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b
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a
;
THE 1998 CANADI
CANADIAN
AN MA
MATHEMA
THEMATICAL
TICAL
OLYMPIAD
1. Determine the number of real solutions a to the equation
1
1
1
a +
a +
a =a.
2
3
5
Here, if x is a real number, then [ x ] denotes the greatest integer that is less than or
equal to x.
2. Find all real numbers x such that
x=
x−
1
x
1/2
+ 1−
1
x
1/2
.
3. Let n be a natural number such that n ≥ 2. Show that
1
1
1
1+ + ···+
n+1
3
2n − 1
1 1
>
n
1
1
.
+ +···+
2 4
2n
4. Let ABC be a triangle with ∠BAC = 40 and ∠ABC = 60 . Let D and E be
the points lying on the sides AC and AB , respectively, such that ∠C BD = 40 and
∠BC E = 7
700 . Let F be the point of intersection of the lines BD and C E . Show that
the line AF is perpendicular to the line BC .
◦
◦
◦
◦
5. Let m be a posit
positiv
ivee integ
integer.
er. Defi
Define
ne the seq
sequen
uence
ce a0 , a1 , a2 , . . . by a0 = 0, a1 = m,
2
Provee that an orde
ordere
red
d pa
pair
ir (a, b) of
and an+1 = m an − an 1 for n = 1, 2, 3, . . . . Prov
non-negative integers, with a ≤ b, gives a solution to the equation
−
a2 + b2
= m2
ab + 1
if and only if (a, b) is of the form (an , an+1 ) for some n ≥ 0.
THE 1999 CANADI
CANADIAN
AN MA
MATHEMA
THEMATICAL
TICAL
OLYMPIAD
1. Find all real solutions to the equation 4 x2 − 40[x] + 51 = 0.
Here, if x is a real number, then [x] denotes the greatest integer that is less than
or equal to x.
2. Let ABC be an equil
equilate
ateral
ral tria
triangl
nglee of altit
altitude
ude 1. A cir
circle
cle wit
with
h rad
radius
ius 1 and
center on the same side of AB as C rolls along the segment AB . Pro
Prove
ve tha
thatt
the arc of the circle that is inside the triangle always has the same length.
3. Dete
Determine
rmine all positiv
positivee intege
integers
rs n with the property that n = (d(n))2 . Here d (n)
denotes the number of positive divisors of n.
4. Suppose a1 , a2 , . . . , a8 are eight distinct integers from {1, 2, . . . , 16, 17}. Sh
Shoow
that there is an integer k > 0 such that the equation ai − aj = k has at least
three
thr
ee diffe
differen
rentt sol
soluti
utions.
ons. Als
Also,
o, find a spec
specific
ific set of 7 dis
distin
tinct
ct inte
integer
gerss fro
from
m
{1, 2, . . . , 16, 17} such that the equation a i − aj = k does not have three distinct
solutions for any k > 0.
5. Let x, y , and z be non-negative real numbers satisfying x + y + z = 1. Sh
Show
ow
that
4
x2 y + y 2 z + z 2x ≤
,
27
and find when equality occurs.
THE 2000 CANADI
CANADIAN
AN MA
MATHEMA
THEMATICAL
TICAL
OLYMPIAD
1. At 12:00 noon, Anne, Beth and Carmen begin running laps around a circular
track of length three hundred meters, all starting from the same point on the
track
tra
ck.. Eac
Each
h jog
jogger
ger main
maintai
tains
ns a con
consta
stant
nt speed in one of the tw
twoo poss
possibl
iblee directions for an indefinite period of time. Show that if Anne’s speed is different
from the other two speeds, then at some later time Anne will be at least one
hundred
hun
dred meter
meterss from eac
each
h of the other runner
runners.
s. (Here
(Here,, dista
distance
nce is measu
measured
red
along the shorter of the two arcs separating two runners.)
2. A permutation of the integers 1901, 1902, . . . , 2000 is a sequence a1 , a2 , . . . , a100
in which each of those integers appears exactly once. Given such a permutation,
we form the sequence of partial sums
s1 = a 1 , s2 = a 1 + a2 , s3 = a 1 + a2 + a3 , . . . , s100 = a 1 + a2 + · · · + a100.
How many of these permutations will have no terms of the sequence s 1 , . . . , s100
divisible by three?
3. Let A = (a1, a2 , . . . , a2000 ) be a sequence of integers each lying in the interval [−1000, 1000
1000].
]. Su
Suppo
ppose
se tha
thatt the
the en
entr
trie
iess in A sum
sum to 1. Sh
Show
ow th
that
at som
somee
nonempty subsequence of A sums to zero.
4. Let ABCD be a convex quadrilateral with
= 2 ADB,
ABD = 2 C DB
AB = C B.
C BD
and
Prove that AD = C D.
5. Suppose that the re
real
al num
numbers
bers a1 , a2, . . . , a100 satisfy
a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0,
a1 + a2 ≤ 100
and
a3 + a4 + · · · + a100 ≤ 100.
2
Determine the maximum possible value of a21 + a22 + · · · + a 100
, and find all
possible sequences a1 , a2 , . . . , a100 which achieve this maximum.
THE 2001 CANADIAN MATHEMATICAL OLYMPIAD
1. Randy: “Hi Rachel, that’s an int
interest
eresting
ing quadratic equati
equation
on you hav
havee writ
written
ten down. What
are its roots?”
Rachel: “The roots are two positive integers. One of the roots is my age, and the other root
is the age of my younger brother, Jimmy.”
Randy: “That is very neat! Let me see if I can figure out how old you and Jimmy are. That
shouldn’t
shoul
dn’t be too difficu
difficult
lt since all of your coefficien
coefficients
ts are int
integer
egers.
s. By the wa
way
y, I notice that
the sum of the three coefficients is a prime number.”
Rachel: “Interesting. Now figure out how old I am.”
Randy: “Instead, I will guess your age and substitute it for x in your quadratic equation
. . . da
darn
rn,, tha
thatt giv
gives
es me −55, and not 0.”
Rachel: “Oh, leave me alone!”
(a) Pro
Prove
ve that Jimm
Jimmy
y is two
two years old.
(b) Dete
Determine
rmine Rac
Rachel’s
hel’s age.
2. Ther
Theree is a board num
numbered
bered −10 to 10 as shown. Each square is coloured either red or white,
and the sum of the numbers on the red squares is n.
n. Maureen starts with a token on the square
labeled
0. She
thenright.
tossesEvery
a fair time
coin ten
Every
shethe
flipstoken
heads,
she
movestothe
one square
to the
she times.
flips tails,
shetime
moves
one
square
thetoken
left.
At the end of the ten flips, the probability that the token finishes on a red square is a rational
number of the form ab . Given
Given th
that
at a + b = 2001, determine the largest possible value for n.
-10
-9
-8
-8
--7
7
--6
6
--5
5
--4
4
--3
3
--2
2
--1
1
0
1
2
3
4
5
6
7
8
9
10
3. Let ABC be a triangle with AC > AB.
AB . Let P be the intersection point of the perpendicular bisector of BC and the internal angle bisector of A. Con
Constr
struct
uct poi
point
ntss X on AB
(extended) and Y on AC such that P X is perpendicular to AB and P Y is perpendicular
to AC . Let Z be the intersection point of X Y and BC . Det
Determ
ermine
ine the v
valu
aluee of BZ/ZC .
C
Y
Z
M
A
B
P
X
4. Let n be a positive integer. Nancy is given a rectangular table in which each entry is a positive
integer. She is permitted to make either of the following two moves:
(a) selec
selectt a row and multi
multiply
ply each ent
entry
ry in this row by n.
n.
(b) select a column and subtract n from each entry in this column.
Find all possible values of n for which the following statement is true:
Given any rectangular table, it is possible for Nancy to perform a finite sequence of
moves to create a table in which each entry is 0.
5. Let P0 , P1 , P2 be three points on the circumference of a circle with radius 1, where P1 P2 =
t < 2. For each i ≥ 3, define P i to be the centre of the circumcircle of Pi−1 Pi−2 Pi−3 .
(a) Pro
Prove
ve that the point
pointss P 1 , P5 , P9 , P13, . . . are collinear.
(b) Let x be the distance from P1 to P1001
, and let y be the distance from P1001 to P2001.
Determine all values of t for which
x/y is an integer.
500
THE 2002 CANADIAN MATHEMATICAL OLYMPIAD
1. Let S be a subset of 1, 2, . . . , 9 , such that the sums formed by adding each unordered pair of
distinct numbers from S are all different. For example, the subset 1, 2, 3, 5 has this property
property,,
but 1, 2, 3, 4, 5 does not, since the pairs 1, 4 and 2, 3 have the same sum, namely 5.
{
{
}
}
{ }
{ }
{
}
What is the maximum number of elements that S can contain?
2. Cal
Calll a posi
positiv
tivee integ
integer
er n practical if every positive integer less than or equal to n can be
written as the sum of distinct divisors of n.
For example, the divisors of 6 are 1, 2, 3, and 6. Since
1=1, 2=2, 3=3, 4=1+3, 5=2+ 3,
6=6,
we see that 6 is practical.
Prove that the product of two practical numbers is also practical.
3. Pro
Prove
ve that for all p
positiv
ositivee real numbers a, b, and c,
a3
b3
c3
+ +
bc
ca
ab
≥ a + b + c,
and determine when equality occurs.
√
4. Let Γ be a circle with radius r. Let A and B be distinct points on Γ such that AB < 3r.
Let the circle with centre B and radius AB meet Γ again at C . Let P be the point inside Γ
such that triangle ABP is equilateral. Finally, let the line C P meet Γ again at Q.
Prove that P Q = r .
N such that
5. Let N = 0, 1, 2, . . . . Determine all functions f : N
xf (y ) + yf (x) = (x + y )f (x2 + y 2 )
{
}
for all x and y in N .
→
The Canadian Mathematical Olympiad - 2003
1. Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let m be an integer, with 1 ≤ m ≤ 720. At precisely m minutes after 12:00, the
angle made by the hour hand and minute hand is exactly 1 . Det
Determ
ermine
ine all possibl
possiblee
values of m.
m.
◦
2001
2002
2. Find the last three digits of the number 2003
.
3. Find all real positive solutions (if any) to
x3 + y3 + z 3 = x + y + z, and
x2 + y2 + z 2 = xyz.
4. Pro
Prove
ve that when three circl
circles
es share the same cho
chord
rd AB
AB,, every line through A different
from AB determines the same ratio X Y : Y Z , where X is an arbitrary point different
from B on the first circle while Y and Z are the points where AX intersects the other
two circles (labelled so that Y is between X and Z ).
l
A
X
Y
Z
B
5. Let S be a set of n points in the plane such that any two points of S are at least 1
unit apart. Prove there √
is a subset T of S with at least n/
n/77 points such that any two
points of T are at least 3 units apart.
36th Canadian Mathematical Olympiad
Wednesday, March 31, 2004
1. Find all ordered triples ( x,y,z ) of real numbers which satisfy the following system of
equations:
xy = z − x − y
xz = y − x − z
yz = x − y − z
2. How many ways can 8 mutually non-attacking rooks be placed
on the 9 × 9 chessboard (shown here) so that all 8 rooks are on
squares of the same colour?
[Two rooks are said to be attacking each other if they are placed
in the same row or column of the board.]
3. Let A
A,, B,C,D be four points on a circle (occurring in clockwise order), with AB < AD
and BC > CD . Let the bisector of angle BAD meet the circle at X and the bisector
of angle BC D meet the circle at Y . Consider the hexagon formed by these six points
on the circle. If four of the six sides of the hexagon have equal length, prove that BD
must be a diameter of the circle.
4. Let p be an odd prime. Prove that
p−1
k=1
k 2p−1
≡
p(p + 1)
2
(mod p2 ).
[Note that a ≡ b (mod m) means that a − b is divisible by m.]
5. Let T be the set of all positive integer divisors of 2004 100. What is the largest possible
number of elements that a subset S of T can have if no element of S is an integer
multiple of any other element of S ?
37th Canadian Mathematical Olympiad
Wednesday, March 30, 2005
1. Consi
Consider
der an equilate
equilateral
ral triangl
trianglee of side length n, which is divided into unit triangles, as shown. Let
f (n) be the number of paths from the triangle in the top row to the middle triangle in the bottom
row, such that adjacent triangles in our path share a common edge and the path never travels up
(from a lower row to a higher row) or revisits a triangle. An example of one such path is illustrated
below for n = 5. Determine the value of f (2005).
2. Let (a,b,c
(a,b,c)) be a Pythagorean triple, i.e., a triplet of positive integers with a 2 + b2 = c2 .
a) Prove that (c/a
(c/a + c/b
c/b))2 > 8.
b) Pro
Prove
ve that ther
theree does not exist any int
integer
eger n for which we can find a Pythagorean triple (a,b,c
(a,b,c))
2
satisfying (c/a
(c/a + c/b
c/b)) = n.
n.
3. Let S be a set of n
≥
3 points in the interior of a circle.
a) Show that there are three distinct points a,b,c ∈ S and three distinct points A,B,C on the
circle such that a is (strictly) closer to A than any other point in S , b is closer to B than any
other point in S and c is closer to C than any other point in S .
b) Show that for no value of n can four such points in S (and corresponding points on the circle)
be guaranteed.
4. Let ABC be a triangle with circumradius R, perimeter P and area K . Det
Determ
ermine
ine the maxi
maximu
mum
m
3
value of KP/R .
5. Let’
Let’ss say that an order
ordered
ed triple of positiv
positivee intege
integers
rs ((a,b,c
a,b,c)) is nn -powerful if a ≤ b ≤ c, gcd(a,b,c
gcd(a,b,c)) = 1,
and a + b + c is divisible by a + b + c. For example, (1,
(1, 2, 2) is 5-powerful.
n
n
n
a) Dete
Determine
rmine all ordered trip
triples
les (if any) which are n-powerful
n -powerful for all n ≥ 1.
b) Determine all ordered triples (if any
any)) which are 2004-powe
2004-powerful
rful and 2005-powerful, but not 2007powerful.
[Note that gcd(a,b,c
gcd(a,b,c)) is the greatest common divisor of a,
a , b and c.]
c .]
39th Canadian Mathematical Olympiad
Wednesday, March 28,2007
111000001110
1. What is the maximum
maximum number of non-over
non-overlapp
lapping
ing 2 1 dominoes that can be placed on a 8 9 checkerboard if six of
them are placed as shown? Each domino must be placed horizontally or vertically so as to cover two adjacent squares of
the board.
×
×
2. You are given a pair of tria
triangles
ngles for which
which
(a) two
two sides of one triangle
triangle are equal in length to two sides of the secon
second
d triangle,
triangle, and
(b) the triangles
triangles are similar,
similar, but not necessar
necessarily
ily congrue
congruent.
nt.
Prove that the ratio of the sides that correspond under the similarity is a number between
1
√ −
√
1) and ( 5 − 1).
( 5
2
1
2
3. Suppose that f is a real-valued function for which
f (xy) + f (y
− x) ≥ f (y + x)
for all real numbers x and y .
(a) Give a nonconstant polynomi
polynomial
al that satisfies the condition.
(b) Prove
Prove that f (x) 0 for all real x .
≥
4. For two real numbers
numbers a , b , with ab = 1, define the
∗ operation by
a + b − 2ab
a∗b=
.
1 − ab
Start with a list of n 2 real numbers whose entries x all satisfy 0 < x < 1. Select any two numbers a and b in the list;
remove them and put the number a b at the end of the list
list,, thereby redu
reducing
cing its length by one. Repeat this procedure
procedure
until a single number remains.
≥
∗
(a) Prov
Provee that this single number
number is the same regardl
regardless
ess of the choic
choicee of pair at each stage.
(b) Suppose that the condition
condition on the num
numbers
bers x in S is weakened to 0 < x
1?
≤ 1. What happens if S contains exactly one
ABC
C touch sides B
BC
C , C A and AB at D, E and F , respectively. Let Γ,Γ1 ,Γ2 and Γ3 denote the
5. Let the incircle
incircle of triangle
triangle AB
circumcircles of triangle AB
ABC
C , AE
AEF
F , B DF and C DE respectively. Let Γ and Γ 1 intersect at A and P , Γ and Γ2 intersect
at B and Q , and Γ and Γ3 intersect at C and R .
(a) Prov
Provee that the circles
circles Γ 1 , Γ2 and Γ3 intersect in a common point.
(b) Show that P D, QE and RF are concurrent.
40th Canadian Mathematical Olympiad
Wednesday, March 26, 2008
1. ABCD is a convex quadrilateral for which AB is the longest side. Poin
Points
ts M and N are located on
sides AB and BC respectively, so that each of the segments AN and C M divides the quadrilateral
into two parts of equal area. Prove that the segment M N bisects the diagonal BD .
2. Dete
Determine
rmine all funct
functions
ions f defined on the set of rational numbers that take rational values for which
f (2f (x) + f (y )) = 2x + y ,
for each x and y .
3. Let a, b, c be positive real numbers for which a + b + c = 1. Prove that
a − bc
b − ca
c − ab
3
≤
.
+
+
a + bc
b + ca
c + ab
2
4. Dete
Determine
rmine al
alll funct
functions
ions f defined on the natural numbers that take v
values
alues among the natural numbers
for which
mod f (p)
(f (n))p ≡ n
for all n ∈ N and all prime numbers p.
5. A self-avoiding rook walk on a chessboard (a rectangular grid of unit squares) is a path traced by
a sequence of moves parallel to an edge of the board from one unit square to another, such that
each begins where the previous move ended and such that no move ever crosses a square that has
previously been crossed, i.e., the rook’s path is non-se
non-self-in
lf-interse
tersecting
cting..
Let R(m, n) be the number of self-avoiding rook walks on an m × n ( m rows, n columns) chessboard
which begin at the lower-left corner and end at the upper-left corner. For example, R(m, 1) = 1 for
all natural numbers m; R(2, 2) = 2; R(3, 2) = 4; R(3, 3) = 11. Fin
Find
d a for
formu
mula
la for R(3, n) for each
natural number n.
41st Canadian Mathematical Olympiad
Wednesday, March 25, 2009
1. Given an m × n grid with squares coloured either black or white, we say
that a black square in the grid is stranded if there is some square to its left in the same
row that is white and there is some square above it in the same column that is white (see
Figure 1.).
Problem
Figure
Figure 1.
A 4 × 5 grid with no stranded black squares
Find a closed formula for the number of 2 × n grids with no stranded black squares.
2. Two circles of different radii are cut out of cardboard. Each circle is subdivided
vid
ed into 200 equa
equall sect
sectors.
ors. On each circl
circlee 100 sector
sectorss are pain
painted
ted white and the othe
otherr
100 are painted black. The smaller circle is then placed on top of the larger circle, so that
Problem
their cen
their
centers
ters coin
coincid
cide.
e. Sho
Show
w that one can rota
rotate
te the sma
small
ll circl
circlee so that the sec
sectors
tors on
the two circles line up and at least 100 sectors on the small circle lie over sectors of the
same color on the big circle.
Problem
3. Define
(xy + yz + zx )(x + y + z )
.
(x + y )(x + z )(y + z )
Determine the set of real numbers r for which there exists a triplet (x,y ,z ) of positive
real num
numbers
bers satisf
satisfying
ying f (x,y ,z) = r .
f (x,y ,z ) =
a
b
4. Find all ordered pairs (a, b) such that a and b are integers and 3 + 7 is a
perfect square.
Problem
Problem
5. A set of points is marked on the plane, with the property that any three
marked points can be cove
marked
covered
red with a dis
disk
k of radi
radius
us 1. Pro
Prove
ve that the set of all marke
marked
d
points can be covered with a disk of radius 1.
nd
42
Canadian Mathematical Olympiad
Wednesday, March 24, 2010
(1) For a positive integer n, an n-staircase is a figure consisting of unit squares, with one
square in the first row, two squares in the second row, and so on, up to n squares in
the nth row
row,, such
such that all the left-most
left-most squares
squares in each row are aligned vertically
vertically.. For
example, the 5-staircase is shown below.
Let f (n) denote the minimum number of square tiles required to tile the n
n-staircase,
-staircase,
where
whe
re the side
side lengths
lengths of the square
square tiles
tiles can be an
any
y positiv
positivee in
integ
teger.
er. For example,
example,
f (2) = 3 and f (4) = 7.
n..
(a) Find all n such that f (n) = n
(b) Find all n such that f (n) = n + 1.
(2) Let A
A,, B,P be three points
points on a cir
circle
cle.. Prove
Prove that
that if a and b are the distances from P
to the tangents at A and B and c is the distance from P to the chord AB
AB,, then c 2 = ab
ab..
(3) Three
Three speed
speed skate
skaters
rs have a friend
friendly
ly “rac
“race”
e” on a skati
skating
ng ov
oval.
al. They
They all start from the
same point and skate
skate in the same direction,
direction, but with different
different speeds that they maintain
throughout
through
out the race. The slowest
slowest skater
skater does 1 lap a minute,
minute, the fastest one does 3.
3 .14
laps a minute, and the middle one does L laps a minute for some 1 < L < 3.
3 .14. The race
ends at the moment when all three skaters again come together to the same point on
the oval
oval (which may differ from the starting
starting point.) Find how many different
different choices
choices for
L are there such that exactly 117 passings occur before the end of the race. (A passing
is defined
defined when
when one skate
skaterr passes
passes another
another one. The beginning
beginning and the end of the race
race
when all three skaters are together are not counted as passings.)
(4) Each vertex of a finite graph can be coloured either black or white. Initially all vertices
are black.
black. We are allowe
allowed
d to pick a ve
verte
rtex
x P and change
change the colour
colour of P and all of its
1
neighbours. Is it possible to change the colour of every vertex from black to white by a
sequence of operations of this type?
(A finite graph consists of a finite set of vertices and a finite set of edges between
vertices. If there is an edge between vertex A and vertex B, then B is called a neighbour
of A.)
(5) Let P (x) and Q(
Q (x) be polynomials with integer coefficients. Let a = n
n!! + n. Show that
if P (a )/Q
/Q((a ) is an integer for every n
n,, then P (n)/Q
/Q((n) is an integer for every integer
n such that Q(
Q (n)
= 0.
n
n
n
2
43rd Canadian Mathematical Olympiad
Wednesday, March 23, 2011
(1) Consider 70-digit numbers n , with the property that each of the digits 1 , 2, 3, . . . , 7
appears in the decimal expansion of n ten times (and 8, 9, and 0 do not appear).
Show that no number of this form can divide another number of this form.
(2) Let ABCD be a cyclic quadrilateral whose opposite sides are not parallel, X the
intersection of AB and C D, and Y the intersection of AD and BC . Let tth
he
AXD
D intersect AD,BC at E, F respectively and let the angle
angle bisector of ∠AX
bisector of ∠AY B intersect AB,CD at G, H respective
respectively
ly.. Prov
Provee that EGFH is
a paralle
parallelogram.
logram.
(3) Amy has divided a square up into finitely many white and red rectangles, each
with
wit
h sid
sides
es parall
parallel
el to the sides of the squ
square.
are. Wit
Within
hin each whit
whitee rect
rectangl
angle,
e, she
writes down its width divided by its height. Within each red rectangle, she writes
down its height divided
divided by its width. Finall
Finally
y, she calcula
calculates
tes x, the sum of these
numbers. If the total area of the white rectangles equals the total area of the red
rectangles, what is the smallest possible value of x?
(4) Show that there exists a positive integer N such that for all integers a > N , there
exists a contiguous substring of the decimal expansion of a that is divisible by 2011.
(For instance, if a = 153204, then 15, 532, and 0 are all contiguous substrings of
a. Note that 0 is divisible by 2011.)
(5) Let d be a positive integer. Show that for every integer S , there exists an integer
n > 0 and a sequence 1 , 2 , . . . , n , where for any k , k = 1 or k = − 1, such that
S = 1 (1 + d)2 + 2 (1 + 2d)2 + 3 (1 + 3d)2 + · · · + n (1 + nd)2 .
1
44th CANADIAN MATHEMATICAL OLYMPIAD
44ème OLYMPIADE MATHÉMATIQUE DU CANADA
1. Let , and be positive real numbers. Show that
4 4.
Soit , et trois nombres réels positifs. Démontrez que 4 4.
2. For aany
ny positive
positive integers and , let ,
, be the least common multiple of the consecutive integers
, 1, …., 1. Show that for any integer , there exist integers and such that
, 1 , .
Soit
,, le plus petit commun multiple de la suite des k entiers consécutifs , 1 , … . , 1 ,
où et sont deux entiers positifs quelconques. Montrez que pour tout entier , il existe des nombres entiers
et tels que ,
1 ,
.
3. Let be a convex quadrilateral and let be the point of intersection of and . Suppose that
. Prove that the internal angle bisectors of
,,
,, and meet at a common point.
Soit un quadrilatère convexe et le point d’intersection des droites et . Supposons que
. Démontrez que les bissectrices des angles internes de
,,
,, et se coupent
en un point.
4. A number of robots are place
placed
d on the squares of a finite, rectangular grid of squa
squares.
res. A square can hold any number of
robots. Every edge of the grid is
i s classified as either passable or impassable
impassable.. All edges on the boundary of the grid are
impassable.
You can give any of the commands up
up,, down
down,, left, or right. All of the robots then simultaneously try to move in the
specified direction. If the edge adjacent to a robot in that direction is passable, the robot moves across the edge and
into the next square. Otherwise, the robot remains on its current square. You can then give another command of up
up,,
down, left, or right, then another, for as long as you want.
down,
Suppose that for any individual robot, and any square on the grid, there is a finite sequence of commands that will
move that robot to that square. Prove that you can also give a finite sequence of commands such that all of the robots
end up on the same square at the
t he same time.
Un certain nombre de robots sont placés sur les carrés composant une grille rectangulaire de dimension finie. Chaque
carré peut contenir un nombre quelconque de robots. Les bords des carrés de la grille sont classés comme
franchissablee ou infranchissable
franchissabl
infranchissable.. Les côtés qui forment le pourtour de la grille sont infranchissables.
Vous pouvez donner n’importe laquelle des commandes suivantes : en haut, en bas
bas,, à gauche ou à droite
droite.. Tous les
robots tentent alors de se déplacer simultanément dans la direction précisée. Si le bord adjacent au carré vers lequel se
déplace un robot est franchissable, le robot le franchit et se place dans le carré suivant. Sinon, le robot reste dans le
carré où il se trouve. Vous pouvez ensuite lancer une autre commande de déplacement vers le haut, le bas, la gauche
ou la droite
l e désirez.
droite,, et encore une autre, aussi longtemps que vous le
Supposons que pour chaque robot, et ce, pour n’importe quel carré, il
i l existe une suite finie de commandes qui
amèneront
robotfiniront
au carrépar
donné.
Démontrez
que vous pouvez
lancer
une suite finie de commandes de sorte
que
tous lescerobots
se retrouver
simultanément
dans leaussi
même
carré.
44th CANADIAN MATHEMATICAL OLYMPIAD
44ème OLYMPIADE MATHÉMATIQUE DU CANADA
5. A bookshelf contains n volumes,
volumes, labelled 1 to in some order. The librarian wishes to put them in the correct order as
follows. The librarian selects a volume that is too far to the right, say the volume with label , takes it out, and inserts
it so that it is in the -th place. For example, if the boo
bookshelf
kshelf contains the volumes 1, 3, 2, 4 in that order, the
librarian could take out volume 2 and place it in the second position. The books will then be in the correct order
1, 2, 3, 4.
(a) Show that if this process is repeated, then, however the librarian makes the selections, all the volumes will
eventually be in the correct order.
(b) What is the largest number of steps that this process can take?
Une étagère contient n volumes étiquetés de 1 à , rangés dans un certain ordre. Le bibliothécaire souhaite les mettre
dans le bon ordre de la façon suivante : il choisit un volume qui se trouve trop à droite, par exemple le volume étiqueté
, le retire de son emplacement et l’insère à la –ième place. Par exemple, si les volumes sont rangés dans l’ordre
1, 3, 2, 4, le bibliothécaire peut prendre le volume 2 et le mettre à la deuxième place. Les livres sont alors rangés dans
le bon ordre, soit 1, 2, 3, 4.
a) Démontrez que si l’on répète ce processus, tous les volumes finiront par être dans le bon ordre, et ce, qu’elle que
soit la manière dont le bibliothécaire les range.
b) Quel est le plus grand nombre d’étapes que peut exiger un tel processus?
45th Canadian Mathematical Olympiad
Wednesday, March 27, 2013
1. Determine all polynomials P (x) with real coefficients such that
(x + 1)P (x − 1) − (x − 1)P (x)
is a constant polynomial.
2. The sequence a1 , a2 , . . . , an consists of the numbers 1, 2, . . . , n in some order. For
which positive integers n is it possible that the n +1 numbers 0, a1 , a1 + a2, a 1 + a2 + a3 ,
. . ., a1 + a2 + · · · + an all have different remainders when divided by n + 1?
3. Let G be the centroid of a right-angled triangle ABC with ∠BC A = 90 . Let P
be the point on ray AG such that ∠C P A = ∠C AB , and let Q be the point on ray
BG such that ∠C QB = ∠ABC . Prove that the circumcircles of triangles AQG and
BP G meet at a point on side AB .
◦
4. Let n be a positiv
positivee integer. For any positive integ
integer
er j and positive real number
r , define fj (r) and gj (r ) by
fj (r) = min(jr, n) + min
j
,n ,
r
and gj (r) = min(jr , n) + min
j
,n ,
r
where x denotes the smallest integer greater than or equal to x. Prove that
n
n
j =1
2
fj (r ) ≤ n + n
≤
gj (r )
j =1
for all positive real numbers r.
5. Let O denote the circumcentre of an acute-angled triangle ABC . Let point P on
BOP
P = ∠ABC , and let point Q on side AC be such that
side AB be such that ∠BO
= ∠AC
ACB
B . Prove that the reflection of BC in the line P Q is tangent to the
circumcircle of triangle AP Q.
∠C OQ
46th Canadian Mathematical Olympiad
Wednesday, April 2, 2014
1. Let a 1 , a2 , . . . , an be positive real numbers whose product is 1. Show that the sum
a1
a2
a3
an
+
+
+· · ·+
1 + a1 (1 + a1 )(1 + a2 ) (1 + a1 )(1 + a2 )(1 + a3 )
(1 + a1 )(1 + a2 ) · · · (1 + an )
2n − 1
is greater than or equal to
.
