21st IMO 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.

Solution　We show first that the A_i 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 A₁B_i. 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 A₁B_i, A₁B_j, A₁B_k. Then considering the triangles A₁A₂B_i, A₁A₂B_j, A₁A₂B_k, the three edges A₂B_i, A₂B_j, A₂B_k must all be red. Two of B_i, B_j, B_k 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 A_i are all the same color. Similarly, the B_i are all the same color. It remains to show that they are the same color. Suppose otherwise, so that the A_i are green and the B_i are red.
Now we argue as before that 3 of the 5 edges A₁B_i must be the same color. If it is red, then as before 2 of the 3 B_i must be adjacent and that gives an all red triangle with A₁. So 3 of the 5 edges A₁B_i must be green. Similarly, 3 of the 5 edges A₂B_i must be green. But there must be a B_i featuring in both sets and it forms an all green triangle with A₁ and A₂. Contradiction. So the A_i and the B_i are all the same color.

Solution　Let the circles have centers C, D and let their other point of intersection be B. Take E on the circle center C and F on the circle center D so that EF is perpendicular to AB and meets it at A. Let the point on the circle center C be at P and the point on the other circle be at Q. Then angle ABP = half angle ACP = half angle ADQ = angle AFQ. Hence angle ABQ = 180 - angle AFQ = 180 - angle ABP, so PQ passes through B. Also angle BPE = 180 - angle BAE = 90 and angle BQE = 180 - angle BAF = 90. In other words the perpendiculars to PQ at its endpoints meet EF at its endpoints. Hence the perpendicular to PQ at its midpoint meets EF at its midpoint. So P and Q are equidistant from the midpoint of EF (which remains fixed as P and Q move).

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 angle RQS = 90. Thus there is a single point maximising the ratio.
If P = X, then we still require angle RQS = 90, but R is no longer restricted to a line, so it can be anywhere on a circle center P.

Solution　Take a² x 1st equ - 2a x 2nd equ + 3rd equ. The rhs is 0. On the lhs the coefficient of x_n 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 x_n = 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, x_n = n and the other x_i are zero.]

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 a_2n-1 = 0.
We derive a recurrence relation for a_2n. This is not easy to do directly, so we introduce b_n which is the number of paths length n from C to E. Then we have immediately:
　 a_2n = 2a_2n-2 + 2b_2n-2 for n > 1
　 b_2n = 2b_2n-2 + a_2n-2 for n > 1
Hence, using the first equation: a_2n - 2a_2n-2 = 2a_2n-2 - 4a_2n-4 + 2b_2n-2 - 4b_2n-4 for n > 2. Using the second equation, this leads to: a_2n = 4a_2n-2 - 2a_2n-4 for n > 2. This is a linear recurrence relation with the general solution: a_2n = a(2 + Ö2)^n-1 + b(2 - Ö2)^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.