A
P
M
O
1989 - 2022
COLLECTION OF
MATHEMATICS OLYMPIAD
THE 1989 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Let x1 , x2 , . . . , xn be positive real numbers, and let
S = x1 + x2 + · · · + xn .
Prove that
(1 + x1 )(1 + x2 ) · · · (1 + xn ) ≤ 1 + S +
S2 S3
Sn
+
+ ··· +
.
2!
3!
n!
Question 2
Prove that the equation
6(6a2 + 3b2 + c2 ) = 5n2
has no solutions in integers except a = b = c = n = 0.
Question 3
Let A1 , A2 , A3 be three points in the plane, and for convenience, let A4 = A1 , A5 = A2 .
For n = 1, 2, and 3, suppose that Bn is the midpoint of An An+1 , and suppose that Cn is
the midpoint of An Bn . Suppose that An Cn+1 and Bn An+2 meet at Dn , and that An Bn+1
and Cn An+2 meet at En . Calculate the ratio of the area of triangle D1 D2 D3 to the area of
triangle E1 E2 E3 .
Question 4
Let S be a set consisting of m pairs (a, b) of positive integers with the property that
1 ≤ a < b ≤ n. Show that there are at least
2
(m − n4 )
4m ·
3n
triples (a, b, c) such that (a, b), (a, c), and (b, c) belong to S.
Question 5
Determine all functions f from the reals to the reals for which
(1) f (x) is strictly increasing,
(2) f (x) + g(x) = 2x for all real x,
where g(x) is the composition inverse function to f (x). (Note: f and g are said to be
composition inverses if f (g(x)) = x and g(f (x)) = x for all real x.)
THE 1990 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Given triagnle ABC, let D, E, F be the midpoints of BC, AC, AB respectively and let G
be the centroid of the triangle.
For each value of ∠BAC, how many non-similar triangles are there in which AEGF is a
cyclic quadrilateral?
Question 2
Let a1 , a2 , . . . , an be positive real numbers, and let Sk be the sum of the products of a1 , a2 ,
. . . , an taken k at a time. Show that
µ ¶2
n
Sk Sn−k ≥
a1 a2 · · · an
k
for k = 1, 2, . . . , n − 1.
Question 3
Consider all the triangles ABC which have a fixed base AB and whose altitude from C is a
constant h. For which of these triangles is the product of its altitudes a maximum?
Question 4
A set of 1990 persons is divided into non-intersecting subsets in such a way that
1. No one in a subset knows all the others in the subset,
2. Among any three persons in a subset, there are always at least two who do not know each
other, and
3. For any two persons in a subset who do not know each other, there is exactly one person
in the same subset knowing both of them.
(a) Prove that within each subset, every person has the same number of acquaintances.
(b) Determine the maximum possible number of subsets.
Note: It is understood that if a person A knows person B, then person B will know person
A; an acquaintance is someone who is known. Every person is assumed to know one’s self.
Question 5
Show that for every integer n ≥ 6, there exists a convex hexagon which can be dissected into
exactly n congruent triangles.
THE 1991 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Let G be the centroid of triangle ABC and M be the midpoint of BC. Let X be on AB
and Y on AC such that the points X, Y , and G are collinear and XY and BC are parallel.
Suppose that XC and GB intersect at Q and Y B and GC intersect at P . Show that triangle
M P Q is similar to triangle ABC.
Question 2
Suppose there are 997 points given in a plane. If every two points are joined by a line
segment with its midpoint coloured in red, show that there are at least 1991 red points in
the plane. Can you find a special case with exactly 1991 red points?
Question 3
Let a1 , a2 , . . . , an , b1 , b2 , . . . , bn be positive real numbers such that a1 + a2 + · · · + an =
b1 + b2 + · · · + bn . Show that
a21
a22
a2n
a1 + a2 + · · · + an
+
+ ··· +
≥
.
a1 + b1 a2 + b2
an + bn
2
Question 4
During a break, n children at school sit in a circle around their teacher to play a game.
The teacher walks clockwise close to the children and hands out candies to some of them
according to the following rule. He selects one child and gives him a candy, then he skips the
next child and gives a candy to the next one, then he skips 2 and gives a candy to the next
one, then he skips 3, and so on. Determine the values of n for which eventually, perhaps
after many rounds, all children will have at least one candy each.
Question 5
Given are two tangent circles and a point P on their common tangent perpendicular to the
lines joining their centres. Construct with ruler and compass all the circles that are tangent
to these two circles and pass through the point P .
THE 1992 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
A triangle with sides a, b, and c is given. Denote by s the semiperimeter, that is s =
(a + b + c)/2. Construct a triangle with sides s − a, s − b, and s − c. This process is repeated
until a triangle can no longer be constructed with the side lengths given.
For which original triangles can this process be repeated indefinitely?
Question 2
In a circle C with centre O and radius r, let C1 , C2 be two circles with centres O1 , O2 and
radii r1 , r2 respectively, so that each circle Ci is internally tangent to C at Ai and so that
C1 , C2 are externally tangent to each other at A.
Prove that the three lines OA, O1 A2 , and O2 A1 are concurrent.
Question 3
Let n be an integer such that n > 3. Suppose that we choose three numbers from the set
{1, 2, . . . , n}. Using each of these three numbers only once and using addition, multiplication,
and parenthesis, let us form all possible combinations.
(a) Show that if we choose all three numbers greater than n/2, then the values of these
combinations are all distinct.
√
(b) Let p be a prime number such that p ≤ n. Show that the number of ways of choosing
three numbers so that the smallest one is p and the values of the combinations are not all
distinct is precisely the number of positive divisors of p − 1.
Question 4
Determine all pairs (h, s) of positive integers with the following property:
If one draws h horizontal lines and another s lines which satisfy
(i) they are not horizontal,
(ii) no two of them are parallel,
(iii) no three of the h + s lines are concurrent,
then the number of regions formed by these h + s lines is 1992.
Question 5
Find a sequence of maximal length consisting of non-zero integers in which the sum of any
seven consecutive terms is positive and that of any eleven consecutive terms is negative.
THE 1993 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Let ABCD be a quadrilateral such that all sides have equal length and angle ABC is 60 deg.
Let l be a line passing through D and not intersecting the quadrilateral (except at D). Let
E and F be the points of intersection of l with AB and BC respectively. Let M be the point
of intersection of CE and AF .
Prove that CA2 = CM × CE.
Question 2
Find the total number of different integer values the function
f (x) = [x] + [2x] + [
5x
] + [3x] + [4x]
3
takes for real numbers x with 0 ≤ x ≤ 100.
Question 3
Let
f (x) = an xn + an−1 xn−1 + · · · + a0 and
g(x) = cn+1 xn+1 + cn xn + · · · + c0
be non-zero polynomials with real coefficients such that g(x) = (x + r)f (x) for some real
number r. If a = max(|an |, . . . , |a0 |) and c = max(|cn+1 |, . . . , |c0 |), prove that ac ≤ n + 1.
Question 4
Determine all positive integers n for which the equation
xn + (2 + x)n + (2 − x)n = 0
has an integer as a solution.
Question 5
Let P1 , P2 , . . . , P1993 = P0 be distinct points in the xy-plane with the following properties:
(i) both coordinates of Pi are integers, for i = 1, 2, . . . , 1993;
(ii) there is no point other than Pi and Pi+1 on the line segment joining Pi with Pi+1 whose
coordinates are both integers, for i = 0, 1, . . . , 1992.
Prove that for some i, 0 ≤ i ≤ 1992, there exists a point Q with coordinates (qx , qy ) on the
line segment joining Pi with Pi+1 such that both 2qx and 2qy are odd integers.
THE 1994 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Let f : R → R be a function such that
(i) For all x, y ∈ R,
f (x) + f (y) + 1 ≥ f (x + y) ≥ f (x) + f (y),
(ii) For all x ∈ [0, 1), f (0) ≥ f (x),
(iii) −f (−1) = f (1) = 1.
Find all such functions f .
Question 2
Given a nondegenerate triangle ABC, with circumcentre O, orthocentre H, and circumradius
R, prove that |OH| < 3R.
Question 3
Let n be an integer of the√form a2 + b2 , where a and b are relatively prime integers and such
that if p is a prime, p ≤ n, then p divides ab. Determine all such n.
Question 4
Is there an infinite set of points in the plane such that no three points are collinear, and the
distance between any two points is rational?
Question 5
You are given three lists A, B, and C. List A contains the numbers of the form 10k in base
10, with k any integer greater than or equal to 1. Lists B and C contain the same numbers
translated into base 2 and 5 respectively:
A
10
100
1000
..
.
B
C
1010
20
1100100
400
1111101000 13000
..
..
.
.
Prove that for every integer n > 1, there is exactly one number in exactly one of the lists B
or C that has exactly n digits.
THE 1995 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Determine all sequences of real numbers a1 , a2 , . . . , a1995 which satisfy:
p
2 an − (n − 1) ≥ an+1 − (n − 1), for n = 1, 2, . . . 1994,
and
√
2 a1995 − 1994 ≥ a1 + 1.
Question 2
Let a1 , a2 , . . . , an be a sequence of integers with values between 2 and 1995 such that:
(i) Any two of the ai ’s are realtively prime,
(ii) Each ai is either a prime or a product of primes.
Determine the smallest possible values of n to make sure that the sequence will contain a
prime number.
Question 3
Let P QRS be a cyclic quadrilateral such that the segments P Q and RS are not parallel. Consider the set of circles through P and Q, and the set of circles through R and S.
Determine the set A of points of tangency of circles in these two sets.
Question 4
Let C be a circle with radius R and centre O, and S a fixed point in the interior of C. Let
AA0 and BB 0 be perpendicular chords through S. Consider the rectangles SAM B, SBN 0 A0 ,
SA0 M 0 B 0 , and SB 0 N A. Find the set of all points M , N 0 , M 0 , and N when A moves around
the whole circle.
Question 5
Find the minimum positive integer k such that there exists a function f from the set Z of
all integers to {1, 2, . . . k} with the property that f (x) 6= f (y) whenever |x − y| ∈ {5, 7, 12}.
THE 1996 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Let ABCD be a quadrilateral AB = BC = CD = DA. Let M N and P Q be two segments
perpendicular to the diagonal BD and such that the distance between them is d > BD/2,
with M ∈ AD, N ∈ DC, P ∈ AB, and Q ∈ BC. Show that the perimeter of hexagon
AM N CQP does not depend on the position of M N and P Q so long as the distance between
them remains constant.
Question 2
Let m and n be positive integers such that n ≤ m. Prove that
2n n! ≤
(m + n)!
≤ (m2 + m)n .
(m − n)!
Question 3
Let P1 , P2 , P3 , P4 be four points on a circle, and let I1 be the incentre of the triangle P2 P3 P4 ;
I2 be the incentre of the triangle P1 P3 P4 ; I3 be the incentre of the triangle P1 P2 P4 ; I4 be the
incentre of the triangle P1 P2 P3 . Prove that I1 , I2 , I3 , I4 are the vertices of a rectangle.
Question 4
The National Marriage Council wishes to invite n couples to form 17 discussion groups under
the following conditions:
1. All members of a group must be of the same sex; i.e. they are either all male or all
female.
2. The difference in the size of any two groups is 0 or 1.
3. All groups have at least 1 member.
4. Each person must belong to one and only one group.
Find all values of n, n ≤ 1996, for which this is possible. Justify your answer.
Question 5
Let a, b, c be the lengths of the sides of a triangle. Prove that
√
√
√
√
√
√
a+b−c+ b+c−a+ c+a−b≤ a+ b+ c ,
and determine when equality occurs.
THE 1997 ASIAN PACIFIC MATHEMATICAL OLYMPIAD
Time allowed: 4 hours
NO calculators are to be used.
Each question is worth seven points.
Question 1
Given
1
1
1
,
1 +
1
1 + ··· +
1
1
1
1+ 3
1+ 3 + 6
1 + 3 + 6 + · · · + 1993006
where the denominators contain partial sums of the sequence of reciprocals of triangular
numbers (i.e. k = n(n + 1)/2 for n = 1, 2, . . . , 1996). Prove that S > 1001.
S =1+
Question 2
Find an integer n, where 100 ≤ n ≤ 1997, such that
2n + 2
n
is also an integer.
Question 3
Let ABC be a triangle inscribed in a circle and let
mb
mc
ma
, lb =
, lc =
,
la =
Ma
Mb
Mc
where ma , mb , mc are the lengths of the angle bisectors (internal to the triangle) and Ma ,
Mb , Mc are the lengths of the angle bisectors extended until they meet the circle. Prove that
la
lb
lc
+
+
≥ 3,
2
2
sin A sin B sin2 C
and that equality holds iff ABC is an equilateral triangle.
Question 4
Triangle A1 A2 A3 has a right angle at A3 . A sequence of points is now defined by the following
iterative process, where n is a positive integer. From An (n ≥ 3), a perpendicular line is
drawn to meet An−2 An−1 at An+1 .
(a) Prove that if this process is continued indefinitely, then one and only one point P is
interior to every triangle An−2 An−1 An , n ≥ 3.
(b) Let A1 and A3 be fixed points. By considering all possible locations of A2 on the plane,
find the locus of P .
Question 5
Suppose that n people A1 , A2 , . . ., An , (n ≥ 3) are seated in a circle and that Ai has ai
objects such that
a1 + a2 + · · · + an = nN,
where N is a positive integer. In order that each person has the same number of objects, each
person Ai is to give or to receive a certain number of objects to or from its two neighbours
Ai−1 and Ai+1 . (Here An+1 means A1 and An means A0 .) How should this redistribution be
performed so that the total number of objects transferred is minimum?
10th Asian Pacific Mathematics Olympiad
March 1998
Time allowed: 4 hours.
No calculators to be used.
Each question is worth 7 points.
1.
Let F be the set of all n-tuples ( A1 , A2 , …, An ) where each Ai , i = 1, 2, …, n is a subset of
{1, 2, …, 1998}. Let | A | denote the number of elements of the set A.
Find the number
∑| A ∪ A
1
( A1 , A2 , , An )
2
∪ ∪ An | .
2.
Show that for any positive integers a and b, (36a + b)(a + 36b) cannot be a power of 2.
3.
 a  b  c   a + b + c 
Let a, b, c be positive real numbers. Prove that 1 + 1 + 1 +  ≥ 21 + 3
 .
abc 
 b  c  a  
4.
Let ABC be a triangle and D the foot of the altitude from A. Let E and F be on a line through D
such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from
D. Let M and N be the midpoints of the line segments BC and EF, respectively. Prove that AN
is perpendicular to NM.
5.
Determine the largest of all integers n with the property that n is divisible by all positive
integers that are less than 3 n .
END OF PAPER
11th Asian Pacific Mathematical Olympiad
March, 1999
1. Find the smallest positive integer n with the following property: there does not exist
an arithmetic progression of 1999 real numbers containing exactly n integers.
2. Let a1 , a2 , . . . be a sequence of real numbers satisfying ai+j ≤ ai +aj for all i, j = 1, 2, . . ..
Prove that
a2 a3
an
a1 +
+
+ ··· +
≥ an
2
3
n
for each positive integer n.
3. Let Γ1 and Γ2 be two circles intersecting at P and Q. The common tangent, closer to
P , of Γ1 and Γ2 touches Γ1 at A and Γ2 at B. The tangent of Γ1 at P meets Γ2 at C,
which is different from P , and the extension of AP meets BC at R. Prove that the
circumcircle of triangle P QR is tangent to BP and BR.
4. Determine all pairs (a, b) of integers with the property that the numbers a2 + 4b and
b2 + 4a are both perfect squares.
5. Let S be a set of 2n + 1 points in the plane such that no three are collinear and no
four concyclic. A circle will be called good if it has 3 points of S on its circumference,
n − 1 points in its interior and n − 1 points in its exterior. Prove that the number of
good circles has the same parity as n.
12th Asian Pacific Mathematics Olympiad
March 2000
Time allowed: 4 hours.
No calculators to be used.
Each question is worth 7 points.
101
3
xi
i
for xi =
.
2
101
i = 0 1 − 3 xi + 3 xi
1.
Compute the sum S = ∑
2.
Given the following triangular arrangement of circles:
Each of the numbers 1, 2, …, 9 is to be written into one of these circles, so that each circle
contains exactly one of these numbers and
(i)
the sums of the four numbers on each side of the triangle are equal;
(ii) the sums of the squares of the four numbers on each side of the triangle are equal.
Find all ways in which this can be done.
3.
Let ABC be a triangle. Let M and N be the points in which the median and the angle bisector,
respectively, at A meet the side BC. Let Q and P be the points in which the perpendicular at N
to NA meets MA and BA, respectively, and O the point in which the perpendicular at P to BA
meets AN produced. Prove that QO is perpendicular to BC.
4.
Let n, k be given positive integers with n > k. Prove that
1
nn
n!
nn
⋅ k
<
<
.
n + 1 k (n − k )n −k k !(n − k )! k k (n − k ) n −k
5.
Given a permutation (a0 , a1 , , an ) of the sequence 0, 1, …, n. A transposition of ai with
a j is called legal if ai = 0 for i > 0 , and ai −1 + 1 = a j . The permutation (a0 , a1 , , an ) is
called regular if after a number of legal transpositions it becomes (1, 2, , n, 0) . For which
numbers n is the permutation (1, n, n − 1, , 3, 2, 0) regular?
END OF PAPER
XIII Asian Pacific Mathematics Olympiad
March, 2001
Time allowed: 4 hours
No calculators to be used
Each question is worth 7 points
Problem 1.
For a positive integer n let S(n) be the sum of digits in the decimal representation of n. Any positive
integer obtained by removing several (at least one) digits from the right-hand end of the decimal
representation of n is called a stump of n. Let T (n) be the sum of all stumps of n. Prove that
n = S(n) + 9T (n).
Problem 2.
Find the largest positive integer N so that the number of integers in the set {1, 2, . . . , N } which are
divisible by 3 is equal to the number of integers which are divisible by 5 or 7 (or both).
Problem 3.
Let two equal regular n-gons S and T be located in the plane such that their intersection is a 2n-gon
(n ≥ 3). The sides of the polygon S are coloured in red and the sides of T in blue.
Prove that the sum of the lengths of the blue sides of the polygon S ∩ T is equal to the sum of the
lengths of its red sides.
Problem 4.
A point in the plane with a cartesian coordinate system is called a mixed point if one of its coordinates
is rational and the other one is irrational. Find all polynomials with real coefficients such that their
graphs do not contain any mixed point.
Problem 5.
Find the greatest integer n, such that there are n + 4 points A, B, C, D, X1 , . . . , Xn in the plane with
AB 6= CD that satisfy the following condition: for each i = 1, 2, . . . , n triangles ABXi and CDXi are
equal.
XIV Asian Pacific Mathematics Olympiad
March 2002
Time allowed: 4 hours
No calculators are to be used
Each question is worth 7 points
Problem 1.
Let a1 , a2 , a3 , . . . , an be a sequence of non-negative integers, where n is a positive integer. Let
An =
a1 + a2 + · · · + an
.
n
Prove that
n
a1 !a2 ! . . . an ! ≥ (bAn c!) ,
where bAn c is the greatest integer less than or equal to An , and a! = 1 × 2 × · · · × a for a ≥ 1 (and 0! = 1).
When does equality hold?
Problem 2.
Find all positive integers a and b such that
a2 + b
b2 − a
and
b2 + a
a2 − b
are both integers.
Problem 3.
Let ABC be an equilateral triangle. Let P be a point on the side AC and Q be a point on the side AB so that
both triangles ABP and ACQ are acute. Let R be the orthocentre of triangle ABP and S be the orthocentre
of triangle ACQ. Let T be the point common to the segments BP and CQ. Find all possible values of 6 CBP
and 6 BCQ such that triangle T RS is equilateral.
Problem 4.
Let x, y, z be positive numbers such that
Show that
√
x + yz +
√
1
1 1
+ + = 1.
x y z
y + zx +
√
z + xy ≥
√
xyz +
√
x+
√
y+
√
z.
Problem 5.
Let R denote the set of all real numbers. Find all functions f from R to R satisfying:
(i) there are only finitely many s in R such that f (s) = 0, and
(ii) f (x4 + y) = x3 f (x) + f (f (y)) for all x, y in R.
XV Asian Pacific Mathematics Olympiad
March 2003
Time allowed: 4 hours
No calculators are to be used
Each question is worth 7 points
Problem 1.
Let a, b, c, d, e, f be real numbers such that the polynomial
p(x) = x8 − 4x7 + 7x6 + ax5 + bx4 + cx3 + dx2 + ex + f
factorises into eight linear factors x − xi , with xi > 0 for i = 1, 2, . . . , 8. Determine all possible values of f .
Problem 2.
Suppose ABCD is a square piece of cardboard with side length a. On a plane are two parallel lines `1 and `2 ,
which are also a units apart. The square ABCD is placed on the plane so that sides AB and AD intersect `1
at E and F respectively. Also, sides CB and CD intersect `2 at G and H respectively. Let the perimeters of
4AEF and 4CGH be m1 and m2 respectively. Prove that no matter how the square was placed, m1 + m2
remains constant.
Problem 3.
Let k ≥ 14 be an integer, and let pk be the largest prime number which is strictly less than k. You may assume
that pk ≥ 3k/4. Let n be a composite integer. Prove:
(a) if n = 2pk , then n does not divide (n − k)! ;
(b) if n > 2pk , then n divides (n − k)! .
Problem 4.
Let a, b, c be the sides of a triangle, with a + b + c = 1, and let n ≥ 2 be an integer. Show that
√
n
√
√
√
2
n
n
an + bn + bn + cn + n cn + an < 1 +
.
2
Problem 5.
Given two positive integers m and n, find the smallest positive integer k such that among any k people, either
there are 2m of them who form m pairs of mutually acquainted people or there are 2n of them forming n pairs
of mutually unacquainted people.
XVI Asian Pacific Mathematics Olympiad
March 2004
Time allowed: 4 hours
No calculators are to be used
Each question is worth 7 points
Problem 1.
Determine all finite nonempty sets S of positive integers satisfying
i+j
(i, j)
is an element of S for all i, j in S,
where (i, j) is the greatest common divisor of i and j.
Problem 2.
Let O be the circumcentre and H the orthocentre of an acute triangle ABC. Prove that the area of one of the
triangles AOH, BOH and COH is equal to the sum of the areas of the other two.
Problem 3.
Let a set S of 2004 points in the plane be given, no three of which are collinear. Let L denote the set of all lines
(extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible
to colour the points of S with at most two colours, such that for any points p, q of S, the number of lines in L
which separate p from q is odd if and only if p and q have the same colour.
Note: A line ` separates two points p and q if p and q lie on opposite sides of ` with neither point on `.
Problem 4.
For a real number x, let bxc stand for the largest integer that is less than or equal to x. Prove that
¹
º
(n − 1)!
n(n + 1)
is even for every positive integer n.
Problem 5.
Prove that
for all real numbers a, b, c > 0.
(a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca)
XVII Asian Pacific Mathematics Olympiad
Time allowed : 4 hours
Each problem is worth 7 points
∗ The contest problems are to be kept confidential until they are posted on the official APMO
website. Please do not disclose nor discuss the problems over the internet until that date.
No calculators are to be used during the contest.
Problem 1. Prove that for every irrational real number a, there are irrational real numbers
b and b0 so that a + b and ab0 are both rational while ab and a + b0 are both irrational.
Problem 2. Let a, b and c be positive real numbers such that abc = 8. Prove that
b2
c2
4
a2
p
+p
+p
≥ .
3
(1 + a3 )(1 + b3 )
(1 + b3 )(1 + c3 )
(1 + c3 )(1 + a3 )
Problem 3. Prove that there exists a triangle which can be cut into 2005 congruent
triangles.
Problem 4. In a small town, there are n × n houses indexed by (i, j) for 1 ≤ i, j ≤ n with
(1, 1) being the house at the top left corner, where i and j are the row and column indices,
respectively. At time 0, a fire breaks out at the house indexed by (1, c), where c ≤ n2 .
During each subsequent time interval [t, t + 1], the fire fighters defend a house which is not
yet on fire while the fire spreads to all undefended neighbors of each house which was on
fire at time t. Once a house is defended, it remains so all the time. The process ends when
the fire can no longer spread. At most how many houses can be saved by the fire fighters?
A house indexed by (i, j) is a neighbor of a house indexed by (k, `) if |i − k| + |j − `| = 1.
Problem 5. In a triangle ABC, points M and N are on sides AB and AC, respectively,
such that M B = BC = CN . Let R and r denote the circumradius and the inradius of the
triangle ABC, respectively. Express the ratio M N/BC in terms of R and r.
XVIII Asian Pacific Mathematics Olympiad
Time allowed : 4 hours
Each problem is worth 7 points
∗ The contest problems are to be kept confidential until they are posted on the official APMO
website. Please do not disclose nor discuss the problems over the internet until that date.
No calculators are to be used during the contest.
Problem 1. Let n be a positive integer. Find the largest nonnegative real number f (n)
(depending on n) with the following property: whenever a1 , a2 , . . . , an are real numbers
such that a1 + a2 + · · · + an is an integer, there exists some i such that | ai − 21 | ≥ f (n).
Problem 2. Prove that every positive integer
can be written as a finite sum of distinct
√
integral powers of the golden mean τ = 1+2 5 . Here, an integral power of τ is of the form
τ i , where i is an integer (not necessarily positive).
Problem 3. Let p ≥ 5 be a prime and let r be the number of ways of placing p checkers
on a p × p checkerboard so that not all checkers are in the same row (but they may all be
in the same column). Show that r is divisible by p 5 . Here, we assume that all the checkers
are identical.
Problem 4. Let A, B be two distinct points on a given circle O and let P be the midpoint
of the line segment AB. Let O1 be the circle tangent to the line AB at P and tangent to
the circle O. Let ` be the tangent line, different from the line AB, to O1 passing through
A. Let C be the intersection point, different from A, of ` and O. Let Q be the midpoint
of the line segment BC and O2 be the circle tangent to the line BC at Q and tangent to
the line segment AC. Prove that the circle O2 is tangent to the circle O.
Problem 5. In a circus, there are n clowns who dress and paint themselves up using a
selection of 12 distinct colours. Each clown is required to use at least five different colours.
One day, the ringmaster of the circus orders that no two clowns have exactly the same set
of colours and no more than 20 clowns may use any one particular colour. Find the largest
number n of clowns so as to make the ringmaster’s order possible.
XIX Asian Pacific Mathematics Olympiad
G
Time allowed : 4 hours
Each problem is worth 7 points
∗ The contest problems are to be kept confidential until they are posted on the official APMO
website. Please do not disclose nor discuss the problems over the internet until that date.
No calculators are to be used during the contest.
Problem 1. Let S be a set of 9 distinct integers all of whose prime factors are at most 3.
Prove that S contains 3 distinct integers such that their product is a perfect cube.
Problem 2. Let ABC be an acute angled triangle with ∠BAC = 60◦ and AB > AC. Let
I be the incenter, and H the orthocenter of the triangle ABC. Prove that
2∠AHI = 3∠ABC.
Problem 3. Consider n disks C1 , C2 , . . . , Cn in a plane such that for each 1 ≤ i < n, the
center of Ci is on the circumference of Ci+1 , and the center of Cn is on the circumference
of C1 . Define the score of such an arrangement of n disks to be the number of pairs (i, j)
for which Ci properly contains Cj . Determine the maximum possible score.
√
√
√
Problem 4. Let x, y and z be positive real numbers such that x + y + z = 1. Prove
that
x2 + yz
y 2 + zx
z 2 + xy
p
+p
+p
≥ 1.
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
Problem 5. A regular (5 × 5)-array of lights is defective, so that toggling the switch for
one light causes each adjacent light in the same row and in the same column as well as
the light itself to change state, from on to off, or from off to on. Initially all the lights are
switched off. After a certain number of toggles, exactly one light is switched on. Find all
the possible positions of this light.
1
XX Asian Pacific Mathematics Olympiad
March, 2008
Time allowed : 4 hours
Each problem is worth 7 points
∗ The contest problems are to be kept confidential until they are posted on the official APMO
website. Please do not disclose nor discuss the problems over the internet until that date. No
calculators are to be used during the contest.
Problem 1. Let ABC be a triangle with ∠A < 60◦ . Let X and Y be the points on the sides
AB and AC, respectively, such that CA + AX = CB + BX and BA + AY = BC + CY . Let
P be the point in the plane such that the lines P X and P Y are perpendicular to AB and
AC, respectively. Prove that ∠BP C < 120◦ .
Problem 2. Students in a class form groups each of which contains exactly three members
such that any two distinct groups have at most one member in common. Prove that, when
the class size is 46, there is a set of 10 students in which no group is properly contained.
Problem 3. Let Γ be the circumcircle of a triangle ABC. A circle passing through points
A and C meets the sides BC and BA at D and E, respectively. The lines AD and CE meet
Γ again at G and H, respectively. The tangent lines of Γ at A and C meet the line DE at L
and M , respectively. Prove that the lines LH and M G meet at Γ.
Problem 4. Consider the function f : N0 → N0 , where N0 is the set of all non-negative
integers, defined by the following conditions :
(i) f (0) = 0, (ii) f (2n) = 2f (n) and (iii) f (2n + 1) = n + 2f (n) for all n ≥ 0.
(a) Determine the three sets L := { n | f (n) < f (n + 1) }, E := { n | f (n) = f (n + 1) }, and
G := { n | f (n) > f (n + 1) }.
(b) For each k ≥ 0, find a formula for ak := max{f (n) : 0 ≤ n ≤ 2k } in terms of k.
Problem 5. Let a, b, c be integers satisfying 0 < a < c − 1 and 1 < b < c. For each k,
0 ≤ k ≤ a, let rk , 0 ≤ rk < c, be the remainder of kb when divided by c. Prove that the two
sets {r0 , r1 , r2 , . . . , ra } and {0, 1, 2, . . . , a} are different.
XXI Asian Pacific Mathematics Olympiad
March, 2009
Time allowed : 4 hours
Each problem is worth 7 points
∗ The contest problems are to be kept confidential until they are posted on the official APMO
website (http://www.kms.or.kr/Competitions/APMO). Please do not disclose nor discuss the
problems over the internet until that date. Calculators are not allowed to use.
Problem 1. Consider the following operation on positive real numbers written on a blackboard: Choose a number r written on the blackboard, erase that number, and then write a
pair of positive real numbers a and b satisfying the condition 2r2 = ab on the board.
Assume that you start out with just one positive real number r on the blackboard, and
apply this operation k 2 − 1 times to end up with k 2 positive real numbers, not necessarily
distinct. Show that there exists a number on the board which does not exceed kr.
Problem 2. Let a1 , a2 , a3 , a4 , a5 be real numbers satisfying the following equations:
a2
a3
a4
a5
1
a1
+ 2
+ 2
+ 2
+ 2
= 2 for k = 1, 2, 3, 4, 5.
+1 k +2 k +3 k +4 k +5
k
k2
Find the value of
a2
a3
a4
a5
a1
+
+
+
+
. (Express the value in a single fraction.)
37 38 39 40 41
Problem 3. Let three circles Γ1 , Γ2 , Γ3 , which are non-overlapping and mutually external,
be given in the plane. For each point P in the plane, outside the three circles, construct
six points A1 , B1 , A2 , B2 , A3 , B3 as follows: For each i = 1, 2, 3, Ai , Bi are distinct points on
the circle Γi such that the lines P Ai and P Bi are both tangents to Γi . Call the point P
exceptional if, from the construction, three lines A1 B1 , A2 B2 , A3 B3 are concurrent. Show
that every exceptional point of the plane, if exists, lies on the same circle.
Problem 4. Prove that for any positive integer k, there exists an arithmetic sequence
a1
a2
ak
,
, ... ,
b1
b2
bk
of rational numbers, where ai , bi are relatively prime positive integers for each i = 1, 2, . . . , k,
such that the positive integers a1 , b1 , a2 , b2 , . . . , ak , bk are all distinct.
Problem 5. Larry and Rob are two robots travelling in one car from Argovia to Zillis. Both
robots have control over the steering and steer according to the following algorithm: Larry
makes a 90◦ left turn after every ` kilometer driving from start; Rob makes a 90◦ right turn
after every r kilometer driving from start, where ` and r are relatively prime positive integers.
In the event of both turns occurring simultaneously, the car will keep going without changing
direction. Assume that the ground is flat and the car can move in any direction.
Let the car start from Argovia facing towards Zillis. For which choices of the pair (`, r) is
the car guaranteed to reach Zillis, regardless of how far it is from Argovia?
XXII Asian Pacific Mathematics Olympiad
G
March, 2010
Time allowed: 4 hours
Each problem is worth 7 points
*The contest problems are to be kept confidential until they are posted on the official APMO
website (http://www.mmjp.or.jp/competitions/APMO). Please do not disclose nor discuss the
problems over the internet until that date. Calculators are not allowed to use.
Problem 1. Let ABC be a triangle with 6 BAC 6= 90◦ . Let O be the circumcenter of the
triangle ABC and let Γ be the circumcircle of the triangle BOC. Suppose that Γ intersects
the line segment AB at P different from B, and the line segment AC at Q different from C.
Let ON be a diameter of the circle Γ. Prove that the quadrilateral AP N Q is a parallelogram.
Problem 2. For a positive integer k, call an integer a pure k-th power if it can be represented
as mk for some integer m. Show that for every positive integer n there exist n distinct positive
integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power.
Problem 3. Let n be a positive integer. n people take part in a certain party. For any pair
of the participants, either the two are acquainted with each other or they are not. What is
the maximum possible number of the pairs for which the two are not acquainted but have a
common acquaintance among the participants?
Problem 4.
Let ABC be an acute triangle satisfying the condition AB > BC and
AC > BC. Denote by O and H the circumcenter and the orthocenter, respectively, of the
triangle ABC. Suppose that the circumcircle of the triangle AHC intersects the line AB at
M different from A, and that the circumcircle of the triangle AHB intersects the line AC at
N different from A. Prove that the circumcenter of the triangle M N H lies on the line OH.
Problem 5. Find all functions f from the set R of real numbers into R which satisfy for
all x, y, z ∈ R the identity
f (f (x) + f (y) + f (z)) = f (f (x) − f (y)) + f (2xy + f (z)) + 2f (xz − yz).
2011 APMO PROBLEMS
Time allowed: 4 hours
Each problem is worth 7 points
*The contest problems are to be kept confidential until they are posted on the official APMO website (http://www.mmjp.or.jp/competitions/APMO). Please do not
disclose nor discuss the problems over the internet until that date. Calculators are
not allowed to use.
Problem 1. Let a, b, c be positive integers. Prove that it is impossible to have
all of the three numbers a2 + b + c, b2 + c + a, c2 + a + b to be perfect squares.
Problem 2. Five points A1 , A2 , A3 , A4 , A5 lie on a plane in such a way that no
three among them lie on a same straight line. Determine the maximum possible
value that the minimum value for the angles ∠Ai Aj Ak can take where i, j, k are
distinct integers between 1 and 5.
Problem 3. Let ABC be an acute triangle with ∠BAC = 30◦ . The internal and
external angle bisectors of ∠ABC meet the line AC at B1 and B2 , respectively, and
the internal and external angle bisectors of ∠ACB meet the line AB at C1 and C2 ,
respectively. Suppose that the circles with diameters B1 B2 and C1 C2 meet inside
the triangle ABC at point P . Prove that ∠BP C = 90◦ .
Problem 4. Let n be a fixed positive odd integer. Take m + 2 distinct points
P0 , P1 , · · · , Pm+1 (where m is a non-negative integer) on the coordinate plane in
such a way that the following 3 conditions are satisfied:
(1) P0 = (0, 1), Pm+1 = (n + 1, n), and for each integer i, 1 ≤ i ≤ m, both x- and
y- coordinates of Pi are integers lying in between 1 and n (1 and n inclusive).
(2) For each integer i, 0 ≤ i ≤ m, Pi Pi+1 is parallel to the x-axis if i is even, and
is parallel to the y-axis if i is odd.
(3) For each pair i, j with 0 ≤ i < j ≤ m, line segments Pi Pi+1 and Pj Pj+1 share
at most 1 point.
Determine the maximum possible value that m can take.
Problem 5. Determine all functions f : R → R, where R is the set of all real
numbers, satisfying the following 2 conditions:
(1) There exists a real number M such that for every real number x, f (x) < M is
satisfied.
(2) For every pair of real numbers x and y,
f (xf (y)) + yf (x) = xf (y) + f (xy)
is satisfied.
1
2012 APMO PROBLEMS
Time allowed: 4 hours
Each problem is worth 7 points
*The contest problems are to be kept confidential until they are posted on the official APMO website (http://www.mmjp.or.jp/competitions/APMO). Please do not
disclose nor discuss the problems over the internet until that date. Calculators are
not allowed to use.
Problem 1. Let P be a point in the interior of a triangle ABC, and let D, E, F
be the point of intersection of the line AP and the side BC of the triangle, of the
line BP and the side CA, and of the line CP and the side AB, respectively. Prove
that the area of the triangle ABC must be 6 if the area of each of the triangles
P F A, P DB and P EC is 1.
Problem 2. Into each box of a 2012 × 2012 square grid, a real number greater
than or equal to 0 and less than or equal to 1 is inserted. Consider splitting the grid
into 2 non-empty rectangles consisting of boxes of the grid by drawing a line parallel
either to the horizontal or the vertical side of the grid. Suppose that for at least one
of the resulting rectangles the sum of the numbers in the boxes within the rectangle
is less than or equal to 1, no matter how the grid is split into 2 such rectangles.
Determine the maximum possible value for the sum of all the 2012 × 2012 numbers
inserted into the boxes.
Problem 3. Determine all the pairs (p, n) of a prime number p and a positive
p
+1
integer n for which npn +1
is an integer.
Problem 4.
Let ABC be an acute triangle. Denote by D the foot of the
perpendicular line drawn from the point A to the side BC, by M the midpoint of
BC, and by H the orthocenter of ABC. Let E be the point of intersection of the
circumcircle Γ of the triangle ABC and the half line M H, and F be the point of
AB
intersection (other than E) of the line ED and the circle Γ. Prove that BF
CF = AC
must hold.
Here we denote by XY the length of the line segment XY .
Problem 5. Let n be an integer greater than or equal to 2. Prove that if the
real numbers a1 , a2 , · · · , an satisfy a21 + a22 + · · · + a2n = n, then
X
n
1
≤
n − ai aj
2
1≤i<j≤n
must hold.
1
XXV Asian Pacific Mathematics Olympiad
Time allowed: 4 hours
Each problem if worth 7 points
Problem 1. Let ABC be an acute triangle with altitudes AD, BE and CF , and let O
be the center of its circumcircle. Show that the segments OA, OF, OB, OD, OC, OE
dissect the triangle ABC into three pairs of triangles that have equal areas.
n2 + 1
is an integer. Here
Problem 2. Determine all positive integers n for which √ 2
[ n] + 2
[r] denotes the greatest integer less than or equal to r.
Problem 3. For 2k real numbers a1 , a2 , . . . , ak , b1 , b2 , . . . , bk define the sequence of
numbers Xn by
k
X
Xn =
[ai n + bi ] (n = 1, 2, . . .).
i=1
P
If the sequence Xn forms an arithmetic progression, show that ki=1 ai must be an
integer. Here [r] denotes the greatest integer less than or equal to r.
Problem 4. Let a and b be positive integers, and let A and B be finite sets of
integers satisfying:
(i) A and B are disjoint;
(ii) if an integer i belongs either to A or to B, then i + a belongs to A or i − b
belongs to B.
Prove that a|A| = b|B|. (Here |X| denotes the number of elements in the set X.)
Problem 5. Let ABCD be a quadrilateral inscribed in a circle ω, and let P be a
point on the extension of AC such that P B and P D are tangent to ω. The tangent at
C intersects P D at Q and the line AD at R. Let E be the second point of intersection
between AQ and ω. Prove that B, E, R are collinear.
XXVI Asian Pacific Mathematics Olympiad
Time allowed: 4 hours
Each problem if worth 7 points
Problem 1. For a positive integer m denote by S(m) and P (m) the sum and
product, respectively, of the digits of m. Show that for each positive integer n, there
exist positive integers a1 , a2 , . . . , an satisfying the following conditions:
S(a1 ) < S(a2 ) < · · · < S(an ) and S(ai ) = P (ai+1 ) (i = 1, 2, . . . , n).
(We let an+1 = a1 .)
(Proposed by the Problem Committee of
the Japan Mathematical Olympiad Foundation)
Problem 2. Let S = {1, 2, . . . , 2014}. For each non-empty subset T ⊆ S, one
of its members is chosen as its representative. Find the number of ways to assign
representatives to all non-empty subsets of S so that if a subset D ⊆ S is a disjoint
union of non-empty subsets A, B, C ⊆ S, then the representative of D is also the
representative of at least one of A, B, C.
(Proposed by Warut Suksompong, Thailand)
Problem 3. Find all positive integers n such that for any integer k there exists an
integer a for which a3 + a − k is divisible by n.
(Proposed by Warut Suksompong, Thailand)
Problem 4. Let n and b be positive integers. We say n is b-discerning if there exists
a set consisting of n different positive integers less than b that has no two different
subsets U and V such that the sum of all elements in U equals the sum of all elements
in V .
(a) Prove that 8 is a 100-discerning.
(b) Prove that 9 is not 100–discerning.
(Proposed by the Senior Problems Committee of
the Australian Mathematical Olympiad Committee)
Problem 5. Circles ω and Ω meet at points A and B. Let M be the midpoint of
the arc AB of circle ω (M lies inside Ω). A chord M P of circle ω intersects Ω at Q
(Q lies inside ω). Let `P be the tangent line to ω at P , and let `Q be the tangent
line to Ω at Q. Prove that the circumcircle of the triangle formed by the lines `P , `Q ,
and AB is tangent to Ω.
(Proposed by Ilya Bogdanov, Russia and
Medeubek Kungozhin, Kazakhstan)
XXVII Asian Pacific Mathematics Olympiad
Time allowed: 4 hours
Each problem if worth 7 points
Problem 1. Let ABC be a triangle, and let D be a point on side BC. A line
through D intersects side AB at X and ray AC at Y . The circumcircle of triangle
BXD intersects the circumcircle ω of triangle ABC again at point Z 6= B. The lines
ZD and ZY intersect ω again at V and W , respectively. Prove that AB = V W .
Proposed by Warut Suksompong, Thailand
Problem 2. Let S = {2, 3, 4, . . .} denote the set of integers that are greater than or
equal to 2. Does there exist a function f : S → S such that
f (a)f (b) = f (a2 b2 ) for all a, b ∈ S with a 6= b?
Proposed by Angelo Di Pasquale, Australia
Problem 3. A sequence of real numbers a0 , a1 , . . . is said to be good if the following
three conditions hold.
(i) The value of a0 is a positive integer.
(ii) For each non-negative integer i we have ai+1 = 2ai + 1 or ai+1 =
ai
.
ai + 2
(iii) There exists a positive integer k such that ak = 2014.
Find the smallest positive integer n such that there exists a good sequence a0 , a1 , . . .
of real numbers with the property that an = 2014.
Proposed by Wang Wei Hua, Hong Kong
Problem 4. Let n be a positive integer. Consider 2n distinct lines on the plane, no
two of which are parallel. Of the 2n lines, n are colored blue, the other n are colored
red. Let B be the set of all points on the plane that lie on at least one blue line,
and R the set of all points on the plane that lie on at least one red line. Prove that
there exists a circle that intersects B in exactly 2n − 1 points, and also intersects R
in exactly 2n − 1 points.
Proposed by Pakawut Jiradilok and Warut Suksompong, Thailand
Problem 5. Determine all sequences a0 , a1 , a2 , . . . of positive integers with a0 ≥ 2015
such that for all integers n ≥ 1:
(i) an+2 is divisible by an ;
(ii) |sn+1 − (n + 1)an | = 1, where sn+1 = an+1 − an + an−1 − · · · + (−1)n+1 a0 .
Proposed by Pakawut Jiradilok and Warut Suksompong, Thailand
XXVIII Asian Pacific Mathematics Olympiad
March, 2016
Time allowed: 4 hours
Each problem is worth 7 points
The contest problems are to be kept confidential until they are posted on the official
APMO website http://apmo.ommenlinea.org.
Please do not disclose nor discuss the problems over online until that date. The use
of calculators is not allowed.
Problem 1. We say that a triangle ABC is great if the following holds: for any
point D on the side BC, if P and Q are the feet of the perpendiculars from D to the
lines AB and AC, respectively, then the reflection of D in the line P Q lies on the
circumcircle of the triangle ABC.
Prove that triangle ABC is great if and only if ∠A = 90◦ and AB = AC.
Problem 2. A positive integer is called fancy if it can be expressed in the form
2a1 + 2a2 + · · · + 2a100 ,
where a1 , a2 , . . . , a100 are non-negative integers that are not necessarily distinct.
Find the smallest positive integer n such that no multiple of n is a fancy number.
Problem 3. Let AB and AC be two distinct rays not lying on the same line, and let
ω be a circle with center O that is tangent to ray AC at E and ray AB at F . Let R
be a point on segment EF . The line through O parallel to EF intersects line AB at
P . Let N be the intersection of lines P R and AC, and let M be the intersection of
line AB and the line through R parallel to AC. Prove that line M N is tangent to ω.
Problem 4. The country Dreamland consists of 2016 cities. The airline Starways
wants to establish some one-way flights between pairs of cities in such a way that each
city has exactly one flight out of it. Find the smallest positive integer k such that no
matter how Starways establishes its flights, the cities can always be partitioned into
k groups so that from any city it is not possible to reach another city in the same
group by using at most 28 flights.
Problem 5. Find all functions f : R+ → R+ such that
(z + 1)f (x + y) = f (xf (z) + y) + f (yf (z) + x),
for all positive real numbers x, y, z.
XXIX Asian Pacific Mathematics Olympiad
March, 2017
Time allowed: 4 hours
Each problem is worth 7 points
The contest problems are to be kept confidential until they are posted on the official
APMO website http://apmo.ommenlinea.org.
Please do not disclose nor discuss the problems over online until that date. The use
of calculators is not allowed.
Problem 1. We call a 5-tuple of integers arrangeable if its elements can be labeled
a, b, c, d, e in some order so that a − b + c − d + e = 29. Determine all 2017-tuples
of integers n1 , n2 , . . . , n2017 such that if we place them in a circle in clockwise order,
then any 5-tuple of numbers in consecutive positions on the circle is arrangeable.
Proposed by Warut Suksompong, Thailand
Problem 2. Let ABC be a triangle with AB < AC. Let D be the intersection point
of the internal bisector of angle BAC and the circumcircle of ABC. Let Z be the
intersection point of the perpendicular bisector of AC with the external bisector of
angle ∠BAC. Prove that the midpoint of the segment AB lies on the circumcircle of
triangle ADZ.
Proposed by Equipo Nicaragua, Nicaragua
Problem 3. Let A(n) denote the number of sequences a1 ≥ a2 ≥ . . . ≥ ak of positive
integers for which a1 + · · · + ak = n and each ai + 1 is a power of two (i = 1, 2, . . . , k).
Let B(n) denote the number of sequences b1 ≥ b2 ≥ . . . ≥ bm of positive integers for
which b1 + · · · + bm = n and each inequality bj ≥ 2bj+1 holds (j = 1, 2, . . . , m − 1).
Prove that A(n) = B(n) for every positive integer n.
Proposed by Senior Problems Committee of the Australian Mathematical Olympiad
Committee
k
Problem 4. Call a rational number r powerful if r can be expressed in the form pq
for some relatively prime positive integers p, q and some integer k > 1. Let a, b, c be
positive rational numbers such that abc = 1. Suppose there exist positive integers
x, y, z such that ax + by + cz is an integer. Prove that a, b, c are all powerful.
Proposed by Jeck Lim, Singapur
Problem 5. Let n be a positive integer. A pair of n-tuples (a1 , . . . , an ) and
(b1 , . . . , bn ) with integer entries is called an exquisite pair if
|a1 b1 + · · · + an bn | ≤ 1.
Determine the maximum number of distinct n-tuples with integer entries such
that any two of them form an exquisite pair.
Proposed by Pakawut Jiradilok and Warut Suksompong, Thailand
XXX Asian Pacific Mathematics Olympiad
March, 2018
Time allowed: 4 hours
Each problem is worth 7 points
The contest problems are to be kept confidential until they are posted on the official
APMO website http://apmo.ommenlinea.org.
Please do not disclose nor discuss the problems over online until that date. The use
of calculators is not allowed.
Problem 1. Let H be the orthocenter of the triangle ABC. Let M and N be
the midpoints of the sides AB and AC, respectively. Assume that H lies inside the
quadrilateral BM N C and that the circumcircles of triangles BM H and CN H are
tangent to each other. The line through H parallel to BC intersects the circumcircles
of the triangles BM H and CN H in the points K and L, respectively. Let F be the
intersection point of M K and N L and let J be the incenter of triangle M HN . Prove
that F J = F A.
Proposed by Mahdi Etesamifard, Iran
Problem 2. Let f (x) and g(x) be given by
f (x) =
and
g(x) =
1
1
1
1
+
+
+ ··· +
x x−2 x−4
x − 2018
1
1
1
1
+
+
+ ··· +
.
x−1 x−3 x−5
x − 2017
Prove that
|f (x) − g(x)| > 2
for any non-integer real number x satisfying 0 < x < 2018.
Proposed by Senior Problems Committee of the Australian Mathematical Olympiad
Committee
Problem 3. A collection of n squares on the plane is called tri-connected if the
following criteria are satisfied:
(i) All the squares are congruent.
(ii) If two squares have a point P in common, then P is a vertex of each of the
squares.
(iii) Each square touches exactly three other squares.
How many positive integers n are there with 2018 ≤ n ≤ 3018, such that there exists
a collection of n squares that is tri-connected?
Proposed by Senior Problems Committee of the Australian Mathematical Olympiad
Committee
Problem 4. Let ABC be an equilateral triangle. From the vertex A we draw a
ray towards the interior of the triangle such that the ray reaches one of the sides of
the triangle. When the ray reaches a side, it then bounces off following the law of
reflection, that is, if it arrives with a directed angle α, it leaves with a directed angle
180◦ − α. After n bounces, the ray returns to A without ever landing on any of the
other two vertices. Find all possible values of n.
Proposed by Daniel Perales and Jorge Garza, Mexico
Problem 5. Find all polynomials P (x) with integer coefficients such that for all real
numbers s and t, if P (s) and P (t) are both integers, then P (st) is also an integer.
Proposed by William Ting-Wei Chao, Taiwan
XXXI Asian Pacific Mathematics Olympiad
March, 2019
Time allowed: 4 hours
Each problem is worth 7 points
The contest problems are to be kept confidential until they are posted on the official
APMO website http://apmo.ommenlinea.org.
Please do not disclose nor discuss the problems over online until that date. The use
of calculators is not allowed.
Problem 1. Let Z+ be the set of positive integers. Determine all functions f : Z+ →
Z+ such that a2 + f (a)f (b) is divisible by f (a) + b for all positive integers a and b.
Problem 2. Let m be a fixed positive integer. The infinite sequence {an }n≥1 is
defined in the following way: a1 is a positive integer, and for every integer n ≥ 1 we
have
(
a2n + 2m if an < 2m
an+1 =
an /2
if an ≥ 2m .
For each m, determine all possible values of a1 such that every term in the sequence
is an integer.
Problem 3. Let ABC be a scalene triangle with circumcircle Γ. Let M be the
midpoint of BC. A variable point P is selected in the line segment AM . The
circumcircles of triangles BP M and CP M intersect Γ again at points D and E,
respectively. The lines DP and EP intersect (a second time) the circumcircles to
triangles CP M and BP M at X and Y , respectively. Prove that as P varies, the
circumcircle of 4AXY passes through a fixed point T distinct from A.
Problem 4. Consider a 2018 × 2019 board with integers in each unit square. Two
unit squares are said to be neighbours if they share a common edge. In each turn,
you choose some unit squares. Then for each chosen unit square the average of all
its neighbours is calculated. Finally, after these calculations are done, the number in
each chosen unit square is replaced by the corresponding average. Is it always possible
to make the numbers in all squares become the same after finitely many turns?
Problem 5. Determine all the functions f : R → R such that
f (x2 + f (y)) = f (f (x)) + f (y 2 ) + 2f (xy)
for all real numbers x and y.
XXXII Asian Pacific Mathematics Olympiad
March, 2020
Time allowed: 4 hours
Each problem is worth 7 points
The contest problems are to be kept confidential until they are posted on the official
APMO website http://apmo-official.org.
Please do not disclose nor discuss the problems online until that date. The use of
calculators is not allowed.
Problem 1. Let Γ be the circumcircle of ∆ABC. Let D be a point on the side BC.
The tangent to Γ at A intersects the parallel line to BA through D at point E. The
segment CE intersects Γ again at F . Suppose B, D, F, E are concyclic. Prove that
AC, BF, DE are concurrent.
Problem 2. Show that r = 2 is the largest real number r which satisfies the following
condition:
If a sequence a1 , a2 , . . . of positive integers fulfills the inequalities
p
an ≤ an+2 ≤ a2n + ran+1
for every positive integer n, then there exists a positive integer M such that an+2 = an
for every n ≥ M .
Problem 3. Determine all positive integers k for which there exist a positive integer
m and a set S of positive integers such that any integer n > m can be written as a
sum of distinct elements of S in exactly k ways.
Problem 4. Let Z denote the set of all integers. Find all polynomials P (x) with
integer coefficients that satisfy the following property:
For any infinite sequence a1 , a2 , . . . of integers in which each integer in Z appears
exactly once, there exist indices i < j and an integer k such that ai + ai+1 + · · · + aj =
P (k).
Problem 5. Let n ≥ 3 be a fixed integer. The number 1 is written n times on a
blackboard. Below the blackboard, there are two buckets that are initially empty.
A move consists of erasing two of the numbers a and b, replacing them with the
numbers 1 and a + b, then adding one stone to the first bucket and gcd(a, b) stones to
the second bucket. After some finite number of moves, there are s stones in the first
bucket and t stones in the second bucket, where s and t are positive integers. Find
t
all possible values of the ratio .
s
XXXIII Asian Pacific Mathematics Olympiad
March, 2021
Time allowed: 4 hours
Each problem is worth 7 points
The contest problems are to be kept confidential until they are posted on the official
APMO website http://apmo-official.org.
