6th Australian Mathematical Olympiad Problems 1985



A1.  Find the sum of the first n terms of 0, 1, 1, 2, 2, 3, 3, 4, 4, ... (each positive integer occurs twice). Show that the sum of the first m + n terms is mn larger than the sum of the first m - n terms.
A2.  Show that any real values satisfying x + y + z = 5, xy + yz + zx = 3 lie between -1 and 13/3.
A3.  A graph has 9 points and 36 edges. Each edge is colored red or black, so that every triangle has at least one red side. Show that there are four points with all edges between them red.
B1.  A triangle ABC has all angles smaller than 120o. An equilateral triangle is constructed on the outside of each side by constructing three new vertices D, E, F. Show that the three lines joining the new vertex of each equilateral triangle to the opposite vertex of ABC meet at a point X. Show that XD + XE + XF = 2(XA + XB + XC).
B2.  Find all possible positive integers which have at least seven positive divisors and equal one less than the sum of the squares of their sixth and seventh divisors, when the divisors are listed in order of increasing size.
B3.  Find all real polynomials p(x) such that p(x2 + x + 1) = p(x) p(x + 1). 

Solutions

Problem A1
Find the sum of the first n terms of 0, 1, 1, 2, 2, 3, 3, 4, 4, ... (each positive integer occurs twice). Show that the sum of the first m + n terms is mn larger than the sum of the first m - n terms.
Answer
n2/4 for n even, (n2-1)/4 for n odd. One could also express that as [n2/4] for all n.
Solution
We know that 1 + 2 + ... + n = n(n+1)/2 (reverse the order and add and we get n pairs of terms with sum n+1 each). So the sum of the first 2n+1 terms is n(n+1). The sum of the first 2n terms is n less or n2. Suppose m+n is even. Then m-n is also even, so the sum of the first m+n less the sum of the first m-n is ¼(m+n)2 - ¼(m-n)2 = mn. Similarly, if m+n is odd, the difference is ¼(m+n)2 - ¼ - ¼(m-n)2 + ¼ = mn. 

Problem A2
Show that any real values satisfying x + y + z = 5, xy + yz + zx = 3 lie between -1 and 13/3.
Solution
We have the solution -1, 3, 3, so -1 is attained. We also have the solution 13/3, 1/3, 1/3, so 13/3 is attained.
We have (x+y)2 = (5-z)2, xy = 3 - z(x+y) = 3 - z(5-z), so (x-y)2 = (x+y)2 - 4xy = 25-10z+z2 - 4(3-5z+z2) = 13 + 10z -3z2 = (13 - 3z)(z + 1). But (x-y)2 ≥ 0, so -1 ≤ z ≤ 13/3. 

Problem A3
A graph has 9 points and 36 edges. Each edge is colored red or black, so that every triangle has at least one red side. Show that there are four points with all edges between them red.
Solution
We show first that given any 6 of the points there is a red triangle amongst their edges. Take any point X of the six. If three points are Ai of the six are joined to X by blue edges, then considering the triangles XAiAj, each of the edges AiAj must be red and we have a red triangle. So we can assume that each point of the six has at least three red edges to other points of the six. Take the points to be A, B, C, D, E, F, with AB, AC, AD all red. Now B has at least two other red edges to the other six. If BC is red, then ABC is red. Similarly if BD is red. So BE and BF are red. Now one of AE, AF, EF must be red, but that gives red triangles ABE, ABF, BEF respectively.
Return to considering all 9 points. Take any point X. If there are 4 points Ai not joined to X by a red edge, then since every triangle XAiAj has a red edge, all the edges AiAj must be red and we are home. So we can assume that every point has at least 5 red edges.
If every point has exactly 5 red edges, then there are 9·5/2 red edges in total, which is impossible, so some point X must have red edges to 6 other points. But there is a red triangle ABC amongst those points (shown above), so that XABC has all edges red. 

Problem B1
A triangle ABC has all angles smaller than 120o. An equilateral triangle is constructed on the outside of each side by constructing three new vertices D, E, F. Show that the three lines joining the new vertex of each equilateral triangle to the opposite vertex of ABC meet at a point X. Show that XD + XE + XF = 2(XA + XB + XC).
Solution
A rotation of 60o about B takes A to F and D to C. So the angle between AD and CF is 60o. Suppose they meet at X. So ∠BAF = ∠BXF = 60o, so AFBX is cyclic, so ∠BXF = ∠BAF = 60o. But a rotation of 60o about A takes FC to BE, so BX and BE make the same angle with CF, so the must be the same line.
A rotation of 60o about F takes B to A and X to X'. Evidently FXX' is equilateral. Since ∠AXF = 60o, A must lie on XX'. But AX' = BX (because of the rotation), so XF = XX' = XA + AX' = XA + XB. Adding the two similar relations gives XD + XE + XF = 2(XA + XB + XC).