2n
2. Let m and n be odd positive integers. Each square of an m by n board is coloured
red or blue. A row is said to be redred-domi
dominate
nated
d if ther
theree are more red squares tha
than
n
blue squares
squares in the ro
row.
w. A column is said to be blu
blue-do
e-domin
minated
ated if there are more
blue squares than red squares in the column. Determi
Determine
ne the maxim
maximum
um possible valu
valuee
of the number of red-dominated rows plus the number of blue-dominated columns.
Express your answer in terms of m and n.
3. Let p be a fixed odd prime. A p-tuple (a1 , a2 , a3, . . . , ap ) of integers is said to be
good if
(i) 0 ≤ ai
≤p−1
for all i, and
(ii) a1 + a2 + a3 + · · · + ap is not divisible by p, and
(iii) a1 a2 + a2 a3 + a3 a4 + · · · + ap a1 is divisible by p.
Determine the number of good p-tuples.
4. The quadrilateral ABCD is inscribed in a circle. The point P lies in the interior
of ABCD , and ∠P AB = ∠P BC = ∠P C D = ∠P DA. The lines AD and BC meet
at Q, and the lines AB and C D meet at R. Prove that the lines P Q and P R form
the same angle as the diagonals of ABCD .
2.. A li
list
st ooff n integers is written in a row on a
5. Fix positive integers n and k ≥ 2
blackboard. You can choose a contiguous block of integers, and I will either add 1 to
all of them or subtract 1 from all of them. You can repeat this step as often as you
like, possibly adapting your selections based on what I do. Prove that after a finite
number of steps, you can reach a state where at least n − k + 2 of the numbers on
the blackboard are all simultaneously divisible by k .
2015 Canadian Mathematical Olympiad
[version of January 28, 2015]
Notation: If V and W are two points, then V W denotes the line segment
Notation:
with endpoints V and W as well as the length of this segment.
N
. . . be the set of positive
1. Let
1, 2, 3,on
positive integ
integers
ers.. Find
Find all1)funcfun
2 c<
tions f =
, defined
N and taking values in N, such that (n
2
f (n)f (f (n)) < n + n for every positive integer n.
{
}
−
2. Let ABC be an acute-angled triangle with altitudes AD, BE , and
2. Let ABC be an acute angled triangle with altitudes AD, BE , and
CF . Let H be the orthocentre, that is, the point where the altitudes
meet. Prove that
·
·
·
·
·
·
AB AC + BC BA + CA CB
AH AD + BH BE + CH C F
≤
2.
3. On a (4n + 2) (4 n + 2) square
square grid, a turtle
turtle can move
move between
between
squares
squar
es sharing a side. The turtle begins in a corner
corner square of the grid
and enters each square exactly once, ending in the square where she
started. In terms of n , what is the largest positive integer k such that
there must be a row or column that the turtle has entered at least k
distinct times?
×
4. Let ABC be an acute-angled triangle with circumcenter O. Let Γ be
a circle with centre on the altitude from A in ABC , passing through
vertex A and points P and Q on sides AB and AC . As
Assu
sume
me tha
thatt
BP CQ = AP AQ. Prove
Prove that Γ is tangent
tangent to the cir
circum
cumcir
circle
cle of
triangle BOC .
·
·
a,, b, c
5. Let p be a prime number for which p−2 1 is also pr
prime
ime,, and let a
be integers not divisible by p. Prove
Prove that
that there are at most 1 + 2p
positive integers n such that n < p and p divides an + bn + cn .
√
1
Problems for 2016 CMO – as of Feb 12, 2016
1. The integers 1 , 2, 3, . . . , 2016 are written
written on a board.
board. You can choose
any two numbers on the board and replace them with their average.
For example, you can replace 1 and 2 with 1 .5, or you can replace 1
and 3 with a second copy of 2. After 2015 replacemen
replacements
ts of this kind,
the board will have only one number left on it.
(a) Prove
Prove that there is a sequence of replacement
replacementss that will make
make the
final number equal to 2.
(b) Prove
Prove that there is a sequence of replacement
replacementss that will make
make the
final number equal to 1000.
2. Consid
Consider
er the followin
followingg system
system of 10 equati
equations
ons in 10 rea
reall vari
ariabl
ables
es
v1 , . . . , v10:
vi = 1 +
6 vi2
2
v12 + v22 + · · · + v10
(i = 1, . . . , 10).
Find all 10-tuples (v1 , v2 , . . . , v10) that are solutions of this system.
3. Find all polynomials
polynomials P (x) with integer coefficients such that P (P (n) +
n) is a prime number for infinitely many integers n.
4. Lavaman
Lavaman versu
versuss the Flea. Let A, B , and F be positive integers, and
assume A < B < 2A. A flea is at the num
number
ber 0 on the numbe
numberr line.
The flea can move by jumping to the right by A or by B . Befo
Before
re
the flea starts jumping, Lavaman chooses finitely many intervals { m +
1, m + 2, . . . , m + A} consisting of A consecutive positive integers, and
placess lava at all of the integers
place
integers in the interv
intervals.
als. The inter
interv
vals must
be chosen so that:
(i ) any two distinct intervals are disjoint and not adjacent;
(ii ) there are at least F positive integers with no lava between any two
intervals; and
(iii ) no lava is placed at any integer less than F .
1
Prove that the smallest F for which the flea can jump over all the
inter
interv
vals and av
avoid
oid all the lava,
lava, regard
regardles
lesss of what
what Lav
Lavaman does,
does,
is F = (n − 1)A + B , wher
wheree n is the positiv
positivee in
integ
teger
er such
such tha
thatt
A
n+1
≤B−A <
A
.
n
5. Let ABC be an acute-angled triangle with altitudes AD and BE
meeting at H . Let M be the midpoint of segment AB , and suppose
DEM
M and ABH meet at points P and
that the circumcircles of DE
Q with P on the same side of CH as A. Pro
Prove
ve that
that the lines
lines ED ,
P H , and M Q all pass through a single point on the circumcircle of
ABC .
2
CMO 2017
https://cms.math.ca/Competitions/CMO/
Problem
Problem Set
2017 Canadian
Mathematical Olympiad
Official Problem Set
1. Let a
a,, b, and c be non-negative real numbers, no two of which are equal. Prove that
c2
b2
a2
> 2.
2.
+
+
(b − c)2 (c − a)2 (a − b)2
2. Let f be a function from the set of positive integers to itself such that, for every n, the
number of positive integer divisors of n is equal to f (f (n)). For example,
example, f (f (6)) = 4 and
f (f (25)) = 3. Prove that if p is prime then f (p
p)) is also prime.
3. Let n be a positive integer, and define S = { 1, 2, . . . , n}. Consider a non-empty subset T of
S . We say that T is balanced if the median of T is equal to the average of T . For example,
for n = 9, each of the subsets {7}, {2, 5}, {2, 3, 4}, {5, 6, 8, 9}, and {1, 4, 5, 7, 8} is balanced;
n
n
however, the subsets {2, 4, 5} and {1, 2, 3, 5} ar
aree not balanced.
balanced. For each n ≥ 1, prove that
the number of balanced subsets of S is odd.
(To define the median of a set of k numbers, first put the numbers in increasing order; then
the median is the middle number if k is odd, and the average of the two middle numbers if
k is even. For example, the median of {1, 3, 4, 8, 9} is 4, and the median of {1, 3, 4, 7, 8, 9} is
(4 + 7)/
7)/2 = 5.5.)
n
4. Points
Points P and Q lie inside parallelogram ABCD and are such that triangles AB P and B CQ
are equilateral.
equilateral. Prove
Prove that the line through P perpendicular to DP and the line through Q
perpendicular
perpendi
cular to DQ
D Q meet on the altitude from B in triangle ABC .
5. One hundred
hundred circles
circles of radius one are positioned in the plane so that the area of any triangle
triangle
formed
for
med by the cent
centres
res of three of these circles
circles is at most 2017. Prove
Prove that there
there is a lin
linee
intersecting at least three of these circles.
c 2017 Canadian Mathematical Society
Page 1/ 1
CMO 1996
SOLUTIONS
QUESTION 1
Solution .
If f (x) = x3
x
1 = (x
− −
α)(x
−
β )(x
−
γ ) has roots α
α,β
,β , γ standard results about roots of
−
polynomials give α + β + γ = 0, αβ + αγ + βγ =
Then
S=
−1, and αβγ = 11..
1+α 1 +β 1 +γ
+
+
=
1 α 1 β 1 γ
(1
−
−
N
−
− α)(1 − β)(1 − γ)
where the numerator simplifies to
− (α + β + γ ) − (αβ + αγ + βγ
β γ ) + 3 αβγ
= 3 − (0) − (−1) + 3(1)
N =3
= 7.
The denominator is f (1) =
1 so the required sum is
−
7.
−
QUESTION 2
Solution 1.
2
For any t, 0
than 1.
2
≤ 4t
< 1 + 4t2 , so 0
≤ 1 +4t4t
2
< 1. Thus x, y and z must be non-negative and less
Observe that if one of x y or z is 0, then x = y = z = 0.
If two of the variables are equal, say x = y , then the first equation becomes
4x2
= x.
1 + 4 x2
1
1
which gives x = y = z = .
2
2
This has the solution x = 0, which gives x = y = z = 0 and x =
Finally, assume
Finally,
assume that x, y and z are non-zero
non-zero and distinc
distinct.
t. Without
Without loss of generalit
generality
y we may
assume that either 0 < x < y < z < 1 or 0 < x < z < y < 1. The two proofs are similar, so we do
only the first case.
We will need the fact that f (t) =
4t2
is increasing on the interval (0, 1).
1 + 4 t2
To prove this, if 0 < s < t < 1 then
2
2
f (t)
− f (s) = 1 +4t4t − 1 +4s4s
4t − 4s
=
2
2
2
2
(1 + 4s2 )(1 + 4t2 )
> 0.
So 0 < x < y < z
⇒ f (x) = y < f (y) = z < f (z) = x, a contradiction.
Hence x = y = z = 0 and x = y = z =
1
are the only real solutions.
2
Solution 2.
Notice that x, y and z are non-negative. Adding the three equations gives
x+y+z=
4y2
4x2
4z 2
+
+
.
1 + 4 z 2 1 + 4 x2 1 + 4 y 2
This can be rearranged to give
x(2x
− 1)
1 + 4 x2
2
+
y (2y
2
− 1)
1 + 4y2
+
z (2z
− 1)
1 + 4z2
2
= 0.
1
Since each term is non-negative, each term must be 0, and hence each
1 variable is either 0 or 2 . The
original equations then show that x = y = z = 0 and x = y = z = are the only two solutions.
2
2
Solution 3.
Notice that x, y, and z are non-negative. Multiply both sides of the inequality
y
1 + 4 y2
by (2y
≥0
2
− 1) , and rearrange to obtain
y
≥
and hence that y z . Similarly
Similarly,, z
two solutions follow.
−
4y 2
1 + 4 y2
≥ x, and x ≥ y.
≥ 0,
Hence,
Hence, x = y = z and, as in Solution 1, the
Solution 4.
As for solution 1, note that x = y = z = 0 is a solution and any other solution will have each of
x, y and z positive.
√
1 + 4x2
1 4x2 = 2x
2
1
4x2
2
=
,
with
equality
if
and
only
if
1
=
4
–
that
is,
=
and hence x
. Similarly, y z
y
x
x
1 + 4 x2
2
1
1
with equality if and only if y = and z x with equality if and only if z = . Adding x y, y z
2
2
and z x gives x + y + x x + y + z . Thus equality must occur in each inequality, so x = y = z = 1 .
2
The arithmetic-geometric mean inequality (or direct computation) shows that
≥
≥
≥
·
≥
≥
≥
≥
3
QUESTION 3
≥
Solution.
Let a1 , a2 , . . . , an be a permutation of 1, 2, . . . , n with properties (i) and (ii).
A crucial observ
observation, needed
needed in Case II (b) is the following:
following: If a k and a k+1 are consecutive integers
(i.e. ak+1 = a k 1), then the terms to the right of ak+1 (also to the left of ak ) are either all less
than both a k and ak+1 or all greater than both ak and ak+1 .
±
Since a1 = 1, by (ii) a2 is either 2 or 3.
CASE I: Suppose a2 = 2. Then a3 , a4 , . . . , an is a permutation of 3, 4, . . . , n. Thus a2 , a3 , . . . , an
is a permutation of 2, 3, . . . , n with a2 = 2 an
and
d pr
prope
opert
rty
y (i
(ii).
i). Clea
Clearl
rly
y th
ther
eree are
are f (n 1) such
−
permutations.
CASE II: Suppose a2 = 3.
3.
(a) Suppo
Suppose
se a3 = 2. Then a4 , a5 , . . . , an is a permutation of 4 , 5, . . . , n with a4 = 4 and property
(ii). There are f (n 3) such permutations.
−
(b)
(b) Suppo
Suppose
se a3
≥ 4.4.
If ak+1 is the first even number in the permutation then, because of (ii),
a1 , a2 , . . . , ak must be 1, 3, 5, . . . , 2k 1 (in that order). Then ak+1 is either 2k or 2k 2, so
−
−
ak that
and aakk+1
are consecutive integers. Applying the crucial observation made above, we
that
deduce
+2 , . . . , an are all either greater than or smaller than a k and a k+1 . But 2 must
be to the right of ak+1 . Henc
Hencee ak+2 , . . . , an are the even integers less than ak+1 . The
The only
only
possibility then, is
1, 3, 5, . . . , ak
1
−
, ak , . . . , 6, 4, 2.
Cases I and II show that
f (n) = f (n
− 1) + f (n − 3) + 1, n ≥ 4.
∗
( )
Calculating the first few values of f (n) directly gives
f (1) = 1, f (2) = 1, f (3) = 2, f (4) = 4, f (5) = 6.
Calculating a few more f (n)’s using (*) and mod 3 arithmetic, f (1) = 1, f (2) = 1, f (3) =
2, f (4) = 1, f (5) = 0, f (6) = 0, f (7) = 2, f (8) = 0, f (9) = 1, f (10) = 1, f (11
(11)) = 2. Si
Sinc
ncee
f (1) = f (9), f (2) = f (10) and f (3) = f (11) mod 3, (*) shows that f (a) = f (a mod 8) , mod 3, a
1.
≥
Hence f (1996)
≡ f (4) ≡ 1
(mod 3) so 3 does not divid
dividee f (1996).
4
QUESTION 4
Solution 1.
Let BE = B
BD
D with E on B C , so that AD = E C :
A
D
4x
2x
x
x
4x
2x
B
C
E
By a standard theorem,
CE D
and
CAB
AB
AD
=
;
CB
DC
so in
we have
have a common angle and
CE = AD = AB = CA .
CD
CD
CB
CB
Thus
CE D
Hence ∠BDE =
Thus ∠A = 180
◦
∼ CAB ,
so that ∠ C DE = ∠ DC E = ∠ ABC = 2x .
∠ BE
BED
D = 4x , whence 9x = 180 so x = 20 .
4x = 100 .
−
◦
◦
◦
Solution 2.
Apply the law of sines to
ABD and BDC to get
sin x
AD
=
BD
sin4x
and
1+
sin3x
AD
BC
=
=
.
BD
BD
sin2x
Now massage the resulting trigonometric equation with standard identities to get
sin2x (sin4x + sin x) = sin
sin 2x (sin5x + sin x) .
Since 0 < 2 x < 90 , we get
◦
5x
− 90
◦
= 90
◦
− 4x ,
so that ∠A = 100 .
◦
5
QUESTION 5
Solution.
Let
m
[r n]
f (n) = n −
k
k=1
m
m
=n
r − [r n]
{r n − [r n]}.
=
k
k
k=1
k=1
m
k
k
k=1
Now 0
≤ x − [x] < 1, and if c is an integer, (c + x) − [c + x] = x − [x].
m
1 = m. Because f (n) is an integer, 0 ≤ f (n) ≤ m − 1.
Hence 0 ≤ f (n) <
k=1
To show that f (n) can achieve these bounds for n > 0, we assume that rk =
integers; ak < bk .
Then, if n = b 1 b2 . . . bm , (rk n) [rk n] = 0 , k = 1, 2, . . . , m and thus f (n) = 0.
Letting n = b 1 b2 . . . bn 1, then
−
−
− 1)
= r {(b b . . . b − b ) + b − 1)}
= integer + r (b − 1).
rk n = r k (b1 b2 . . . bm
k
m
1 2
k
k
This gives
rk n
k
k
− [r n] = r (b − 1) − [r (b − 1)]
a
a
= (b − 1) −
(b − 1)
b
b
k
k
k
k
k
k
k
k
k
k
k
= ak
k
a
f (n) =
k
k
k
− b − (a − 1)
k
k
− ab
k
=1
k
Hence
k
k
= ak
=1
− ab − a − ab
m
k=1
(1
−r .
k
− r ) = m − 1.
6
1997
k
ak
where ak , bk are
bk
SOLUTIONS
Problem 1 – Deepee Khosla, Lisgar Collegiate Institute, Ottawa, ON
Let p 1 , . . . , p12 denote, in increasing order, the primes from 7 to 47. Then
0
5! = 23 · 31 · 51 · p01 · p02 . . . p12
and
b
50! = 2a · 3a · 5a · pb1 · p2b . . . p12
.
1
2
1
3
2
12
Note that 24 , 32 , 52 , p1 , . . . , p12 all divide 50!, so all its prime powers differ from those of 5!
Since x, y|50!, they are of the form
n
x = 2n · 3n · . . . p12
1
m1
y = 2
15
2
m2
·3
m15
· ...p
12
.
Then max(n
max(ni , mi ) is the ith prime power in 50!
and min(n
min(ni , mi ) is the i th prime power in 5!
Since, by the above note, the prime powers for p12 and under
under differ
differ in 5! and 50!, there
there are 215
choices for x, only half of which will be less than y . (Si
(Since
nce for each
each choice
choice of x, y is forced and
15
14
either x < y or y < x.) So the number of pairs is 2 /2 = 2 .
Problem 2 – Byung Kuy Chun, Harry Ainlay Composite High School, Edmonton, AB
Look at the first point of each
each given
given unit in
inter
terv
val. This
This point
point unique
uniquely
ly defines
defines the given
given unit
interval.
Lemma. In any interval [x,
[ x, x + 1) there must be at least one of these first points (0 ≤ x ≤ 49).
Proof. Suppose
Proof.
Suppose the opposit
opposite.
e. The last
last first point
point before x must be x − ε for some ε > 0.
0. Th
Thee
corresponding unit interval ends at x − ε + 1 < x + 1. Howev
However,
er, the next given
given unit interv
interval
al cannot
begin until at least x + 1.
This implies that points (x
( x − ε + 1,
1, x + 1) are not in set A, a contradiction.
∴
There must be a first point in [x,
[x, x + 1).
Note that for two first points in intervals [x,
[ x, x +1) and [x
[x + 2, x +3) respectively, the corresponding
unit intervals are disjoint since the intervals are in the range [x,
[ x, x +2) and [x
[x + 2, x +4) respectively.
∴
We can choose a given unit interval that begins in each of
[0,
[0, 1)[2
1)[2,, 3) . . . [2k,
[2k, 2k + 1) . . . [48
[48,, 49)
49)..
Since there are 25 of these intervals, we can find 25 points which correspond to 25 disjoint unit
intervals.
Problem 2 – Colin Percival, Burnaby Central Secondary School, Burnaby, BC
I prove the more general result, that if [0,
[0, 2n] =
that A ai
Ai , |Ai | = 1, Ai are intervals then ∃a1 . . . an , such
i
Aaj = ∅ .
Let 0 < ε ≤
2
and let bi = (i − 1)(2 + ε), i = 1 . . . n.
n. Then
n−1
2
( n − 1) 2 +
min{bi } = 0, max{bi } = (n − 1)(2 + ε) ≤ (n
n−1
= (n − 1)
2n
n−1
= 2n.
So all the b i are in [0,
[0, 2n].
Let a i be such that b i ∈ Aai . Since
Ai = [0,
[0, 2n], this is possible.
j )(2 + ε)
ε) ≥ 2 + ε > 2, and the Ai are intervals of length 1, min Aai −
Then since (b
(bi − bj ) = (i − j)(2
max Aaj > 2 − 1 − 1 = 0, so A ai Aaj = ∅ .
Substituting n = 25, we get the required result. Q.E.D.
2
Problem 3 – Mihaela Enachescu,
Enachescu, Dawson College, Montr´
Montré
eal,
al, PQ
Let P =
...
1 3
1997
1
1
3
3
· · ... ·
. Then > because 2 < 3, > because 4 < 5,
5 , . . .,
.,
2 4
1998
2
3
4
5
1997
1997
>
because 1998 < 1999 .
1998
1999
So
P >
1 3
1997
1
· · ... ·
=
.
3 5
1999
1999
Also
1
2
3
4
< because 1 · 3 < 2 · 2, < because 3 · 5 < 4 · 4, . . .
2
3
4
5
(1)
1997
1998
<
because 1997 · 1999 = 19982 − 1 < 19982 .
1998
1999
2 4
1998
So P < · · . . . ·
=
3 5
1999
2 4 6
1998
· · · ... ·
1 3 5
1997
1
P
1
.
1999
1
1
1
1
<
= 2 and P <
.
1999
1936
44
44
Hence P 2 <
Then (1) and (2) give
1
<P <
1
(2)
(q.e.d.)
44
1999
Problem 4 – Joel Kamnitzer, Earl Haig Secondary School, North York, ON
O’
A
B
O
D
C
Consider a translation which maps D to A. It wil
willl map
map 0 → 0 with OO = DA
DA,, and C will be
mapped to B because CB = DA
DA..
◦
This translation keeps angles invariant, so ∠AO B = ∠DOC = 180 − ∠AOB
AOB..
∴
AOBO is a cyclic quadrilateral.
∴ ∠ODC
= ∠O AB = ∠O OB
but, since O O is parallel to B C ,
∠O
OB =
∴ ∠ODC
=
∠OBC
∠OBC.
3
Problem 4 – Adrian Chan, Upper Canada College, Toronto, ON
A
B
θ
180−α
α
Ο
180−θ
D
C
◦
◦
Let ∠AOB = θ and ∠BOC
BO C = α.
α . Then ∠COD
CO D = 180 − θ and ∠AOD = 180 − α.
◦
Since AB = CD (parallelogram) and sin θ = sin(180 − θ),
θ ), the sine law on OC
OCD
D and OAB
gives
sin ∠CD
CDO
O
sin(180 − θ)
sin θ
sin ∠ABO
=
=
=
OC
CD
AB
OA
so
OA
sin ∠ABO
=
.
(1)
OC
sin ∠CD
CDO
O
◦
Similarily, the sine law on OB
OBC
C and OAD gives
sin ∠CB O
OC
so
=
sin α
◦
=
sin(180 − α)
=
BC
AD
OA
sin ∠ADO
=
.
OC
sin ∠CB O
sin ∠ADO
OA
(2)
Equations (1) and (2) show that sin ∠ABO · sin ∠CB O = sin ∠ADO · sin ∠CD
CDO
O hence
1
1
[cos(∠ABO + ∠CB O) − cos(∠ABO − ∠CB O)] = [cos(∠ADO + ∠CD
CDO
O) − cos(∠ADO − ∠C DO
DO)]
)]..
2
2
Since ∠ADC = ∠ABC (parallelogram) and ∠ADO + ∠CD
CDO
O = ∠ADC and
∠ABC it follows that cos(∠ABO − ∠CB O ) = cos(∠ADO − ∠CD
CDO
O).
∠ABO
+ ∠C BO =
There are two cases to consider.
∠
∠
∠
∠
Case (i): ABO − CB O = ADO − CD
CDO
O.
Since ∠ABO + ∠ CB O = ∠ADO + ∠CD
CDO
O, subtracting gives 2 ∠CB O = 2 ∠CDO
CD O
∠CDO, and we
we are done.
Case (ii):
∠ABO
so
∠C BO
=
∠CD
CDO
O
so
− ∠CB O = ∠CDO
CD O − ∠ADO
ADO..
Since we know that ∠ABO + ∠ C BO =
∠ABO = ∠CD
CDO
O and ∠CB O = ∠ADO
ADO..
∠CD
CDO
O
+ ∠ ADO
ADO,, adding gives 2
∠ABO
=2
Substituting this into (1), it follows that OA
O A = OC .
Also, ∠ADO + ∠ABO = ∠CB O + ∠ABO = ∠ABC .
◦
Now, ∠ABC = 180 − ∠BAD since ABCD is a parallelogram.
◦
◦
Hence ∠BAD + ∠ADO + ∠ABO = 180 so ∠DOB = 180 and D,O ,B are collinear.
We now have the diagram
4
A
B
θ
O
180−θ
D
∠
∠
◦
C
∠
∠
Then CO
COD
D + BOC = 180 , so BO
BOC
C = θ = AOB
AOB..
AOB is congruent to C OB (SAS, OB is common, ∠AOB = ∠CO
COB
B and AO = C O), so
∠ABO = ∠CB O . Since also ∠ABO = ∠C DO we conclude that ∠CB O = ∠CD
CDO
O.
Since it is true in both cases, then
∠CB O
= ∠CDO
CD O.
Q.E.D.
5
Problem 5 – Sabin Cautis, Earl Haig Secondary School, North York, ON
We first note that
k 3 + 9k
9k2 + 26k
26k + 24 = (k
(k + 2)(k
2)(k + 3)(k
3)(k + 4).
4) .
(−1)k
n
Let S (n) =
Then
n
k
k2 + 9k
9k2 + 26k
26k + 24
k=0
.
n
S (n) =
(−1)k n!
k!(n
!(n − k )!(k
)!(k + 2)(k
2)(k + 3)(k
3)(k + 4)
k=0
n
=
k=0
(−1)k (n + 4)!
(k + 4)!(n
4)!(n − k)!
×
k+1
.
(n + 1)(n
1)(n + 2)(n
2)(n + 3)(n
3)(n + 4)
Let
n
n+4
(−1)k
T (n) = (n + 1)(n
1)(n + 2)(n
2)(n + 3)(n
3)(n + 4)S
4)S (n) =
k+4
k=0
Now, for n ≥ 1,
n
(−1)i
i=0
n
i
(k + 1) .
(∗)
=0
since
(1 − 1)n =
n
0
n
1
−
n
2
+
+ . . . + (−1)n
n
n
= 0.
Also
n
(−1)
i=0
i
n
i
n
i =
(−1)i
i · n!
0 · n!
+ (−1)0 ·
i! · (n − i)!
0! · n!
(−1)i
n!
(i − 1)!(
1)!(n
n − i)!
i=1
n
=
i=1
n
=
(−1)i n
i=1
n
(−1)i
= n
i=1
n
= −n
(−1)
i=1
6
Substituting j = i − 1, (*) shows that
n−1
i−1
n−1
i−1
i−1
n−1
i−1
.
n
(−1)i
i=0
n
i
n−1
i = −n
(−1)j
j =0
n−1
j
= 0.
(∗∗)
Hence
n
T (n) =
n+4
k+4
(−1)k
k=0
n
=
n+4
k+4
(−1)k+4
k=0
n
=
(−1)k+4
k=−4
Substituting j = k + 4,
4,
n+4
=
(−1)
n+4
j
j =0 (−1)
(k + 1)
n+4
k+4
n+4
j
(k + 1) − −3 + 2(n
2(n + 4) −
n+4
j
j − 3 j =0 (−1)
n+4
j
The first two terms are zero because of results (*) and (**) so
T (n) =
n2 + 3n
3n + 2
.
2
Then
S (n) =
=
=
=
(−1)k
n
∴
k=0
3
2
n
k
k + 9k
9k + 26k
26k + 24
=
n+4
2
(n + 4)(n
4)(n + 3)
(j − 3) − 2n + 8 − 3 −
2
n+4
j
j
j =0
=
(k + 1)
T (n)
(n + 1)(n
1)(n + 2)(n
2)(n + 3)(n
3)(n + 4)
n2 + 3n
3n + 2
2(n
2(n + 1)(n
1)(n + 2)(n
2)(n + 3)(n
3)(n + 4)
(n + 1)(n
1)(n + 2)
2(n
2(n + 1)(n
1)(n + 2)(n
2)(n + 3)(n
3)(n + 4)
1
.
2(n
2(n + 3)(n
3)(n + 4)
1
2(n
2(n + 3)(n
3)(n + 4)
7
1998
1
.
2
7n − 12)
(4n + 10 − n − 7n
− 2 (4n
SOLUTIONS
The solutions to the problems of the 1998 CMO presented below are taken from students papers.
Some minor editing has been done - unnecesary steps have been eliminated and some wording has
been changed to make the proofs clearer. But for the most part, the proofs are as submitted.
Solution to Problem 1 – David Arthur, Upper Canada College, Toronto, ON
Let a = 30k + r, where k is an integer and r is a real number between 0 and 29 inclusive.
− − −
− − −
− − −
1
1
1
1
r
r
r
Then
a =
(30k + r) = 15k +
. Similarly
a = 10k +
and
a = 6k +
.
2
2
2
3
3
5
5
Now,
1
1
1
r
a +
a +
a = a , so 15k +
2
3
5
2
hence k = r
r
r
r
2
3
5
r
+ 6k +
3
r
5
= 30k + r and
.
Clearly, r has to be an integer, or r
equal k .
+ 10k +
r
2
r
r will not be an integer, and therefore, cannot
3
5
On the other hand, if r is an integer, then r
exactly one solution for k.
r
r
r
2
3
5
will also be an integer, giving
≤ ≤
For each r (0 r 29), a = 30k + r will have a different remainder mod 30, so no two different
values of r give the same result for a.
Since there are 30 possible values for r(0, 1, 2, . . . , 29), there are then 30 solutions for a .
1
Solution to Problem 2 – Jimmy Chui, Earl Haig S.S., North York, ON
−
1
Since x
1/2
≥ 0 and
x
1
Note that x = 0. Else,
−
1
1
1/2
≥ 0, then 0
x
≤ −
−
1/2
1
x
1
+ 1
x
1/2
=x .
x
would not be defined so x > 0.
x
Squaring both sides gives,
1
2
1
1
1
− − − −
−
− −
x
=
x2
= x+1
x
x
+ 1
2
x
x
+2 x
+2
1
1
x
x
+
1
x
1
x2
x
.