Please do not disclose nor discuss the problems online until that date. The use of
calculators is not allowed.
Problem 1. Prove that for each real number r > 2, there are exactly two or three
positive real numbers x satisfying the equation x2 = rbxc.
Note: bxc denotes the largest integer less than or equal to x.
Problem 2. For a polynomial P and a positive integer n, define Pn as the number
of positive integer pairs (a, b) such that a < b ≤ n and |P (a)| − |P (b)| is divisible by
n. Determine all polynomial P with integer coefficients such that Pn ≤ 2021 for all
positive integers n.
Problem 3. Let ABCD be a cyclic convex quadrilateral and Γ be its circumcircle.
Let E be the intersection of the diagonals AC and BD, let L be the center of the
circle tangent to sides AB, BC, and CD, and let M be the midpoint of the arc BC
of Γ not containing A and D. Prove that the excenter of triangle BCE opposite E
lies on the line LM .
Problem 4. Given a 32 × 32 table, we put a mouse (facing up) at the bottom left
cell and a piece of cheese at several other cells. The mouse then starts moving. It
moves forward except that when it reaches a piece of cheese, it eats a part of it, turns
right, and continues moving forward. We say that a subset of cells containing cheese
is good if, during this process, the mouse tastes each piece of cheese exactly once and
then falls off the table. Show that:
(a) No good subset consists of 888 cells.
(b) There exists a good subset consisting of at least 666 cells.
Problem 5. Determine all functions f : Z → Z such that f (f (a) − b) + bf (2a) is a
perfect square for all integers a and b.
XXXIV Asian Pacific Mathematics Olympiad
March, 2022
Time allowed: 4 hours
Each problem is worth 7 points
The contest problems are to be kept confidential until they are posted on the official
APMO website http://apmo-official.org.
Please do not disclose nor discuss the problems online until that date. The use of
calculators is not allowed.
Problem 1. Find all pairs (a, b) of positive integers such that a3 is a multiple of b2
and b − 1 is a multiple of a − 1.
Note: An integer n is said to be a multiple of an integer m if there is an integer k
such that n = km.
Problem 2. Let ABC be a right triangle with ∠B = 90◦ . Point D lies on the line
CB such that B is between D and C. Let E be the midpoint of AD and let F be
the second intersection point of the circumcircle of 4ACD and the circumcircle of
4BDE. Prove that as D varies, the line EF passes through a fixed point.
Problem 3. Find all positive integers k < 202 for which there exists a positive
integer n such that
n n o 2n kn
k
+
+ ··· +
= ,
202
202
202
2
where {x} denote the fractional part of x.
Note: The fractional part of a real number x is defined as the real number k with
0 ≤ k < 1 such that x − k is an integer.
Problem 4. Let n and k be positive integers. Cathy is playing the following game.
There are n marbles and k boxes, with the marbles labelled 1 to n. Initially, all
marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the
marbles with the smallest label, say i, to either any empty box or the box containing
marble i + 1. Cathy wins if at any point there is a box containing only marble n.
Determine all pairs of integers (n, k) such that Cathy can win this game.
Problem 5. Let a, b, c, d be real numbers such that a2 + b2 + c2 + d2 = 1. Determine
the minimum value of (a − b)(b − c)(c − d)(d − a) and determine all values of (a, b, c, d)
such that the minimum value is achieved.
1st APMO 1989
Problem 1
ai are positive reals. s = a1 + ... + an. Prove that for any integer n > 1 we have (1 + a1) ...
(1 + an) < 1 + s + s2/2! + ... + sn/n! .
Solution
We use induction on n. For n = 2 the rhs is 1 + a1 + a2 + a1a2 + (a12 + a22)/2 > lhs.
Assume the result is true for n. We note that, by the binomial theorem, for s and t
positive we have sm+1 + (m+1) t sm < (s + t)m+1, and hence sm+1/(m+1)! + t sm/m! < (s +
t)m+1/(m+1)! . Summing from m = 1 to n+1 we get (s + t) + (s2/2! + t s/1!) + (s3/3! + t
s2/2!) + ... + (sn+1/(n+1)! + t sn/n!) < (s + t) + (s + t)2/2! + ... + (s + t)n+1/(n+1)! . Adding 1
to each side gives that (1 + t)(1 + s + s2/2! + ... + sn/n!) < (1 + (s+t) + ... + (s+t)n+1/(n+1)!
. Finally putting t = an+1 and using the the result for n gives the result for n+1.
Problem 2
Prove that 5n2 = 36a2 + 18b2 + 6c2 has no integer solutions except a = b = c = n = 0.
Solution
The rhs is divisible by 3, so 3 must divide n. So 5n2 - 36a2 - 18b2 is divisible by 9, so 3
must divide c. We can now divide out the factor 9 to get: 5m2 = 4a2 + 2b2 + 6d2. Now
take m, a, b, d to be the solution with the smallest m, and consider residues mod 16.
Squares = 0, 1, 4, or 9 mod 16. Clearly m is even so 5m2 = 0 or 4 mod 16. Similarly,
4a2 = 0 or 4 mod 16. Hence 2b2 + 6d2 = 0, 4 or 12 mod 16. But 2b2 = 0, 2 or 8 mod 16
and 6d2 = 0, 6 or 8 mod 16. Hence 2b2 + 6d2 = 0, 2, 6, 8, 10 or 14 mod 16. So it must be
0. So b and d are both even. So a cannot be even, otherwise m/2, a/2, b/2, d/2 would be a
solution with smaller m/2 < m.
So we can divide out the factor 4 and get: 5k2 = a2 + 2e2 + 6f2 with a odd. Hence k is
also odd. So 5k2 - a2 = 4 or 12 mod 16. But we have just seen that 2e2 + 6 f2 cannot be 4
or 12 mod 16. So there are no solutions.
Problem 3
ABC is a triangle. X lies on the segment AB so that AX/AB = 1/4. CX intersects the
median from A at A' and the median from B at B''. Points B', C', A'', C'' are defined
similarly. Find the area of the triangle A''B''C'' divided by the area of the triangle A'B'C'.
Solution
Answer: 25/49.
Let M be the midpoint of AB. We use vectors center G. Take GA = A, GB = B, GC = C.
Then GM = A/2 + B/2 and GX = 3/4 A + 1/4 C. Hence GA' = 2/5 A (showing it lies on
GA) = 4/5 (3/4 A + 1/4 B) + 1/5 C, since A + B + C = 0 (which shows it lies on CX).
Similarly, GB'' = 4/7 (1/2 A + 1/2 C) (showing it lies on the median through B) =
2/7 A + 2/7 C = 5/7 (2/5 A) + 2/7 C (showing it lies on CA' and hence on CX). Hence
GB'' = -2/7 B. So we have shown that GB'' is parallel to GB' and 5/7 the length. The
same applies to the distances from the centroid to the other vertices. Hence triangle
A''B''C'' is similar to triangle A'B'C' and its area is 25/49 times the area of A'B'C'.
Problem 4
Show that a graph with n vertices and k edges has at least k(4k - n2)/3n triangles.
Solution
Label the points 1, 2, ... , n and let point i have degree di (no. of edges). Then if i and j
are joined they have at least di + dj - 2 other edges between them, and these edges join
them to n - 2 other points. So there must be at least di + dj - n triangles which have i and j
as two vertices. Hence the total number of triangles must be at least ∑edges ij (di + dj n)/3. But ∑edges ij (di + dj) = ∑ di2, because each point i occurs in just di terms. Thus the
total number of triangles is at least (∑ di2)/3 - nk/3. But ∑ di2 ≥ (∑ di) 2/n (a special case
of Chebyshev's inequality) = 4k2/n. Hence result.
Problem 5
f is a strictly increasing real-valued function on the reals. It has inverse f-1. Find all
possible f such that f(x) + f-1(x) = 2x for all x.
Solution
Answer: f(x) = x + b for some fixed real b.
Suppose for some a we have f(a) ≠ a. Then for some b ≠ 0 we have f(a) = a + b. Hence
f(a + b) = a + 2b (because f( f(a) ) + f-1( f(a) ) = 2 f(a), so f(a + b) + a = 2a + 2b ) and by
two easy inductions, f(a + nb) = a + (n+1)b for all integers n (positive or negative).
Now take any x between a and a + b. Suppose f(x) = x + c. The same argument shows
that f(x + nc) = x + (n+1)c. Since f is strictly increasing x + c must lie between f(a) = a +
b and f(a+b) = a + 2b. So by a simple induction x + nc must lie between a + nb and a +
(n+1)b. So c lies between b + (x-a)/n and b + (a+b-x)/n or all n. Hence c = b. Hence f(x)
= x + b for all x.
If there is no a for which f(a) ≠ a, then we have f(x) = x for all x.
2nd APMO 1990
Problem 1
Given θ in the range (0, π) how many (incongruent) triangles ABC have angle A = θ, BC
= 1, and the following four points concyclic: A, the centroid, the midpoint of AB and the
midpoint of AC?
Solution
Answer: 1 for θ ≤ 60 deg. Otherwise none.
Let O be the circumcenter of ABC and R the circumradius, let M be the midpoint of BC,
and let G be the centroid. We may regard A as free to move on the circumcircle, whilst
O, B and C remain fixed. Let X be the point on MO such that MX/MO = 1/3. An
expansion by a factor 3, center M, takes G to A and X to O, so G must lie on the circle
center X radius R/3.
The circle on diameter OA contains the midpoints of AB and AC (since if Z is one of the
midpoints OZ is perpendicular to the corresponding side). So if G also lies on this circle
then angle OGA = 90 deg and hence angle MGO = 90 deg, so G must also lie on the
circle diameter OM. Clearly the two circles for G either do not intersect in which case no
triangle is possible which satisfies the condition or they intersect in one or two points.
But if they intersect in two points, then corresponding triangles are obviously congruent
(they just interchange B and C). So we have to find when the two circle intersect.
Let the circle center X meet the line OXM at P and Q with P on the same side of X as M.
Now OM = R cos θ, so XM = 1/3 R cos θ < 1/3 R = XP, so M always lies inside PQ.
Now XO = 2/3 OM = 1/3 R (2 cos θ), so XQ = 1/3 R > XO iff 2 cos θ < 1 or θ > π/3.
Thus if θ > π/3, then XQ > XO and so the circle diameter OM lies entirely inside the
circle center X radius R/3 and so they cannot intersect. If θ = π/3, then the circles touch
at O, giving the equilateral triangle as a solution. If θ < π/3, then the circles intersect
giving one incongruent triangle satisfying the condition.
Problem 2
x1, ... , xn are positive reals. sk is the sum of all products of k of the xi (for example, if n =
3, s1 = x1 + x2 + x3, s2 = x1x2 + x2x3 + x3x1, s3 = x1x2x3). Show that sksn-k ≥ (nCk)2 sn for 0
< k < n.
Solution
Each of sk and sn-khave nCk terms. So we may multiply out the product sksn-k to get a
sum of (nCk)2 terms. We now apply the arithmetic/geometric mean result. The product
of all the terms must be a power of sn by symmetry and hence must be sn to the power of
(nCk)2. So the geometric mean of the terms is just sn. Hence result.
Problem 3
A triangle ABC has base AB = 1 and the altitude from C length h. What is the maximum
possible product of the three altitudes? For which triangles is it achieved?
Solution
Answer: for h ≤ 1/2, maximum product is h2, achieved by a triangle with right-angle at
C; for h > 1/2, the maximum product is h3/(h2 + 1/4), achieved by the isosceles triangle
(AC = BC).
Solution by David Krumm
Let AC = b, BC = a, let the altitude from A have length x and the altitude from B have
length y. Then ax = by = h, so hxy = h3/ab. But h = a sin B and b/sin B = 1/sin C, so h =
ab sin C and the product hxy = h2 sin C.
The locus of possible positions for C is the line parallel to AB and a distance h from it.
[Or strictly the pair of such lines.] If h ≤ 1/2, then there is a point on that line with angle
ACB = 90 deg, so in this case we can obtain hxy = h2 by taking angle ACB = 90 deg and
that is clearly the best possible.
If h > 1/2, then there is no point on the line with angle ACB = 90 deg. Let L be the
perpendicular bisector of AB and let L meet the locus at C. Then C is the point on the
locus with the angle C a maximum. For if D is any other point of the line then the
circumcircle of ABD also passes through the corresponding point D' on the other side of
C and hence C lies inside the circumcircle. If L meets the circumcircle at C', then angle
ADB = angle AC'B > angle ACB. Evidently sin C = 2 sin C/2 cos C/2 = h/(h2 + 1/4), so
the maximum value of hxy is h3/(h2 + 1/4).
My original, less elegant, solution is as follows.
Take AP perpendicular to AB and length h. Take Q to be on the line parallel to AB
through P so that BQ is perpendicular to AB. Then C must lie on the line PQ (or on the
corresponding line on the other side of AB). Let a(A) be the length of the altitude from
A to BC and a(B) the length of the altitude from B to AC. If C maximises the product h
a(A) a(B), then it must lie on the segment PQ, for if angle ABC is obtuse, then both a(A)
and a(B) are shorter than for ABQ. Similarly if BAC is obtuse. So suppose PC = x with
0 ≤ x ≤ 1. Then AC = √(x2 + h2), so a(B) = h/√(x2 + h2). Similarly, a(A) = h/√( (1-x)2 +
h2). So we wish to minimise f(x) = (x2 + h2)( (1-x)2 + h2) = x4 - 2x3 + (2h2 + 1)x2 - 2h2x +
h4 + h2. We have f '(x) = 2(2x-1)(x2 - x + h2), which has roots x = 1/2, 1/2 ± √(1/4 - h2).
Thus for h >= 1/2, the minimum is at x = 1/2, in which case CA = CB. For h < 1/2, the
minimum is at x = 1/2 ± √(1/4 - h2). But if M is the midpoint of AB and D is the point on
AB with AD = 1/2 ± √(1/4 - h2), then DM = √(1/4 - h2). But DC = h, and angle CDM =
90, so MC = 1/2 and hence angle ACB = 90.
(DK's solution added 6 Jun 2002)
Problem 4
A graph with n points satisfies the following conditions: (1) no point has edges to all the
other points, (2) there are no triangles, (3) given any two points A, B such that there is
no edge AB, there is exactly one point C such that there are edges AC and BC. Prove
that each point has the same number of edges. Find the smallest possible n.
Solution
Answer: 5.
We say A and B are joined if there is an edge AB. For any point X we write deg X for
the number of points joined to X. Take any point A. Suppose deg A = m. So there are m
points B1, B2, ... , Bm joined to A. No Bi, Bj can be joined for i ≠ j, by (2), and a point C
≠ A cannot be joined to Bi and Bj for i ≠ j, by (3). Hence there are deg Bi - 1 points
Cij joined to Bi and all the Cij are distinct.
Now the only points that can be joined to Cij, apart from Bi, are other Chk, for by (3) any
point of the graph is connected to A by a path of length 1 or 2. But Cij cannot be joined
to Cik, by (2), and it cannot be joined to two distinct points Ckh and Ckh' by (3), so it is
joined to at most one point Ckh for each k ≠ i. But by (3) there must be a point X joined
to both Bk and Cij (for k ≠ i), and the only points joined to Bk are A and Ckh. Hence
Cij must be joined to at least one point Ckh for each k ≠ i. Hence deg Cij = m.
But now if we started with Bi instead of A and repeated the whole argument we would
establish that deg Bi is the same as the deg Chk, where Chk is one of the points joined to
Ci1. Thus all the points have the same degree.
Suppose the degree of each point is m. Then with the notation above there is 1 point A,
m points Bi and m(m-1) points Cjk or m2 + 1 in all. So n = m2 + 1. The smallest possible
m is 1, but that does not yield a valid graph because if does not satisfy (1). The next
smallest possibility is m = 2, giving 5 points. It is easy to check that the pentagon
satisfies all the conditions.
Problem 5
Show that for any n ≥ 6 we can find a convex hexagon which can be divided into n
congruent triangles.
Solution
We use an isosceles trianglea as the unit. The diagram shows n = 4 and n = 5. We can
get any n ≥ 4 by adding additional rhombi in the middle.
3rd APMO 1991
Problem 1
ABC is a triangle. G is the centroid. The line parallel to BC through G meets AB at B'
and AC at C'. Let A'' be the midpoint of BC, C'' the intersection of B'C and BG, and B''
the intersection of C'B and CG. Prove that A''B''C'' is similar to ABC.
Solution
Let M be the midpoint of AB and N the midpoint of AC. Let A''M meet BG at X. Then
X must be the midpoint of A''M (an expansion by a factor 2 center B takes A''M to CA
and X to N). Also BX/BN = 1/2 and BG/BN = 2/3, so XG = BX/3. Let the ray CX meet
AB at Z. Then ZX = CX/3. (There must be a neat geometric argument for this, but if we
take vectors origin B, then BX = BN/2 = BA/4 + BC/4, so BZ= BA/3 and so XZ = 1/3
(BA/4 - 3BC/4) = CX/3.) So now triangles BXC and ZXG are similar, so ZG is parallel
to BC, so Z is B' and X is C''. But A''X is parallel to AC and 1/4 its length, so A''C'' is
parallel to AC and 1/4 its length. Similarly A''B'' is parallel to AB and 1/4 its length.
Hence A''B''C'' is similar to ABC.
Problem 2
There are 997 points in the plane. Show that they have at least 1991 distinct midpoints.
Is it possible to have exactly 1991 midpoints?
Solution
Answer: yes. Take the 997 points collinear at coordinates x = 1, 3, ... , 1993. The
midpoints are 2, 3, 4, ... , 1992.
Take two points A and B which are the maximum distance apart. Now consider the
following midpoints: M, the midpoint of AB, the midpoint of each AX for any other X
in the set (not A or B), and the midpoint of each BX. We claim that all these are distinct.
Suppose X and Y are two other points (apart from A and B). Clearly the midpoints of
AX and AY must be distinct (otherwise X and Y would coincide). Similarly the
midpoints of BX and BY must be distinct. Equally, the midpoint of AX cannot be M (or
X would coincide with B), nor can the midpoint of BX be M. Suppose, finally, that N is
the midpoint of AX and BY. Then AYXB is a parallelogram and either AX or BY must
exceed AB, contradicting the maximality of AB. So we have found 1991 distinct
midpoints. The example above shows that there can be exactly 1991 midpoints.
Problem 3
xi and yi are positive reals with ∑1n xi = ∑1n yi. Show that ∑1n xi2/(xi + yi) ≥ (∑1n xi)/2.
Solution
We use Cauchy-Schwartz: ∑ (x/√(x+y) )2 ∑ (√(x+y) )2 ≥ (∑ x )2. So ∑ x2/(x+y) >= (∑
x)2/(∑(x+y) = 1/2 ∑ x.
Problem 4
A sequence of values in the range 0, 1, 2, ... , k-1 is defined as follows: a1 = 1, an = an-1 +
n (mod k). For which k does the sequence assume all k possible values?
Solution
Let f(n) = n(n+1)/2, so an = f(n) mod k. If k is odd, then f(n+k) = f(n) mod k, so the
sequence can only assume all possible values if a1, ... , ak are all distinct. But f(k-n) =
f(n) mod k, so there are at most (k+1)/2 distinct values. Thus k odd does not work.
If k is even, then f(n+2k) = f(n) mod k, so we need only look at the first 2k values. But
f((2k-1-n) = f(n) mod k and f(2k-1) = 0 mod k, so the sequence assumes all values iff a1,
a2, ... , ak-1 assume all the values 1, 2, ... , k-1.
Checking the first few, we find k = 2, 4, 8, 16 work and k = 6, 10, 12, 14 do not. So this
suggests that k must be a power of 2. Suppose k is a power of 2. If f(r) = f(s) mod k for
some 0 < r, s < k, then (r - s)(r + s + 1) = 0 mod k. But each factor is < k, so neither can
be divisible by k. Hence both must be even. But that is impossible (because their sum is
2r+1 which is odd), so each of f(1), f(2), ... , f(k-1) must be distinct residues mod k.
Obviously none can be 0 mod k (since 2k cannot divide r(r+1) for 0 < r < k and so k
cannot divide f(r) ). Thus they must include all the residues 1, 2, ... k-1. So k a power of
2 does work.
Now suppose h divides k and k works. If f(n) = a mod k, then f(n) = a mod h, so h must
also work. Since odd numbers do not work, that implies that k cannot have any odd
factors. So if k works it must be a power of 2.
Problem 5
Circles C and C' both touch the line AB at B. Show how to construct all possible circles
which touch C and C' and pass through A.
Solution
Take a common tangent touching C' at Q' and C at Q. Let the line from Q to A meet C
again at P. Let the line from Q' to A meet C' again at P'. Let the C have center O and C'
have center O'. Let the lines OP and O'P' meet at X. Take X as the center of the required
circle. There are two common tangents, so this gives two circles, one enclosing C and C'
and one not.
To see that this construction works, invert wrt the circle on center A through B. C and C'
go to themselves under the inversion. The common tangent goes to a circle through A
touching C and C'. Hence the point at which it touches C must be P and the point at
which it touches C' must be P'.
4th APMO 1992
Problem 1
A triangle has sides a, b, c. Construct another triangle sides (-a + b + c)/2, (a - b + c)/2,
(a + b - c)/2. For which triangles can this process be repeated arbitrarily many times?
Solution
Answer: equilateral.
We may ignore the factor 1/2, since clearly a triangle with sides x, y, z can be
constructed iff a triangle with sides 2x, 2y, 2z can be constructed.
The advantage of considering the process as generating (-a + b + c), (a - b + c), (a + b c) from a, b, c is that the sum of the sides remains unchanged at a + b + c, so we can
focus on just one of the three sides. Thus we are looking at the sequence a, (a + b + c) 2a, a + b + c - 2(-a + b + c), ... . Let d = 2a - b - c. We show that the process generates
the sequence a, a - d, a + d, a - 3d, a + 5d, a - 11d, a + 21d, ... . Let the nth term be a + (1)nand. We claim that an+1 = 2an + (-1)n. This is an easy induction, for we have a + (1)n+1an+1d = a + b + c - 2(a + (-1)nand) and hence (-1)n+1an+1d = -d - 2(-1)nand, and hence
an+1= 2an + (-1)n. But this shows that an is unbounded. Hence if d is non-zero then the
process ultimately generates a negative number. Thus a necessary condition for the
process to generate triangles indefinitely is that 2a = b + c. Similarly, 2b = c + a is a
necessary condition. But these two equations imply (subtracting) a = b and hence a = c.
So a necessary condition is that the triangle is equilateral. But this is obviously also
sufficient.
Problem 2
Given a circle C centre O. A circle C' has centre X inside C and touches C at A. Another
circle has centre Y inside C and touches C at B and touches C' at Z. Prove that the lines
XB, YA and OZ are concurrent.
Solution
We need Ceva's theorem, which states that given points D, E, F on the lines BC, CA,
AB, the lines AD, BE, CF are concurrent iff (BD/DC) (CE/EA) (AF/FB) = 1 (where we
pay attention to the signs of BD etc, so that BD is negative if D lies on the opposite side
of B to C). Here we look at the triangle OXY, and the points A on OX, B on OY and Z
on XY (it is obvious that Z does lie on XY). We need to consider (OA/AX) (XZ/ZY)
(YB/BO). AX and BY are negative and the other distances positive, so the sign is plus.
Also OA = OB, AX = XZ, and ZY = YB (ignoring signs), so the expression is 1. Hence
AY, XB and OZ are concurrent as required.
Problem 3
Given three positive integers a, b, c, we can derive 8 numbers using one addition and
one multiplication and using each number just once: a+b+c, a+bc, b+ac, c+ab, (a+b)c,
(b+c)a, (c+a)b, abc. Show that if a, b, c are distinct positive integers such that n/2 < a, b,
c, <= n, then the 8 derived numbers are all different. Show that if p is prime and n ≥ p2,
then there are just d(p-1) ways of choosing two distinct numbers b, c from {p+1, p+2, ...
, n} so that the 8 numbers derived from p, b, c are not all distinct, where d(p-1) is the
number of positive divisors of p-1.
Solution
If 1 < a < b < c, we have a + b + c < ab + c < b + ac < a + bc and (b+c)a < (a+c)b <
(a+b)c < abc. We also have b + ac < (a+c)b. So we just have to consider whether a + bc
= (b+c)a. But if a > c/2, which is certainly the case if n/2 < a, b, c ≤ n, then a(b + c - 1) >
c/2 (b + b) = bc, so a + bc < a(b + c) and all 8 numbers are different.
The numbers are not all distinct iff p + bc = (b + c)p. Put b = p + d. Then c = p(p-1)/d +
p. Now we are assuming that b < c, so p + d < p(p-1)/d + p, hence d2 < p(p-1), so d < p.
But p is prime so d cannot divide p, so it must divide p-1. So we get exactly d(p-1)
solutions provided that all the c ≤ n. The largest c is that corresponding to d = 1 and is
p(p-1) + p = p2 ≤ n.
Problem 4
Find all possible pairs of positive integers (m, n) so that if you draw n lines which
intersect in n(n-1)/2 distinct points and m parallel lines which meet the n lines in a
further mn points (distinct from each other and from the first n(n-1)/2) points, then you
form exactly 1992 regions.
Solution
Answer: (1, 995), (10, 176), (21, 80).
n lines in general position divide the plane into n(n+1)/2 + 1 regions and each of the m
parallel lines adds a further n+1 regions. So we require n(n+1)/2 + 1 + m(n+1) = 1992 or
(n+1)(2m+n) = 3982 = 2·11·181. So n+1 must divide 3982, also (n+1)n < 3982, so n ≤
62. We are also told that n is positive Thus n = 0 is disallowed. The remaining
possibilities are n+1 = 2, 11, 2·11. These give the three solutions shown above.
Problem 5
a1, a2, a3, ... an is a sequence of non-zero integers such that the sum of any 7 consecutive
terms is positive and the sum of any 11 consecutive terms is negative. What is the largest
possible value for n?
Solution
Answer: 16.
We cannot have 17 terms, because then:
a1 +
a2 +
a3 +
...
a7 +
a2 + ... + a11 < 0
a3 + ... + a12 < 0
a4 + ... + a13 < 0
a8 + ... + a17 < 0
So if we add the inequalities we get that an expression is negative. But notice that each
column is positive. Contradiction.
On the other hand, a valid sequence of 16 terms is: -5, -5, 13, -5, -5, -5, 13, -5, -5, 13, -5,
-5, -5, 13, -5, -5. Any run of 7 terms has two 13s and five -5s, so sums to 1. Any run of
11 terms has three 13s and eight -5s, so sums to -1.
5th APMO 1993
Problem 1
A, B, C is a triangle. X, Y, Z lie on the sides BC, CA, AB respectively, so that AYZ and
XYZ are equilateral. BY and CZ meet at K. Prove that YZ2 = YK·YB.
Solution
Use vectors. Take A as the origin. Let AZ = b, AY = c. We may take the equilateral
triangles to have side 1, so b2 = c2 = 1 and b.c = 1/2. Take AB to be k b. AX is b + c,
so AC must be k/(k-1) c (then AX = 1/k (k b) + (1 - 1/k) ( k/(k-1) c), which shows that X
lies on BC).
Hence AK = k/(k2 - k + 1) (b + (k-1) c). Writing this as (k2-k)/(k2-k+1) c + 1/(k2-k+1)
(k b) shows that it lies on BY and writing it as k/(k2-k+1) b + (k2-2k+1) ( k/(k-1) c)
shows that it lies on CZ. Hence YK.YB = YK.YB= ( k/(k2-k+1) b - 1/(k2-k+1) c) . (
k b - c) = (k b - c)2/(k2-k+1) = 1 = YZ2.
Thank to Achilleas Porfyriadis for the following geometric proof
BZX and XYC are similar (sides parallel), so BZ/ZX = XY/YC. But XYZ is equilateral,
so BZ/ZY = ZY/YC. Also ∠BZY = ∠ZYC = 120o, so BZY and ZYC are similar. Hence
∠ZBY = ∠YZC. Hence YZ is tangent to the circle ZBK. Hence YZ2 = YK·YB
Problem 2
How many different values are taken by the expression [x] + [2x] + [5x/3] + [3x]+ [4x]
for real x in the range 0 ≤ x ≤ 100?
Solution
Answer: 734.
Let f(x) = [x] + [2x] + [3x] + [4x] and g(x) = f(x) + [5x/3]. Since [y+n] = n + [y] for any
integer n and real y, we have that f(x+1) = f(x) + 10. So for f it is sufficient to look at the
half-open interval [0, 1). f is obviously monotonic increasing and its value jumps at x =
0, 1/4, 1/3, 1/2, 2/3, 3/4. Thus f(x) takes 6 different values on [0, 1).
g(x+3) = g(x), so for g we need to look at the half-open interval [0, 3). g jumps at the
points at which f jumps plus 4 additional points: 3/5, 1 1/5, 1 4/5, 2 2/5. So on [0, 3),
g(x) takes 3 x 6 + 4 = 22 different values. Hence on [0, 99), g(x) takes 33 x 22 = 726
different values. Then on [99, 100] it takes a further 6 + 1 + 1 (namely g(99), g(99 1/4),
g(99 1/3), g(99 1/2), g(99 3/5), g(99 2/3), g(99 3/4), g(100) ). Thus in total g takes 726 +
8 = 734 different values.
Problem 3
p(x) = (x + a) q(x) is a real polynomial of degree n. The largest absolute value of the
coefficients of p(x) is h and the largest absolute value of the coefficients of q(x) is k.
Prove that k ≤ hn.
Solution
Let p(x) = p0 + p1x + ... + pnxn, q(x) = q0 + q1x + ... + qn-1xn-1, so h = max |pi|, k = max
|qi|.
If a = 0, then the result is trivial. So assume a is non-zero. We have pn = qn-1, pn-1 = qn-2 +
aqn-1, pn-2 = qn-3 + aqn-2, ... , p1 = q0 + aq1, p0 = aq0.
We consider two cases. Suppose first that |a| ≥ 1. Then we show by induction that |qi| ≤
(i+1) h. We have q0 = p0/a, so |q0| ≤ h, which establishes the result for i = 0. Suppose it is
true for i. We have qi+1 = (pi+1 - qi)/a, so |qi+1| ≤ |pi+1| + |qi| ≤ h + (i+1)h = (i+2)h, so it is
true for i+1. Hence it is true for all i < n. So k ≤ max(h, 2h, ... , nh) = nh.
The remaining possibility is 0 < |a| < 1. In this case we show by induction that |qn-i| ≤ ih.
We have qn-1 = pn, so |qn-1| ≤ |pn| ≤ h, which establishes the result for i = 1. Suppose it is
true for i. We have qn-i-1 = pn-i - aqn-i, so |qn-i-1n-i| + |qn-i| ≤ h + ih = (i+1)h, so it is true for
i+1. Hence it is true for all 1 ≤ i ≤ n. Hence k ≤ max(h, 2h, ... , nh) = nh.
Problem 4
Find all positive integers n for which xn + (x+2)n + (2-x)n = 0 has an integral solution.
Solution
Answer: n = 1.
There are obviously no solutions for even n, because then all terms are non-negative and
at least one is positive. x = -4 is a solution for n = 1. So suppose n is odd n and > 3.
If x is positive, then xn + (x+2)n > (x+2)n > (x-2)n, so xn + (x+2)n + (2-x)n > 0. Hence any
solution x must be negative. Put x = -y. Clearly x = -1 is not a solution for any n, so if x
= -y is a solution then (x+2) = -(y-2) ≤ 0 we have (y+2)n = yn + (y-2)n. Now 4 = ( (y+2) (y-2) ) divides (y+2)n - (y-2)n. Hence 2 divides y. Put y = 2z, then we have (z+1)n = zn +
(z-1)n. Now 2 divides (z+1)n - (z-1)n so 2 divides z, so z+1 and z-1 are both odd. But an bn = (a - b)(an-1n-2b + an-3b2 + ... + bn-1). If a and b are both odd, then each term in (an-1n2b + an-3b2 + ... + bn-1) is odd and since n is odd there are an odd number of terms, so (an1
n-2b + an-3b2 + ... + bn-1) is odd. Hence, putting a=z+1, b=z-1, we see that (z+1)n - (z1)n = 2(an-1n-2b + an-3b2 + ... + bn-1) is not divisible by 4. But it equals zn with z even.
Hence n must be 1.
Problem 5
C is a 1993-gon of lattice points in the plane (not necessarily convex). Each side of C
has no lattice points except the two vertices. Prove that at least one side contains a point
(x, y) with 2x and 2y both odd integers.
Solution
We consider the midpoint of each side. We say that a vertex (x, y) is pure if x and y have
the same parity and impure if x and y have opposite parity. Since the total number of
vertices is odd, there must be two adjacent pure vertices P and Q or two adjacent impure
vertices P and Q. But in either case the midpoint of P and Q either has both coordinates
integers, which we are told does not happen, or as both coordinates of the form an
integer plus half, which therefore must occur.
6th APMO 1994
Problem 1
Find all real-valued functions f on the reals such that (1) f(1) = 1, (2) f(-1) = -1, (3) f(x) ≤
f(0) for 0 < x < 1, (4) f(x + y) ≥ f(x) + f(y) for all x, y, (5) f(x + y) ≤ f(x) + f(y) + 1 for all
x, y.
Solution
Answer: f(x) = [x].
f(x+1) >= f(x) + f(1) = f(x) + 1 by (4) and (1). But f(x) ≥ f(x+1) + f(-1) = f(x+1) - 1 by
(4) and (2). Hence f(x+1) = f(x) + 1.
In particular, 1 = f(1) = f(0+1) = f(0) + 1, so f(0) = 0. Hence, by (3), f(x) ≤ 0 for 0 < x <
1. But, by (5), 1 = f(1) = f(x + 1-x) ≤ f(x) + f(1-x) + 1, so f(x) + f(1-x) ≥ 0. But if 0 < x <
1, then also 0 < 1-x < 1, so f(x) = f(1-x) = 0.
Thus we have established that f(x) = 0 for 0 ≤ x < 1, and f(x+1) = f(x) + 1. It follows that
f(x) = [x] for all x.
Problem 2
ABC is a triangle and A, B, C are not collinear. Prove that the distance between the
orthocenter and the circumcenter is less than three times the circumradius.
Solution
We use vectors. It is well-known that the circumcenter O, the centroid G and the
orthocenter H lie on the Euler line and that OH = 3 OG. Hence taking vectors with origin
O, OH = 3 OG = OA + OB + OC. Hence |OH| ≤ |OA| + |OB| + |OC| = 3 x circumradius.
We could have equality only if ABC were collinear, but that is impossible, because ABC
would not then be a triangle.
Problem 3
Find all positive integers n such that n = a2 + b2, where a and b are relatively prime
positive integers, and every prime not exceeding √n divides ab.
Solution
Answer: 2 = 12 + 12, 5 = 12 + 22, 13 = 22 + 32.
The key is to use the fact that a and b are relatively prime. We show in fact that they
must differ by 1 (or 0). Suppose first that a = b. Then since they are relatively prime they
must both be 1. That gives the first answer above. So we may take a > b. Then (a - b)2 <
a2 + b2 = n, so if a - b is not 1, it must have a prime factor which divides ab. But then it
must divide a or b and hence both. Contradiction. So a = b + 1.
Now (b - 1)2 < b2 < n, so any prime factor p of b - 1 must divide ab = b(b + 1). It cannot
divide b (or it would divide b and b - 1 and hence 1), so it must divide b + 1 and hence
must be 2. But if 4 divides b - 1, then 4 does not divide b(b - 1), so b - 1 must be 0, 1 or
2. But it is now easy to check the cases a, b = (4, 3), (3, 2), (2, 1).
Problem 4
Can you find infinitely many points in the plane such that the distance between any two
is rational and no three are collinear?
Solution
Answer: yes.
Let θ = cos-13/5. Take a circle center O radius 1 and a point X on the circle. Take Pn on
the circle such that angle XOPn = 2nθ. We establish (A) that the Pn are all distinct and
(B) that the distances PmPn are all rational.
We establish first that we can express 2 cos nx as a polynomial of degree n in (2 cos x)
with integer coefficients and leading coefficient 1. For example, 2 cos 2x = (2 cos x)2 1, 2 cos 3x = (2 cos x)3 - 3 (2 cos x). We proceed by induction. Let the polynomial be
pn(2 cos x). We have that p1(y) = y and p2(y) = y2 - 1. Suppose we have found pm for m ≤
n. Now cos(n+1)x = cos nx cos x - sin nx sin x, and cos(n-1)x = cos nx cos x + sin nx sin
x, so cos(n+1)x = 2 cos x cos nx - cos(n-1)x. Hence pn+1(y) = y pn(y) - pn-1(y). Hence the
result is also true for n+1.
It follows that (1) if cos x is rational, then so is cos nx, and (2) if cos x is rational, then
x/π is irrational. To see (2), suppose that x/π = m/n, with m and n integers. Then nx is a
multiple of π and hence cos nx = 0, so pn(2 cos x) = 0. Now we may write pn(y) = yn +
an-1yn-1 + ... + a0. Now if also cos x = r/s, with r and s relatively prime integers, then we
have, pn(2 cos x) = rn + an-1s rn-1 + ... + a0sn = 0. But now s divides all terms except the
first. Contradiction.
Thus we cannot have cos mθ = cos nθ for any distinct integers m, n, for then θ/π would
be rational as well as cos θ. So we have established (A).
We have also established that all cos nθ are rational. But since sin(n+1)x = sin nx cos x
+ cos nx sin x and sin θ = 4/5, it is a trivial induction that all sin nθ are also rational.
Now PmPn = 2 |sin(m - n)θ|, so all the distances PmPn are rational, thus establishing (B).
Problem 5
Prove that for any n > 1 there is either a power of 10 with n digits in base 2 or a power of
10 with n digits in base 5, but not both.
Solution
10k has n digits in base 5 iff 5n-1 < 10k < 5n. Similarly, 10h has n digits in base 2 iff 2n-1 <
10h < 2n. So if we can find both 10k with n digits in base 5 and 10h with n digits in base
2, then, multiplying the two inequalities, we have 10n-1 < 10h+k < 10n, which is clearly
impossible. This establishes the "but not both" part.
Let S be the set of all positive powers of 2 or 5. Order the members of S in the usual way
and let an be the n-1th member of the set. We claim that if an = 2k, then 10k has n digits
in base 5, and if an = 5h, then 10h has n digits in base 2. We use induction on n.
a2 = 21, a3 = 22, a4 = 51, a5 = 23, ... . Thus the claim is certainly true for n = 2. Suppose it
is true for n.
Note that 10k has n digits in base 5 iff 5n-k-1 < 2k < 5n-k. Similarly, 10h has n digits in base
2 iff 2n-h-1 < 5h < 2n-h. There are 3 cases. Case (1). an = 2k and an+1 = 2k+1. Hence 10k+1 has
n+1 digits in base 5. Case (2). an = 2k and an+1 is a power of 5. Hence an+1 must be 5n-k.
Hence 2k < 5n-k < 2k+1. Hence 2n < 10n-k < 2n+1. So 10n-k has n+1 digits in base 2. Case
(3). an = 5h. Since there is always a power of 2 between two powers of 5, an+1 must be a
power of 2. Hence it must be 2n-h. So we have 5h < 2n-h < 5h+1. So 5n < 10n-h < 5n+1 and
hence 10n-h has n+1 digits in base 5.
Jacob Tsimerman pointed out that the second part can be done in a similar way to the
first - which is neater than the above:
If no power of 10 has n digits in base 2 or 5, then for some h, k: 10h < 2n-1 < 2n <
10h+1 and 10k < 5n-1 < 5n < 10k+1. Hence 10h+k < 10n-1 < 10n < 10h+k+2. But there is only
one power of 10 between h+k and h+k+2.
7th APMO 1995
Problem 1
Find all real sequences x1, x2, ... , x1995 which satisfy 2√(xn - n + 1) ≥ xn+1 - n + 1 for n =
1, 2, ... , 1994, and 2√(x1995 - 1994) ≥ x1 + 1.
Solution
Answer: the only such sequence is 1, 2, 3, ... , 1995.
Put x1995 = 1995 + k. We show by induction (moving downwards from 1995) that xn ≥ n
+ k. For suppose xn+1 ≥ n + k + 1, then 4(xn - n + 1) ≥ (xn+1- n + 1)2 ≥ (k+2)2 ≥ 4k + 4, so
xn ≥ n + k. So the result is true for all n ≥ 1. In particular, x1 ≥ 1 + k. Hence 4(x1995 1994) = 4(1 + k) ≥ (2 + k)2 = 4 + 4k + k2, so k2 ≤ 0, so k = 0.
Hence also xn ≥ n for n = 1, 2, ... , 1994. But now if xn = n + k, with k > 0, for some n <
1995, then the same argument shows that x1 ≥ 1 + k and hence 4 = 4(x1995 - 1994) ≥
(x1 + 1)2 ≥ (2 + k)2 = 4 + 4k + k2> 4. Contradiction. Hence xn = n for all n ≤ 1995.
Problem 2
Find the smallest n such that any sequence a1, a2, ... , an whose values are relatively
prime square-free integers between 2 and 1995 must contain a prime. [An integer is
square-free if it is not divisible by any square except 1.]
Solution
Answer: n = 14.
We can exhibit a sequence with 13 terms which does not contain a prime: 2·101 = 202,
3·97 = 291, 5·89 = 445, 7·83 = 581, 11·79 = 869, 13·73 = 949, 17·71 = 1207, 19·67 =
1273, 23·61 = 1403, 29·59 = 1711, 31·53 = 1643, 37·47 = 1739, 41·43 = 1763. So
certainly n ≥ 14.
If there is a sequence with n ≥ 14 not containing any primes, then since there are only 13
primes not exceeding 41, at least one member of the sequence must have at least two
prime factors exceeding 41. Hence it must be at least 43·47 = 2021 which exceeds 1995.
So n =14 is impossible.
Problem 3
ABCD is a fixed cyclic quadrilateral with AB not parallel to CD. Find the locus of
points P for which we can find circles through AB and CD touching at P.
Solution
Answer: Let the lines AB and CD meet at X. Let R be the length of a tangent from X to
the circle ABCD. The locus is the circle center X radius R. [Strictly you must exclude
four points unless you allow the degenerate straight line circles.]
Let X be the intersection of the lines AB and CD. Let R be the length of a tangent from
X to the circle ABCD. Let C0 be the circle center X radius R. Take any point P on C0.
Then considering the original circle ABCD, we have that R2 = XA·XB = XC·XD, and
hence XP2 = XA·XB = XC·XD.
If C1 is the circle through C, D and P, then XC.XD = XP2, so XP is tangent to the circle
C1. Similarly, the circle C2 through A, B and P is tangent to XP. Hence C1 and C2 are
tangent to each other at P. Note that if P is one of the 4 points on AB or CD and C0, then
this construction does not work unless we allow the degenerate straight line circles AB
and CD.
So we have established that all (or all but 4) points of C0 lie on the locus. But for any
given circle through C, D, there are only two circles through A, B which touch it (this is
clear if you consider how the circle through A, B changes as its center moves along the
perpendicular bisector of AB), so there are at most 2 points on the locus lying on a given
circle through C, D. But these are just the two points of intersection of the circle with C0.
So there are no points on the locus not on C0.
Problem 4
Take a fixed point P inside a fixed circle. Take a pair of perpendicular chords AC, BD
through P. Take Q to be one of the four points such that ASCQ, ASDQ, BSCQ or BSDQ
is a rectangle. Find the locus of all possible Q for all possible such chords.
Solution
Let O be the center of the fixed circle and let X be the center of the rectangle ASCQ. By
the cosine rule we have OQ2 = OX2 + XQ2 - 2·OX·XQ cos θ and OP2 = OX2 + XP2 2·OX·XP cos(θ+π), where θ is the angle OXQ. But cos(θ+π) = -cos θ, so OQ2 + OP2=
2OX2 + 2XQ2. But since X is the center of the rectangle XQ = XC and since X is the
midpoint of AC, OX is perpendicular to AC and hence XO2 + XC2 = OC2. So OQ2 =
2OC2 - OP2. But this quantity is fixed, so Q must lie on the circle center O radius √(2R2 OP2), where R is the radius of the circle.
Conversely, it is easy to see that all points on this circle can be reached. For given a
point Q on the circle radius √(2R2 - OP2) let X be the midpoint of PQ. Then take the
chord AC to have X as its midpoint.
Problem 5
f is a function from the integers to {1, 2, 3, ... , n} such that f(A) and f(B) are unequal
whenever A and B differ by 5, 7 or 12. What is the smallest possible n?
Solution
Answer: n = 4.
Each pair of 0, 5, 12 differ by 5, 7 or 12, so f(0), f(5), f(12) must all be different, so n ≥
3.
We can exhibit an f with n = 4. Define f(m) = 1 for m = 1, 3, 5, 7, 9, 11 (mod 24), f(m) =
2 for m = 2, 4, 6, 8, 10, 12 (mod 24), f(m) = 3 for m = 13, 15, 17, 19, 21, 23 (mod 24),
f(m) = 4 for m = 14, 16, 18, 20, 22, 0 (mod 24).
8th APMO 1996
Problem 1
ABCD is a fixed rhombus. P lies on AB and Q on BC, so that PQ is perpendicular to
BD. Similarly P' lies on AD and Q' on CD, so that P'Q' is perpendicular to BD. The
distance between PQ and P'Q' is more than BD/2. Show that the perimeter of the
hexagon APQCQ'P' depends only on the distance between PQ and P'Q'.
Solution
BPQ and DQ'P' are similar. Let PQ meet BD at X and P'Q' meet BD at Y. XY is fixed,
so BX + DY is fixed. Hence also, BP + DQ' and BQ + DP' and PQ + P'Q' are fixed. So
PQ + P'Q' - BP - BQ - DP' - DQ' is fixed, so PQ + P'Q' + (AB - BP) + (BC - BQ) + (CD
- DP') + (DA - DQ') is fixed, and that is the perimeter of the hexagon.
Problem 2
Prove that (n+1)mnm ≥ (n+m)!/(n-m)! ≥ 2mm! for all positive integers n, m with n ≥ m.
Solution
For any integer k ≥ 1, we have (n + k)(n - k + 1) = n2 + n - k2 + k ≤ n(n + 1). Taking the
product from k = 1 to m we get (n + m)!/(n - m)! ≤ (n + 1)mnm.
For k = 1, 2, ... , m, we have n ≥ k and hence n + k ≥ 2k. Taking the product from k = 1
to m, we get (n + m)!/(n - m)! ≥ 2mm! .
Problem 3
Given four concyclic points. For each subset of three points take the incenter. Show that
the four incenters from a rectangle.
Solution
Take the points as A, B, C, D in that order. Let I be the incenter of ABC. The ray CI
bisects the angle ACB, so it passes through M, the midpoint of the arc AB. Now ∠MBI
= ∠MBA + ∠IBA = ∠MCA + ∠IBA = (∠ACB + ∠ABC)/2 = 90o - (∠CAB) /2 = 90o ∠CMB/2 = 90o - ∠IMB/2. So the bisector of ∠IMB is perpendicular to IB. Hence MB =
MI. Let J be the incenter of ABD. Then similarly MA = MJ. But MA = MB, so the four
points A, B, I, J are concyclic (they lie on the circle center M). Hence ∠BIJ = 180o ∠BAJ = 180o - ∠BAD/2.
Similarly, if K is the incenter of ADC, then ∠BJK = 180o - ∠BDC/2. Hence ∠IJK =
360o - ∠BIJ - ∠BJK = (180o - ∠BIJ) + (180o - ∠BJK) = (∠BAD + ∠BDC)/2 = 90o.
Similarly, the other angles of the incenter quadrilateral are 90o, so it is a rectangle.
Problem 4
For which n in the range 1 to 1996 is it possible to divide n married couples into exactly
17 single sex groups, so that the size of any two groups differs by at most one.
Solution
Answer: 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 27, 28, 29, 30, 31, 32,
36, 37, 38, 39, 40, 45, 46, 47, 48, 54, 55, 56, 63, 64, 72.
If n = 17k, then the group size must be 2k. Hence no arrangement is possible, because
one sex has at most 8 groups and 8.2k < n.
If 2n = 17k+h with 0 < h < 17, then the group size must be k or k+1. One sex has at most
8 groups, so 8(k+1) ≥ n. Hence 16k + 16 ≥ 17k + h, so 16 - h ≥ k (*). We also require
that 9k ≤ n. Hence 18k < 2n = 17k + h, so k ≤ h (**). With (*) this implies that k ≤ 8. So
n ≤ 75.
Each group has at least one person, so we certainly require n ≥ 9 and hence k ≥ 1. It is
now easiest to enumerate. For k = 1, we can have h = 1, 3, ... 15, giving n = 9-16. For k
= 2, we can have h = 2, 4, ... 14, giving n = 18-24. For k = 3, we can have h = 3, 5, ... 13,
giving n = 27-32. For k = 4, we can have h = 4, 6, ... 12, giving n = 36-40. For k = 5 we
can have h = 5, 7, 9, 11, giving n = 45-48. For k = 6, we can have h = 6, 8, 10, giving n =
54, 55, 56. For k = 7, we can have h = 7, 9, giving n = 63, 64. For k = 8, we can have h =
8, giving n = 72.
Problem 5
A triangle has side lengths a, b, c. Prove that √(a + b - c) + √(b + c - a) + √(c + a - b) ≤
√a + √b + √c. When do you have equality?
Solution
Let A2 = b + c - a, B2 = c + a - b, C2 = a + b - c. Then A2 + B2 = 2c. Also A = B iff a = b.
We have (A - B)2 ≥ 0, with equality iff A = B. Hence A2 + B2 ≥ 2AB and so 2(A2 + B2)
≥ (A + B)2 or 4c ≥ (A + B)2 or 2√c ≥ A + B, with equality iff A = B. Adding the two
similar relations we get the desired inequality, with equality iff the triangle is equilateral.
9th APMO 1997
Problem 1
Let Tn = 1 + 2 + ... + n = n(n+1)/2. Let Sn= 1/T1 + 1/T2 + ... + 1/Tn. Prove that 1/S1 +
1/S2 + ... + 1/S1996 > 1001.
Solution
1/Tm = 2(1/m - 1/(m+1) ). Hence Sn/2 = 1 - 1/(n+1). So 1/Sn = (1 + 1/n)/2. Hence 1/S1 +
1/S2 + ... + 1/Sn = 1996/2 + (1+ 1/2 + 1/3 + ... + 1/1996)/2.
Now 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + ... + 1/16) + (1/17 + ... +
1/32) + (1/33 + ... + 1/64) + (1/65 + ... + 1/128) + (1/129 + ... + 1/256) + (1/257 + ... +
1/512) + (1/513 + ... + 1/1024) > 1 + 1/2 + 1/2 + ... + 1/2 = 6. So 1/S1 + 1/S2 + ... +
1/Sn = 1996/2 + 6/2 = 998 + 3 = 1001.
Problem 2
Find an n in the range 100, 101, ... , 1997 such that n divides 2n + 2.
Solution
Answer: the only such number is 946.
We have 2p-1 = 1 mod p for any prime p, so if we can find h in {1, 2, ... , p-2} for which
2h = -2 mod p, then 2k= -2 mod p for any h = k mod p. Thus we find that 2k = -2 mod 5
for k = 3 mod 4, and 2k = -2 mod 11 for k = 6 mod 10. So we might then hope that 5·11
= 3 mod 4 and = 6 mod 10. Unfortunately, it does not! But we try searching for more
examples.
The simplest would be to look at pq. Suppose first that p and q are both odd, so that pq is
odd. If k = h mod p-1, then we need h to be odd (otherwise pq would have to be even).
So the first step is to get a list of primes p with 2h = -2 mod p for some odd h < p. We
always have 2p-1 = 1 mod p, so we sometimes have 2(p-1)/2 = -1 mod p and hence 2(p+1)/2 =
-2 mod p. If (p+1)/2 is to be odd then p =1 mod 4. So searching such primes we find 3
mod 5, 7 mod 13, 15 mod 29, 19 mod 37, 27 mod 53, 31 mod 61. We require pq to lie in
the range 100-1997, so we check 5·29 (not = 3 mod 4), 5·37 (not = 3 mod 4), 5·53 (not =
3 mod 4), 5·61 (not = 3 mod 4), 13·29 (not = 7 mod 12), 13·37 (not = 7 mod 12), 13.53
(not = 7 mod 12), 13·61 (not = 7 mod 12), 29·37 (not = 15 mod 28), 29·53 (not = 15
mod 28), 29·61 (not = 15 mod 28), 37·53 (not = 19 mod 36). So that does not advance
matters much!
2p will not work (at least with h = (p+1)/2) because we cannot have 2p = (p+1)/2 mod p1. So we try looking at 2pq. This requires that p and q = 3 mod 4. So searching for
suitable p we find 6 mod 11, 10 mod 19, 22 mod 43, 30 mod 59, 34 mod 67, 42 mod 83.
So we look at 2·11·43 = 946, which works.
Proving that it is unique is harder. The easiest way is to use a computer to search (approx
5 min to write a Maple program or similar and a few seconds to run it).
Problem 3
ABC is a triangle. The bisector of A meets the segment BC at X and the circumcircle at
Y. Let rA = AX/AY. Define rB and rC similarly. Prove that rA/sin2A + rB/sin2B + rC/sin2C
>= 3 with equality iff the triangle is equilateral.
Solution
AX/AB = sin B/sin AXB = sin B/sin(180 - B - A/2) =sin B/sin(B + A/2). Similarly,
AB/AY = sin AYB/sin ABY = sin C/sin(B + CBY) = sin C/sin(B + A/2). So AX/AY =
sin B sin C/sin2(B + A/2). Hence rA/sin2A = sA/sin2(B + A/2), where sA = sin B sin
C/sin2A. Similarly for rB and rC. Now sAsBsC = 1, so the arithmetic/geometric mean
result gives sA + sB + sC ≥ 3. But 1/sin k ≥ 1 for any k, so rA/sin2A + rB/sin2B + rC/sin2C
≥ 3.
A necessary condition for equality is that sin2(B + A/2) = sin2(B + A/2) = sin2(B + A/2)
= 1 and hence A = B = C. But it is easily checked that this is also sufficient.