 Problem B2
Find all possible positive integers which have at least seven positive divisors and equal one less than the sum of the squares of their sixth and seventh divisors, when the divisors are listed in order of increasing size.
Answer
144, 1984
Solution
Let the number be n and its divisors be 1 = d1 < d2 < ... , so we have n + 1 = d62 + d72.
Suppose d6 = ab, d7 = cd with 1 < a < b, 1 < c < d. Note that d6 and d7 must be coprime, because if d divides d6 and d7, then it also divides n and hence 1 = n - d62 - d72. So a, b, c, d are all distinct. If we must have a < d (or d6 = ab > a2 > d2 > cd = d7), hence ac < ad = d7. But ac ≠ d6 and ac divides n (since a and c are coprime and both divide n), so ac < d6. Hence we have 6 divisors of n all < d6 (namely 1, a, b, c, d, ac). Contradiction. So at least one of d6, d7 is a prime or a prime squared. Neither can be 2 or 4 (because there cannnot be 5 divisors of n less than 4).
But note that d6 divides n - d62 = d72 - 1 = (d7 - 1)(d7 + 1). (d7 ± 1) are coprime except possibly for a factor 2, so if d6 is an odd prime or odd prime squared, then it divides d7 - 1 or d7 + 1. Similarly, if d7 is an odd prime or odd prime squared, then it divides d6 - 1 or d6 + 1.
Suppose first that d6 is the odd prime or odd prime squared. If d6 divides d7 - 1, then d7 = kd6 + 1. Hence n = d62 + k2d62 + 2kd6 = d6(k2d6 + d6 + 2k). So d7 (coprime to d6) must divide k2d6 + d6 + 2k and hence also (k2d6 + d6 + 2k) - k(kd6 + 1) = d6 + k. So 0 ≤ d6 + k - kd6 - 1 = - (k - 1)(d6 - 1). Hence k = 1.
If d6 divides d7 + 1, then d7 = kd6 - 1 for some k > 1. Hence n = d6(k2d6 + d6 - 2k), so d7 divides k2d6 + d6 - 2k and hence also d6 - k. If d6 - k > 0, then d6 - k > kd6 - 1, so (k - 1)(d6 + 1) < 0. Contradiction. So d6 - k < 0. But d7 > |d6 - k|, so k > 2d6. But d6 is a prime or a prime square, so some di < d6 is coprime to d6. Hence n has another divisor between d6 and d62 (namely did6), so k ≤ d6. Contradiction. So d6 cannot divide d7 + 1.
The other case is much easier. If d7 divides d6 ± 1, then it must equal d6 + 1. So we have established that in all cases d7 = d6 + 1.
Put d6 = d. Then d and d+1 are coprime and both divide n, so n = kd(d+1) for some k. But n + 1 = d2 + (d+1)2, so k = 2 and n = 2d(d+1).
We can still get more out of the fact that d or d+1 is a prime or prime squared. Suppose d is a prime. Then d1, d2, d3, d4, d5, d7, 2d7 are precisely the factors of 2(d+1). But paqb ... has (a+1)(b+1)... factors, so 2(d+1) = p6 for some prime p which must be 2. Hence d+1 = 32, d = 31. That gives n = 1984 and it is easy to check that d1 = 1, d2 = 2, d3 = 4, d4 = 8, d5 = 16, d6 = 31, d7 = 32, and 1984 = 312 + 322 - 1.
Similarly, if d+1 is a prime, then d1, d2, d3, d4, d5, d6, 2d6 are precisely the factors of 2d. So 2d = 26 and d = 32, but then d+1 is not prime. Contradiction.
If d is a prime squared, put d = p2. If p = 3, then d+1 = 10, and n = 180, but then d1 = 1, d2 = 2, d3 = 3, d4 = 4, d5 = 5, d6 = 6 (not 9). Contradiction. So p > 3. d+1 is even, so we certainly have the factors 1, 2, 4, p, 2p, 4p less than p2. Contradiction.
The final case is d+1 a prime squared, so put d+1 = p2. If p = 3, then d = 8 and n = 144. Then d1 = 1, d2 = 2, d3 = 3, d4 = 4, d5 = 6, d6 = 8, d7 = 9 and 144 = 82 + 92 - 1, which is a solution. If p > 3, then d is even, so we have the factors 1, 2, 4, p, 2p, 4p less than p2 = d+1, so 4p = d6 = p2 - 1. Contradiction.
Problem B3
Find all real polynomials p(x) such that p(x2 + x + 1) = p(x) p(x + 1).
Answer
0, 1, (x2 + 1)n for n = 1, 2, 3, ...
Solution
Suppose p(x) has a real root. Then take k to be the largest real root. But k2 + k + 1 > k and is also a real root. Contradiction. So p(x) has no real roots. So p(x) must have even degree. Note also that it is immediate from the relation that the leading coefficient is 1.
It is easy to check that x2 + 1 is a solution and the only quadratic solution. Also notice that if q(x) and r(x) are solutions, then so is q(x)r(x), so we might conjecture that the solutions are (x2 + 1)n.
Let p(x) be a solution with degree 2n. Put q(x) = p(x) - (x2 + 1)n. Suppose q(x) ≠ 0. It has no term in x2n, so let deg q = m < 2n. Hence q(x2 + x + 1) - q(x)q(x+1) has degree < 4n. But it equals p(x)((x+1)2+1)n + p(x+1)(x2+1)n, which has degree 4n. Contradiction. So we must have q(x) = 0, which shows that the only solution of degree 2n is (x2 + 1)n.
Finally notice that in addition to the solution 1 of degree 0, we also have the solution 0.



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