Multiplying both sides by x and rearranging, we get
2
x3
3
(x
−x −x+2
− x − x + 1) − 2 x − x − x + 1 + 1
( x − x − x + 1 − 1)
2
3
2
= 0
−x −x+1
x −x −x+1
x(x − x − 1)
x −x−1
= 1
x
2
3
2
2
√5
±
Thus x =
2
√
1+ 5
Let α =
2
2
−x −x+1
2
3
1
= 0
2
2
3
= 2 x3
= 1
= 0
= 0
since x = 0 .
. We must
must chec
check
k to see
see if these
these are
are indeed
indeed solutions
solutions..
, β=
1
− √5 . No
Note
te that
that α + β = 1,
2
αβ =
−1 and α > 0 > β.
Since β < 0 , β is not a solution.
Now, if x = α , then
− −
α
1
α
1/2
+ 1
1
α
1/2
= (α + β )1/2 + (1 + β )1/2
(since αβ
= 11/2 + (β 2 )1/2
(since α + β = 1 and β 2 = β + 1)
= 1
−β
− −1)
(since β < 0)
= α
(since α + β = 1).
So x = α is the unique solution to the equation.
2
Solution 1 to Problem 3 – Chen He, Columbia International Collegiate, Hamilton, ON
1
1
1 1 1 1
1
+ ... +
= + + + +...
3
2n 1
2 2 3 5
2n 1
1+
−
(1)
−
Since
1
1 1
1
1
1
> ,
> , ... ,
>
,
3
4 5
6
2n 1
2n
−
(1) gives
1
1
1 1 1 1
1
1
1+ +...+
= +
> + + + +...+
3
2n 1
2 2 4 6
2n
2
−
1 1 1
1
+ + + ... +
.
2 4 6
2n
(2)
Since
1
1 1
1 1
1
1
1
> ,
> ,
> , ... ,
>
2
4 2
6 2
8
2
2n
then
n
2
so
1
=
2
+
1
2
+
1
+...+
>
2n
1
1
>
2
n
1
2
1
2
+
1
4
+
1
+ ...+
6
1
2n
1 1
1
1
+ + + ... +
.
2 4 6
2n
(3)
Then (1), (2) and (3) show
1
1
1+ + ... +
3
2n 1
−
>
=
=
1
1 1 1
1
+ + + ...+
+
4 6
2n
n 2
1
1 1
1
1+
+ + ... +
n
2 4
2n
n+1 1 1
1
+ + ... +
.
n
2 4
2n
1
1
1
Therefore
1+ + ... +
3
2n 1
n+1
−
>
1
n
1 1
1
+ +...+
2 4
2n
1 1 1
1
+ + +...+
2 4 6
2n
for all n
∈ N and n ≥ 2.
3
Solution 2 to Problem 3 – Yin Lei, Vincent Massey S.S., Windsor, ON
Since n
≥ 2,
n(n + 1)
≥ 0. Therefore the given inequality is equivalent to
1
1
n 1+ + ...+
3
2n 1
−
≥
1 1
1
(n + 1)
+ +...+
.
2 4
2n
We shall use mathematical induction to prove this.
1
1
For n = 2, obviously 3 1 + 3
4
1
= 9>2
1 1
2+ 4
3
= 8.
Suppose that the inequality stands for n = k , i.e.
1
1
k 1+ +...+
3
2k 1
−
1
1 1
+ +...+
.
> ( k + 1)
2 4
2k
(1)
Now we have to prove it for n = k + 1.
We know
−
− −− −
1
1
1+ + ...+
3
2k 1
1 1
1
+ + ... +
2 4
2k
1 + 1 1 + 1 1 + ... +
1
2
3 4
5 6
2k 1
1
1
1
1
=
+
+
+ ...+
.
1 2 3 4 5 6
(2k 1)(2k )
= 1
×
×
×
− − 21k
−
Since
1
× 2 < 3 × 4 < 5 × 6 < . . . < (2k − 1)(2k) < (2k + 1)(2k + 2)
then
1
1
2
×
hence
1+
+
1
3
4
+ ...+
1
k
>
1)(2k )
(2k + 1)(2k + 2)
(2k
×
−
1
1
1 1
1
k
+ ... +
> + + ... +
+
.
3
2k 1
2 4
2k (2k + 1)(2k + 2)
(2)
−
Also
k+1
2k + 1
−
k+2
2k2 + 2k + 2k + 2 2k 2 4k
=
2k + 2
(2k + 1)(2k + 2)
−
− −k−2 =−
k
(2k + 1)(2k + 2)
therefore
k+1
k+2
=
2k + 1
2k + 2
k
− (2k + 1)(2
.
k + 2)
4
Adding 1, 2 and 3:
(3)
−
−
1
1
1
1
k+1
k 1+ + ...+
+ 1+ +...+
+
3
2k 1
3
2k 1
2k + 1
> ( k + 1)
1 1
1
+ + ... +
2 4
2k
+
1 1
1
+ + ... +
2 4
2k
+
k
(2k + 1)(2k + 2)
+
k+2
2k + 2
k
− (2k + 1)(2
k + 2)
Rearrange both sides to get
(k + 1) 1 + 13 + . . . + 2k 1+ 1
> ( k + 2)
12 + 14 + . . . + 2k 1+ 2 .
Proving the induction.
5
Solution 1 to Problem 4 – Keon Choi, A.Y. Jackson S.S., North York, ON
Suppose H is the foot of the perpendicular line from A to B C ; construct equilateral ABG, with
C on BG . I will pro
prove
ve that
that if F is the point where AH meets BD , then F CB = 70◦ . (Be
(Becau
cause
se
that means AH , and the given lines B D and C E meet at one point and that proves the question.)
Suppose B D extended meets AG at I .
A
E
I
D
F
B
C
H
G
200◦ . Also IF G = F BG + F GB =
Now B
BF
F = GF and F BG = F GB = 40◦ so that IGF = 2
80◦ , so that
F IG
= 180◦
= 180◦
◦
− IF G −
− 80 − 20
◦
I GF
◦
= 80 .
Therefore
GIF is an isoceles triangle, so
GI = GF = B
BF.
F.
But
BGI and ABC are congruent, since BG = AB
AB,,
(1)
GBI
= BAC,
BGI
= ABC .
Therefore
GI = B
BC.
C.
(2)
From (1) and (2) we get
BC = B
BF.
F.
So in
BC F ,
BC F
=
180◦
−
F BC
2
=
180◦
◦
− 40
2
= 70◦ .
Thus F CB = 70◦ and that proves that the given lines C E and BD and the perpendicular line
AH meet at one point.
6
Solution 2 to Problem 4 – Adrian Birka, Lakeshore Catholic H.S., Port Colborne, ON
First we prove the following lemma:
In
ABC, AA , BB , CC
intersect if-f
sin α1 sin β1 sin γ1
= 1,
sin α2 sin β2 sin γ2
·
·
where α1 , α2 , β1 , β2 , γ1 , γ2 are as shown in the diagram just below.
(Editor: This is a known variant of Ceva’s Theorem.)
B
c1
β1 β2 a2
A’
C’
D
c2
a1
α2
α1
A
b1
γ2
b2
B’
Proof: Let BB C = x , then BB A = 180◦
γ1
C
− x. Using the sine law in BB C yields
b2
a
=
.
sin β2
sin x
Similarly using the sine law in
(1)
BB A yields
b1
c
=
sin β1
sin(180◦
− x)
=
c
sin x
.
(2)
Hence,
b1 : b 2 = c sin β1
a sin β2
(3)
(from (1),(2)). (Editor: Do you recognize this when β1 = β 2 ?)
Similarly,
a1 : a 2 =
b sin α1
,
c sin α2
a sin γ1
.
b sin γ2
c1 : c 2 =
(4)
By Ceva’s
Ceva’s theore
theorem,
m, the necess
necessary
ary and sufficie
sufficient
nt conditi
condition
on for AA , BB , CC to intersect is:
(a1 : a 2 ) (b1 : b 2 ) (c1 : c 2 ) = 1. Using
Using (3), (4) on this yields:
yields:
·
·
b sin α1
c sin α2
·
γ c sin β
· ab · sin
· ·
sin γ a sin β
1
1
2
2
7
so
=1
sin α1 sin β1 sin γ1
= 1.
sin α2 sin β2 sin γ2
·
·
(5)
This is just what we needed to show, therefore the lemma is proved.
Now, in our original question, give BAC = 40◦ ,
Since CB D = 40◦ ,
ABD
= ABC
DBC
ABC
= 6600◦ . It follows that ACB = 80◦ .
= 20◦ . Similarly, EC A = 20◦ .
−
B
K
E
F
A
C
D
Now let us show that F AD = 10◦ . Suppose otherwise. Let F be such that F,
F , F are in the same
side of AC and DAF = 10◦ . Then BAF = BAC DAF = 30◦ .
−
Thus
sin ABD sin BC E sin C AF
sin DBC sin EC A sin F AB
·
·
sin 10◦
sin 70◦ sin
sin20◦ sin
sin 30◦
sin40◦ sin10◦ sin
◦
cos20◦
sin20
=
2sin20◦ cos20◦ sin30◦
1
=
= 1.
2sin30◦
=
·
·
·
By the lemma above, AF passes through C E BD = F . Therefore AF = AF , and F AD = 10◦ ,
contrary
con
trary to assumpti
assumption.
on. Thus
Thus F AD must be 10◦ . Now let
let AF BC = K . Si
Sinc
ncee KAC =
10◦ , KC A = 80◦ , it follows that AKC = 90◦ . Therefore AK BC
AF BC as needed.
∩
⊥
8
∩
⇒
⊥
Solution to Problem 5
Adrian Chan, Upper Canada College, Toronto, ON
Let us first prove by induction that
a2n + a2n+1
= m 2 for all n
an an+1 + 1
·
≥ 0.
Proof:
Base Case (n = 0) :
a20 + a21
0 + m2
= m2 .
=
0+1
a0 a1 + 1
·
Now, let us assume that it is true for n = k,
k, k
≥ 0. Then,
a2k + a2k+1
ak · ak
+1
+1
2
ak + a2k+1
a2k+1 + m4 a2k+1
2
− 2m · ak · ak
2
2
+1
ak+1 + ( m ak+1
+ a2k
− ak )
2
2
2
ak+1 + ak+2
So
a2k+1 + a2k+2
·
ak+1 ak+2 + 1
= m2
= m2 ak ak+1 + m2
= m2 + m4 a2k+1 m2 ak ak+1
· ·
2
= m
2
= m
−
· ·
+ m ak (m ak − ak )
+ m · ak · ak .
2
2
+1
+1
2
+1
+2
= m 2,
proving the induction. Hence (an , an+1 ) is a solution to
Now, consider the equation
Then
a2 + b 2
= m 2 for all n
ab + 1
≥ 0.
a2 + b2
= m 2 and suppose (a, b) = (x, y) is a solution with 0
ab + 1
≤ x ≤ y.
x2 + y 2
= m2.
xy + 1
(1)
If x = 0 then it is easily seen that y = m , so (x, y) = (a0 , a1 ). Since
Since we are given
given x
now that x > 0.
Let us show that y
≥ 0, suppose
2
≤ m x.
Proof by contradiction: Assume that y > m2 x. Then y = m 2 x + k where k
≥ 1.
Substituting into (1) we get
x2 + (m2 x + k)2
(x)(m2 x + k ) + 1
= m2
x2 + m4 x2 + 2m2 xk + k2
(x2 + k2 ) + m2 (kx
= m4 x2 + m2 kx + m2
− 1)
= 0.
− 1) ≥ 0 since kx ≥ 1 and x + k ≥ x + 1 ≥ 1 so (x
Thus we have a contradiction, so y ≤ m x if x > 0.
2
Now, m2 (kx
2
2
2
9
2
2
2
+ k 2 ) + m2 (kx
− 1) = 0.
Now substitute y
m x
x1 , where 0
≤x
1
< m x, into (1).
We have
x2 + (m2 x x1 )2
x(m2 x x1 ) + 1
−
2
4
2
x +m x
−
− 2m x · x
2
1
= m2
x2 + x21
2
2
1
x +x
·
2
+ x21 = m4 x2
=
−m x·x
m (x · x + 1)
2
1
+ m2
1
= m2 .
(2)
1
x x +1
If x1 = 0, then x2 = m 2 . Hence x = m and (x1 , x) = (0, m) = (a0 , a1 ). But y = m 2 x
(x, y) = (a1 , a2 ). Thus suppose x1 > 0.
−x
1
= a 2 , so
Let us now show that x1 < x.
Proof by contradiction: Assume x 1
Then m2 x
−y ≥x
since y = m 2 x
(x, y) is a solut
solution
ion to
to
3
2
2
≥ x.
− x , and
1
x2 + y 2
xy + 1
x
− y ≥ x since
a2 + b2
= m2.
ab + 1
≥
2
3
2
≥
≥
−
x y + x + y which is a contradiction since y x > 0.
x y + xy + x + y , hence x
So x + xy
2
With the same proof that y m x, we have x m2 x1 . So the subst
substitut
itution
ion x = m2 x1 x2 with
x2 0 is valid.
≤
≥
2
Substituting x = m x1
−
≤
x21 + x22
= m2 .
x2 into (2) gives
x1 x2 + 1
·
2
−
If x2 = 0, then we continue with the substitution xi = m x +1 xi+2 (*) until we get
i
and xj +1 = 0. (The sequence xi is decreasing, nonnegative and integer.)
x2j + x2j +1
·
xj xj +1 + 1
So, if xj +1 = 0, then x2j = m 2 so xj = m and (xj +1 , xj ) = (0, m) = (a0 , a1 ).
Then (xj , xj −1 ) = (a1 , a2 ) sinc
sincee xj −1 = m 2 xj
xj +1 (from (*)).
−
Continuing, we have (x1 , x) = (an−1 , an ) for
for some
some n. Then (x, y) = (an , an+1 ).
2
2
a +b
= m 2 has solutions (a, b) if and only if (a, b) = (an , an+1 ) for some n.
Hence
ab + 1
10
= m2
GRADERS REPORT
REPORT
Each question was worth a maximum of 7 marks. Every solution on every paper was graded by two
different markers. If the two marks differed by more than one point, the solution was reconsidered
un
until
til the differe
difference
nce resolv
resolved.
ed. If the tw
twoo marks
marks differed
differed by one point,
point, the averag
averagee wa
wass use
used
d in
computing the total score.
The various grades assigned each solution are displayed below, as a percentage.
MARKS
#1
#2
#3
#4
#5
0
1
2
3
4
5
6
7
7.6
14.1
10.9
6.5
3.3
6.0
16.3
35.3
9.8
27
27.7
16.8
16.3
2.2
14.1
6.0
7.1
40
40.2
7.1
16.8
3.8
1.6
4.3
7.1
19.0
31
31.0
27
27.7
21.7
1.6
2.2
3.8
2.2
9.8
73
73.9
9.2
12.0
1.1
0.5
0.0
1.1
2.2
PROBLEM 1
This question
question was well done. 47 studen
students
ts received
received 6 or 7 and only 6 students
students receive
received
d no marks.
marks.
Many students came up with a proof similar to David Arthur’s proof. Another common approach
was to find bounds for a (either 0
a < 60 or 0
a < 90) and to then check which of these a
satisfy
satisf
y the equation.
equation.
≤
≤
PROBLEM 2
Although most students
Although
students attempted this problem, there were
were only 6 perfect
perfect solutions. A further 6
solutions earned a mark of 6/7 and 13 solutions earned a mark of 5/7.
The most common approach was to square both sides of the equation, rearrange the terms to isolate
the radical, and to then square both sides again. This resulted in the polynomial x6 2x5 x4 +
2x3 + x 2 = 0. Many
Many student
studentss we
were
re unable
unable to factor
factor this polynomi
polynomial,
al, and so earned
earned only 2 or 3
points.
−
√
−
−√
−√
1+ 5
1
5
The polynomial has three distinct roots: 0,
, and
. Most stud
students
ents recognized
recognized that
2
2
1
5
0 is extraneous
extraneous.. One point was deducted
deducted for not finding that
is extra
extraneous
neous,, aand
nd a ffurthe
urtherr
2
1+ 5
point was deducted for not checking that
is a solution.
solution. (It’s not
not obvious
obvious that the
the equation
equation
2
has any solution
solutions.)
s.) Failing
ailing to ch
chec
eck
k for extrane
extraneous
ous roots is consid
considere
ered
d to be a major erro
error.
r. The
graders should, perhaps, have deducted more points for this mistake.
√
The solution included here avoids the 6th degree polynomial, thus avoiding the difficult factoring.
11
However, the solutions must still be checked.
PROBLEM 3
There were 17 perfect solutions and eleven more contestants earned either 5 or 6 points.
1 1
1
The most elegant
elegant solution
solution uses two
two simple observ
observations:
ations: that 1 = + and that is greater than
2 2
2
1
1 1
the average of 2 , 4 , . . . , 2n . A telescope argument also works, adding the first and last terms
from each side, and so on. The key to a successful proof by induction is to be careful with algebra
and to avoid
avoid the temptation
temptation to use inequalities.
inequalities. For example, many students
students used the induction
hypothesis to deduce that
1
n+2
then used
1
1
1+ +...+
3
2n + 1
n+1
1
1
1+ + ... +
>
(n + 2)n
2
2n
+
1
(n + 2)(2n + 1)
n+1
1
, which is too sloppy for a successful induction proof.
>
(n + 2)n
n+1
PROBLEM 4
Many contes
Many
contestan
tants
ts attempt
attempted
ed this
this questi
question,
on, though
though few got beyond
beyond labelin
labelingg the most
most appare
apparent
nt
angles.. Nine students
angles
students successfully
successfully completed
completed the problem,
problem, while another six made a significan
significantt
attempt.
Most of these efforts employed trigonometry or coordinates to set up a trigonometric equation for
an unknown
unknown angle. This
This yields
yields to an assault
assault by identiti
identities.
es. Adrian
Adrian Birka
Birka produce
produced
d a ve
very
ry clean
solution of this nature.
Only Keon Choi managed to complete
complete a (very
(very pretty) synthetic
synthetic solution.
solution. One other contestan
contestantt
made significant progress with the same idea.
PROBLEM 5
{ }
Many students were successful in finding the expression for the terms of the sequence an by a
variet
ariety
y of methods
methods:: produci
producing
ng an explici
explicitt formu
formula,
la, by means of a genera
generating
ting function
function and as a
sum of binomial coefficients involving parameter m. Unfor
Unfortunate
tunately
ly this does not help solving the
problem. Nevertheless seventeen contestants were able to prove by induction that the terms of the
sequence satisfy the required relation.
To prove the ”only if” part one should employ the method of descent which technically is the same
calcul
cal
culatio
ation
n as in the direct
direct part
part of the problem
problem.. Thr
Three
ee students
students succee
succeeded
ded in this,
this, but only two
obtained a complete solution by showing that the sequence constructed by descent is decreasing
and must have m and 0 as the last two terms.
12
1999
SOLUTIONS
Most of the solutions to the problems of the 1999 CMO presented below are taken from students’
papers.
pape
rs. Some
Some min
minor
or editin
editingg has been done
done - unnece
unnecessa
ssary
ry steps have
have been eliminat
eliminated
ed and some
wordi
wo
rding
ng has been ch
chang
anged
ed to make
make the proofs
proofs clearer.
clearer. But for the most part, the proofs
proofs are as
submitted.
Solution to Problem 1 – Adrian Chan, Upper Canada College, Toronto, ON
Rearranging the equation we get 4x
4 x2 + 51 = 40[x
40[ x]. It is known that x
4x2 + 51 = 40[x
40[x] > 40(
40(x
x
2
− 40
40x
x + 91
(2x
(2x − 13)(2
13)(2x
x − 7)
4x
> 0
≥ [x] > x − 1, so
− 1)
> 0
Hence x > 13/
13 /2 or x < 77/
/2. Also,
4x2 + 51 = 40[x
40[ x]
2
−−
4x
40
40x
x + 51
(2x
(2x 17)(2x
17)(2x 3)
−
≤
≤≤
40x
40x
0
0
≤ x ≤ 17
17/
/2. Combining these inequalities gives 3/
3 /2 ≤ x < 7/
7 /2 or 13/
13/2 < x ≤ 17/
17/2 .
CASE 1: 3/2 ≤ x < 7
7/
/2.
Hence 3/
3/2
For this case, the possible values for [x
[ x] are 1, 2 and 3.
If [x] = 1 then 4x
4 x2 + 51 = 40 1 so 44x
x2 =
·
−11, which has no real solutions.
If [x] = 2 then 4x
4 x2 + 51 = 40 2 so 44x
x2 = 29 and x =
2 < x < 3 and [x
[x] = 2.
·
If [x] = 3 then 4x
4 x2 + 51 = 40 3 and x =
·
CASE 2: 13
13/
/2 < x
√
√69
69/
/2. But
69
2
√29
2
>
. No
Notic
ticee that
that
√64
2
√16
2
<
√29
2
<
√36
2
so
= 4. So, this solution is re
rejecte
jected.
d.
≤
17
17/
/2.
For this case, the possible values for [x
[ x] are 6, 7 and 8.
If [x] = 6 then 44x
x2 + 51 = 40 6 so x =
[x] = 6.
·
If [x] = 7 then 44x
x2 + 51 = 40 7 so x =
[x] = 7.
·
If [x] = 8 then 44x
x2 + 51 = 40 8 so x =
[x] = 8.
The solutions are x =
√29
√189
2
√229
2
. Notice
Notice that
that
. Notice
Notice that
that
√144
2
√196
2
√269
√256
2
2
·
. Notice
Notice that
that
√189 √229 √269
,
,
,
.
2
2
2
2
(Editor: Adrian then checks these four solutions.)
1
<
<
<
√189
2
√229
2
√269
2
<
<
<
√196
2
√256
2
√324
2
so 6 < x < 7 and
so 7 < x < 8 and
so 8 < x < 9 and
Solution 1 to Problem 2 – Keon Choi, A.Y. Jackson S.S., North York, ON
E
Let D and E be the intersections of BC
B C and extended
AC respectively with the circle.
||
Since C O AB (because both the altitude and the radius are 1) BC O = 60◦ and therefore
therefore EC O =
180◦ ACB BC 0 = 60◦ .
−
C O
−
Since a circle is always symmetric in its diameter and
line CE is reflection of line CB in CO,
CO , line segment
CE is reflection of line segment CB .
Therefore CE = C
CD
D.
Therefore
D
A
B
CE D is an isosceles.
Therefore CE D = CDE
CD E and C ED + CD
CDE
E = ACB = 60◦ .
CE D = 30◦ regardless of the position of centre 0. Since CE D is also the angle subtended from
the arc inside the triangle, if C ED is constant, the arc length is also constant.
Editor’s Note:
This proof has had no editing.
2
Solution 2 to Problem 2 – Jimmy Chui, Earl Haig S.S., North York, ON
√
1
Place C at the origin,
origin, point
point A at
− √
1
point B at
, 1 . Then
3
with altitude of length 1.
3
,1
and
y
B
ABC is equilateral
A
A’
B’
Let O be th
thee ce
cen
nte
terr of the circle.
circle. Be
Beca
caus
usee th
thee
circle has radius 1, and since it touches line AB,
AB ,
the locus of O is on the line through C parallel to
AB (since C is length 1 away from AB ), i.e., the
locus of O is on the x-axis.
Let point O be at (a,
(a, 0)
0).. Then
Then
along AB .
1
1
−√ ≤ a ≤ √
3
3
x
C
O(a,0)
since we have the restriction that the circle rolls
Now, let A and B be the intersection of the circle with CA and CB respectiv
respectively
ely.. The equation
1
1
√3 , of CB is y =
of CA is y = 3 x, 0
x
3 x, √3
x
0, and of the circle is
(x a)2 + y 2 = 1.
√
−√
≤ ≤
−
− ≤ ≤
√
√
a ± 4 − 3a2
2
2
We solve for A by substituting y = 3 x into (x
(x − a) + y = 1 to get x =
.
4
Visually, we can see that solutions represent the intersection of AC extended and the circle, but
we are only concerned with the greater x-value – this is the solution that is on AC , not on AC
extended. Therefore
x=
a+
√4 − 3a
2
,
4
y=
√
3
a+
√4 − 3a
2
4
.
Likewise we solve for B , but we take the lesser x
x-value
-value to get
x=
a
− √4 − 3a
4
2
,
y=
−
√4 − 3a
√
a+
√4 − 3a − − √3 a − √4 − 3a
4
4
3
a+
2
4
.
Let us find the length of A B :
|AB|2 =
=
a+
4
− 3a
√4 − 3a − a − √4 − 3a
4
4
2
2
4
2
√
2
+
3
2
2
a2
+3
4
= 1
which is independent of a
a..
Consider the points 0,
0, A and B .
Therefore A 0B =
π
3
0AB is an equilateral triangle (because AB = 0A = 0B = 1).
and arc A B =
π
3
, a constant.
3
2
Solution to Problem 3 – Masoud Kamgarpour, Carson S.S., North Vancouver, BC
α
Note that n = 1 is a solution
solution.. For n > 1 write n in the form n = P1α P2α ...Pm
where the Pi ’s,
1 i m, are distinct prime numbers and αi > 0. Since d(n) is an integer, n is a perfect square,
so αi = 2βi for integers β i > 0.
1
2
m
≤ ≤
Using the formula for the number of divisors of n,
d(n) = (2β
(2β1 + 1)(2β
1)(2β2 + 1) . . . (2β
(2βm + 1)
2
which
whi
ch is an odd nu
numbe
mber.
r. Now
Now because
because d(n) is odd, (d
(d(n)) is odd, therefore n is odd as well, so
Pi 3, 1 i m. We get
≥
≤ ≤
P1α
1
or using α i = 2βi
·P
α2
2
α
. . . Pm
= [(α
[(α1 + 1)(α
1)(α2 + 1) . . . (αm + 1)]2
m
β
P1β P2β . . . Pm
= (2
( 2β1 + 1)(2β
1)(2β2 + 1) . . . (2β
(2βm + 1).
1) .
1
2
m
Now we prove a lemma:
Lemma: P t
≥ 2t + 1 for positive integers t and P ≥ 3, with equality only when P=3 and t=1.
Proof: We use mathematical induction on tt.. The statement is true for t = 1 because P ≥ 3. Now
suppose P ≥ 2k + 1,
1 , k ≥ 1; then we have
k
k
k+1
Thus P t
≥
k
k
· ≥
≥
= P P P (1 + 2) > P + 2 (2k
(2k + 1) + 2 = 2(k
2(k + 1) + 1
P
2t + 1 and equality occurs only when P = 3 and t = 1.
β
Let’s say n has a prime factor Pk > 3; then (by the lemma) Pkβ > 2β
2 βk +1 and we have P1β . . . Pm
>
(2
(2β
β1 + 1) . . . (2
(2β
βm + 1), a contradiction.
1
k
m
Therefore, the only prime factor of n is P = 3 and we have 3 α = 2α + 1. By the lemma α = 1.
The only positive integer solutions are 1 and 9.
Solution 1 to Problem 4 – David Nicholson, Fenelon Falls S.S., Fenelon Falls, ON
Without loss of generality let a 1 < a2 < a3 . . . < a8 .
Assume that there is no such integer k
k.. Let’s just look at the seven differences d i = ai+1
amongst the di there can be at most two 1s, two 2s, and two 3s, which totals to 12.
Now 16 = 17
−1 ≥a −a
8
1
− a . Then
i
= d 1 + d2 + . . . + d7 so the seven differences must be 1,1,2,2,3,3,4.
Now let’s think of arranging
arranging the differences
differences 1,1,2,2,3,3,4.
1,1,2,2,3,3,4. Note that the sum of consecutiv
consecutivee differences is another difference. (Eg d 1 + d2 = a 3 a1 , d1 + d2 + d3 = a 4 a1 )
−
−
We can
can’t
’t place the two 1s side
side by side beca
because
use that will give
give us ano
anothe
therr differe
difference
nce of 2. The 1s
can’t be beside a 2 because then we have three 3s. They can’t both be beside a 3 because then we
have three 4s! So we must have either 1,
1, 4, , , , 3, 1 or 1,
1, 4, 1, 3, , , (or their reflections).
−−−
−−−
In either case we have
have a 3,1 giving a difference
difference of 4 so we can’t put the 2s beside each other.
other. Also we
can’t have
have 2,3,2 because then (with the 1,4) we will have three
three 5s. So all cases give
give a cont
contradic
radiction.
tion.
Therefore there will always be three differences equal.
{
}
A set of seven numbers satisfying the criteria are 1, 2, 4, 7, 11,
11, 16,
16, 17 . (Edito
(Editor:
r: There
There are
are many
many
such sets)
4
Solution 2 to Problem 4 – The CMO committee
−
−
Consider all the consecutive differences (ie, di above) as well as the differences bi = a i+2 ai , i =
1 . . . 6. Then the sum of these thirteen differences is 2 (a8 a1 )+ (a7 a2 ) 2(17 1)+(16 2) = 46.
Now if no difference occurs more than twice, the smallest the sum of the thirteen differences can
be is 2 (1 + 2 + 3 + 4 + 5 + 6) + 7 = 49, giving a contradic
contradiction.
tion.
· −
− ≤
−
·
Solution 1 to Problem 5 – The CMO committee
Let f (x,y,z
x,y,z)) = x2 y + y 2 z + z 2 x. We wish to determ
determine
ine where
where f is maximal
maximal.. Since
Since f is cyclic,
without loss of generality we may assume that x y, z . Since
≥
f (x,y ,z)
,z)
− f (x,z ,y)
,y)
= x2 y + y2 z + z 2 x
=
≥ z. Then
f (x + z,y, 0) − f (x,y,z
x,y,z))
2
2
2
−x z−z y−y x
(y − z )(x
)(x − y )(
)(x
x − z ),
we may also assume y
= (x + z )2 y
=
2
2
2
−x y−y z−z x
z y + yz
yz((x − y) + xz(
xz (y − z) ≥ 0,
2
so we may now assume z = 0. The rest follows from the arithmetic-geometric mean inequality:
2x2 y
f (x,y, 0) =
2
≤
1
2
x + x + 2y
2y
3
3
=
4
27
Equality occurs when x = 2y, hence at (x,y
(x,y ,z)
,z) = ( 23 , 13 , 0). (As well as (0,
(0, 32 , 31 ) and ( 13 , 0, 32 ).