Problem 4
P1 and P3 are fixed points. P2 lies on the line perpendicular to P1P3 through P3. The
sequence P4, P5, P6, ... is defined inductively as follows: Pn+1 is the foot of the
perpendicular from Pn to Pn-1Pn-2. Show that the sequence converges to a point P (whose
position depends on P2). What is the locus of P as P2 varies?
Solution
PnPn+1Pn+2 lies inside Pn-1PnPn+1. So we have sequence of nested triangles whose size
shrinks to zero. Each triangle is a closed set, so there is just one point P in the
intersection of all the triangles and it is clear that the sequence Pn converges to it.
Obviously all the triangles PnPn+1Pn+2 are similar (but not necessarily with the vertices in
that order). So P must lie in the same position relative to each triangle and we must be
able to obtain one triangle from another by rotation and expansion about P. In particular,
P5P4P6 is similar (with the vertices in that order) to P1P2P3, and P4P5 is parallel to P1P2,
so the rotation to get one from the other must be through π and P must lie on P1P5.
Similarly P3P4P5 must be obtained from P1P2P3 by rotation about P through π/2 and
expansion. But this takes P1P5 into a perpendicular line through P3. Hence P1P is
perpendicular to P3P. Hence P lies on the circle diameter P1P3.
However, not all points on this circle are points of the locus. P3P5 = P3P4 cos P1 =
P3P1 sin P1 cos P2 = 1/2 P3P1 sin 2P1, so we can obtain all values of P3P5 up to P1P3/2.
[Of course, P2, and hence P5, can be on either side of P3.]. Thus the locus is an arc XP3Y
of the circle with XP3 = YP3 and ∠XP1Y = 2 tan-11/2. If O is the midpoint of P1P3, then
O is the center of the circle and ∠XOY = 4 tan-11/2 (about 106o).
Problem 5
n people are seated in a circle. A total of nk coins are distributed amongst the people, but
not necessarily equally. A move is the transfer of a single coin between two adjacent
people. Find an algorithm for making the minimum number of moves which result in
everyone ending up with the same number of coins?
Solution
Label the people from 1 to n, with person i next to person i+1, and person n next to
person 1. Let person i initially hold ci coins. Let di = ci - k.
It is not obvious how many moves are needed. Clearly at least 1/2 ∑ |di| are needed. But
one may need more. For example, suppose the starting values of di are 0, 1, 0, -1, 0.
Then one needs at least 2 moves, not 1.
Obviously ∑ di = 0, so not all di can be negative. Relabel if necessary so that d1 ≥= 0.
Now consider X = |d1| + |d1 + d2| + |d1 + d2 + d3| + ... + |d1 + d2 + ... + dn-1|. Note first that
X is zero iff all di are zero. Any move between i and i+1, except one between n and 1,
changes X by 1, because only the term |d1 + d2 + ... + di| is affected. Thus if we do not
make any moves between n and 1, then we need at least X moves to reach the desired
final position (with all di zero).
Assume X > 1. We show how to find a move which reduces X by 1. This requires a little
care to avoid specifying a move which might require a person with no coins to transfer
one. We are assuming that d1≥ 0. Take the first i for which di+1 < 0. There must be such
an i, otherwise all di would be non-negative, but they sum to 0, so they would all have to
be zero, contradicting X > 0. If d1 + ... + di > 0, then we take the move to be a transfer
from i to i+1. This will reduce |d1 + ... + di| by 1 and leave the other terms in X
unchanged, so it will reduce X by 1. If d1 + ... + di is not strictly positive, then by the
minimality of i we must have d1 = d2 = ... = di = 0. We know that di+1 < 0. Now find the
first j > i+1 such that dj ≥ 0. There must be such a j, otherwise we would have ∑ dm < 0.
We have d1 + ... + dj-1 < 0, so a transfer from j to j-1 will reduce |d1 + ... + dj-1| and hence
reduce X. Finally note that the move we have chosen leaves d1 ≥ 0. Thus we can repeat
the process and reduce X to zero in X moves.
We have proved that this procedure minimises the number of moves if we accept the
restriction that we do not make any transfers between 1 and n. Thus the full algorithm is:
calculate the effect of the transfers from 1 to n and from n to 1 on X. If either of these
transfers reduces X by more than 1, then take the move with the larger reduction;
otherwise, find a move as above which reduces X by 1; repeat.
10th APMO 1998
Problem 1
S is the set of all possible n-tuples (X1, X2, ... , Xn) where each Xi is a subset of {1, 2, ... ,
1998}. For each member k of S let f(k) be the number of elements in the union of its n
elements. Find the sum of f(k) over all k in S.
Solution
Answer: 1998(21998n - 21997n).
Let s(n, m) be the sum where each Xi is a subset of {1, 2, ... , m}. There are 2m possible
Xi and hence 2mn possible n-tuples. We have s(n, m) = 2ns(n, m-1) + (2n - 1)2n(m-1) (*).
For given any n-tuple {X1, ... , Xn} of subsets of {1, 2, ... , m-1} we can choose to add m
or not (2 choices) to each Xi. So we derive 2n n-tuples of subsets of {1, 2, ... , m}. All but
1 of these have f(k) incremented by 1. The first term in (*) gives the sum for m-1 over
the increased number of terms and the second term gives the increments to the f(k) due
to the additional element.
Evidently s(n, 1) = 2n - 1. It is now an easy induction to show that s(n, m) = m(2nm - 2n(m1)
).
Putting m = 1998 we get that the required sum is 1998(21998n - 21997n).
Problem 2
Show that (36m + n)(m + 36n) is not a power of 2 for any positive integers m, n.
Solution
Assume there is a solution. Take m ≤ n and the smallest possible m. Now (36m + n) and
(m + 36n) must each be powers of 2. Hence 4 divides n and 4 divides m. So m/2 and n/2
is a smaller solution with m/2 < m. Contradiction.
Problem 3
Prove that (1 + x/y)(1 + y/z)(1 + z/x) ≥ 2 + 2(x + y + z)/w for all positive reals x, y, z,
where w is the cube root of xyz.
Solution
(1 + x/y)(1 + y/z)(1 + z/x) = 1 + x/y + y/x + y/z + z/y + z/x + x/z = (x + y + z)(1/x + 1/y
+ 1/z) - 1 ≥ 3(x + y + z)/w - 1, by the arithmetic geometric mean inequality,
= 2(x + y + z)/w + (x + y + z)/w - 1 ≥ 2(x + y + z) + 3 - 1, by the arithmetic geometric
mean inequality.
Problem 4
ABC is a triangle. AD is an altitude. X lies on the circle ABD and Y lies on the circle
ACD. X, D and Y are collinear. M is the midpoint of XY and M' is the midpoint of BC.
Prove that MM' is perpendicular to AM
Solution
Take P, Q so that PADB, AQCD are rectangles. Let N be the midpoint of PQ. Then PD
is a diameter of the circumcircle of ABC, so PX is perpendicular to XY. Similarly, QY
is perpendicular to XY. N is the midpoint of PQ and M' the midpoint of XY, so NM is
parallel to PX and hence perpendicular to XY. NADM' is a rectangle, so ND is a
diameter of its circumcircle and M must lie on the circumcircle. But AM' is also a
diameter, so ∠AMM' = 90o.
Thanks to Michael Lipnowski for the above. My original solution is below.
Let P be the circumcenter of ABD and Q the circumcenter of ADC. Let R be the
midpoint of AM'. P and Q both lie on the perpendicular bisector of AD, which is parallel
to BC and hence also passes through R. We show first that R is the midpoint of PQ.
Let the feet of the perpendiculars from P, Q, R to BC to P', Q', R' respectively. It is
sufficient to show that . BP' = BD/2. BR' = BM' + M'R' = (BD + DC)/2 + M'D/2 = (BD
+ DC)/2 + ( (BD + DC)/2 - DC)/2 = 3BD/4 + DC/4, so P'R' = (BD + DC)/4. Q'C = DC/2,
so BQ' = BD + DC/2 and P'Q' = (BD + DC)/2 = 2P'R'.
Now the circumcircle centre P meets XY in X and D, and the circumcircle centre Q
meets XY in D and Y. Without loss of generality we may take XD >= DY. Put XD = 4x,
DY = 4y. The circle center R through A, M' and D meets XY in a second point, say M''.
Let the feet of the perpendiculars from P, Q, R to XY be P'', Q'', R'' respectively. So on
XY we have, in order, X, P'', M'', R'', D, Q'', Y. Since R is the midpoint of PQ, R'' is the
midpoint of P''Q''. Now P'' is the midpoint of XD and Q'' is the midpoint of DY, so P''Q''
= XY/2 = 2(x+y), so R''Q'' = x+y. But DQ'' = 2y, so R''D = x-y. R'' is the midpoint of
M''D, so M''D = 2(x-y) and hence M''Y = M''D + DY = 2(x-y) + 4y = 2(x+y) = XY/2. So
M'' is just M the midpoint of XY. Now AM' is a diameter of the circle center R, so AM
is perpendicular to MM'.
Problem 5
What is the largest integer divisible by all positive integers less than its cube root.
Solution
Answer: 420.
Let N be a positive integer satisfying the condition and let n be the largest integer not
exceeding its cube root. If n = 7, then 3·4·5·7 = 420 must divide N. But N cannot exceed
83 - 1 = 511, so the largest such N is 420.
If n ≥ 8, then 3·8·5·7 = 840 divides N, so N > 729 = 93. Hence 9 divides N, and hence
3·840 = 2520 divides N. But we show that no N > 2000 can satisfy the condition.
Note that 2(x - 1)3 > x3 for any x > 4. Hence [x]3 > x3/2 for x > 4. So certainly if N >
2000, we have n3 > N/2. Now let pk be the highest power of k which does not exceed n.
Then pk > n/k. Hence p2p3p5 > n3/30 > N/60. But since N > 2000, we have 7 < 11 < n
and hence p2, p3, p5, 7, 11 are all ≤ n. But 77 p2p3p5 > N, so N cannot satisfy the
condition.
11th APMO 1999
Problem 1
Find the smallest positive integer n such that no arithmetic progression of 1999 reals
contains just n integers.
Solution
Answer: 70.
Using a difference of 1/n , where n does not divide 1999, we can get a progression of
1999 terms with m = [1998/n] or m = [1998/n] - 1 integers. Thus {0, 1/n, 2/n, ... ,
1998/n} has m+1 integers, and {1/n, 2/n, ... , 1999/n} has m integers. So we are ok until
n gets so large that the gap between [1998/n] and [1998/(n+1)] is 3 or more. This could
happen for 1998/n - 1998/(n+1) just over 2 or n > 31. So checking, we find [1998/31] =
64, [1998/30] = 66, [1998/29] = 68, [1998/28] = 71.
We can get 68 integers with {1/29, 2/29, ... , 1999/29} and 69 with {0, 1/29, 2/29, ... ,
1998/29}. We can get 71 with {1/28, 2/28, ... , 1999/28}, but we cannot get 70. Note that
a progression with irrational difference gives at most 1 integer. A progression with
difference a/b, where a and b are coprime integers, gives the same number of integers as
a progression with difference 1/b. So it does not help to widen the class of progressions
we are looking at.
Problem 2
The real numbers x1, x2, x3, ... satisfy xi+j <= xi + xj for all i, j. Prove that x1 + x2/2 + ... +
xn/n >= xn.
Solution
We use induction. Suppose the result is true for n. We have:
x1 >= x1
x1 + x2/2 >= x2
...
x1 + x2/2 + ... + xn/n >= xn
Also: x1 + 2x2/2 + ... + nxn/n = x1 + ... + xn
Adding: (n+1) x1 + (n+1)x2/2 + ... + (n+1)xn/n >= 2(x1 + ... + xn). But rhs = (x1 + xn) +
(x2 + xn-1) + ... + (xn + x1) >= n xn+1. Hence result.
Problem 3
Two circles touch the line AB at A and B and intersect each other at X and Y with X
nearer to the line AB. The tangent to the circle AXY at X meets the circle BXY at W.
The ray AX meets BW at Z. Show that BW and BX are tangents to the circle XYZ.
Solution
Let angle ZXW =  and angle ZWX = . XW is tangent to circle AXY at X, so angle
AYX = . AB is tangent to circle AXY at A, so angle BAX = . AB is tangent to circle
BXY at B, so angle ABX = . Thus, considering triangle ABX, angle BXZ = .
Considering triangle ZXW, angle BZX = .
BXYW is cyclic, so angle BYX = angle BWX = . Hence angle AYB = angle AYX +
angle XYB =  = angle AZB. So AYZB is cyclic. Hence angle BYZ = angle BAZ
= . So angle XYZ = angle XYB + angle BYZ = . Hence angle BZX = angle XYZ,
so BZ is tangent to circle XYZ at Z. Similarly angle BXY = angle XYZ, so BX is
tangent to circle XYZ at X.
Problem 4
Find all pairs of integers m, n such that m2 + 4n and n2 +4m are both squares.
Solution
Answer: (m, n) or (n, m) = (0, a2), (-5, -6), (-4, -4), (a+1, -a) where a is a non-negative
integer.
Clearly if one of m, n is zero, then the other must be a square and that is a solution.
If both are positive, then m2 + 4n must be (m + 2k)2 for some positive k, so n = km +
k2 > m. But similarly m > n. Contradiction. So there are no solutions with m and n
positive.
Suppose both are negative. Put m = -M, n = -N, so M and N are both positive. Assume
M >= N. M2 - 4N is a square, so it must be (M - 2k)2 for some k, so N = Mk - k2. If M =
N, then M(k-1) = k2, so k-1 divides k2 and hence k2 - (k-1)(k+1) = 1, so k = 2 and M = 4,
giving the solution (m, n) = (-4, -4). So we may assume M > N and hence M > Mk - k2 >
0. But that implies that k = 1 or M-1 and hence N = M-1. [If M > Mk - k2, then (k-1)M <
k2. Hence k = 1 or M < k+2. But Mk - k2 > 0, so M > k. Hence k = 1 or M = k+1.].
But N2 - 4M is also a square, so (M-1)2 - 4M = M2 - 6M + 1 is a square. But (M-3)2 >
M2 - 6M + 1 and (M-4)2 < M2 - 6M + 1 for M >= 8, so the only possible solutions are M
= 1, 2, ... , 7. Checking, we find that only M = 6 gives M2 - 6M + 1 a square. This gives
the soluton (m, n) = (-6, -5). Obviously, there is also the solution (-5, -6).
Finally, consider the case of opposite signs. Suppose m = M > 0, n = -N < 0. Then N2 +
4M is a square, so by the argument above M > N. But M2 - 4N is a square and so the
argument above gives N = M-1. Now we can easily check that (m, n) = (M, -(M-1) ) is a
solution for any positive M.
Problem 5
A set of 2n+1 points in the plane has no three collinear and no four concyclic. A circle is
said to divide the set if it passes through 3 of the points and has exactly n - 1 points
inside it. Show that the number of circles which divide the set is even iff n is even.
Solution
Take two of the points, A and B, and consider the 2n-1 circles through A and B. We will
show that the number of these circles which divide the set is odd. The result then follows
almost immediately, because the number of pairs A, B is (2n+1)2n/2 = N, say. The total
number of circles which divide the set is a sum of N odd numbers divided by 3 (because
each such circle will be counted three times). If n is even, then N is even, so a sum of N
odd numbers is even. If n is odd, then N is odd, so a sum of N odd numbers is odd.
Dividing by 3 does not change the parity.
Their centers all lie on the perpendicular bisector of AB. Label them C1, C2, ... , C2n-1,
where the center of Ci lies to the left of Cj on the bisector iff i < j. We call the two halfplanes created by AB the left-hand half-plane L and the right-hand half-plane R
correspondingly. Let the third point of the set on Ci be Xi. Suppose i < j. Then
Ci contains all points of Cj that lie in L and Cj contains all points of Ci that lie R. So
Xi lies inside Cj iff Xi lies in R and Xj lies inside Ci iff Xj lies in L
Now plot f(i), the number of points in the set that lie inside Ci, as a function of i. If
Xi and Xi+1 are on opposite sides of AB, then f(i+1) = f(i). If they are both in L, then
f(i+1) = f(i) - 1, and if they are both in R, then f(i+1) = f(i) + 1. Suppose m of the Xi lie
in L and 2n-1-m lie in R. Now suppose f(i) = n-2, f(i+1) = f(i+2) = ... = f(i+j) = n-1,
f(i+j+1) = n. Then j must be odd. For Xi and Xi+1 must lie in R. Then the points must
alternate, so Xi+2 lies in L, Xi+3 lies in R etc. But Xi+j and Xi+j+1 must lie in R. Hence j
must be odd. On the other hand, if f(i+j+1) = n-2, then j must be even. So the parity of
the number of C1, C2, ... , Ci which divide the set only changes when f crosses the line n-
1 from one side to the other. We now want to say that f starts on one side of the line n-1
and ends on the other, so the final parity must be odd. Suppose there are m points in L
and hence 2n-1-m in R. Without loss of generality we may take m <= n-1. The first
circle C1 contains all the points in L except X1 if it is in L. So f(1) = m or m-1. Similarly
the last circle C2n-1 contains all the points in R except X2n-1 if it is in R. So f(2n-1) = 2n1-m or 2n-2-m. Hence if m < n-1, then f(1) = m or m-1, so f(1) < n-1. But 2n-1-m >=
n+1, so f(2n-1) > n-1. So in this case we are done.
However, there are complications if m = n-1. We have to consider 4 cases. Case (1): m =
n-1, X1 lies in R, X2n-1 lies in L. Hence f(1) = n-1, f(2n-1) = n > n-1. So f starts on the
line n-1. If it first leaves it downwards, then for the previous point i, Xi is in L and hence
there were an even number of points up to i on the line. So the parity is the same as if
f(1) was below the line. f(2n-1) is above the line, so we get an odd number of points on
the line. If f first leaves the line upwards, then for the previous point i, Xi is in R and
hence there were an odd number of points up to i on the line. So again the parity is the
same as if f(1) was below the line.
Case (2): m = n-1, X1 lies in R, X2n-1 lies in R. Hence f(1) = f(2n-1) = n-1. As in case (1)
the parity is the same as if f(1) was below the line. If the last point j with f(j) not on the
line has f(j) < n-1, then (since X2n-1 lies in R) there are an odd number of points above j,
so the parity is the same as if f(2n-1) was above the line. Similarly if f(j) > n-1, then
there are an even number of points above j, so again the parity is the same as if f(2n-1)
was above the line.
Case (3): m = n-1, X1 lies in L, X2n-1 lies in L. Hence f(1) = n-2, f(2n-1) = n. So case has
already been covered.
Case (4): m=n-1, X1 lies in L, Xn-1 lies in R. So f(1) = n-2, f(2n-1) = n-1. As in case (2),
the parity is the same as if f(2n-1) was above the line.
12th APMO 2000
Problem 1
Find a13/(1 - 3a1 + 3a12) + a23/(1 - 3a2 + 3a22) + ... + a1013/(1 - 3a101 + 3a1012), where an =
n/101.
Solution
Answer: 51.
The nth term is an3/(1 - 3an + 3an2) = an3/( (1 - an)3 + an3) = n3/( (101 - n)3 + n3). Hence
the sum of the nth and (101-n)th terms is 1. Thus the sum from n = 1 to 100 is 50. The
last term is 1, so the total sum is 51.
Problem 2
Find all permutations a1, a2, ... , a9 of 1, 2, ... , 9 such that a1 + a2 + a3 + a4 = a4 + a5 +
a6 + a7 = a7 + a8 + a9 + a1 and a12 + a22 + a32 + a42 = a42 + a52 + a62 + a72 = a72 + a82 + a92 +
a12.
Solution
We may start by assuming that a1 < a4 < a7 and that a2 < a3, a5 < a6, a8 < a9.
Note that 1 + ... + 9 = 45 and 12 + ... + 92 = 285. Adding the three square equations
together we get (a12 + ... + a92) + a12 + a42 + a72 = 285 + a12 + a42 + a72. The total must be
a multiple of 3. But 285 is a multiple of 3, so a12 + a42 + a72 must be a multiple of 3. Now
32, 62 and 92 are all congruent to 0 mod 3 and the other squares are all congruent to 1
mod 3. Hence either a1, a4 and a7 are all multiples of 3, or none of them are. Since 45 is
also a multiple of three a similar argument with the three linear equations shows that a1 +
a4 + a7 is a multiple of 3. So if none of a1, a4, a7 are multiples of 3, then they are all
congruent to 1 mod 3 or all congruent to 2 mod 3. Thus we have three cases: (1) a1 = 3,
a4 = 6, a7 = 9, (2) a1 = 1, a4 = 4, a7 = 7, and (3) a1 = 2, a4 = 5, a7 = 8.
In case (1), we have that each sum of squares equals 137. Hence a82 + a92 = 47. But 47 is
not a sum of two squares, so this case gives no solutions.
In case (2), we have that each sum of squares is 117. Hence a52 + a62 = 52. But the only
way of writing 52 as a sum of two squares is 42 + 62 and 4 is already taken by a4, so this
case gives no solutions.
In case (3), we have that each sum of squares is 126 and each linear sum 20. We quickly
find that the only solution is 2, 4, 9, 5, 1, 6, 8, 3, 7.
Obviously, this generates a large number of equivalent solutions. We can interchange
a2 and a3, or a5 and a6, or a8 and a9. We can also permute a1, a4 and a7. So we get a total
of 2 x 2 x 2 x 6 =48 solutions.
Problem 3
ABC is a triangle. The angle bisector at A meets the side BC at X. The perpendicular to
AX at X meets AB at Y. The perpendicular to AB at Y meets the ray AX at R. XY meets
the median from A at S. Prove that RS is perpendicular to BC.
Solution
Let the line through C parallel to AX meet the ray BA at C'. Let the perpendicular from
B meet the ray C'C at T and the ray AX at U. Let the line from C parallel to BT meet BA
at V and let the perpendicular from V meet BT at W. So CVWT is a rectangle.
AU bisects ∠CAV and CV is perpendicular to AU, so U is the midpoint of WT. Hence
the intersection N of AU and CW is the center of the rectangle and, in particular, the
midpoint of CW. Let M be the midpoint of BC. Then since M, N are the midpoints of
the sides CB and CW of the triangle CBW, MN = BW/2.
Since CC' is parallel to AX, ∠CC'A = ∠BAX = ∠CAX = ∠C'CA, so AC' = AC. Let A'
be the midpoint of CC'. Then AU = C'T - C'A'. But N is the center of the rectangle
CTWV, so NU = CT/2 and AN = AU - NU = C'T - C'A' - CT/2 = C'T/2. Hence MN/AN
= BW/C'T. But MN is parallel to BW and XY, so SX/AX = MN/AN = BW/C'T.
Now AX is parallel to VW and XY is parallel to BW, so AXY and VWB are similar and
AX/XY = VW/BW = CT/BW. Hence SX/XY = (SX/AX) (AX/XY) = CT/C'T.
YX is an altitude of the right-angled triangle AXR, so AXY and YXR are similar. Hence
XY/XR = XA/XY. But AXY and C'TB are similar, so XA/XY = C'T/BT. Hence SX/XR
= (SX/XY) (XY/XR) = (CT/C'T) (C'T/BT) = CT/BT. But angles CTB and SXR are both
right angles, so SXR and CTB are similar. But XR is perpendicular to BT, so SR is
perpendicular to BC.
Problem 4
If m < n are positive integers prove that nn/(mm (n-m)n-m) > n!/( m! (n-m)! ) > nn/(
mm(n+1) (n-m)n-m).
Solution
The key is to consider the binomial expansion (m + n-m)n. This is a sum of positive
terms, one of which is nCm mm(n-m)n-m, where nCm is the binomial coefficient n!/( m!
(n-m)! ). Hence nCm mm(n-m)n-m < nn, which is one of the required inequalities.
We will show that nCm mm(n-m)n-m is the largest term in the binomial expansion. It then
follows that (n+1) nCm mm(n-m)n-m > nn, which is the other required inequality.
Comparing the rth term nCr mr(n-m)n-r with the r+1th term nCr+1 mr+1(n-m)n-r-1 we see
that the rth term is strictly larger for r ≥ m and smaller for r < m. Hence the mth term is
larger than the succeeding terms and also larger than the preceding terms.
Problem 5
Given a permutation s0, s2, ... , sn of 0, 1, 2, .... , n, we may transform it if we can find i, j
such that si = 0 and sj = si-1 + 1. The new permutation is obtained by transposing si and sj.
For which n can we obtain (1, 2, ... , n, 0) by repeated transformations starting with (1, n,
n-1, .. , 3, 2, 0)?
Solution
Experimentation shows that we can do it for n=1 (already there), n = 2 (already there), 3,
7, 15, but not for n = 4, 5, 6, 8, 9, 10, 11, 12, 13, 14. So we conjecture that it is possible
just for n = 2m - 1 and for n = 2.
Notice that there is at most one transformation possible. If n = 2m, then we find that
after m-1 transformations we reach
1
n
0
n-2
n-1
n-4
n-3 ... 4
5
2
3
and we can go no further. So n even fails for n > 2.
If n = 15 we get successively:
1 15
1
0
1
2
moves
1
2
moves
1
2
moves
14
14
3
13
15
0
12
12
12
11
13
13
10
10
14
9
11
15
8
8
8
7
9
9
6
6
10
5
7
11
4
4
4
3
5
5
2
2
6
0
3
7
start
after 7 moves
after 8 more
3
4
5
6
7
0
8
9
10
11
12
13
14
15
after 8 more
3
4
5
6
7
8
9
10
11
12
13
14
15
0
after 8 more
This pattern is general. Suppose n = 2m - 1. Let P0 be the starting position and Pr be the
position:
1
2
3 ... R-1
0,
... , R R+1 ... 2R-1
n-R+1
n-R+2
n-R+3 ... n,
n-2R+1
n-2R+2 ... n-R,
Here R denotes 2r and the commas highlight that, after the initial 1 2 ... R-1 0, we have
increasing runs of R terms. If we start from Pr, then the 0 is transposed successively with
R, 3R, 5R, ... , n-R+1, then with R+1, 3R+1, ... , n-R+2, and so on up to 2R-1, 4R-1, ... ,
n. But that gives Pr+1. It is also easy to check that P0 leads to P1 and that Pm is the
required finishing position. Thus the case n = 2m - 1 works.
Now suppose n is odd but not of the form 2m - 1. Then we can write n = (2a + 1)2b - 1
(just take 2b as the highest power of 2 dividing n + 1). We can now define P0, P1, ... ,
Pb as before. As before we will reach Pb:
1 2 ¼ B-1 0, 2aB 2aB+1¼ (2a+1)B-1, (2a-1)B ¼ 2aB-1, ¼ , 3B, 3B+1, ¼ 4B-1, 2B,
2B+1, ¼ , 3B-1, B, B+1, ¼ , 2B-1
where B = 2b - 1. But then the 0 is transposed successively with B, 3B, 5B, ... , (2a-1)B,
which puts it immediately to the right of (2a+1)B-1 = n, so no further transformations
are possible and n = (2a+1)2b - 1 fails.
13th APMO 2001
Problem 1
If n is a positive integer, let d+1 be the number of digits in n (in base 10) and s be the
sum of the digits. Let n(k) be the number formed by deleting the last k digits of n. Prove
that n = s + 9 n(1) + 9 n(2) + ... + 9 n(d).
Solution
Let the digits of n be ad, ad-1, ... , a0, so that n = ad 10d + ... + a0. Then n(k) = ad 10d-k + ad1
10d-k-1 + ... + ak. Obviously s = ad + ... + a0. Hence s + 9 n(1) + ... + 9 n(d) = ad(9.10d-1 +
9.10d-2 + ... + 9 + 1) + ad-1(9.10d-2 + ... + 9 + 1) + ... + ad-k(9.10d-k-1 + ... + 9 + 1) + a1(9 +
1) + a0 = ad 10d + ... + a0 = n.
Problem 2
Find the largest n so that the number of integers less than or equal to n and divisible by 3
equals the number divisible by 5 or 7 (or both).
Solution
Answer: 65.
Let f(n) = [n/3] - [n/5] - [n/7] + [n/35]. We are looking for the largest n with f(n) = 0.
Now [n/5] + [n/7} <= [n/5 + n/7] = [12n/35] = [n/3 + n/105]. So for [n/5] + [n/7] to
exceed [n/3] we certainly require n/105 ≥ 1/3 or n ≥ 35. Hence f(n) ≥ 0 for n ≤ 35. But
f(n+35) = [n/3 + 11 + 2/3] - [n/5 + 7] - [n/7 + 5] + [n/35 + 1] = [n/3 + 2/3] - [n/5] - [n/7]
+ [n/35] ≥ f(n) (*). Hence f(n) ≥ 0 for all n. But f(n+105) = [n/3 + 35] - [n/5 + 21] - [n/7
+ 15] + [n/35 + 3] = f(n) + 2. Hence f(n) ≥ 2 for all n ≥ 105.
Referring back to (*) we see that f(n+35) > f(n), and hence f(n+35) > 0, unless n is a
multiple of 3. But if n is a multiple of 3, then n + 35 is not and hence f(n+70) > f(n+35)
> 0. So f(n) > 0 for all n ≥ 70.
f(70) = 1. So f(69) = 2 (we have lost 70, a multiple of 7). So f(68) = f(67) = f(66) = 1
(we have lost 69, a multiple of 3). Hence f(65) = 0 (we have lost 66, a multiple of 3).
Problem 3
Two equal-sized regular n-gons intersect to form a 2n-gon C. Prove that the sum of the
sides of C which form part of one n-gon equals half the perimeter of C.
Solution
Let one regular n-gon have vertices P1, P2, ... , Pn and the other have vertices Q1, Q2, ... ,
Qn. Each side of the 2n-gon forms a triangle with one of the Pi or Qi. Note that Pi and
Qj must alternate as we go around the 2n-gon. For convenience assume that the order is
P1, Q1, P2, Q2, ... , Pn, Qn. Let the length of the side which forms a triangle with Pi be pi,
and the length of the side which forms a triangle with Qibe qi. Each of these triangles has
one angle (180o - 360o/n). Adjacent triangles have one of their other angles equal
(alternate angles), so all the triangles are similar. If the sides of the triangle vertex
Pi have lengths ai, bi, pi, then the side PiPi+1 is bi + qi + ai+1, and ai/ai+1 = bi/bi+1 = pi/pi+1.
Similarly, if the sides of the triangle vertex Qi have lengths ci, di, qi, then the side
QiQi+1 is di + pi+1 + ci+1 and ci/ci+1 = di/di+1 = qi/qi+1. But ai/bi = di/ci (not ci/di), because
the triangles alternate in orientation.
Put ai/pi = h, bi/pi = k. Note that ai + bi > pi, so h + k - 1 > 0. We have also ci/qi = k,
di/qi = h. Adding the expressions for PiPi+1 we get perimeter Pi = ∑(bi + qi + ai+1) = k ∑
pi + ∑ qi + h ∑ pi. Similarly, perimeter Qi = (h + k) ∑ qi + ∑ pi. The two n-gons are
equal, so (h + k - 1) ∑ pi = (h + k - 1) ∑ qi. Hence ∑ pi = ∑ qi, which is the required
result.
Problem 4
Find all real polynomials p(x) such that x is rational iff p(x) is rational.
Solution
It is necessary for all the coefficients of x to be rational. This is an easy induction on the
number of coefficients. For p(0) must be rational, hence ( p(x) - p(0) )/x must be rational
for all rational x and so on.
Clearly this condition is also sufficient for polynomials of degree 0 or 1.
There are obviously no quadratics, for if p(x) = ax2 + bx + c, with a, b, c rational, then
p(√2 - b/2a) = 2a - b2/4a + c, which is rational.
We prove that if there are also no higher degree polynomials. The idea is to show that
there is a rational value k which must be taken for some real x, but which cannot be
taken by any rational x.
Suppose p(x) has degree n > 1. Multiplying through by the lcm of the denominators of
the coefficients, we get p(x) = (a xn + b xn-1 + ... + u x + v)/w, where a, b, ... , w are all
integers. Put x = r/s, where r and s are coprime integers, then p(r/s) = (a rn + b rn-1s + ... +
u r sn-1 + v sn)/( w sn). Let q be any prime which does not divide a or w. Consider first a
> 0. p(x) must assume all sufficiently large positive values. So it must in particular take
the value k = m + 1/q, where m is a sufficiently large integer. So k = (mq + 1)/q. The
denominator is divisible by q, but not q2 and the numerator is not divisible by q. Suppose
p(r/s) = k for some integers r, s. The denominator of p(r/s) is w sn. We know that w is not
divisible by q, so q must divide s. But n > 1, so q2 divides w sn. The numerator of p(r/s)
has the form a rn + h s. Neither a nor r is divisible by q, so the numerator is not divisible
by q. Thus no cancellation is possible and we cannot have p(r/s) = k. Thus there must be
some irrational x such that p(x) = k.
If a < 0, then the same argument works except that we take k = m + 1/q, where m is a
sufficiently large negative integer.
Problem 5
What is the largest n for which we can find n + 4 points in the plane, A, B, C, D, X1, ... ,
Xn, so that AB is not equal to CD, but for each i the two triangles ABXi and CDXi are
congruent?
Answer
4
Solution
Many thanks to Allen Zhang for completing the proof
Assume AB = a, CD = b with a > b. If ABX and CDX are congruent, then either AX or
BX = CD = b, so X lies either on the circle SA center A radius b, or on the circle
SBcenter B radius b. Similarly, CX or DX = AB = a, so X lies either on the circle
SC center C radius a, or on the circle SD center D radius a. Thus we have four pairs of
circles, (SA, SC), (SA, SD), (SB, SC), (SB, SD) each with at most 2 points of intersection. X
must be one of these 8 points.
However, we show that if two points of intersection of (SA, SC) are included,
then no points of (SB, SD) can be included. The same applies to each pair of circles, so at
most 4 points X are possible. Finally, we will give an example of n = 4, showing that the
maximum of 4 is achieved.
So suppose (SA, SC) intersect at X and Y. We must have BX = DX and BY = DY, so X
and Y both lie on the perpendicular bisector of BD. In other words, XY is the
perpendicular bisector of BD, so D is the reflection of B in the line XY. There is no loss
of generality in taking B (and D) to be on the same side of AC as X. Let A' be the
reflection of A in the line XY. Since B lies on the circle center A radius a, D must lie on
the circle center A' radius A. Thus the triangles A'XC and CDA' are congruent.
(Note that A and C can be on the same side of XY or opposite sides.) Hence D is the
same height above AC as X, so DX is perpendicular to XY. Hence X is the midpoint of
BD. Also ∠A'CD = ∠CA'X = 180o - ∠CAX, so AX and CD are parallel. They are also
equal, so ACDX is a parallelogram and hence AC = DX = BD/2. In the second
configuration above, both A and C are on the same side of XY as D, so the midpoint M
of AC lies on the same side of XY as D. In the first configuration, since AX = b < a =
CX, M lies to the right of XY.
Now suppose there is a solution for the configuration (SB, SD). Thus there is a point Z
such that ABZ and ZDC are congruent. Then AZ = CZ, so Z lies on the perpendicular
bisector of AC and hence on the same side of XY as D. But it is also a distance a from D
and a distance b from B, and a > b, so it must lie on the same side of XY as B.
Contradiction. So there are no solutions for the configuration (SB, SD), as required. That
completes the proof that n ≤ 4.
For an example with n = 4, take a regular hexagon ACDBX3X2. Extend the side X2X3 to
X1X4, with X1, X2, X3, X4 equally spaced in that order, so that X1AX2 and X3BX4 are
equilateral. Then ABX1 and CX1D are congruent, ABX2 and DX2C are congruent,
ABX3 and X3CD are congruent, and ABX4 and X4DC are congruent.
Last corrected/updated 8 Feb 04
14th APMO 2002
Problem 1
xi are non-negative integers. Prove that x1! x2! ... xn! ≥ ( [(x1 + ... + xn)/n] ! )n (where [y]
denotes the largest integer not exceeding y). When do you have equality?
Solution
Answer: Equality iff all xi equal.
For given 2m the largest binomial coefficient is (2m)!/(m! m!) and for 2m+1 the largest
binomial coefficient is (2m+1)!/( m! (m+1)!). Hence for fixed xi + xj the smallest value
of xi! xj! is for xi and xj as nearly equal as possible.
If x1 + x2 + ... + xn = qn + r, where 0 < r < n, then we can reduce one or more xi to reduce
the sum to qn. This will not affect the rhs of the inequality in the question, but will
reduce the lhs. Equalising the xiwill not increase the lhs (by the result just proved). So it
is sufficient to prove the inequality for all xi equal. But in this case it is trivial since k! =
k! .
Problem 2
Find all pairs m, n of positive integers such that m2 - n divides m + n2 and n2 - m divides
m2 + n.
Solution
Assume n ≥ m.
(m+1)2 - m = (m+1) + m2. Clearly n2 increases faster than n, so n2 - m > n + m2 for n >
m+1 and hence there are no solutions with n > m+1. It remains to consider the two cases
n = m and n = m+1.
Suppose n = m. Then we require that n2 - n divides n2 + n. If n > 3, then n2 > 3n, so
2(n2 - n) > n2 + n. Obviously n2 - n < n2 + n, so if n > 3, then n2 - n cannot divide n2 + n.
It is easy to check that the only solutions (with n = m) less than 3 are n = 2 and n = 3.
Finally suppose n = m+1. We require m2 - m - 1 divides m2 + 3m + 1. If m >= 6, then
m(m - 5) > 3, so 2(m2 - m - 1) > m2 + 3m + 1. Obviously m2 - m - 1 < m2 + 3m + 1, so
m2 - m - 1 cannot divide m2 + 3m + 1 for m >= 6. Checking the smaller values, we find
the solutions less than 6 are m = 1 and m = 2.
Summarising, the only solutions are: (n, m) = (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2).
Problem 3
ABC is an equilateral triangle. M is the midpoint of AC and N is the midpoint of AB. P
lies on the segment MC, and Q lies on the segment NB. R is the orthocenter of ABP and
S is the orthocenter of ACQ. The lines BP and CQ meet at T. Find all possible values for
angle BCQ such that RST is equilateral.
Answer
15o.
Solution
Suppose CP < BQ. Since R is the intersection of BM and the perpendicular from P to
AB, and S is the intersection of CN and the perpendicular from Q to AC, we have MR >
NS. Hence (treating BC as horizontal), R is below S. But T must be to the right of the
midpoint of BC. Hence T is to the right of the perpendicular bisector of RS, so RST
cannot be equilateral. Contradiction. Similarly if CP > BQ. So CP = BQ.
Let L be the midpoint of BC. Put ∠CBP = x and ∠RAM = y. So RM = AM tan y, TL =
BL tan x = AM tan x. But ∠APB = 60o + x, so y = 30o - x. So if x ≠ 15o, then TL ≠ RM.
However, RST equilateral implies TL = RM, so x and hence also ∠BCQ = 15o.
Problem 4
The positive reals a, b, c satisfy 1/a + 1/b + 1/c = 1. Prove that √(a + bc) + √(b + ca) +
√(c + ab) ≥ √(abc) + √a + √b + √c.
Solution
Thanks to Suat Namli for this.
Multiplying by √(abc), we have √(abc) = √(ab/c) + √(bc/a) + √(ca/b). So it is sufficient
to prove that √(c + ab) ≥ √c + √(ab/c).
Squaring, this is equivalent to c + ab ≥ c + ab/c + 2√(ab) or c + ab ≥ c + ab(1 - 1/a - 1/b)
+ 2√(ab) or a + b >= 2√(ab) or (√a - √b)2 ≥ 0.
Problem 5
Find all real-valued functions f on the reals which have at most finitely many zeros and
satisfy f(x4 + y) = x3f(x) + f(f(y)) for all x, y.
Solution
Putting x = 0, we get f(y) = f(f(y)) for all y (*).
Putting x = 1, y = 0, we get f(0) = 0. Hence putting y = 0, f(x4) = x3f(x) (**).
If f(1) = 0, then f(2) = f(1 + 1) = f(1) + f(1) = 0, so f(3) = f(1 + 2) = f(1) + f(2) = 0 and so
on. Contradiction (only finitely many zeros). So f(1) = k for some non-zero k. By (*),
f(k) = k.
Suppose f(h) = 0 for some h not 0 or 1. Then f(h4) = h3f(h) = 0, so f(x) = 0 for any of the
distinct values x = h, h4, h16, h64, ... . Contradiction (only finitely many zeros). So f(h) is
not 0 for any non-zero h.
Given any x, put z = f(x4) - x4. Then f(x4) = f(f(x4)) = f(x4 + z) = x3f(x) + f(z). But f(x4) =
x3f(x), so f(z) = 0. Hence z = 0. So for any positive x we have f(x) = x. But f(x4) = f( (x)4 ) = - x3 f(-x). Hence f(-x) = - f(x). So f(x) = x for all x.
15th APMO 2003
Problem 1
The polynomial a8x8 +a7x7 + ... + a0 has a8 = 1, a7 = -4, a6 = 7 and all its roots positive
and real. Find the possible values for a0.
Answer
1/28
Solution
Thanks to Jonathan Ramachandran
Let the roots be xi. We have Sum xi2 = 42 - 2·7 = 2. By Cauchy we have (x1·1 + ... +
x8·1) ≤ (x12 + ... + x82)1/2(12 + ... + 12)1/2 with equality iff all xi are equal. Hence all xi are
equal. So they must all be 1/2.
Problem 2
A unit square lies across two parallel lines a unit distance apart, so that two triangular
areas of the square lie outside the lines. Show that the sum of the perimeters of these two
triangles is independent of how the square is placed.
Solution
Let the lines be L, L'. Let the square be ABA'B', with A, A' the two vertices not between
L and L'. Let L meet AB at X and AB' at Y. Let L' meet A'B' at X' and A'B at Y'. So
AXY and A'X'Y' are similar. Suppose angle AXY = x. If we move L towards A by a
distance d perpendicular to itself, then AX is shortened by d cosec x. If L' remains a
distance 1 from L, then A'X' is lengthened by d cosec x. The new triangle AXY is
similar to the old. Suppose that perimeter AXY = k·AX, then perimeter AXY is
increased by kd cosec x. Since AXY and A'X'Y' are similar, perimeter A'X'Y' is
shortened by kd cosec x, so the sum of their perimeters is unchanged. It remains to show
that the sum of the perimeters does not depend on the angle x.
Let us move L towards A until L' passes through A', at which point the perimeter of
A'X'Y' is zero. Now if h is the height of AXY (from the base XY), then 1 + h = AA'
sin(45o + x) = sin x + cos x. The perimeter of AXY is h/sin x + h/cos x + h/(sin x cos x)
= h(sin x + cos x + 1)/(sin x cos x) = (sin x + cos x - 1)(sin x + cos x + 1)/(sin x cos x) =
2, which is independent of x.
Problem 3
k > 14 is an integer and p is the largest prime smaller than k. k is chosen so that p ≥ 3k/4.
Prove that 2p does not divide (2p - k)!, but that n does divide (n - k)! for composite n >
2p.
Solution
Since k > p, we have p > 2p-k and hence p does not divide any of 1, 2, 3, ... , 2p-k. But p
is prime, so it does not divide (2p - k)! So 2p does not either.
Now consider composite n > 2p. Note that k > 14, so p ≥ 13, so n > 26. Take q to be the
largest prime divisor of n and put n = qr. We now have three cases.
(1) q > r ≥ 3. Then n > 2p ≥ 3k/2, so 2n/3 > k. Hence n-k > n-2n/3 = n/3 ≥ n/r = q > r. So
q and r are distinct integers < n-k. Hence n = qr divides (n-k)!.
(2) r = 2. Then n = 2q > 2p. But p is the largest prime < k. Hence q ≥ k, so n-k ≥ n-q = q.
Obviously q > 2 (since n > 26), so q and 2 are distinct integers < n-k. Hence n = 2q
divides (n-k)! .
(3) The final case is n = q2. We show that 2q ≤ n-k. Suppose not. Then 2q ≥ n-k+1 ≥
(2p+1)-k+1 ≥ 3k/2 + 1 - k + 1 = k/2 + 2. So q ≥ k/4 + 1. Also since 2q ≥ n-k+1, we have
k ≥ n-2q+1 = (q-1)2 ≥ k2/16. If k > 16, this is a contradiction. If k = 15 or 16, then p = 13
and k ≥ (q-1)2 gives q ≤ 5, so n = q2 ≤ 25. But n > 2p = 26. Contradiction. So we have
established that 2q ≤ n-k. But that means that q and 2q are distinct integers ≤ n-k and so
their product 2n divides (n-k)!.
Thanks to Gerhard Woeginger for this.
Problem 4
Show that (an + bn)1/n + (bn + cn)1/n + (cn + an)1/n < 1 + (21/n)/2, where n > 1 is an integer
and a, b, c are the sides of a triangle with unit perimeter.
Solution
Thanks to Olena Bormashenko
We may take a ≥ b ≥ c. Since a + b + c = 1 and a < b+c, we have b ≤ a < 1/2. Hence (an +
bn)1/n < 21/n/2 (*).
We have (b + c/2)n = bn + n/2 c bn-1 + other positive terms > bn + cn. Hence (bn + cn)1/n <
b + c/2. Similarly, (cn + an)1/n < a + c/2. Adding we get (bn + cn)1/n + (cn + an)1/n < a+b+c
= 1. Adding to (*) gives the required result.
Problem 5
Find the smallest positive integer k such that among any k people, either there are 2m
who can be divided into m pairs of people who know each other, or there are 2n who can
be divided into n pairs of people who do not know each other.
Answer
k = m + n + max(m, n) - 1
Solution
We have to find the smallest k so that any graph with k points has either m disjoint edges
or n disjoint pairs of points with each pair not joined by an edge. Let the smallest k be
f(m, n). For any graph G, there is another graph G' with the same points and the
complementary set of edges (so that A, B are joined by an edge in G' iff they are not
joined in G). This shows that f(m, n) = f(n, m), so there is no loss of generality in
assuming that m ≤ n.
Consider the graph G with m+2n-2 points defined as follows. Let A be a subset of m-1
points and B the remaining 2n-1 points. Every point of A is joined to every other point
(in A or B) and there are no other edges. Now since all points in A are joined to every
other point in G, if two points are not joined then they must both be in B. So there can be
at most n-1 pairs of unjoined points. If two points are joined then at least one of them
must be in A (since if both are in B they are unjoined). But A has less than m points, so
if there are m pairs of joined points, then they must overlap. Thus G shows that f(m, n) >
m+2n-2.
We now prove by induction on m that any graph with m-1+2n points (and m <= n) has
either m joined pairs or n unjoined pairs. It is trivial for m = 1 (if the graph has an edge,
then we have the pair of joined points, if not then we can divide the 2n points arbitrarily
into n pairs of unjoined points). So suppose it is true for m-1. We now use induction on
n. We suppose that a graph G with m-1+2n points does not have m pairs of joined points
or n pairs of unjoined points. We seek a contradiction.
Ignore for a moment the case n = m. Suppose that n > m and we have already established
the result for n-1. Since m-1+2(n-1) < m-1+2n, and G does not have m pairs of joined
points it must have n-1 pairs of unjoined points.
So let B be a set of n-1 pairs of unjoined points. Let A be the remaining m+1 points.
Take any two points P and Q in A and any pair P', Q' in B. If P is not joined to P' and Q
is not joined to Q', then we have a contradiction, because we could remove the pair P', Q'
from B and replace it by the two pairs P, P' and Q, Q', thus getting n pairs of unjoined
points. So we may assume that P is joined to P'. Now remove P from A, leaving m points
and mark the pair P', Q' in B as used. We now repeat. Take any two points in the reduced
A and any unmarked pair in B and conclude that one of the pair in A is joined to one of
the pair in B. Remove the point from A and mark the pair in B. Since n-1 >= m we can
continue in this way until we obtain m pairs of joined points (one of each pair in A and
the other in B). Contradiction. So the result is true for n also.
It remains to consider the case n = m.
To get the n-1 pairs we use the induction on m. G has m-1+2m = 3m-1 points and does
not have either m pairs of joined points or m pairs of unjoined points. We know by
induction that f(m-1, m) = m-1+2m-1 = 3m-2, so f(m, m-1) = 3m-2 also. Since m-1+2m
> 3m-2, G has either m pairs of joined points or m-1 points of unjoined points. It does
not have the former (by assumption) so it must have the latter.
We now proceed as before. So B is a set of m-1 pairs of unjoined points and A the set of
the remaining m+1 points. We get m-1 pairs of joined points (one of each pair in A and
the other in B). But now we have run out of unmarked pairs in B. However, there are
still two points in A. If they are not joined, then we would have a contradiction, because
with the pairs in B they would give n pairs of unjoined points. So they are joined and
hence give us an mth joined pair. Contradiction.
Comment. This all seems complicated, and certainly much harder work than questions
1-3. So maybe I am missing something. Does anyone have a simpler solution?
16th APMO 2004
Problem 1
Find all non-empty finite sets S of positive integers such that if m,n ∈ S, then
(m+n)/gcd(m,n) ∈ S.
Answer
{2}
Solution
Let k ∈ S. Then (k+k)/gcd(k,k) = 2 ∈ S. Let M be the largest odd element of S. Then
(M+2)/gcd(M,2) = M+2 ∈ S. Contradiction. So all elements of S are even.
Let m = 2n be the smallest element of S greater than 2. Then (m+2)/2 = n+1 ∈ S. But n
must be > 1 (or m = 2), so 2n > n+1. Hence 2n = 2 (by minimality of m), so n = 1.
Contradiction. So S has no elements apart from 2.
Last corrected/updated 22 Mar 04
Problem 2
ABC is an acute-angled triangle with circumcenter O and orthocenter H (and O ≠ H).
Show that one of area AOH, area BOH, area COH is the sum of the other two.
Solution
Let G be the centroid. Note that OH is the Euler line, so G lies on OH. wlog A is on one
side of the line OH, and B, C are on the other side. Let M be the midpoint of BC. Then
the length of the perpendicular from M to the line OH is the mean of the lengths of the
perpendiculars from B and C. Thus if ∠MGO = θ, then area BOH + area COH =
OH·GM sin θ = (OH·AG/2) sin θ = area AOH.
Problem 3
2004 points are in the plane, no three collinear. S is the set of lines through any two of
the points. Show that the points can be colored with two colors so that any two of the
points have the same color iff there are an odd number of lines in S which separate them
(a line separates them if they are on opposite sides of it).
Solution
Let us denote by dXY the number of points separating the points X and Y. If the result is
true, then the coloring is effectively determined: take a point X and color it blue. Then
for every other point Y, color it blue iff dXY is odd. This will work provided that given
any three points A, B, C, we have dAB + dBC + dCA is odd. (For then if Y and Z are the
same color, dXY and dXZ have th same parity, so dYZ is odd, which is correct. Similarly, if
Y and Z are opposite colors, then dYZ is even, which is correct.)
We are interested in lines which pass through an interior point of one or more of AB,
BC, CA. Lines cannot pass through all three. If they pass through two, then they do not
affect the parity of dAB + dBC + dCA. So we are interested in lines which pass through A
and the (interior of) BC, and similarly for B, C. Let n1, n2, ... , n7 be the number of points
(excluding A, B, C) in the various regions (as shown). The number of lines through A
and BC is n1 + n2 + n3. So dAB + dBC + dCA = (n1 + n2 + n3) + (n1 + n4 + n5) + (n1 + n6 +
n7) = n1 + n2 + n3 + n4 + n5 + n6 + n7 = (2004 - 3) = 1 mod 2.
Thanks to Dinu Razvan
Problem 4
Show that [(n-1)!/(n2+n)] is even for any positive integer n.
Solution
Thanks to Juan Ignacio Restrepo
For n = 1, 2, 3, 4, 5 we have [(n-1)!/(n2+n)] = 0, which is even. So assume n ≥ 6.
If n and n+1 are composite, then they must divide (n-1)!. They are coprime, so their
product must divide (n-1)!. Note that just one of n, n+1 is even. For m ≥ 6, (m-2)! is
divisible by more powers of 2 than m, so (n-1)!/(n2+n) is even. It remains to consider the
two cases n+1 = p, a prime, and n = p a prime.
If n+1 = p, then p-1 is composite, so p-1 divides (p-2)!. Let k = (p-2)!/(p-1). By Wilson's
theorem we have (p-2)! = 1 mod p, so k(p-1) = 1 mod p and hence k = -1 mod p. So
(k+1)/p is an integer. But k is even, so k+1 is odd and hence (k+1)/p is odd. Now [k/p] =
(k+1)/p - 1, so [k/p] = [(n-1)!/(n2+n)] is even.
If n = p, then k = (p-1)!/(p+1) is an even integer, so k+1 is an odd integer. By Wilson's
theorem, k(p+1) = -1 mod p, so (k+1)/p is an integer and hence an odd integer. Hence
[(n-1)!/(n2+n)] = [k/p] = (k+1)/p - 1 is even.
Wilson's theorem states that p is prime iff (p-1)! = -1 mod p. If p is composite, then it has
a factor ≤ p-1, which divides (p-1)! and so does not divide (p-1)! + 1. Hence (p-1)! ≠ -1
mod p. Now suppose p is prime. If p = 2, then (p-1)! = 1 = -1 mod 2. So assume p is odd.
Note first that if a2 = 1 mod p and 0 < a < p, then a = 1 or p-1. For p divides (a+1)(a-1)
and so it divides either a+1, giving a = p-1, or it divides a-1, giving a = 1. Now for each
0 < a < p, there is a unique a' such that aa' = 1 mod p. We have just shown that a = a'
iff a = 1 or p-1. So we can divide 2, 3, ... , p-2 into pairs with the product of each pair
being 1 mod p. Hence (p-2)! = 1 mod p, as required.
On the powers of 2, note that 2, 22, 23, ... , 2k-1 < 2k and their product is 2k(k-1)/2. If 2k < n
< 2k+1, then n is divisible by at most 2k-1. So for n ≥ 16, for example, (n-8)!/n is even. For
n = 8, 9, ... , 15, (n-2)! is divisible by 2·4·6 and n is divisible by at most 8, so (n-2)!/n is
even. Finally, we can easily check n = 6, 7.
