Sample Part II Problems and Solutions
These sample problems are being provided to give participants an indication of the
kinds of questions that might be included in Part II of the Kennesaw State University
Mathematics Competition. Like this sample set, the actual competition will consist of
five questions to be completed in a total time of two hours. However, the questions in
the sample set are not necessarily representative of the level of difficulty of each
question on the actual competition. Some may be more challenging, some less.
1.
How many ordered triples of positive integers (x, y, z) are there for which the product
xyz = 10 6 ? Justify your answer.
C
2.
A
The figure at the right is a regular nonagon (nine sides).
Prove that the sum of the length of side AB and the length
of diagonal BC is equal to the length of diagonal PC .
B
P
3.
The triangular numbers, so named for the pattern
at the right, are the numbers 1, 3, 6, 10, ….
a) Prove that the nth triangular number is
1
n ( n  1) for all positive integers n.
2

1

 
3

 
  
   
10

 
  
6
b) Find, with proof, the smallest positive integer P
such that Pt + 1 is a triangular number whenever
t is a triangular number.
4.
Let a 1 , a 2 , a 3 , ... , a 7 be distinct integers. Prove that the polynomial
P(x) = ( x  a 1 )( x  a 2 )( x  a 3 )    ( x  a 7 )  1
cannot be expressed as the product of two polynomials, each of degree less than 7,
with integer coefficients.
B
5.
In equilateral triangle ABC, shown, P is the midpoint of altitude BD .
Compute, with proof, the ratio of AP to PQ.
Q
P
A
D
C
Sample Part II Problems and Solutions
SOLUTIONS
1. The number 10 6 has the prime factors 2 and 5. If we let a and b denote the number
of times 2 appears as a factor of x and of y, respectively, then 6 – (a + b) is the
number of times 2 appears as a factor of z. Since a  0 and b  0 and
6 – (a +b)  0, we count the number of pairs of nonnegative integers (a, b) for which
a + b  6. For each choice of a from 0 to 6, inclusive, there are 7 – a different
choices for b. These pairs of values for a and b are listed in the following table.
a
b
0000000
0123456
111111
012345
22222
01234
3333
0123
444
012
55
01
6
0
Thus there are 28 different ways to allot the six factors 2 among the three numbers
x, y, z. The same discussion applies to the prime 5. Since the distribution of twos
among x, y, z is independent of the distribution of fives, the total number of ordered
triples (x, y, z) is (28)(28) = 784.
C
A
2. Inscribe the regular nonagon in a circle. Each side
360
of the nonagon intercepts an arc of
= 40˚.
9
Extend diagonals CB and PQ until they intersect at
B
point R. Angle RCP and angle RPC are inscribed angles
and each intercepts an arc with measure 3(40) = 120˚.
1
Therefore, m<RCP = m<RPC = (120) = 60˚.
Q
2
P
Hence, triangle RPC is equilateral. Since
BQ // PC (they are chords that intercept congruent arcs),
triangle RBQ is also equilateral. Noting that AB = BQ = BR, and RC = PC, we have
AB + BC = BR + BC = PC.
R
Sample Part II Problems and Solutions
3. a) Proof by mathematical induction:
(i)
For n = 1, obviously 1 =
1
(1)(1  1)
2
1
(k )( k  1)
2
Prove for n = k + 1. That is, prove the (k + 1)st triangular number is
1
(k  1)( k  2) .
2
By definition, the (k + 1)st triangular number is k + 1 more than the kth triangular
number. Therefore the (k + 1)st triangular number is equal to
(ii)
Assume for n = k. That is, the kth triangular number is
1
1
1
3
1
1
(k )( k  1)  (k  1)  (k  1)( k  1)  k 2  k  1  (k 2  3k  2)  (k  1)( k  2) .
2
2
2
2
2
2
Therefore, the formula is true for all triangular numbers n.
1
(n )( n  1) , where n is a positive integer.
2
Since it is required that Pt 1  1  t 2 , where t 1 and t 2 are triangular numbers, we have
1
1

P  (r )( r  1)  1  (s)(s  1) , where r and s are positive integers.
2
2

2
Therefore, Pr  Pr  2  s(s  1) . Let s = A + 1. Then this last equation becomes
b) By part (a) a triangular number t is of the form
Pr 2  Pr  2  (A  1)( A  2) or Pr 2  Pr  2  A 2  3A  2 .
A s 1
From this last equation, Pr 2  A 2 and Pr  3A . Therefore, r  
.
3
3
Since the smallest value of P is required, choose r = 1. Then s = 4 and P = 9.
4. Suppose P(x) = ( x  a 1 )( x  a 2 )( x  a 3 )    ( x  a 7 )  1 = g(x)h(x) where g(x) and h(x) are
polynomials each with degree less than 7 and integer coefficients. One of these
polynomials, say g(x), must have degree  3. For each a i (i = 1, 2, 3,…, 7)
P( a i ) = 1 = g( a i )h( a i ).
Since g(x) and h(x) have integer coefficients, and each a i is an integer, it follows that
g( a i ) and h( a i ) are integers for each a i .
Therefore, g( a i ) = h( a i ) = 1 or g( a i ) = h( a i ) = –1 for each i = 1, 2, 3,…, 7. Thus we
have one of the following situations
(1) g( a i ) = h( a i ) = 1 for at least four distinct integers a i . If this is the case, then
the equation g(x) – 1 = 0 has at least four solutions. This is impossible, since the
degree of g(x) is  3.
(2) g( a i ) = h( a i ) = –1 for at least four distinct integers a i . If this is the case, then
the equation g(x) + 1 = 0 has at least four solutions. This is also impossible,
since the degree of g(x) is  3.
Therefore, P(x) is not expressible as a product of two polynomials, each of degree less
than 7, with integer coefficients.
Sample Part II Problems and Solutions
B
5. Note that BD bisects ABC. An angle bisector of a triangle divides
a
the side to which it is drawn into two segments whose lengths are
Q
proportional to the lengths of the other two sides of the triangle.
s
P
AB AP

Therefore, in ABQ,
.
BQ PQ
Let AB = BC = AC = s and let BQ = a.
AB s

It suffices to find the ratio
 .
A
BQ a
D
s
Represent PAD as  . Since ABD is a 30-60-90 triangle, AD =
and
2
1
1s 3 s 3

PD = BD 
2
2 2
4
C
3
s 3 4
3
2
3
1

, cos =
, and sin =
. Also, mBAQ = 60 – tan
and
2
s2
2
7
7
3
3
1
1
mBQA = 180 – [60 + (60 – tan
)] = 60 + tan
.
2
2
and tan =
Using the Law of Sines on triangle ABQ, and the formulas for sin(A  B)
3
3
3
sin( 60  tan 1
)
sin( 60  tan 1
) sin( 60  tan 1
)
s
2 
2
2
or
=

s
a
a
3
1
sin( 60  tan
)
2
3 2
1 3
3
3
3

7
sin 60cos(tan 1
)  cos 60sin(tan 1
)
2
2 = 2 7 2 7  2
 3.
1
3 2
1 3
3
1 3
1
7

sin 60cos(tan
)  cos 60sin(tan
)
2
2 7 2 7
2
2
Therefore, the ratio of AP to PQ is 3:1.
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