Solution 2 to Problem 5 – The CMO committee
With f as above, and x
≥ y, z
2
−
f x + z, y + z, 0
2
2
f (x,y ,z)
,z) = yz
yz((x
−
y) + xz (x
2
so we may assume that z = 0. The rest follows as for solution 1.
5
−
3
z) + z y + z
8
4
GRADERS’ REPORT
REPORT
Each question was worth a maximum of 7 marks. Every solution on every paper was graded by two
different markers. If the two marks differed by more than one point, the solution was reconsidered
un
until
til the differe
difference
nce resolv
resolved.
ed. If the tw
twoo marks
marks differed
differed by one point,
point, the averag
averagee wa
wass use
used
d in
computing the total score.
The various grades assigned to each solution are displayed below, as a percentage.
MARKS
#1
#2
#3
#4
#5
0
1
2
3
4
5
6
7
7.6
13.9
12.0
5.7
4.4
8.9
8.9
39.9
45
4 5.6
15
1 5.8
12.7
2.5
1.3
2.5
0.0
20.9
18
1 8.4
43
4 3.0
15.2
5.1
8.9
3.2
1.3
6.3
38
3 8.6
0.0
41.8
7.6
4.4
5.7
1.9
1.3
51
5 1.3
32
3 2.3
13.9
2.5
0.0
1.3
0.0
0.0
PROBLEM 1 The aim of the question was to give the competitors an encouraging start (it was
not a give away!). Over half of the students had good scores of 5, 6 or 7.
The general approach was to find bounds for x and then to find the exact value for x by substituting
in the resulting possible values of [x
[x]. Depending on how the bounds were determined, this meant
checking 6 - 10 different cases.
Points were lost for not adequately verifying the bounds on x.
x . For example, 2 points were deducted
2
for assuming, without proof, that 4x
4x + 51 > 40[
40[x
x] for x 9.
≥
PROBLEM 2 Many
Many competitors saw that the key here is to prove
prove that the angle subtended
subtended by
the arc at its centre is constant, namely π/3.
π/3. In all, 16 students managed a complete proof. Most
attempted an analytic solution – indeed, the problem is nearly routine if one chooses coordinates
wisely
wis
ely and later on not
notes
es that two such
such x-coord
x-coordina
inates
tes are roots of the same quadrat
quadratic.
ic. A few
students used trigonometry, namely the law of sines on a couple of useful triangles. Two students
found essentially the same synthetic solution, which is very elegant.
PROBLEM
PR
OBLEM 3 Most competitors determined by direct calculation that n = 1 and n = 9 are
solutions. The difficulty
solutions.
difficulty was to show that these are the only solutions,
solutions, which boils down to proving
k
that p
22k
k + 1 for all primes p > 2 and all k > 0 with equality only for k = 1 and p = 3. This
can be done by induction or by calculus. Only 5 students obtained perfect marks.
≥
PROBLEM 4
Many students found a specific set of seven integers such that the equation did not have three
different
differ
ent solutions.
solutions. This earned two
two point
p oints.
s. (One student
student found such a set with maximum value
value
14. A maximum value of 13 is not possible.)
6
Only eight competitors received high marks on the question (5, 6, or 7), and only one student scored
a perfect 7. All of the successful solvers considered differences of consecutive integers, showing that
they must be 1, 1, 2, 2, 3, 3, and 4, and then showed that every ordering of these differences led
to at least three repetitions of the same value. Most competitors recognized that the 1s could not
be togethe
together,
r, nor could
could they
they be beside
beside a 2. They
They then proceed
proceed by conside
considerin
ringg all such
such possible
possible
arrangements, which often resulted in close to a dozen cases (depending on how the the cases were
handled.) David Nicholson was the most efficient at pruning the cases. (See Solution 1 to Problem
4.) Most students failed to consider one or two (easily dismissed) cases, hence lost 1 or 2 points.
A number of the contestants attempted to solve the problem by examining the odd-even character
of the set of eight integers, counting how many of the differences were odd or even, and using the
pigeon-hole
pigeon
-hole principle.
principle. Although
Although this approach looked promising,
promising, no one was able to handle the
case that 3 of the integers were of one parity, and 5 were of the other parity.
PROBLEM 5 No students
students received
received full marks for this problem. One student
student received
received 5 marks
for a proof
proof that had minor
minor errors
errors.. This
This proof was by Calculus.
Calculus. The comm
committe
itteee wa
wass aware
aware that
the proble
problem
m could
could be solve
solved
d using
using Calculu
Calculuss but (erron
(erroneou
eously
sly)) though
thoughtt it unlik
unlikely
ely high
high school
school
studen
stu
dents
ts would
would attemp
attemptt such
such a solutio
solution.
n. Many
Many studen
students
ts receiv
received
ed 1 poin
pointt for “guessin
“guessing”
g” that
1
2
2
1
2
1
, 3, 0 ,
0, 3 , 3 and 3 , 0 , 3 are where
where equalit
equality
y occurre
occurred.
d. Some
Some studen
students
ts receiv
received
ed
3
a further point for verifying the inequality on the boundary of the region.
7
2000 Canadian Mathematics Olympiad Solutions
Chair: Luis Goddyn, Simon Fraser
Fraser University, goddyn@math.sfu.ca
The Year 2000 Canadian Mathematics Olympiad was written on Wednesday April 2, by 98 high
school students
students across Canada. A correct and well
well presented solution to any
any of the ve questions
was awarded seven points. This year’s exam was a somewhat harder than usual, with the mean
score being 8.37 out of 35. The top few scores were: 30, 28, 27, 22, 20, 20, 20. The rst, second and
third prizes are awarded to: Daniel Brox (Sentinel Secondary BC), David Arth
Arthur
ur (Upper Canada
College ON), and
a nd David Pritchard (Woburn
(Woburn Collegiate Institute ON).
1. At 12:00 noon, Anne, Beth and Carmen begin running laps around a circular track of length
three hundred meters, all starting from the same point on the track. Each jogger maintains
a constant speed in one of the two possible directions
directions for an inde nite period of time. Show
that if Anne’s speed is di
di erent from the other two speeds, then at some later time Anne
Anne will
be at least one hundred meters from each of the other runners. (Here, distance is measured
along the shorter of the two arcs separating two runners.)
runners.)
Comment: We were surprised by the difficulty of this question, having awarded an average
grade of 1 .43 out of 7.
7 . We pr
present
esent two
two solutions; only the rst appear
appeared
ed among the graded
graded
papers.
Solution 1: By rotating the frame of reference we may assume that Anne has speed zero, that
Beth runs at least as fast as Carmen, and that Carmen’s
Carmen’s speed
speed is positive. If Beth is no more
than twice as fast as Carmen, then both are at least 100 meters from Anne when Carmen has
run 100 meters. If Beth runs more that twice
twice as fast as Carmen, then Beth runs a stretch of
more than 200 meters during the time Carmen
C armen runs between 100 and 200 meters. Some part
of this stretch lies more than 100 meters from Anne, at which time both Beth and Carmen
are at least (in fact, more than) 100 meters away from Anne.
Solution 2: By rotating the frame of reference we may assume Anne’s speed to equal zero,
and that the other two runners have non-zero speed.
speed. We may assume that Beth is running at
least as fast as Carmen. Suppose that it takes t seconds for Beth to run 200 meters. Then at
all times in the in nite set T = {t, 2t, 4t, 8t , . . .}, Beth is exactly 100 meters from Anne. At
time t, Carmen has traveled exactly d meters where 0 < d 200
200.. Let k be the least
l east integer
k
k
kd
200,, so at time 2 t ∈ T both Beth and
100.. Then k 0 and 100
100
2 d 200
such that 2
Carmen are at least 100 meters from Anne.
2. A permutatio
permutation
n ooff the integers 1901 , 1902, . . . , 2000 is a sequence a1, a2, . . . , a100 in which each
of those integers appears exactly once. Given such a permutation, we form the sequence of
partial sums
s1 = a 1, s2 = a1 + a2 , s3 = a 1 + a2 + a3 , . . . , s100 = a 1 + a2 +
+ a100.
How many of these permutations will
wil l have no terms of the sequence s1 , . . . , s100 divisible by
three?
Comment: This question was the easiest and most straight forward,
forward, with an average grade
grade of
3.07
07..
Solution: Let {1901, 1902, . . . , 2000} = R0 ∪ R 1 ∪ R 2 where each integer in Ri is congruent
to i modulo 3. We not
notee that
that |R0| = |R1| = 33 and |R2| = 34.
34. Each pe
permut
rmutation
ation S =
(a1, a2, . . . , a100) can be uniquely speci
speci ed by describing a sequence S = (a1, a2, . . . , a100) of
residues modulo
modulo 3 (containing exactly 33 zeros, 33 ones and 34 twos), and three permutations
(one each of R0 , R1 , and R2). Note that the number of permutations
permutations of Ri is exactly |Ri|! =
1 2 |Ri |.
The condition on the partial sums of S depends only on the sequence of residues S . In
order to avoid a partial sum divisible by three, the subsequence formed by the 67 ones and
twos in S must equal either 1, 1, 2, 1, 2, . . . , 1, 2 or 2, 2, 1, 2, 1, . . . , 2, 1. Since |R2| = |R1| + 1,
only the second pattern is possible. The 33 zero entries in S may appear anywhere among
a1 , a2 , . . . , a100 provided that a1
= 0.
0. There ar
aree 99
= 33!99!66! ways to choose which entries in
33
S equal zero. Thus there are exactly 99
sequences S whose partial sums are not divisible
33
by three. Therefore the total number of permutations S satisfyin
satisfyingg this requirement
requirement is exactly
99
33
99! 33! 34!
.
66!
33! 33! 34! =
Incidently, this number equals approximately 4.4 10138.
3. Let A = (a1, a2, . . . , a2000) be a sequence of integers each lying in the interv
i nterval
al [ −1000, 1000].
Suppose that the entries in A sum to 1. Show that some nonempty subsequence of A sums
to zero.
Comment: This students found this question to be the most difficult, with an average grade
of 0.51
51,, and only one perfect
perfect solution among 100 papers.
Solution: We may assume no entry of A is zero, for otherwise we are
are done. We sort A into
a new list B = (b1, . . . , b2000) by selecting elements from A one at a time in such a way that
b1 > 0
0,, b2 < 0 and, for each i = 2, 3, . . . , 2000
2000,, the sign of bi is opposite to that of the partial
sum
si 1 = b 1 + b2 +
+ bi 1 .
−
−
(We can assume that each si i
= 0 for otherwise we are done.) At each step of the selection
process a candidate for bi is guaranteed
guaranteed to exist, since the condition a1 + a2 +
+ a2000 = 1
implies that the sum of unselected entries in A is either zero or has sign
s ign opposite to si 1 .
−
−
Fr
From
om the way they were de ned, each of s1 , s2, . . . , s2000 is one of the 1999 nonzero integers
in the interval [ −999, 1000]
1000].. By the Pigeon Hole Principle, sj = sk for some j, k satisfying
1 j < k 2000
2000.. Thus bj +1 + bj +2 +
+ bk = 0 and we are done.
4. Let ABCD be a convex
convex quadrilateral with
and
CBD
CB
D
= 2 ADB,
ABD
CD B
= 2 CDB
AB
CB .
= CB.
C D.
Prove that AD = CD
Comment: There are
are several di erent solutions to this, including some using purely
purely trigonometric arguments
arguments (involving the law of sines and standard
s tandard angle sum formulas). We present
here two prettier
prettier geometric
geometric arguments
arguments (with diagrams).
diagrams). The rst solution is perhaps the more
more
attractive
attr
active of the two. Average grade:
grade: 1 .84 out of 7.
7.
DB
P
A
1
Solution
1 (fromatcontestant
Extend
to aand
pointAP Don=the circle
CP DChoi):
CBD
CB
D = ADB
ABD through
CD B,
B . Then Keon
=
= CDB
and C centered
2
1
2
2
so APCD is a parallelogram. Now P D bisects AC so
s o BD is an angle bisector of isosceles
triangle ABC . We have
have
ADB =
1
1
CBD
CB D =
ABD = CDB
2
2
ADC
C , this triangle
so DB
D B is the angle bisector of ADC . As DB bisects the base of triangle AD
must be isosceles and AD = C D.
CB D meet CD
Solution 2: Let the bisector of ABD meet AD at E . Let the bisector of CBD
at F . Then F BD = BDE and EB D = BDF , which imply BE F D and BF E D.
Thus BEDF is a parallelogram whence
BD intersects EF at its midpoint
midpoint M .
(1)
AB
AE
On the other hand since BE is an angle bisector, we have BD
= ED
. Simila
Similarly
rly we have
have
CB
CF
AE
CF
DEF
F
= F D . By assumption
assumption AB = C B so ED = F D which implies E F AC . Thus DE
BD
and DAC are similar, which implies by (1) that B D intersects AC at its midpoint N . Since
ABC is isosceles, this implies AC ⊥ BD . Thus N AD and N CD are right triangles
with equal legs and hence are congruent.
congruent. Thus AD = C D.
P
B
B
A
A
C
C
N
F
E
E
M
F
D
D
Diagram for Solution 1
Diagram for Solution 2
5. Suppose that the real numbers a1, a2, . . . , a100 satisfy
a1
a2
a1 + a2
a3 + a4 +
a100
0
100
+ a100
Determine the maximum possible value of a 21 + a22 +
a1 , a2 , . . . , a100 which achieve this maximum.
100.
2
, and nd all possible sequences
+ a100
Comment: All of the correct solutions involved a sequence of adjustments to the variables,
2
each of which increase a 21 + a22 + + a100
while satisfying the constraints, eventually arriving
at the two optimal sequences: 100 , 0, 0, . . . , 0 and 50 , 50, 50, 50, 0, 0, . . . , 0. We present
present here a
sharper proof,
proof, which might be arrived at after guessing that the optimal value is 1002. Average
grade: 1 .52 out of 7.
7.
3
Solution: We have a1 + a2 +
a21 + a22 +
2
+ a100
+ a100
200,, so
200
(100 − a2)2 + a22 + a23 +
2
+ a100
2
= 1002 − 200a2 + 2a22 + a23 +
+ a100
1002 − (a1 + a2 +
+ a100)a2 + 2a22 + a23 +
= 1002 + (a22 − a1a2 ) + (a23 − a3 a2 ) + (a24 − a4 a2 ) +
2
+ (a100
− a100a2)
= 1002 + (a2 − a1)a2 + (a3 − a2 )a3 + (a4 − a2 )a4 +
+ (a100 − a2)a100
a100 0, none of the terms (ai − aj )ai is positive. Thus a21 + a22 +
10, 000 with equality
equality holding if and only if
Since a 1
2
a100
2
+ a100
a2
a1 = 100 − a2
a1 + a2 +
and
+
+ a100 = 200
and each of the products
(a2 − a1 )a2 , (a3 − a2 )a3, (a4 − a2)a4 ,
, (a100 − a2 )a100
equals zero. Since a1 a2 a3
a100 0, the last condition holds if and only if for
some i 1 we have a1 = a 2 =
= a i and ai+1 =
= a 100 = 0.
0. If i = 1,
1, then we get the
solution 100 , 0, 0, . . . , 0. If i 2, then from a1 + a2 = 100,
100, we get that i = 4 and the second
second
optimal solution 50 , 50, 50, 50, 0, 0, . . . , 0.
4
2001
SOLUTIONS
Several solutions are edited versions of solutions submitted by the contestants whose names appear
in italics..
1. (Daniel Brox)
Let R be Rachel’s age, and let J be Jimmy’s age. Rachel’s quadratic is
)(x − J ) = ax 2 − a(R + J )x + aRJ.
a(x − R)(x
for some number a. We are
are given
given that
that th
thee coeffic
coefficie
ient
nt a is an inte
intege
ger.
r. The
The sum
sum of the
the
coefficients is
a − a(R + J ) + aRJ = a(
a (R − 1)(
1)(J
J − 1).
1).
Since this is a prime number, two of the three integers a, R − 1, J − 1 multiply to 1. We are
given that R > J > 0, so we must have that a = 1, J = 2, R − 1 is prime, and the quadratic
is
(x − R)(x
)(x − 2).
2).
We are told that this quadratic takes the value −55 = −5 · 11 for some positive integer x.
Since R > 2, the first factor, (x
( x − R), must be the negative one. We have four cases:
x − R = −55 and x − 2 = 1, which implies x = 3, R = 58.
x − R = −11 and x − 2 = 5, which implies x = 7, R = 18.
x − R = −5 and x − 2 = 11, which implies x = 13, R = 18.
x − R = −1 and x − 2 = 53, which implies x = 57, R = 58.
Since R − 1 is prime, the first and last cases are rejected, so R = 18 and J = 2.
2. (Lino Demasi)
After ten coin flips, the token finishes on the square numbered 2k
2 k − 10, where k is the number
10
of heads obtained. Of the 2 = 1024 possible results of ten coin flips, there are exactly 10
k
ways to obtain exactly k heads, so the probability of finishing on the square labeled 2 k − 10
equals 10
/1024.
k
The probability of landing on a red square equals c/1024
c/ 1024 where c is the sum of a selection of
the numbers from the list
10
10
10
10
,
,
,...,
= 1, 10
10,, 45,
45, 120
120,, 210
210,, 252
252,, 210
210,, 120
120,, 45,
45, 10,
10, 1.
0
1
2
10
(1)
We are given that for some integers a, b satisfying a + b = 2001,
a/b = c/1024
c/ 1024..
If we assume (as most contestan
contestants
ts did!) that a and b are relatively prime, then the solution
proceedss as follows.
proceed
follows. Since 0 ≤ a/b ≤ 1 and a + b = 2001, we have 1001 ≤ b ≤ 20
2001.
01. Also
Also b
divides 1024, so we have b = 1024. Thus a = c = 2001 − 1024 = 977. There is only one way
to select terms from (1) so that the sum equals 977.
977 = 10 + 10 + 45 + 120 + 120 + 210 + 210 + 252.
252.
(2)
(This is easy to check, since the remaining terms in (1) must add to 1024 − 977 = 47, and
47 = 45 + 1 + 1 is the only possibility for this.)
1
In order to maximize n, we must
must colour
colour the strip as fol
follo
lows.
ws. Odd numbere
numbered
d square
squaress are
10
red if positiv
positive,
e, and white if negativ
negative.
e. Since
Since 252 = 5 is in the sum, the square labeled
2 · 5 − 10 = 0 is red. For k = 0, 1, 2, 3, 4, if 10
appears twice in the sum (2), then both 2k
2 k − 10
k
10
and 10 − 2k ar
aree coloured
coloured red.
red. If k does not appear in the sum, then both 2k
2k − 10 and
10
10 − 2k are coloured white. If k appears once in the sum, then 10 − 2k is red and 2k
2 k − 10 is
white. Thus
Thus the maximum value
value of n is obtained when the red squares are those numbered
{1, 3, 5, 7, 9, −8, 8, −4, 4, −2, 2, 0, 6} giving n = 31.
(Alternatively) If we do not assume a and b are relatively prime, then there are several more
possibilities
possibil
ities to consider.
consider. The greatest
greatest common divisor
divisor of a and b divides a + b = 2001, so
gcd(a,
gcd(
a, b) is one of
1, 3, 23
23,, 29
29,, 3 · 23
23,, 3 · 29,
29, 23 · 29,
29, 3 · 23 · 29.
29.
Since a/b = c/1024,
c/ 1024, dividing
dividing b by gcd(a,
gcd(a, b) results in a power of 2. Thus the prime factorization of b is one of the following, for some integer k.
2k , 3 · 2k , 23 · 2k , 29 · 2k , 69 · 2k , 87 · 2k , 667 · 2k , 2001
2001..
Again we have 1001 ≤ b ≤ 2001, so b must be one of the following numbers.
1024, 3 · 512 = 1536,
1024,
1536, 23 · 64 = 1472,
1472, 29 · 64 = 1856,
1856, 69 · 16 = 1104,
1104, 87 · 16 = 1392,
1392, 667 · 2 = 1334,
1334, 2001.
2001.
Thus a/b = (2001 − b)/b is one of the following fractions.
977 , 465 , 529 , 145 , 897 , 609 , 667 , 0 .
1024 1536 1472 1856 1104 1392 1334 2001
Thus c = 1024a/b
1024a/b is one of the following integers.
977,, 310,
977
310, 368,
368, 80
80,, 832,
832, 448
448,, 512
512,, 0.
After some (rather tedious) checking, one finds that only the following sums with terms from
(1) can add to a possible value of c.
c.
977 = 10 + 10
10 + 45 + 120 + 120
120 + 2210
10 + 2210
10 + 2252
52
310
310 = 10 + 45 + 45 + 210
210
51
5122 = 10 + 10 + 120 + 120 + 252
252
51
5122 = 1 + 1 + 45 + 45 + 210 + 210
210
0 = 0.
Again only those terms appearing exactly once in a sum can affect maximum value of n.
n . We
make the following table.
c
977
310
512
512
0
Terms appearing
appearing once in sum
{45
45,, 252}
{10
10,, 210}
{252}
∅
∅
Correspondin
Correspondingg red squares
squares
{6, 0}
{8, 2}
{0}
∅
∅
Evidently, the maximum possible value of n is obtained when c = 310 =
10
2
+
10
6
+
10
8
+
10
9
,
the
red465/
squares
, 3, 5,=
7, 155/
9, −/6512,
, 6, 2and
, 8}, nth
the
pr
prob
obab
abili
ility
ty of landi
landing
ng on a re
red
d squa
square
re is
a/b =
465
/1536 =are
310{/11024
310/
155
=e 35.
2
3. Solution 1: (Daniel
(Daniel Brox)
Set O be the centre of the circumcircle of ABC . Let the angle bisector of BAC meet this
circumcircle at R
R.. We have
BO
BOR
R
= 2 BAR = 2 CAR = CO
COR
R
Thus BR = CR and R lies on the perpendicular bisector of BC . Th
Thus
us R = P and ABCP
are concyclic. The points X , Y , M are the bases of the three perpendiculars dropped from P
onto the sides of ABC . Thus by Simson’s rule, X,Y,
X,Y, M are collinear. Thus we have M = Z
and BZ/ZC = BM/MC = 1.
Note: X Y Z is called a Simson line, Wallace line or pedal line for ABC . To prove Simson’s
rule, we note that BMPX are concyclic, as are A
AY
Y P X , thus
BX M
= BP M = 90 − P BC = 90 − P AC = AP Y = AX
AXY
Y
Solution 2: (Kenneth Ho)
Since P AX = P AY and P X A = P Y A = 90, triangles P AX and P AY are congruent, so AX = AY and P X = P Y . As P is on the perpendicular bisector of B C , we have
P C = P B . Thus P Y C and P X B are congruent right triangles, which implies CY
C Y = BX
BX.
Since X , Y and Z are collinear, we have by Menelaus’ Theorem
AY CZ BX
= −1.
Y C ZB XA
Applying AX = AY and CY = BX
B X , this is equivalent to BZ/ZC = 11..
4. We shall see that the only solution is n = 2. Fi
Firs
rstt we show
show that if n
= 2, then the table
1
can not be ch
chang
anged
ed into a table
table containi
containing
ng two zeros.
zeros. For n = 1, this is
T0 =
n−1
a
, let d(T ) be the quantity b − a
b
(mod n − 1). We shall show that neither of the two permitted moves can change the value of
d(T ). If we subtract n from both elements in T , then b − a does not change. If we multiply
the first row by n, then the element a changes to na,
na, for a difference of (n
( n − 1)a
1)a, which is
co
cong
ngru
ruen
entt to
to 0 (mod
(mod n − 1). Similarly
Similarly,, multiplyin
multiplyingg the second row by n does not change
d(T ). Since d
d((T0 ) = (n − 1) − 1 ≡ −1 (mo d n − 1), we can never obtain the table with two
zeros by starting with T 0 , because 0 − 0 is not congruent to − 1 modulo n − 1.
ve
very
ry easy to see. Suppose
Suppose n ≥ 3. For any table
table T =
For n = 2, and any table of positive integers, the following procedure will always result in a
table of zeros. We shall begin by converting the first column into a column of zeros as follows.
We repeatedly subtract 2 from all entries in the first column until at least one of the entries
equals either 1 or 2. Now we repeat the following sequence of three steps:
(a) multip
multiply
ly by 2 all rows
rows with 1 in the first column
column
(b) now multiply by 2 every row having a 2 in the first column (there is at least one such
row)
(c) subtract
subtract 2 from all entries in the first column.
column.
Each iteration of the three steps decreases the sum of those entries in the first column which
are greater than 2. Thus
Thus the first column eventu
eventually
ally consists
consists entirely
entirely of ones and twos,
twos, at
which time we apply (a) and (c) once again to obtain a column of zeros. We now repeat the
3
above procedure for each successive
above
successive column of the table. The procedure does not affect any
column which has already been set to zero, so we eventually obtain a table with all entries
zero.
5. (Daniel Brox)
Let P1 P3 P2 = 2α. As P1 P2 P3 is isosceles, we have that
t = P 1 P2 = 2 sin
sin α.
The line P 3 P4 is the perpendicular bisector of P 1 P2 . Since P2 P3 P4 is isosceles, we calculate
its length,
P2 P3 /2
1
P3 P4 =
=
.
cos α
2cos α
As P5 is the circumcentre of P2 P3 P4 , we have P3 P5 P4 = 2 P3 P2 P4 = 2 P2 P3 P4 = 2α.
The isosceles triangle P3 P4 P5 is therefore similar to P1 P2 P3 . As P3 P4 ⊥ P1 P2 , we have
P1 P3 P5 = 90. Furthermore, the ratio P 3 P5 : P 1 P3 equals r where
r=
P3 P4
1
1
=
=
.
P1 P2
(2sin α)(2cos α)
2 sin(
sin(22α)
P
P
0
3
α
P
2
P
1
P
5
P
4
By the same argument, we see that each Pi Pi+2 Pi+4 is a right angle with Pi+2 Pi+4 : P i Pi+2 =
r . Thus the points P1 , P3 , P5 , . . . lie on a logarithmic spiral of ratio r and period four as shown
below. It follows that P 1 , P5 , P9 , . . . are collinear, proving part (a).
4
P
3
1
r
P1
P
11
P9
P
13
r3
r2
P
5
P
7
By the self-similarity of the spiral, we have that P1 P1001 = r 500 P1001 P2001 , so
500
x/y = 1/r = 2 sin(
sin(22α).
This is an in
This
inte
tege
gerr when
when sin(
sin(22α) ∈ {0, ±1/2, ±1}. Since 0 < α < 90, th
this
is is equi
equiv
valent to α ∈ {15
15,, 45
45,, 75}. Thus
x/y is an in
integ
teger
er exactl
exactly
y when
when t belongs to the set
{2sin15
2sin15,, 2sin45
2sin45,, 2sin75}. This answers part (b).
500
5
GRADERS’ REPORT
REPORT
Eighty four of the eighty five eligible students
Eighty
students submitted
submitted an examination paper. Each
Each paper contained proposed solutions the some or all of the five examination questions. Each correct and well
presented
prese
nted solution
solution was awarded
awarded seven marks for a maxim
maximum
um total score of 35. The mean score was
10.8/35. The top three scores were
were 28, 27, and 22, thus
thus special scrutiny
scrutiny was required
required to separate
separate
the top two papers.
Each
Eac
h soluti
solution
on was
was indepen
independen
dently
tly marked
marked by two graders.
graders. If the two marks differed,
differed, then
then the
solution
solutio
n was
was recons
reconside
idered
red until
until the difference
difference was resolv
resolved.
ed. The top tw
twen
enty
ty papers were then
then
carefully regraded by the chair to ensure that nothing was amiss.
The grade distribu
distribution
tion and av
avera
erage
ge mark
mark for each question
question appears
appears in the followin
followingg tabl
table.
e. For
example, 13.1% of students were awarded 3 marks for question #1.
Marks
#1
#2
#3
#4
#5
0
1
2
3
4
5
6
7
Ave.
Mark
10.7
8.3
8.3
13.1
10.7
9.5
20.2
19.0
11.9
8.3
6.0
9.5
17
17.9
32.1
13
13.1
1.2
45.2
26
26.2
4.8
3.6
0.0
3.6
0.0
16
16.7
60.7
10
10.7
8.3
9.5
4.8
2.4
1.2
2.4
90.5
3.6
1.2
1.2
1.2
2.4
0.0
0.0
4.05
3.64
1.79
1.09
.26
Ninety
Ninety five perce
p ercent
nt of students
students found the correct solution,
solution, although a surprising
surprising
number arrived at a solution through trial and error or by guessing and verifying a solution. Many
assumed without proof that the leading coefficient equals one, which resulted in a two-point penalty.
Another common error was not to consider all four possibilities for the pair (x
( x − R), (x − 2).
PROBLEM
PR
OBLEM 1
There was
was a flaw in question
question 2. The proposers
proposers intended
intended that the integers
integers a, b be
relatively prime. This was made explicit in an early draft, but somehow was lost with the ambiguous
phrase “of the form a/b.”
a/b.” Without
Without this assumption, the problem is much
much more tedious to solve.
solve.
Remarkably, one student (Lino Demasi) considered more (but not all) possible values for gcd( a, b)
and obtained the correct solution n = 35. All other students assumed implicitly (and in two cases,
explicitly) that gcd(a,
gcd(a, b) = 1. Solutions to both problems are presented in this publication.
PROBLEM
PR
OBLEM 2
Most students
students either completely solved
solved or wer
weree baffled by this basic geometry
problem. There were at least four types of solutions: one trigonometric, one using basic geometry,
and two
two which
which refer
refer to standard
standard theorems
theorems relatin
relatingg to the trian
triangle
gle.. The first two
two tended
tended to be
lengthy
length
y or cumbersome
cumbersome,, and the last two
two are presented
presented here.
here. There were
were complaints
complaints from some
participants regarding the inaccurate angles appearing in the diagram supplied with the question.