Problem 5
Show that (x2 + 2)(y2 + 2)(z2 + 2) ≥ 9(xy + yz + zx) for any positive reals x, y, z.
Solution
Expanding lhs we get x2y2z2 + 2(x2y2 + y2z2 + z2x2) + 4(x2 + y2 + z2) + 8.
By AM/GM we have 3(x2 + y2 + z2) ≥ 3(xy + yz + zx) and (2x2y2 + 2) + (2y2z2 + 2) +
(2z2x2 + 2) ≥ 4(xy + yz + zx). That leaves us needing x2y2z2 + x2 + y2 + z2 + 2 ≥ 2(xy +
yz + zx) (*). That is hard, because it is not clear how to deal with the xyz on the lhs.
By AM/GM we have (a+b)(b+c)(c+a) ≥ 8abc. So if (u+v-w), (u-v+w), (-u+v+w) are all
non-negative, we can put 2a = -u+v+w, 2b = u-v+w, 2c = u+v-w and get uvw ≥ (u+v+w)(u-v+w)(u+v-w). If u, v, w are positive, then at most one of (u+v-w), (u-v+w), (u+v+w) is negative. In that case uvw ≥ (-u+v+w)(u-v+w)(u+v-w) is trivially true. So it
holds for all positive u, v, w. Expanding, we get u3 + v3 + w3 + 3uvw ≥ uv(u+v) +
vw(v+w) + wu(w+u), which is a fairly well-known inequality. Applying AM/GM to u+v
etc we get, u3 + v3 + w3 + 3uvw ≥ 2(uv)3/2 + 2(vw)3/2 + 2(wu)3/2. Finally, putting u = x2/3,
v = y2/3, w = z2/3, we get x2 + y2 + z2 + 3(xyz)2/3 ≥ 2(xy + yz + zx) (**).
Thus we have x2y2z2 + x2 + y2 + z2 + 2 ≥ ≥ (x2 + y2 + z2) + 3(xyz)2/3, by applying
AM/GM to x2y2z2, 1, 1, and now (**) gives the required (*).
Thanks to Jacob Tsimerman
XVII APMO - March, 2005
Problems and Solutions
Problem 1. Prove that for every irrational real number a, there are irrational real numbers
b and b0 so that a + b and ab0 are both rational while ab and a + b0 are both irrational.
(Solution) Let a be an irrational number. If a2 is irrational, we let b = −a. Then,
a + b = 0 is rational and ab = −a2 is irrational. If a2 is rational, we let b = a2 − a. Then,
a + b = a2 is rational and ab = a2 (a − 1). Since
a=
ab
+1
a2
is irrational, so is ab.
Now, we let b0 =
1
2
or b0 = . Then ab0 = 1 or 2, which is rational. Note that
a
a
a + b0 =
Since,
a2 + 1
a
or a + b0 =
a2 + 2 a2 + 1
1
−
= ,
a
a
a
at least one of them is irrational.
1
a2 + 2
.
a
Problem 2. Let a, b and c be positive real numbers such that abc = 8. Prove that
a2
b2
c2
4
p
+p
+p
≥ .
3
(1 + a3 )(1 + b3 )
(1 + b3 )(1 + c3 )
(1 + c3 )(1 + a3 )
(Solution) Observe that
√
1
2
≥
.
3
2 + x2
1+x
(1)
In fact, this is equivalent to (2 + x2 )2 ≥ 4(1 + x3 ), or x2 (x − 2)2 ≥ 0. Notice that equality
holds in (1) if and only if x = 2.
We substitute x by a, b, c in (1), respectively, to find
c2
+p
(1 + a3 )(1 + b3 )
(1 + b3 )(1 + c3 )
(1 + c3 )(1 + a3 )
4a2
4b2
4c2
≥
+
+
.
(2 + a2 )(2 + b2 ) (2 + b2 )(2 + c2 ) (2 + c2 )(2 + a2 )
p
a2
b2
+p
(2)
We combine the terms on the right hand side of (2) to obtain
Left hand side of (2) ≥
2
2S(a, b, c)
=
,
36 + S(a, b, c)
1 + 36/S(a, b, c)
(3)
where S(a, b, c) := 2(a2 + b2 + c2 ) + (ab)2 + (bc)2 + (ca)2 . By AM-GM inequality, we have
p
3
a2 + b2 + c2 ≥ 3p
(abc)2 = 12 ,
(ab)2 + (bc)2 + (ca)2 ≥ 3 3 (abc)4 = 48 .
Note that the equalities holds if and only if a = b = c = 2. The above inequalities yield
Therefore
S(a, b, c) = 2(a2 + b2 + c2 ) + (ab)2 + (bc)2 + (ca)2 ≥ 72 .
(4)
2
4
2
≥
= ,
1 + 36/S(a, b, c)
1 + 36/72
3
(5)
which is the required inequality.
2
Problem 3. Prove that there exists a triangle which can be cut into 2005 congruent
triangles.
(Solution) Suppose that one side of a triangle has length n. Then it can be cut into n2
congruent triangles which are similar to the original one and whose corresponding sides to
the side of length n have lengths 1.
Since 2005 = 5 × 401 where 5 and 401 are primes and both primes are of the type
4k + 1, it is representable as a sum of two integer squares. Indeed, it is easy to see that
2005 = 5 × 401 = (22 + 1)(202 + 1)
= 402 + 202 + 22 + 1
= (40 − 1)2 + 2 × 40 + 202 + 22
= 392 + 222 .
Let ABC be a right-angled triangle with the legs AB and BC having lengths 39 and
22, respectively. We draw the altitude BK, which divides ABC into two similar triangles.
Now we divide ABK into 392 congruent triangles as described above and BCK into 222
congruent triangles. Since ABK is similar to BKC, all 2005 triangles will be congruent.
3
Problem 4. In a small town, there are n × n houses indexed by (i, j) for 1 ≤ i, j ≤ n with
(1, 1) being the house at the top left corner, where i and j are the row and column indices,
respectively. At time 0, a fire breaks out at the house indexed by (1, c), where c ≤ n2 .
During each subsequent time interval [t, t + 1], the fire fighters defend a house which is not
yet on fire while the fire spreads to all undefended neighbors of each house which was on
fire at time t. Once a house is defended, it remains so all the time. The process ends when
the fire can no longer spread. At most how many houses can be saved by the fire fighters?
A house indexed by (i, j) is a neighbor of a house indexed by (k, `) if |i − k| + |j − `| = 1.
(Solution) At most n2 + c2 − nc − c houses can be saved. This can be achieved under the
following order of defending:
(2, c), (2, c + 1); (3, c − 1), (3, c + 2); (4, c − 2), (4, c + 3); . . .
(c + 1, 1), (c + 1, 2c); (c + 1, 2c + 1), . . . , (c + 1, n).
(6)
Under this strategy, there are
2 columns (column numbers c, c + 1) at which n − 1 houses are saved
2 columns (column numbers c − 1, c + 2) at which n − 2 houses are saved
···
2 columns (column numbers 1, 2c) at which n − c houses are saved
n − 2c columns (column numbers n − 2c + 1, . . . , n) at which n − c houses are saved
Adding all these we obtain :
2[(n − 1) + (n − 2) + · · · + (n − c)] + (n − 2c)(n − c) = n2 + c2 − cn − c.
(7)
We say that a house indexed by (i, j) is at level t if |i − 1| + |j − c| = t. Let d(t) be the
number of houses at level t defended by time t, and p(t) be the number of houses at levels
greater than t defended by time t. It is clear that
p(t) +
t
X
d(i) ≤ t and p(t + 1) + d(t + 1) ≤ p(t) + 1.
i=1
Let s(t) be the number of houses at level t which are not burning at time t. We prove that
s(t) ≤ t − p(t) ≤ t
for 1 ≤ t ≤ n − 1 by induction. It is obvious when t = 1. Assume that it is true for
t = k. The union of the neighbors of any k − p(k) + 1 houses at level k + 1 contains at
least k − p(k) + 1 vertices at level k. Since s(k) ≤ k − p(k), one of these houses at level
k is burning. Therefore, at most k − p(k) houses at level k + 1 have no neighbor burning.
Hence we have
s(k + 1) ≤ k − p(k) + d(k + 1)
= (k + 1) − (p(k) + 1 − d(k + 1))
≤ (k + 1) − p(k + 1).
4
We now prove that the strategy given above is optimal. Since
n−1
X
t=1
µ ¶
n
s(t) ≤
,
2
the maximum number of houses
¡n¢at levels less than or equal to n − 1, that can be saved
under any strategy is at most 2 , which is realized by the strategy above. Moreover, at
levels bigger than n − 1, every house is saved under the strategy above.
The following is an example
when n = 11 and c = 4. The houses with ° mark are
N
burned. The houses with
mark are blocked ones and hence those and the houses below
them are saved.
J
J
J
J
•
J
J
J
J
J
J
J
J
J
J
N
N
J
J
J
J
J
J
J
J
N
N
J
J
J
J
J
J
N
N
J
J
J
J
N
N
N
N
N
5
Problem 5. In a triangle ABC, points M and N are on sides AB and AC, respectively,
such that M B = BC = CN . Let R and r denote the circumradius and the inradius of the
triangle ABC, respectively. Express the ratio M N/BC in terms of R and r.
(Solution) Let ω, O and I be the circumcircle, the circumcenter and the incenter of ABC,
respectively. Let D be the point of intersection of the line BI and the circle ω such that
D 6= B. Then D is the midpoint of the arc AC. Hence OD ⊥ CN and OD = R.
We first show that triangles M N C and IOD are similar. Because BC = BM , the line
BI (the bisector of ∠M BC) is perpendicular to the line CM . Because OD ⊥ CN and
ID ⊥ M C, it follows that
∠ODI = ∠N CM
(8)
Let ∠ABC = 2β. In the triangle BCM , we have
CM
CM
=
= 2 sin β
NC
BC
(9)
Since ∠DIC = ∠DCI, we have ID = CD = AD. Let E be the point of intersection
of the line DO and the circle ω such that E 6= D. Then DE is a diameter of ω and
∠DEC = ∠DBC = β. Thus we have
DI
CD
2R sin β
=
=
= 2 sin β.
OD
OD
R
(10)
Combining equations (8), (9), and (10) shows that triangles M N C and IOD are similar.
It follows that
MN
MN
IO
IO
=
=
=
.
(11)
BC
NC
OD
R
The well-known Euler’s formula states that
OI 2 = R2 − 2Rr.
Therefore,
MN
=
BC
r
1−
2r
.
R
(12)
(13)
(Alternative Solution) Let a (resp., b, c) be the length of BC (resp., AC, AB). Let α
(resp., β, γ) denote the angle ∠BAC (resp., ∠ABC, ∠ACB). By introducing coordinates
B = (0, 0), C = (a, 0), it is immediate that the coordinates of M and N are
M = (a cos β, a sin β), N = (a − a cos γ, a sin γ),
6
(14)
respectively. Therefore,
(M N/BC)2 = [(a − a cos γ − a cos β)2 + (a sin γ − a sin β)2 ]/a2
= (1 − cos γ − cos β)2 + (sin γ − sin β)2
= 3 − 2 cos γ − 2 cos β + 2(cos γ cos β − sin γ sin β)
= 3 − 2 cos γ − 2 cos β + 2 cos(γ + β)
= 3 − 2 cos γ − 2 cos β − 2 cos α
= 3 − 2(cos γ + cos β + cos α).
Now we claim
cos γ + cos β + cos α =
From
(15)
r
+ 1.
R
(16)
a = b cos γ + c cos β
b = c cos α + a cos γ
c = a cos β + b cos α
(17)
we get
a(1 + cos α) + b(1 + cos β) + c(1 + cos γ) = (a + b + c)(cos α + cos β + cos γ).
(18)
Thus
cos α + cos β + cos γ
1
=
(a(1 + cos α) + b(1 + cos β) + c(1 + cos γ))
a+b+c
µ µ
¶
µ
¶
µ
¶¶
1
b2 + c2 − a2
a2 + c2 − b2
a2 + b2 − c2
=
a 1+
+b 1+
+c 1+
a+b+c
2bc
2ac
2ab
1
=
a+b+c
=1+
µ
a2 (b2 + c2 − a2 ) + b2 (a2 + c2 − b2 ) + c2 (a2 + b2 − c2 )
a+b+c+
2abc
¶
2a2 b2 + 2b2 c2 + 2c2 a2 − a4 − b4 − c4
.
2abc(a + b + c)
(19)
a
it follows that
On the other hand, from R =
2 sin α
R2 =
=
a2
=
4(1 − cos2 α)
Ã
4 1−
µ
a2
b2 + c2 − a2
2bc
a2 b2 c2
.
2a2 b2 + 2b2 c2 + 2c2 a2 − a4 − b4 − c4
7
¶2 !
(20)
Also from
1
1
(a + b + c)r = bc sin α, it follows that
2
2
Ã
r2 =
=
2 2
2
b c (1 − cos α)
=
(a + b + c)2
µ
b2 c2 1 −
b2 + c2 − a2
2bc
¶2 !
(a + b + c)2
(21)
2a2 b2 + 2b2 c2 + 2c2 a2 − a4 − b4 − c4
.
4(a + b + c)2
Combining (19), (20) and (21), we get (16) as desired.
Finally, by (15) and (16) we have
MN
=
BC
r
1−
2r
.
R
(22)
Another proof of (16) from R.A. Johnson’s “Advanced Euclidean Geometry”1 :
Construct the perpendicular bisectors OD, OE, OF , where D, E, F are the midpoints
of BC, CA, AB, respectively. By Ptolemy’s Theorem applied to the cyclic quadrilateral
OEAF , we get
a
b
c
· R = · OF + · OE.
2
2
2
Similarly
b
c
a
c
a
b
· R = · OD + · OF,
· R = · OE + · OD.
2
2
2
2
2
2
Adding, we get
sR = OD ·
b+c
c+a
a+b
+ OE ·
+ OF ·
,
2
2
2
(23)
a
where s is the semiperimeter. But also, the area of triangle OBC is OD · , and adding
2
similar formulas for the areas of triangles OCA and OAB gives
b
c
a
+ OE · + OF ·
2
2
2
Adding (23) and (24) gives s(R + r) = s(OD + OE + OF ), or
rs = 4ABC = OD ·
OD + OE + OF = R + r.
Since OD = R cos A etc., (16) follows.
1
This proof was introduced to the coordinating country by Professor Bill Sands of Canada.
8
(24)
Problem 1. Let n be a positive integer. Find the largest nonnegative real number f (n)
(depending on n) with the following property: whenever a1 , a2 , . . . , an are real numbers
such that a1 + a2 + · · · + an is an integer, there exists some i such that | ai − 21 | ≥ f (n).
(Solution) The answer is
½
f (n) =
0
1
2n
if n is even,
if n is odd.
First, assume that n is even. If ai = 12 for all i, then the sum a1 + a2 + · · · + an is an
integer. Since | ai − 12 |= 0 for all i, we may conclude f (n) = 0 for any even n.
1
1
PnNow assume that n is odd. Suppose that | ai − 2 |< 2n for all 1 ≤ i ≤ n. Then, since
i=1 ai is an integer,
¯ n
¯
¯
n ¯
1 ¯¯
n ¯¯ X ¯¯
1
1 ¯¯X
1
ai − ¯ ≤
ai − ¯ <
≤¯
·n= ,
¯
2 ¯ i=1
2 ¯ i=1
2
2n
2
1
a contradiction. Thus |ai − 12 | ≥ 2n
for some
P i, as required. On the other hand, putting
m
n = 2m + 1 and ai = 2m+1 for all i gives
ai = m, while
¯
¯
¯
¯
1
1
¯ai − 1 ¯ = 1 − m =
=
¯
¯
2
2 2m + 1
2(2m + 1)
2n
for all i. Therefore, f (n) =
1
2n
is the best possible for any odd n.
Problem 2. Prove that every positive integer
can be written as a finite sum of distinct
√
1+ 5
integral powers of the golden mean τ = 2 . Here, an integral power of τ is of the form
τ i , where i is an integer (not necessarily positive).
(Solution) We will prove this statement by induction using the equality
τ 2 = τ + 1.
If n = 1, then 1 = τ 0 . Suppose that n − 1 can be written as a finite sum of integral powers
of τ , say
k
X
n−1=
ai τ i ,
(1)
i=−k
where ai ∈ {0, 1} and n ≥ 2. We will write (1) as
n − 1 = ak · · · a1 a0 .a−1 a−2 · · · a−k .
For example,
1 = 1.0 = 0.11 = 0.1011 = 0.101011.
(2)
Firstly, we will prove that we may assume that in (2) we have ai ai+1 = 0 for all i with
−k ≤ i ≤ k − 1. Indeed, if we have several occurrences of 11, then we take the leftmost
such occurrence. Since we may assume that it is preceded by a 0, we can replace 011
with 100 using the identity τ i+1 + τ i = τ i+2 . By doing so repeatedly, if necessary, we will
eliminate all occurrences of two 1’s standing together. Now we have the representation
n−1=
K
X
bi τ i ,
(3)
i=−K
where bi ∈ {0, 1} and bi bi+1 = 0.
If b0 = 0 in (3), then we just add 1 = τ 0 to both sides of (3) and we are done.
Suppose now that there is 1 in the unit position of (3), that is b0 = 1. If there are two
0’s to the right of it, i.e.
n − 1 = · · · 1.00 · · · ,
then we can replace 1.00 with 0.11 because 1 = τ −1 + τ −2 , and we are done because we
obtain 0 in the unit position. Thus we may assume that
n − 1 = · · · 1.010 · · · .
Again, if we have n − 1 = · · · 1.0100 · · · , we may rewrite it as
n − 1 = · · · 1.0100 · · · = · · · 1.0011 · · · = · · · 0.1111 · · ·
and obtain 0 in the unit position. Therefore, we may assume that
n − 1 = · · · 1.01010 · · · .
Since the number of 1’s is finite, eventually we will obtain an occurrence of 100 at the end,
i.e.
n − 1 = · · · 1.01010 · · · 100.
Then we can shift all 1’s to the right to obtain 0 in the unit position, i.e.
n − 1 = · · · 0.11 · · · 11,
and we are done.
Problem 3. Let p ≥ 5 be a prime and let r be the number of ways of placing p checkers
on a p × p checkerboard so that not all checkers are in the same row (but they may all be
in the same column). Show that r is divisible by p 5 . Here, we assume that all the checkers
are identical.
µ 2¶
p
− p . Hence, it suffices to show that
(Solution) Note that r =
p
(p 2 − 1)(p 2 − 2) · · · (p 2 − (p − 1)) − (p − 1)! ≡ 0 (mod p 4 ).
(1)
Now, let
f (x) := (x − 1)(x − 2) · · · (x − (p − 1)) = xp−1 + sp−2 xp−2 + · · · + s1 x + s0 .
(2)
Then the congruence equation (1) is same as f (p 2 )−s0 ≡ 0 (mod p 4 ). Therefore, it suffices
to show that s1 p 2 ≡ 0 (mod p 4 ) or s1 ≡ 0 (mod p 2 ).
Since ap−1 ≡ 1 (mod p) for all 1 ≤ a ≤ p − 1, we can factor
xp−1 − 1 ≡ (x − 1)(x − 2) · · · (x − (p − 1))
(mod p).
(3)
Comparing the coefficients of the left hand side of (3) with those of the right hand side
of (2), we obtain p | si for all 1 ≤ i ≤ p − 2 and s0 ≡ −1 (mod p). On the other hand,
plugging p for x in (2), we get
f (p) = (p − 1)! = s0 = p p−1 + sp−2 p p−2 + · · · + s1 p + s0 ,
which implies
p p−1 + sp−2 p p−2 + · · · + s2 p 2 = −s1 p.
Since p ≥ 5, p | s2 and hence s1 ≡ 0 (mod p 2 ) as desired.
Problem 4. Let A, B be two distinct points on a given circle O and let P be the midpoint
of the line segment AB. Let O1 be the circle tangent to the line AB at P and tangent to
the circle O. Let ` be the tangent line, different from the line AB, to O1 passing through
A. Let C be the intersection point, different from A, of ` and O. Let Q be the midpoint
of the line segment BC and O2 be the circle tangent to the line BC at Q and tangent to
the line segment AC. Prove that the circle O2 is tangent to the circle O.
(Solution) Let S be the tangent point of the circles O and O1 and let T be the intersection
point, different from S, of the circle O and the line SP . Let X be the tangent point of
` to O1 and let M be the midpoint of the line segment XP . Since ∠T BP = ∠ASP , the
triangle T BP is similar to the triangle ASP . Therefore,
PA
PT
=
.
PB
PS
Since the line ` is tangent to the circle O1 at X, we have
∠SP X = 90◦ − ∠XSP = 90◦ − ∠AP M = ∠P AM
which implies that the triangle P AM is similar to the triangle SP X. Consequently,
MP
XP
XS
=
=
XP
MA
2M A
and
XP
MA
=
.
PS
AP
From this and the above observation follows
XS P T
XP P A
XP M A
1
·
=
·
=
·
= .
XP P B
2M A P S
2M A XP
2
(1)
Let A0 be the intersection point of the circle O and the perpendicular bisector of the chord
BC such that A, A0 are on the same side of the line BC, and N be the intersection point
of the lines A0 Q and CT . Since
∠N CQ = ∠T CB = ∠T CA = ∠T BA = ∠T BP
and
∠CAB
∠XAP
=
= ∠P AM = ∠SP X,
2
2
the triangle N CQ is similar to the triangle T BP and the triangle CA0 Q is similar to the
triangle SP X. Therefore
∠CA0 Q =
QN
PT
=
QC
PB
and
QC
XS
=
.
0
QA
XP
and hence QA0 = 2QN by (1). This implies that N is the midpoint of the line segment
QA0 . Let the circle O2 touch the line segment AC at Y . Since
∠ACN = ∠ACT = ∠BCT = ∠QCN
and | CY | = | CQ|, the triangles Y CN and QCN are congruent and hence N Y ⊥ AC and
N Y = N Q = N A0 . Therefore, N is the center of the circle O2 , which completes the proof.
Remark : Analytic solutions are possible : For example, one can prove for a triangle ABC
inscribed in a circle O that AB = k(2 + 2t), AC = k(1 + 2t), BC = k(1 + 4t) for some
positive numbers k, t if and only if there exists a circle O1 such that O1 is tangent to the
side AB at its midpoint, the side AC and the circle O.
One obtains AB = k 0 (1 + 4t0 ), AC = k 0 (1 + 2t0 ), BC = k 0 (2 + 2t0 ) by substituting
t = 1/4t0 and k = 2k 0 t0 . So, there exists a circle O2 such that O2 is tangent to the side BC
at its midpoint, the side AC and the circle O.
tan α
In the above, t = tan2 α and k = (1+tan24R
, where R is the radius of O and
α)(1+4 tan2 α)
4R
tan
γ
∠A = 2α. Furthermore, t0 = tan2 γ and k 0 = (1+tan2 γ)(1+4 tan2 γ) , where ∠C = 2γ. Observe
√
T
XS
that tt0 = tan α · tan γ = XP
· PP B
= 12 , which implies tt0 = 41 . It is now routine easy to
check that k = 2k 0 t0 .
Problem 5. In a circus, there are n clowns who dress and paint themselves up using a
selection of 12 distinct colours. Each clown is required to use at least five different colours.
One day, the ringmaster of the circus orders that no two clowns have exactly the same set
of colours and no more than 20 clowns may use any one particular colour. Find the largest
number n of clowns so as to make the ringmaster’s order possible.
(Solution) Let C be the set of n clowns. Label the colours 1, 2, 3, . . . , 12. For each
i = 1, 2, . . . , 12, let Ei denote the set of clowns who use colour i. For each subset S of
{1, 2, . . . , 12}, let ES be the set of clowns who use exactly those colours in S. Since S 6= S 0
implies ES ∩ ES 0 = ∅, we have
X
|ES | = |C| = n,
S
where S runs over all subsets of {1, 2, . . . , 12}. Now for each i,
ES ⊆ Ei
if and only if i ∈ S,
and hence
|Ei | =
X
|ES |.
i∈S
By assumption, we know that |Ei | ≤ 20 and that if ES 6= ∅, then |S| ≥ 5. From this we
obtain
Ã
!
12
12
X
X
X
X
20 × 12 ≥
|Ei | =
|ES | ≥ 5
|ES | = 5n.
i=1
i=1
i∈S
S
Therefore n ≤ 48.
Now, define a sequence {ci }52
i=1 of colours in the following way:
1
4
3
2
2
1
4
3
3
2
1
4
4
3
2
1
|
|
|
|
5
8
7
6
6
5
8
7
7
6
5
8
8
7
6
5
|
|
|
|
9 10 11 12
12 9 10 11
11 12 9 10
10 11 12 9
|
|
|
|1234
The first row lists c1 , . . . , c12 in order, the second row lists c13 , . . . , c24 in order, the third
row lists c25 , . . . , c36 in order, and finally the last row lists c37 , . . . , c52 in order. For each
j, 1 ≤ j ≤ 48, assign colours cj , cj+1 , cj+2 , cj+3 , cj+4 to the j-th clown. It is easy to check
that this assignment satisfies all conditions given above. So, 48 is the largest for n.
Remark : The fact that n ≤ 48 can be obtained in a much simpler observation that
5n ≤ 12 × 20 = 240.
There are many other ways of constructing 48 distinct sets consisting of 5 colours. For
example, consider the sets
{1, 2, 3, 4, 5, 6},
{3, 4, 5, 6, 7, 8}, {5, 6, 7, 8, 9, 10}, {7, 8, 9, 10, 11, 12},
{9, 10, 11, 12, 1, 2}, {11, 12, 1, 2, 3, 4}, {1, 2, 5, 6, 9, 10}, {3, 4, 7, 8, 11, 12}.
Each of the above 8 sets has 6 distinct subsets consisting of exactly 5 colours. It is easy to
check that the 48 subsets obtained in this manner are all distinct.
XIX Asian Pacific Mathematics Olympiad
Problem 1. Let S be a set of 9 distinct integers all of whose prime factors are at most 3.
Prove that S contains 3 distinct integers such that their product is a perfect cube.
Solution. Without loss of generality, we may assume that S contains only positive integers.
Let
S = {2ai 3bi | ai , bi ∈ Z, ai , bi ≥ 0, 1 ≤ i ≤ 9}.
It suffices to show that there are 1 ≤ i1 , i2 , i3 ≤ 9 such that
ai1 + ai2 + ai3 ≡ bi1 + bi2 + bi3 ≡ 0 (mod 3).
(†)
For n = 2a 3b ∈ S, let’s call (a (mod 3), b (mod 3)) the type of n. Then there are 9 possible
types :
(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2).
Let N (i, j) be the number of integers in S of type (i, j). We obtain 3 distinct integers
whose product is a perfect cube when
(1) N (i, j) ≥ 3 for some i, j, or
(2) N (i, 0)N (i, 1)N (i, 2) 6= 0 for some i = 0, 1, 2, or
(3) N (0, j)N (1, j)N (2, j) 6= 0 for some j = 0, 1, 2, or
(4) N (i1 , j1 )N (i2 , j2 )N (i3 , j3 ) 6= 0, where {i1 , i2 , i3 } = {j1 , j2 , j3 } = {0, 1, 2}.
Assume that none of the conditions (1)∼(3) holds. Since N (i, j) ≤ 2 for all (i, j), there
are at least five N (i, j)’s that are nonzero. Furthermore, among those nonzero N (i, j)’s, no
three have the same i nor the same j. Using these facts, one may easily conclude that the
condition (4) should hold. (For example, if one places each nonzero N (i, j) in the (i, j)-th
box of a regular 3 × 3 array of boxes whose rows and columns are indexed by 0,1 and 2,
then one can always find three boxes, occupied by at least one nonzero N (i, j), whose rows
and columns are all distinct. This implies (4).)
Second solution. Up to (†), we do the same as above and get 9 possible types :
(a (mod 3), b (mod 3)) = (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)
for n = 2a 3b ∈ S.
Note that (i) among any 5 integers, there exist 3 whose sum is 0 (mod 3), and that (ii)
if i, j, k ∈ {0, 1, 2}, then i + j + k ≡ 0 (mod 3) if and only if i = j = k or {i, j, k} = {0, 1, 2}.
Let’s define
T : the set of types of the integers in S ;
N (i) : the number of integers in S of the type (i, ·) ;
M (i) : the number of integers j ∈ {0, 1, 2} such that (i, j) ∈ T .
If N (i) ≥ 5 for some i, the result follows from (i). Otherwise, for some permutation (i, j, k)
of (0, 1, 2),
N (i) ≥ 3, N (j) ≥ 3, N (k) ≥ 1.
If M (i) or M (j) is 1 or 3, the result follows from (ii). Otherwise M (i) = M (j) = 2. Then
either
(i, x), (i, y), (j, x), (j, y) ∈ T or (i, x), (i, y), (j, x), (j, z) ∈ T
for some permutation (x, y, z) of (0, 1, 2). Since N (k) ≥ 1, at least one of (k, x), (k, y) and
(k, z) contained in T . Therefore, in any case, the result follows from (ii). (For example, if
(k, y) ∈ T , then take (i, y), (j, y), (k, y) or (i, x), (j, z), (k, y) from T .)
Problem 2. Let ABC be an acute angled triangle with ∠BAC = 60◦ and AB > AC. Let
I be the incenter, and H the orthocenter of the triangle ABC. Prove that
2∠AHI = 3∠ABC.
Solution. Let D be the intersection point of the lines AH and BC. Let K be the
intersection point of the circumcircle O of the triangle ABC and the line AH. Let the line
through I perpendicular to BC meet BC and the minor arc BC of the circumcircle O at
E and N , respectively. We have
1
1
∠BIC = 180◦ − (∠IBC + ∠ICB) = 180◦ − (∠ABC + ∠ACB) = 90◦ + ∠BAC = 120◦
2
2
and also ∠BN C = 180◦ − ∠BAC = 120◦ = ∠BIC. Since IN ⊥ BC, the quadrilateral
BICN is a kite and thus IE = EN .
Now, since H is the orthocenter of the triangle ABC, HD = DK. Also because
ED ⊥ IN and ED ⊥ HK, we conclude that IHKN is an isosceles trapezoid with
IH = N K.
Hence
∠AHI = 180◦ − ∠IHK = 180◦ − ∠AKN = ∠ABN.
Since IE = EN and BE ⊥ IN , the triangles IBE and N BE are congruent. Therefore
1
∠N BE = ∠IBE = ∠IBC = ∠IBA = ∠ABC
2
and thus
3
∠AHI = ∠ABN = ∠ABC.
2
Second solution. Let P, Q and R be the intersection points of BH, CH and AH with
AC, AB and BC, respectively. Then we have ∠IBH = ∠ICH. Indeed,
1
∠IBH = ∠ABP − ∠ABI = 30◦ − ∠ABC
2
and
1
1
∠ICH = ∠ACI − ∠ACH = ∠ACB − 30◦ = 30◦ − ∠ABC,
2
2
◦
◦
because ∠ABH = ∠ACH = 30 and ∠ACB+∠ABC = 120 . (Note that ∠ABP > ∠ABI
and ∠ACI > ∠ACH because AB is the longest side of the triangle ABC under the given
conditions.) Therefore BIHC is a cyclic quadrilateral and thus
1
∠BHI = ∠BCI = ∠ACB.
2
On the other hand,
∠BHR = 90◦ − ∠HBR = 90◦ − (∠ABC − ∠ABH) = 120◦ − ∠ABC.
Therefore,
1
∠AHI = 180◦ − ∠BHI − ∠BHR = 60◦ − ∠ACB + ∠ABC
2
3
1
◦
◦
= 60 − (120 − ∠ABC) + ∠ABC = ∠ABC.
2
2
Problem 3. Consider n disks C1 , C2 , . . . , Cn in a plane such that for each 1 ≤ i < n, the
center of Ci is on the circumference of Ci+1 , and the center of Cn is on the circumference
of C1 . Define the score of such an arrangement of n disks to be the number of pairs (i, j)
for which Ci properly contains Cj . Determine the maximum possible score.
Solution. The answer is (n − 1)(n − 2)/2.
Let’s call a set of n disks satisfying the given conditions an n-configuration. For an nconfiguration C = {C1 , . . . , Cn }, let SC = {(i, j) | Ci properly contains Cj }. So, the score
of an n-configuration C is |SC |.
We’ll show that (i) there is an n-configuration C for which |SC | = (n − 1)(n − 2)/2, and
that (ii) |SC | ≤ (n − 1)(n − 2)/2 for any n-configuration C.
Let C1 be any disk. Then for i = 2, . . . , n − 1, take Ci inside Ci−1 so that the circumference of Ci contains the center of Ci−1 . Finally, let Cn be a disk whose center is on
the circumference of C1 and whose circumference contains the center of Cn−1 . This gives
SC = {(i, j) | 1 ≤ i < j ≤ n − 1} of size (n − 1)(n − 2)/2, which proves (i).
For any n-configuration C, SC must satisfy the following properties:
(1) (i, i) 6∈ SC ,
(2) (i + 1, i) 6∈ SC , (1, n) 6∈ SC ,
(3) if (i, j), (j, k) ∈ SC , then (i, k) ∈ SC ,
(4) if (i, j) ∈ SC , then (j, i) 6∈ SC .
Now we show that a set G of ordered pairs of integers between 1 and n, satisfying the
conditions (1)∼(4), can have no more than (n − 1)(n − 2)/2 elements. Suppose that there
exists a set G that satisfies the conditions (1)∼(4), and has more than (n − 1)(n − 2)/2
elements. Let n be the least positive integer with which there exists such a set G. Note
that G must have (i, i + 1) for some 1 ≤ i ≤ n or (n, 1), since otherwise G can have at
most
µ ¶
n
n(n − 3)
(n − 1)(n − 2)
−n=
<
2
2
2
elements. Without loss of generality we may assume that (n, 1) ∈ G. Then (1, n − 1) 6∈ G,
since otherwise the condition (3) yields (n, n−1) ∈ G contradicting the condition (2). Now
let G0 = {(i, j) ∈ G | 1 ≤ i, j ≤ n − 1}, then G0 satisfies the conditions (1)∼(4), with n − 1.
We now claim that |G − G0 | ≤ n − 2 :
Suppose that |G − G0 | > n − 2, then |G − G0 | = n − 1 and hence for each 1 ≤ i ≤ n − 1,
either (i, n) or (n, i) must be in G. We already know that (n, 1) ∈ G and (n − 1, n) ∈ G
(because (n, n − 1) 6∈ G) and this implies that (n, n − 2) 6∈ G and (n − 2, n) ∈ G. If we
keep doing this process, we obtain (1, n) ∈ G, which is a contradiction.
Since |G − G0 | ≤ n − 2, we obtain
|G0 | ≥
(n − 1)(n − 2)
(n − 2)(n − 3)
− (n − 2) =
.
2
2
This, however, contradicts the minimality of n, and hence proves (ii).
√
√
√
Problem 4. Let x, y and z be positive real numbers such that x + y + z = 1. Prove
that
x2 + yz
y 2 + zx
z 2 + xy
p
+p
+p
≥ 1.
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
Solution. We first note that
x2 − x(y + z) + yz
x(y + z)
p
+p
2x2 (y + z)
2x2 (y + z)
2x2 (y + z)
r
(x − y)(x − z)
y+z
+
= p
2
2
2x (y + z)
√
√
y+ z
(x − y)(x − z)
≥ p
+
.
2
2x2 (y + z)
p
x2 + yz
=
(1)
Similarly, we have
√
√
(y − z)(y − x)
z+ x
p
≥ p
+
,
2
2y 2 (z + x)
2y 2 (z + x)
√
√
x+ y
z 2 + xy
(z − x)(z − y)
p
≥ p
+
.
2
2z 2 (x + y)
2z 2 (x + y)
y 2 + zx
(2)
(3)
We now add (1)∼(3) to get
x2 + yz
y 2 + zx
z 2 + xy
p
+p
+p
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
√
(x − y)(x − z) (y − z)(y − x) (z − x)(z − y) √
√
≥ p
+ p
+ p
+ x+ y+ z
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
(x − y)(x − z) (y − z)(y − x) (z − x)(z − y)
= p
+ p
+ p
+ 1.
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
Thus, it suffices to show that
(x − y)(x − z) (y − z)(y − x) (z − x)(z − y)
p
+ p
+ p
≥ 0.
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
Now, assume without loss of generality, that x ≥ y ≥ z. Then we have
(x − y)(x − z)
p
≥0
2x2 (y + z)
(4)
and
(z − x)(z − y) (y − z)(y − x)
(y − z)(x − z) (y − z)(x − y)
p
+ p
= p
− p
2z 2 (x + y)
2y 2 (z + x)
2z 2 (x + y) Ã 2y 2 (z + x)
!
(y − z)(x − y) (y − z)(x − y)
1
1
≥ p
− p
= (y − z)(x − y) p
−p
.
2z 2 (x + y)
2y 2 (z + x)
2z 2 (x + y)
2y 2 (z + x)
The last quantity is non-negative due to the fact that
y 2 (z + x) = y 2 z + y 2 x ≥ yz 2 + z 2 x = z 2 (x + y).
This completes the proof.
Second solution. By Cauchy-Schwarz inequality,
Ã
!
x2
y2
z2
p
+p
+p
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
p
p
p
√
√
√
× ( 2(y + z) + 2(z + x) + 2(x + y)) ≥ ( x + y + z)2 = 1,
(5)
and
Ã
yz
zx
xy
!
p
+p
+p
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
r
µr
¶2
r
p
p
p
yz
zx
xy
× ( 2(y + z) + 2(z + x) + 2(x + y)) ≥
+
+
.
x
y
z
(6)
We now combine (5) and (6) to find
Ã
!
x2 + yz
y 2 + zx
z 2 + xy
p
+p
+p
2x2 (y + z)
2y 2 (z + x)
2z 2 (x + y)
p
p
p
× ( 2(x + y) + 2(y + z) + 2(z + x))
r
r
¶2
µr
¶
µr
r
r
yz
zx
xy
yz
zx
xy
+
+
≥2
+
+
.
≥1+
x
y
z
x
y
z
Thus, it suffices to show that
r
µr
¶
r
p
p
p
yz
zx
xy
+
+
2
≥ 2(y + z) + 2(z + x) + 2(x + y) .
x
y
z
Consider the following inequality using AM-GM inequality
r
r µ r
r
·r
µ r
¶¸2
¶
yz
1 zx 1 xy
yz 1 zx 1 xy
+
+
≥4
+
= 2(y + z),
x
2 y
2 z
x 2 y
2 z
(7)
or equivalently
r
yz
+
x
r
µ r
¶
p
1 zx 1 xy
+
≥ 2(y + z) .
2 y
2 z
Similarly, we have
r
¶
µ r
p
zx
1 xy 1 yz
+
+
≥ 2(z + x) ,
2 z
2 x
ry
¶
µ r
r
p
xy
1 yz 1 zx
≥ 2(x + y) .
+
+
z
2 x
2 y
r
Adding the last three inequalities, we get
r
¶
µr
r
p
p
p
yz
zx
xy
≥ 2(y + z) + 2(z + x) + 2(x + y) .
2
+
+
x
y
z
This completes the proof.
Problem 5. A regular (5 × 5)-array of lights is defective, so that toggling the switch for
one light causes each adjacent light in the same row and in the same column as well as
the light itself to change state, from on to off, or from off to on. Initially all the lights are
switched off. After a certain number of toggles, exactly one light is switched on. Find all
the possible positions of this light.
Solution. We assign the following first labels to the 25 positions of the lights:
1
0
1
0
1
1
0
1
0
1
0
0
0
0
0
1
0
1
0
1
1
0
1
0
1
For each on-off combination of lights in the array, define its first value to be the sum
of the first labels of those positions at which the lights are switched on. It is easy to
check that toggling any switch always leads to an on-off combination of lights whose first
value has the same parity(the remainder when divided by 2) as that of the previous on-off
combination.
The 90◦ rotation of the first labels gives us another labels (let us call it the second
labels) which also makes the parity of the second value(the sum of the second labels of
those positions at which the lights are switched on) invariant under toggling.
1
1
0
1
1
0
0
0
0
0
1
1
0
1
1
0
0
0
0
0
1
1
0
1
1
Since the parity of the first and the second values of the initial status is 0, after certain
number of toggles the parity must remain unchanged with respect to the first labels and
the second labels as well. Therefore, if exactly one light is on after some number of toggles,
the label of that position must be 0 with respect to both labels. Hence according to the
above pictures, the possible positions are the ones marked with ∗i ’s in the following picture:
∗2
∗1
∗0
∗3
∗4
Now we demonstrate that all five positions are possible :
Toggling the positions checked by t (the order of toggling is irrelevant) in the first
picture makes the center(∗0 ) the only position with light on and the second picture makes
the position ∗1 the only position with light on. The other ∗i ’s can be obtained by rotating
the second picture appropriately.
t t
t
t t
t
t
t t
t
t
t
t t
t
t
t t
t
t t
t
XX Asian Pacific Mathematics Olympiad
March, 2008
Problem 1. Let ABC be a triangle with ∠A < 60◦ . Let X and Y be the points on the sides
AB and AC, respectively, such that CA + AX = CB + BX and BA + AY = BC + CY . Let
P be the point in the plane such that the lines P X and P Y are perpendicular to AB and
AC, respectively. Prove that ∠BP C < 120◦ .
(Solution) Let I be the incenter of △ABC, and let the feet of the perpendiculars from I
to AB and to AC be D and E, respectively. (Without loss of generality, we may assume
that AC is the longest side. Then X lies on the line segment AD. Although P may or
may not lie inside △ABC, the proof below works for both cases. Note that P is on the line
perpendicular to AB passing through X.) Let O be the midpoint of IP , and let the feet of
the perpendiculars from O to AB and to AC be M and N , respectively. Then M and N are
the midpoints of DX and EY , respectively.
1
The conditions on the points X and Y yield the equations
AX =
From AD = AE =
AB + BC − CA
2
and
AY =
BC + CA − AB
.
2
CA + AB − BC
, we obtain
2
BD = AB − AD = AB −
AB + BC − CA
CA + AB − BC
=
= AX.
2
2
Since M is the midpoint of DX, it follows that M is the midpoint of AB. Similarly, N is the
midpoint of AC. Therefore, the perpendicular bisectors of AB and AC meet at O, that is,
O is the circumcenter of △ABC. Since ∠BAC < 60◦ , O lies on the same side of BC as the
point A and
∠BOC = 2∠BAC.
We can compute ∠BIC as follows :
1
1
∠BIC = 180◦ − ∠IBC − ∠ICB = 180◦ − ∠ABC − ∠ACB
2
2
1
1
1
= 180◦ − (∠ABC + ∠ACB) = 180◦ − (180◦ − ∠BAC) = 90◦ + ∠BAC
2
2
2
It follows from ∠BAC < 60◦ that
1
2∠BAC < 90◦ + ∠BAC,
2
i.e., ∠BOC < ∠BIC.
From this it follows that I lies inside the circumcircle of the isosceles triangle BOC because
O and I lie on the same side of BC. However, as O is the midpoint of IP , P must lie outside
the circumcircle of triangle BOC and on the same side of BC as O. Therefore
∠BP C < ∠BOC = 2∠BAC < 120◦ .
Remark. If one assumes that ∠A is smaller than the other two, then it is clear that the
line P X (or the line perpendicular to AB at X if P = X) runs through the excenter IC
of the excircle tangent to the side AB. Since 2∠ACIC = ∠ACB and BC < AC, we have
2∠P CB > ∠C. Similarly, 2∠P BC > ∠B. Therefore,
)
(
∠A
∠B + ∠C
◦
◦
= 90 +
< 120◦ .
∠BP C = 180 − (∠P BC + ∠P CB) < 180 −
2
2
In this way, a special case of the problem can be easily proved.
2
Problem 2. Students in a class form groups each of which contains exactly three members
such that any two distinct groups have at most one member in common. Prove that, when
the class size is 46, there is a set of 10 students in which no group is properly contained.
(Solution) We let C be the set of all 46 students in the class and let
s := max{ |S| : S ⊆ C such that S contains no group properly }.
Then it suffices to prove that s ≥ 10. (If |S| = s > 10, we may choose a subset of S consisting
of 10 students.)
Suppose that s ≤ 9 and let S be a set of size s in which no group is properly contained.
Take any student, say v, from outside S. Because of the maximality of s, there should be a
group containing the student v and two other students in S. The number of ways to choose
two students from S is
( ) ( )
s
9
≤
= 36.
2
2
On the other hand, there are at least 37 = 46 − 9 students outside of S. Thus, among those
37 students outside, there is at least one student, say u, who does not belong to any group
containing two students in S and one outside. This is because no two distinct groups have two
members in common. But then, S can be enlarged by including u, which is a contradiction.
Remark. One may choose a subset S of C that contains no group properly. Then, assuming
|S| < 10, prove that there is a student outside S, say u, who does not belong to any group
containing two students in S. After enlarging S by including u, prove that the enlarged S
still contains no group properly.
3
Problem 3. Let Γ be the circumcircle of a triangle ABC. A circle passing through points
A and C meets the sides BC and BA at D and E, respectively. The lines AD and CE meet
Γ again at G and H, respectively. The tangent lines of Γ at A and C meet the line DE at L
and M , respectively. Prove that the lines LH and M G meet at Γ.
(Solution) Let M G meet Γ at P . Since ∠M CD = ∠CAE and ∠M DC = ∠CAE, we have
M C = M D. Thus
M D2 = M C 2 = M G · M P
and hence M D is tangent to the circumcircle of △DGP . Therefore ∠DGP = ∠EDP .
Let Γ′ be the circumcircle of △BDE. If B = P , then, since ∠BGD = ∠BDE, the tangent
lines of Γ′ and Γ at B should coincide, that is Γ′ is tangent to Γ from inside. Let B ̸= P .
If P lies in the same side of the line BC as G, then we have
∠EDP + ∠ABP = 180◦
because ∠DGP + ∠ABP = 180◦ . That is, the quadrilateral BP DE is cyclic, and hence P is
on the intersection of Γ′ with Γ.
4
Otherwise,
∠EDP = ∠DGP = ∠AGP = ∠ABP = ∠EBP.
Therefore the quadrilateral P BDE is cyclic, and hence P again is on the intersection of Γ′
with Γ.
Similarly, if LH meets Γ at Q, we either have Q = B, in which case Γ′ is tangent to Γ
from inside, or Q ̸= B. In the latter case, Q is on the intersection of Γ′ with Γ. In either
case, we have P = Q.
5
Problem 4. Consider the function f : N0 → N0 , where N0 is the set of all non-negative
integers, defined by the following conditions :
(i) f (0) = 0, (ii) f (2n) = 2f (n) and (iii) f (2n + 1) = n + 2f (n) for all n ≥ 0.
(a) Determine the three sets L := { n | f (n) < f (n + 1) }, E := { n | f (n) = f (n + 1) }, and
G := { n | f (n) > f (n + 1) }.
(b) For each k ≥ 0, find a formula for ak := max{f (n) : 0 ≤ n ≤ 2k } in terms of k.
(Solution) (a) Let
L1 := {2k : k > 0}, E1 := {0} ∪ {4k + 1 : k ≥ 0}, and G1 := {4k + 3 : k ≥ 0}.
We will show that L1 = L, E1 = E, and G1 = G. It suffices to verify that L1 ⊆ E, E1 ⊆ E,
and G1 ⊆ G because L1 , E1 , and G1 are mutually disjoint and L1 ∪ E1 ∪ G1 = N0 .
Firstly, if k > 0, then f (2k) − f (2k + 1) = −k < 0 and therefore L1 ⊆ L.
Secondly, f (0) = 0 and
f (4k + 1) = 2k + 2f (2k) = 2k + 4f (k)
f (4k + 2) = 2f (2k + 1) = 2(k + 2f (k)) = 2k + 4f (k)
for all k ≥ 0. Thus, E1 ⊆ E.
Lastly, in order to prove G1 ⊂ G, we claim that f (n + 1) − f (n) ≤ n for all n. (In fact,
one can prove a stronger inequality : f (n + 1) − f (n) ≤ n/2.) This is clearly true for even n
from the definition since for n = 2t,
f (2t + 1) − f (2t) = t ≤ n.
If n = 2t + 1 is odd, then (assuming inductively that the result holds for all nonnegative
m < n), we have
f (n + 1) − f (n) = f (2t + 2) − f (2t + 1) = 2f (t + 1) − t − 2f (t)
= 2(f (t + 1) − f (t)) − t ≤ 2t − t = t < n.
For all k ≥ 0,
f (4k + 4) − f (4k + 3) = f (2(2k + 2)) − f (2(2k + 1) + 1)
= 4f (k + 1) − (2k + 1 + 2f (2k + 1)) = 4f (k + 1) − (2k + 1 + 2k + 4f (k))
= 4(f (k + 1) − f (k)) − (4k + 1) ≤ 4k − (4k + 1) < 0.
This proves G1 ⊆ G.
(b) Note that a0 = a1 = f (1) = 0. Let k ≥ 2 and let Nk = {0, 1, 2, . . . , 2k }. First we claim
that the maximum ak occurs at the largest number in G ∩ Nk , that is, ak = f (2k − 1). We
use mathematical induction on k to prove the claim. Note that a2 = f (3) = f (22 − 1).
Now let k ≥ 3. For every even number 2t with 2k−1 + 1 < 2t ≤ 2k ,
f (2t) = 2f (t) ≤ 2ak−1 = 2f (2k−1 − 1)
(†)
by induction hypothesis. For every odd number 2t + 1 with 2k−1 + 1 ≤ 2t + 1 < 2k ,
f (2t + 1) = t + 2f (t) ≤ 2k−1 − 1 + 2f (t)
≤ 2k−1 − 1 + 2ak−1 = 2k−1 − 1 + 2f (2k−1 − 1)
6
(‡)
again by induction hypothesis. Combining (†), (‡) and
f (2k − 1) = f (2(2k−1 − 1) + 1) = 2k−1 − 1 + 2f (2k−1 − 1),
we may conclude that ak = f (2k − 1) as desired.
Furthermore, we obtain
ak = 2ak−1 + 2k−1 − 1
for all k ≥ 3. Note that this recursive formula for ak also holds for k ≥ 0, 1 and 2. Unwinding
this recursive formula, we finally get
ak = 2ak−1 + 2k−1 − 1 = 2(2ak−2 + 2k−2 − 1) + 2k−1 − 1
= 22 ak−2 + 2 · 2k−1 − 2 − 1 = 22 (2ak−3 + 2k−3 − 1) + 2 · 2k−1 − 2 − 1
= 23 ak−3 + 3 · 2k−1 − 22 − 2 − 1
..
.
= 2k a0 + k2k−1 − 2k−1 − 2k−2 − ... − 2 − 1
= k2k−1 − 2k + 1 for all k ≥ 0.
7
Problem 5. Let a, b, c be integers satisfying 0 < a < c − 1 and 1 < b < c. For each k,
0 ≤ k ≤ a, let rk , 0 ≤ rk < c, be the remainder of kb when divided by c. Prove that the two
sets {r0 , r1 , r2 , . . . , ra } and {0, 1, 2, . . . , a} are different.
(Solution) Suppose that two sets are equal. Then gcd(b, c) = 1 and the polynomial
f (x) := (1 + xb + x2b + · · · + xab ) − (1 + x + x2 + · · · + xa−1 + xa )
is divisible by xc − 1. (This is because : m = n + cq =⇒ xm − xn = xn+cq − xn = xn (xcq − 1)
and (xcq − 1) = (xc − 1)((xc )q−1 + (xc )q−2 + · · · + 1).) From
f (x) =
x(a+1)b − 1 xa+1 − 1
F (x)
−
=
,
b
x −1
x−1
(x − 1)(xb − 1)
where F (x) = xab+b+1 + xb + xa+1 − xab+b − xa+b+1 − x , we have
F (x) ≡ 0
(mod xc − 1) .
Since xc ≡ 1 (mod xc − 1), we may conclude that
{ab + b + 1, b, a + 1} ≡ {ab + b, a + b + 1, 1} (mod c).
(†)
Thus,
b ≡ ab + b, a + b + 1 or 1 (mod c).
But neither b ≡ 1 (mod c) nor b ≡ a + b + 1 (mod c) are possible by the given conditions.
Therefore, b ≡ ab + b (mod c). But this is also impossible because gcd(b, c) = 1.
8
XXI Asian Pacific Mathematics Olympiad
March, 2009
Problem 1. Consider the following operation on positive real numbers written on a blackboard:
Choose a number r written on the blackboard, erase that number, and then write a
pair of positive real numbers a and b satisfying the condition 2r2 = ab on the board.
Assume that you start out with just one positive real number r on the blackboard, and apply
this operation k 2 − 1 times to end up with k 2 positive real numbers, not necessarily distinct.
Show that there exists a number on the board which does not exceed kr.
(Solution) Using AM-GM inequality, we obtain
1
2ab
2
a2 + b2
1
1
=
=
≤
≤ 2+ 2.
2
2
2
2
2
r
ab
a b
a b
a
b
(∗)
Consequently, if we let Sℓ be the sum of the squares of the reciprocals of the numbers written
on the board after ℓ operations, then Sℓ increases as ℓ increases, that is,
S0 ≤ S1 ≤ · · · ≤ Sk2 −1 .
(∗∗)
Therefore if we let s be the smallest real number written on the board after k 2 − 1 operations,
1
1
then 2 ≥ 2 for any number t among k 2 numbers on the board and hence
s
t
k2 ×
1
1
≥ Sk2 −1 ≥ S0 = 2 ,
s2
r
which implies that s ≤ kr as desired.
Remark. The nature of the problem does not change at all if the numbers on the board
are restricted to be positive integers. But that may mislead some contestants to think the
problem is a number theoretic problem rather than a combinatorial problem.
1
Problem 2. Let a1 , a2 , a3 , a4 , a5 be real numbers satisfying the following equations:
a2
a3
a4
a5
1
a1
+ 2
+ 2
+ 2
+ 2
= 2 for k = 1, 2, 3, 4, 5.
+1 k +2 k +3 k +4 k +5
k
k2
Find the value of
a1
a2
a3
a4
a5
+
+
+
+
. (Express the value in a single fraction.)
