A1. Let m and n be positive integers such that: m/n = 1 - 1/2 + 1/3 - 1/4 + ... - 1/1318 + 1/1319.
Prove that m is divisible by 1979.
A2. A prism with pentagons A1A2A3A4A5 and B1B2B3B4B5 as the top and bottom faces is given. Each side of the two pentagons and each of the 25 segments AiBj is colored red or green. Every triangle whose vertices are vertices of the prism and whose sides have all been colored has two sides of a different color. Prove that all 10 sides of the top and bottom faces have the same color.
A3. Two circles in a plane intersect. A is one of the points of intersection. Starting simultaneously from A two points move with constant speed, each traveling along its own circle in the same sense. The two points return to A simultaneously after one revolution. Prove that there is a fixed point P in the plane such that the two points are always equidistant from P.
B1. Given a plane k, a point P in the plane and a point Q not in the plane, find all points R in k such that the ratio (QP + PR)/QR is a maximum.
B2. Find all real numbers a for which there exist non-negative real numbers x1, x2, x3, x4, x5 satisfying: x1 + 2x2 + 3x3 + 4x4 + 5x5 = a,
x1 + 23x2 + 33x3 + 43x4 + 53x5 = a2,
x1 + 25x2 + 35x3 + 45x4 + 55x5 = a3.
B3. Let A and E be opposite vertices of an octagon. A frog starts at vertex A. From any vertex except E it jumps to one of the two adjacent vertices. When it reaches E it stops. Let an be the number of distinct paths of exactly n jumps ending at E. Prove that: a2n-1 = 0
a2n = (2 + √2)n-1/√2 - (2 - √2)n-1/√2.
Solutions
Problem A1
Let m and n be positive integers such that:
m/m = 1 - 1/2 + 1/3 - 1/4 + ... - 1/1318 + 1/1319.
Prove that m is divisible by 1979.
Solution
This is difficult.
The obvious step of combining adjacent terms to give 1/(n(n+1) is unhelpful. The trick is to separate out the negative terms:
1 - 1/2 + 1/3 - 1/4 + ... - 1/1318 + 1/1319 = 1 + 1/2 + 1/3 + ... + 1/1319 - 2(1/2 + 1/4 + ... + 1/1318) = 1/660 + 1/661 + ... + 1/1319.
and to notice that 660 + 1319 = 1979. Combine terms in pairs from the outside:
1/660 + 1/1319 = 1979/(660.1319); 1/661 + 1/1318 = 1979/(661.1318) etc.
There are an even number of terms, so this gives us a sum of terms 1979/m with m not divisible by 1979 (since 1979 is prime and so does not divide any product of smaller numbers). Hence the sum of the 1/m gives a rational number with denominator not divisible by 1979 and we are done.
Problem A2
A prism with pentagons A1A2A3A4A5 and B1B2B3B4B5 as the top and bottom faces is given. Each side of the two pentagons and each of the 25 segments AiBj is colored red or green. Every triangle whose vertices are vertices of the prism and whose sides have all been colored has two sides of a different color. Prove that all 10 sides of the top and bottom faces have the same color.
Solution
We show first that the Ai are all the same color. If not then, there is a vertex, call it A1, with edges A1A2, A1A5 of opposite color. Now consider the five edges A1Bi. At least three of them must be the same color. Suppose it is green and that A1A2 is also green. Take the three edges to be A1Bi, A1Bj, A1Bk. Then considering the triangles A1A2Bi, A1A2Bj, A1A2Bk, the three edges A2Bi, A2Bj, A2Bk must all be red. Two of Bi, Bj, Bk must be adjacent, but if the resulting edge is red then we have an all red triangle with A2, whilst if it is green we have an all green triangle with A1. Contradiction. So the Ai are all the same color. Similarly, the Bi are all the same color. It remains to show that they are the same color. Suppose otherwise, so that the Ai are green and the Bi are red.
Now we argue as before that 3 of the 5 edges A1Bi must be the same color. If it is red, then as before 2 of the 3 Bi must be adjacent and that gives an all red triangle with A1. So 3 of the 5 edges A1Bi must be green. Similarly, 3 of the 5 edges A2Bi must be green. But there must be a Bi featuring in both sets and it forms an all green triangle with A1 and A2. Contradiction. So the Ai and the Bi are all the same color.