The inaccuracy was intensional, since the key observation M = Z would have otherwise been given
PROBLEM
PR
OBLEM 3
6
away.. Unfortunate
away
Unfortunately
ly,, this caused some students
students to doubt their own proofs that BZ : Z C = 1; as
the ratio appears to be closer to 2 in the misleading diagram!
This problem
problem was left unanswered
unanswered by about 60% of students.
students. Several
Several solutions
solutions
consisted only of a proof that n = 1 is not possible. About 25% described a procedure which works
when n = 2. Indeed
Indeed the proced
procedure
ure for n = 2 se
seem
emss to be un
uniq
ique
ue.. About
About 10% prov
proved that for
no other value of n was possible, and all of the proofs explicitly or implicitly involved considering
PROBLEM
PR
OBLEM 4
residues modulo n − 1.
This problem proved
proved to be very
very difficult. Only two students
students completely answered
answered
part (a), and no students correctly answered part (b). Of the students receiving more than 0 marks,
only two were were not among the top 15 This suggests that the question effectively resolved the
ranking of the strongest participants, which is arguably the purpose of Problem 5.
PROBLEM 5
7
2002 Canadian Mathematical Olympiad
Solutions
{
}
1. Let S be a subset of 1, 2, . . . , 9 , such that the sums formed by adding each unordered pair of
distinct numbers from S are all different. For example, the subset 1, 2, 3, 5 has this property,
property,
but 1, 2, 3, 4, 5 does not, since the pairs 1, 4 and 2, 3 have the same sum, namely 5.
{
}
{ }
{
{ }
}
What is the maximum number of elements that S can contain?
Solution
Solu
tion 1
{
}
It can be checked that all the sums of pairs for the set 1, 2, 3, 5, 8 are different.
{
}
Suppose, for a contradiction, that S is a subset of 1, . . . , 9 containing 6 elements such that
all the sums of pairs are different. Now the smallest possible sum for two numbers from S is
1 + 2 = 3 and the largest
largest possible
possible sum is 8 + 9 = 17. That
That gives
gives 15 poss
possibl
iblee sums: 3, . . . , 17.
6
Also there are
= 15 pairs from S . Thus
Thus,, each
each of 3 , . . . , 17 is the sum of exactly one
2
pair. The only pair from 1, . . . , 9 that adds to 3 is 1, 2 and to 17 is 8, 9 . Thus 1, 2, 8, 9
are in S . But then 1+9 = 2+8, giving a contradiction. It follows that the maximum number
of elements that S can contain is 5.
{
}
{ }
{ }
Solution
Solu
tion 2.
It can be checked that all the sums of pairs for the set 1, 2, 3, 5, 8 are different.
Suppose, for a contradiction, that S is a subset of 1, . . . 9 such that all the sums of pairs
are different and that a1 < a2 < . . . < a6 are the members of S .
{
−
− −
{
}
}
−
−
Similar
ilarly
ly a6 a5 = a 4
Since a1 + a6 = a2 + a5 , it follows that a6 a5 = a2 a1 . Sim
a4 a3 = a 2 a1 . These three differences must be distinct positive integers, so,
(a6
Similarly a3
−a
− a ) + (a − a ) + (a − a ) ≥ 1 + 2 + 3 = 6 .
5
4
3
2
1
− a = a − a , so
2
5
4
(a3
− a ) + (a − a ) ≥ 1 + 2 = 3 .
2
5
4
Adding the above 2 inequalities yields
−a +a −a + a −a + a −a + a −a ≥6 + 3 = 9,
− a ≥ 9. This is impossible since the numbers in S are between 1 and 9.
a6
and hence a6
1
5
5
4
4
3
3
2
2
1
3
and
2. Call a positive integer n practical if every positive integer less than or equal to n can be
written as the sum of distinct divisors of n.
For example, the divisors of 6 are 1, 2, 3, and 6 . Since
1=1, 2=2, 3=3, 4=1+3, 5=2+ 3,
6=6,
we see that 6 is practical.
Prove that the product of two practical numbers is also practical.
Solution
≤ pq , we can write
k = aq + b wi
with
th 0 ≤ a ≤ p, 0 ≤ b < q.
Let p aand
nd q be practical. For any k
Since p and q are practical, we can write
a = c 1 + . . . + c m , b = d 1 + . . . + dn
where the ci ’s are distinct divisors of p and the dj ’s are distinct divisors of q . Now
k
= (c1 + . . . + cm )q + ( d1 + . . . + dn )
= c1 q + . . . + cm q + d1 + . . . + dn .
Each of c i q and d j divides pq . Since d j < q
and we conclude that pq is practical.
≤ c q for any i, j, the c q’s and d ’s are all distinct,
i
i
j
3. Prove
Prove that for all positiv
p ositivee real numbers
numbers a , b, and c,
c3
b3
a3
+
+
ab
ca
bc
≥ a + b + c,
and determine when equality occurs.
Each of the inequalities used in the solutions below has the property that equality holds if
and only if a = b = c . Thus equality holds for the given inequality if and only if a = b = c .
Solution 1.
4
4
4
4
4
4
Note that a + b + c = (a +
2 a ) . Applying the arithmetic-geometric
2 c ) + (c +
2 b ) + (b +
mean inequality to each term, we see that the right side is greater than or equal to
4
4
4
a2 b2 + b2 c2 + c2 a2 .
We can rewrite this as
a2 (b2 + c2 )
2
+
b2 (c2 + a2 )
2
+
c2 (a2 + b2 )
2
.
Applying the arithmetic mean-geometric mean inequality again we obtain a4 + b 4 + c 4
a2 bc + b2 ca + c2 ab. Dividing both sides by abc (which is positive) the result follows.
≥
Solution 2.
a,, b, c are replaced by ka,kb,kc, k > 0 we
Notice the inequalit
inequality
y is homogeneous.
homogeneous. That is, if a
get the original
original inequality
inequality. Thus
Thus we can assume, without loss of generality
generality,, that abc = 1.
Then
c3
b3
a3
+
+
ab
ca
bc
= abc
c3
b3
a3
+
+
ab
ca
bc
= a4 + b 4 + c 4 .
So we need prove that a4 + b4 + c4
By the Power Mean Inequality,
≥ a + b + c.
a4 + b4 + c4
3
≥
a+b+c
3
4
,
3
so a4 + b4 + c4
b + c)
≥ (a + b + c) · (a + 27
.
By the arithmetic mean-geometric mean inequality,
4
4
Hence, a + b + c
4
≥
( a + b + c)3
( a + b + c)
27
·
≥
a+b+c
3
≥
√
3
abc = 1, so a + b + c
≥ 3.
33
( a + b + c) = a + b + c.
27
Solution 3.
Rather than using the Power-Mean inequality to prove a4 + b4 + c4
the Cauchy-Schwartz-Bunjakovsky inequality can be used twice:
(a4 + b4 + c4 )(12 + 12 + 12 )
(a2 + b2 + c2 )(12 + 12 + 12 )
So
a4 + b4 + c4
3
≥ (a
2
+ b 2 + c 2 )2
9
4
b + c)
≥ (a + 81
≥ a + b + c in Proof 2,
(a2 + b2 + c2 )2
≥
(a + b + c)2
. Contin
Continue
ue as in Proo
Prooff 2.
√
4. Let Γ be a circle with radius r. Let A and B be distinct points on Γ such that AB < 3r .
Let the circle with centre B and radius AB meet Γ again at C . Le
Lett P be the point inside
Γ such that triangle ABP is equ
equilat
ilatera
eral.
l. Finally
Finally,, let CP meet Γ again at Q. Prov
Provee that
that
P Q = r.
Γ
O
P
Q
C
B
A
Solution 1.
B C , let θ = BP C = BC P .
Let the center of Γ be O, the radius r. Since BP = BC
Quadrilateral QABC is cyclic, so BAQ = 180
Also AP Q = 180
◦
−
AP B
−
BP C
= 120
◦
Again because quadrilateral QABC is cyclic,
◦
θ and hence
P AQ
= 120
◦
θ.
− −θ, so P Q = AQ and AQP =−2 θ − 60 .
ABC = 180 − AQC = 240 − 2θ .
◦
◦
◦
OC B are congruent, since OA = OB = O
OC
C = r and AB = B C .
Triangles OAB and OCB
1
Thus ABO = CB O = ABC = 120
θ.
2
We have now shown that in triangles AQP and AOB , P AQ = BAO = AP Q = ABO .
OB
B = r.
Also AP = AB , so AQP = AOB . Hence QP = O
◦
−
∼
Solution 2.
Let the center of Γ be O, the radius r. Sinc
Sincee A, P and C lie on a circle centered at B ,
60 = ABP = 2ACP , so ACP = ACQ = 30 .
◦
◦
Since Q, A, and C lie on Γ, QOA = 2QCA = 60 .
◦
So QA = r since if a chord of a circle subtends an angle of 60 at the center, its length is the
radius of the circle.
◦
B C , so BP C = BC P = ACB + 30 .
Now BP = BC
◦
Thus AP Q = 180
◦
−
AP B
−
BP C
= 90
◦
−
ACB .
Q, A, B a
BC,
C, AQP = AQC = AQB + BQC = 2ACB .
Since Q,
and
nd C lie on Γ and AB = B
Finally, QAP = 180 AQP AP Q = 90 ACB .
−
−
So P AQ = AP Q hence P Q = AQ = r .
−
5. Let
N
{
}
= 0, 1, 2, . . . . Determine all functions f : N
→
N
such that
xf (y ) + yf (x) = (x + y )f (x2 + y 2 )
for all x and y in N.
Solution 1.
We claim that f is a constant funct
function.
ion. Suppose,
Suppose, for a contradiction,
contradiction, that there exist x and
a nd
y with f (x) < f (y ); choose x, y such that f (y ) f (x) > 0 is minimal. Then
f (x) =
−
xf (x) + yf (x)
xf (y ) + yf (x)
xf (y ) + yf (y )
<
<
= f (y )
x+y
x+y
x+y
so f (x) < f (x2 + y 2 ) < f (y ) and 0 < f (x2 + y 2 ) f (x) < f (y ) f (x), contradicting the
choice of x and y . Th
Thus,
us, f is a constant function.
function. Since f (0) is in N, the constant must be
from N.
−
Also, for any c in
to the equation.
N,
−
xc + yc = (x + y )c for all x and y , so f (x) = c, c
∈
N
are the solutions
Solution 2.
−
We claim f is a constant function. Define g (x) = f (x) f (0). Then g (0) = 0, g (x)
and
xg (y ) + yg (x) = (x + y )g (x2 + y 2 )
≥ −f (0)
for all x, y in N.
Letting y = 0 shows g (x2 ) = 0 (in particular, g (1) = g (4) = 0), and letting x = y = 1 shows
g (2) = 0. Also, if x, y and z in N satisfy x2 + y 2 = z 2 , then
−xy g(x).
g (y ) =
∗
∗
( )
Letting x = 4 and y = 3, ( ) shows that g (3) = 0.
For any even number x = 2n > 4, let y = n 2 1. Then y > x and x2 + y 2 = (n2 + 1) 2 . For
any odd number x = 2n + 1 > 3, let y = 2(n + 1)n. Then y > x and x 2 + y 2 = ((n + 1)2 + n2 )2 .
Thus for every x > 4 there is y > x such that ( ) is satisfied.
−
∗
Suppose for a contradiction, that there is x > 4 withxig+1
(x)g >
0. The
Then
n we can constru
construct
ct a
sequence x = x0 < x1 < x2 < . . . where g (xi+1 ) =
followss that g (xi+1 ) >
(xi ). It follow
−
xi
|g(x )| and the signs of g (x ) alternate. Since g (x) is always an integer, |g(x
Thus for some sufficiently large value of i, g (x ) < −f (0), a contradiction.
i+1
i
i
i
|
|
)| ≥ |g (x )| + 1.
i
As for Proof 1, we now conclude that the functions that satisfy the given functional equation
are f (x) = c, c N.
∈
Solution 3.
Suppose that W is the set of nonnegative integers and that f : W
xf (y ) + yf (x) = (x + y )f (x2 + y 2 ).
→ W satisfies:
(∗)
We will show that f is a constant function.
Let f (0) = k , and set S = x f (x) = k .
Letting y = 0 in ( ) shows that f (x2 ) = k
∗
{ |
}
x2
∀
x > 0, and so
∈S ∀
x>0
(1)
In particular, 1
∈ S.
Suppose x 2 + y 2 = z 2 . Then yf (x) + xf (y ) = (x + y )f (z 2 ) = (x + y )k . Thus,
x
∈S
iff y
∈ S.
(2)
whenever x2 + y 2 is a perfect square.
For a contradiction, let n be the smallest non-negative integer such that f (2n ) = k . By (l) n
−
−
n 1
n 1
1
n
1
n
2
2
k.. Letting x = y = 2
must be odd, so 2 is an in
integ
teger.
er. Now 2 < n so f (2
)=k
in ( ) shows f (2n ) = k , a contradiction. Thus every power of 2 is an element of S .
−
∗
For each integer n
−
≥ 2 define p(n) to be the largest prime such that p (n) | n.
For any integer n > 1 that is not a power of 2, there exists a sequence of integers
x1 , x2 , . . . , xr such that the following conditions hold:
Claim:
a) x1 = n .
b) x2i + x2i+1 is a perfect square for each i = 1, 2, 3, . . . , r
c) p(x1 )
≥ p(x ) ≥ . . . ≥ p(x ) = 2.
2
r
n
Proof:
− 1.
p n
p x
p x
a
Since
is nota and
a power
of 2,p (b( ) )<=2 m(+11.
)
b(2m + 1)
b , where
, for some
≥ 3.3.
m
Le
Lett ( 1 ) = 2
n
+ 1,
1, so
x
=
1
=
Case 1: a = 1. Since (2m+1 , 2m2 +2m, 2m2 +2m +1) is a Pythagorean Triple, if x2 = b (2m2 +
2m), then x21 + x22 = b 2 (2m2 + 2m + 1) 2 is a perfect square. Furthermore, x2 = 2bm(m + 1),
and so p(x2 ) < 2 m + 1 = p (x1 ).
Case 2: a > 1. If n = x1 = (2m + 1)a b , let x2 = (2m + 1)a 1 b (2 m2 + 2m), x3 =
(2m + 1)a 2 b (2m2 + 2m)2 , . . ., x a+1 = (2m + 1)0 b (2m2 + 2m)a = b 2a ma (m + 1)a . Note
that for 1 i a , x 2i + x2i+1 is a perfect square and also note that p (xa+1 ) < 2 m + 1 = p (x1 ).
·
· ·
≤ ≤
−
−
· ·
· ·
·
If xa+1 is not a power of 2, we extend the sequence xi using the same procedure described
above. We keep doing this until p(xr ) = 2, for some integer r.
∈
∈
−
∈
∈
By (2), xi S iff xi+1 S for i = 1, 2, 3, . . . , r 1. Thus,
Thus, n = x1 S iff xr S . But xr is
a power of 2 because p(xr ) = 2, and we earlier proved that powers of 2 are in S. Therefore,
n S , proving the claim.
∈
≥
We have proven that every integer n
1 is an element of S , and so we have proven that
f (n) = k = f (0), for each n
1. Therefore, f is constant, Q.E.D.
≥
Soluti
Sol
utions
ons to the 200
2003
3 CMO
written March 26, 2003
1. Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let m be an integer, with 1 m 720. At precisely m minutes after 12:00, the
angle made by the hour hand and minute hand is exactly 1 . Deter
Determine
mine all pos
possible
sible
≤ ≤
◦
values of m.
m.
Solution
The minute hand makes a full revolution of 360 every 60 minutes, so after m minutes
it has swept through 360
m = 6m degrees. The hour hand makes a full revolution every
60
360
12 hours (720 minutes), so after m minutes it has swept through 720
m = m/
m/22 degrees.
Since both hands started in the same position at 12:00, the angle between the two
hands will be 1 if 6m m/
m/22 = 1 + 360k
360k for some integer k. Solvi
Solving
ng this equa
equation
tion
we get
720
720k
k 2
5k
5k 2
m=
= 65
65k
k+
.
11
11
Since 1 m 720, we have 1 k 11. Since m is an integer, 5k
5k 2 must be divisible
◦
◦
−
±
±
≤ ≤
≤ ≤
by 11, say 5k
5k ± 2 = 11q
11q . Then
5k = 11q
11q ± 2 ⇒
±
±
k = 2q +
q
± 2.
5
If is now clear that only q = 2 and q = 3 sat
satisf
isfy
y all the cond
conditi
itions
ons.. Th
Thus
us k = 4 or
k = 7 and substituting these values into the expression for m we find that the only
possible values of m are 262 and 458.
2001
2. Find the last three digits of the number 2003 2002
.
Solution
2001
We must find the remainder when 20032002
is divided by 1000, which will be the
2002
same as the remainder when 3
is divided by 1000, since 2003 3 (mod 1000).
To do this we will first find a positive integer n such that 3n 1 (mod 1000) and then
2001
≡
2001
in the form nk + r , so that
try to express 2002
2001
20032002
Since 32 = 10
32m = (10
nk +r
≡3
m
r
n k
r
k
r
≡ (3 ) · 3 ≡ 1 · 3 ≡ 3
− 1, we can evaluate 3
− 1)
≡
2m
using the binom
binomial
ial theor
theorem:
em:
= ( 1)m + 10m
10m( 1)m
−
−
(mod 1000).
1000).
1
−
+ 100
− −
m(m 1)
( 1)m
2
2
−
+
· · · + 10
m
.
After the first 3 terms of this expansion, all remaining terms are divisible by 1000, so
letting m = 2q , we have that
34q
≡ 1 − 20q
20q + 100q
100q (2q
(2q − 1) (mod 1000).
1000).
(1)
≡ 1 (mod 1000) and now we wish to find the
we can
can che
hecck that
that 3
100
Usingg th
Usin
this
is,,
remainder when 20022001 is divided by 100.
Now 20022001 22001 (mod 100) 4 21999 (mod 4 25), so we’ll investigate powers of
2 modulo 25. Noting that 210 = 1024
1 (mod 25), we have
≡
≡ ·
21999 = (210 )199
·
≡−
· 2 ≡ (−1) · 512 ≡ −12 ≡ 13 (mod 25).
25).
9
199
4 13 = 52 (m
(mod
od 100
100).
). The
Theref
refore
ore 20022001 can be written in the form
Thus 22001
100k
100
k + 52 for some integer k,
k , so
≡ ·
2001
20032002
52
≡3
(mod 1000)
≡ 1 − 20 · 13 + 1300 · 25 ≡ 241 (mod 1000)
2001
using equation (1). So the last 3 digits of 20032002
are 241.
3. Find all real positive solutions (if any) to
x3 + y 3 + z 3 = x + y + z, and
x2 + y 2 + z 2 = xyz.
Solution
Solut
ion 1
abovee is equiva
equivalen
lentt
(z 3 z ). The first equation abov
( y 3 y ) + (z
Let f (x,y,z ) = (x3 x) + (y
to f (x,y,z ) = 0. If x,
x, y,z 1, then f (x,y,z ) 0 with equality only if x = y = z = 1.
But if x = y = z = 1, then the second equation is not satisfied.
satisfied. So in any solution to
the system
system of equat
equation
ions,
s, at least one of the vari
ariabl
ables
es is les
lesss tha
than
n 1. Wit
Withou
houtt los
losss of
generality, suppose that x < 1. Then
−
≥
−
−
≥
x2 + y 2 + z 2 > y 2 + z 2
≥ 2yz > yz > xyz.
Therefore the system has no real positive solutions.
Solution
Solut
ion 2
We will show that the system has no real positive solution. Assume otherwise.
The second equation can be written x2 (yz )x + (y
( y 2 + z 2 ). Since this quadratic in x
has a real solution by hypothesis, its discrimant is nonnegative. Hence
−
y2z2
2
2
− 4y − 4z ≥ 0.
Dividing through by 4y
4y 2 z 2 yields
1
4
Hence y 2
4 and so y
≥
≥ y1
2
+
1
z2
≥ y1 .
2
2, y being positive.
positive. A similar argum
argument
ent yields x,y,z
≥
≥
But the first equation can be written as
x(x2
contradicting x
x,, y,z
2
2.
2
− 1) + y(y − 1) + z(z − 1) = 0,0,
≥ 2. Hence, a real positive solution cannot exist.
Solution
Solut
ion 3
Applying the arithmetic-geometric mean inequality and the Power Mean Inequalities
to x
x,, y,z we have
√xyz ≤ x + y + z ≤
3
3
x + y + z x + y + z
2
2
2
≤
3
3
3
3
3
3
.
Letting S = x + y + z = x3 + y 3 + z 3 and P = xy
xyzz = x2 + y 2 + z 2 , this inequality can
be written
S
P
S
P
.
3
3
3
√
3
≤ ≤
≤
3
√ ≤
Now P
implies P ≤ P /27, so P ≥ 27. Also ≤
implies S /27 ≤ S/3,
S/3,
√
≤ 1 which is inconsistent with √P ≤ .
so S ≤ 33.. Bu
Butt th
then
en P ≥ 3 and
3
P
2
S
3
3
3
3
3
S
3
Therefore the system cannot have a real positive solution.
3
S
3
3
3
3
S
3
4. Pro
Prove
ve that when three circles share the same chor
chord
d AB
AB,, ever
every
y line throug
through
h A different
from AB determines the same ratio X Y : Y Z , where X is an arbitrary point different
from B on the first circle while Y and Z are the points where AX intersects the other
two circles (labelled so that Y is between X and Z ).
l
A
α
X
β
Y
γ
Z
B
Solution
Solut
ion 1
Let l be a line through A different from AB and join B to A, X , Y and Z as in the
above diagram. No matter how l is chosen, the angles AX
AXB
B , AY B and AZB always
subtend the chord AB
AB.. For this reason the angles in the tria
triangles
ngles BX
B X Y and BX
B X Z are
the same for all such l.
l . Thus the ratio X Y : Y Z remains constant by similar triangles.
Note that this is true no matter how X , Y and Z lie in relation to A.
A. Suppose X , Y and
Z all lie on the same side of A (as in the diagram) and that AX
AXB
B = α,
α , AY B = β
β and
α, BY X = β , BY Z = 180
and AZB = γ . Then BX Y = 180
BZ Y = γ . Now suppose l is chosen so that X is now on the opposite side of A from
α, but
Y and Z . Now since X is on the other side of the chord AB
AB,, AX
AXB
B = 180
α and all other angles in the two pertinent
it is still the case that BX Y = 180
◦
−
◦
◦
◦
−
−
−
tria
triangles
ngles to
rremain
emain
unc
unchang
hanged.
ed.
chosen
so that
identical
then
lher
isr
tangent
the first
circle
andIfit lisisstill
the case
thatXis
Al
l ot
othe
BX
Y = 180with αA. , All
cases can be checked in a similar manner.
◦
−
l
A
X
P
Q R
O1 O2
Y
Z
O3
B
Solution
Solut
ion 2
Let m be the perpendicular bisector of AB and let O1 , O2 , O3 be the centres of the
three circles. Since AB is a chord common to all three circles, O 1 , O 2 , O 3 all lie on m.
m.
Let l be a line through A different from AB and suppose that X , Y , Z all lie on the
same
AB,, as in theSince
AB
above
Letthe
perpendiculars
from bisects
O 1 , O 2 ,any
O 3 chord,
meet l
at P , side
a diagram.
line through
centre of a circle
Q,
Q , Rof, respectively.
AX = 2AP,
AY = 22AQ
AQ and AZ = 2AR.
Now
− AX = 2(AQ
2(AQ − AP ) = 22P
P Q and, similarly
similarly,, Y Z = 22QR.
QR.
Therefore X Y : Y Z = P Q : QR
QR.. But O P || O Q || O R, so P Q : QR = O O : O O .
X Y = AY
1
2
3
1
2
2
3
Since the centres of the circles are fixed, the ratio X Y : Y Z = O 1 O2 : O 2 O3 does not
depend on the choice of l.
l.
If X , Y , Z do not all lie on the same side of AB
AB,, we can obtain the same result with
a similar proof. For instance, if X and Y are opposite sides of AB
AB,, then we will have
X Y = AY + AX , but since in this case P Q = AQ + AP , it is still the case that
X Y = 22P
P Q and result still follows, etc.
5. Let S be a set of n points in the plane such that any two points of S are at least 1
unit apart. Prove there is a subset T of S with at least n/7
n/7 points such that any two
points of T are at least 3 units apart.
√
Solution
We will construct the set T in the follo
following
wing wa
way:
y: Assum
Assumee the point
pointss of S are in the
xy
xy-plane
-plane and let P be a point in S with maximum y-coordinate.
y -coordinate. This point P wil
willl be a
member of the set T and now, from S , we will remove P and all points in S which are
less than 3 units from P . From the remaining points we choose one with maximum
y -coordinate to be a member of T and remove from S all points at distance less than
3 units from this new point
point.. We con
contin
tinue
ue in this wa
way
y, unt
until
il all the points of S are
exhausted.
exhau
sted. Clear
Clearly
ly any two point
pointss in T are at least 3 units apart. To show that T
has at least n/
n/77 points, we must prove that at each stage no more than 6 other points
are removed along with P .
√
√
√
At a typical stage in this process, we’ve selected a point P with maximum y-coordinate,
y -coordinate,
so any points at distance less than 3 from P must lie inside the semicircular region
of radius 3 centred at P shown in the first diagram below.
below. Since point
pointss of S are at
least 1 unit apart, these points must lie outside (or on) the semicircle of radius 1. (So
they lie in the shaded region of the first diagram.) Now divide this shaded region into
6 congruent regions R1 , R2 , . . . , R6 as shown in this diagram.
√
√
We will show that each of these regions contains at most one point of S . Sinc
Sincee all 6
regions are congruent, consider one of them as depicted in the second diagram below.
The distance between any two points in this shaded region must be less than the length
of the line segment AB
AB.. The le
lengt
ngths
hs of P A and P B are 3 and 1, respectively, and
angle AP B = 30 . If we constr
construct
uct a perpendicular fro
from
m B to P A at C , then the length
of P C is cos30 = 3/2. Thus
Thus BC is a perpendicular bisector of P A and therefore
AB = P B = 1. So the distance betw
between
een any two point
pointss in this regio
region
n is less than 1.
Therefore each of R 1 , . . . , R6 can contain at most one point of S , which completes the
proof.
√
◦
◦
√
1
√
P
3
R1
R6
R2
R5
R3
R4
P
C
◦
30
B
A
Solu
So
luti
tion
ons
s to th
the
e 20
2004
04 CM
CMO
O
written March 31, 2004
1. Find all ordered triples (x,y,z) of real numbers which satisfy the following system of
equations:
xy = z − x − y
Solution
Solut
ion 1
xz = y − x − z
yz = x − y − z
Subtracting the second equation from the first gives xy − xz = 2z − 2y . Factoring y − z
from each side and rearranging gives
(x + 2)(y − z ) = 0 ,
so either x = −2 or z = y .
If x = −2, the first equation becomes −2y = z + 2 − y , or y + z = −2. Subst
Substituti
ituting
ng
x = −2, y + z = −2 into the third equation gives y
yzz = −2 − (−2) = 0. Hence either y
or z is 0, so if x = −2, the only solutions are (−2, 0, −2) and (−2, −2, 0).
If z = y the first equation becomes xy = −x, or x(y + 1) = 0. If x = 0 and z = y ,
the third equation becomes y 2 = −2y which gives y = 0 or y = −2. If y = −1 and
z = y = −1, the third equation gives x = −1. So if
if y = z , the only solutions are
(0, 0, 0), (0, −2, −2) and (−1, −1, −1).
In summary
summary, ther
theree are 5 solut
solutions:
ions: (−2, 0, −2), (−2, −2, 0), (0, 0, 0), (0, −2, −2) and
(−1, −1, −1).
Solution
Solut
ion 2
Adding x to both sides of the first equation gives
x(y + 1) = z − y = (z + 1) − (y + 1) ⇒ (x + 1)(y + 1) = z + 1 .
Similarly
thesystem
other two
equations
letting a = x + 1, b = y + 1,
c = z + 1, manipulating
we can write the
in the
followingand
way.
ab = c
ac = b
bc = a
If any one of a,b,c is 0, th
then
en it’s
it’s clear
clear th
that
at all three
three are 0. So ( a,b,c) = (0, 0, 0) is
one solution and now suppose that a
a,, b, c are all nonzero.
nonzero. Subst
Substitut
ituting
ing c = ab into the
2
2
second and third equations gives a b = b and b a = a, respec
respective
tively
ly.. Hence a2 = 1,
b2 = 1 (since a, b nonze
nonzero).
ro). This gives
gives 4 more solutions:
solutions: (a,b,c) = (1, 1, 1), (1, −1, −1),
x,y,
,y, z , we obtain the 5 ordered
(−1, 1, −1) or (−1, −1, 1). Ree
Reexpr
xpress
essing
ing in terms
terms of x
triples listed in Solution 1.
2. How many ways can 8 mutually non-attacking rooks be placed
on the 9 × 9 chessboard (shown here) so that all 8 rooks are on
squares of the same colour?
[Two rooks are said to be attacking each other if they are placed
in the same row or column of the board.]
Solution
Solut
ion 1
We will first count the number of ways of placing 8 mutually non-attacking rooks on
black squares and then count the number of ways of placing them on white squares.
Suppose that the rows of the board have been numbered 1 to 9 from top to bottom.
First notice that a rook placed on a black square in an odd O O O O O
E
E
E
E
numbered row cannot attack a rook on a black square in an even
O
O
O
O
O
numbered row. This effectively partitions the black squares into
E
E
E
E
a 5 × 5 board and a 4 × 4 board (squares labelled O and E O O O O O
respectively, in the diagram at right) and rooks can be placed O E O E O E O E O
independent
indepen
dently
ly on these two
two boards. There are 5! ways to place
E
E
E
E
O
O
O
O
O
5 non-attacking rooks on the squares labelled O and 4! ways to
place 4 non-attacking rooks on the squares labelled E .
This gives 5!4! ways to place 9 mutually non-attacking rooks on black squares and
removing
remo
ving any one of these 9 rooks give
givess one of the desired configurat
configurations
ions.. Thu
Thuss there
are 9 · 5!4! ways to place 8 mutually non-attacking rooks on black squares.