37 38 39 40 41
a1
a2
a3
a4
a5
+ 2
+ 2
+ 2
+ 2
. Then R(±1) = 1,
+1
x +2
x +3
x +4
x +5
1
1
1
1
R(±2) = , R(±3) = , R(±4) =
, R(±5) =
and R(6) is the value to be found.
4
9
16
25
2
2
2
2
2
Let’s put P (x) := (x + 1)(x + 2)(x + 3)(x + 4)(x + 5) and Q(x) := R(x)P (x). Then for
P (k)
, that is, P (k) − k 2 Q(k) = 0. Since
k = ±1, ±2, ±3, ±4, ±5, we get Q(k) = R(k)P (k) =
k2
P (x) − x2 Q(x) is a polynomial of degree 10 with roots ±1, ±2, ±3, ±4, ±5, we get
(Solution) Let R(x) :=
x2
P (x) − x2 Q(x) = A(x2 − 1)(x2 − 4)(x2 − 9)(x2 − 16)(x2 − 25).
Putting x = 0, we get A =
(∗)
1
P (0)
=−
. Finally, dividing both sides
(−1)(−4)(−9)(−16)(−25)
120
of (∗) by P (x) yields
1 − x2 R(x) = 1 − x2
Q(x)
1 (x2 − 1)(x2 − 4)(x2 − 9)(x2 − 16)(x2 − 25)
=−
·
P (x)
120 (x2 + 1)(x2 + 2)(x2 + 3)(x2 + 4)(x2 + 5)
and hence that
1 − 36R(6) = −
35 × 32 × 27 × 20 × 11
3 × 7 × 11
231
=−
=−
,
120 × 37 × 38 × 39 × 40 × 41
13 × 19 × 37 × 41
374699
which implies R(6) =
187465
.
6744582
1105
2673
1862
1885
1323
, a2 = −
, a3 =
, a4 = −
, a5 =
by solving
72
40
15
18
40
the given system of linear equations, which is extremely messy and takes a lot of time.
Remark. We can get a1 =
2
Problem 3. Let three circles
1,
2,
3,
which are non-overlapping and mutually external,
be given in the plane. For each point P in the plane, outside the three circles, construct
six points A1 , B1 , A2 , B2 , A3 , B3 as follows: For each i = 1, 2, 3, Ai , Bi are distinct points
on the circle
i
such that the lines P Ai and P Bi are both tangents to
i.
Call the point
P exceptional if, from the construction, three lines A1 B1 , A2 B2 , A3 B3 are concurrent. Show
that every exceptional point of the plane, if exists, lies on the same circle.
(Solution) Let Oi be the center and ri the radius of circle
i
for each i = 1, 2, 3. Let P be
an exceptional point, and let the three corresponding lines A1 B1 , A2 B2 ,A3 B3 concur at Q.
Construct the circle with diameter P Q. Call the circle , its center O and its radius r. We
now claim that all exceptional points lie on .
3
Let P O1 intersect A1 B1 in X1 . As P O1 ⊥ A1 B1 , we see that X1 lies on
tangent to
1,
. As P A1 is a
triangle P A1 O1 is right-angled and similar to triangle A1 X1 O1 . It follows that
O1 X1
O1 A1
=
,
O1 A1
O1 P
i.e., O1 X1 · O1 P = O1 A1 2 = r1 2 .
On the other hand, O1 X1 · O1 P is also the power of O1 with respect to , so that
r12 = O1 X1 · O1 P = (O1 O − r)(O1 O + r) = O1 O2 − r2 ,
(∗)
and hence
r2 = OO12 − r12 = (OO1 − r1 )(OO1 + r1 ).
Thus, r2 is the power of O with respect to
O with respect to
2
and
3.
1.
By the same token, r2 is also the power of
Hence O must be the radical center of the three given circles.
Since r, as the square root of the power of O with respect to the three given circles, does not
depend on P , it follows that all exceptional points lie on .
Remark. In the event of the radical point being at in nity (and hence the three radical
axes being parallel), there are no exceptional points in the plane, which is consistent with the
statement of the problem.
4
Problem 4. Prove that for any positive integer k, there exists an arithmetic sequence
a2
,
b2
a1
,
b1
ak
bk
... ,
of rational numbers, where ai , bi are relatively prime positive integers for each i = 1, 2, . . . , k,
such that the positive integers a1 , b1 , a2 , b2 , . . . , ak , bk are all distinct.
(Solution) For k = 1, there is nothing to prove. Henceforth assume k ≥ 2.
Let p1 , p2 , . . . , pk be k distinct primes such that
k < p k < · · · < p 2 < p1
and let N = p1 p2 · · · pk . By Chinese Remainder Theorem, there exists a positive integer x
satisfying
x ≡ − i (mod pi )
for all i = 1, 2, . . . , k and x > N 2 . Consider the following sequence :
x+1
,
N
x+2
,
N
,... ,
x+k
.
N
This sequence is obviously an arithmetic sequence of positive rational numbers of length k.
For each i = 1, 2, . . . , k, the numerator x + i is divisible by pi but not by pj for j ̸= i, for
otherwise pj divides |i − j|, which is not possible because pj > k > |i − j|. Let
ai :=
x+i
,
pi
bi :=
N
pi
for all i = 1, 2, . . . , k.
Then
x+i
ai
=
,
N
bi
gcd(ai , bi ) = 1
for all i = 1, 2, . . . , k,
and all bi ’s are distinct from each other. Moreover, x > N 2 implies
ai =
x+i
N2
N
>
>N >
= bj
pi
pi
pj
for all i, j = 1, 2, . . . , k
and hence all ai ’s are distinct from bi ’s. It only remains to show that all ai ’s are distinct from
each other. This follows from
aj =
x+j
x+i
x+i
>
>
= ai
pj
pj
pi
for all i < j
by our choice of p1 , p2 , . . . , pk . Thus, the arithmetic sequence
a1
,
b1
a2
,
b2
... ,
ak
bk
of positive rational numbers satis es the conditions of the problem.
5
Remark. Here is a much easier solution :
For any positive integer k ≥ 2, consider the sequence
(k!)2 + 1 (k!)2 + 2
(k!)2 + k
,
,...,
.
k!
k!
k!
Note that gcd(k!, (k!)2 + i) = i for all i = 1, 2, . . . , k. So, taking
ai :=
(k!)2 + i
,
i
bi :=
k!
i
for all i = 1, 2, . . . , k,
we have gcd(ai , bi ) = 1 and
ai =
(k!)2 + i
(k!)2 + j
k!
k!
> aj =
> bi =
> bj =
i
j
i
j
for any 1 ≤ i < j ≤ k. Therefore this sequence satis es every condition given in the problem.
6
Problem 5. Larry and Rob are two robots travelling in one car from Argovia to Zillis. Both
robots have control over the steering and steer according to the following algorithm: Larry
makes a 90◦ left turn after every ℓ kilometer driving from start; Rob makes a 90◦ right turn
after every r kilometer driving from start, where ℓ and r are relatively prime positive integers.
In the event of both turns occurring simultaneously, the car will keep going without changing
direction. Assume that the ground is flat and the car can move in any direction.
Let the car start from Argovia facing towards Zillis. For which choices of the pair (ℓ, r) is the
car guaranteed to reach Zillis, regardless of how far it is from Argovia?
(Solution) Let Zillis be d kilometers away from Argovia, where d is a positive real number.
For simplicity, we will position Argovia at (0, 0) and Zillis at (d, 0), so that the car starts
out facing east. We will investigate how the car moves around in the period of travelling the
rst ℓr kilometers, the second ℓr kilometers, . . . , and so on. We call each period of travelling
ℓr kilometers a section. It is clear that the car will have identical behavior in every section
except the direction of the car at the beginning.
Case 1: ℓ − r ≡ 2 (mod 4) . After the rst section, the car has made ℓ − 1 right turns and
r − 1 left turns, which is a net of 2(≡ ℓ − r (mod 4)) right turns. Let the displacement vector
for the rst section be (x, y). Since the car has rotated 180◦ , the displacement vector for
the second section will be (−x, −y), which will take the car back to (0, 0) facing east again.
We now have our original situation, and the car has certainly never travelled further than ℓr
kilometers from Argovia. So, the car cannot reach Zillis if it is further apart from Argovia.
Case 2: ℓ − r ≡ 1 (mod 4) . After the rst section, the car has made a net of 1 right turn.
Let the displacement vector for the rst section again be (x, y). This time the car has rotated
90◦ clockwise. We can see that the displacements for the second, third and fourth section
will be (y, −x), (−x, −y) and (−y, x), respectively, so after four sections the car is back at
(0, 0) facing east. Since the car has certainly never travelled further than 2ℓr kilometers from
Argovia, the car cannot reach Zillis if it is further apart from Argovia.
Case 3: ℓ − r ≡ 3 (mod 4) . An argument similar to that in Case 2 (switching the roles of
left and right) shows that the car cannot reach Zillis if it is further apart from Argovia.
Case 4: ℓ ≡ r (mod 4) . The car makes a net turn of 0◦ after each section, so it must be
facing east. We are going to show that, after traversing the rst section, the car will be at
(1, 0). It will be useful to interpret the Cartesian plane as the complex plane, i.e. writing
√
x + iy for (x, y), where i = −1. We will denote the k-th kilometer of movement by mk−1 ,
7
which takes values from the set {1, i, −1, −i}, depending on the direction. We then just have
to show that
ℓ∑
r−1
mk = 1,
k=0
which implies that the car will get to Zillis no matter how far it is apart from Argovia.
Case 4a: ℓ ≡ r ≡ 1 (mod 4) . First note that for k = 0, 1, . . . , ℓr − 1,
mk = i⌊k/ℓ⌋ (−i)⌊k/r⌋
since ⌊k/ℓ⌋ and ⌊k/r⌋ are the exact numbers of left and right turns before the (k + 1)st
kilometer, respectively. Let ak (≡ k (mod ℓ)) and bk (≡ k (mod r)) be the remainders of k
when divided by ℓ and r, respectively. Then, since
⌊ ⌋
⌊ ⌋
⌊ ⌋
⌊ ⌋
k
k
k
k
ak = k −
ℓ≡k−
(mod 4) and bk = k −
r≡k−
ℓ
ℓ
r
r
(mod 4),
we have ⌊k/ℓ⌋ ≡ k − ak (mod 4) and ⌊k/r⌋ ≡ k − bk (mod 4). We therefore have
mk = ik−ak (−i)k−bk = (−i2 )k i−ak (−i)−bk = (−i)ak ibk .
As ℓ and r are relatively prime, by Chinese Remainder Theorem, there is a bijection between
pairs (ak , bk ) = (k(mod ℓ), k(mod r)) and the numbers k = 0, 1, 2, . . . , ℓr − 1. Hence
) ( r−1 )
( ℓ−1
ℓ∑
r−1
ℓ∑
r−1
∑
∑
mk =
(−i)ak ibk =
(−i)ak
ibk = 1 × 1 = 1
k=0
k=0
k=0
k=0
as required because ℓ ≡ r ≡ 1 (mod 4).
Case 4b: ℓ ≡ r ≡ 3 (mod 4) . In this case, we get
mk = iak (−i)bk ,
where ak (≡ k (mod ℓ)) and bk (≡ k (mod r)) for k = 0, 1, . . . , ℓr − 1. Then we can proceed
analogously to Case 4a to obtain
ℓ∑
r−1
k=0
mk =
ℓ∑
r−1
(−i)ak ibk
( ℓ−1
) ( r−1 )
∑
∑
=
(−i)ak
ibk = i × (−i) = 1
k=0
k=0
k=0
as required because ℓ ≡ r ≡ 3 (mod 4).
Now clearly the car traverses through all points between (0, 0) and (1, 0) during the
rst
section and, in fact, covers all points between (n − 1, 0) and (n, 0) during the n-th section.
Hence it will eventually reach (d, 0) for any positive d.
8
To summarize: (ℓ, r) satis es the required conditions if and only if
ℓ≡r≡1
or ℓ ≡ r ≡ 3
(mod 4).
Remark. In case gcd(ℓ, r) = d ̸= 1, the answer is :
ℓ
r
≡ ≡1
d
d
or
ℓ
r
≡ ≡ 3 (mod 4).
d
d
9
SOLUTIONS FOR 2010 APMO PROBLEMS
Problem 1. Let ABC be a triangle with ∠BAC 6= 90◦ . Let O be the circumcenter of the
triangle ABC and let Γ be the circumcircle of the triangle BOC. Suppose that Γ intersects
the line segment AB at P different from B, and the line segment AC at Q different from C.
Let ON be a diameter of the circle Γ. Prove that the quadrilateral AP N Q is a parallelogram.
Solution: From the assumption that the circle Γ intersects both of the line segments AB
and AC, it follows that the 4 points N, C, Q, O are located on Γ in the order of N, C, Q, O
or in the order of N, C, O, Q. The following argument for the proof of the assertion of the
problem is valid in either case. Since ∠N QC and ∠N OC are subtended by the same arc
_
N C of Γ at the points Q and O, respectively, on Γ, we have ∠N QC = ∠N OC. We also
_
have ∠BOC = 2∠BAC, since ∠BOC and ∠BAC are subtended by the same arc BC of the
circum-circle of the triangle ABC at the center O of the circle and at the point A on the
circle, respectively. From OB = OC and the fact that ON is a diameter of Γ, it follows that
the triangles OBN and OCN are congruent, and therefore we obtain 2∠N OC = ∠BOC.
Consequently, we have ∠N QC = 21 ∠BOC = ∠BAC, which shows that the 2 lines AP, QN
are parallel.
In the same manner, we can show that the 2 lines AQ, P N are also parallel. Thus, the
quadrilateral AP N Q is a parallelogram.
Problem 2. For a positive integer k, call an integer a pure k-th power if it can be represented
as mk for some integer m. Show that for every positive integer n there exist n distinct positive
integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power.
Solution: For the sake of simplicity, let us set k = 2009.
First of all, choose n distinct positive integers b1 , · · · , bn suitably so that their product is a
pure k +1-th power (for example, let bi = ik+1 for i = 1, · · · , n). Then we have b1 · · · bn = tk+1
for some positive integer t. Set b1 + · · · + bn = s.
2
Now we set ai = bi sk −1 for i = 1, · · · , n, and show that a1 , · · · , an satisfy the required
conditions. Since b1 , · · · , bn are distinct positive integers, it is clear that so are a1 , · · · , an .
From
a1 + · · · + an = sk
2 −1
a1 · · · an = (sk
2
(b1 + · · · + bn ) = sk = (sk )2009 ,
2 −1
)n b1 · · · bn = (sk
2 −1
)n tk+1 = (s(k−1)n t)2010
we can see that a1 , · · · , an satisfy the conditions on the sum and the product as well. This
ends the proof of the assertion.
Remark: We can find the appropriate exponent k 2 − 1 needed for the construction of the
ai ’s by solving the simultaneous congruence relations: x ≡ 0 (mod k + 1), x ≡ −1 (mod k).
Problem 3. Let n be a positive integer. n people take part in a certain party. For any
pair of the participants, either the two are acquainted with each other or they are not. What
is the maximum possible number of the pairs for which the two are not acquainted but have
a common acquaintance among the participants?
1
Solution: When 1 participant, say the person A, is mutually acquainted with each of the
remaining n − 1 participants, and if there are no other acquaintance relationships among
the participants, then for any pair of participants not involving A, the two are not mutual
acquaintances, but they have a common acquaintance, namely A, so any such pair satisfies
2
the requirement. Thus, the number desired in this case is (n−1)(n−2)
= n −3n+2
.
2
2
n2 −3n+2
Let us show that
is the maximum possible number of the pairs satisfying the
2
requirement of the problem. First, let us observe that in the process of trying to find the
maximum possible number of such pairs, if we split the participants into two non-empty
subsets T and S which are disjoint, we may assume that there is a pair consisting of one
person chosen from T and the other chosen from S who are mutual acquaintances. This is
so, since if there are no such pair for some splitting T and S, then among the pairs consisting
of one person chosen from T and the other chosen from S, there is no pair for which the two
have a common acquaintance among participants, and therefore, if we arbitrarily choose a
person A ∈ T and B ∈ S and declare that A and B are mutual acquaintances, the number of
the pairs satisfying the requirement of the problem does not decrease.
Let us now call a set of participants a group if it satisfies the following 2 conditions:
• One can connect any person in the set with any other person in the set by tracing a
chain of mutually acquainted pairs. More precisely, for any pair of people A, B in the set
there exists a sequence of people A0 , A1 , · · · , An for which A0 = A, An = B and, for each
i : 0 ≤ i ≤ n − 1, Ai and Ai+1 are mutual acquaintances.
• No person in this set can be connected with a person not belonging to this set by tracing
a chain of mutually acquainted pairs.
In view of the discussions made above, we may assume that the set of all the participants
to the party forms a group of n people. Let us next consider the following lemma.
Lemma. In a group of n people, there are at least n − 1 pairs of mutual acquaintances.
Proof: If you choose a mutually acquainted pair in a group and declare the two in the pair
are not mutually acquainted, then either the group stays the same or splits into 2 groups.
This means that by changing the status of a mutually acquainted pair in a group to that
of a non-acquainted pair, one can increase the number of groups at most by 1. Now if in a
group of n people you change the status of all of the mutually acquainted pairs to that of
non-acquainted pairs, then obviously, the number of groups increases from 1 to n. Therefore,
there must be at least n − 1 pairs of mutually acquainted pairs in a group consisting of n
people.
2
− (n − 1) = n −3n+2
pairs satisfying
The lemma implies that there are at most n(n−1)
2
2
the condition of the problem. Thus the desired maximum number of pairs satisfying the
2
requirement of the problem is n −3n+2
.
2
Remark: One can give a somewhat different proof by separating into 2 cases depending on
whether there are at least n − 1 mutually acquainted pairs, or at most n − 2 such pairs. In the
former case, one can argue in the same way as the proof above, while in the latter case, the
Lemma above implies that there would be 2 or more groups to start with, but then, in view
of the comment made before the definition of a group above, these groups can be combined
to form one group, thereby one can reduce the argument to the former case.
Alternate Solution 1: The construction of an example for the case for which the number
n2 −3n+2
appears, and the argument for the case where there is only 1 group would be the
2
same as in the preceding proof.
Suppose, then, n participants are separated into k(k ≥ 2) groups, and the number of people
in each group is given by ai , i = 1, · · · , k. In such a case, the number of pairs for which paired
P
people are not mutually acquainted but have a common acquaintance is at most ki=1 ai C2 ,
where we set 1 C2 = 0 for convenience. Since a C2 + b C2 ≤ a+b C2 holds for any pair of
P
positive integers a, b, we have ki=1 ai C2 ≤ a1 C2 + n−a1 C2 . From
n2 − n
n 2 n2 − 2n
= a1 −
+
2
2
4
it follows that a1 C2 + n−a1 C2 takes its maximum value when a1 = 1, n − 1. Therefore, we
P
have ki=1 ai C2 ≤ n−1 C2 , which shows that in the case where the number of groups are 2
or more, the number of the pairs for which paired people are not mutually acquainted but
2
have a common acquaintance is at most n−1 C2 = n −3n+2
, and hence the desired maximum
2
n2 −3n+2
.
number of the pairs satisfying the requirement is
2
a1 C2
+
n−a1 C2
= a21 − na1 +
Alternate Solution 2: Construction of an example would be the same as the preceding
proof.
For a participant, say A, call another participant, say B, a familiar face if A and B are
not mutually acquainted but they have a common acquaintance among the participants, and
in this case call the pair A, B a familiar pair.
Suppose there is a participant P who is mutually acquainted with d participants. Denote
by S the set of these d participants, and by T the set of participants different from P and
not belonging to the set S. Suppose there are e pairs formed by a person in S and a person
in T who are mutually acquainted.
Then the number of participants who are familiar faces to P is at most e. The number of
pairs formed by two people belonging to the set S and are mutually acquainted is at most
d C2 . The number of familiar pairs formed by two people belonging to the set T is at most
n−d−1 C2 . Since there are e pairs formed by a person in the set S and a person in the set
T who are mutually acquainted (and so the pairs are not familiar pairs), we have at most
d(n − 1 − d) − e familiar pairs formed by a person chosen from S and a person chosen from T.
Putting these together we conclude that there are at most e+ d C2 + n−1−d C2 +d(n−1−d)−e
familiar pairs. Since
e + d C2 +
the number we seek is at most
n−1−d C2 + d(n − 1 − d) − e =
n2 −3n+2
,
2
n2 − 3n + 2
,
2
and hence this is the desired solution to the problem.
Problem 4.
Let ABC be an acute triangle satisfying the condition AB > BC and
AC > BC. Denote by O and H the circumcenter and the orthocenter, respectively, of the
triangle ABC. Suppose that the circumcircle of the triangle AHC intersects the line AB at
M different from A, and that the circumcircle of the triangle AHB intersects the line AC at
N different from A. Prove that the circumcenter of the triangle M N H lies on the line OH.
Solution: In the sequel, we denote ∠BAC = α, ∠CBA = β, ∠ACB = γ. Let O0 be the
circumcenter of the triangle M N H. The lengths of line segments starting from the point H
will be treated as signed quantities.
Let us denote by M 0 , N 0 the point of intersection of CH, BH, respectively, with the circumcircle of the triangle ABC (distinct from C, B, respectively.) From the fact that 4
points A, M, H, C lie on the same circle, we see that ∠M HM 0 = α holds. Furthermore,
_
∠BM 0 C, ∠BN 0 C and α are all subtended by the same arc BC of the circumcircle of the
triangle ABC at points on the circle, and therefore, we have ∠BM 0 C = α, and ∠BN 0 C = α
_
as well. We also have ∠ABH = ∠ACN 0 as they are subtended by the same arc AN 0 of the
circumcircle of the triangle ABC at points on the circle. Since HM 0 ⊥ BM, HN 0 ⊥ AC, we
conclude that
∠M 0 HB = 90◦ − ∠ABH = 90◦ − ∠ACN 0 = α
is valid as well. Putting these facts together, we obtain the fact that the quadrilateral
HBM 0 M is a rhombus. In a similar manner, we can conclude that the quadrilateral HCN 0 N
is also a rhombus. Since both of these rhombuses are made up of 4 right triangles with an
angle of magnitude α, we also see that these rhombuses are similar.
Let us denote by P, Q the feet of the perpendicular lines on HM and HN , respectively,
drawn from the point O0 . Since O0 is the circumcenter of the triangle M N H, P, Q are respectively, the midpoints of the line segments HM, HN. Furthermore, if we denote by R, S
the feet of the perpendicular lines on HM and HN , respectively, drawn from the point O,
then since O is the circumcenter of both the triangle M 0 BC and the triangle N 0 BC, we see
that R is the intersection point of HM and the perpendicular bisector of BM 0 , and S is the
intersection point of HN and the perpendicular bisector of CN 0 .
We note that the similarity map φ between the rhombuses HBM 0 M and HCN 0 N carries
the perpendicular bisector of BM 0 onto the perpendicular bisector of CN 0 , and straight line
HM onto the straight line HN, and hence φ maps R onto S, and P onto Q. Therefore, we get
HP : HR = HQ : HS. If we now denote by X, Y the intersection points of the line HO0 with
the line through R and perpendicular to HP , and with the line through S and perpendicular
to HQ, respectively, then we get
HO0 : HX = HP : HR = HQ : HS = HO0 : HY
so that we must have HX = HY , and therefore, X = Y. But it is obvious that the point
of intersection of the line through R and perpendicular to HP with the line through S and
perpendicular to HQ must be O, and therefore, we conclude that X = Y = O and that the
points H, O0 , O are collinear.
Alternate Solution: Deduction of the fact that both of the quadrilaterals HBM 0 M and
HCN 0 N are rhombuses is carried out in the same way as in the preceding proof.
We then see that the point M is located in a symmetric position with the point B with
respect to the line CH, we conclude that we have ∠CM B = β. Similarly, we have ∠CN B = γ.
If we now put x = ∠AHO0 , then we get
∠O0 = β − α − x, ∠M N H = 90◦ − β − α + x,
from which it follows that
∠AN M = 180◦ − ∠M N H − (90◦ − α) = β − x.
Similarly, we get
∠N M A = γ + x.
Using the laws of sines, we then get
sin(γ + x)
AN
AC AB AN
=
=
·
·
sin(β − x)
AM
AM AC AB
sin β
sin γ sin(γ − α)
sin(γ − α)
=
·
·
=
.
sin(β − α) sin β
sin γ
sin(β − α)
On the other hand, if we let y = ∠AHO, we then get
∠OHB = 180◦ − γ − y, ∠CHO = 180◦ − β + y,
and since
∠HBO = γ − α, ∠OCH = β − α,
using the laws of sines and observing that OB = OC, we get
sin(180◦ − γ − y) · OH
sin ∠HBO
sin(γ − α)
OB
=
=
sin(β − α)
sin ∠OCH
sin(180◦ − β + y) · OH
OC
sin(180◦ − γ − y)
sin(γ + y)
=
=
.
◦
sin(180 − β + y)
sin(β − y)
We then get sin(γ + x) sin(β − y) = sin(β − x) sin(γ + y). Expanding both sides of the last
identity by using the addition formula for the sine function and after factoring and using
again the addition formula we obtain that sin(x − y) sin(β + γ) = 0. This implies that x − y
must be an integral multiple of 180◦ , and hence we conclude that H, O, O0 are collinear.
Problem 5. Find all functions f from the set R of real numbers into R which satisfy for
all x, y, z ∈ R the identity
f (f (x) + f (y) + f (z)) = f (f (x) − f (y)) + f (2xy + f (z)) + 2f (xz − yz).
Solution: It is clear that if f is a constant function which satisfies the given equation,
then the constant must be 0. Conversely, f (x) = 0 clearly satisfies the given equation, so,
the identically 0 function is a solution. In the sequel, we consider the case where f is not a
constant function.
Let t ∈ R and substitute (x, y, z) = (t, 0, 0) and (x, y, z) = (0, t, 0) into the given functional
equation. Then, we obtain, respectively,
f (f (t) + 2f (0)) = f (f (t) − f (0)) + f (f (0)) + 2f (0),
f (f (t) + 2f (0)) = f (f (0) − f (t)) + f (f (0)) + 2f (0),
from which we conclude that f (f (t)−f (0)) = f (f (0)−f (t)) holds for all t ∈ R. Now, suppose
for some pair u1 , u2 , f (u1 ) = f (u2 ) is satisfied. Then by substituting (x, y, z) = (s, 0, u1 ) and
(x, y, z) = (s, 0, u2 ) into the functional equation and comparing the resulting identities, we
can easily conclude that
f (su1 ) = f (su2 )
(∗)
holds for all s ∈ R. Since f is not a constant function there exists an s0 such that f (s0 )−f (0) 6=
0. If we put u1 = f (s0 ) − f (0), u2 = −u1 , then f (u1 ) = f (u2 ), so we have by (∗)
f (su1 ) = f (su2 ) = f (−su1 )
for all s ∈ R. Since u1 6= 0, we conclude that
f (x) = f (−x)
holds for all x ∈ R.
Next, if f (u) = f (0) for some u 6= 0, then by (∗), we have f (su) = f (s0) = f (0) for all
s, which implies that f is a constant function, contradicting our assumption. Therefore, we
must have f (s) 6= f (0) whenever s 6= 0.
We will now show that if f (x) = f (y) holds, then either x = y or x = −y must hold.
Suppose on the contrary that f (x0 ) = f (y0 ) holds for some pair of non-zero numbers x0 , y0
for which x0 6= y0 , x0 6= −y0 . Since f (−y0 ) = f (y0 ), we may assume, by replacing y0 by −y0
if necessary, that x0 and y0 have the same sign. In view of (∗), we see that f (sx0 ) = f (sy0 )
holds for all s, and therefore, there exists some r > 0, r 6= 1 such that
f (x) = f (rx)
holds for all x. Replacing x by rx and y by ry in the given functional equation, we obtain
f (f (rx) + f (ry) + f (z)) = f (f (rx) − f (ry)) + f (2r2 xy + f (z)) + 2f (r(x − y)z)
(i),
and replacing x by r2 x in the functional equation, we get
f (f (r2 x) + f (y) + f (z)) = f (f (r2 x) − f (y)) + f (2r2 xy + f (z)) + 2f ((r2 x − y)z)
(ii).
Since f (rx) = f (x) holds for all x ∈ R, we see that except for the last term on the right-hand
side, all the corresponding terms appearing in the identities (i) and (ii) above are equal, and
hence we conclude that
f (r(x − y)z) = f ((r2 x − y)z))
(iii)
must hold for arbitrary choice of x, y, z ∈ R. For arbitrarily fixed pair u, v ∈ R, substitute
v−r2 u
(x, y, z) = ( rv−u
2 −1 , r 2 −1 , 1) into the identity (iii). Then we obtain f (v) = f (ru) = f (u), since
x − y = u, r2 x − y = v, z = 1. But this implies that the function f is a constant, contradicting
our assumption. Thus we conclude that if f (x) = f (y) then either x = y or x = −y must
hold.
By substituting z = 0 in the functional equation, we get
f (f (x) + f (y) + f (0)) = f (f (x) − f (y) + f (0)) = f ((f (x) − f (y)) + f (2xy + f (0)) + 2f (0).
Changing y to −y in the identity above and using the fact that f (y) = f (−y), we see that
all the terms except the second term on the right-hand side in the identity above remain the
same. Thus we conclude that f (2xy + f (0)) = f (−2xy + f (0)), from which we get either
2xy + f (0) = −2xy + f (0) or 2xy + f (0) = 2xy − f (0) for all x, y ∈ R. The first of these
alternatives says that 4xy = 0, which is impossible if xy 6= 0. Therefore the second alternative
must be valid and we get that f (0) = 0.
Finally, let us show that if f satisfies the given functional equation and is not a constant
function, then f (x) = x2 . Let x = y in the functional equation, then since f (0) = 0, we get
f (2f (x) + f (z)) = f (2x2 + f (z)),
from which we conclude that either 2f (x) + f (z) = 2x2 + f (z) or 2f (x) + f (z) = −2x2 − f (z)
must hold. Suppose there exists x0 for which f (x0 ) 6= x20 , then from the second alternative,
we see that f (z) = −f (x0 ) − x20 must hold for all z, which means that f must be a constant
function, contrary to our assumption. Therefore, the first alternative above must hold, and
we have f (x) = x2 for all x, establishing our claim.
It is easy to check that f (x) = x2 does satisfy the given functional equation, so we conclude
that f (x) = 0 and f (x) = x2 are the only functions that satisfy the requirement.
SOLUTIONS FOR 2011 APMO PROBLEMS
Problem 1.
Solution: Suppose all of the 3 numbers a2 + b + c, b2 + c + a and c2 + a + b are
perfect squares. Then from the fact that a2 + b + c is a perfect square bigger than
a2 it follows that a2 + b + c ≥ (a + 1)2 , and therefore, b + c ≥ 2a + 1. Similarly we
obtain c + a ≥ 2b + 1 and a + b ≥ 2c + 1.
Adding the corresponding sides of the preceding 3 inequalities, we obtain
2(a + b + c) ≥ 2(a + b + c) + 3, a contradiction. This proves that it is impossible to have all the 3 given numbers to be perfect squares.
Alternate Solution: Since the given conditions of the problem are symmetric in
a, b, c, we may assume that a ≥ b ≥ c holds. From the assumption that a2 +b+c is a
perfect square, we can deduce as in the solution above the inequality b + c ≥ 2a + 1.
But then we have
2a ≥ b + c ≥ 2a + 1,
a contradiction, which proves the assertion of the problem.
Problem 2.
Solution: We will show that 36◦ is the desired answer for the problem.
First, we observe that if the given 5 points form a regular pentagon, then the
minimum of the angles formed by any triple among the five vertices is 36◦ , and
therefore, the answer we seek must be bigger than or equal to 36◦ .
Next, we show that for any configuration of 5 points satisfying the condition of
the problem, there must exist an angle smaller than or equal to 36◦ formed by a
triple chosen from the given 5 points. For this purpose, let us start with any 5
points, say A1 , A2 , A3 , A4 , A5 , on the plane satisfying the condition of the problem,
and consider the smallest convex subset, call it Γ, in the plane containing all of the
5 points. Since this convex subset Γ must be either a triangle or a quadrilateral
or a pentagon, it must have an interior angle with 108◦ or less. We may assume
without loss of generality that this angle is ∠A1 A2 A3 . By the definition of Γ it is
clear that the remaining 2 points A4 and A5 lie in the interior of the angular region
determined by ∠A1 A2 A3 , and therefore, there must be an angle smaller than or
1
equal to · 108◦ = 36◦ , which is formed by a triple chosen from the given 5 points,
3
and this proves that 36◦ is the desired maximum.
1
2
Problem 3.
Solution:
Since ∠B1 BB2 = 90◦ , the circle having B1 B2 as its diameter goes
through the points B, B1 , B2 . From B1 A : B1 C = B2 A : B2 C = BA : BC, it
follows that this circle is the Apolonius circle with the ratio of the distances from
the points A and C being BA : BC. Since the point P lies on this circle, we have
P A : P C = BA : BC = sin C : sin A,
from which it follows that P A sin A = P C sin C. Similarly, we have P A sin A =
P B sin B, and therefore, P A sin A = P B sin B = P C sin C.
Let us denote by D, E, F the foot of the perpendicular line drawn from P to the
line segment BC, CA and AB, respectively. Since the points E, F lie on a circle
having P A as its diameter, we have by the law of sines EF = P A sin A. Similarly,
we have F D = P B sin B and DE = P C sin C. Consequently, we conclude that
DEF is an equilateral triangle. Furthermore, we have ∠CP E = ∠CDE, since the
quadrilateral CDP E is cyclic. Similarly, we have ∠F P B = ∠F DB. Putting these
together, we get
∠BP C = 360◦ − (∠CP E + ∠F P B + ∠EP F )
= 360◦ − {(∠CDE + ∠F DB) + (180◦ − ∠F AE)}
= 360◦ − (120◦ + 150◦ ) = 90◦ ,
which proves the assertion of the problem.
Alternate Solution: Let O be the midpoint of the line segment B1 B2 . Then
the points B and P lie on the circle with center at O and going through the point
B1 . From
∠OBC = ∠OBB1 − ∠CBB1 = ∠OB1 B − ∠B1 BA = ∠BAC
it follows that the triangles OCD and OBA are similar, and therefore we have that
OC · OA = OB 2 = OP 2 . Thus we conclude that the triangles OCP and OP A are
similar, and therefore, we have ∠OP C = ∠P AC. Using this fact, we obtain
∠P BC − ∠P BA = (∠B1 BC + ∠P BB1 ) − (∠ABB1 − ∠P BB1 )
= 2∠P BB1 = ∠P OB1 = ∠P CA − ∠OP C
= ∠P CA − ∠P AC,
from which we conclude that ∠P AC + ∠P BC = ∠P BA + ∠P CA. Similarly, we
get ∠P AB + ∠P CB = ∠P BA + ∠P CA. Putting these facts together and taking
into account the fact that
(∠P AC + ∠P BC) + (∠P AB + ∠P CB) + (∠P BA + ∠P CA) = 180◦ ,
we conclude that ∠P BA + ∠P CA = 60◦ , and finally that
∠BP C = (∠P BA+∠P AB)+(∠P CA+∠P AC) = ∠BAC+(∠P BA+∠P CA) = 90◦ ,
proving the assertion of the problem.
Problem 4.
Solution: We will show that the desired maximum value for m is n(n − 1).
First, let us show that m ≤ n(n−1) always holds for any sequence P0 , P1 , · · · , Pm+1
satisfying the conditions of the problem.
3
Call a point a turning point if it coincides with Pi for some i with 1 ≤ i ≤ m.
Let us say also that 2 points {P, Q} are adjacent if {P, Q} = {Pi−1 , Pi } for some
i with 1 ≤ i ≤ m, and vertically adjacent if, in addition, P Q is parallel to the
y-axis.
Any turning point is vertically adjacent to exactly one other turning point.
Therefore, the set of all turning points is partitioned into a set of pairs of points
using the relation of ”vertical adjacency”. Thus we can conclude that if we fix
k ∈ {1, 2, · · · , n}, the number of turning points having the x-coordinate k must be
even, and hence it is less than or equal to n − 1. Therefore, altogether there are
less than or equal to n(n − 1) turning points, and this shows that m ≤ n(n − 1)
must be satisfied.
It remains now to show that for any positive odd number n one can choose a
sequence for which m = n(n − 1). We will show this by using the mathematical
induction on n. For n = 1, this is clear. For n = 3, choose
P0 = (0, 1),
P1 = (1, 1),
P2 = (1, 2),
P3 = (2, 2),
P4 = (2, 1),
P5 = (3, 1),
P6 = (3, 3),
P7 = (4, 3).
It is easy to see that these points satisfy the requirements (See fig. 1 below).
figure 1
Let n be an odd integer ≥ 5, and suppose there exists a sequence satisfying the
desired conditions for n−4. Then, it is possible to construct a sequence which gives
a configuration indicated in the following diagram (fig. 2), where the configuration
inside of the dotted square is given by the induction hypothesis:
figure 2
By the induction hypothesis, there are exactly (n − 4)(n − 5) turning points for
the configuration inside of the dotted square in the figure 2 above, and all of the
lattice points in the figure 2 lying outside of the dotted square except for the 4
points (n, 2), (n − 1, n − 2), (2, 3), (1, n − 1) are turning points. Therefore, the total
4
number of turning points in this configuration is
(n − 4)(n − 5) + (n2 − (n − 4)2 − 4) = n(n − 1),
showing that for this n there exists a sequence satisfying the desired properties,
and thus completing the induction process.
Problem 5.
Solution: By substituting x = 1 and y = 1 into the given identity we obtain
f (f (1)) = f (1). Next, by substituting x = 1 and y = f (1) into the given identity
and using f (f (1)) = f (1), we get f (1)2 = f (1), from which we conclude that either
f (1) = 0 or f (1) = 1. But if f (1) = 1, then substituting y = 1 into the given
identity, we get f (x) = x for all x, which contradicts the condition (1). Therefore,
we must have f (1) = 0.
By substituting x = 1 into the given identity and using the fact f (1) = 0, we
then obtain f (f (y)) = 2f (y) for all y. This means that if a number t belongs to the
range of the function f , then so does 2t, and by induction we can conclude that for
any non-negative integer n, 2n t belongs to the range of f if t does. Now suppose
that there exists a real number a for which f (a) > 0, then for any non-negative
integer n 2n f (a) must belong to the range of f , which leads to a contradiction to
the condition (1). Thus we conclude that f (x) ≤ 0 for any real number x.
By substituting x2 for x and f (y) for y in the given identity and using the fact
that f (f (y)) = 2f (y), we obtain
x
x
f (xf (y)) + f (y)f
f (y) ,
= xf (y) + f
2
2
x
from which it follows that xf (y) − f (xf (y)) = f (y)f 2 − f x2 f (y) ≥ 0, since the
values of f are non-positive. Combining this with the given identity, we conclude
1
that yf (x) ≥ f (xy). When x > 0, by letting y to be
and using the fact that
x
f (1) = 0, we get f (x) ≥ 0. Since f (x) ≤ 0 for any real number x, we conclude that
f (x) = 0 for any positive real number x. We also have f (0) = f (f (1)) = 2f (1) = 0.
If f is identically 0, i.e., f (x) = 0 for all x, then clearly, this f satisfies the given
identity. If f satisfies the given identity but not identically 0, then there exists a
b < 0 for which f (b) < 0. If we set c = f (b), then we have f (c) = f (f (b)) = 2f (b) =
2c. For any negative real number x, we have cx > 0 so that f (cx) = f (2cx) = 0,
and by substituting y = c into the given identity, we get
f (2cx) + cf (x) = 2cx + f (cx),
from which it follows that f (x) = 2x for any negative real x.
We therefore conclude that if f (
satisfies the given identity and is not identically
0
if x ≥ 0
0, then f is of the form f (x) =
Finally, let us show that the
2x
if x < 0.
function f of the form shown above does satisfy the conditions of the problem.
Clearly, it satisfies the condition (1). We can check that f satisfies the condition
(2) as well by separating into the following 4 cases depending on whether x, y are
non-negative or negative.
• when both x and y are non-negative, both sides of the given identity are 0.
• when x is non-negative and y is negative, we have xy ≤ 0 and both sides
of the given identity are 4xy.
5
• when x is negative and y is non-negative, we have xy ≤ 0 and both sides
of the given identity are 2xy.
• when both x and y are negative, we have xy > 0 and both sides of the given
identity are 2xy.
Summarizing the arguments above, we conclude that the functions f satisfying the
conditions of the problem are
(
0
if x ≥ 0
f (x) = 0
and
f (x) =
2x
if x < 0.
SOLUTIONS FOR 2012 APMO PROBLEMS
Problem 1.
Solution:
Let us denote by 4XY Z the area of the triangle XY Z.
x = 4P AB, y = 4P BC and z = 4P CA.
From
Let
y : z = 4BCP : 4ACP = BF : AF = 4BP F : 4AP F = (x − 1) : 1
follows that z(x − 1) = y, which yields (z + 1)x = x + y + z. Similarly, we get
(x + 1)y = x + y + z and (y + 1)z = x + y + z. Thus, we obtain (x + 1)y = (y + 1)z =
(z + 1)x.
We may assume without loss of generality that x ≤ y, z. If we assume that y > z
holds, then we get (y + 1)z > (z + 1)x, which is a contradiction. Similarly, we see
that y < z leads to a contradiction (x + 1)y < (y + 1)z. Therefore, we must have
y = z. Then, we also get from (y + 1)z = (z + 1)x that x = z must hold. We now
obtain from (x − 1) : 1 = y : z = 1 : 1 that x = y = z = 2 holds. Therefore, we
conclude that the area of the triangle ABC equals x + y + z = 6.
Problem 2.
Solution: If we insert numbers as in the figure below (00 s are to be inserted in the
remaining blank boxes), then we see that the condition of the problem is satisfied
and the total number of all the numbers inserted is 5.
0
1
0
1
1
1
0
1
0
We will show that the sum of all the numbers to be inserted in the boxes of the
given grid cannot be more than 5 if the distribution of the numbers has to satisfy
the requirement of the problem. Let n = 2012. Let us say that the row number
(the column number) of a box in the given grid is i (j, respectively) if the box lies
on the i-th row and the j-th column. For a pair of positive integers x and y, denote
by R(x, y) the sum of the numbers inserted in all of the boxes whose row number is
greater than or equal to x and less than or equal to y (assign the value 0 if x > y).
First let a be the largest integer satisfying 1 ≤ a ≤ n and R(1, a − 1) ≤ 1, and
then choose the smallest integer c satisfying a ≤ c ≤ n and R(c + 1, n) ≤ 1. It is
possible to choose such a pair a, c since R(1, 0) = 0 and R(n + 1, n) = 0. If a < c,
then we have a < n and so, by the maximality of a, we must have R(1, a) > 1,
while from the minimality of c, we must have R(a + 1, n) > 1. Then by splitting the
grid into 2 rectangles by means of the horizontal line bordering the a-th row and
the a + 1-th row, we get the splitting contradicting the requirement of the problem.
Thus, we must have a = c.
Similarly, if for any pair of integers x, y we define C(x, y) to be the sum of the
numbers inserted in all of the boxes whose column number is greater than or equal
to x and less than or equal to y (C(x, y) = 0 if x > y), then we get a number b for
which
C(1, b − 1) ≤ 1, C(b + 1, n) ≤ 1, 1 ≤ b ≤ n.
1
2
If we let r be the number inserted in the box whose row number is a and the column
number is b, then since r ≤ 1, we conclude that the sum of the numbers inserted
into all of the boxes is
≤ R(1, a − 1) + R(a + 1, n) + C(1, b − 1) + C(b + 1, n) + r ≤ 5.
Problem 3.
Solution
For integers a, b and a positive integer m, let us write a ≡ b (mod m) if a − b
p
+1
must be a positive integer, we see that pn ≤ np must
is divisible by m. Since npn +1
hold. This means that if p = 2, then 2n ≤ n2 must hold. As it is easy to show by
induction that 2n > n2 holds if n ≥ 5, we conclude that if p = 2, then n ≤ 4 must
be satisfied. And we can check that (p, n) = (2, 2), (2, 4) satisfy the condition of
the problem, while (2, 3) does not.
Next, we consider the case where p ≥ 3.
Suppose s is an integer satisfying s ≥ p. If sp ≤ ps for such an s, then we have
p
p
1
1
≤ ps 1 +
(s + 1)p = sp 1 +
s
p
p
p
X
X
1
1
s
= ps
p Cr r < p
p
r!
r=0
r=0
p
X
1
≤ ps 1 +
2r−1
r=1
< ps (1 + 2) ≤ ps+1
Thus we have (s + 1)p < ps+1 , and by induction on n, we can conclude that if
n > p, then np < pn . This implies that we must have n ≤ p in order to satisfy our
requirement pn ≤ np .
We note that since pn + 1 is even, so is np + 1, which, in turn implies that n must
be odd and therefore, pn + 1 is divisible by p + 1, and np + 1 is also divisible by
p + 1. Thus we have np ≡ −1 (mod (p + 1)), and therefore, n2p ≡ 1 (mod (p + 1)).
Now, let e be the smallest positive integer for which ne ≡ 1 (mod (p+1)). Then,
we can write 2p = ex + y, where x, y are non-negative integers and 0 ≤ y < e, and
we have
1 ≡ n2p = (ne )x · ny ≡ ny
(mod (p + 1)),
which implies, because of the minimality of e, that y = 0 must hold. This means
that 2p is an integral multiple of e, and therefore, e must equal one of the numbers
1, 2, p, 2p.
Now, if e = 1, p, then we get np ≡ 1 (mod (p + 1)), which contradicts the
fact that p is an odd prime. Since n and p + 1 are relatively prime, we have by
Euler’s Theorem that nϕ(p+1) ≡ 1 (mod (p+1)), where ϕ(m) denotes the number of
integers j(1 ≤ j ≤ m) which are relatively prime with m. From ϕ(p+1) < p+1 < 2p
and the minimality of e, we can then conclude that e = 2 must hold.
From n2 ≡ 1 (mod (p + 1)), we get
−1 ≡ np = n(2·
p−1
2 +1)
≡ n (mod (p + 1)),
3
which implies that p + 1 divides n + 1. Therefore, we must have p ≤ n, which,
together with the fact n ≤ p, show that p = n must hold.
It is clear that the pair (p, p) for any prime p ≥ 3 satisfies the condition of the
problem, and thus, we conclude that the pairs (p, n) which satisfy the condition of
the problem must be of the form (2, 4) and (p, p) with any prime p.
Alternate Solution. Let us consider the case where p ≥ 3. As we saw in the
preceding solution, n must be odd if the pair (p, n) satisfy the condition of the
problem. Now, let q be a prime factor of p + 1. Then, since p + 1 divides pn + 1, q
must be a prime factor of pn + 1 and of np + 1 as well. Suppose q ≥ 3. Then, from
np ≡ −1 (mod q), it follows that n2p ≡ 1 (mod q) holds. If we let e be the smallest
positive integer satisfying ne ≡ 1 (mod q), then by using the same argument as
we used in the preceding solution, we can conclude that e must equal one of the
numbers 1, 2, p, 2p. If e = 1, p, then we get np ≡ 1 (mod q), which contradicts the
assumption q ≥ 3. Since n is not a multiple of q, by Fermat’s Little Theorem we
get nq−1 ≡ 1 (mod q), and therefore, we get by the minimality of e that e = 2 must
hold. From n2 ≡ 1 (mod q), we also get
np = n(2·
p−1
2 +1)
≡ n (mod q),
p
and since n ≡ −1 (mod q), we have n ≡ −1 (mod q) as well.
Now, if q = 2 then since n is odd, we have n ≡ −1 (mod q) as well. Thus, we
conclude that for an arbitrary prime factor q of p + 1, n ≡ −1 (mod q) must hold.
Suppose, for a prime q, q k for some positive integer k is a factor of p + 1. Then
k
q must be a factor of np + 1 as well. But since
np + 1 = (n + 1)(np−1 − np−2 + · · · − n + 1)
p−1
p−2
p−1
and
p−2
n
−n
+ · · · − n + 1 ≡ (−1)
− (−1)
+ · · · − (−1) + 1 6≡ 0 (mod q),
k
we see that q must divide n + 1. By applying the argument above for each prime
factor q of p + 1, we can then conclude that n + 1 must be divisible by p + 1, and
as we did in the preceding proof, we can conclude that n = p must hold.
Problem 4.
Solution: If AB = AC, then we get BF = CF and the conclusion of the problem
is clearly satisfied. So, we assume that AB 6= AC in the sequel.
Due to symmetry, we may suppose without loss of generality that AB > AC.
Let K be the point on the circle Γ such that AK is a diameter of this circle. Then,
we get
∠BCK = ∠ACK − ∠ACB = 90◦ − ∠ACB = ∠CBH
and
∠CBK = ∠ABK − ∠ABC = 90◦ − ∠ABC = ∠BCH,
from which we conclude that the triangles BCK and CBH are congruent. Therefore, the quadrilateral BKCH is a parallelogram, and its diagonal HK passes
through the center M of the other diagonal BC. Therefore, the 3 points H, M, K
lie on the same straight line, and we have ∠AEM = ∠AEK = 90◦ .
From ∠AED = 90◦ = ∠ADM , we see that the 4 points A, E, D, M lie on
the circumference of the same circle, from which we obtain ∠AM B = ∠AED =
∠AEF = ∠ACF . Putting this fact together with the fact that ∠ABM = ∠AF C,
AC
AM
=F
we conclude that the triangles ABM and AF C are similar, and we get BM
C.
By a similar argument, we get that the triangles ACM and AF B are similar, and
4
AM
AB
AC
AB
therefore, that CM
= F
B holds. Noting that BM = CM , we also get F C = F B ,
BF
AB
from which we can conclude that CF = AC , proving the assertion of the problem.
Problem 5.
Solution:
Let us note first that if i 6= j, then since ai aj ≤
n − ai aj ≥ n −
a2i +a2j
,
2
we have
a2i + a2j
n
n
≥ n − = > 0.
2
2
2
If we set bi = |ai | (i = 1, 2, . . . , n), then we get b21 + b22 + · · · + b2n = n and n−a1i aj ≤
1
n−bi bj , which shows that it is enough to prove the assertion of the problem in the
case where all of a1 , a2 , · · · , an are non-negative. Hence, we assume from now on
that a1 , a2 , · · · , an are all non-negative.
By multiplying by n the both sides of the desired inequality we get the inequality:
n2
n
≤
n − ai aj
2
X
1≤i<j≤n
and since
n(n−1)
2
n
n−ai aj
a a
i j
= 1 + n−a
, we obtain from the inequality above by subtracting
i aj
from both sides the following inequality:
X
(i)
1≤i<j≤n
ai aj
n
≤
n − ai aj
2
We will show that this inequality (i) holds.
If for some i the equality a2i = n is valid, then aj = 0 must hold for all j 6= i and
the inequality (i) is trivially satisfied. So, we assume from now on that a2i < n is
valid for each i.
2
a2 +a2
ai +aj
Let us assume that i 6= j from now on. Since 0 ≤ ai aj ≤
≤ i2 j
2
holds, we have
(ii)
ai +aj
2
ai aj
ai aj
≤
≤
a2 +a2
n − ai aj
n− i2 j
n−
2
a2i +a2j
2
=
1
(ai + aj )2
·
.
2 (n − a2i ) + (n − a2j )
Since n − a2i > 0, n − a2j > 0, we also get from the Cauchy-Schwarz inequality that
a2j
a2i
+
n − a2i
n − a2j
!
((n − a2i ) + (n − a2j )) ≥ (ai + aj )2 ,
from which it follows that
(iii)
(ai + aj )2
≤
(n − a2i ) + (n − a2j )
a2j
a2i
+
n − a2i
n − a2j
!
5
holds. Combining the inequalities (ii) and (iii), we get
X
1≤i<j≤n
ai aj
1
≤
n − ai aj
2
=
=
1≤i<j≤n
1X
2
a2j
a2i
+
2
n − ai
n − a2j
X
i6=j
a2j
n − a2i
n
1 X n − a2
i
2 i=1 n − a2i
n
= ,
2
which establishes the desired inequality (i).
!
Solutions of APMO 2013
Problem 1. Let ABC be an acute triangle with altitudes AD, BE and CF , and let O
be the center of its circumcircle. Show that the segments OA, OF, OB, OD, OC, OE dissect
the triangle ABC into three pairs of triangles that have equal areas.
Solution. Let M and N be midpoints of sides BC and AC, respectively. Notice that
∠M OC = 21 ∠BOC = ∠EAB, ∠OM C = 90◦ = ∠AEB, so triangles OM C and AEB are
OA
= OC
. For triangles ON A and BDA we also have ON
= BA
. Then
similar and we get OM
AE
AB
BD
OM
ON
= BD or BD · OM = AE · ON.
AE
Denote by S(Φ) the area of the figure Φ. So, we see that S(OBD) = 12 BD · OM =
1
AE · ON = S(OAE). Analogously, S(OCD) = S(OAF ) and S(OCE) = S(OBF ).
2
Alternative solution. Let R be the circumradius of triangle ABC, and as usual write
A, B, C for angles ∠CAB, ∠ABC, ∠BCA respectively, and a, b, c for sides BC, CA, AB
respectively. Then the area of triangle OCD is
S(OCD) =
1
2
· OC · CD · sin(∠OCD) = 21 R · CD · sin(∠OCD).
Now CD = b cos C, and
180◦ − 2A
= 90◦ − A
∠OCD =
2
(since triangle OBC is isosceles, and ∠BOC = 2A). So
S(OCD) = 21 Rb cos C sin(90◦ − A) = 21 Rb cos C cos A.
A similar calculation gives
S(OAF ) = 21 OA · AF · sin(∠OAF )
= 12 R · (b cos A) sin(90◦ − C)
= 21 Rb cos A cos C,
so OCD and OAF have the same area. In the same way we find that OBD and OAE have
the same area, as do OCE and OBF .
Problem 2. Determine all positive integers n for which
denotes the greatest integer less than or equal to r.