Problem A3
Two circles in a plane intersect. A is one of the points of intersection. Starting simultaneously from A two points move with constant speed, each traveling along its own circle in the same sense. The two points return to A simultaneously after one revolution. Prove that there is a fixed point P in the plane such that the two points are always equidistant from P.
Solution
Let the circles have centers O, O' and let the moving points by X, X. Let P be the reflection of A in the perpendicular bisector of OO'. We show that triangles POX, X'O'P are congruent. We have OX = OA (pts on circle) = O'P (reflection). Also OP = O'A (reflection) = O'X' (pts on circle). Also ∠AOX = ∠AO'X' (X and X' circle at same rate), and ∠AOP = ∠AO'P (reflection), so ∠POX = ∠PO'X'. So the triangles are congruent. Hence PX = PX'.
Another approach is to show that XX' passes through the other point of intersection of the two circles, but that involves looking at many different cases depending on the relative positions of the moving points.
Problem B1
Given a plane k, a point P in the plane and a point Q not in the plane, find all points R in k such that the ratio (QP + PR)/QR is a maximum.
Solution
Consider the points R on a circle center P. Let X be the foot of the perpendicular from Q to k. Assume P is distinct from X, then we minimise QR (and hence maximise (QP + PR)/QR) for points R on the circle by taking R on the line PX. Moreover, R must lie on the same side of P as X. Hence if we allow R to vary over k, the points maximising (QP + PR)/QR must lie on the ray PX. Take S on the line PX on the opposite side of P from X so that PS = PQ. Then for points R on the ray PX we have (QP + PR)/QR = SR/QR = sin RQS/sin QSR. But sin QSR is fixed for points on the ray, so we maximise the ratio by taking ∠RQS = 90o. Thus there is a single point maximising the ratio.
If P = X, then we still require ∠RQS = 90o, but R is no longer restricted to a line, so it can be anywhere on a circle center P.
Problem B2
Find all real numbers a for which there exist non-negative real numbers x1, x2, x3, x4, x5 satisfying:
x1 + 2x2 + 3x3 + 4x4 + 5x5 = a,
x1 + 23x2 + 33x3 + 43x4 + 53x5 = a2,
x1 + 25x2 + 35x3 + 45x4 + 55x5 = a3.
x1 + 23x2 + 33x3 + 43x4 + 53x5 = a2,
x1 + 25x2 + 35x3 + 45x4 + 55x5 = a3.
Solution
Take a2 x 1st equ - 2a x 2nd equ + 3rd equ. The rhs is 0. On the lhs the coefficient of xn is a2n - 2an3 + n5 = n(a - n2)2. So the lhs is a sum of non-negative terms. Hence each term must be zero separately, so for each n either xn = 0 or a = n2. So there are just 5 solutions, corresponding to a = 1, 4, 9, 16, 25. We can check that each of these gives a solution. [For a = n2, xn = n and the other xi are zero.]
Problem B3
Let A and E be opposite vertices of an octagon. A frog starts at vertex A. From any vertex except E it jumps to one of the two adjacent vertices. When it reaches E it stops. Let an be the number of distinct paths of exactly n jumps ending at E. Prove that:
a2n-1 = 0
a2n = (2 + √2)n-1/√2 - (2 - √2)n-1/√2.
a2n = (2 + √2)n-1/√2 - (2 - √2)n-1/√2.
Solution
Each jump changes the parity of the shortest distance to E. The parity is initially even, so an odd number of jumps cannot end at E. Hence a2n-1 = 0.
We derive a recurrence relation for a2n. This is not easy to do directly, so we introduce bn which is the number of paths length n from C to E. Then we have immediately:
a2n = 2a2n-2 + 2b2n-2 for n > 1
b2n = 2b2n-2 + a2n-2 for n > 1
b2n = 2b2n-2 + a2n-2 for n > 1
Hence, using the first equation: a2n - 2a2n-2 = 2a2n-2 - 4a2n-4 + 2b2n-2 - 4b2n-4 for n > 2. Using the second equation, this leads to: a2n = 4a2n-2 - 2a2n-4 for n > 2. This is a linear recurrence relation with the general solution: a2n = a(2 + √2)n-1 + b(2 - √2)n-1. But we easily see directly that a4 = 2, a6 = 8 and we can now solve for the coefficients to get the solution given.
Read more
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985
, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 
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