Using very similar reasoning we can partition the white squares
as shown
shown in the dia
diagra
gram
m at right.
right. The white
white squ
square
aress are parpartitioned into two 5 × 4 boards such that no rook on a square
marked O can attack a rook on a square mark E . At mos
mostt 4
non-attacking rooks can be placed on a 5 × 4 board and they
can be placed in 5 · 4 · 3 · 2 = 5! ways. Thus there are (5!) 2 ways
to place 8 mutually non-attacking rooks on white squares.
O
E
O
E
O
E
O
E
O
E
E
E
E
E
E
O
E
O
E
O
E
O
O
O
E
O
O
E
O
O
O
E
O
E
E
O
E
O
E
O
In total there are 9 · 5!4! + (5!)2 = (9 + 5)5!4! = 14 · 5!4! = 40320 ways to place 8
mutually non-attacking rooks on squares of the same colour.
Solution
Solut
ion 2
Consider rooks on black squares first. We have 8 rooks and 9 rows, so exactly one row
will be without rooks. There are two cases: either the empty row has 5 black squares
or it has 4 black squares. By permutation these rows can be made either last or second
last. In each case we’ll count the possible number of ways of placing the rooks on the
board as we proceed row by row.
In the first case we have 5 choices for the empty row, then we can place a rook on any
of the black squares in row 1 (5 possibilities) and any of the black squares in row 2 (4
possibiliti
possib
ilities).
es). When we attempt to place a rook in row 3, we mus
mustt avoid the column
containing the rook that was placed in row 1, so we have 4 possibilities. Using similar
reasoning, we can place the rook on any of 3 possible black squares in row 4, etc. The
total number of possibilities for the first case is 5 · 5 · 4 · 4 · 3 · 3 · 2 · 2 · 1 = (5!)2 . In the
second case, we have 4 choices for the empty row (but assume it’s the second last row).
We now place rooks as before and using similar logic, we get that the total number of
possibilities for the second case is 4 · 5 · 4 · 4 · 3 · 3 · 2 · 1 · 1 = 4(5!4!).
Now, do the same
Now,
same for the white
white squares.
squares. If a ro
row
w wit
with
h 4 white square
squaress is empty
empty (5
2
ways to choose it), then the total number of possibilities is (5!) . It’
It’ss impossi
impossible
ble to
have a row with 5 white squares empty, so the total number of ways to place rooks is
(5!)2 + 4(5!4!) + (5!)2 = (5 + 4 + 5)5!4! = 14(5!4!) .
3. Let A
A,, B,C,D be four points on a circle (occurring in clockwise order), with AB < AD
and BC > CD. Let the bisector of angle BAD meet the circle at X and the bisector
of angle BC D meet the circle at Y . Consider the hexagon formed by these six points
on the circle. If four of the six sides of the hexagon have equal length, prove that BD
must be a diameter of the circle.
B
A
X
α
α
γ
γ
Y
C
D
Solution
Solut
ion 1
We’re given that AB < AD. Since C Y bisects BC D, BY = Y D, so Y lies between
D and A on the circle, as in the diagram above, and D
DY
Y > Y A, D
DY
Y > AB
AB . Similar
reasoning confirms that X lies between B and C and BX > XC , BX > CD. So if
ABXCDY has 4 equal sides, then it must be that Y A = AB = X C = C D.
Let
DCY
Y = γ . Sinc
= DAX = α and let BC Y = DC
Sincee ABCD is cyclic,
A+C = 180 , which implies
implies that α+γ = 90 . The ffact
act tthat
hat Y A = AB = X C = C D
means that the arc from Y to B (which is subtended by Y C B ) is equal to the arc from
X to D (which is subtended by X AD). He
Henc
ncee Y C B = X AD, so α = γ = 45 .
Finally, B D is subtended by BAD = 2α = 90 . There
Therefore
fore BD is a diameter of the
BAX
◦
◦
◦
◦
circle.
Solution
Solut
ion 2
We’re given that AB < AD. Since C Y bisects BC D, BY = Y D, so Y lies between
D and A on the circle, as in the diagram above, and D
DY
Y > Y A, D
DY
Y > AB
AB . Similar
reasoning confirms that X lies between B and C and B X > X C , B X > C D . So
if ABXCDY has 4 equal sides, then it must be that Y A = AB = X C = C D.
This implies that the arc from Y to B is equal to the arc from X to D and hence
that Y B = X D. Si
Sinc
ncee BAX = X AD, BX = X D and since DC Y = Y C B ,
DY = Y B . There
Therefore
fore BXDY is a square and its diagonal, BD , must be a diameter
of the circle.
4. Let p be an odd prime. Prove that
p−1
k 2p
1
−
≡
p(p + 1)
(mod p2 ).
2
k=1
[Note that a ≡ b (mod m) means that a − b is divisible by m .]
Solution
Since p − 1 is even, we can pair up the terms in the summation in the following way
(first term with last, 2nd term with 2nd last, etc.):
p
p−1
2
k
2p−1
k =1
Expanding (p − k )2p
1
−
(p − k)2p
1
−
−1
=
k
2p−1
+ (p − k )
2p−1
k =1
.
with the binomial theorem, we get
= p2p
1
−
2p − 1
−···−
p2 k 2p
3
−
2p − 1
+
pk 2p
2
−
− k 2p 1 ,
−
2
1
where every term on the right-hand side is divisible by p2 except the last two. Therefore
k 2p
1
−
+ (p − k )2p
1
−
≡ k 2p
1
−
+
2 1
p−
1
pk 2p
2
−
− k 2p
1
−
≡ (2p − 1)pk 2p
2
−
(mod p2 ).
For 1 ≤ k < p, k is not divisible by p, so k p 1 ≡ 1 (mod p), by Fermat’s
Fermat’s Little Theorem.
So (2p − 1)k2p 2 ≡ (2 p − 1)(12 ) ≡ −1 (mod p), say (2p − 1)k 2p 2 = mp − 1 for some
integer m . Then
(2p − 1)pk2p 2 = mp2 − p ≡ −p (mod p2 ).
−
−
−
−
Finally,
p
p−1
−1
2
k
2p−1
≡
k =1
(
−p) ≡
k =1
≡
p − p2
2
1 (
p−
+ p2 ≡
2
−p) (mod p2 )
p(p + 1)
2
(mod p2 ).
5. Let T be the set of all positive
positive integer
integer divisors
divisors of 2004100 . What is the largest possible
number of elements that a subset S of T can have if no element of S is an integer
multiple of any other element of S ?
Solution
Assume throughout that a
a,, b, c are nonnegative integers. Since the prime factorization
2
of 2004 is 2004 = 2 · 3 · 167,
T
Let
= 2 3 167 0
200 0
= 2
3 167 0
a b
S
c
≤a≤
200−b−c
b
,
c
100
100
≤ b, c ≤
≤ b, c ≤
.
.
For any 0 ≤ b, c ≤ 100, we have 0 ≤ 200 − b − c ≤ 200, so S is a subset of T . Since there
are 101 possible values for b and 101 possible values for c, S contains 1012 elements.
We will show that no element of S is a multiple of another and that no larger subset
of T satisfies this condition.
Suppose 2200
b c b
− −
3 167c is an integer multiple of 2 200
j −k j
−
200 − b − c ≥ 200 − j − k,
b ≥ j,
3 167k . Then
c ≥ k.
But this first inequality implies b + c ≤ j + k , which together with b ≥ j , c ≥ k gives
b = j and c = k . Hence no element of S is an integer multiple of another element of S .
Let U be a subset of T with more than 1012 elements. Since there are only 1012 distinct
pairs (b, c) with 0 ≤ b, c ≤ 100, then (by the pigeonhole principle) U must contain two
elements u 1 = 2a 3b 167c and u 2 = 2a 3b 167c , with b 1 = b 2 and c1 = c 2, but a 1
= a 2.
If a 1 > a2, then u 1 is a multiple of u 2 and if a 1 < a2 , then u 2 is a multiple of u 1 . Hence
U does not satisfy the desired condition.
1
1
1
2
2
2
Therefore the largest possible number of elements that such a subset of T can have is
1012 = 10201.
Solut
So
lution
ions
s to the 200
2005
5 CM
CMO
O
written March 30, 2005
1. Consider an equilateral
equilateral triangle of side length
length n
n,, which is divided into unit triangles, as
shown. Let f (n) be the number of paths from the triangle in the top row to the middle
triangle in the bottom row, such that adjacent triangles in our path share a common
edge and the path never travels up (from a lower row to a higher row) or revisits a
triangle.
trian
gle. An example of one such path is illustrated
illustrated below for n = 5. Det
Determ
ermine
ine the
value of f (2005).
Solution
We shall show that f (n) = (n
(n
− 1)!.
Label the horizontal line segments in the triangle l1 , l2, . . . as in th
thee di
diag
agra
ram
m be
belo
low.
w.
Since the path goes from the top triangle to a triangle in the bottom row and never
travels up, the path must cross each of l 1 , l2 , . . . , ln−1 exactly once. The diagonal lines
in the triangle divide l k into k unit line segments and the path must cross exactly one
of these k segments for each k . (In the diagram below, these line segments have been
highli
hig
hligh
ghted
ted.)
.) The path is com
comple
pletel
tely
y det
determ
ermine
ined
d by the set of n 1 line segments
whic
wh
ich
h ar
aree cr
cros
osse
sed.
d. So as th
thee pat
path
h mo
mov
ves from the k th row to the (k
(k + 1)st row,
there are k possible line segments where the path could cross lk . Si
Sinc
ncee ther
theree are
are
1 2 3 (n 1) = (n
(n 1)! ways that the path could cross the n 1 horizontal lines,
and each one corresponds to a unique path, we get f (n) = (n 1)!.
−
· · ··· −
−
−
−
Therefore f (2005) = (2004)!.
l1
l2
l3
l4
2. Let (a,b,c
(a,b,c)) be
b e a Pytha
Pythagorea
gorean
n tripl
triple,
e, i.e., a triplet of positive integers with a2 + b2 = c2 .
a) Prove that (c/a
(c/a + c/b
c/b))2 > 8.
b) Prov
Provee that there does not exist any
any integer n for which we can find a Pythagorean
triple (a,b,c
(a,b,c)) satisfying (c/a
(c/a + c/b
c/b))2 = n
n..
Solution
Solutio
n 1
a)
Let (a,b,c
(a,b,c)) be a Pythagor
Pythagorean
ean triple
triple.. Vie
View
w a, b as lengths of the legs of a right
angled triangle with hypotenuse of length cc;; let θ be the angle determined by the
sides with lengths a and cc.. Then
2
c c 2
1
1
sin2 θ + cos2 θ + 2 si
sin
n θ cos θ
+
=
+
=
a b
cos θ sin θ
(sin θ cos θ)2
1 + sin
sin 2θ
4
4
=
+
sin2θθ
sin2 2θ
sin2 2θ sin2
Note that because 0 < θ < 90◦ , we have 0 < sin2
sin2θθ
1, with equality only if
◦
θ = 45 . But then a = b and we obtain 2 = c/a
c/a,, contradicting a, c both being
integers. Thus, 0 < si
sin
n 2θ < 1 which gives (c/a
(c/a + c/b
c/b))2 > 8.
= 4
≤
√
Solution
Solut
ion 2
Defining θ as in Solution 1, we have c/a + c/b = sec θ + csc θ. By the
the AM-G
AM-GM
M
inequality, we have (sec θ + csc θ)/2
sec θ csc θ. So
≥
√
√
√
√
≥ 2 2.
√
Since a
a,, b, c are integers, we have c/a + c/b > 2 2 which gives (c/a
(c/a + c/b
c/b))
c/a + c/b
2
=
sin θ cos θ
≥√
2 2
sin2θθ
sin2
2
> 8.
Solution
Solut
ion 3
By simplifying and using the AM-GM inequality,
c c
+
a b
2
= c2
2
a+b
ab
(a2 + b2 )(
)(a
a + b) 2
=
a2 b2
≥
√
√
2 a2b2 (2 ab
ab))2
= 8,
a2 b2
with equality only if a = bb.. By using the same argumen
argumentt as in Solution 1, a cannot
equal b and the inequality is strict.
Solution
Solut
ion 4
c c
+
a b
2
c2 c 2 22cc2
b2 a 2
2(a
2(
a2 + b2)
= 2+ 2+
=1 + 2 + 2 +1+
a
b
ab
a
b
ab
2
a b
2
= 2+
+2+
(a b)2 + 2ab
2ab
b a
ab
= 4+
−
−
a
b
b
a
2
+
2(a
2(
a
− b)
ab
−
2
+4
≥ 8,
with equality only if a = bb,, which (as argued previously) cannot occur.
b)
Solution
Solutio
n 1
Since c/a + c/b is rational, (c/a
(c/a + c/b
c/b))2 can only be an integer if c/a + c/b is an
integer
int
eger.. Suppose c/a + c/b = m. We may assume
assume that gcd(
gcd(a, b) = 1. (If not
not,,
divide the common factor from (a,b,c
(a,b,c),
), leaving m unchanged.)
Since cc((a + b) = mab and gcd(a,
gcd(a, a + b) = 1, a must divide cc,, say c = ak
ak.. This gives
2
2
2 2
2
2
2
a + b = a k which implies b = (k
1)a
1)
a . But then a divides b contradicting
the fact that gcd(a,
gcd(a, b) = 1. Therefore (c/a
(c/a + c/b
c/b))2 is not equal to any integer n
n..
−
Solution
Solut
ion 2
We begin as in Solution 1, supposing that c/a + c/b = m with gcd(a,
gcd(a, b) = 1.
Hence a and b are not
not both even.
even. It is also the case that
that a and b are not both
odd, for then c2 = a2 + b 2
2 (mod 4), and perfect squares are congruent to
eith
ei
ther
er 0 or 1 mod
modul
uloo 4. So one of a, b is odd and the other
other is even.
even. The
There
refor
foree
c must be odd.
Now c/a + c/b = m implies cc((a + b) = mab
mab,, which cannot be true because cc((a + b)
is odd and mab is even.
≡
3. Let S be a set of n
≥ 3 points in the interior of a circle.
a) Show that there are three distinct points a,b,c ∈ S and three distinct points
A,B,C on the circle such that a is (strictly) closer to A than any other point in
S , b is closer to B than any other point in S and c is closer to C than any other
point in S .
b) Show that for no value of n can four such points in S (and corresponding points
on the circle) be guaranteed.
Solution
Solut
ion 1
a) Let H be the smallest convex set of points in the plane which contains S .† Take
3 points a,b,c S which lie on the boundary of H . (Th
(There
ere must
must alway
alwayss be at
least 3 (but not necessarily 4) such points.)
Since a lies on the boundary of the convex region H , we can construct a chord L
such that no two points of H lie on opposite sides of L
L.. Of the two points where
the perpendicular to L at a meets the circle, choose one which is on a side of L
not containing any points of H and call this point A. Cert
Certainly
ainly A is closer to a
than to any other point on L or on the other side of L. He
Henc
ncee A is closer to a
∈
than to any other point of S . We can find the requir
required
ed points
points B and C in an
analogous way and the proof is complete.
[Note that this argument still holds if all the points of S lie on a line.]
P
A
√
3
r
2
a
a
c
r
b
L
H
r
b
Q
(a)
c
R
(b)
b) Let P QR be an equilateral triangle inscribed in the circle and let a
a,, b, c be midpoints of the three sides of P QR
QR.. If r is the radius of the circle, then every
point on the circle is within ( 3/2)
2)rr of one of a
a,, b or cc.. (See figure (b) above.)
Now 3/2 < 99//10, so if S consists of a
a,, b, c and a cluster of points within rr//10 of
the centre of the circle, then we cannot select 4 points from S (and corresponding
points on the circle) having the desired property.
√
†
√
By the way, H is called the convex hull of S . If the points of S lie on a line, then H will be the shortest
line segment containing the points of S . Other
Otherwise,
wise, H is a polygon whose vertices are all elements of S and
such that all other points in S lie inside or on this polygon.
Solution
Solut
ion 2
a) If all the points of S lie on a line L, then choose any 3 of them to be a
a,, b, c. Let
A be a point on the circle which meets the perpendicular to L at a. Clearly A is
closer to a than to any other point on L
L,, and hence closer than other other point
in S . We find B and C in an analogous way.
Otherwise, choose a
a,, b, c from S so that the triangle formed by these points has
maximal area. Construct the altitude from the side bc to the point a and extend
this line until it meets the circle at A
A.. We claim that A is closer to a than to any
other point in S .
Suppose not. Let x be a point in S for which the distance from A to x is less than
the distance from A to a. Then the perpendicular distance from x to the line bc
must be greater than the perpendicular distance from a to the line bc
bc.. But then
the triangle formed by the points x
x,, b, c has greater area than the triangle formed
by a
a,, b, c, contradicting the original choice of these 3 points. Therefore A is closer
to a than to any other point in S .
The points B and C are found by constructing similar altitudes through b and cc,,
respectively.
b) See Solution 1.
4. Let AB
ABC
C be a triangle with circumradius R
R,, perime
perimeter
ter P and area K . Determine the
maximum value of KP/R3 .
Solution
Solut
ion 1
Since similar triangles give the same value of KP/R3 , we can fix R = 1 and maximize
KP over all triangles inscribed in the unit circle. Fix points A and B on the unit circle.
The locus of points C with a given perimeter P is an ellipse that meets the circle in at
most four points.
points. The area
area K is maximized (for a fixed P ) when C is chosen on the
perpendicular bisector of AB
AB,, so we get a maximum value for K P if C is where the
perpendicular bisector of AB meet
meetss the circle. Thu
Thuss the maximum value
value of K P for
a given AB occurs when ABC is an isosceles triangle. Repeating this argument with
BC fixed, we have that the maximum occurs when ABC is an equilateral triangle.
Consider an equilateral triangle with side length a
a.. It has P = 3a
3 a. It has height equal
2 R = a/ sin(60) giving
to a 3/2 giving K = a2 3/4. ¿From the extended law of sines, 2R
R = a/ 3. Therefore the maximum value we seek is
√
√
√
√
KP/R3 =
a2 3
4
(3a
(3
a)
√3
a
3
=
Solution
Solut
ion 2
27
.
4
From the extended law of sines, the lengths of the sides of the triangle are 2R
2 R sin A,
2R sin B and 2R
2R sin C . So
1
P = 2R
2 R(sin A + sin B + sin C ) and K = (2
(2R
R sin A)(2
)(2R
R sin B )(sin C ),
2
giving
KP
= 4sin A sin B sin C (sin A + sin B + sin C ).
R3
We wish to find the maximum value of this expression over all A + B + C = 180◦ .
Using well-known identities for sums and products of sine functions, we can write
cos(B
cos(B
KP
=
4sin
A
R3
2
− C ) − cos(
cos(B
B + C)
2
sin A + 2 si
sin
n
−
B+C
2
cos
B
C
2
.
If we first consider A to be fixed, then B + C is fixed also and this expression takes
its maximum value when cos(B
cos(B C ) and cos B −2 C equal 1; i.e. when B = C . In a
similar way, one can show that for any fixed value of B , KP/R3 is maximized when
A = C . The
Theref
refore
ore the maximum
maximum value
value of KP/R3 occurs when A = B = C = 60◦ ,
and it is now an easy task to substitute this into the above expression to obtain the
maximum value of 27/
27/4.
−
Solution
Solut
ion 3
As in Solution 2, we obtain
KP
= 4sin A sin B sin C (sin A + sin B + sin C ).
R3
From the AM-GM inequality, we have
sin A sin B sin C
giving
KP
R3
≤
sin A + sin B + sin C
3
3
,
≤ 274 (sin A + sin B + sin C ) ,
4
with equality when sin A = sin B = sin C . Sin
Since
ce the sine functio
function
n is concave
concave on the
interval from 0 to π
π,, Jensen’s inequality gives
sin A + sin B + sin C
3
≤
√
A+B+C
sin
3
π
3
= sin =
.
3
2
Since equality occurs here when sin A = sin B = sin C also, we can conclude that the
3
maximum value of KP/R is
4
27
√
3
2
3
4
= 27
27//4.
5. Let’s say that an ordered triple of positive integers (a,b,c
( a,b,c)) is n-powerful if a b c,
n
n
n
gcd(a,b,c
gcd(
a,b,c)) = 1, and a + b + c is divisible by a + b + cc.. For examp
example,
le, (1,
(1, 2, 2) is
5-powerful.
≤ ≤
a) Dete
Determine
rmine all ordered
ordered triples (if any) which
which are n
n-powerful
-powerful for all n
≥ 1.
b) Determine all ordered triples (if any)
any) which are 2004-powerful
2004-powerful and 2005-powerful,
but not 2007-powerful.
[Note that gcd(a,b,c
gcd(a,b,c)) is the greatest common divisor of a
a,, b and c.]
Solution
Solut
ion 1
Let T n = a n + bn + cn and consider the polynomial
P (x) = (x a)(
)(x
x b)(
)(x
x c) = x 3 (a + b + c)x2 + (ab
(ab + ac + bc
bc))x abc.
Since P (a) = 0, we get a 3 = (a + b + c)a2 (ab + ac + bc
bc))a + abc and multiplying both
n−3
n
n−1
sides by a
we obtain a = (a + b + c)a
(ab + ac + bc
bc))an−2 + (abc
abc))an−3 . Applying
the same reasoning, we can obtain similar expressions for bn and cn and adding the
three identities we get that T n satisfies the following 3-term recurrence:
Tn = (a + b + c)Tn 1 (ab + ac + bc
bc))Tn 2 + (abc
(abc))Tn 3 , for all n 3.
−
−
¿From this we see that if T−
n−2 and T n−3 are divisible by a + b + c, then so is T n . This
immediately resolves part (b)—there are no ordered triples which are 2004-powerful
and 2005-powerful, but not 2007-powerful—and reduces the number of cases to be
considere
consi
dered
d in part (a): since all tripl
triples
es are 1-pow
1-powerfu
erful,
l, the recurrence
recurrence implies that any
ordered triple which is both 2-powerful and 3-powerful is n
-powerful for all n 1.
n-powerful
−
−
−
−
−
−
−
−
≥
≥
Putting n = 3 in the recurrence, we have
a3 + b3 + c3 = (a
( a + b + c)(
)(a
a2 + b2 + c2 ) (ab + ac + bc
bc)(
)(a
a + b + c) + 3abc
3abc
which implies that (a,b,c
(a,b,c)) is 3-powerful if and only if 3abc
3 abc is divisible by a + b + cc..
Since
a2 + b2 + c2 = (a
( a + b + c)2 2(
2(ab
ab + ac + bc
bc)),
(a,b,c
2(ab + ac + bc
a,b,c)) is 2-powerful if and only if 2(ab
bc)) is divisible by a + b + c.
−
−
≥
Suppose a prime p 5 divides a + b + c. Then p divides abc
abc.. Since gcd(a,b,c
gcd(a,b,c)) = 1, p
divides exactly one of a, b or c; but then p doesn’t divide 2(ab
2(ab + ac + bc
bc).
).
Suppose 32 divides a + b + c. Then 3 divides abc
abc,, implying 3 divides exactly one of a,
b or c. But then 3 doesn’t divide 2(ab
2(ab + ac + bc
bc).
).
Suppose 22 divides a + b + cc.. The
Then
n 4 divides
divides abc
abc.. Sin
Since
ce gcd(
gcd(a,b,c
a,b,c)) = 1, at most one
of a
a,, b or c is even, implying one of a
a,, b, c is divisible by 4 and the others are odd. But
then ab + ac + bc is odd and 4 doesn’t divide 2(ab
2( ab + ac + bc
bc).
).
So if (a,b,c
(a,b,c)) is 2- and 3-powerful, then a + b + c is not divisible by 4 or 9 or any prime
greate
gre
aterr than 3. Sin
Since
ce a + b + c is at least 3, a + b + c is eith
either
er 3 or 6. It is now
now a
simple matter to check the possibilities and conclude that the only triples which are
n-powerful for all n 1 are (1,
(1, 1, 1) and (1,
(1, 1, 4).
≥
Solution
Solut
ion 2
Let p be a prime. By Fermat’s Little Theorem,
a −1 ≡
p
1 (mod pp)), if p doesn’t divide a
a;;
0 (mod pp)), if p divides a.
a.
1
1
1
≡
Since gcd(a,b,c
gcd(a,b,c)) = 1, we
havep−that
ap−p−1 + bp− + cp−
1, 2 or 3 (mod pp).
). Therefore if
p−1
1
p is a prime divisor
divisor of a + b + c , then p equals 2 or 3. So if (a,b,c
(a,b,c)) is n
n-powerful
-powerful
for all n 1, then the only primes which can divide a + b + c are 2 or 3.
≥
We can proceed in a similar fashion to show that a + b + c is not divisible by 4 or 9.
Since
a2
≡
0 (mod 4),
4), if p is even;
1 (mod 4),
4), if p is odd
and a
a,, b, c aren’t all even, we have that a 2 + b2 + c2
≡ 1, 2 or 3 (mod 4).
By expanding (3k
(3k )3 , (3
(3k
k + 1)3 and (3k
(3k + 2)3 , we find that a3 is congruent to 0, 1 or
1 modulo 9. Hence
−
a6
≡
01 (mod
9)
dividess a
divide
a;
; ea
(mod 9),
9),, if
9),
if 33 doesn’t
divide
divid
a..
Since a
a,, b, c aren’t all divisible by 3, we have that a 6 + b6 + c6
≡ 1, 2 or 3 (mod 9).
So a2 + b2 + c2 is not divisible by 4 and a 6 + b6 + c6 is not divisible by 9. Thus if (a,b,c
(a,b,c))
is n
n-powerful
-powerful for all n 1, then a + b + c is not divisible by 4 or 9. Therefore a + b + c
is either 3 or 6 and checking all possibilities, we conclude that the only triples which
are n-powerful for all n 1 are (1,
(1, 1, 1) and (1,
(1, 1, 4).
≥
≥
See Solution 1 for the (b) part.
38th Canadian Mathematical Olympiad
Wednesday, March 29, 2006
Solutions to the 2006 CMO paper
1. Let f (n, k ) be the number of ways of distributing k candies to n children so that each child receives at most 2 candies.
For example, if n = 3, then f (3
(3,, 7) = 0, f (3
(3,, 6) = 1 and f (3
(3,, 4) = 6.
Determine the value of
f (2006
(2006,, 1) + f (2006
(2006,, 4) + f (2006
(2006,, 7) + · · · + f (2006
(2006,, 1000) + f (2006
(2006,, 1003) .
Comment. Unfortunat
Unfortunately
ely,, there was
was an error
error in the statement
statement of this problem. It was intended
intended that the sum should
continue to f (2006
(2006,, 4012).
Solution 1. The number of ways of distributing k candies to 2006 children is equal to the number of ways of distributing
0 to a particular child and k to the rest
rest,, plus
plus the number
number of ways
ways of dis
distri
tribut
buting
ing 1 to the particu
particular
lar child
child and k − 1
to the rest, plus the number of ways of distributing 2 to the particular child and k − 2 to the rest
rest.. Thus
Thus f (2006
(2006,, k ) =
f (2005
(2005,, k ) + f (2005
(2005,, k − 1) + f (2005
(2005,, k − 2), so that the required sum is
1003
1+
f (2005
(2005,, k ) .
k=1
In evaluating f (n, k ), suppose that there are r children who receive 2 candies; these r children can be chosen in
Then there are k − 2r candies from which at most one is given to each of n − r children. Hence
k/2
n
r
ways.
∞
n n − r n n − r
f (n, k) =
=
,
with
k − 2r
r
r =0
r=0
k − 2r
r
x
y
= 0 when x < y and when y < 0. The answer is
20052005 − r = 2005 2005 − r .
1003
∞
∞
r
k=0 r =0
k − 2r
1003
r
r =0
k=0
k − 2r
The desired number is the sum of the coefficients of the terms of degree not exceeding 1003 in the expansion
of (1 + x + x )
, which is equal to the coefficient of x 1003 in the expansion of
Solution 2.
2 2005
(1 + x + x2 )2005 (1 + x + · · · + x1003 ) = [(1 − x3 )2005 (1 − x)−2005 ](1 − x1004 )(1 − x)−1
= (1 − x3 )2005 (1 − x)−2006 − (1 − x3)2005 (1 − x)−2006 x1004 .
Since the degree of every term in the expansion of the second member on the right exceeds 1003, we are looking for the
coefficient of x 1003 in the expansion of the first member:
∞
2005
3 2005
(1 − x )
−2006
(1 − x)
=
(−1) 2005x (−1) −2006x
i
i
i=0
1
j
3i
j =0
j
j
2005
=
∞
(−1) 20052005 + jx
i
j
(−1) 20052005 + k − 3ix
i
3i+j
i=0 j =0
∞
=
3008−3i
2005
(Note that
i
i=1
334
334
(−1) 20053008 − 3i = (−1)
i
i
i
i=1
k
i
k =0
The desired number is
2005
2005
i=1
2005
.
(3008 − 3i)!
.
i!(2005 − i)!(1003 − 3i)!
= 0 when i ≥ 335.)
2. Le
Lett ABC be an acute-angled
acute-angled triangle.
triangle. Inscribe
Inscribe a rectangle
rectangle DEFG in this triangle so that D is on AB
AB,, E is on AC
and both F and G are on BC . Des
Descri
cribe
be the locus
locus of ( i.e., the curve occupied by) the intersections of the diagonals of all
possible rectangles DEFG
DEFG..
Solution. The locus is the line segment joining the midpoint M of B C to the midpoint K of the altitude AH . Note that
a segment DE with D on AB and E on AC determines an inscribed rectangle; the midpoint F of DE lies on the median
AM , while the midpoint of the perpendicular from F to BC is the centre
centre of the rectangle.
rectangle. This lies on the median M K of
the triangle AMH
AM H .
Conversely
Conve
rsely,, any point
p oint P on M K is the centre of a rectangle with base along BC
B C whose height is double the distance from
K to B C .
3. In a rectangular array of nonnegative real numbers with m rows and n columns, each row and each column contains at
least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements
are the same. Prove that m = n.
n.