2
√n +1
[ n]2 +2
is an integer. Here [r]
Solution. We will show that there are no positive integers n satisfying the condition of
the problem. √
Let m = [ n] and a = n−m2 . We have m ≥ 1 since n ≥ 1. From n2 +1 = (m2 +a)2 +1 ≡
(a − 2)2 + 1 (mod (m2 + 2)), it follows that the condition of the problem is equivalent to the
fact that (a − 2)2 + 1 is divisible by m2 + 2. Since we have
0 < (a − 2)2 + 1 ≤ max{22 , (2m − 2)2 } + 1 ≤ 4m2 + 1 < 4(m2 + 2),
1
we see that (a − 2)2 + 1 = k(m2 + 2) must hold with k = 1, 2 or 3. We will show that none
of these can occur.
Case 1. When k = 1. We get (a − 2)2 − m2 = 1, and this implies that a − 2 = ±1, m = 0
must hold, but this contradicts with fact m ≥ 1.
Case 2. When k = 2. We have (a − 2)2 + 1 = 2(m2 + 2) in this case, but any perfect
square is congruent to 0, 1, 4 mod 8, and therefore, we have (a − 2)2 + 1 ≡ 1, 2, 5 (mod 8),
while 2(m2 + 2) ≡ 4, 6 (mod 8). Thus, this case cannot occur either.
Case 3. When k = 3. We have (a − 2)2 + 1 = 3(m2 + 2) in this case. Since any perfect
square is congruent to 0 or 1 mod 3, we have (a − 2)2 + 1 ≡ 1, 2 (mod 3), while 3(m2 + 2) ≡ 0
(mod 3), which shows that this case cannot occur either.
Problem 3. For 2k real numbers a1 , a2 , . . . , ak , b1 , b2 , . . . , bk define the sequence of
numbers Xn by
k
X
Xn =
[ai n + bi ] (n = 1, 2, . . .).
i=1
If the sequence Xn forms an arithmetic progression, show that
Here [r] denotes the greatest integer less than or equal to r.
Pk
i=1
ai must be an integer.
P
P
Solution. Let us write A = ki=1 ai and B = ki=1 bi . Summing the corresponding terms
of the following inequalities over i,
ai n + bi − 1 < [ai n + bi ] ≤ ai n + bi ,
we obtain An + B − k < Xn < An + B. Now suppose that {Xn } is an arithmetic progression
with the common difference d, then we have nd = Xn+1 − X1 and A + B − k < X1 ≤ A + B
Combining with the inequalities obtained above, we get
A(n + 1) + B − k < nd + X1 < A(n + 1) + B,
or
An − k ≤ An + (A + B − X1 ) − k < nd < An + (A + B − X1 ) < An + k,
from which we conclude that |A − d| < nk must hold. Since this inequality holds for any
positive integer n, we must have A = d. Since {Xn } is a sequence of integers, d must be an
integer also, and thus we conclude that A is also an integer.
Problem 4. Let a and b be positive integers, and let A and B be finite sets of integers
satisfying:
(i) A and B are disjoint;
(ii) if an integer i belongs either to A or to B, then i + a belongs to A or i − b belongs
to B.
Prove that a|A| = b|B|. (Here |X| denotes the number of elements in the set X.)
Solution. Let A∗ = {n − a : n ∈ A} and B ∗ = {n + b : n ∈ B}. Then, by (ii),
A ∪ B ⊆ A∗ ∪ B ∗ and by (i),
|A ∪ B| ≤ |A∗ ∪ B ∗ | ≤ |A∗ | + |B ∗ | = |A| + |B| = |A ∪ B|.
2
(1)
Thus, A ∪ B =P
A∗ ∪ B ∗ P
and A∗ and B ∗ have no element in common. For each finite set X
of integers, let (X) = x∈X x. Then
X
X
X
(A) +
(B) =
(A ∪ B)
X
X
X
=
(A∗ ∪ B ∗ ) =
(A∗ ) +
(B ∗ )
X
X
=
(A) − a|A| +
(B) + b|B|,
(2)
which implies a|A| = b|B|.
Alternative solution. Let us construct a directed graph whose vertices are labelled by
the members of A ∪ B and such that there is an edge from i to j iff j ∈ A and j = i + a or
j ∈ B and j = i − b. From (ii), each vertex has out-degree ≥ 1 and, from (i), each vertex has
in-degree ≤ 1. Since the sum of the out-degrees equals the sum of the in-degrees, each vertex
has in-degree and out-degree equal to 1. This is only possible if the graph is the union of
disjoint cycles, say G1 , G2 , . . . , Gn . Let |Ak | be the number of elements of A in Gk and |Bk |
be the number of elements of B in Gk . The cycle Gk will involve increasing vertex labels by
a a total of |Ak | times and decreasing them by b a total of |Bk | times. Since it is a cycle, we
have a|Ak | = b|Bk |. Summing over all cycles gives the result.
Problem 5. Let ABCD be a quadrilateral inscribed in a circle ω, and let P be a point
on the extension of AC such that P B and P D are tangent to ω. The tangent at C intersects
P D at Q and the line AD at R. Let E be the second point of intersection between AQ and
ω. Prove that B, E, R are collinear.
Solution. To show B, E, R are collinear, it is equivalent to show the lines AD, BE, CQ
are concurrent. Let CQ intersect AD at R and BE intersect AD at R0 . We shall show
RD/RA = R0 D/R0 A so that R = R0 .
Since 4P AD is similar to 4P DC and 4P AB is similar to 4P BC, we have AD/DC =
P A/P D = P A/P B = AB/BC. Hence, AB · DC = BC · AD. By Ptolemy’s theorem,
AB · DC = BC · AD = 21 CA · DB. Similarly CA · ED = CE · AD = 12 AE · DC.
Thus
DB
2DC
=
,
(3)
AB
CA
and
DC
2ED
=
.
(4)
CA
AE
3
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A
ω
D
B
R(R0 )
E
Q
C
P
Since the triangles RDC and RCA are similar, we have
RD · RA
RD
=
=
RA
RA2
RC
RA
2
=
DC
CA
RD
RC
2
=
DC
CA
=
2ED
AE
2
=
RC
.
RA
.
Thus using (4)
(5)
Using the similar triangles ABR0 and EDR0 , we have R0 D/R0 B = ED/AB. Using the
similar triangles DBR0 and EAR0 we have R0 A/R0 B = EA/DB. Thus using (3) and (4),
2
ED · DB
2ED
R0 D
.
(6)
=
=
R0 A
EA · AB
AE
It follows from (5) and (6) that R = R0 .
4
Solutions of APMO 2014
Problem 1. For a positive integer m denote by S(m) and P (m) the sum and product,
respectively, of the digits of m. Show that for each positive integer n, there exist positive
integers a1 , a2 , . . . , an satisfying the following conditions:
S(a1 ) < S(a2 ) < · · · < S(an ) and S(ai ) = P (ai+1 ) (i = 1, 2, . . . , n).
(We let an+1 = a1 .) (Problem Committee of the Japan Mathematical Olympiad Foundation)
Solution. Let k be a sufficiently large positive integer. Choose for each i = 2, 3, . . . , n,
ai to be a positive integer among whose digits the number 2 appears exactly k + i − 2 times
and the number 1 appears exactly 2k+i−1 − 2(k + i − 2) times, and nothing else. Then, we
have S(ai ) = 2k+i−1 and P (ai ) = 2k+i−2 for each i, 2 ≤ i ≤ n. Then, we let a1 be a positive
integer among whose digits the number 2 appears exactly k + n − 1 times and the number
1 appears exactly 2k − 2(k + n − 1) times, and nothing else. Then, we see that a1 satisfies
S(a1 ) = 2k and P (a1 ) = 2k+n−1 . Such a choice of a1 is possible if we take k to be large
enough to satisfy 2k > 2(k + n − 1) and we see that the numbers a1 , . . . , an chosen this way
satisfy the given requirements.
Problem 2. Let S = {1, 2, . . . , 2014}. For each non-empty subset T ⊆ S, one of its
members is chosen as its representative. Find the number of ways to assign representatives
to all non-empty subsets of S so that if a subset D ⊆ S is a disjoint union of non-empty
subsets A, B, C ⊆ S, then the representative of D is also the representative of at least one
of A, B, C. (Warut Suksompong, Thailand)
Solution. Answer: 108 · 2014!.
For any subset X let r(X) denotes the representative of X. Suppose that x1 = r(S).
First, we prove the following fact:
If x1 ∈ X and X ⊆ S, then x1 = r(X).
If |X| ≤ 2012, then we can write S as a disjoint union of X and two other subsets of S,
which gives that x1 = r(X). If |X| = 2013, then let y ∈ X and y 6= x1 . We can write X as
a disjoint union of {x1 , y} and two other subsets. We already proved that r({x1 , y}) = x1
(since |{x1 , y}| = 2 < 2012) and it follows that y 6= r(X) for every y ∈ X except x1 . We
have proved the fact.
Note that this fact is true and can be proved similarly, if the ground set S would contain
at least 5 elements.
There are 2014 ways to choose x1 = r(S) and for x1 ∈ X ⊆ S we have r(X) = x1 . Let
S1 = S \ {x1 }. Analogously, we can state that there are 2013 ways to choose x2 = r(S1 )
and for x2 ∈ X ⊆ S1 we have r(X) = x2 . Proceeding similarly (or by induction), there
are 2014 · 2013 · · · 5 ways to choose x1 , x2 , . . . , x2010 ∈ S so that for all i = 1, 2 . . . , 2010,
xi = r(X) for each X ⊆ S \ {x1 , . . . , xi−1 } and xi ∈ X.
We are now left with four elements Y = {y1 , y2 , y3 , y4 }. There are 4 ways to choose r(Y ).
Suppose that y1 = r(Y ). Then we clearly have y1 = r({y1 , y2 }) = r({y1 , y3 }) = r({y1 , y4 }).
The only subsets whose representative has not been assigned yet are {y1 , y2 , y3 }, {y1 , y2 , y4 },
{y1 , y3 , y4 }, {y2 , y3 , y4 }, {y2 , y3 }, {y2 , y4 }, {y3 , y4 }. These subsets can be assigned in any way,
hence giving 34 · 23 more choices.
1
In conclusion, the total number of assignments is 2014 · 2013 · · · 4 · 34 · 23 = 108 · 2014!.
Problem 3. Find all positive integers n such that for any integer k there exists an
integer a for which a3 + a − k is divisible by n. (Warut Suksompong, Thailand)
Solution. Answer: All integers n = 3b , where b is a nonnegative integer.
We are looking for integers n such that the set A = {a3 + a | a ∈ Z} is a complete residue
system by modulo n. Let us call this property by (*). It is not hard to see that n = 1
satisfies (*) and n = 2 does not.
If a ≡ b (mod n), then a3 + a ≡ b3 + b (mod n). So n satisfies (*) iff there are no
a, b ∈ {0, . . . , n − 1} with a 6= b and a3 + a ≡ b3 + b (mod n).
First, let us prove that 3j satisfies (*) for all j ≥ 1. Suppose that a3 + a ≡ b3 + b (mod 3j )
for a 6= b. Then (a − b)(a2 + ab + b2 + 1) ≡ 0 (mod 3j ). We can easily check mod 3 that
a2 + ab + b2 + 1 is not divisible by 3.
Next note that if A is not a complete residue system modulo integer r, then it is also not
a complete residue system modulo any multiple of r. Hence it remains to prove that any
prime p > 3 does not satisfy (*).
If p ≡ 1 (mod 4), there exists b such that b2 ≡ −1 (mod p). We then take a = 0 to
obtain the congruence a3 + a ≡ b3 + b (mod p).
Suppose now that p ≡ 3 (mod 4). We will prove that there are integers a, b 6≡ 0 (mod p)
such that a2 + ab + b2 ≡ −1 (mod p). Note that we may suppose that a 6≡ b (mod p), since
otherwise if a ≡ b (mod p) satisfies a2 + ab + b2 + 1 ≡ 0 (mod p), then (2a)2 + (2a)(−a) +
a2 + 1 ≡ 0 (mod p) and 2a 6≡ −a (mod p). Letting c be the inverse of b modulo p (i.e.
bc ≡ 1 (mod p)), the relation is equivalent to (ac)2 + ac + 1 ≡ −c2 (mod p). Note that −c2
can take on the values of all non-quadratic residues modulo p. If we can find an integer x
such that x2 + x + 1 is a non-quadratic residue modulo p, the values of a and c will follow
immediately. Hence we focus on this latter task.
Note that if x, y ∈ {0, . . . , p − 1} = B, then x2 + x + 1 ≡ y 2 + y + 1 (mod p) iff p divides
x + y + 1. We can deduce that x2 + x + 1 takes on (p + 1)/2 values as x varies in B. Since
there are (p − 1)/2 non-quadratic residues modulo p, the (p + 1)/2 values that x2 + x + 1
take on must be 0 and all the quadratic residues.
Let C be the set of quadratic residues modulo p and 0, and let y ∈ C. Suppose that
y ≡ z 2 (mod p) and let z ≡ 2w + 1 (mod p) (we can always choose such w). Then y + 3 ≡
4(w2 + w + 1) (mod p). From the previous paragraph, we know that 4(w2 + w + 1) ∈ C.
This means that y ∈ C =⇒ y + 3 ∈ C. Unless p = 3, the relation implies that all elements
of B are in C, a contradiction. This concludes the proof.
Problem 4. Let n and b be positive integers. We say n is b-discerning if there exists a
set consisting of n different positive integers less than b that has no two different subsets U
and V such that the sum of all elements in U equals the sum of all elements in V .
(a) Prove that 8 is a 100-discerning.
(b) Prove that 9 is not 100–discerning.
(Senior Problems Committee of the Australian Mathematical Olympiad Committee)
Solution.
(a) Take S = {3, 6, 12, 24, 48, 95, 96, 97}, i.e.
S = {3 · 2k : 0 ≤ k ≤ 5} ∪ {3 · 25 − 1, 3 · 25 + 1}.
2
As k ranges between 0 to 5, the sums obtained from the numbers 3 · 2k are 3t, where
1 ≤ t ≤ 63. These are 63 numbers that are divisible by 3 and are at most 3 · 63 = 189.
Sums of elements of S are also the numbers 95 + 97 = 192 and all the numbers that
are sums of 192 and sums obtained from the numbers 3 · 2k with 0 ≤ k ≤ 5. These are 64
numbers that are all divisible by 3 and at least equal to 192. In addition, sums of elements
of S are the numbers 95 and all the numbers that are sums of 95 and sums obtained from
the numbers 3 · 2k with 0 ≤ k ≤ 5. These are 64 numbers that are all congruent to −1 mod
3.
Finally, sums of elements of S are the numbers 97 and all the numbers that are sums of
97 and sums obtained from the numbers 3 · 2k with 0 ≤ k ≤ 5. These are 64 numbers that
are all congruent to 1 mod 3.
Hence there are at least 63 + 64 + 64 + 64 = 255 different sums from elements of S. On
the other hand, S has 28 − 1 = 255 non-empty subsets. Therefore S has no two different
subsets with equal sums of elements. Therefore, 8 is 100-discerning.
(b) Suppose that 9 is 100-discerning. Then there is a set S = {s1 , . . . , s9 }, si < 100 that
has no two different subsets with equal sums of elements. Assume that 0 < s1 < · · · < s9 <
100.
Let X be the set of all subsets of S having at least 3 and at most 6 elements and let Y
be the set of all subsets of S having exactly 2 or 3 or 4 elements greater than s3 .
The set X consists of
9
9
9
9
+
+
+
= 84 + 126 + 126 + 84 = 420
3
4
5
6
subsets of S. The set in X with the largest sums of elements is {s4 , . . . , s9 } and the smallest
sums is in {s1 , s2 , s3 }. Thus the sum of the elements of each of the 420 sets in X is at least
s1 + s2 + s3 and at most s4 + · · · + s9 , which is one of (s4 + · · · + s9 ) − (s1 + s2 + s3 ) + 1 integers.
From the pigeonhole principle it follows that (s4 + · · · + s9 ) − (s1 + s2 + s3 ) + 1 ≥ 420, i.e.,
(s4 + · · · + s9 ) − (s1 + s2 + s3 ) ≥ 419.
(1)
6
Now let us calculate
the number of subsets6in Y . Observe that {s4 , . . . , s9 } has 2
6
2-element subsets, 3 3-element subsets and 4 4-element subsets, while {s1 , s2 , s3 } has
exactly 8 subsets. Hence the number of subsets of S in Y equals
6
6
6
8
+
+
= 8(15 + 20 + 15) = 400.
2
3
4
The set in Y with the largest sum of elements is {s1 , s2 , s3 , s6 , s7 , s8 , s9 } and the smallest
sum is in {s4 , s5 }. Again, by the pigeonhole principle it follows that (s1 + s2 + s3 + s6 + s7 +
s8 + s9 ) − (s4 + s5 ) + 1 ≥ 400, i.e.,
(s1 + s2 + s3 + s6 + s7 + s8 + s9 ) − (s4 + s5 ) ≥ 399.
(2)
Adding (1) and (2) yields 2(s6 + s7 + s8 + s9 ) ≥ 818, so that s9 + 98 + 97 + 96 ≥
s9 + s8 + s7 + s6 ≥ 409, i.e. s9 ≥ 118, a contradiction with s9 < 100. Therefore, 9 is not
100-discerning.
3
Problem 5. Circles ω and Ω meet at points A and B. Let M be the midpoint of the
arc AB of circle ω (M lies inside Ω). A chord M P of circle ω intersects Ω at Q (Q lies inside
ω). Let `P be the tangent line to ω at P , and let `Q be the tangent line to Ω at Q. Prove
that the circumcircle of the triangle formed by the lines `P , `Q , and AB is tangent to Ω.
(Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan)
Solution. Denote X = AB ∩ `P , Y = AB ∩ `Q , and Z = `P ∩ `Q . Without loss of
generality we have AX < BX. Let F = M P ∩ AB.
`Q
`P
Z
X
YY
YY
D
D
D
A
A
A
A
P
S
Q
F
M
Ω
ω
R
B
B
B
T
Denote by R the second point of intersection of P Q and Ω; by S the point of Ω such that
SR k AB; and by T the point of Ω such that RT k `P . Since M is the midpoint of arc AB,
the tangent `M at M to ω is parallel to AB, so ∠(AB, P M ) = ∠(P M, `P ). Therefore we
have ∠P RT = ∠M P X = ∠P F X = ∠P RS. Thus the point Q is the midpoint of the
arc T QS of Ω, hence ST k `Q . So the corresponding sides of the triangles RST and XY Z
are parallel, and there exist a homothety h mapping RST to XY Z.
Let D be the second point of intersection of XR and Ω. We claim that D is the center of
the homothety h; since D ∈ Ω, this implies that the circumcircles of triangles RST and XY Z
are tangent, as required. So, it remains to prove this claim. In order to do this, it suffices to
show that D ∈ SY .
By ∠P F X = ∠XP F we have XF 2 = XP 2 = XA · XB = XD · XR. Therefore,
XF
XR
= XF
, so the triangles XDF and XF R are similar, hence ∠DF X = ∠XRF = ∠DRQ =
XD
∠DQY ; thus the points D, Y , Q, and F are concyclic. It follows that ∠Y DQ = ∠Y F Q =
∠SRQ = 180◦ − ∠SDQ which means exactly that the points Y , D, and S are collinear, with
D between S and Y .
4
Solutions of APMO 2015
Problem 1. Let ABC be a triangle, and let D be a point on side BC. A line through
D intersects side AB at X and ray AC at Y . The circumcircle of triangle BXD intersects the
circumcircle ω of triangle ABC again at point Z 6= B. The lines ZD and ZY intersect ω again
at V and W , respectively. Prove that AB = V W .
Solution. Suppose XY intersects ω at points P and Q, where Q lies between X and Y . We
will show that V and W are the reflections of A and B with respect to the perpendicular bisector
of P Q. From this, it follows that AV W B is an isosceles trapezoid and hence AB = V W .
First, note that
∠BZD = ∠AXY = ∠AP Q + ∠BAP = ∠AP Q + ∠BZP,
so ∠AP Q = ∠P ZV = ∠P QV , and hence V is the reflection of A with respect to the perpendicular bisector of P Q.
Now, suppose W 0 is the reflection of B with respect to the perpendicular bisector of P Q,
and let Z 0 be the intersection of Y W 0 and ω. It suffices to show that B, X, D, Z 0 are concyclic.
Note that
∠Y DC = ∠P DB = ∠P CB + ∠QP C = ∠W 0 P Q + ∠QP C = ∠W 0 P C = ∠Y Z 0 C.
So D, C, Y , Z 0 are concyclic. Next, ∠BZ 0 D = ∠CZ 0 B − ∠CZ 0 D = 180◦ − ∠BXD and due to
the previous concyclicity we are done.
Alternative solution 1. Using cyclic quadrilaterals BXDZ and ABZV in turn, we have
∠ZDY = ∠ZBA = ∠ZCY. So ZDCY is cyclic.
Using cyclic quadrilaterals ABZC and ZDCY in turn, we have ∠AZB = ∠ACB = ∠W ZV
(or 180◦ − ∠W ZV if Z lies between W and C).
So AB = V W because they subtend equal (or supplementary) angles in ω.
Alternative solution 2. Using cyclic quadrilaterals BXDZ and ABZV in turn, we have
∠ZDY = ∠ZBA = ∠ZCY. So ZDCY is cyclic.
Using cyclic quadrilaterals BXDZ and ABZV in turn, we have ∠DXA = ∠V ZB =
180◦ − BAV. So XD k AV .
Using cyclic quadrilaterals ZDCY and BCW Z in turn, we have ∠Y DC = ∠Y ZC =
∠W BC. So XD k BW .
Hence BW k AV which implies that AV W B is an isosceles trapezium with AB = V W . Problem 2. Let S = {2, 3, 4, . . .} denote the set of integers that are greater than or equal
to 2. Does there exist a function f : S → S such that
f (a)f (b) = f (a2 b2 ) for all a, b ∈ S with a 6= b?
Solution. We prove that there is no such function. For arbitrary elements a and b of S,
choose an integer c that is greater than both of them. Since bc > a and c > b, we have
f (a4 b4 c4 ) = f (a2 )f (b2 c2 ) = f (a2 )f (b)f (c).
Furthermore, since ac > b and c > a, we have
f (a4 b4 c4 ) = f (b2 )f (a2 c2 ) = f (b2 )f (a)f (c).
Comparing these two equations, we find that for all elements a and b of S,
f (a2 )f (b) = f (b2 )f (a)
=⇒
1
f (b2 )
f (a2 )
=
.
f (a)
f (b)
It follows that there exists a positive rational number k such that
f (a2 ) = kf (a),
for all a ∈ S.
(1)
Substituting this into the functional equation yields
f (ab) =
f (a)f (b)
,
k
for all a, b ∈ S with a 6= b.
(2)
Now combine the functional equation with equations (1) and (2) to obtain
f (a)f (a2 ) = f (a6 ) =
f (a)f (a)f (a4 )
f (a)f (a)f (a2 )
f (a)f (a5 )
=
,
=
k
k2
k
for all a ∈ S.
It follows that f (a) = k for all a ∈ S. Substituting a = 2 and b = 3 into the functional equation
yields k = 1, however 1 6∈ S and hence we have no solutions.
Problem 3. A sequence of real numbers a0 , a1 , . . . is said to be good if the following three
conditions hold.
(i) The value of a0 is a positive integer.
(ii) For each non-negative integer i we have ai+1 = 2ai + 1 or ai+1 =
ai
.
ai + 2
(iii) There exists a positive integer k such that ak = 2014.
Find the smallest positive integer n such that there exists a good sequence a0 , a1 , . . . of real
numbers with the property that an = 2014.
Answer: 60.
Solution. Note that
ai+1 + 1 = 2(ai + 1) or ai+1 + 1 =
Hence
ai + ai + 2
2(ai + 1)
=
.
ai + 2
ai + 2
1
1
1
ai + 2
1
1
1
1
= ·
or
=
= ·
+ .
ai+1 + 1
2 ai + 1
ai+1 + 1
2(ai + 1)
2 ai + 1 2
Therefore,
k
X εi
1
1
1
= k·
+
,
ak + 1
2 a0 + 1 i=1 2k−i+1
(1)
where εi = 0 or 1. Multiplying both sides by 2k (ak + 1) and putting ak = 2014, we get
!
k
X
2015
2k =
+ 2015 ·
εi · 2i−1 ,
a0 + 1
i=1
where εi = 0 or 1. Since gcd(2, 2015) = 1, we have a0 + 1 = 2015 and a0 = 2014. Therefore,
!
k
X
2k − 1 = 2015 ·
εi · 2i−1 ,
i=1
where εi = 0 or 1. We now need to find the smallest k such that 2015|2k − 1. Since 2015 =
5 · 13 · 31, from the Fermat little theorem we obtain 5|24 − 1, 13|212 − 1 and 31|230 − 1. We also
have lcm[4, 12, 30] = 60, hence 5|260 − 1, 13|260 − 1 and 31|260 − 1, which gives 2015|260 − 1.
2
But 5 - 230 − 1 and so k = 60 is the smallest positive integer such that 2015|2k − 1. To conclude,
the smallest positive integer k such that ak = 2014 is when k = 60.
Alternative solution 1. Clearly all members of the sequence are positive rational numbers.
2ai+1
ai+1 − 1
or ai =
. Since ai > 0 we deduce
For each positive integer i, we have ai =
2
1 − ai+1
that


 ai+1 − 1 if ai+1 > 1

2
ai =
2a
i+1


if ai+1 < 1.

1 − ai+1
Thus ai is uniquely determined from ai+1 . Hence starting from ak = 2014, we simply run the
sequence backwards until we reach a positive integer. We compute as follows.
2014 2013 2011 2007 1999 1983 1951 1887 1759 1503 991 1982 1949 1883 1751 1487 959 1918 1821 1627
, 2 , 4 , 8 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 , 33 , 66 , 132 , 264 , 528 , 1056 , 97 , 194 , 388 ,
1
1239 463 926 1852 1689 1363 711 1422 829 1658 1301 587 1174 333 666 1332 649 1298 581 1162
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
776 1552 1089 163 326 652 1304 593 1186 357 714 1428 841 1682 1349 683 1366 717 1434 853
309 618 1236 457 914 1828 1641 1267 519 1038 61
,
,
,
,
,
,
,
,
,
,
, 122 , 244 , 488 , 976 , 1952 , 1889
, 1763 , 1511 , 1007 , 2014
.
1706 1397 779 1558 1101 187 374 748 1496 977 1954 1893 1771 1527 1039 63
126 252 504 1008
1
There are 61 terms in the above list. Thus k = 60.
Alternative solution 1 is quite computationally intensive. Calculating the first few terms
indicates some patterns that are easy to prove. This is shown in the next solution.
m0
where m0 = 2014 and n0 = 1 as in alternative
Alternative solution 2. Start with ak =
n0
mi
solution 1. By inverting the sequence as in alternative solution 1, we have ak−i =
for i ≥ 0
ni
where
(mi − ni , 2ni ) if mi > ni
(mi+1 , ni+1 ) =
(2mi , ni − mi ) if mi < ni .
Easy inductions show that mi + ni = 2015, 1 ≤ mi , ni ≤ 2014 and gcd(mi , ni ) = 1 for
i ≥ 0. Since a0 ∈ N+ and gcd(mk , nk ) = 1, we require nk = 1. An easy induction shows that
(mi , ni ) ≡ (−2i , 2i ) (mod 2015) for i = 0, 1, . . . , k.
Thus 2k ≡ 1 (mod 2015). As in the official solution, the smallest such k is k = 60. This
yields nk ≡ 1 (mod 2015). But since 1 ≤ nk , mk ≤ 2014, it follows that a0 is an integer.
Problem 4. Let n be a positive integer. Consider 2n distinct lines on the plane, no two
of which are parallel. Of the 2n lines, n are colored blue, the other n are colored red. Let B be
the set of all points on the plane that lie on at least one blue line, and R the set of all points
on the plane that lie on at least one red line. Prove that there exists a circle that intersects B
in exactly 2n − 1 points, and also intersects R in exactly 2n − 1 points.
Solution. Consider a line ` on the plane and a point P on it such that ` is not parallel to
any of the 2n lines. Rotate ` about P counterclockwise until it is parallel to one of the 2n lines.
Take note of that line and keep rotating until all the 2n lines are met. The 2n lines are now
ordered according to which line is met before or after. Say the lines are in order `1 , . . . , `2n .
Clearly there must be k ∈ {1, . . . , 2n − 1} such that `k and `k+1 are of different colors.
Now we set up a system of X– and Y – axes on the plane. Consider the two angular bisectors
of `k and `k+1 . If we rotate `k+1 counterclockwise, the line will be parallel to one of the bisectors
before the other. Let the bisector that is parallel to the rotation of `k+1 first be the X–axis,
and the other the Y –axis. From now on, we will be using the directed angle notation: for lines
s and s0 , we define ∠(s, s0 ) to be a real number in [0, π) denoting the angle in radians such
that when s is rotated counterclockwise by ∠(s, s0 ) radian, it becomes parallel to s0 . Using this
3
notation, we notice that there is no i = 1, . . . , 2n such that ∠(X, li ) is between ∠(X, `k ) and
∠(X, `k+1 ).
Because the 2n lines are distinct, the set S of all the intersections between `i and `j (i 6= j)
is a finite set of points. Consider a rectangle with two opposite vertices lying on `k and the
other two lying on `k+1 . With respect to the origin (the intersection of `k and `k+1 ), we can
enlarge the rectangle as much as we want, while all the vertices remain on the lines. Thus,
there is one of these rectangles R which contains all the points in S in its interior. Since each
side of R is parallel to either X– or Y – axis, R is a part of the four lines x = ±a, y = ±b.
where a, b > 0.
Y
lk+1
lk
b
N
M
a
−a
X
−b
Consider the circle C tangent to the right of the x = a side of the rectangle, and to both `k
and `k+1 . We claim that this circle intersects B in exactly 2n − 1 points, and also intersects R
in exactly 2n − 1 points. Since C is tangent to both `k and `k+1 and the two lines have different
colors, it is enough to show that C intersects with each of the other 2n − 2 lines in exactly 2
points. Note that no two lines intersect on the circle because all the intersections between lines
are in S which is in the interior of R.
Consider any line L among these 2n − 2 lines. Let L intersect with `k and `k+1 at the points
M and N , respectively (M and N are not necessarily distinct). Notice that both M and N
must be inside R. There are two cases:
(i) L intersects R on the x = −a side once and another time on x = a side;
(ii) L intersects y = −b and y = b sides.
However, if (ii) happens, ∠(`k , L) and ∠(L, `k+1 ) would be both positive, and then ∠(X, L)
would be between ∠(X, `k ) and ∠(X, `k+1 ), a contradiction. Thus, only (i) can happen. Then
L intersects C in exactly two points, and we are done.
Alternative solution. By rotating the diagram we can ensure that no line is vertical. Let
`1 , `2 , . . . , `2n be the lines listed in order of increasing gradient. Then there is a k such that lines
`k and `k+1 are oppositely coloured. By rotating our coordinate system and cyclicly relabelling
our lines we can ensure that `1 , `2 , . . . , `2n are listed in order of increasing gradient, `1 and `2n
are oppositely coloured, and no line is vertical.
Let D be a circle centred at the origin and of sufficiently large radius so that
• All intersection points of all pairs of lines lie strictly inside D; and
• Each line `i intersects D in two points Ai and Bi say, such that Ai is on the right semicircle
(the part of the circle in the positive x half plane) and Bi is on the left semicircle.
Note that the anticlockwise order of the points Ai , Bi around D is A1 , A2 , . . . , An , B1 , B2 , . . . , Bn .
4
(If Ai+1 occurred before Ai then rays ri and ri+1 (as defined below) would intersect outside D.)
r2n
r2n−1
D
A2n
B1
A2n−1
B2
B2n−1 A2
B2n A1
C
r2
r1
For each i, let ri be the ray that is the part of the line `i starting from point Ai and that
extends to the right. Let C be any circle tangent to r1 and r2n , that lies entirely to the right
of D. Then C intersects each of r2 , r3 , . . . , r2n−1 twice and is tangent to r1 and r2n . Thus C has
the required properties.
Problem 5. Determine all sequences a0 , a1 , a2 , . . . of positive integers with a0 ≥ 2015
such that for all integers n ≥ 1:
(i) an+2 is divisible by an ;
(ii) |sn+1 − (n + 1)an | = 1, where sn+1 = an+1 − an + an−1 − · · · + (−1)n+1 a0 .
Answer: There are two families of answers:
(a) an = c(n + 2)n! for all n ≥ 1 and a0 = c + 1 for some integer c ≥ 2014, and
(b) an = c(n + 2)n! for all n ≥ 1 and a0 = c − 1 for some integer c ≥ 2016.
Solution. Let {an }∞
n=0 be a sequence of positive integers satisfying the given conditions.
We can rewrite (ii) as sn+1 = (n + 1)an + hn , where hn ∈ {−1, 1}. Substituting n with n − 1
yields sn = nan−1 + hn−1 , where hn−1 ∈ {−1, 1}. Note that an+1 = sn+1 + sn , therefore there
exists δn ∈ {−2, 0, 2} such that
an+1 = (n + 1)an + nan−1 + δn .
(1)
We also have |s2 − 2a1 | = 1, which yields a0 = 3a1 − a2 ± 1 ≤ 3a1 , and therefore a1 ≥ a30 ≥ 671.
Substituting n = 2 in (1), we find that a3 = 3a2 + 2a1 + δ2 . Since a1 |a3 , we have a1 |3a2 + δ2 ,
and therefore a2 ≥ 223. Using (1), we obtain that an ≥ 223 for all n ≥ 0.
Lemma 1: For n ≥ 4, we have an+2 = (n + 1)(n + 4)an .
Proof. For n ≥ 3 we have
an = nan−1 + (n − 1)an−2 + δn−1 > nan−1 + 3.
(2)
By applying (2) with n substituted by n − 1 we have for n ≥ 4,
an = nan−1 + (n − 1)an−2 + δn−1 < nan−1 + (an−1 − 3) + δn−1 < (n + 1)an−1 .
Using (1) to write an+2 in terms of an and an−1 along with (2), we obtain that for n ≥ 3,
an+2 = (n + 3)(n + 1)an + (n + 2)nan−1 + (n + 2)δn + δn+1
< (n + 3)(n + 1)an + (n + 2)nan−1 + 3(n + 2)
< (n2 + 5n + 5)an .
5
(3)
Also for n ≥ 4,
an+2 = (n + 3)(n + 1)an + (n + 2)nan−1 + (n + 2)δn + δn+1
> (n + 3)(n + 1)an + nan
= (n2 + 5n + 3)an .
Since an |an+2 , we obtain that an+2 = (n2 + 5n + 4)an = (n + 1)(n + 4)an , as desired. Lemma 2: For n ≥ 4, we have an+1 =
(n + 1)(n + 3)
an .
n+2
Proof. Using the recurrence an+3 = (n + 3)an+2 + (n + 2)an+1 + δn+2 and writing an+3 ,
an+2 in terms of an+1 , an according to Lemma 1 we obtain
(n + 2)(n + 4)an+1 = (n + 3)(n + 1)(n + 4)an + δn+2 .
(n + 1)(n + 3)
an , as desired. n+2
(n + 1)(n + 3)
Suppose there exists n ≥ 1 such that an+1 6=
an . By Lemma 2, there exist a
n+2
(m + 2)(m + 4)
greatest integer 1 ≤ m ≤ 3 with this property. Then am+2 =
am+1 . If δm+1 = 0,
m+3
(m + 1)(m + 3)
we have am+1 =
am , which contradicts our choice of m. Thus δm+1 6= 0.
m+2
Hence n + 4|δn+2 , which yields δn+2 = 0 and an+1 =
Clearly m + 3|am+1 . Write am+1 = (m + 3)k and am+2 = (m + 2)(m + 4)k. Then (m +
1)am + δm+1 = am+2 − (m + 2)am+1 = (m + 2)k. So, am |(m + 2)k − δm+1 . But am also divides
am+2 = (m + 2)(m + 4)k. Combining the two divisibility conditions, we obtain am |(m + 4)δm+1 .
Since δm+1 6= 0, we have am |2m + 8 ≤ 14, which contradicts the previous result that an ≥ 223
for all nonnegative integers n.
(n + 1)(n + 3)
an for n ≥ 1. Substituting n = 1 yields 3|a1 . Letting a1 = 3c,
So, an+1 =
n+2
we have by induction that an = n!(n + 2)c for n ≥ 1. Since |s2 − 2a1 | = 1, we then get
a0 = c ± 1, yielding the two families of solutions. By noting that (n + 2)n! = n! + (n + 1)!,
we have sn+1 = c(n + 2)! + (−1)n (c − a0 ). Hence both families of solutions satisfy the given
conditions.
6
Solutions of APMO 2016
Problem 1. We say that a triangle ABC is great if the following holds: for any point
D on the side BC, if P and Q are the feet of the perpendiculars from D to the lines AB and
AC, respectively, then the reflection of D in the line P Q lies on the circumcircle of the triangle
ABC.
Prove that triangle ABC is great if and only if ∠A = 90◦ and AB = AC.
Solution. For every point D on the side BC, let D0 be the reflection of D in the line P Q.
We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled
at A.
Choose D to be the point where the angle bisector from A meets BC. Note that P and Q
lie on the rays AB and AC respectively. Furthermore, P and Q are reflections of each other
in the line AD, from which it follows that P Q ⊥ AD. Therefore, D0 lies on the line AD and
we may deduce that either D0 = A or D0 is the second point of the angle bisector at A and
the circumcircle of ABC. However, since AP DQ is a cyclic quadrilateral, the segment P Q
intersects the segment AD. Therefore, D0 lies on the ray DA and therefore D0 = A. By angle
chasing we obtain
∠P D0 Q = ∠P DQ = 180◦ − ∠BAC,
and since D0 = A we also know ∠P D0 Q = ∠BAC. This implies that ∠BAC = 90◦ .
Now we choose D to be the midpoint of BC. Since ∠BAC = 90◦ , we can deduce that
DQP is the medial triangle of triangle ABC. Therefore, P Q||BC from which it follows that
DD0 ⊥ BC. But the distance from D0 to BC is equal to both the circumradius of triangle
ABC and to the distance from A to BC. This can only happen if A = D0 . This implies that
ABC is isosceles and right-angled at A.
We will now prove that if ABC is isosceles and right-angled at A then the required property
in the problem holds. Let D be any point on side BC. Then D0 P = DP and we also have
DP = BP . Hence, D0 P = BP and similarly D0 Q = CQ. Note that AP DQD0 is cyclic
with diameter P Q. Therefore, ∠AP D0 = ∠AQD0 , from which we obtain ∠BP D0 = ∠CQD0 .
So triangles D0 P B and D0 QC are similar. It follows that ∠P D0 Q = ∠P D0 C + ∠CD0 Q =
0
0
∠P D0 C + ∠BD0 P = ∠BD0 C and D0 P = D 0 B . So we also obtain that triangles D0 P Q and
DQ
DC
D0 BC are similar. But since DP Q and D0 P Q are congruent, we may deduce that ∠BD0 C =
∠P D0 Q = ∠P DQ = 90◦ . Therefore, D0 lies on the circle with diameter BC, which is the
circumcircle of triangle ABC.
Problem 2. A positive integer is called fancy if it can be expressed in the form
1
2a1 + 2a2 + · · · + 2a100 ,
where a1 , a2 , . . . , a100 are non-negative integers that are not necessarily distinct.
Find the smallest positive integer n such that no multiple of n is a fancy number.
Answer: The answer is n = 2101 − 1.
Solution. Let k be any positive integer less than 2101 − 1. Then k can be expressed
in binary notation using at most 100 ones, and therefore there exists a positive integer r and
non-negative integers a1 , a2 , . . . , ar such that r ≤ 100 and k = 2a1 + · · · + 2ar . Notice that for
a positive integer s we have:
2s k = 2a1 +s + 2a2 +s + · · · + 2ar−1 +s + (1 + 1 + 2 + · · · + 2s−1 )2ar
= 2a1 +s + 2a2 +s + · · · + 2ar−1 +s + 2ar + 2ar + · · · + 2ar +s−1 .
This shows that k has a multiple that is a sum of r + s powers of two. In particular, we
may take s = 100 − r ≥ 0, which shows that k has a multiple that is a fancy number.
We will now prove that no multiple of n = 2101 − 1 is a fancy number. In fact we will prove
a stronger statement, namely, that no multiple of n can be expressed as the sum of at most
100 powers of 2.
For the sake of contradiction, suppose that there exists a positive integer c such that cn is
the sum of at most 100 powers of 2. We may assume that c is the smallest such integer. By
repeatedly merging equal powers of two in the representation of cn we may assume that
cn = 2a1 + 2a2 + · · · + 2ar
where r ≤ 100 and a1 < a2 < . . . < ar are distinct non-negative integers. Consider the following
two cases:
• If ar ≥ 101, then 2ar − 2ar −101 = 2ar −101 n. It follows that 2a1 + 2a2 + · · · + 2ar−1 + 2ar −101
would be a multiple of n that is smaller than cn. This contradicts the minimality of c.
• If ar ≤ 100, then {a1 , . . . , ar } is a proper subset of {0, 1, . . . , 100}. Then
n ≤ cn < 20 + 21 + · · · + 2100 = n.
This is also a contradiction.
From these contradictions we conclude that it is impossible for cn to be the sum of at most
100 powers of 2. In particular, no multiple of n is a fancy number.
Problem 3. Let AB and AC be two distinct rays not lying on the same line, and let ω
be a circle with center O that is tangent to ray AC at E and ray AB at F . Let R be a point
on segment EF . The line through O parallel to EF intersects line AB at P . Let N be the
intersection of lines P R and AC, and let M be the intersection of line AB and the line through
R parallel to AC. Prove that line M N is tangent to ω.
Solution. We present two approaches. The first one introduces an auxiliary point and
studies similarities in the figure. The second one reduces the problem to computations involving
a particular exradius of a triangle. The second approach has two variants.
Solution 1.
2
Let the line through N tangent to ω at point X 6= E intersect AB at point M 0 . It suffices
to show that M 0 R k AC, since this would yield M 0 = M .
Suppose that the line P O intersects AC at Q and the circumcircle of AM 0 O at Y , respectively. Then
∠AY M 0 = ∠AOM 0 = 90◦ − ∠M 0 OP.
By angle chasing we have ∠EOQ = ∠F OP = 90◦ − ∠AOF = ∠M 0 AO = ∠M 0 Y P and by
symmetry ∠EQO = ∠M 0 P Y . Therefore 4M 0 Y P ∼ 4EOQ.
On the other hand, we have
1
∠M 0 OP = ∠M 0 OF + ∠F OP = (∠F OX + ∠F OP + ∠EOQ) =
2
1 180◦ − ∠XOE
∠XOE
=
= 90◦ −
.
2
2
2
Since we know that ∠AY M 0 and ∠M 0 OP are complementary this implies
∠XOE
= ∠N OE.
2
Therefore, ∠AY M 0 and ∠N OE are congruent angles, and this means that A and N are
corresponding points in the similarity of triangles 4M 0 Y P and 4EOQ. It follows that
∠AY M 0 =
AM 0
NE
NR
=
=
.
M 0P
EQ
RP
We conclude that M 0 R k AC, as desired.
Solution 2a.
3
PR = PM0 .
As in Solution 1, we introduce point M 0 and reduce the problem to proving RN
M 0A
Menelaus theorem in triangle AN P with transversal line F RE yields
P R N E AF
·
·
= 1.
RN EA F P
F P = P R , so that it suffices to prove
Since AF = EA, we have N
E
RN
FP
P M0
= 0 .
(1)
NE
MA
This is a computation regarding the triangle AM 0 N and its excircle opposite A. Indeed,
setting a = M 0 N , b = N A, c = M 0 A, s = a + 2b + c , x = s − a, y = s − b and z = s − c, then
r2
AE = AF = s, M 0 F = z and N E = y. From 4OF P ∼ 4AF O we have F P = sa , where
ra = OF is the exradius opposite A. Combining the following two standard formulas for the
area of a triangle
|AM 0 N |2 = xyzs (Heron’s formula) and |AM 0 N | = ra (s − a),
yz
we have ra2 = yzs
x . Therefore, F P = x . We can now write everything in (1) in terms of x, y, z.
We conclude that we have to verify
yz
x
y
=
z + yz
x
x+y
,
which is easily seen to be true.
Note: Antoher approach using Menalaus theorem is to construct the tangent from M to
create a point N 0 in AC and then prove, using the theorem, that P , R and N 0 are collinear.
This also reduces to an algebraic identity.
Solution 2b.
As in Solution 1, we introduce point M 0 . Let the line through M 0 and parallel to AN
intersect EF at R0 . Let P 0 be the intersection of lines N R0 and AM . It suffices to show that
P 0 O k F E, since this would yield P = P 0 , and then R = R0 and M = M 0 . Hence it is enough
to prove that
AD
AF
=
,
(2)
0
FP
DO
where D is the intersection of AO and EF . Once again, this reduces to a computation regarding
the triangle AM 0 N and its excircle opposite A.
Let u = P 0 F and x, y, z, s as in Solution 2a. Note that since AE = AF and M 0 R0 k AE, we
0 R0
0
0
have M 0 R0 = M 0 F = z. Since M 0 R0 k AN , we have P M
=M
0
N A , that is,
PA
u+z
z
=
.
u+x+y+z
x+z
AF
xs
From this last equation we obtain u = yz
x . Hence F P 0 = yz . Also, as in Solution 2a, we have
ra2 = yzs
x .
Finally, using similar triangles ODF , F DA and OF A, and the above equalities, we have
AD
AD DF
AF AF
s2
s2
xs
AF
=
·
=
·
= 2 = yzs =
=
,
0
DO
DF DO
OF OF
ra
yz
F
P
x
as required.
4
Problem 4. The country Dreamland consists of 2016 cities. The airline Starways wants to
establish some one-way flights between pairs of cities in such a way that each city has exactly
one flight out of it. Find the smallest positive integer k such that no matter how Starways
establishes its flights, the cities can always be partitioned into k groups so that from any city
it is not possible to reach another city in the same group by using at most 28 flights.
Answer: 57
Solution. The flights established by Starways yield a directed graph G on 2016 vertices
in which each vertex has out-degree equal to 1.
We first show that we need at least 57 groups. For this, suppose that G has a directed cycle
of length 57. Then, for any two cities in the cycle, one is reachable from the other using at
most 28 flights. So no two cities in the cycle can belong to the same group. Hence, we need at
least 57 groups.
We will now show that 57 groups are enough. Consider another auxiliary directed graph H
in which the vertices are the cities of Dreamland and there is an arrow from city u to city v
if u can be reached from v using at most 28 flights. Each city has out-degree at most 28. We
will be done if we can split the cities of H in at most 57 groups such that there are no arrows
between vertices of the same group. We prove the following stronger statement.
Lemma: Suppose we have a directed graph on n ≥ 1 vertices such that each vertex has
out-degree at most 28. Then the vertices can be partitioned into 57 groups in such a way that
no vertices in the same group are connected by an arrow.
Proof: We apply induction. The result is clear for 1 vertex. Now suppose we have more
than one vertex. Since the out-degree of each vertex is at most 28, there is a vertex, say v, with
in-degree at most 28. If we remove the vertex v we obtain a graph with fewer vertices which
still satifies the conditions, so by inductive hypothesis we may split it into at most 57 groups
with no adjacent vertices in the same group. Since v has in-degree and out-degree at most 28,
it has at most 56 neighboors in the original directed graph. Therefore, we may add v back and
place it in a group in which it has no neighbors. This completes the inductive step.
Problem 5. Find all functions f : R+ → R+ such that
(z + 1)f (x + y) = f (xf (z) + y) + f (yf (z) + x),
(3)
for all positive real numbers x, y, z.
Answer: The only solution is f (x) = x for all positive real numbers x.
Solution. The identity function f (x) = x clearly satisfies the functional equation. Now,
let f be a function satisfying the functional equation. Plugging x = y = 1 into (3) we get
2f (f (z) + 1) = (z + 1)(f (2)) for all z ∈ R+ . Hence, f is not bounded above.
Lemma. Let a, b, c be positive real numbers. If c is greater than 1, a/b and b/a, then the
system of linear equations
cu + v = a u + cv = b
has a positive real solution u, v.
Proof. The solution is
cb − a
ca − b
v= 2
.
2
c −1
c −1
The numbers u and v are positive if the conditions on c above are satisfied.
u=
5
We will now prove that
f (a) + f (b) = f (c) + f (d) for all a, b, c, d ∈ R+ with a + b = c + d.
(4)
Consider a, b, c, d ∈ R+ such that a + b = c + d. Since f is not bounded above, we can
choose a positive number e such that f (e) is greater than 1, a/b, b/a, c/d and d/c. Using the
above lemma, we can find u, v, w, t ∈ R+ satisfying
f (e)u + v = a,
f (e)w + t = c,
u + f (e)v = b
w + f (e)t = d.
Note that u + v = w + t since (u + v)(f (e) + 1) = a + b and (w + t)(f (e) + 1) = c + d.
Plugging x = u, y = v and z = e into (3) yields f (a) + f (b) = (e + 1)f (u + v). Similarly, we
have f (c) + f (d) = (e + 1)f (w + t). The claim follows immediately.
We then have
yf (x) = f (xf (y)) for all x, y ∈ R+
(5)
since by (3) and (4),
x
x
x
f (y) +
+f
f (y) +
= f (xf (y)) + f (x).
(y + 1)f (x) = f
2
2
2
2
Now, let a = f (1/f (1)). Plugging x = 1 and y = 1/f (1) into (5) yields f (a) = 1. Hence
a = af (a) and f (af (a)) = f (a) = 1. Since af (a) = f (af (a)) by (5), we have f (1) = a = 1. It
follows from (5) that
x
f (f (y)) = y
for all y ∈ R+ .
(6)
Using (4) we have for all x, y ∈ R+ that
f (x + y) + f (1) = f (x) + f (y + 1),
f (y + 1) + f (1) = f (y) + f (2).
and
Therefore
f (x + y) = f (x) + f (y) + b for all x, y ∈ R+ ,
(7)
where b = f (2) − 2f (1) = f (2) − 2. Using (5), (7) and (6), we get
4 + 2b = 2f (2) = f (2f (2)) = f (f (2) + f (2)) = f (f (2)) + f (f (2)) + b = 4 + b.
This shows that b = 0 and thus
f (x + y) = f (x) + f (y) for all x, y ∈ R+ .
In particular, f is strictly increasing.
We conclude as follows. Take any positive real number x. If f (x) > x, then f (f (x)) >
f (x) > x = f (f (x)), a contradiction. Similarly, it is not possible that f (x) < x. This shows
that f (x) = x for all positive real numbers x.
Marking Scheme:
6
• (2pt) Showing that f (a) + f (b) = f (c) + f (d) when a + b = c + d.
• (1pt) Showing that yf (x) = f (xf (y)).
• (1pt) Showing that f (f (y)) = y.
• (2pt) Showing that f (x + y) = f (x) + f (y).
• (1pt) Conclusion
Note: Since f (x) = x clearly satisfies the functional equation, no points will be awarded
or deducted for statements , or lack thereof, related to this fact.
7
Solutions of APMO 2017
Problem 1. We call a 5-tuple of integers arrangeable if its elements can be labeled a,
b, c, d, e in some order so that a − b + c − d + e = 29. Determine all 2017-tuples of integers
n1 , n2 , . . . , n2017 such that if we place them in a circle in clockwise order, then any 5-tuple of
numbers in consecutive positions on the circle is arrangeable.
Answer: n1 = · · · = n2017 = 29.
Solution. A valid 2017-tuple is n1 = · · · = n2017 = 29. We will show that it is the only
solution.
We first replace each number ni in the circle by mi := ni − 29. Since the condition a − b +
c − d + e = 29 can be rewritten as (a − 29) − (b − 29) + (c − 29) − (d − 29) + (e − 29) = 0, we
have that any five consecutive replaced integers in the circle can be labeled a, b, c, d, e in such
a way that a − b + c − d + e = 0. We claim that this is possible only when all of the mi ’s are 0
(and thus all of the original ni ’s are 29).
We work with indexes modulo 2017. Notice that for every i, mi and mi+5 have the same
parity. Indeed, this follows from mi ≡ mi+1 + mi+2 + mi+3 + mi+4 ≡ mi+5 (mod 2). Since
gcd(5, 2017) = 1, this implies that all mi ’s are of the same parity. Since m1 +m2 +m3 +m4 +m5
is even, all mi ’s must be even as well.
Suppose for the sake of contradiction that not all mi ’s are zero. Then our condition still
holds when we divide each number in the circle by 2. However, by performing repeated divisions,
we eventually reach a point where some mi is odd. This is a contradiction.
Problem 2. Let ABC be a triangle with AB < AC. Let D be the intersection point of
the internal bisector of angle BAC and the circumcircle of ABC. Let Z be the intersection
point of the perpendicular bisector of AC with the external bisector of angle ∠BAC. Prove
that the midpoint of the segment AB lies on the circumcircle of triangle ADZ.
Solution 1. Let N be the midpoint of AC. Let M be the intersection point of the
circumcircle of triangle ADZ and the segment AB. We will show that M is the midpoint of
AB. To do this, let D0 the reflection of D with respect to M . It suffices to show that ADBD0
is a parallelogram.
The internal and external bisectors of an angle in a triangle are perpendicular. This implies
that ZD is a diameter of the circumcircle of AZD and thus ∠ZM D = 90◦ . This means that
ZM is the perpendicular bisector of D0 D and thus ZD0 = ZD. By construction, Z is in the
perpendicular bisector of AC and thus ZA = ZC.
Now, let α be the angle ∠BAD = ∠DAC. In the cyclic quadrilateral AZDM we get
∠M ZD = ∠M AD = α, and thus ∠D0 ZD = 2α. By angle chasing we get
∠AZN = 90◦ − ∠ZAN = ∠DAC = α,
which implies that ∠AZC = 2α. Therefore,
∠D0 ZA = ∠D0 ZD − ∠AZD = 2α − ∠AZD = ∠AZC − ∠AZD = ∠DZC.
Combining ∠D0 ZA = ∠DZC, ZD0 = ZD and ZA = ZC, we obtain by the SAS criterion
that the triangles D0 ZA and DZC are congruent. In particular, D0 A = DC and ∠D0 AZ =
∠DCZ. From here DB = DC = D0 A.
Finally, let β = ∠ABC = ∠ADC. We get the first of the following equalities by the sum of
angles around point A and the second one by the sum of internal angles of quadrilateral AZCD
1
360◦ = ∠D0 AZ + ∠ZAD + ∠DAB + ∠BAD0 = ∠D0 AZ + 90◦ + α + ∠BAD0
360◦ = ∠DCZ + ∠ZAD + ∠CZA + ∠ADC = ∠DCZ + 90◦ + 2α + β.
By canceling equal terms we conclude that ∠BAD0 = α + β. Also, ∠ABD = α + β.
Therefore, the segments D0 A and DB are parallel and have the same length. We conclude that
ADBD0 is a parallelogram. As the diagonals of a parallelogram intersect at their midpoints,
we obtain that M is the midpoint of AB as desired.
Variant of solution. The solution above is indirect in the sense that it assumes that M is
in the circumcircle of AZD and then shows that M is the midpoint of AB. We point out that
the same ideas in the solution can be used to give a direct solution. Here we present a sketch
on how to proceed in this manner.
Now we know that M is the midpoint of the side AB. We construct the point D0 in the
same way. Now we have directly that ADBD0 is a parallelogram and thus D0 A = DB = DC.
By construction ZA = ZC. Also, the two sums of angles equal to 360◦ in the previous solution
let us conclude that ∠D0 AZ = ∠DCZ. Once again, we use (differently) the SAS criterion and
obtain that the triangles D0 AZ and DCZ are congruent. Thus, D0 Z = DZ.
We finish the problem by noting that ZM is a median of the isosceles triangle D0 ZD, so it
is also a perpendicular bisector. This shows that ∠DM Z = 90◦ = ∠DAZ, and therefore M
lies in the circumcircle of DAZ.
Solution 2. We proceed directly. As above, we name
α = ∠DAC = ∠AZN = ∠CZN = ∠DCB.
Let L be the midpoint of the segment BC. Since M and N are midpoints of AB and AC, we
have that M N = CL and that the segment M N is parallel to BC. Thus, ∠AN M = ∠ACB.
2
Therefore, ∠ZN M = ∠ACB + 90◦ . Also, ∠DCZ = ∠DCB + ∠ACB + ∠ZCA = ∠ACB + 90◦ .
We conclude that ∠ZN M = ∠DCZ.
Now, the triangles ZN C and CLD are similar since ∠DCL = 90◦ = ∠CN Z and ∠DCL =
α = ∠CZN . We use this fact to obtain the following chain of equalities
CD
CZ
CD
=
=
.
MN
CL
ZN
We combine the equality above with ∠ZN M = ∠DCZ to conclude that the triangles
ZD
ZN M and ZCD are similar. In particular, ∠M ZN = ∠DZC and ZM
ZN = ZC . Since we
also have ∠M ZD = ∠M ZN + ∠N ZD = ∠DZC + ∠N ZD = ∠N ZC, we conclude that the
triangles M ZD and N ZC are similar. This yields that ∠ZM D = 90◦ and therefore M is in
the circumcircle of triangle DAZ.
Solution 3. Let m be the perpendicular bisector of AD; thus m passes through the center
O of the circumcircle of triangle ABC. Since AD is the internal angle bisector of A and OM
and ON are perpendicular to AB and AC respectively, we obtain that OM and ON form equal
angles with AD. This implies that they are symmetric with respect to M .
Therefore, the line M O passes through the point Z 0 symmetrical to Z with respect to m.
Since ∠DAZ = 90◦ , then also ∠ADZ 0 = 90◦ . Moreover, since AZ = DZ 0 , we have that
∠AZZ 0 = ∠DZ 0 Z = 90◦ , so AZZ 0 D is a rectangle. Since ∠AM Z 0 = ∠AM O = 90◦ , we
conclude that M is in the circle with diameter AZ 0 , which is the circumcircle of the rectangle.
Thus M lies on the circumcircle of the triangle ADZ.
Problem 3. Let A(n) denote the number of sequences a1 ≥ a2 ≥ . . . ≥ ak of positive
integers for which a1 + · · · + ak = n and each ai + 1 is a power of two (i = 1, 2, . . . , k).
Let B(n) denote the number of sequences b1 ≥ b2 ≥ . . . ≥ bm of positive integers for which
b1 + · · · + bm = n and each inequality bj ≥ 2bj+1 holds (j = 1, 2, . . . , m − 1).
Prove that A(n) = B(n) for every positive integer n.
Solution. We say that a sequence a1 ≥ a2 ≥ . . . ≥ ak of positive integers has type A if
ai + 1 is a power of two for i = 1, 2, . . . , k. We say that a sequence b1 ≥ b2 ≥ . . . ≥ bm of
positive integer has type B if bj ≥ 2bj+1 for j = 1, 2, . . . , m − 1.
Recall that the binary representation of a positive integer expresses it as a sum of distinct
powers of two in a unique way. Furthermore, we have the following formula for every positive
integer N
2N − 1 = 2N −1 + 2N −2 + · · · + 21 + 20 .
Given a sequence a1 ≥ a2 ≥ . . . ≥ ak of type A, use the preceding formula to express each
term as a sum of powers of two. Write these powers of two in left-aligned rows, in decreasing
order of size. By construction, ai is the sum of the numbers in the ith row. For example, we
obtain the following array when we start with the type A sequence 15, 15, 7, 3, 3, 3, 1.
8
8
4
2
2
2
1
27
4
4
2
1
1
1
2
2
1
1
1
13
5
2
3
Define the sequence b1 , b2 , . . . , bn by setting bj to be the sum of the numbers in the jth
column of the array. For example, we obtain the sequence 27, 13, 5, 2 from the array above.
We now show that this new sequence has type B. This is clear from the fact that each column
in the array contains at least as many entries as the column to the right of it and that each
number larger than 1 in the array is twice the number to the right of it. Furthermore, it is
clear that a1 + a2 + · · · + ak = b1 + b2 + · · · + bm , since both are equal to the sum of all the
entries in the array.
We now show that we can do this operation backwards. Suppose that we are given a type
B sequence b1 ≥ b2 ≥ . . . ≥ bm . We construct an array inductively as follows:
• We fill bm left-aligned rows with the numbers 2m−1 , 2m−2 , . . . , 21 , 20 .
• Then we fill bm−1 − 2bm left aligned rows with the numbers 2m−2 , 2m−3 , . . . , 21 , 20 .
• Then we fill bm−2 − 2bm−1 left aligned rows with the numbers 2m−3 , 2m−4 , . . . , 21 , 20 , and
so on.
• In the last step we fill b1 − 2b2 left-aligned rows with the number 1.
For example, if we start with the type B sequence 27, 13, 5, 2, we obtain once again the array
above. We define the sequence a1 , a2 , . . . , ak by setting ai to be the sum of the numbers in the
ith row of the array. By construction, this sequence has type A. Furthermore, it is clear that
a1 + · · · + ak = b1 + · · · + bm , since once again both sums are equal to the sum of all the entries
in the array.
We have defined an operation that starts with a sequence of type A, produces an array
whose row sums are given by the sequence, and outputs a sequence of type B corresponding
to the column sums. We have also defined an operation that starts with a sequence of type
B, produces an array whose column sums are given by the sequence, and outputs a sequence
of type A corresponding to the row sums. The arrays produced in both cases comprise leftaligned rows of the form 2N −1 , 2N −2 , . . . , 21 , 20 , with non-increasing lengths. Let us refer to
arrays obeying these properties as marvelous.
To show that these two operations are inverses of each other, it then suffices to prove that
marvelous arrays are uniquely defined by either their row sums or their column sums. The
former is obviously true and the latter arises from the observation that each step in the above
inductive algorithm was forced in order to create a marvelous array with the prescribed column
sums.
Thus, we have produced a bijection between the sequences of type A with sum n and the
sequences of type B with sum n. So we can conclude that A(n) = B(n) for every positive
integer n.
Remark The solution above provides a bijection between type A and type B sequences via
an algorithm. There are alternative ways to provide such a bijection. For example, given the
numbers a1 ≥ . . . ≥ ak we may define the bi ’s as
X ai + 1 .
bj =
2j
i
Conversely, given the numbers b1 ≥ . . . ≥ bm , one may define the ai ’s by taking, as in the
solution, bm numbers equal to 2m − 1, bm−1 − 2bm numbers equal to 2m−1 − 1, . . ., and b1 − 2b2
numbers equal to 21 − 1. One now needs to verify that these maps are mutually inverse.
k
Problem 4.
Call a rational number r powerful if r can be expressed in the form pq
for some relatively prime positive integers p, q and some integer k > 1. Let a, b, c be positive
4
rational numbers such that abc = 1. Suppose there exist positive integers x, y, z such that
ax + by + cz is an integer. Prove that a, b, c are all powerful.
Solution. Let a = ab 1 , b = ab 2 , where gcd(a1 , b1 ) = gcd(a2 , b2 ) = 1. Then c = ab11 ba22 . The
1
2
condition that ax + by + cz is an integer becomes
x+z y+z
x
a1x+z az2 by2 + az1 ay+z
2 b1 + b1 b 2
∈ Z,
az1 az2 bx1 by2
which can be restated as
z y
z y+z x
x+z y+z
az1 az2 bx1 by2 | ax+z
1 a2 b 2 + a1 a2 b 1 + b 1 b 2 .
(1)
In particular, az1 divides the right-hand side. Since it divides the first and second terms in the
y+z
z
sum, we conclude that az1 | b1x+z by+z
2 . Since gcd(a1 , b1 ) = 1, we have a1 | b2 .
Let p be a prime that divides a1 . Let m, n ≥ 1 be integers such that pn ka1 (i.e. pn |a1
but pn+1 - a1 ) and pm kb2 . The fact that az1 | b2y+z implies nz ≤ m(y + z). Since gcd(a1 , b1 ) =
gcd(a2 , b2 ) = 1, we have p does not divide b1 and does not divide a2 . Thus
x
m(y+z) x+z y+z
pnz kaz1 ay+z
kb1 b2 .
2 b1 and p
(2)
On the other hand, (1) implies that
y+z
pnz+my | az1 a2y+z bx1 + bx+z
1 b2 .
(3)
y+z
If nz < m(y + z), then (2) gives pnz kaz1 a2y+z bx1 + bx+z
1 b2 , which contradicts (3). Thus
y+z
nz = m(y + z) so n is divisible by k := gcd(z,
> 1. Thus each exponent in the prime
y + z)
decomposition of a1 must be divisible by k. Hence a1 is a perfect k-power which means a is
powerful. Similarly, b and c are also powerful.
Problem 5. Let n be a positive integer. A pair of n-tuples (a1 , . . . , an ) and (b1 , . . . , bn )
with integer entries is called an exquisite pair if
|a1 b1 + · · · + an bn | ≤ 1.
Determine the maximum number of distinct n-tuples with integer entries such that any two
of them form an exquisite pair.
Answer: The maximum is n2 + n + 1.
Solution. First, we construct an example with n2 + n + 1 n-tuples, each two of them
forming an exquisite pair. In the following list, ∗ represents any number of zeros as long as the
total number of entries is n.
• (∗)
• (∗, 1, ∗)
• (∗, −1, ∗)
• (∗, 1, ∗, 1, ∗)
• (∗, 1, ∗, −1, ∗)
5
For example, for n = 2 we have the tuples (0, 0), (0,
1), (1,
0), (0, −1), (−1, 0), (1, 1), (1, −1).
The total number of such tuples is 1 + n + n + n2 + n2 = n2 + n + 1. For any two of them,
at most two of the products ai bi are non-zero. The only case in which two of them are non-zero
is when we take a sequence (∗, 1, ∗, 1, ∗) and a sequence (∗, 1, ∗, −1, ∗) with zero entries in the
same places. But in this case one ai bi is 1 and the other −1. This shows that any two of these
sequences form an exquisite pair.
Next, we claim that among any n2 + n + 2 tuples, some two of them do not form an exquisite
pair. We begin with lemma.
Lemma. Given 2n + 1 distinct non-zero n-tuples of real numbers, some two of them
(a1 , . . . , an ) and (b1 , . . . , bn ) satisfy a1 b1 + · · · + an bn > 0.
Proof of Lemma. We proceed by induction. The statement is easy for n = 1 since for every
three non-zero numbers there are two of them with the same sign. Assume that the statement
is true for n − 1 and consider 2n + 1 tuples with n entries. Since we are working with tuples of
real numbers, we claim that we may assume that one of the tuples is a = (0, 0, . . . , 0, −1). Let
us postpone the proof of this claim for the moment.
If one of the remaining tuples b has a negative last entry, then a and b satisfy the desired
condition. So we may assume all the remaining tuples has a non-negative last entry. Now, from
each tuple remove the last number. If two n-tuples b and c yield the same (n − 1)-tuple, then
b1 c1 + · · · + bn−1 cn−1 + bn cn = b21 + · · · + b2n−1 + bn cn > 0,
and we are done. The remaining case is that all the n-tuples yield distinct (n − 1)-tuples. Then
at most one of them is the zero (n − 1)-tuple, and thus we can use the inductive hypothesis on
2n − 1 of them. So we find b and c for which
(b1 c1 + · · · + bn−1 cn−1 ) + bn cn > 0 + bn cn > 0.
The only thing that we are left to prove is that in the inductive step we may assume that
one of the tuples is a = (0, 0, . . . , 0, −1). Fix one of the tuples x = (x1 , . . . , xn ). Set a real
x1 . Change each tuple a = (a , a , . . . , a ) (including x), to the
number ϕ for which tan ϕ = x
1 2
n
2
tuple
(a1 cos ϕ − a2 sin ϕ, a1 sin ϕ + a2 cos ϕ, a3 , a4 , . . . , an ).
A straightforward calculation shows that the first coordinate of the tuple x becomes 0, and
that all the expressions of the form a1 b1 + · · · + an bn are preserved. We may iterate this process
until all the entries of x except for the last one are equal to 0. We finish by multiplying all the
entries in all the tuples by a suitable constant that makes the last entry of x equal to −1. This
preserves the sign of all the expressions of the form a1 b1 + · · · + an bn .
We proceed to the proof of our claim. Let A be a set of non-zero tuples among which any
two form an exquisite pair. It suffices to prove that |A| ≤ n2 + n. We can write A as a disjoint
union of subsets A1 ∪ A2 ∪ . . . ∪ An , where Ai is the set of tuples in A whose last non-zero
entry appears in the ith position. We will show that |Ai | ≤ 2i, which will finish our proof since
2 + 4 + · · · + 2n = n2 + n.
Proceeding by contradiction, suppose that |Ai | ≥ 2i + 1. If Ai has three or more tuples
whose only non-zero entry is in the ith position, then for two of them this entry has the same
sign.
P Since the tuples are different and their entries are integers, this yields two tuples for which
| ai bi | ≥ 2, a contradiction. So there are at most two such tuples. We remove them from Ai .
Now, for each of the remaining tuples a, if it has a positive ith coordinate, we keep a as
it is. If it has a negative ith coordinate, we replace it with the opposite tuple −a with entries
with opposite signs. This does not changes the exquisite pairs condition.
6
After making the necessary changes, we have two cases. The first case is that there are two
tuples a and b that have the same first i − 1 coordinates and thus
a1 b1 + · · · + ai−1 bi−1 = a21 + · · · + a2i−1 > 0,
and thus is at least 1 (the entries are integers). The second case is that no two tuples have the
same first i − 1 coordinates, but then by the Lemma we find two tuples a and b for which
a1 b1 + · · · + ai−1 bi−1 ≥ 1.
In any case, we obtain
a1 b1 + · · · + ai−1 bi−1 + ai bi ≥ 2.
This yields a final contradiction to the exquisite pair hypothesis.
7
Solutions of APMO 2018
Problem 1.
Let H be the orthocenter of the triangle ABC. Let M and N be the
midpoints of the sides AB and AC, respectively. Assume that H lies inside the quadrilateral
BM N C and that the circumcircles of triangles BM H and CN H are tangent to each other.
The line through H parallel to BC intersects the circumcircles of the triangles BM H and CN H
in the points K and L, respectively. Let F be the intersection point of M K and N L and let J
be the incenter of triangle M HN . Prove that F J = F A.
Solution.
Lemma 1. In a triangle ABC, let D be the intersection of the interior angle bisector at A
with the circumcircle of ABC, and let I be the incenter of 4ABC. Then
DI = DB = DC.
Proof.
∠DBI =
b
∠BAC B
+ = ∠DIB
2
2
⇒
DI = DB.
Analogously DI = DC.
We start solving the problem. First we state some position considerations. Since there is
an arc of the circumcircle of BHM outside the triangle ABC, it must happen that K and N
lie on opposite sides of AM . Similarly, L and M lie on opposite sides of AN . Also, K and L
lie on the same side of M N , and opposite to A. Therefore, F lies inside the triangle AM N .
Now, since H is the orthocenter of 4ABC and the circumcircles of BM H and CN H are
tangent we have
∠ABH = 90◦ − ∠BAC = ∠ACH
⇒
∠M HN = ∠M BH + ∠N CH = 180◦ − 2∠BAC. (1)
So ∠M BH = ∠M KH = ∠N CH = ∠N LH = 90◦ − ∠BAC and, since M N kKL, we have
∠F M N = ∠F N M = 90◦ − ∠BAC
⇒
∠M F N = 2∠BAC.
(2)
The relations (1) and (2) yield that the quadrilateral M F N H is cyclic, with the vertices
in this order around the circumference. Since F M = F N , ∠M F N = 2∠BAC and F is the
correct side of M N we have that the point F is the circumcenter of triangle AM N , and thus
F A = F M = F N.
1
Since the quadrilateral M F N H is cyclic, F M = F N and H lies on the correct side of
M N , we have that H, J and F are collinear. According to Lemma 1, F J = F M = F N . So
F J = F A.
Solution 2: According to Solution 1, we have ∠M HN = 180◦ − 2∠BAC and since the
point J is the incenter of 4M HN , we have ∠M JN = 90◦ + 12 ∠M HN = 180◦ − ∠BAC. So
the quadrilateral AM JN is cyclic.
According to Solution 1, the point F is the circumcenter of 4AM N . So F J = F A.
Problem 2. Let f (x) and g(x) be given by
f (x) =
and
g(x) =
1
1
1
1
+
+
+ ··· +
x x−2 x−4
x − 2018
1
1
1
1
+
+
+ ··· +
.
x−1 x−3 x−5
x − 2017
Prove that
|f (x) − g(x)| > 2
for any non-integer real number x satisfying 0 < x < 2018.
Solution 1 There are two cases: 2n − 1 < x < 2n and 2n < x < 2n + 1. Note that
f (2018 − x) = −f (x) and g(2018 − x) = −g(x), that is, a half turn about the point (1009, 0)
preserves the graphs of f and g. So it suffices to consider only the case 2n − 1 < x < 2n.
Let d(x) = g(x) − f (x). We will show that d(x) > 2 whenever 2n − 1 < x < 2n and
n ∈ {1, 2, . . . , 1009}.
For any non-integer x with 0 < x < 2018, we have
1
1
1
1
d(x + 2) − d(x) =
−
+
−
> 0 + 0 = 0.
x+1 x+2
x − 2018 x − 2017
1
Hence it suffices to prove d(x) > 2 for 1 < x < 2. Since x < 2, it follows that x − 2i
−1 >
1 for i = 2, 3, . . . , 1008. We also have
1
x − 2018 < 0. Hence it suffices to prove the following
x − 2i
2
for 1 < x < 2.
1
1
1
1
+
− −
>2
x − 1 x − 3 x x − 2
1
1
1
1
⇔
+
+
−
>2
x−1 2−x
x−3 x
1
3
⇔
+
> 2.
(x − 1)(2 − x) x(x − 3)
By the GM − HM inequality (alternatively, by considering the maximum of the quadratic
(x − 1)(2 − x)) we have
1
1
·
>
x−1 2−x
2
(x − 1) + (2 − x)
2
= 4.
To find a lower bound for x(x 3− 3) , note that x(x − 3) < 0 for 1 < x < 2. So we seek an upper
bound for x(x − 3). From the shape of the quadratic, this occurs at x = 1 or x = 2, both of
which yield x(x 3− 3) > − 32 .
It follows that d(x) > 4 − 32 > 2, as desired.
Solution 2
As in Solution 1, we may assume 2n − 1 < x < 2n for some 1 ≤ n ≤ 1009. Let d(x) =
f (x) − g(x), and note that
1009
1
1 X
d(x) = +
x m=1 (x − 2m)(x − 2m + 1)
We split the sum into three parts: the terms before m = n, after m = n, and the term m = n.
The first two are
0≤
n−1
X
1
(x − 2m)(x − 2m + 1)
m=1
n−1
X
n−1
1008
X
X 1
1
1
1
≤
=
≤
− ,
(2n − 1 − 2m)(2n − 2m)
(2i)(2i − 1)
2i − 1 2i
m=1
i=1
i=1
0≤
1009
X
1
(2m − x)(2m − 1 − x)
m=n+1
1009
X
1009−n
1008
X
X
1
1
1
1
≤
=
≤
−
.
(2m − 2n + 1)(2m − 2n)
(2i + 1)(2i)
2i 2i + 1
m=n+1
i=1
i=1
When we add the two sums the terms telescope and we are left with
0≤
1
1
≤1−
< 1,
(x − 2m)(x − 2m + 1)
2017
1≤m≤1009,m6=n
X
For the term m = n, we write
0 < −(x − 2n)(x − 2n + 1) = 0.25 − (x − 2n + 0.5)2 ≤ 0.25,
3
whence
−4 ≥
1
.
(x − 2n)(x − 2n + 1)
Finally, x1 < 1 since x > 2n − 1 ≥ 1. Combining these we get
1009
d(x) =
1 X
1
+
< 1 + 1 − 4 < −2.
x m=1 (x − 2m)(x − 2m + 1)
Solution 3
First notice that
f (x) − g(x) =
1
1
1
1
1
−
+
− ··· −
+
.
x x−1 x−2
x − 2017 x − 2018
As in Solution 1, we may deal only with the case 2n < x < 2n + 1. Then x − 2k + 1 and
x − 2k never differ in sign for any integer k. Then
−
1
1
1
+
=
> 0 for k = 1, 2, . . . , n − 1, n + 2, . . . , 1009.
x − 2k + 1 x − 2k
(x − 2k + 1)(x − 2k)
2
1
1
1
2
= 4,
−
=
≥
x − 2n x − 2n − 1
(x − 2n)(2n + 1 − x)
x − 2n + 2n + 1 − x
Therefore, summing all inequalities and collecting the remaining terms we find f (x)−g(x) >
1 > 4 − 1 = 3 for 0 < x < 1 and, for n > 0,
4+ x−
2
1
1
1
−
+4+
x x − 2n + 1
x − 2n − 2
1
1
1
= −
+4−
x x − 2n + 1
2n + 2 − x
1
1
1
> −
+4−
x 2n − 2n + 1
2n + 2 − 2n − 1
1
= 2 + > 2.
x
f (x) − g(x) >
Problem 3. A collection of n squares on the plane is called tri-connected if the following
criteria are satisfied:
(i) All the squares are congruent.
(ii) If two squares have a point P in common, then P is a vertex of each of the squares.
(iii) Each square touches exactly three other squares.
How many positive integers n are there with 2018 ≤ n ≤ 3018, such that there exists a collection
of n squares that is tri-connected?
Answer: 501
Solution. We will prove that there is no tri-connected collection if n is odd, and that
tri-connected collections exist for all even n ≥ 38. Since there are 501 even numbers in the
range from 2018 to 3018, this yields 501 as the answer.
4
For any two different squares A and B, let us write A ∼ B to mean that square A touches
square B. Since each square touches exactly three other squares, and there are n squares in
total, the total number of instances of A ∼ B is 3n. But A ∼ B if and only if B ∼ A. Hence
the total number of instances of A ∼ B is even. Thus 3n and hence also n is even.
We now construct tri-connected collections for each even n in the range. We show two
Construction 1 The idea is to use the following two configurations. Observe that in each
configuration every square is related to three squares except for the leftmost and rightmost
squares which are related to two squares. Note that the configuration on the left is of variable
length. Also observe that multiple copies of the configuration on the right can be chained
together to end around corners.
Putting the above two types of configurations together as in the following figure yields a
tri-connected collection for every even n ≥ 38.
Construction 2 Consider a regular 4n−gon A1 A2 · · · A4n , and make 4n squares on the
outside of the 4n−gon with one side being on the 4n−gon. Reflect squares sharing sides
A4m+2 A4m+3 , A4m+3 A4m+4 across line A4m+2 A4m+4 , for 0 ≤ m ≤ n − 1. This will produce a
tri-connected set of 6n squares, as long as the squares inside the 4n−gon do not intersect.
When n ≥ 4, this will be true. The picture for n = 24 is as follows:
To treat the other cases, consider the following gadget
5
Y
X
Two squares touch 3 other squares, and the squares containing X, Y touch 2 other squares.
Take the 4n−gon from above, and break it into two along the line A1 A2n , moving the two
parts away from that line. Do so until the gaps can be exactly filled by inserting two copies of
the above figure, so that the vertices X, Y touch the two vertices which used to be A1 in one
instance, and the two vertices which used to be A2n in the other.
This gives us a valid configuration for 6n + 8 squares, n ≥ 4. Finally, if we had instead spread
the two parts out more and inserted two copies of the above figure into each gap, we would get
6n + 16 for n ≥ 4, which finishes the proof for all even numbers at least 36.
Problem 4.
Let ABC be an equilateral triangle. From the vertex A we draw a ray
towards the interior of the triangle such that the ray reaches one of the sides of the triangle.
When the ray reaches a side, it then bounces off following the law of reflection, that is, if it
arrives with a directed angle α, it leaves with a directed angle 180◦ − α. After n bounces, the
ray returns to A without ever landing on any of the other two vertices. Find all possible values
of n.
Answer: All n ≡ 1, 5 mod 6 with the exception of 5 and 17
Solution. Consider an equilateral triangle AA1 A2 of side length m and triangulate it with
unitary triangles. See the figure. To each of the vertices that remain after the triangulation we
can assign a pair of coordinates (a, b) where a, b are non-negative integers, a is the number of
edges we travel in the AA1 direction and b is the number of edges we travel in the AA2 direction
to arrive to the vertex, (we have A = (0, 0), A1 = (m, 0) and A2 = (0, m)). The unitary triangle
with vertex A will be our triangle ABC, (B = (1, 0), C = (0, 1)). We can obtain every unitary
triangle by starting with ABC and performing reflections with respect to a side (the vertex
(1, 1) is the reflection of A with respect to BC, the vertex (0, 2) is the reflection of B = (1, 0)
with respect to the side formed by C = (1, 0) and (1, 1), and so on).
When we reflect a vertex (a, b) with respect to a side of one of the triangles, the congruence
of a−b is preserved modulo 3. Furthermore, an induction argument shows that any two vertices
(a, b) and (a0 , b0 ) with a − b ≡ a0 − b0 mod 3 can be obtained from each other by a series of such
reflections. Therefore, the set of vertices V that result from the reflections of A will be those
of the form (a, b) satisfying a ≡ b mod 3. See the green vertices in the figure.
Now, let U be the set of vertices u that satisfy that the line segment between u and A
does not pass through any other vertex. A pair (a, b) is in U if and only if gcd(a, b) = 1, since
otherwise for d = gcd(a, b) we have that the vertex (a/d, b/d) also lies on the line segment
between u and A.
Observe that the rays that come out from A and eventually return to A are those that come
out towards a vertex in V ∩ U (they would be in V to be able to come back to A and in U
so that they do not reach a vertex beforehand). In the diagram, a ray toward one such vertex
(a, b) will intersect exactly (a − 1) + (b − 1) + (a + b − 1) = 2(a + b) − 3 lines: a − 1 of them
parallel to AB, b − 1 parallel to AC and a + b − 1 parallel to BC. Therefore, in the triangle
ABC the ray will bounce 2(a + b) − 3 times before returning to A. So we want to find all
6
n = 2(a + b) − 3 where a ≡ b mod 3 and gcd(a, b) = 1.
If a + b is a multiple of 3 then we cannot satisfy both conditions simultaneously, therefore n
is not a multiple of 3. We also know that n is odd. Therefore n ≡ 1, 5, 7, 11 mod 12. Note that
the pair (1, 3k + 1) satisfies the conditions and we can create n = 2(3k + 2) − 3 = 6k + 1 for
all k ≥ 0 (this settles the question for n ≡ 1, 7 mod 12). For n ≡ 5 mod 12 consider the pair
(3k − 1, 3k + 5) when k is even or (3k − 4, 3k + 8) when k is odd. This gives us all the integers
of the form 12k + 5 for k ≥ 2. For 11 mod 12, take the pairs (3k − 1, 3k + 2) (with k ≥ 1),
which yield all positive integers of the form 12k − 1.
Finally, to discard 5 and 17 note that the only pairs (a, b) that are solutions to 2(a+b)−3 = 5
or 2(a + b) − 3 = 17 with the same residue mod 3 in this range are the non-relatively prime
pairs (2, 2), (2, 8) and (5, 5).
Problem 5. Find all polynomials P (x) with integer coefficients such that for all real
numbers s and t, if P (s) and P (t) are both integers, then P (st) is also an integer.
Answer: P (x) = xn + k, −xn + k for n a non-negative integer and k an integer.
Solution 1: P (x) = xn + k, −xn + k for n a non-negative integer and k an integer.
Notice that if P (x) is a solution, then so is P (x) + k and −P (x) + k for any integer k, so
we may assume that P
the leading coefficient of P (x) is positive and that P (0) = 0, i.e., we can
assume that P (x) = ni=1 ai xi with an >P
0. We are going to prove that P (x) = xn in this case.
Let p be a large prime such that p > ni=1 |ai |. Because P has a positive leading coefficient
and p is large enough, we can find t ∈ R such that P (t) = p. Denote the greatest common
divisor of the polynomial P (x) − p and P (2x) − P (2t) as f (x), and t is a root of it, so f is a
non-constant polynomial. Notice that P (2t) is an integer by using the hypothesis for s = 2 and
t. Since P (x) − p and P (2x) − P (2t) are polynomials with integer coefficients, f can be chosen
as a polynomial with rational coefficients.
In the following, we will prove that f is the same as P (x) − p up to a constant multiplier.
Say P (x) − p = f (x)g(x), where f and g are non-constant polynomials. By Gauss’s lemma, we
can get f1 , g1 with P (x) − p = f1 (x)g1 (x) where f1 is a scalar multiple of f and g1 is a scalar
multiple of g and one of f1 , g1 has constant term ±1 (this is because −p = P (0) − p = f (0)g(0)
with p prime). So P (x) − p has at least one root r with absolute value not greater than 1 (using
7
Vieta, the product of the roots of the polynomial with constant term ±1 is ±1), but
|P (r) − p| =
n
X
i
ai r − p > p −
i=1
n
X
|ai | > 0,
i=1
hence we get a contradiction!
Therefore f is a constant multiple of P (x) − p, so P (2x) − P (2t) is a constant multiple
of P (x) − p because they both have the same degree. By comparing leading coefficients we
get that P (2x) − P (2t) = 2n (P (x) − p). Comparing the rest of the coefficients we get that
P (x) = an xn . If we let a = b = (1/an )1/n , then P (a) = P (b) = 1, so P (ab) must also be an
integer. But P (ab) = a1n . Therefore an = 1 and the proof is complete.
Solution 2: Assume P (x) =
Pn
i=0
ai xi . Consider the following system of equations
a0 = P (0)
an t + an−1 t
+ · · · + a0 = P (t)
n
n
n−1
n−1
2 an t + 2 an−1 t
+ · · · + a0 = P (2t)
..
.
n
n
n−1
n−1
n an t + n an−1 t
+ · · · + a0 = P (nt).
n
n−1
viewing ak tk as variables. Note that if P (t) is an integer, then by the hypothesis all the terms
on the right hand side of the equations are integers as well. By using Cramer’s rule, we can get
that ak tk = D/M , where D is an integer and M is the following determinant
1
1
1
..
.
0
1
2
..
.
0
1
4
..
.
···
···
···
1 n n2 · · ·
0
1
2n 6= 0.
..
.
nn
Thus, if we let r be the smallest positive index such that ar 6= 0, we can express each t ∈ R
with P (t) ∈ Z in the form ( m0 )1/r for some integer m, and where M 0 = M × ar is a constant.
M
We can choose L large enough such that P |R≥L is injective, and for any larger N , the growth
order of the number of values in the form ( m0 )1/r is N r , while the growth order of the number
M
of integers in [P (L), P (N )] is N n , so r = n. Therefore P (x) is of the form an xn + k. The
problem can be finished as in Solution 1.
8
Solutions of APMO 2019
Problem 1. Let Z+ be the set of positive integers. Determine all functions f : Z+ → Z+
such that a2 + f (a)f (b) is divisible by f (a) + b for all positive integers a and b.
Answer: The answer is f (n) = n for all positive integers n.
Clearly, f (n) = n for all n ∈ Z+ satisfies the original relation. We show some possible
approaches to prove that this is the only possible function.
Solution. First we perform the following substitutions on the original relation:
1. With a = b = 1, we find that f (1) + 1 | f (1)2 + 1, which implies f (1) = 1.
2. With a = 1, we find that b + 1 | f (b) + 1. In particular, b ≤ f (b) for all b ∈ Z+ .
3. With b = 1, we find that f (a) + 1 | a2 + f (a), and thus f (a) + 1 | a2 − 1. In particular,
f (a) ≤ a2 − 2 for all a ≥ 2.
Now, let p be any odd prime. Substituting a = p and b = f (p) in the original relation, we
find that 2f (p)|p2 + f (p)f (f (p)). Therefore, f (p)|p2 . Hence the possible values of f (p) are 1, p
and p2 . By (2) above, f (p) ≥ p and by (3) above f (p) ≤ p2 − 2. So f (p) = p for all primes p.
Substituting a = p into the original relation, we find that b + p | p2 + pf (b). However, since
(b + p)(f (b) + p − b) = p2 − b2 + bf (b) + pf (b), we have b + p | bf (b) − b2 . Thus, for any fixed b
this holds for arbitrarily large primes p and therefore we must have bf (b) − b2 = 0, or f (b) = b,
as desired.
Solution 2: As above, we have relations (1)-(3). In (2) and (3), for b = 2 we have 3|f (2)+1
and f (2) + 1|3. These imply f (2) = 2.
Now, using a = 2 we get 2 + b|4 + 2f (b). Let f (b) = x. We have
1+x≡0
4 + 2x ≡ 0
(mod b + 1)
(mod b + 2).
From the first equation x ≡ b (mod b + 1) so x = b + (b + 1)t for some integer t ≥ 0. Then
0 ≡ 4 + 2x ≡ 4 + 2(b + (b + 1)t) ≡ 4 + 2(−2 − t) ≡ −2t (mod b + 2).
Also t ≤ b − 2 because 1 + x|b2 − 1 by (3).
If b + 2 is odd, then t ≡ 0 (mod b + 2). Then t = 0, which implies f (b) = b.
If b + 2 is even, then t ≡ 0 (mod (b + 2)/2). Then t = 0 or t = (b + 2)/2. But if t 6= 0,
then by definition (b + 4)/2 = (1 + t) = (x + 1)/(b + 1) and since x + 1|b2 − 1, then (b + 4)/2
divides b − 1. Therefore b + 4|10 and the only possibility is b = 6. So for even b, b 6= 6 we have
f (b) = b.
Finally, by (2) and (3), for b = 6 we have 7|f (6) + 1 and f (6) + 1|35. This means f (6) = 6
or f (6) = 34. The later is discarded as, for a = 5, b = 6, we have by the original equation that
11|5(5 + f (6)). Therefore f (n) = n for every positive integer n.
Solution 3: We proceed by induction. As in Solution 1, we have f (1) = 1. Suppose that
f (n − 1) = n − 1 for some integer n ≥ 2.
With the substitution a = n and b = n − 1 in the original relation we obtain that f (n) +
n − 1|n2 + f (n)(n − 1). Since f (n) + n − 1|(n − 1)(f (n) + n − 1), then f (n) + n − 1|2n − 1.
With the substitution a = n − 1 and b = n in the original relation we obtain that 2n −
1|(n − 1)2 + (n − 1)f (n) = (n − 1)(n − 1 + f (n)). Since (2n − 1, n − 1) = 1, we deduce that
2n − 1|f (n) + n − 1.
1
Therefore, f (n) + n − 1 = 2n − 1, which implies the desired f (n) = n.
Problem 2. Let m be a fixed positive integer. The infinite sequence {an }n≥1 is defined in
the following way: a1 is a positive integer, and for every integer n ≥ 1 we have
(
a2n + 2m if an < 2m
an+1 =
an /2
if an ≥ 2m .
For each m, determine all possible values of a1 such that every term in the sequence is an
integer.
Answer: The only value of m for which valid values of a1 exist is m = 2. In that case, the
only solutions are a1 = 2` for ` ≥ 1.
Solution. Suppose that for integers m and a1 all the terms of the sequence are integers.
For each i ≥ 1, write the ith term of the sequence as ai = bi 2ci where bi is the largest odd
divisor of ai (the “odd part” of ai ) and ci is a nonnegative integer.
Lemma 1. The sequence b1 , b2 , . . . is bounded above by 2m .
Proof. Suppose this is not the case and take an index i for which bi > 2m and for which ci is
minimal. Since ai ≥ bi > 2m , we are in the second case of the recursion. Therefore, ai+1 = ai /2
and thus bi+1 = bi > 2m and ci+1 = ci − 1 < ci . This contradicts the minimality of ci .
Lemma 2. The sequence b1 , b2 , . . . is nondecreasing.
Proof. If ai ≥ 2m , then ai+1 = ai /2 and thus bi+1 = bi . On the other hand, if ai < 2m , then
ai+1 = a2i + 2m = b2i 22ci + 2m ,
and we have the following cases:
• If 2ci > m, then ai+1 = 2m (b2i 22ci −m + 1), so bi+1 = b2i 22ci −m + 1 > bi .
• If 2ci < m, then ai+1 = 22ci (b2i + 2m−2ci ), so bi+1 = b2i + 2m−2ci > bi .
• If 2ci = m, then ai+1 = 2m+1 ·
b2i +1
,
2
so bi+1 = (b2i + 1)/2 ≥ bi since b2i + 1 ≡ 2 (mod 4).
By combining these two lemmas we obtain that the sequence b1 , b2 , . . . is eventually constant.
Fix an index j such that bk = bj for all k ≥ j. Since an descends to an /2 whenever an ≥ 2m ,
there are infinitely many terms which are smaller than 2m . Thus, we can choose an i > j such
that ai < 2m . From the proof of Lemma 2, ai < 2m and bi+1 = bi can happen simultaneously
only when 2ci = m and bi+1 = bi = 1. By Lemma 2, the sequence b1 , b2 , . . . is constantly 1 and
thus a1 , a2 , . . . are all powers of two. Tracing the sequence starting from ai = 2ci = 2m/2 < 2m ,
2m/2 → 2m+1 → 2m → 2m−1 → 22m−2 + 2m .
Note that this last term is a power of two if and only if 2m − 2 = m. This implies that m
must be equal to 2. When m = 2 and a1 = 2` for ` ≥ 1 the sequence eventually cycles through
2, 8, 4, 2, . . .. When m = 2 and a1 = 1 the sequence fails as the first terms are 1, 5, 5/2.
Solution 2: Let m be a positive integer and suppose that {an } consists only of positive
integers. Call a number small if it is smaller than 2m and large otherwise. By the recursion,
2
after a small number we have a large one and after a large one we successively divide by 2 until
we get a small one.
First, we note that {an } is bounded. Indeed, a1 turns into a small number after a finite
number of steps. After this point, each small number is smaller than 2m , so each large number
is smaller than 22m + 2m . Now, since {an } is bounded and consists only of positive integers, it
is eventually periodic. We focus only on the cycle.
Any small number an in the cycle can be writen as a/2 for a large, so an ≥ 2m−1 , then
an+1 ≥ 22m−2 + 2m = 2m−2 (4 + 2m ), so we have to divide an+1 at least m − 1 times by 2 until
we get a small number. This means that an+m = (a2n + 2m )/2m−1 , so 2m−1 |a2n , and therefore
2d(m−1)/2e | an for any small number an in the cycle. On the other hand, an ≤ 2m − 1, so
an+1 ≤ 22m − 2m+1 + 1 + 2m ≤ 2m (2m − 1), so we have to divide an+1 at most m times by
two until we get a small number. This means that after an , the next small number is either
N = am+n = (a2n /2m−1 ) + 2 or am+n+1 = N/2. In any case, 2d(m−1)/2e divides N .
If m is odd, then x2 ≡ −2 (mod 2d(m−1)/2e ) has a solution x = an /2(m−1)/2 . If (m − 1)/2 ≥
2 ⇐⇒ m ≥ 5 then x2 ≡ −2 (mod 4), which has no solution. So if m is odd, then m ≤ 3.
If m is even, then 2m−1 | a2n =⇒ 2d(m−1)/2e | an ⇐⇒ 2m/2 | an . Then if an = 2m/2 x,
2
2x ≡ −2 (mod 2m/2 ) ⇐⇒ x2 ≡ −1 (mod 2(m/2)−1 ), which is not possible for m ≥ 6. So if m
is even, then m ≤ 4.
The cases m = 1, 2, 3, 4 are handed manually, checking the possible small numbers in the
cycle, which have to be in the interval [2m−1 , 2m ) and be divisible by 2d(m−1)/2e :
• For m = 1, the only small number is 1, which leads to 5, then 5/2.
• For m = 2, the only eligible small number is 2, which gives the cycle (2, 8, 4). The only
way to get to 2 is by dividing 4 by 2, so the starting numbers greater than 2 are all
numbers that lead to 4, which are the powers of 2.
• For m = 3, the eligible small numbers are 4 and 6; we then obtain 4, 24, 12, 6, 44, 22, 11, 11/2.
• For m = 4, the eligible small numbers are 8 and 12; we then obtain 8, 80, 40, 20, 10, . . . or
12, 160, 80, 40, 20, 10, . . ., but in either case 10 is not an elegible small number.
Problem 3. Let ABC
be a scalene triangle with circumcircle Γ. Let M be the midpoint of BC. A variable point P
is selected in the line segment AM . The circumcircles of triangles BP M and CP M intersect
Γ again at points D and E, respectively. The lines DP and EP intersect (a second time) the
circumcircles to triangles CP M and BP M at X and Y , respectively. Prove that as P varies,
the circumcircle of 4AXY passes through a fixed point T distinct from A.
Solution. Let N be the radical center of the circumcircles of triangles ABC, BM P and
CM P . The pairwise radical axes of these circles are BD, CE and P M , and hence they concur
at N . Now, note that in directed angles:
∠M CE = ∠M P E = ∠M P Y = ∠M BY.
3
It follows that BY is parallel to CE, and analogously that CX is parallel to BD. Then, if L
is the intersection of BY and CX, it follows that BN CL is a parallelogram. Since BM = M C
we deduce that L is the reflection of N with respect to M , and therefore L ∈ AM . Using power
of a point from L to the circumcircles of triangles BP M and CP M , we have
LY · LB = LP · LM = LX · LC.
Hence, BY XC is cyclic. Using the cyclic quadrilateral we find in directed angles:
∠LXY = ∠LBC = ∠BCN = ∠N DE.
Since CX k BN , it follows that XY k DE.
Let Q and R be two points in Γ such that CQ, BR, and AM are all parallel. Then in
directed angles:
∠QDB = ∠QCB = ∠AM B = ∠P M B = ∠P DB.
Then D, P, Q are collinear. Analogously E, P, R are collinear. From here we get ∠P RQ =
∠P DE = ∠P XY , since XY and DE are parallel. Therefore QRY X is cyclic. Let S be the
radical center of the circumcircle of triangle ABC and the circles BCY X and QRY X. This
point lies in the lines BC, QR and XY because these are the radical axes of the circles. Let
T be the second intersection of AS with Γ. By power of a point from S to the circumcircle of
ABC and the circle BCXY we have
SX · SY = SB · SC = ST · SA.
4
Therefore T is in the circumcircle of triangle AXY . Since Q and R are fixed regardless of
the choice of P , then S is also fixed, since it is the intersection of QR and BC. This implies
T is also fixed, and therefore, the circumcircle of triangle AXY goes through T 6= A for any
choice of P .
Now we show an alternative way to prove that BCXY and QRXT are cyclic.
Solution 2. Let the lines DP and EP meet the circumcircle of ABC again at Q and R,
respectively. Then ∠DQC∠DBC = ∠DP M , so QC k P M . Similarly, RB k P M .
Now, ∠QCB = ∠P M B = ∠P XC = ∠(QX, CX), which is half of the arc QC in the
circumcircle ωC of QXC. So ωC is tangent to BS; analogously, ωB , the circumcicle of RY B,
is also tangent to BC. Since BR k CQ, the inscribed trapezoid BRQC is isosceles, and by
symmetry QR is also tangent to both circles, and the common perpendicular bisector of BR
and CQ passes through the centers of ωB and ωC . Since M B = M C and P M k BR k CQ, the
line P M is the radical axis of ωB and ωC .
However, P M is also the radical axis of the circumcircles γB of P M B and γC of P M C.
Let CX and P M meet at Z. Let p(K, ω) denote the power of a point K with respect to a
circumference ω. We have
p(Z, γB ) = p(Z, γC ) = ZX · ZC = p(Z, ωB ) = p(Z, ωC ).
Point Z is thus the radical center of γB , γC , ωB , ωC . Thus, the radical axes BY, CX, P M meet
at Z. From here,
ZY · ZB = ZC · ZX ⇒ BCXY cyclic
P Y · P R = P X · P Q ⇒ QRXT cyclic.
We may now finish as in Solution 1.
Problem 4. Consider a 2018 × 2019 board with integers in each unit square. Two unit
squares are said to be neighbours if they share a common edge. In each turn, you choose some
unit squares. Then for each chosen unit square the average of all its neighbours is calculated.
Finally, after these calculations are done, the number in each chosen unit square is replaced by
the corresponding average. Is it always possible to make the numbers in all squares become
the same after finitely many turns?
Answer: No
5
Solution. Let n be a positive integer relatively prime to 2 and 3. We may study the whole
process modulo n by replacing divisions by 2, 3, 4 with multiplications by the corresponding
inverses modulo n. If at some point the original process makes all the numbers equal, then the
process modulo n will also have all the numbers equal. Our aim is to choose n and an initial
configuration modulo n for which no process modulo n reaches a board with all numbers equal
modulo n. We split this goal into two lemmas.
Lemma 1. There is a 2 × 3 board that stays constant modulo 5 and whose entries are not
all equal.
Proof. Here is one such a board:
The fact that the board remains constant regardless of the choice of squares can be checked
square by square.
Lemma 2. If there is an r × s board with r ≥ 2, s ≥ 2, that stays constant modulo 5, then
there is also a kr × ls board with the same property.
Proof. We prove by a case by case analysis that repeateadly reflecting the r × s with respect
to an edge preserves the property:
• If a cell had 4 neighbors, after reflections it still has the same neighbors.
• If a cell with a had 3 neighbors b, c, d, we have by hypothesis that a ≡ 3−1 (b + c + d) ≡
2(b + c + d) (mod 5). A reflection may add a as a neighbor of the cell and now
4−1 (a + b + c + d) ≡ 4(a + b + c + d) ≡ 4a + 2a ≡ a (mod 5)
• If a cell with a had 2 neighbors b, c, we have by hypothesis that a ≡ 2−1 (b + c) ≡ 3(b + c)
(mod 5). If the reflections add one a as neighbor, now
3−1 (a + b + c) ≡ 2(3(b + c) + b + c) ≡ 8(b + c) ≡ 3(b + c) ≡ a (mod 5)
• If a cell with a had 2 neighbors b, c, we have by hypothesis that a ≡ 2−1 (b + c) (mod 5).
If the reflections add two a’s as neighbors, now
4−1 (2a + b + c) ≡ (2−1 a + 2−1 a) ≡ a (mod 5)
In the three cases, any cell is still preserved modulo 5 after an operation. Hence we can fill
in the kr × ls board by k × l copies by reflection.
Since 2|2018 and 3|2019, we can get through reflections the following board:
6
By the lemmas above, the board is invariant modulo 5, so the answer is no.
Problem 5. Determine all the functions f : R → R such that
f (x2 + f (y)) = f (f (x)) + f (y 2 ) + 2f (xy)
for all real number x and y.
Answer: The possible functions are f (x) = 0 for all x and f (x) = x2 for all x.
Solution. By substituting x = y = 0 in the given equation of the problem, we obtain that
f (0) = 0. Also, by substituting y = 0, we get f (x2 ) = f (f (x)) for any x.
Furthermore, by letting y = 1 and simplifying, we get
2f (x) = f (x2 + f (1)) − f (x2 ) − f (1),
from which it follows that f (−x) = f (x) must hold for every x.
Suppose now that f (a) = f (b) holds for some pair of numbers a, b. Then, by letting y = a
and y = b in the given equation, comparing the two resulting identities and using the fact that
f (a2 ) = f (f (a)) = f (f (b)) = f (b2 ) also holds under the assumption, we get the fact that
f (a) = f (b) ⇒ f (ax) = f (bx) for any real number x.
(1)
Consequently, if for some a 6= 0, f (a) = 0, then we see that, for any x, f (x) = f (a · xa ) =
f (0 · xa ) = f (0) = 0, which gives a trivial solution to the problem.
In the sequel, we shall try to find a non-trivial solution for the problem. So, let us assume
from now on that if a 6= 0 then f (a) 6= 0 must hold. We first note that since f (f (x)) = f (x2 )
for all x, the right-hand side of the given equation equals f (x2 ) + f (y 2 ) + 2f (xy), which is
invariant if we interchange x and y. Therefore, we have
f (x2 ) + f (y 2 ) + 2f (xy) = f (x2 + f (y)) = f (y 2 + f (x)) for every pair x, y.
(2)
Next, let us show that for any x, f (x) ≥ 0 must hold. Suppose, on the contrary, f (s) = −t2
holds for some pair s, t of non-zero real numbers. By setting x = s, y = t in the right hand
side of (2), we get f (s2 + f (t)) = f (t2 + f (s)) = f (0) = 0,√ so f (t) = −s2 . We also have
f (t2 ) = f (−t2 ) = f (f (s)) = f (s2 ). By applying (2) with x = s2 + t2 and y = s, we obtain
√
f (s2 + t2 ) + 2f (s · s2 + t2 ) = 0,
√
and similarly, by applying (2) with x = s2 + t2 and y = t, we obtain
√
f (s2 + t2 ) + 2f (t · s2 + t2 ) = 0.
Consequently, we obtain
√
√
f (s · s2 + t2 ) = f (t · s2 + t2 ).
√
√
√
By applying (1) with a = s s2 + t2 , b = t s2 + t2 and x = 1/ s2 + t2 , we obtain f (s) =
f (t) = −s2 , from which it follows that
0 = f (s2 + f (s)) = f (s2 ) + f (s2 ) + 2f (s2 ) = 4f (s2 ),
a contradiction to the fact s2 > 0. Thus we conclude that for all x 6= 0, f (x) > 0 must be
satisfied.
Now, we show the following fact
k > 0, f (k) = 1 ⇔ k = 1.
7
(3)
Let k > 0 for which f (k) = 1. We have f (k 2 ) = f (f√
(k)) = f (1), so by (1), f (1/k) = f (k) =
1, so we may assume k ≥ 1. By applying (2) with x = k 2 − 1 and y = k, and using f (x) ≥ 0,
we get
√
f (k 2 − 1 + f (k)) = f (k 2 − 1) + f (k 2 ) + 2f (k k 2 − 1) ≥ f (k 2 − 1) + f (k 2 ).
This simplifies to 0 ≥ f (k 2 − 1) ≥ 0, so k 2 − 1 = 0 and thus k = 1.
Next we focus√ on showing f (1) = 1. If f (1) = m ≤ 1, then we may proceed as above
by setting x = 1 − m and y = 1 to get m = 1. If f (1) = m ≥ 1, now we √
note that
2
2
f (m) = f (f (1)) = f (1 ) = f (1) = m ≤ m . We may then proceed as above with x = m2 − m
and y = 1 to show m2 = m and thus m = 1.
We are now ready to finish. Let x > 0 and m = f (x). Since f (f (x)) = f (x2 ), then f (x2 ) =
f (m). But by (1), f (m/x2 ) = 1. Therefore m = x2 . For x < 0, we have f (x) = f (−x) = f (x2 )
as well. Therefore, for all x, f (x) = x2 .
Solution 2 After proving that f (x) > 0 for x 6= 0 as in the previous solution, we may also
proceed as follows. We claim that f is injective on the positive real numbers. Suppose that
a > b > 0 satisfy f (a) = f (b). Then by setting x = 1/b in (1) we have f (a/b) = f (1). Now,
by induction on n and iteratively setting x = a/b in (1) we get f ((a/b)n ) = 1 for any positive
integer n.
n
p Now, let m = f (1) and n be a positive integer such that (a/b) > m. By setting x =
n
(a/b) − m and y = 1 in (2) we obtain that
p
f ((a/b)n − m + f (1)) = f ((a/b)n − m) + f (12 ) + 2f ( (a/b)n − m)) ≥ f ((a/b)n − m) + f (1).
Since f ((a/b)n ) = f (1), this last equation simplifies to f ((a/b)n − m) ≤ 0 and thus m =
(a/b)n . But this is impossible since m is constant and a/b > 1. Thus, f is injective on the
positive real numbers. Since f (f (x)) = f (x2 ), we obtain that f (x) = x2 for any real value x.
8
APMO 2020 Solution
1. Let Γ be the circumcircle of ∆ABC. Let D be a point on the side BC. The tangent to Γ at A intersects
the parallel line to BA through D at point E. The segment CE intersects Γ again at F . Suppose
B, D, F, E are concyclic. Prove that AC, BF, DE are concurrent.
Solution 1 From the conditions, we have
∠CBA = 180◦ − ∠EDB = 180◦ − ∠EF B
= 180◦ − ∠EF A − ∠AF B
= 180◦ − ∠CBA − ∠ACB = ∠BAC.