Solution
Solut
ion 1. Consider first the case where all the rows have the same positive sum s; this covers the particular situation
in which m = 1. Then each column,
column, sharing a positive
positive element
element with some row,
row, must also have
have the sum s. Then the sum of
all the entries in the matrix is ms = ns
ns,, whence m = n.
n.
We prove the general case by induction on m
m.. The case m = 1 is already covered. Suppose that we have an m × n array
not all of whose rows
rows have the same sum. Let r < m of the rows have the sum s, and each of the of the other rows have a
different
differ
ent sum. Then every
every column sharing a positiv
positivee entry with one of these rows must also have sum s, and these are the
only columns with the sum ss.. Suppose there are c columns with sum s.
s . The situation is essentially unchanged if we permute
the rows and then the column so that the first r rows have the sum s and the first c columns have the sum s. Si
Sinc
ncee all
all
the entries of the first r rows not in the first c columns and in the first c columns not in the first r rows must be 0, we can
partition the array into a r × c array in which all rows and columns have sum s and which satisfies the hypothesis of the
problem, two rectangular arrays of zeros in the upper right and lower left and a rectangular (m
( m − r) × (n − c) array in the
lower right that satisfies the conditions of the problem. By the induction hypothesis, we see that r = c and so m = n.
n.
Solution 2. [Y. Zhao] Let the term in the ith
i th row and the j th column of the array be denoted by a ij , and let S = {(i, j ) :
aij > 0 }. Sup
Suppose
pose that
that ri is the sum of the ith row and cj the sum of the j th column.
column. Then
Then ri = cj whenever (i,
(i, j ) ∈ S .
Then we have that
aij
aij
: (i, j ) ∈ S } .
: (i, j ) ∈ S } =
{
{
cj
ri
We evaluate the sums on either side independently.
{ a
ij
ri
{ a
: (i, j ) ∈ S } =
aij
: (i, j ) ∈ S } =
{
cj
Hence m = n
n..
ij
ri
aij
: 1 ≤ i ≤ m, 1 ≤ j ≤ n} =
{
cj
m
n
m
m
1
1
: 1 ≤ i ≤ m, 1 ≤ j ≤ n} =
a =
r =
1=m .
i=1
n
j =1
ri
1
cj
ij
j =1
i=1
m
n
aij =
i=1
j =1
ri
1
cj
i
i=1
n
cj =
1=n .
j =1
2
The second solution can be made cleaner and more elegant by defining u ij = a ij /ri for all (i,
(i, j ). When a ij = 0,
= 0. When a ij > 0, then, by hypothesis, u ij = a ij /cj , a relation that in fact holds for all (i,
(i, j ). We find that
Comment.
then u ij
n
n
u
=1
ij
u
and
ij
=1
i=1
j =1
for 1 ≤ i ≤ m and 1 ≤ j ≤ n, so that (u
(uij ) is an m × n array whose row sums and column sums are all equal to 1. Hence
m
n
n
=
m=
u
{u
ij
ij
i=1
ij
j =1
j =1
m
=n
: 1 ≤ i ≤ m, 1 ≤ j ≤ n} =
u
i=1
(being the sum of all the entries in the array).
4. Consider a round-robin tournament with 2n
2 n + 1 teams, where each team plays each other team exactly once. We say
that three teams X , Y and Z , form a cycle triplet if X beats Y , Y beats Z , and Z beats X . There are no ties.
(a) Determine the minimum number of cycle triplets possible.
(b) Determine the maximum number of cycle triplets possible.
Solution 1.
(a) The minimum is 0, which is achieved by a tournament in which team T i beats Tj if and only if i > j .
(b) Any set of three teams constitutes either a cycle triplet or a “dominated triplet” in which one team beats the other
two; let there be c of the former and d of the latter. Then c + d = 2n3+1 . Suppose that team T i beats xi other teams; then
n
it is the winning team in exactly x2 dominated triples. Observe that 2i=1+1 xi = 2n2+1 , the total number of games. Hence
x 1 1 2n + 1
x −
d=
=
.
2
2
2
2
x ≥ ( x ) = n (2(2nn + 1) , whence
By the Cauchy-Schwarz Inequality, (2n
(2 n + 1)
2n + 1 x 2n + 1 n (2(2nn + 1) 1 2n + 1 n(n + 1)(2n
1)(2n + 1)
i
2n+1
2n+1
i
2
i
i=1
i=1
2n+1
i=1
2n+1
i=1
2
i
2n+1
c=
3
i
−
i=1
2
i
2
2
2
2
≤
3
−
2
+
2
2
=
6
.
To realize the upper bound, let the teams be T1 = T2n+2, T2 = T2n+3. · · ·, Ti = T2n+1+i , · · ·, T2n+1 = T4n+2. For
each i, let team Ti beat Ti+1, Ti+2, · · · , Ti+n and lose to Ti+n+1, · · · , Ti+2n . We need to chec
check
k that
that this is a consisten
consistentt
assign
ass
ignmen
mentt of wins and losses,
losses, since
since the result
result for each
each pai
pairr of te
teams
ams is defined
defined twice.
twice. This
This can be seen
seen by noting
noting that
that
(2n
(2n + 1 + i)
i) − (i + j)
j ) = 2n + 1 − j ≥ n + 1 for 1 ≤ j ≤ n . The cycle triplets are (T
( Ti , Ti+j , Ti+j +k ) where 1 ≤ j ≤ n and
(2n
(2n + 1 + i) − (i + j + k) ≤ n, i.e., when 1 ≤ j ≤ n and n + 1 − j ≤ k ≤ n. For each i,
i , this
this cou
coun
nts 1 + 2 + · · · + n = 21 n(n + 1)
cycle triplets. When we range over all i,
i , each cycle triplet gets counted three times, so the number of cycle triplets is
2n + 1 n(n + 1)
3
2
=
n(n + 1)(2n
1)(2n + 1)
.
6
[S. Eastwood] (b) Let t be the number of cycle triplets and u be the number of ordered triplets of teams
(X , Y , Z ) where X beats Y and Y beats Z . Each
Each cycle triplet
triplet generates
generates three ordered triplets
triplets while other triplets
triplets generate
exactly one. The total number of triplets is
n(4
(4n
n2 − 1)
2n + 1
.
=
3
3
Solution
Solut
ion 2.
The number of triples that are not cycle is
n(4
(4n
n2 − 1)
−t .
3
Hence
2
u = 3t +
n(4(4nn − 1)
3
− t =⇒
3
t=
n (n + 1)(2n
1)(2n + 1)
u − (2
(2n
n + 1)n
1)n2 n(
3u − n(4
(4n
n2 − 1)
.
+
=
6
2
6
If team Y beats a teams and loses to b teams, then the number of ordered triples with Y as the central element is ab.
ab.
Since a + b = 2n, by the Arithmetic-Geometric Means Inequality, we have that ab ≤ n2. Hence u ≤ (2
(2n
n + 1)n
1)n2, so that
t≤
n(n + 1)(2n
1)(2n + 1)
.
6
The maximum is attainable when u = (2
(2n
n + 1)
1)n
n2, which can occur when we arrange all the teams in a circle with each team
beating exactly the n teams in the clockwise direction.
Comment.
Interestingly enough, the maximum is
n
2
i=1 i ;
is there a nice argument that gives the answer in this form?
5. The vertices
vertices of a right
right triangle ABC inscribed in a circle
circle divide the circumfere
circumference
nce into three arcs. The right angle is
at A,
A , so that the opposite arc BC
B C is a semicircle while arc AB and arc AC are supplementary. To each of the three arcs, we
draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended
lines AB and AC . More precisely, the point D on arc BC
B C is the midpoint of the segment joining the points D and D where
the tangent at D intersects the extended lines AB and AC . Similarly for E on arc AC and F on arc AB
AB..
Prove that triangle DEF is equilateral.
Solution
Solut
ion 1.
and F Fwhere
= FF
A prime
a tangent
meets
meetred
s AB
andwathat
double
AC . It is giv
givas
enisthat
th
at
DD = DD , EE
= EEindicates
. It is required
requi
to show
sho
arc prime
EF is where
a thirditofmeets
the circumference
arc
DBF .
AF is the median to the hypotenuse of right triangle AF F , so that F F = F A and therefore
arc AF = 2∠F F A = 2(∠F F A + ∠F AF ) = 4∠F AF = 4 ∠F AB = 2 arc B F ,
whence arc F A = (2/3) arc BF A. Similarly
Similarly,, arc AE = (2/3) arc AEC . Therefore,
Therefore, arc F E is 2/3 of the semicircle, or 1/3
of the circumfer
circumference
ence as desired.
desired.
As for arc DBF , arc BD = 2∠BAD = ∠BAD + ∠BD D = ∠ADD = (1/2) arc ACD
ACD.. But, arc BF = (1/2) arc AF ,
so arc D
DBF
BF = (1/2) arc F AED
AED. That is, arc D
DBF
BF is 1/3 the circumference and the proof is complete.
Solution 2. Since AE E is a right triangle,
that ∠CD
CDD
D = ∠CAD = ∠CD D. As EADC
180◦
=
∠EAD
AE = E E = EE
E E so that ∠C AE = ∠C E E . Also AD = D D = D
DD
D , so
is a concyclic quadrilateral,
+ ∠EC D
= ∠DAC + ∠CAE + ∠EC A + ∠ACD
= ∠DAC + ∠CAE + ∠CE E + ∠CE E + ∠CD
CDD
D + ∠CD D
= ∠DAC + ∠CAE + ∠CAE + ∠CAE + ∠CAD + ∠CAD
= 3(∠DAC + ∠DAE ) = 3(∠DAE )
Hence
∠DF E
=
= 60 ◦ . Similarly,
∠DAE
= 60◦ . It follows that triangle D
DEF
EF is equilateral.
∠DEF
4
Report - Fortieth Canadian Mathematical Olympiad 2008
40th Canadian Mathematical Olympiad
Wednesday, March 26, 2008
Solutio
Solu
tions
ns - CMO 2008
1. ABCD is a convex quadrilateral in which AB is the longest side. Points M and N are located on sides
AB and B C respectively, so that each of the segments AN and C M divides the quadrilateral into two
parts of equal area. Prove that the segment M N bisects the diagonal B D .
1
Solution.
ABCD]] = [NADC ], it follows that [AN
[AN C ] = [AM
AMC
C ], so that M N �AC .
Since [MADC
[MADC ] = 2 [ABCD
Let m be a line through D parallel to AC and M N and let BA produced meet m at P and BC produced
meet m at Q.
Q . Then
[M P C ] = [M AC ] + [C
[ C AP ] = [M AC ] + [CAD
[ CAD]] = [MADC ] = [BM C ]
whence BM = M P . Similarly B N = N Q, so that M N is a midline of triangle B P Q and must bisect B D.
2. Determine
Determine all functions
functions f defined on the set of rationals that take rational values for which
f (2f
(2f (x) + f (y )) = 2x
2x + y
for each x and y .
Solution 1. The only solutions are f (x) = x for all rational x and f (x) = −x
− x for all rational x.
x . Both of
these readily check out.
Setting y = x yields f (3f
(3f (x)) = 3x
3x for all rational x.
x . Now replacing x by 3f (x), we find that
f (9x
(9x) = f (3f
(3f (3f
(3f (x)) = 3[3f
3[3f (x)] = 9f
9f (x) ,
for all rational x
x.. Setting x = 0 yields f (0) = 9f
9f (0), whence f (0) = 0.
Setting x = 0 in the given functional equation yields f (f (y )) = y for all rational y . Thus f is one-one
onto. Applying f to the functional equation yields that
2f (x) + f (y ) = f (2x
(2x + y )
for every rational pair (x,
(x, y ).
Setting y = 0 in the functional equation yields f (2f
(2f (x)) = 2x
2x, whence 2f
2f (x) = f (2
(2x
x). Theref
Therefore
ore
f (2x
(2x) + f (y ) = f (2x
(2x + y ) for each rational pair (x,
(x, y ), so that
f (u + v ) = f (u) + f (v)
for each rational pair (u,
(u, v).
6
Report - Fortieth Canadian Mathematical Olympiad 2008
Since 0 = f (0) = f (−1) + f (1), f (−1) = −f
− f (1). By induction, it can be established that for each intger
n and rational x
x,, f (nx
nx)) = nf (x). If k = f (1), we can establish from this that f (n) = nk
nk,, f (1
(1/n
/n)) = k/
k/n
n and
f (m/n
m/n)) = mk/n for each integer pair (m,
(m, n). Thus f (x) = kx
k x for all rational x.
x . Since f (f (x)) = x,
x , we must
have k
k22 = 1. Hence f (x) = x or f (x) = −x
− x. These check out.
Solution 2.
In the functional equation, let
x = y = 2f (z ) + f (w )
to obtain f (x) = f (y ) = 2z + w and
f (6z
(6z + 3w
3 w) = 66ff (z ) + 3f
3 f (w )
for all rational pairs ( z, w). Set (z, w) = (0, 0) to obtain f (0) = 0, w = 0 to obtain f (6z ) = 6f (z ) and z = 0
to obtain f (3w) = 3f (w) for all rationals z and w. Hence f (6z + 3 w) = f (6z ) + f (3w). Replacing (6z, 3w)
by (u, v) yields
f (u + v ) = f (u) + f (v)
f x) = kx where k = f (1) for all rational x . Substitution of this into the
for all rational
pairs with
( u, v).(x,Hence
functional
equation
y ) = (1(, 1)
leads to 3 = f (3f (1)) = f (3k ) = 3k 2, so that k = ±1. It ccan
an be
checked that both f (x) ≡ 1 and f (x) ≡ −1 satisfy the equation.
Acknowledgment. The first solution is due to Man-Duen Choi and the second to Ed Doolittle.
3. Let a , b , c be positive real numbers for which a + b + c = 1. Prove that
3
a − bc
b − ca
c − ab
+
+
≤
.
a + bc
b + ca
c + ab
2
Solution 1. Note that
1−
a − bc
2bc
2bc
=
=
.
a + bc
1 − b − c + bc
(1 − b)(1 − c)
The inequality is equivalent to
3
2ab
2ca
2bc
≥
+
+
.
(1 − b)(1 − c) (1 − c)(1 − a) (1 − a)(1 − b)
2
Manipulation yields the equivalent
4(bc + ca + ab − 3abc) ≥ 3(bc + ca + ab + 1 − a − b − c − abc) .
This simplifies to ab + bc + ca ≥ 9abc or
1
a
+
1
b
+
1
c
≥
9.
This is a consequence of the harmonic-arithmetic means inequality.
Solution 2. Observe that
a + bc = a (a + b + c) + bc = (a + b)(a + c)
and that a + b = 1 − c, with analogous relations for other permutations of the variables. Then
(b + c)(c + a)(a + b) = (1 − a)(1 − b)(1 − c) = ( ab + bc + ca) − abc .
Putting the left side of the desired inequality over a common denominator, we find that it is equal to
7
Report - Fortieth Canadian Mathematical Olympiad 2008
(a − bc)(1 − a) + (b − ac)(1 − b) + (c − ab)(1 − c)
(a + b + c) − (a2 + b2 + c2) − (bc + ca + ab) + 3 abc
=
(b + c)(c + a)(a + b)
(b + c)(c + a)(a + b)
1 − (a + b + c)2 + (bc + ca + ab) + 3 abc
=
(ab + bc + ca) − abc
(bc + ca + ab) + 3 abc
=
(bc + bc + ab) − abc
4abc
= 1+
.
(a + b)(b + c)(c + a)
Using the arithmetic-geometric means inequality, we obtain that
(a + b)(b + c)(c + a) = (a2 b + b2 c + c2 a) + ( ab2 + bc2 + ca2) + 2abc
≥ 3abc + 3 abc + 2 abc = 8abc ,
whence 4abc/
4abc/[(
[(a
a + b)(
b)(bb + c)(
c)(cc + a)]
a)] ≤ 21 . The desired resu
result
lt follows.
follows. Equality
Equality occurs exactly
exactly when a = b =
1
c = 3.
4. Find all functions f defined on the natural numbers that take values among the natural numbers for
which
(f (n))p ≡ n
mod f (p)
p)
for all n ∈ N and all prime numbers p.
p.
The substitution n = p,
p , a prime, yields p ≡ (f (p))
p))p ≡ 0 (mod f (p)),
p)), so that p is divisible by
f (p).
p). Hence, for each prime p,
p , f (p)
p) = 1 or f (p)
p) = p.
p.
Solution.
Let S = {p : p is prime and f (p)
p) = p}. If S is infinite, then f (n)p ≡ n (mod p) for infinitely many
primes p. By the little Fermat theorem, n ≡ f (n)p ≡ f (n), so that f (n) − n is a multiple of p for infinitely
many primes p. This
This can happen
happen only if f (n) = n for all values of n, and it can be verified that this is a
solution.
If S is empty, then f (p)
p) = 1 for all primes p
p,, and any function satisfying this condition is a solution.
Now suppose that S is finite and non-empty. Let q be the largest prime in S . Suppose, if possible, that
q ≥ 3. Theref
Therefore,
ore, for any prime
prime p exceeding q, p ≡ 1 (mod q). Howe
Howeve
ver,
r, this
this is not true.
true. Let Q be the
product of all the odd primes up to q. Th
Then
en Q + 2 must have a prime factor exceeding q and at least one
of them must be incongruent to 1 (mod q). (An alternat
alternative
ive argume
argument
nt notes that Bertrand’s
Bertrand’s postulate can
turn up a prime p between q and 2q
2q which fails to satisfy p ≡ 1 mod q .)
The only remaining case is that S = {2}. Then
Then f (2) = 2 and f (p)
p) = 1 for every odd prime p. Si
Sinc
ncee
f (n)2 ≡ n (mod 2), f (n) and n must have the same parity
parity.. Conversely
Conversely,, any function f for which f (n) ≡ n
(mod 2) for all n
n,, f (2) = 2 and f (p
p)) = 1 for all odd primes p satisfies the condition.
Therefore
There
fore the only solutions are
• f (n) = n for all n ∈ N;
p) = 1 for all primes p
p;;
• any function f with f (p)
• any function for which f (2) = 2, f (p)
p) = 1 for primes p exceeding 2 and f (n) and n have the same
parity.
rook walk
5. A
chessboard
(a rectangular
grid of
squares)
is a pathsuch
traced
byeach
a sequence
of self-avoiding
rook moves parallel
to on
an aedge
of the board
from one unit
square
to another,
that
begins
where the previous move ended and such that no move ever crosses a square that has previously been
crossed, i.e., the rook’s
ro ok’s path is non-se
non-self-in
lf-intersec
tersecting.
ting.
Let R(m, n) be the number of self-avoiding rook walks on an m × n (m rows, n columns) chessboard
which begin at the lower-left corner and end at the upper-left corner. For example, R
R((m, 1) = 1 for all
8
Report - Fortieth Canadian Mathematical Olympiad 2008
natural numbers m
m;; R(2
R (2,, 2) = 2; R(3
R (3,, 2) = 4; R(3
R (3,, 3) = 11. Find a formula for R(3
R (3,, n) for each natural
number n
n..
Solution
Solut
ion 1.
1
2
Let rnthe
= cell
R(3,
(3in
, n).
can
befrom
chec
checked
ked
dire
directly
ctlyand
tha
thattthe
r j=th1column
and r from
= 4. the
Let
Letleft,
1 ≤soi ≤
3 and
1 ≤ j;
j ; let (i,
(i, j ) denote
theItiith
th
row
the bottom
that
the
paths in question go from (1,
(1, 1) to (3,
(3, 1).
Suppose that n ≥ 3. The rook walks fall into exactly one of the following six categories:
(1) One walk given by (1,
(1, 1) → (2,
(2, 1) → (3,
(3, 1).
(2) Walks that avoid the cell (2,
(2 , 1): Any such walk must start with (1,
(1 , 1) → (1
(1,, 2) and finish with (3,
(3, 2) →
(3,
(3, 1); there are r n−1 such walks.
(3) Walks that begin with (1,
(1 , 1) → (2,
(2, 1) → (2,
(2, 2) and never return to the first row: Suc
Such
h walks enter
enter the
third row from (2,
(2, k) for some k with 2 ≤ k ≤ n and then go along the third row leftwards to (3,
(3, 1); there
are n − 1 such walks.
(4) Walks that begin with (1 , 1) → (2, 1) → ·· · → (2, k) → (1, k) → (1, k + 1) and end with (3, k + 1) →
(3, k) → (3, k − 1) → ·· · → (3, 2) → (3, 1) for some k with 2 ≤ k ≤ n − 1; there are r −2 + r −3 + · · · + r1
such walks.
(5) Walks that are the horizontal reflected images of walks in (3) that begin with (1 , 1)
(2 , 1) and never
n
enter the third row until the final cell; there are n − 1 such walks.
(6) Walks that are horizontal reflected images of walks in (5); there are r
+r
−3
n
Thus, r 3 = 1 + r2 + 2(2 + r1 ) = 11 and, for n
r = 1+r
n
n
−1
≥ 3,
+ 2[(n − 1) + r
−2
n
= 2n − 1 + r
−1
n
+ 2( r
−2
n
−2
n
+r
−3
n
+ · · · + r1 ]
+ · · · + r1 ) ,
and
r
+1
n
= 2n + 1 + r + 2( r
n
−1
n
+r
−2
n
+ · · · + r1 ) .
Therefore
r
n
+1
−r
n
= 2+r +r
n
−1
n
=⇒ r
+1
n
= 2 + 2r + r
n
−1
n
Thus
r
+1
n
whence
r +1 =
n
and
r =
n
√1
+ 1 = 2(r + 1) + (r
n
√1
2 2
(1 +
(1 +
√
√
2)
n
2)
+1
+1
n
−1
n
− √1
2 2
− √1
+ 1) ,
(1 −
(1 −
√
2)
√
2)
+1
n
n
+1
,
−1 .
.
n
→
+ ··· +r
1
such walks.
2 2
2 2
Employ the same notation as in Solut
Solution
ion 1. We have that r1 = 1, r 2 = 4 and r 3 = 11. Let
n ≥ 3. Consider the situation that there are r +1 columns. There are basically three types of rook walks.
Solution 2.
n
Type 1. There are four rook walks that enter only the first two columns.
Type
Typ
e 2. There are 3r
rooks walks that do not pass between the second and third columns in the
middle row (in either direction), viz. r −1 of each of the types:
−1
n
n
(1, 1) −→ (1, 2) −→ (1, 3) −→···−→ (3, 3) −→ (3, 2) −→ (3, 1) ;
(1, 1) −→ (2, 1) −→ (2, 2) −→ (1, 2) −→ (1, 3) −→···−→ (3, 3) −→ (3, 2) −→ (3, 1) ;
(1, 1) −→ (1, 2) −→ (1, 3) −→···−→ (3, 3) −→ (3, 2) −→ (2, 2) −→ (2, 1) −→ (3, 1) .
Type 3. Consider the rook walks that pass between the second and third column along the middle row.
9
Report - Fortieth Canadian Mathematical Olympiad 2008
They are of Type 3a:
(1, 1) −→ ∗ −→ (2, 2) −→ (2, 3) −→···−→ (3, 3) −→ (3, 2) −→ (3, 1) ,
or Type 3b:
(1, 1) −→ (1, 2) −→ (1, 3) −→···−→ (2, 3) −→ (2, 2) −→ ∗ −→ (3, 1) ,
where in each case the asterisk stands for one of two possible options.
We can associate in a two-one way the walks of Type 3a to a rook walk on the last n columns, namely
(1, 2) −→ (2, 2) −→ (2, 3) −→···−→ (3, 3) −→ (3, 2)
and the walks of Type 3b to a rook walk on the last n columns, namely
(1, 2) −→ (1, 3) −→···−→ (2, 3) −→ (2, 2) −→ (3, 2) .
The number of rook walks of the latter two types together is r n 1 rn−1 . From the number of rook walks
on the last n columns, we subtract one for (1 , 2)
(2, 2)
(3, 2) and r n−1 for those of the type
→
(1, 2)
(1, 3)
−→
− −
→
(3, 3)
−→···−→
Therefore, the number of rook walks of Type 3 is 2(rn
rn+1 = 4 + 3 rn−1 + 2( rn
(2, 3) .
−→
− 1 − r − ) and we find that
n
1
− 1 − r − ) = 2 + 2r
n
1
n
+ rn−1 .
We can now complete the solution as in Solution 1.
Solution 3. Let S (3, n) be the set of self-avoiding rook walks in which the rook occupies column n but
does not occupy column n +1. Then R (3, n) = S (3, 1) + S (3, 2) + + S (3, n) . Furthermore, topological
considerations allow us to break S (3, n) into three disjoint subsets S 1 (3, n), the set of paths in which corner
(1, n) is not occupied, but there is a path segment (2, n)
(3, n); S 2 (3, n), the set of paths in which corners
(1, n) and (3, n) are both occupied by a path (1, n)
(2, n)
(3, n); and S3 (3, n), the set of paths in
which corner (3, n) is not occupied but there is a path segment (1, n)
(2 , n). Let
Let si (n) = Si (3, n) for
i = 1, 2, 3. Note that s1 (1) = 0, s2 (1) = 1 and s 3 (1) = 0. By symmetry
symmetry,, s1 (n) = s 3 (n) for every positive n.
Furthermore, we can construct paths in S (3, n + 1) by “bulging” paths in S (3, n), from which we obtain
|
| |
−→
−→
−→
| ··· |
−→
−→
−→
−→
s1 (n + 1) = s 1 (n) + s 2 (n) ;
s2 (n + 1) = s 1 (n) + s 2 (n) + s3 (n) ;
s3 (n + 1) = s 2 (n) + s 3 (n) ;
|
|
|
or, upon simplification,
s1 (n + 1) = s 1 (n) + s2 (n) ;
s2 (n + 1) = 2 s1 (n) + s 2 (n) .
Hence,, for n
Hence
≥ 2,
s1 (n + 1) = s 1 (n) + 2 s1 (n
− 1) + s (n − 1)
= s (n) + 2 s (n − 1) + s (n) − s (n − 1)
= 2s (n) + s (n − 1) .
1
1
1
and
s n
2
2
s n
1
1
1
s n
s n
s n
s n
( + 1) = 2 1 ( ) + 2 ( ) = 2 1 (
1) + 2 2 (
1) +
= s 2 (n) s2 (n 1) + 2s2 (n 1) + s2 (n)
−
−
= 2s (n) + s (n − 1) .
2
−−
−
2
( )
2
10
Report - Fortieth Canadian Mathematical Olympiad 2008
We find that
s1 (n) =
√
√
1
1
√
(1 + 2) − − √ (1 − 2) −
2 2
2 2
n
1
s2 (n) = 2 (1 +
n
√2) −
1
n
1
1
+ 2 (1
n
− √2) −
1
1
;
.
Summing a geometric series yields that
� 1 1· ·�· +(1 +( √))2)+ 2(− 1(1) +�· ·1· + 1( �)) (1 − √2) − 1
√2
√2
=
+√
+
− √2
2
2
2
−
�1 √
√
R(3, n) = (s2 (1) +
s2 n
s1 n
s1
n
=
√
2 2
[(1 +
2)n+1
− (1 −
n
2)n+1 ]
−1.
The formula agrees with R(3, 1) = 1, R (3, 2) = 4 and R(3, 3) = 11.
Acknowledgment. The first two solutions are due to Man-Duen Choi, and the third to Ed Doolittle.
11
Report - Forty First Canadian Mathematical Olympiad 2009
CANADIAN MA
MATHEMA
THEMATICAL
TICAL OL
OLYMPIAD
YMPIAD 2009
SOLUTIONS
1. Given an m × n grid with squares coloured either black or white, we say
that a black square in the grid is stranded if there is some square to its left in the same
row that is white and there is some square above it in the same column that is white (see
Figure).
Problem
Figure
Figur
e 1.
A 4 × 5 grid with no stranded black squares
Find a closed formula for the number of 2 × n grids with no stranded black squares.
There is no condition for squares in the first row. A square in the second row
can be black only if the square above it is black or all squares to the left of it are black.
Suppose the first k squares in the second row are black and the ( k + 1)1)-st
st squar
squaree is whi
white
te
or k = n. When k < n then for each of the first k + 1 squares in the first row we have 2
choices, and for each of the remaining n − k − 1 columns we have 3 choices. When k = n ,
there are 2n choices for the first row. The total number of choices is thus:
Solution.
−
n
1
2k+1 3n−k−1 + 2n .
k=0
This expression
expression simpl
simplifies
ifies to
2 · 3n − 2n .
7
Report - Forty First Canadian Mathematical Olympiad 2009
2
CANADIAN MATHEMATICAL OLYMPIAD 2009 SOLUTIONS
2. Two circles of different radii are cut out of cardboard. Each circle is subdivided into 200 equal sectors. On each circle 100 sectors are painted white and the other
100 are painted black.
black. The smalle
smallerr circle is then placed on top of the larger circle, so that
their centers
centers coinc
coincide.
ide. Sho
Show
w that one can rotate the small circle so that the sectors on
the two circles line up and at least 100 sectors on the small circle lie over sectors of the
same color on the big circle.
Problem
Let x 0 , . . . , x 199 be variables. Assign the value of +1 or −1 to x i depending on
whether the (i + 1)st segment of the larger circle (counting counterclockwise) is black or
white,, respec
white
respectiv
tively
ely.. Simil
Similarly
arly,, assign the value of +1 or −1 to the variable yi depending
on whether the (i + 1)t
1)th
h seg
segmen
mentt of the small
smaller
er circl
circlee is bla
black
ck or whi
white.
te. We can now
restate the problem in the following equivalent way: show that
Solution.
200
Sj =
xi yi+j ≥ 0 ,
i=1
for some j = 0, . . . , 199. Here the subscript i + j is understood modulo 200.
Now observe that y0 + · · · + y199 = 0 and thus
199
S0 + · · · + S199 =
xi (y0 + · · · + y199 ) = 0 .
I =0
Thus Sj
≥
0 for some j = 0, . . . , 199, as claimed.
8
Report - Forty First Canadian Mathematical Olympiad 2009
CANADIAN
CANADI
AN MA
MATHEMA
THEMATICAL
TICAL OL
OLYMPIAD
YMPIAD 2009 SOLUTI
SOLUTIONS
ONS
Problem
3. Define
z x)(x + y + z )
(xy + yz + zx
.
(x + y )(x + z )(y + z )
Determine the set of real numbers r for which there exists a triplet (x,y,z ) of positive
real numbers satisfying f (x,y,z ) = r .
f (x,y,z ) =
We prove that 1 < f (x,y,z ) ≤ 89 , and that f (x,y,z ) can take on any value
within the range (1, 89 ].
The expression for f (x,y,z ) can be simplified to
Solution.
f (x,y,z ) = 1 +
xyz
.
(x + y )(x + z )(y + z )
x,, y,z are positive, we get 1 < f (x,y,z ).
Since x
The inequality f (x,y,z ) ≤ 89 can be simplified to
x2 y + x2 z + y 2 x + y 2 z + z 2 x + z 2 y − 6xyz
≥ 0.
Rearrange the left hand side as follows:
x2 y + x2z + y 2 x + y 2 z + z 2 x + z 2y − 6xyz =
x(y 2 + z 2 ) − 2xyz + y (x2 + z 2 ) − 2xyz + z (x2 + y 2 ) − 2xyz =
x(y − z )2 + y (x − z )2 + z (x − y )2 .
x,, y,z are non-negative.
This expression is clearly non-negative when x
To prove that f (x,y,z ) takes any values in the interval (1, 98 ], define
g (t) = f (t, 1, 1) = 1 +
t
2(1 + t)2
.
Then g (1) = 98 and g(t) approaches 1 as t appr
approac
oaches
hes 0. It follows from the con
contin
tinuit
uity
y
9
of g (t) for 0 < t ≤ 1 that it takes all values in the interval (1, 8 ]. (Alt
(Alternat
ernative
ively
ly,, one
can check that the quadratic equation g(t) = r has a solution t for any number r in the
interval (1, 98 ].)
9
Report - Forty First Canadian Mathematical Olympiad 2009
4
CANADIAN MATHEMATICAL OLYMPIAD 2009 SOLUTIONS
4. Find all ordered pairs (a, b) such that a and b are integers and 3a + 7b is a
perfect square.
Problem
It is obvious that a and b must be non-negative.
Suppose that 3a + 7 b = n2 . We can assum
assumee that n is positive.
positive. We first work modulo 4.
a
b
2
Since 3 + 7 = n , it follows that
Solution.
n2 ≡ (−1)a + (−
(−1)b
(mod 4).
Since no square can be congruent to 2 modulo 4, it follows that we have either (i) a is
odd and b is even or (ii) a is even and b is odd.
Case (i): Let b = 2c. Then
3a = (n − 7c )(n + 7c ).
It cannot be the case that 3 divides both n − 7c and n + 7c . But each of these is a power
of 3. It follows that n − 7c = 1, and therefore
3a = 2 · 7c + 1.
If c = 0, then a = 1, and we obtain the solution a = 1, b = 0. So sup
suppos
posee that
that c ≥ 1.
a
Then 3 ≡ 1 (mod 7). This is impossible, since the smallest positive value of a such that
3a ≡ 1 (mod 7) is given by a = 6, and therefore all a such that 3a ≡ 1 (mod 7) are even,
contradicting the fact that a is odd.
Case (ii): Let a = 2c. Then
7b = (n − 3c )(n + 3c ).
Thus each of n − 3c and n + 3c is a pow
power
er of 7. Sin
Since
ce 7 can
cannot
not div
divide
ide both of thes
these,
e, it
c
follows that n − 3 = 1, and therefore
7b = 2 · 3c + 1.
Look first at the case c = 1. Th
Then
en b = 1, and we obtain the solution a = 2, b = 1. So
from now on we may assume that c > 11.. Th
Then
en 7b ≡ 1 (mod 9)
9).. The sma
smalle
llest
st positi
positive
ve
b
integer b such that 7 ≡ 1 (mod 9) is given by b = 3. It follows that b must be a multiple
of 3. Let b = 3d. Note that d is odd, so in particular d ≥ 1.
Let y = 7d . Then y 3 − 1 = 2 · 3c , and therefore
2 · 3c = (y − 1)(y 2 + y + 1) .
It follows that y − 1 = 2 · 3u for some positive u , and that y 2 + y + 1 = 3v for some v ≥ 2.
But since
3y = (y 2 + y + 1) − (y − 1)2 ,
it follows that 3 | y , which is impossible since 3 | (y − 1).
10
Report - Forty First Canadian Mathematical Olympiad 2009
CANADIAN
CANADI
AN MA
MATHEMA
THEMATICAL
TICAL OL
OLYMPIAD
YMPIAD 2009 SOLUTI
SOLUTIONS
ONS
5. A set of points is marked on the plane, with the property that any three
marke
mar
ked
d poin
points
ts can be cove
covered
red wit
with
h a dis
disk
k of radius 1. Pro
Prove
ve that the set of all mark
marked
ed
points can be covered with a disk of radius 1.
Problem
(For a finite set of points only.) Let D be a disk of smallest radius that covers
all marked points. Consider the marked points on the boundary C of this disk. Note that
if all marked points on C lie on an arc smaller than the half circle (ASTTHC for short),
then the disk can be moved a little towards these points on the boundary and its radius
can be decr
decreased
eased.. Since we assume
assumed
d that our disk has minima
minimall radius, the mark
marked
ed points
on its boundary do not lie on an ASTTHC.
If the two endpoints of a diagonal of D are marked, then D is the smallest disk containing
these two points, hence must have radius at most 1.
Solution.
If there are 3 marked points on C that do not lie on an ASTTHC, then D is the smallest
disk
dis
k co
cove
verin
ringg the
these
se 3 poin
points
ts and hen
hence
ce must hav
havee rad
radius
ius at mos
mostt 1. (In this case the
triangle formed by the three points is acute and C is its circumcircle.)
If there are more than 3 marked points on the boundary that do not lie on an ASTTHC,
then we can remove one of them so that the remaining points again do not lie on an
ASTTHC
AST
THC.. By induc
inductio
tion
n thi
thiss lea
leads
ds us to the case of 3 poin
points.
ts. Ind
Indeed
eed,, giv
given
en 4 or more
points on C , choo
hoose
se 3 poi
point
ntss th
that
at lie on a ha
half
lf circ
circle
le.. Th
Then
en the midd
middle
le poin
pointt ca
can
n be
removed.
11
CANADIAN MA
MATHEMA
THEMATICAL
TICAL OL
OLYMPIAD
YMPIAD 2010
PROBLEM
PRO
BLEMS
S AND SOLU
SOLUTIONS
TIONS
(1) For a positive integer n,
n , an n -staircase is a figure consisting of unit squares, with
one square in the first row, two squares in the second row, and so on, up to n
squares in the n th row, such that all the left-most squares in each row are aligned
vertically. For example, the 5-staircase is shown below.
Let f (n) denote the minimum number of square tiles required to tile the nstaircase, where the side lengths of the square tiles can be any positive integer.
For example, f (2) = 3 and f (4) = 7.
(a) Find all n such that f (n) = n.
n.
(b) Find all n such that f (n) = n + 1.
(a) A diagonal square in an n-staircase is a unit square that lies on
the diagonal
diagonal going from the top-left to the bottom-right
bottom-right.. A minimal tiling of an
n-staircase is a tiling consisting of f (n) square tiles.
Solution.
Observe that f (n) ≥ n for all n.
n . There are n diagonal squares in an n-staircase,
n -staircase,
and a square tile can cover at most one diagonal square, so any tiling requires at
least n sq
squar
uaree til
tiles.
es. In other wo
words
rds,, f (n) ≥ n. He
Henc
nce,
e, iiff f (n) = n, then each
square tile covers exactly one diagonal square.
Let n be a positive integer such that f (n) = n,
n , and consider a minimal tiling of
an n-stai
n -staircase
rcase.. The only square tile that can cover
cover the unit squar
squaree in the first row
is the unit square itself.
Now consider the left-most unit square in the second row. The only square tile
that can cover this unit square and a diagonal square is a 2 × 2 square tile.
1
Next, consi
Next,
consider
der the left
left-most
-most unit square in the fourth row. The only square tile
that can cover this unit square and a diagonal square is a 4 × 4 square tile.
Continuing this construction, we see that the side lengths of the square tiles
we encounter will be 1, 2, 4, and so on, up to 2k for some nonnegative integer k.
Therefore, n,
n, the height of the n-st
n-stair
aircas
case,
e, is equ
equal
al to 1+ 2 + 4 + · · · + 2k = 2k+1 − 1.
k
k
Alternatively,
= can
2 −tile
1 fora some
positive integer
k
p(
p (k )tiles
= 2 recursively
− 1.
Conve
Con
versely
rsely,, nwe
p(k)-staircase
with pk.
(k. )Let
square
as
follows:
follo
ws: We have that p(1) = 1, and we can tile a 1-staircase with 1 square tile.
Assume that we can tile a p(k )-staircase with p(k ) square tiles for some positive
integer k.
k.
Consider a p(k + 1)1)-sta
stairc
ircase
ase.. Pla
Place
ce a 2k × 2k square tile in the bottom left
corner. Note that this square tile covers a digaonal square. Then p(
p (k + 1) − 2k =
2k+1 − 1 − 2k = 2k − 1 = p(
p(k), so we are left with two p(k )-staircases.
p(k)
2k
2k
p(k)
Furthermore, these two p(
p(k )-staircases can be tiled with 2 p(
p(k ) square tiles, which
means we use 2p(
p(k) + 1 = p(
p (k + 1) squar
squaree tile
tiles.
s.
k
Therefore, f (n) = n if and only if n = 2 − 1 = p(
p (k) for some positive integer
k . In other words, the binary representation of n consists of all 1s, with no 0s.
(b) Let n be a positive integer such that f (n) = n + 1, and consider a minimal
tiling of an n-st
-stair
aircas
case.
e. Sin
Since
ce the
there
re are n diag
diagonal
onal squar
squares,
es, ev
every
ery squar
squaree tile
except
exc
ept one co
cove
vers
rs a dia
diagon
gonal
al squ
square
are.. We cla
claim
im tha
thatt the squar
squaree tile
tile tha
thatt co
cove
vers
rs
the bottom-left unit square must be the square tile that does not cover a diagonal
square.
If n is even,
thiscannot
fact iscover
obvious,
because square,
the square
tile that
covers
the
bottom-left
unitthen
square
any diagonal
so assume
that
n is odd.
Let n = 2m + 1. We may assume that n > 1, so m ≥ 1. Suppose that the square
tile covering
covering the bottom-le
bottom-left
ft unit squar
squaree also cov
covers
ers a diagon
diagonal
al square. Then the
side length of this square tile must be m + 1.
1. Aft
After
er this ((m
m + 1) × (m + 1) square
tile has been placed, we are left with two m-staircases.
2
m
m+1
m+1
m
Hence, f (n) = 2f (m) + 1. Bu
Butt 22ff (m) + 1 is odd, and n + 1 = 2m
2m + 2 is even,
so f (n) cannot be equal to n + 1, con
contradi
tradiction
ction.. Ther
Therefor
efore,
e, the square tile that
covers the bottom-left unit square is the square tile that does not cover a diagonal
square.
Let t be the side length of the square tile covering the bottom-left unit square.
Then every other square tile must cover a diagonal square, so by the same construction as in part (a), n = 1 + 2 + 4 + · · · + 2k 1 + t = 2k + t − 1 for some positive
−
k
integer
k.
k . way
Furthermore,
the toptiling
p(
p(k) of
= a2 p−
rows of theTherefore,
n-staircase
n -staircasethe
must
be
tiled
the
same
as the minimal
(k1)-staircase.
hor
horizontal
izontal
line between rows p(k ) and p(k ) + 1 does not pas
passs thr
throug
ough
h an
any
y squ
square
are tile
tiles.
s. Let
us call such a line a fault line. Simila
Similarly
rly,, the vertic
vertical
al line betwe
between
en columns t and
t + 1 is also a fault line. These two fault lines partition two p(k )-staircases.
p(k)
t
t
p(k)
If these two p(k )-staircases do not overlap, then t = p(k), so n = 2p(
p(k ). For
example, the minimal tiling for n = 2p(2)
p(2) = 6 is shown below.
Hence, assume that the two p(k )-sta
)-staircas
ircases
es do overlap
overlap.. The inter
intersect
section
ion of the
two
two p(k )-staircases is a [p(
p(k ) − t]-sta
]-staircas
ircase.
e. Since thi
thiss [ p(
p(k ) − t]-staircase is tiled
the same way as the top p(k ) − t rows of a minimal tiling of a p(k )-staircase,
p(
p(k ) − t = p(
p(l) for some positive integer l < k
k,, so t = p(
p(k ) − p(
p(l). Then
n = t + p(
p(k ) = 2p(
p(k ) − p(
p(l).
Since p(0)
p (0) = 0, we can summarize by saying that n must be of the form
n = 2p(
p(k ) − p(
p(l) = 2k+1 − 2l − 1,
3
where k is a positive integer and l is a nonne
nonnegativ
gativee intege
integer.
r. Also, our argumen
argumentt
shows how if n is of this form, then an n-staircase
n -staircase can be tiled with n + 1 square
tiles.
Finally, we observe that n is of this form if and only if the binary representation
of n contains exactly one 0:
2k+1 − 2l − 1 = 11 . . . 1 0 11 . . . 1 .
k − l 1s
l 1s
(2) Let A,B,P be thr
three
ee poin
points
ts on a cir
circle
cle.. Pro
Prove
ve tha
thatt if a and b are the distances
from P to the tangents at A and B and c is the distance from P to the chord AB
AB,,
2
then c = ab
ab..
Let r be the radius of the circle, and let a’ and b’ be the respective
lengths of P A and P B . Since b = 2r sin ∠P AB = 2rc/a , c = a b /(2r
(2r). Let AC
be the diameter of the circle and H the foot of the perpendicular from P to AC .
The similarity of the triangles AC
ACP
P and AP H imply that AH : AP = AP : AC
2
2
or (a ) = 2ra
ra.. Similarly, (b
(b ) = 2rb
rb.. Hence
Solution.
(a )2 (b )2
c =
= ab
2r 2r
2
as desired.
Let E,F,G be the feet of the perpendiculars to the
tangents at A and B and the chord AB
AB,, res
respect
pectiv
ively
ely.. We need to sho
show
w tha
thatt
P E : P G = P G : GF , where G is the foot of the perpendicular from P to AB
AB..
This suggest that we try to prove that the triangles E P G and
a nd GP F are similar.
Since P G is parallel to the bisector of the angle between the two tangents,
Alternate
Alte
rnate Solut
Solution.
ion.
∠EP G
= ∠F P G. Since AEPG and B FPG are concyclic quadrilaterals (having
opposite angles right), ∠P GE = ∠P AE and ∠P F G = ∠P BG.
BG . But ∠P AE =
∠P BA = ∠P BG
BG,, whence ∠P GE = ∠P F G. Therefore triangles E P G and GP F
are similar.
The argument above with concyclic quadrilaterals only works when P lies on
the shorter arc between A and B . The other case can be proved similarly.
(3) Three
speed
skaters
skate
rs skate
have
have ainfri
friend
endly
ly race
on a skati
skating
ngwith
oval.
oval.different
The
They
y allspeeds
start from
the same
point
and
the
same
direction,
but
that
they maintain
maintain through
throughout
out the race
race.. The slow
slowest
est skat
skater
er does 1 lap a minute, the
fastest one does 3.
3.14 laps a minute, and the middle one does L laps a minute for
some 1 < L < 3.14. The race ends at the moment
moment when all thr
three
ee ska
skater
terss agai
again
n
come together to the same point on the oval (which may differ from the starting
4
point.) Find how many diffe
point.)
differen
rentt choices
choices for L are there such that 117 passings
occurr befo
occu
before
re the end of the race
race.. (A passin
passingg is define
defined
d whe
when
n one sk
skate
aterr pas
passes
ses
another one. The beginning and the end of the race when all three skaters are at
together are not counted as a passing.)
As
Assu
sume
me tha
thatt th
thee le
leng
ngth
th of th
thee ov
oval
al is on
onee un
unit
it.. Let
Let x(t) be the
difference of distances that the slowest and the fastest skaters have skated by time
t. Similarly
Similarly,, let y(t) be the difference between the middle skater and the slowest
skate
sk
ater.
r. The pa
path
th ((x
x(t), y(t)) is a straight ray R in R2 , starting from the origin,
with slope depending on L
L.. By assumption, 0 < y(t) < x(t).
One skater passes another one when either x(
x(t) ∈ Z, y(
y (t) ∈ Z or x(
x(t) − y(t) ∈ Z.
The race ends when both x(
x (t), y(t) ∈ Z.
2
Let (a,
(a, b) ∈ Z be the endpoint of the ray R. We need to find the num
number
ber of
Solution.
such
(a) 0points
< b <satisfying:
a
(b) The ray R intersects Z2 at endpoints only.
(c) The ray R crosses 357 times the lines x ∈ Z, y ∈ Z, y − x ∈ Z.
The second condition says that a and b are relatively prime. The ray R crosses
a − 1 of the lines x ∈ Z, b − 1 of the lines y ∈ Z and a − b − 1 of the lines x − y ∈ Z.
Thus, we need (a
(a − 1 ) + (b
(b − 1 ) + (a
(a − b − 1) = 117, or equivalently, 2a
2a − 3 = 117.
That is a = 60.
Now b m
must
ust be a posi
positiv
tivee in
integ
teger
er less than and relati
relative
vely
ly prime to 60. The
number of such b can be found using the Euler’s φ function:
φ(60) = φ(2
φ(22 · 3 · 5) = (2 − 1) · 2 · (3 − 1) · (5 − 1) = 16.
16.
Thus the answer is 16.
Firs
First,
t, let us name our ska
skaters.
ters. From fastest to slowe
slowest,
st,
call them: A, B and C . (Abel, Bernoulli and Cayley?)
Now, it is helpful to consider the race from the viewpoint of C . Relative to C ,
Alternate Solution.
both A and B complete a whole number of laps, since they both start and finish
at C .
Let n be the number of laps completed by A relative to C , and let m be the
number of laps completed by B relative to C . Note that: n > m ∈ Z+
Consider
Consi
der the num
number
ber of minutes requir
required
ed to compl
complete
ete the race
race.. Rela
Relativ
tivee to C ,
A is moving with a speed of 3.
3.14 − 1 = 2.14 laps per minute and completes the
n
race in 2.14 minutes. Also relative to C , B is moving with a speed of (L
( L − 1) laps
m
per minute and completes the race in L 1 minutes. Since A and B finish the race
−
together (when they both meet C ):
n
m
=
⇒
2.14
L−1
L = 2.14
m
n
+ 1.
1.
Hence, there is a one-to-one relation between values of L and values of the postive
proper fraction m
. The fraction should be reduced, that is the pair (m,
(m, n) should
n
5
be relatively prime, or else, with k = gcd(m,
gcd(m, n), the race ends after n/k laps for
A and m/k laps for B when they first meet C together.
It is also helpful to consider the race from the viewpoint of B . In th
this
is fr
fram
amee
of reference, A completes only n − m la
laps.
ps. Hen
Hence
ce A passes B only (n
(n − m) − 1
times,
time
s, since the racers do not ””pass”
pass” at the end of the race (nor at the beginnin
beginning).
g).
Similarily A passes C only n − 1 times and B passes C only m − 1 times
times.. Th
Thee
total number of passings is:
117 = (n
(n − 1) + (m
(m − 1) + (n
(n − m − 1) = 2n
2n − 3
⇒
n = 60
m
Hence the number of values of L equals the number of m for which the fraction 60
is
positive, proper and reduced. That is the number of positive integer values smaller
than and relative
relatively
ly prime to 60. One could simply count:
count: {1,7,11,13,17,...}, but
Euler’s φ function gives this number:
2
φ(60) = φ(2
φ(2 · 3 · 5) = (2 − 1) · 2 · (3 − 1) · (5 − 1) = 16.
16.
Therefore, there are 16 values for L which give the desired number of passings.
Note that the actual values for the speeds of A and C do not affect the result.
They could be any values, rational or irrational, just so long as they are different,
and there will be 16 possible values for the speed of B between them.
(4) Eac
Each
h ve
verte
rtex
x of a fini
finite
te graph can be col
colore
ored
d eit
either
her blac
black
k or white
white.. Ini
Initia
tially
lly all
ve
vertice
rticess are blac
black.
k. We are allo
allowe
wed
d to pic
pick
k a ve
vertex
rtex P and cha
change
nge the color of P
and all of its neigh
neighbours.
bours. Is it possible to chan
change
ge the colour of every ver
vertex
tex from
black to white by a sequence of operations of this type?
Solution.
The
answer
Proof by induction
on suppose
the number
n of
If n = 1, this is
obviou
obvious.
s. Fis
or yes.
the induction
assump
assumption,
tion,
we can
dovertices.
this for
any graph with n − 1 vertices for some n ≥ 2 and let X be a graph with n vertices
which we will denote by P 1 , . . . , Pn+1 .
Let us denote the “basic” operation of changing the color of P and all of its
Let us denote the basic operation of changing the color of Pi and all of its
neighbours by f i . Removing a vertex Pi from X (along with all edges connecting
to Pi ) and applying the induction assumption to the resulting smaller graph, we
see that there exists a sequence of operations g i (obtained by composing some f j ,
with j
= i)
i ) which changes the colour of every vertex in X , except for possibly P i .
If gi it also changes the color of Pi then we are done.
done. So, we may assum
assumee that
gi does not change the colour of P for every i = 1, . . . , n.
n. Now consider two cases.
Case 1: n is even. Then composing g 1 , . . . , gn we will change the color of every
vertex from white to black.
Case 2: n is odd. I claim that in this case X has a vertex with an even number
of neighbours.
Indeed, denote the number of neighbours of Pi (or equivalently, the number of
edges connected to P ) by ki . Then P1 + · · · + Pn+1 = 2e, where e is the number
of edges of X . Thus one of the numbers k i has to be even as claimed.
6
After renumbering the vertices, we may assume that P 1 has 2k
2k neighbours, say
P2 , . . . , P2k+1 . The ccompo
omposit
sition
ion of f1 with g1 , g2 , . . . , g2k+1 will then change the
colour of every vertex, as desired.
(5) Let P (x) and Q(x) be poly
polynom
nomial
ialss wit
with
h in
integ
teger
er coeffici
coefficien
ents.
ts. Let an = n! + n.
n.
Show that if P (an )/Q
/Q((an ) is an integer for every n,
n , then P (n)/Q
/Q((n) is an integer
for every integer n such that Q(
Q (n)
= 0.
Imagine dividing P (x) by Q(
Q (x). We find that
P (x)
R(x)
= A(
A(x) +
,
Q(x)
Q(x)
where A(
A(x) and R(
R (x) are polynomials with rational coefficients, and R(
R (x) is either
identically 0 or has degree less than the degree of Q(x).
By bringing the coefficients of A(
A (x) to their least common multiple, we can find
a polynomial B (x) with integer coefficients, and a positive integer b, such that
Suppose
pose fir
first
st tha
thatt R(x) is not ide
ident
ntica
ically
lly 0. Not
Notee that for any
A(x) = B (x)/b
/b.. Sup
integer k , either A(k ) = 0, or |A(k)| ≥ 1/b
/b.. But w
when
henev
ever
er |k | is large enough,
0 < |R(k )/Q
/Q((k )| < 1/b
/b,, and therefore if n is large enough, P (an )/Q
/Q((an ) cannot
be an integer.
So R(
R (x) is identically 0, and P (x)/Q
/Q((x) = B(
B (x)/b (at least whenever Q(
Q(x)
= 0.)
Now let n be an int
integer.
eger. Then there are infinit
infinitely
ely many inte
integers
gers k such that
n ≡ ak (mod b). Bu
Butt B (ak )/b is an integer, or equivalently b divides B (ak ). It
follows that b divides B (n), and therefore P (n)/Q
/Q((n) is an integer.
Solution.
7
43rd
Canadian Mathematical Olympiad
Wednesday, March 23, 2011
Problems and Solutions
(1) Consid
Consider
er 70-digit
70-digit numbers
numbers n, with
with th
thee pr
prope
opert
rty
y th
that
at each
each of th
thee di
digit
gitss 1, 2, 3, . . . , 7
appears in the decimal expansion of n ten times (and 8, 9, and 0 do not appear). Show
that no number of this form can divide another number of this form.
Assume the contrary: there exist a and b of the prescribed form, such that
b ≥ a and a divides b. Then a divides b − a.
Claim: a is not divisible by 3 but b − a is divisible by 9. Indeed, the sum of the digits
is 10(1 + · · · + 7) = 280, for both a and b. [Here
[Here one needs
needs to know
know or prove
prove that
that an
integer n is equivalent of the sum of its digits modulo 3 and modulo 9.]
We conclude that b − a is divisible by 9 a. But this is impossible, since 9 a has 71 digits
and b has only 70 digits, so 9 a > b > b − a.
Solution.
(2) Let ABCD be a cyclic quadrilateral whose opposite sides are not parallel, X the intersection of AB and C D, and Y the intersection of AD and BC . Let the angle
angle bisector
bisector
of ∠AX
AXD
D intersect AD,BC at E, F respectively and let the angle bisector of ∠AY B
intersect A
AB,
B, CD at G, H respectively. Prove that EGFH is a parallelogram.
Solution.
Since ABCD is cyclic, ∆X AC ∼ ∆X DB and ∆Y AC ∼ ∆Y BD . There-
fore,
XA
XC
AC
YA
YC
=
=
=
=
.
XD
XB
DB
YB
YD
Let s be this ratio. Therefore, by the angle bisector theorem,
AE
XA
XC
CF
=
=
=
= s,
ED
XD
XB
FB
and
YA
YC
CH
AG
=
=
=
= s.
GB
YB
YD
HD
AE
F
=
and ED
= C
FB
EGFH is a parallelogram.
Hence,
AG
GB
DH
.
HC
Therefore,
Therefore, EH ||AC ||GF and EG ||DB ||H F . Hence,
Hence,
1
(3) Amy has divided a square up into finitely many white and red rectangles, each with sides
parallel
square.
eachngle,
white
rectangle,
she writes
downdivided
its width
divided
divid
ed to
bythe
its sides
height.
heighof
t. the
Within
each
eacWithin
h red rectangle,
recta
she
writes down
its height
by
its width. Finally
Finally, she calculates
calculates x, the sum of these numbe
numbers.
rs. If the total area of the
white rectangles equals the total area of the red rectangles, what is the smallest possible
value of x?
Let ai and b i denote the width and height of each white rectangle, and let
ci and di denote the width and height of each red rectangle. Also, let L denote the side
length of the original square.
Solution.
Lemma:
Either
ai ≥ L o
orr
Prooff of lem
Proo
lemma:
ma:
di ≥ L.
Suppose there exists a horizontal line across the square that is
covered
cov
ered
entirely
entirely
rectangles.
rectan
gles. wise,
Then,there
the total
width
of these
rectangles
rectecting
anglesevery
isery
at
least
the with
claimwhite
is proven.
proven
. Otherwise,
Other
is a red
rectangle
rectang
le intersecting
inters
ev
L, and
horizontal line, and hence the total height of these rectangles is at least L.
Now, let us assume without loss of generality that
inequality,
·
ai
bi
But we know
ai bi =
L2
2
ai bi
, so it follows that
ai ≥ L. By the Cauchy-Schwarz
2
≥
ai
≥ L2 .
ai
bi
≥ 2. Furthermore, each ci ≤ L, so
≥ 1 · = 1
di
ci
ci di
L2
2
.
Therefore, x is at least 2.5. Conversely, x = 2.5 can be achieved by making the top half
of the square one colour, and the bottom half the other colour.
(4) Show that there exists a positive integer N such that for all integers a > N , there exists
a contiguous substring of the decimal expansion of a th
that
at is divisi
divisible
ble by 2011. (F
(For
or
instance, if a = 153204, then 15, 532, and 0 are all contiguous substrings of a . Note that
0 is divisible by 2011.)
We claim that if the decimal expansion of a has at least 2012 digits, then
contains
ains the required
required substring.
substring. Let the decimal
decimal expansion
expansion of a b
bee ak ak 1 . . . a0 . For
a cont
i = 0, . . . , 2011, Let bi be the number with decimal expansion ai ai 1 . . . a0 . Then
Then by
pidgenhole principle, bi ≡ bj mod 2011 for some i < j ≤ 20
2011.
11. It foll
follow
owss that 2011
i
Sinc
ncee 2011
2011 and 10 ar
aree
divides bj − bi = c · 10 . Here c is the substring aj . . . ai+1. Si
relatively prime, it follows that 2011 divides c.
Solution.
−
−
2
(5) Let d be a positive integer. Show that for every integer S , there exists an integer n > 0
and a sequence 1, 2, . . . , n , where for any k , k = 1 or k = −1, such that
S = 1 (1 + d)2 + 2 (1 + 2d)2 + 3 (1 + 3d)2 + · · · + n (1 + nd)2 .
calculate Uk+3 − Uk+2 − Uk+1 + U k . This
This turns
turns
Let Uk = (1 + kd )2 . We calculate
2
2
out to be 4d , a constant. Changing signs, we obtain the sum −
−44d .
Thus if we have found an expression for a certain number S0 as a sum of the desired
type, we can obtain an expression of the desired type for S0 + (4 d2 )q , for any integer q .
It remains to show that for any S , there
there exists
exists an in
integ
teger
er S such that S ≡ S
2
(mod 4d ) and S can be expressed in the desired form. Look at the sum
Solution.
(1 + d)2 + (1 + 2d)2 + · · · + (1 + N d)2,
where N is “large.” We can at will choose N so that the sum is odd, or so that the sum
is even.
By changing the sign in front of (1 + kd )2 to a minus sign, we decrease the sum by
2(1+ kd )2 . In particular, if k ≡ 0 (mod 2d), we decrease the sum by 2 (modulo 4d2 ). So
If N is large enough, there are many k < N such that k is a multiple of 2 d. By
switching the sign in front of r of these, we change (“downward”) the congruence class
modulo 4d2 by 2 r . By choosing N so that the original sum is odd, and choosing suitable
r < 2 d2 , we can obtain numbers congruent to all odd numbers modulo 4 d2 . By choosing
N so that the original sum is even, we can obtain numbers congruent to all even numbers
2