Let P be the intersection of AC and BF . Then we have
∠P AE = ∠CBA = ∠BAC = ∠BF C.
This implies A, P, F, E are concyclic. It follows that
∠F P E = ∠F AE = ∠F BA,
and hence AB and EP are parallel. So E, P, D are collinear, and the result follows.
Solution 2
Let E 0 be any point on the extension of EA. From ∠AED = ∠E 0 AB = ∠ACD, points A, D, C, E are
concyclic.
1
Let P be the intersection of BF and DE. From ∠AF P = ∠ACB = ∠AEP , the points A, P, F, E are
concyclic. In addition, from ∠EP A = ∠EF A = ∠DBA, points A, B, D, P are concyclic.
By considering the radical centre of (BDF E), (AP F E) and (BDP A), we find that the lines BD, AP, EF
are concurrent at C. The result follows.
2. Show that r = 2 is the largest real number r which satisfies the following condition:
If a sequence a1 , a2 , . . . of positive integers fulfills the inequalities
p
an ≤ an+2 ≤ a2n + ran+1
for every positive integer n, then there exists a positive integer M such that an+2 = an for every n ≥ M .
Solution 1. First, let us assume that r > 2, and take a positive integer a ≥ 1/(r − 2).
Then, if we let an = a + bn/2c for n = 1, 2, . . ., the sequence an satisfies the inequalities
s
p
p
1
2
2
2
an ≥ an + 1 = an+2 ,
an + ran+1 ≥ an + ran ≥ an + 2 +
a
but since an+2 > an for any n, we see that r does not satisfy the condition given in the problem.
Now we show that r = 2 does satisfy the condition of the problem. Suppose a1 , a2 , . . . is a sequence of
positive integers satisfying the inequalities given in the problem, and there exists a positive integer m
for which am+2 > am is satisfied.
By induction we prove the following assertion:
(†)
am+2k ≤ am+2k−1 = am+1 holds for every positive integer k.
The truth of (†) for k = 1 follows from the inequalities below
2am+2 − 1 = a2m+2 − (am+2 − 1)2 ≤ a2m + 2am+1 − (am+2 − 1)2 ≤ 2am+1 .
Let us assume that (†) holds for some positive integer k. From
a2m+1 ≤ a2m+2k+1 ≤ a2m+2k−1 + 2am+2k ≤ a2m+1 + 2am+1 < (am+1 + 1)2 ,
it follows that am+2k+1 = am+1 must hold. Furthermore, since am+2k ≤ am+1 , we have
a2m+2k+2 ≤ a2m+2k + 2am+2k+1 ≤ a2m+1 + 2am+1 < (am+1 + 1)2 ,
2
from which it follows that am+2k+2 ≤ am+1 , which proves the assertion (†).
We can conclude that for the value of m with which we started our argument above, am+2k+1 = am+1
holds for every positive integer k. Therefore, in order to finish the proof, it is enough to show that
am+2k becomes constant after some value of k. Since every am+2k is a positive integer less than or
equal to am+1 , there exists k = K for which am+2K takes the maximum value. By the monotonicity
of am+2k , it then follows that am+2k = am+2K for all k ≥ K.
Solution 2
We only give an alternative proof of the assertion (†) in solution 1. Let {an } be a sequence satisfying
the inequalities given in the problem. We will use the following key observations:
(a) If an+1 ≤ an for some n ≥ 1, then
an ≤ an+2 ≤
p
p
a2n + 2an+1 < a2n + 2an + 1 = an + 1,
hence an = an+2 .
(b) If an ≤ an+1 for some n ≥ 1, then
an ≤ an+2 ≤
q
p
a2n + 2an+1 < a2n+1 + 2an+1 + 1 = an+1 + 1,
hence an ≤ an+2 ≤ an+1 .
Now let m be a positive integer such that am+2 > am . By the observations above, we must have
am < am+2 ≤ am+1 . Thus the assertion (†) is true for k = 1. Assume that the assertion holds for some
positive integer k. Using observation (a), we get am+2k+1 = am+2k−1 = am+1 . Thus am+2k ≤ am+2k+1 ,
and then using observation (b), we get am+2k+2 ≤ am+2k+1 = am+1 , which proves the assertion (†).
3. Determine all positive integers k for which there exist a positive integer m and a set S of positive
integers such that any integer n > m can be written as a sum of distinct elements of S in exactly k
ways.
Solution:
We claim that k = 2a for all a ≥ 0.
Let A = {1, 2, 4, 8, . . . } and B = N\A. For any set T , let s(T ) denote the sum of the elements of T .
(If T is empty, we let s(T ) = 0.)
We first show that any positive integer k = 2a satisfies the desired property. Let B 0 be a subset of
B with a elements, and let S = A ∪ B 0 . Recall that any nonnegative integer has a unique binary
representation. Hence, for any integer t > s(B 0 ) and any subset B 00 ⊆ B 0 , the number t − s(B 00 ) can
be written as a sum of distinct elements of A in a unique way. This means that t can be written as a
sum of distinct elements of B 0 in exactly 2a ways.
Next, assume that some positive integer k satisfies the desired property for a positive integer m ≥ 2
and a set S. Clearly, S is infinite.
Lemma: For all sufficiently large x ∈ S, the smallest element of S larger than x is 2x.
Proof of Lemma: Let x ∈ S with x > 3m, and let x < y < 2x. We will show that y 6∈ S. Suppose
first that y > x + m. Then y − x can be written as a sum of distinct elements of S not including x in
k ways. If y ∈ S, then y can be written as a sum of distinct elements of S in at least k + 1 ways, a
contradiction. Suppose now that y ≤ x + m. We consider z ∈ (2x − m, 2x). Similarly as before, z − x
can be written as a sum of distinct elements of S not including x or y in k ways. If y ∈ S, then since
m < z − y < x, z − y can be written as a sum of distinct elements of S not including x or y. This
means that z can be written as a sum of distinct elements of S in at least k + 1 ways, a contradiction.
We now show that 2x ∈ S; assume for contradiction that this is not the case. Observe that 2x can be
written as a sum of distinct elements of S including x in exactly k − 1 ways. This means that 2x can
also be written as a sum of distinct elements of S not including x. If this sum includes any number
less than x − m, then removing this number, we can write some number y ∈ (x + m, 2x) as a sum of
distinct elements of S not including x. Now if y = y 0 + x where y 0 ∈ (m, x) then y 0 can be written as
3
a sum of distinct elements of S including x in exactly k ways. Therefore y can be written as a sum of
distinct elements of S in at least k + 1 ways, a contradiction. Hence the sum only includes numbers
in the range [x − m, x). Clearly two numbers do not suffice. On the other hand, three such numbers
sum to at least 3(x − m) > 2x, a contradiction.
From the Lemma, we have that S = T ∪ U , where T is finite and U = {x, 2x, 4x, 8x, . . . } for some
positive integer x. Let y be any positive integer greater than s(T ). For any subset T 0 ⊆ T , if
y − s(T 0 ) ≡ 0 (mod x), then y − s(T 0 ) can be written as a sum of distinct elements of U in a unique
way; otherwise y − s(T 0 ) cannot be written as a sum of distinct elements of U . Hence the number of
ways to write y as a sum of distinct elements of S is equal to the number of subsets T 0 ⊆ T such that
s(T 0 ) ≡ y (mod x). Since this holds for all y, for any 0 ≤ a ≤ x − 1 there are exactly k subsets T 0 ⊆ T
such that s(T 0 ) ≡ a (mod x). This means that there are kx subsets of T in total. But the number of
subsets of T is a power of 2, and therefore k is a power of 2, as claimed.
Solution 2. We give an alternative proof of the first half of the lemma in the Solution 1 above.
Qr
Let s1 < s2 < · · · be the elements of S. For any positive integer r, define Ar (x) = n=1 (1 + xsn ).
For each n such that m ≤ n < sr+1 , all k ways of writing n as a sum of elements of S must only use
s1 , . . . , sr , so the coefficient of xn in Ar (x) is k. Similarly the number of ways of writing sr+1 as a sum
of elements of S without using sr+1 is exactly k − 1. Hence the coefficient of xsr+1 in Ar (x) is k − 1.
Fix a t such that st > 2(m + 1). Write
At−1 (x) = u(x) + k xm+1 + · · · + xst −1 + xst v(x)
for some u(x), v(x) where u(x) is of degree at most m.
Note that
At+1 (x) = At−1 (x) + xst At−1 (x) + xst+1 At−1 (x) + xst +st+1 At−1 (x).
If st+1 + m + 1 < 2st , we can find the term xst+1 +m+1 in xst At−1 (x) and in xst+1 At−1 (x). Hence the
coefficient of xst+1 +m+1 in At+1 (x) is at least 2k, which is impossible. So st+1 ≥ 2st − (m + 1) >
st + m + 1.
Now
At (x) = At−1 (x) + xst u(x) + k(xst +m+1 + · · · x2st −1 ) + x2st v(x).
Recall that the coefficent of xst+1 in At (x) is k − 1. But if st + m + 1 < st+1 < s2t , then the coefficient
of xst+1 in At (x) is at least k, which is a contradiction. Therefore st+1 ≥ 2st .
4. Let Z denote the set of all integers. Find all polynomials P (x) with integer coefficients that satisfy the
following property:
For any infinite sequence a1 , a2 , . . . of integers in which each integer in Z appears exactly once, there
exist indices i < j and an integer k such that ai + ai+1 + · · · + aj = P (k).
Solution:
Part 1: All polynomials with deg P = 1 satisfy the given property.
Suppose P (x) = cx + d, and assume without loss of generality that c > d ≥ 0. Denote si = a1 + a2 +
· · · + ai (mod c). It suffices to show that there exist indices i and j such that j − i ≥ 2 and sj − si ≡ d
(mod c).
Consider c + 1 indices e1 , e2 , . . . , ec+1 > 1 such that ael ≡ d (mod c). By the pigeonhole principle,
among the n + 1 pairs (se1 −1 , se1 ), (se2 −1 , se2 ), . . . , (sen+1 −1 , sen+1 ), some two are equal, say (sm−1 , sm )
and (sn−1 , sn ). We can then take i = m − 1 and j = n.
Part 2: All polynomials with deg P 6= 1 do not satisfy the given property.
Lemma: If deg P 6= 1, then for any positive integers A, B, and C, there exists an integer y with |y| > C
such that no value in the range of P falls within the interval [y − A, y + B].
Proof of Lemma: The claim is immediate when P is constant or when deg P is even since P is bounded
from below. Let P (x) = an xn + · · · + a1 x + a0 be of odd degree greater than 1, and assume without
4
loss of generality that an > 0. Since P (x + 1) − P (x) = an nxn−1 + . . . , and n − 1 > 0, the gap between
P (x) and P (x + 1) grows arbitrarily for large x. The claim follows. Suppose deg P 6= 1. We will inductively construct a sequence {ai } such that for any indices i < j and
any integer k it holds that ai + ai+1 + · · · + aj 6= P (k). Suppose that we have constructed the sequence
up to ai , and m is an integer with smallest magnitude yet to appear in the sequence. We will add two
more terms to the sequence. Take ai+2 = m. Consider all the new sums of at least two consecutive
terms; each of them contains ai+1 . Hence all such sums are in the interval [ai+1 − A, ai+1 + B] for fixed
constants A, B. The lemma allows us to choose ai+1 so that all such sums avoid the range of P .
Alternate Solution for Part 1: Again, suppose P (x) = cx+d, and assume without loss of generality
that c > d ≥ 0. Let Si = {aj + aj+1 + · · · + ai (mod c) | j = 1, 2, . . . , i}. Then Si+1 = {si + ai+1
(mod c) | si ∈ Si } ∪ {ai+1 (mod c)}. Hence |Si+1 | = |Si | or Si+1 = |Si | + 1, with the former occuring
exactly when 0 ∈ Si . Since |Si | ≤ c, the latter can only occur finitely many times, so there exists I
such that 0 ∈ Si for all i ≥ I. Let t > I be an index with at ≡ d (mod c). Then we can find a sum of
at least two consecutive terms ending at at and congruent to d (mod c).
Alternate Construction when P (x) is constant or of even degree
If P (x) is of even degree, then P is bounded from below or from above. In case of P is constant
or bounded from above, then there exists a positive integer c such that P (x) < c. Let {ai } be the
sequence
0, 1, −1, 2, 3, −2, 4, 5, −3, · · ·
which is given by a3n+1 = 2n, a3n+2 = 2n + 1, a3n+3 = −(n + 1) for all n ≥ 0. Notice that for
any i < j we have ai + · · · + aj ≥ 0 . Then for the sequence {bn } defined by bn = an + c, clearly
bi + · · · + bj ≥ (ai + · · · + aj ) + 2c > c which is out side the range of P (x).
Now if P is bounded from below, there exist a positive integer c such that P (x) > −c. In this case,
take bn to be bn = −an − c. Then for all i < j we have bi + · · · + bj ≤ −(a1 + · · · an ) − 2c < −c which
is again out side the range of P (x).
5. Let n ≥ 3 be a fixed integer. The number 1 is written n times on a blackboard. Below the blackboard,
there are two buckets that are initially empty. A move consists of erasing two of the numbers a and b,
replacing them with the numbers 1 and a + b, then adding one stone to the first bucket and gcd(a, b)
stones to the second bucket. After some finite number of moves, there are s stones in the first bucket and
t stones in the second bucket, where s and t are positive integers. Find all possible values of the ratio st .
Solution:
The answer is the set of all rational numbers in the interval [1, n − 1). First, we show that no other
numbers are possible. Clearly the ratio is at least 1, since for every move, at least one stone is added
to the second bucket. Note that the number s of stones in the first bucket is always equal to p − n,
where p is the sum of the numbers on the blackboard. We will assume that the numbers are written
in a row, and whenever two numbers a and b are erased, a + b is written in the place of the number on
the right. Let a1 , a2 , ..., an be the numbers on the blackboard from left to right, and let
q = 0 · a1 + 1 · a2 + · · · + (n − 1)an .
Since each number ai is at least 1, we always have
q ≤ (n − 1)p − (1 + · · · + (n − 1)) = (n − 1)p −
n(n − 1)
n(n − 1)
= (n − 1)s +
.
2
2
Also, if a move changes ai and aj with i < j, then t changes by gcd(ai , aj ) ≤ ai and q increases by
(j − 1)ai − (i − 1)(ai − 1) ≥ iai − (i − 1)(ai − 1) ≥ ai .
Hence q − t never decreases. We may assume without loss of generality that the first move involves the
rightmost 1. Then immediately after this move, q = 0 + 1 + · · · + (n − 2) + (n − 1) · 2 = (n+2)(n−1)
and
2
5
t = 1. So after that move, we always have
(n + 2)(n − 1)
2
n(n − 1) (n + 2)(n − 1)
≤ (n − 1)s +
−
+1
2
2
= (n − 1)s − (n − 2) < (n − 1)s.
t≤q+1−
Hence,
t
s
< n − 1. So
t
s
must be a rational number in [1, n − 1).
After a single move, we have st = 1, so it remains to prove that st can be any rational number in
(1, n − 1). We will now show by induction on n that for any positive integer a, it is possible to reach
a situation where there are n − 1 occurrences of 1 on the board and the number an−1 , with t and s
equal to an−2 (a − 1)(n − 1) and an−1 − 1, respectively. For n = 2, this is clear as there is only one
possible move at each step, so after a − 1 moves s and t will both be equal to a − 1. Now assume
that the claim is true for n − 1, where n > 2. Call the algorithm which creates this situation using
n − 1 numbers algorithm A. Then to reach the situation for size n, we apply algorithm A, to create
the number an−2 . Next, apply algorithm A again and then add the two large numbers, repeat until
we get the number an−1 . Then algorithm A was applied a times and the two larger numbers were
added a − 1 times. Each time the two larger numbers are added, t increases by an−2 and each time
algorithm A is applied, t increases by an−3 (a − 1)(n − 2). Hence, the final value of t is
t = (a − 1)an−2 + a · an−3 (a − 1)(n − 2) = an−2 (a − 1)(n − 1).
This completes the induction.
Now we can choose 1 and the large number b times for any positive integer b, and this will add b stones
to each bucket. At this point we have
an−2 (a − 1)(n − 1) + b
t
=
.
s
an−1 − 1 + b
So we just need to show that for any rational number pq ∈ (1, n − 1), there exist positive integers a and
b such that
p
an−2 (a − 1)(n − 1) + b
=
q
an−1 − 1 + b
Rearranging, we see that this happens if and only if
b=
qan−2 (a − 1)(n − 1) − p(an−1 − 1)
.
p−q
If we choose a ≡ 1 (mod p−q), then this will be an integer, so we just need to check that the numerator
is positive for sufficiently large a.
qan−2 (a − 1)(n − 1) − p(an−1 − 1) > qan−2 (a − 1)(n − 1) − pan−1
= an−2 (a(q(n − 1) − p) − (n − 1)) ,
which is positive for sufficiently large a since q(n − 1) − p > 0.
Alternative solution for the upper bound. Rather than starting with n occurrences of 1, we may
start with infinitely many 1s, but we are restricted to having at most n − 1 numbers which are not
equal to 1 on the board at any time. It is easy to see that this does not change the problem. Note
also that we can ignore the 1 we write on the board each move, so the allowed move is to rub off two
numbers and write their sum. We define the width and score of a number on the board as follows.
Colour that number red, then reverse every move up to that point all the way back to the situation
when the numbers are all 1s. Whenever a red number is split, colour the two replacement numbers
6
red. The width of the original number is equal to the maximum number of red integers greater than 1
which appear on the board at the same time. The score of the number is the number of stones which
were removed from the second bucket during these splits. Then clearly the width of any number is
at most n − 1. Also, t is equal to the sum of the scores of the final numbers. We claim that if a
number p > 1 has a width of at most w, then its score is at most (p − 1)w. We will prove this by
strong induction on p. If p = 1, then clearly p has a score of 0, so the claim is true. If p > 1, then p
was formed by adding two smaller numbers a and b. Clearly a and b both have widths of at most w.
Moreover, if a has a width of w, then at some point in the reversed process there will be w numbers
in the set {2, 3, 4, ...} that have split from a, and hence there can be no such numbers at this point
which have split from b. Between this point and the final situation, there must always be at least one
number in the set {2, 3, 4, ...} that split from a, so the width of b is at most w − 1. Therefore, a and b
cannot both have widths of w, so without loss of generality, a has width at most w and b has width at
most w − 1. Then by the inductive hypothesis, a has score at most (a − 1)w and b has score at most
(b − 1)(w − 1). Hence, the score of p is at most
(a − 1)w + (b − 1)(w − 1) + gcd(a, b) ≤ (a − 1)w + (b − 1)(w − 1) + b
= (p − 1)w + 1 − w
≤ (p − 1)w.
This completes the induction.
Now, since each number p in the final configuration has width at most (n − 1), it has score less than
(n − 1)(p − 1). Hence the number t of stones in the second bucket is less than the sum over the values
of (n − 1)(p − 1), and s is equal to the sum of the the values of (p − 1). Therefore, st < n − 1.
7
APMO 2021 Solution
1. Prove that for each real number r > 2, there are exactly two or three positive real numbers x satisfying
the equation x2 = rbxc.
Note: bxc denotes the largest integer less than or equal to x
Solution Let r > 2 be a real number. Let x be a positive real number such that x2 = rbxc with
bxc = k. Since x > 0 and x2 = rk, we also have k > 0. From k ≤ x < k + 1, we get k 2 ≤ x2 = rk <
k 2 + 2k + 1 ≤ k 2 + 3k, hence k ≤ r < k + 3, or r − 3 < k ≤ r. There are at most three positive integers
in the interval (r − 3, r]. Thus there are at most three possible values for k. Consequently, there are
at most three positive solutions to the given equation.
Now suppose that k is a positive
integer
are at least two such positive
p in the interval [r − 2, r]. There √
√
rb
rkc. We conclude that the
integer. Observe that k ≤ rk ≤ (k + 2)k < k + 1 and so rk =
√
2
equation x = rbxc has at least two positive solutions, namely x = rk with k ∈ [r − 2, r].
2. For a polynomial P and a positive integer n, define Pn as the number of positive integer pairs (a, b)
such that a < b ≤ n and |P (a)| − |P (b)| is divisible by n.
Determine all polynomial P with integer coefficients such that for all positive integers n, Pn ≤ 2021.
Solution There are two possible families of solutions:
• P (x) = x + d, for some integer d ≥ −2022.
• P (x) = −x + d, for some integer d ≤ 2022.
Suppose P satisfies the problem conditions. Clearly P cannot be a constant polynomial. Notice that
a polynomial P satifies the conditions if and only if −P also satisfies them. Hence, we may assume
the leading coefficient of P is positive. Then, there exists positive integer M such that P (x) > 0 for
x ≥ M.
Lemma 1. For any positive integer n, the integers P (1), P (2), . . . , P (n) leave pairwise distinct remainders upon division by n.
Proof. Assume for contradiction that this is not the case. Then, for some 1 ≤ y < z ≤ n, there exists
0 ≤ r ≤ n − 1 such that P (y) ≡ P (z) ≡ r (mod n). Since P (an + b) ≡ P (b) (mod n) for all a, b
integers, we have P (an + y) ≡ P (an + z) ≡ r (mod n) for any integer a. Let A be a positive integer
such that An ≥ M , and let k be a positive integer such that k > 2A + 2021. Each of the 2(k − A)
integers P (An + y), P (An + z), P ((A + 1)n + y), P ((A + 1)n + z), . . . , P ((k − 1)n + y), P ((k − 1)n + z)
leaves one of the k remainders
r, n + r, 2n + r, . . . , (k − 1)n + r
upon division by kn. This implies that at least 2(k − A) − k = k − 2A (possibly overlapping) pairs
leave the same remainder upon division by kn. Since k − 2A > 2021 and all of the 2(k − A) integers
are positive, we find more than 2021 pairs a, b with a < b ≤ kn for which |P (b)| − |P (a)| is divisible by
kn - hence, Pkn > 2021, a contradiction.
1
Next, we show that P is linear. Assume that this is not the case, i.e., deg P ≥ 2. Then we
can find a positive integer k such that P (k) − P (1) ≥ k. This means that among the integers
P (1), P (2), . . . , P (P (k) − P (1)), two of them, namely P (k) and P (1), leave the same remainder upon
division by P (k) − P (1) - contradicting the lemma (by taking n = P (k) − P (1)). Hence, P must be
linear.
We can now write P (x) = cx + d with c > 0. We prove that c = 1 by two ways.
Solution 1 If c ≥ 2, then P (1) and P (2) leave the same remainder upon division by c, contradicting
the Lemma. Hence c = 1.
3
> 2022 and
Solution 2 Suppose c ≥ 2. Let n be a positive integer such that n > 2cM , n 1 −
2c
3n
n
3n
+ i < n, P
+i −P
+ i = n.
2c|n. Notice that for any positive integers i such that
2c
2c
2c
n
3n
Hence,
+ i,
+ i satifies the condition in the question for all positive integers i such that
2c
2c
3n
+ i < n. Hence, Pn > 2021, a contradiction. Then, c = 1.
2c
If d ≤ −2023, then there are at least 2022 pairs a < b such that P (a) = P (b), namely (a, b) =
(1, −2d − 1), (2, −2d − 2), ..., (−d − 1, −d + 1). This implies that d ≥ −2022.
Finally, we verify that P (x) = x + d satisfies the condition for any d ≥ −2022. Fix a positive integer
n. Note that ||P (b)| − |P (a)|| < n for all positive integers a < b ≤ n, so the only pairs a, b for which
|P (b)| − |P (a)| could be divisible by n are those for which |P (a)| = |P (b)|. When d ≥ −2022, there are
indeed at most 2021 such pairs.
3. Let ABCD be a cyclic convex quadrilateral and Γ be its circumcircle. Let E be the intersection of the
diagonals AC and BD, let L be the center of the circle tangent to sides AB, BC, and CD, and let M
be the midpoint of the arc BC of Γ not containing A and D. Prove that the excenter of triangle BCE
opposite E lies on the line LM .
Solution 1
Let L be the intersection of the bisectors of ∠ABC and ∠BCD. Let N be the E-excenter of 4BCE.
Let ∠BAC = ∠BDC = α, ∠DBC = β and ∠ACB = γ.
We have the following:
1
1
1
1
1
∠ABC = 90◦ − α − γ and ∠BCL = 90◦ − α − β,
2
2
2
2
2
1
1
◦
◦
∠CBN = 90 − β and ∠BCN = 90 − γ,
2
2
1
1
◦
◦
∠M BL = ∠M BC + ∠CBL = 90 − γ and ∠M CL = 90 − β,
2
2
1
◦
∠LCN = ∠LBN = 180 − (α + β + γ) .
2
∠CBL =
Applying the sine rule to 4M BL and 4M CL we obtain
MB
MC
sin ∠BLM
sin ∠CLM
=
=
=
.
ML
ML
sin ∠M BL
sin ∠M CL
It follows that
sin ∠BLM
sin ∠M BL
cos(γ/2)
=
=
.
sin ∠CLM
sin ∠M CL
cos(β/2)
2
(1)
Now
sin ∠BLM sin ∠LCN sin ∠N BC
cos(γ/2) sin(90◦ − 12 β)
= 1.
·
·
=
·
sin ∠M LC sin ∠N CB sin ∠N BL
cos(β/2) sin(90◦ − 12 γ)
Hence LM, BN, CN are concurrent and therefore L, M, N are collinear.
Alternative proof
We proceed similarly as above until the equation (1).
We use the following lemma.
Lemma: If π > α, β, γ, δ > 0, α + β = γ + δ < π, and
Proof of Lemma: Let θ = α + β = γ + δ. Then
sin α
sin β
sin(θ−β)
sin β
=
=
sin γ
sin δ ,
then α = γ and β = δ.
sin(θ−δ)
sin δ .
⇐⇒ sin(θ − β) sin δ = sin(θ − δ) sin β
⇐⇒ (sin θ cos β − sin β cos θ) sin δ = (sin θ cos δ − sin δ cos θ) sin β
⇐⇒ sin θ cos β sin δ = sin θ cos δ sin β
⇐⇒ sin θ sin(β − δ) = 0
Since 0 < θ < π, then sin θ 6= 0. Therefore, sin(β − δ) = 0, and we must have β = δ.
Applying the sine rule to 4N BL and 4N CL we obtain
NB
sin ∠BLN
=
,
NL
sin ∠LBN
NC
sin ∠CLN
=
.
NL
sin ∠LCN
Since ∠LBN = ∠LCN , it follows that
sin ∠BLN
NB
sin ∠BCN
cos(γ/2)
sin ∠BLM
=
=
=
=
.
sin ∠CLN
NC
sin ∠CBN
cos(β/2)
sin ∠CLM
By the lemma, it is concluded that ∠BLM = ∠BLN and ∠CLM = ∠CLN . Therefore, L, M, N are
collinear.
Solution 2
Denote by N the excenter of triangle BCE opposite E. Since BL bisects ∠ABC, we have ∠CBL =
∠ABC
1
. Since M is the midpoint of arc BC, we have ∠M BC = (∠M BC + ∠M CB) It follows by
2
2
angle chasing that
1
(∠M BC + ∠M CB + ∠ABC)
2
1
∠BCE
= (∠M BA + ∠M CB) = 90◦ −
= ∠BCN.
2
2
∠M BL = ∠M BC + ∠CBL =
Denote by X and Y the second intersections of lines BM and CM with the circumcircle of BCL,
respectively. Since ∠M BC = ∠M CB, we have BC k XY . It suffices to show that BN k XL and
CN k Y L. Indeed, from this it follows that 4BCN ∼ 4XY L, and therefore a homothety with center
M that maps B to X and C to Y also maps N to L, implying that N lies on the line LM .
By symmetry, it suffices to show that CN k Y L, which is equivalent to showing that ∠BCN = ∠XY L.
But we have ∠BCN = ∠M BL = ∠XBL = ∠XY L, completing the proof.
4. Given a 32 × 32 table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at
several other cells. The mouse then starts moving. It moves forward except that when it reaches a
piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset
of cells containing cheese is good if, during this process, the mouse tastes each piece of cheese exactly
once and then falls off the table. Show that:
3
(a) No good subset consists of 888 cells.
(b) There exists a good subset consisting of at least 666 cells.
Solution.
(a) For the sake of contradiction, assume a good subset consisting of 888 cells exists. We call those
cheese-cells and the other ones gap-cells. Observe that since each cheese-cell is visited once, each
gap-cell is visited at most twice (once vertically and once horizontally). Define a finite sequence s
whose i-th element is C if the i-th step of the mouse was onto a cheese-cell, and G if it was onto a
gap-cell. By assumption, s contains 888 C’s. Note that s does not contain a contiguous block of
4 (or more) C’s. Hence s contains at least 888/3 = 296 such C-blocks and thus at least 295 G’s.
But since each gap-cell is traversed at most twice, this implies there are at least d295/2e = 148
gap-cells, for a total of 888 + 148 = 1036 > 322 cells, a contradiction.
(b) Let Li , Xi be two 2i × 2i tiles that allow the mouse to “turn left” and “cross”, respectively. In
detail, the “turn left” tiles allow the mouse to enter at its bottom left cell facing up and to leave
at its bottom left cell facing left. The “cross” tiles allow the mouse to enter at its top right facing
down and leave at its bottom left facing left, while also to enter at its bottom left facing up and
leave at its top right facing right.
(a) Basic tiles
Li+1
L1
(c) 16 × 16
(b) Inductive construction
X1
Xi+1
Li
Li
Li
Xi
Xi
Li
Xi
Li
Note that given two 2i × 2i tiles Li , Xi we can construct larger 2i+1 × 2i+1 tiles Li+1 , Xi+1
inductively as shown on in (b). The construction works because the path intersects itself (or the
other path) only inside the smaller X-tiles where it works by induction.
For a tile T , let |T | be the number of pieces of cheese in it. By straightforward induction,
|Li | = |Xi | + 1 and |Li+1 | = 4 · |Li | − 1. From the initial condition |L1 | = 3. We now easily
compute |L2 | = 11, |L3 | = 43, |L4 | = 171, and |L5 | = 683. Hence we get the desired subset.
Another proof of (a).
Let XN be the largest possible density of cheese-cells in a good subset on an N ×N table. We will show
that XN ≤ 4/5 + o(1). Specifically, this gives X32 ≤ 817/1024. We look at the (discrete analogue) of
the winding number of the trajectory of the mouse. Since the mouse enters and leaves the table, for
every 4 right turns in its trajectory there has to be a self-crossing. But each self-crossing requires a
different empty square, hence XN ≤ 4/5.
5. Determine all functions f : Z → Z such that f (f (a) − b) + bf (2a) is a perfect square for all integers a
and b.
Solution 1.
There are two families of functions which satisfy the condition:
(
0
if n is even, and
(1) f (n) =
any perfect square
if n is odd
(2) f (n) = n2 , for every integer n.
4
It is straightforward to verify that the two families of functions are indeed solutions. Now, suppose
that f is any function which satisfies the condition that f (f (a) − b) + bf (2a) is a perfect square for
every pair (a, b) of integers. We denote this condition by (*). We will show that f must belong to
either Family (1) or Family (2).
Claim 1. f (0) = 0 and f (n) is a perfect square for every integer n.
Proof. Plugging (a, b) → (0, f (0)) in (*) shows that f (0)(f (0) + 1) = z 2 for some integer z. Thus,
(2f (0) + 1 − 2z)(2f (0) + 1 + 2z) = 1. Therefore, f (0) is either -1 or 0.
Suppose, for sake of contradiction, that f (0) = −1. For any integer a, plugging (a, b) → (a, f (a)) implies
that f (a)f (2a) − 1 is a square. Thus, for each a ∈ Z, there exists x ∈ Z such that f (a)f (2a) = x2 + 1
This implies that any prime divisor of f (a) is either 2 or is congruent to 1 (mod 4), and that 4 - f (a),
for every a ∈ Z.
Plugging (a, b) → (0, 3) in (*) shows that f (−4) − 3 is a square. Thus, there is y ∈ Z such that
f (−4) = y 2 + 3. Since 4 - f (−4), we note that f (−4) is a positive integer congruent to 3 (mod 4),
but any prime dividing f (−4) is either 2 or is congruent to 1 (mod 4). This gives a contradiction.
Therefore, f (0) must be 0.
For every integer n, plugging (a, b) → (0, −n) in (*) shows that f (n) is a square.
Replacing b with f (a) − b, we find that for all integers a and b,
f (b) + (f (a) − b)f (2a) is a square.
(**)
Now, let S be the set of all integers n such that f (n) = 0. We have two cases:
• Case 1: S is unbounded from above.
We claim that f (2n) = 0 for any integer n. Fix some integer n, and let k ∈ S with k > f (n).
Then, plugging (a, b) 7→ (n, k) in (**) gives us that f (k) + (f (n) − k)f (2n) = (f (n) − k)f (2n) is
a square. But f (n) − k < 0 and f (2n) is a square by Claim 1. This is possible only if f (2n) = 0.
In summary, f (n) = 0 whenever n is even and Claim 1 shows that f (n) is a square whenever n is
odd.
• Case 2: S is bounded from above.
Let T be the set of all integers n such that f (n) = n2 . We show that T is unbounded from above.
In fact, we show that p+1
2 ∈ T for all primes p big enough.
Fix a prime number p big enough, and let n = p+1
2 . Plugging (a, b) 7→ (n, 2n) in (**) shows us
that f (2n)(f (n) − 2n + 1) is a square for any integer n. For p big enough, we have 2n 6∈ S, so
f (2n) is a non-zero square. As a result, when p is big enough, f (n) and f (n) − 2n + 1 = f (n) − p
are both squares. Writing f (n) = k 2 and f (n) − p = m2 for some k, m ≥ 0, we have
(k + m)(k − m) = k 2 − m2 = p =⇒ k + m = p, k − m = 1 =⇒ k = n, m = n − 1.
Thus, f (n) = k 2 = n2 , giving us n = p+1
2 ∈ T.
Next, for all k ∈ T and n ∈ Z, plugging (a, b) 7→ (n, k) in (**) shows us that k 2 + (f (n) − k)f (2n)
is a square. But that means (2k − f (2n))2 − (f (2n)2 − 4f (n)f (2n)) = 4(k 2 + (f (n) − k)f (2n)) is
also a square. When k is large enough, we have |f (2n)2 − 4f (n)f (2n)| + 1 < |2k − f (2n)|. As a
result, we must have f (2n)2 = 4f (n)f (2n) and thus f (2n) ∈ {0, 4f (n)} for all integers n.
Finally, we prove that f (n) = n2 for all integers n. Fix n, and take k ∈ T big enough such that
2k 6∈ S. Then, we have f (k) = k 2 and f (2k) = 4f (k) = 4k 2 . Plugging (a, b) 7→ (k, n) to (**)
shows us that f (n) + (k 2 − n)4k 2 = (2k 2 − n)2 + (f (n) − n2 ) is a square. Since T is unbounded
from above, we can take k ∈ T such that 2k 6∈ S and also |2k 2 − n| > |f (n) − n2 |. This forces
f (n) = n2 , giving us the second family of solution.
Another approach of Case 1.
Claim 2. One of the following is true.
5
(i) For every integer n, f (2n) = 0.
(ii) There exists an integer K > 0 such that for every integer n ≥ K, f (n) > 0.
Proof. Suppose that there exists an integer α 6= 0 such that f (2α) > 0. We claim that for every integer
n ≥ f (α) + 1, we have f (n) > 0.
For every n ≥ f (α) + 1, plugging (a, b) → (α, f (α) − n) in (*) shows that f (n) + (f (α) − n)f (2α) is a
square, and in particular, is non-negative. Hence, f (n) ≥ (n − f (α))f (2α) > 0, as desired.
If f belongs to Case (i), Claim 1 shows that f belongs to Family (1).
If f belongs to Case (ii), then S is bounded from above. From Case 2 we get f (n) = n2 .
6
APMO 2022 Solution
1. Find all pairs (a, b) of positive integers such that a3 is a multiple of b2 and b − 1 is a multiple of a − 1.
Note: An integer n is said to be a multiple of an integer m if there is an integer k such that n = km.
Solution
Solution 1.1
By inspection, we see that the pairs (a, b) with a = b are solutions, and so too are the pairs (a, 1). We
will see that these are the only solutions.
• Case 1. Consider the case b < a. Since b − 1 is a multiple of a − 1, it follows that b = 1. This
yields the second set of solutions described above.
• Case 2. This leaves the case b ≥ a. Since the positive integer a3 is a multiple of b2 , there is a
positive integer c such that a3 = b2 c.
Note that a ≡ b ≡ 1 modulo a − 1. So we have
1 ≡ a3 = b2 c ≡ c (mod a − 1).
If c < a, then we must have c = 1, hence, a3 = b2 . So there is a positive integer d such that
a = d2 and b = d3 . Now a − 1 | b − 1 yields d2 − 1 | d3 − 1. This implies that d + 1 | d(d + 1) + 1,
which is impossible.
If c ≥ a, then b2 c ≥ b2 a ≥ a3 = b2 c. So there’s equality throughout, implying a = c = b. This
yields the first set of solutions described above.
Therefore, the solutions described above are the only solutions.
Solution 1.2
We will start by showing that there are positive integers x, c, d such that a = x2 cd and b = x3 c. Let
g = gcd(a, b) so that a = gd and b = gx for some coprime d and x. Then, b2 | a3 is equivalent to
g 2 x2 | g 3 d3 , which is equivalent to x2 | gd3 . Since x and d are coprime, this implies x2 | g. Hence,
g = x2 c for some c, giving a = x2 cd and b = x3 c as required.
Now, it remains to find all positive integers x, c, d satisfying
x2 cd − 1 | x3 c − 1.
That is, x3 c ≡ 1 (mod x2 cd − 1). Assuming that this congruence holds, it follows that d ≡ x3 cd ≡ x
(mod x2 cd − 1). Then, either x = d or x − d ≥ x2 cd − 1 or d − x ≥ x2 cd − 1.
• If x = d then b = a.
• If x − d ≥ x2 cd − 1, then x − d ≥ x2 cd − 1 ≥ x − 1 ≥ x − d. Hence, each of these inequalities must
in fact be an equality. This implies that x = c = d = 1, which implies that a = b = 1.
• If d − x ≥ x2 cd − 1, then d − x ≥ x2 cd − 1 ≥ d − 1 ≥ d − x. Hence, each of these inequalities must
in fact be an equality. This implies that x = c = 1, which implies that b = 1.
Hence the only solutions are the pairs (a, b) such that a = b or b = 1. These pairs can be checked to
satisfy the given conditions.
1
Solution 1.3
All answers are (n, n) and (n, 1) where n is any positive integer. They all clearly work.
To show that these are all solutions, note that we can easily eliminate the case a = 1 or b = 1. Thus,
assume that a, b 6= 1 and a 6= b. By the second divisibility, we see that a − 1 | b − a. However,
gcd(a, b) | b − a and a − 1 is relatively prime to gcd(a, b). This implies that (a − 1) gcd(a, b) | b − a,
b−1
which implies gcd(a, b)
− 1.
a−1
b−1
The last relation implies that gcd(a, b) <
, since the right-hand side are positive. However, due
a−1
to the first divisibility,
gcd(a, b)3 = gcd(a3 , b3 ) ≥ gcd(b2 , b3 ) = b2 .
Combining these two inequalities, we get that
2
b3 <
b−1
b
<2 .
a−1
a
1
3
This implies a < 2b 3 . However, b2 | a3 gives b ≤ a 2 . This forces
√
3 1
a < 2(a 2 ) 3 = 2 a =⇒ a < 4.
Extracting a = 2, 3 by hand yields no additional solution.
2. Let ABC be a right triangle with ∠B = 90◦ . Point D lies on the line CB such that B is between D
and C. Let E be the midpoint of AD and let F be the second intersection point of the circumcircle of
4ACD and the circumcircle of 4BDE. Prove that as D varies, the line EF passes through a fixed
point.
2
Solution
Solution 2.1
Let the line EF intersect the line BC at P and the circumcircle of 4ACD at G distinct from F . We
will prove that P is the fixed point.
First, notice that 4BED is isosceles with EB = ED. This implies ∠EBC = ∠EDP .
Then, ∠DAG = ∠DF G = ∠EBC = ∠EDP which implies AG k DC. Hence, AGCD is an isosceles
trapezoid.
∼ 4DEP and AG = DP .
Also, AG k DC and AE = ED. This implies 4AEG =
Since B is the foot of the perpendicular from A onto the side CD of the isosceles trapezoid AGCD,
we have P B = P D + DB = AG + DB = BC, which does not depend on the choice of D. Hence, the
initial statement is proven.
Solution 2.2
Set up a coordinate system where BC is along the positive x-axis, BA is along the positive y-axis,
and B is the origin. Take A = (0, a), B = (0, 0), C = (c, 0), D = (−d, 0) where a, b, c, d > 0. Then
E = (− d2 , a2 ). The general equation of a circle is
x2 + y 2 + 2f x + 2gy + h = 0
(1)
Substituting the coordinates of A, D, C into (1) and solving for f, g, h, we find that the equation of the
circumcircle of 4ADC is
cd
(2)
x2 + y 2 + (d − c)x + ( − a)y − cd = 0.
a
Similarly, the equation of the circumcircle of 4BDE is
x2 + y 2 + dx + (
a
d2
− )y = 0.
2a 2
(3)
Then (3)−(2) gives the equation of the line DF which is
cx +
a2 + d2 − 2cd
y + cd = 0.
2a
(4)
Solving (3) and (4) simultaneously, we get
F =
2ac(c − d)
c(d2 − a2 − 2cd)
, 2
2
2
a + (d − 2c)
a + (d − 2c)2
,
and the other solution D = (−d, 0).
From this we obtain the equation of the line EF which is ax + (d − 2c)y + ac = 0. It passes through
P (−c, 0) which is independent of d.
3. Find all positive integers k < 202 for which there exists a positive integer n such that
n n o 2n kn
k
+
+ ··· +
= ,
202
202
202
2
where {x} denote the fractional part of x.
Note: {x} denotes the real number k with 0 ≤ k < 1 such that x − k is an integer.
Solution
Denote the equation in the problem statement as (*), and note that it is equivalent
to the
condition
in
that the average of the remainders when dividing n, 2n, . . . , kn by 202 is 101. Since
is invariant
202
in each residue class modulo 202 for each 1 ≤ i ≤ k, it suffices to consider 0 ≤ n < 202.
3
in
If n = 0, so is
, meaning that (*) does not hold for any k. If n = 101, then it can be checked
202
that (*) is satisfied if and only if k = 1. From now on, we will assume that 101 - n.
in
in
in
=
−
. Rewriting (*) and multiplying the equation by
For each 1 ≤ i ≤ k, let ai =
202
202
202
202, we find that
n(1 + 2 + . . . + k) − 202(a1 + a2 + . . . + ak ) = 101k.
Equivalently, letting z = a1 + a2 + . . . + ak ,
nk(k + 1) − 404z = 202k.
Since n is not divisible by 101, which is prime, it follows that 101 | k(k + 1). In particular, 101 | k or
101 | k + 1. This means that k ∈ {100, 101, 201}. We claim that all these values of k work.
• If k = 201, we may choose n = 1. The remainders when dividing n, 2n, . . . , kn by 202 are 1, 2,
. . . , 201, which have an average of 101.
• If k = 100, we may choose n = 2. The remainders when dividing n, 2n, . . . , kn by 202 are 2, 4,
. . . , 200, which have an average of 101.
• If k = 101, we may choose n = 51. To see this, note that the first four remainders are 51, 102, 153,
2, which have an average of 77. The next four remainders (53, 104, 155, 4) are shifted upwards
from the first four remainders by 2 each, and so on, until the 25th set of the remainders (99,
150, 201, 50) which have an average of 125. Hence, the first 100 remainders have an average of
77 + 125
= 101. The 101th remainder is also 101, meaning that the average of all 101 remainders
2
is 101.
In conclusion, all values k ∈ {1, 100, 101, 201} satisfy the initial condition.
4. Let n and k be positive integers. Cathy is playing the following game. There are n marbles and k
boxes, with the marbles labelled 1 to n. Initially, all marbles are placed inside one box. Each turn,
Cathy chooses a box and then moves the marbles with the smallest label, say i, to either any empty
box or the box containing marble i + 1. Cathy wins if at any point there is a box containing only
marble n.
Determine all pairs of integers (n, k) such that Cathy can win this game.
Solution
We claim Cathy can win if and only if n ≤ 2k−1 .
First, note that each non-empty box always contains a consecutive sequence of labeled marbles. This
is true since Cathy is always either removing from or placing in the lowest marble in a box. As a
consequence, every move made is reversible.
Next, we prove by induction that Cathy can win if n = 2k−1 . The base case of n = k = 1 is trivial.
Assume a victory can be obtained for m boxes and 2m−1 marbles. Consider the case of m + 1 boxes
and 2m marbles. Cathy can first perform a sequence of moves so that only marbles 2m−1 , . . . , 2m are
left in the starting box, while keeping one box, say B, empty. Now move the marble 2m−1 to box B,
then reverse all of the initial moves while treating B as the starting box. At the end of that, we will
have marbles 2m−1 + 1, . . . , 2m in the starting box, marbles 1, 2, . . . , 2m−1 in box B, and m − 1 empty
boxes. By repeating the original sequence of moves on marbles 2m−1 + 1, . . . , 2m , using the m boxes
that are not box B, we can reach a state where only marble 2m remains in the starting box. Therefore
4
a victory is possible if n = 2k−1 or smaller.
We now prove by induction that Cathy loses if n = 2k−1 + 1. The base case of n = 2 and k = 1 is
trivial. Assume a victory is impossible for m boxes and 2m−1 +1 marbles. For the sake of contradiction,
suppose that victory is possible for m + 1 boxes and 2m + 1 marbles. In a winning sequence of moves,
consider the last time a marble 2m−1 + 1 leaves the starting box, call this move X. After X, there
cannot be a time when marbles 1, . . . , 2m−1 + 1 are all in the same box. Otherwise, by reversing these
moves after X and deleting marbles greater than 2m−1 + 1, it gives us a winning sequence of moves
for 2m−1 + 1 marbles and m boxes (as the original starting box is not used here), contradicting the
inductive hypothesis. Hence starting from X, marbles 1 will never be in the same box as any marbles
greater than or equal to 2m−1 + 1.
Now delete marbles 2, . . . , 2m−1 and consider the winning moves starting from X. Marble 1 would only
move from one empty box to another, while blocking other marbles from entering its box. Thus we
effectively have a sequence of moves for 2m−1 + 1 marbles, while only able to use m boxes. This again
contradicts the inductive hypothesis. Therefore, a victory is not possible if n = 2k−1 + 1 or greater.
5. Let a, b, c, d be real numbers such that a2 + b2 + c2 + d2 = 1. Determine the minimum value of
(a − b)(b − c)(c − d)(d − a) and determine all values of (a, b, c, d) such that the minimum value is
achieved.
Solution
The minimum value is −
1
. There are eight equality cases in total. The first one is
8
√
√
√
√ !
1
3 1
3 1
3 1
3
+
,− −
, −
,− +
.
4
4
4
4 4
4
4
4
Cyclic shifting all the entries give three more quadruples. Moreover, flipping the sign ((a, b, c, d) →
(−a, −b, −c, −d)) all four entries in each of the four quadruples give four more equality cases.
Solution 5.1
Since the expression is cyclic, we could WLOG a = max{a, b, c, d}. Let
S(a, b, c, d) = (a − b)(b − c)(c − d)(d − a)
Note that we have given (a, b, c, d) such that S(a, b, c, d) = − 81 . Therefore, to prove that S(a, b, c, d) ≥
− 18 , we just need to consider the case where S(a, b, c, d) < 0.
• Exactly 1 of a − b, b − c, c − d, d − a is negative.
Since a = max{a, b, c, d}, then we must have d − a < 0. This forces a > b > c > d. Now, let us
write
S(a, b, c, d) = −(a − b)(b − c)(c − d)(a − d)
Write a − b = y, b − c = x, c − d = w for some positive reals w, x, y > 0. Plugging to the original
condition, we have
(d + w + x + y)2 + (d + w + x)2 + (d + w)2 + d2 − 1 = 0 (∗)
and we want to prove that wxy(w + x + y) ≤ 81 . Consider the expression (∗) as a quadratic in d:
4d2 + d(6w + 4x + 2y) + ((w + x + y)2 + (w + x)2 + w2 − 1) = 0
5
Since d is a real number, then the discriminant of the given equation has to be non-negative, i.e.
we must have
4 ≥ 4((w + x + y)2 + (w + x)2 + w2 ) − (3w + 2x + y)2
= (3w2 + 2wy + 3y 2 ) + 4x(w + x + y)
≥ 8wy + 4x(w + x + y)
= 4(x(w + x + y) + 2wy)
However, AM-GM gives us
1
wxy(w + x + y) ≤
2
x(w + x + y) + 2wy
2
2
≤
1
8
This proves S(a, b, c, d) ≥ − 18 for any a, b, c, d ∈ R such that a > b > c > d. Equality holds if and
only if w = y, x(w + x + y) = 2wy and wxy(w + x + y) = 81 . Solving these equations gives us
1
w4 = 16
which forces w = 12 since w > 0. Solving for x gives us x(x + 1) = 12 , and we will get
√
√
1
x = − 2 + 23 as x > 0. Plugging back gives us d = − 41 − 43 , and this gives us
√
√
√
√ !
1
3 1
3 1
3 1
3
(a, b, c, d) =
+
,− +
, −
,− −
4
4
4
4 4
4
4
4
Thus, any cyclic permutation of the above solution will achieve the minimum equality.
• Exactly 3 of a − b, b − c, c − d, d − a are negative Since a = max{a, b, c, d}, then a − b has to
be positive. So we must have b < c < d < a. Now, note that
S(a, b, c, d) = (a − b)(b − c)(c − d)(d − a)
= (a − d)(d − c)(c − b)(b − a)
= S(a, d, c, b)
Now, note that a > d > c > b. By the previous case, S(a, d, c, b) ≥ − 81 , which implies that
S(a, b, c, d) = S(a, d, c, b) ≥ −
1
8
as well. Equality holds if and only if
(a, b, c, d) =
√
√
√
√ !
1
3 1
3 1
3 1
3
+
,− −
, −
,− +
4
4
4
4 4
4
4
4
and its cyclic permutation.
Solution 5.2
The minimum value is −
1
. There are eight equality cases in total. The first one is
8
√
√
√
√ !
1
3 1
3 1
3 1
3
+
,− −
, −
,− +
.
4
4
4
4 4
4
4
4
Cyclic shifting all the entries give three more quadruples. Moreover, flipping the sign ((a, b, c, d) →
(−a, −b, −c, −d)) all four entries in each of the four quadruples give four more equality cases. We then
begin the proof by the following optimization:
Claim 1. In order to get the minimum value, we must have a + b + c + d = 0.
6
Proof. Assume not, let δ =
a+b+c+d
4
and note that
(a − δ)2 + (b − δ)2 + (c − δ)2 + (d − δ)2 < a2 + b2 + c2 + d2 ,
so by shifting by δ and scaling, we get an even smaller value of (a − b)(b − c)(c − d)(d − a).
The key idea is to substitute the variables
x = ac + bd
y = ab + cd
z = ad + bc,
so that the original expression is just (x − y)(x − z). We also have the conditions x, y, z ≥ −0.5 because
of:
2x + (a2 + b2 + c2 + d2 ) = (a + c)2 + (b + d)2 ≥ 0.
Moreover, notice that
0 = (a + b + c + d)2 = a2 + b2 + c2 + d2 + 2(x + y + z) =⇒ x + y + z =
−1
.
2
Now, we reduce to the following optimization problem.
Claim 2. Let x, y, z ≥ −0.5 such that x + y + z = −0.5. Then, the minimum value of
(x − y)(x − z)
is −1/8. Moreover, the equality case occurs when x = −1/4 and {y, z} = {1/4, −1/2}.
Proof. We notice that
1
1
1
2y + z +
2z + y +
+
2
2
8
1
1
1
2
= (4y + 4z + 1) + y +
z+
≥ 0.
8
2
2
1
(x − y)(x − z) + =
8
1
1
and z + are not less than zero.
2
2
1
1
The equality in the last inequality is attained when either y + = 0 or z + = 0, and 4y + 4z + 1 = 0.
2
2
This system of equations give (y, z) = (1/4, −1/2) or (y, z) = (−1/2, 1/4) as the desired equality cases.
The last inequality is true since both y +
Note: We can also prove (the weakened) Claim 2 by using Lagrange Multiplier, as follows. We first
prove that, in fact, x, y, z ∈ [−0.5, 0.5]. This can be proved by considering that
−2x + (a2 + b2 + c2 + d2 ) = (a − c)2 + (b − d)2 ≥ 0.
We will prove the Claim 2, only that in this case, x, y, z ∈ [−0.5, 0.5]. This is already sufficient to prove
the original question. We already have the bounded domain [−0.5, 0.5]3 , so the global minimum must
occur somewhere. Thus, it suffices to consider two cases:
• If the global minimum lies on the boundary of [−0.5, 0.5]3 . Then, one of x, y, z must be −0.5 or
0.5. By symmetry between y and z, we split to a few more cases.
7
– If x = 0.5, then y = z = −0.5, so (x − y)(x − z) = 1, not the minimum.
– If x = −0.5, then both y and z must be greater or equal to x, so (x − y)(x − z) ≥ 0, not the
minimum.
– If y = 0.5, then x = z = −0.5, so (x − y)(x − z) = 0, not the minimum.
– If y = −0.5, then z = −x, so
(x − y)(x − z) = 2x(x + 0.5),
which obtain the minimum at x = −1/4. This gives the desired equality case.
• If the global minimum lies in the interior (−0.5, 0.5)3 , then we apply Lagrange multiplier:
∂
∂
(x − y)(x − z) = λ (x + y + z) =⇒ 2x − y − z = λ.
∂x
∂x
∂
∂
(x − y)(x − z) = λ (x + y + z) =⇒ z − x = λ.
∂y
∂y
∂
∂
(x − y)(x − z) = λ (x + y + z) =⇒ y − x = λ.
∂z
∂z
Adding the last two equations gives λ = 0, or x = y = z. This gives (x − y)(x − z) = 0, not the
minimum.
Having exhausted all cases, we are done.
8