28th British Mathematical Olympiad 1992 Problems

28th British Mathematical Olympiad 1992 Problems

1.  p is an odd prime. Show that there are unique positive integers m, n such that m² = n(n + p). Find m and n in terms of p.

2.  Show that 12/(w + x + y + z) ≤ 1/(w + x) + 1/(w + y) + 1/(w + z) + 1/(x + y) + 1/(x + z) + 1/(y + z) ≤ 3(1/w + 1/x + 1/y + 1/z)/4 for any positive reals w, x, y, z.

3.  The circumradius R of a triangle with sides a, b, c satisfies a² + b² = c² - R². Find the angles of the triangle.

4.  Each edge of a connected graph with n points is colored red, blue or green. Each point has exactly three edges, one red, one blue and one green. Show that n must be even and that such a colored graph is possible for any even n > 2. X is a subset of 1 < k < n points. In order to isolate X from the other points (so that there is no edge between a point in X and a point not in X) it is necessary and sufficient to delete R red edges, B blue edges and G green edges. Show that R, B, G are all even or all odd. 

Solutions

Problem 1

p is an odd prime. Show that there are unique positive integers m, n such that m² = n(n + p). Find m and n in terms of p.

Solution

We have (k² + k)² = k²(k² + 2k + 1). So if p = 2k + 1, then m = k² + k, n = k² is always a solution (whether or not p is prime).

4m² = 4n² + 4np, so 4m² + p² = (2n+p)². Hence p² = (2n+p+m)(2n+p-m). But p is prime, so the only way of expressing p² as a product of two unequal numbers is p² x 1. Hence 2n+p-m = 1, 2n+p+m = p². Solving for m and n, we get n = (p² - 2p + 1)/4 = (p-1)²/4, m = (p² - 1)/2. 

Problem 2

Show that 12/(w + x + y + z) ≤ 1/(w + x) + 1/(w + y) + 1/(w + z) + 1/(x + y) + 1/(x + z) + 1/(y + z) ≤ 3(1/w + 1/x + 1/y + 1/z)/4 for any positive reals w, x, y, z.

Solution

The left-hand inequality is just the harmonic mean inequality applied to w+x, w+y etc. We have 1/(w + x) + ... ≥ 6²/(w + x + w + y + ... ) = 36/( 3(w + x + y + z) ).

The harmonic mean inequality applied to w and x gives 1/w + 1/x ≥ 4/(w + x). So applying it to the other pairs also and adding we get 3(1/w + 1/x + 1/y + 1/z) ≥ 4(1/(w + x) + ... ), which is the right-hand inequality. 

Problem 3

The circumradius R of a triangle with sides a, b, c satisfies a² + b² = c² - R². Find the angles of the triangle.

Solution

Answer: A = B = 30^o and C = 120^o.

We have a = 2R sin A etc (if O is the circumcenter, then ∠BOC = 2A). So we have that sin²A + sin²B = sin²C - 1/4. We try to simplify this. Put C = 180^o - A - B, then sin C = sin(A + B) = sin A cos B + cos A sin B. Hence we get sin²A + sin²B = sin²A cos²B + cos²A sin²B + 2 sin A cos A sin B cos B - 1/4. Hence 2 sin²A sin²B = 2 sin A cos A sin B cos B - 1/4, or sin A sin B (sin A sin B - cos A cos B) = -1/8 or sin A sin B cos C = -1/8.

It is not clear why this should determine A, B, C. We suspect that maybe sin A sin B cos C >= -1/8 with equality only at the angles we want (that is a common ruse). sin A and sin B must be positive, so cos C must be negative. For fixed A + B, we maximise sin A sin B by taking A = B (for 2 sin A sin B = cos(A - B) - cos(A + B) ). So sin A sin B cos C >= sin²A cos(180^o - 2A) = - sin²A cos 2A = - sin²A (1 - 2 sin²A). To minimise this we must maximise k(1 - 2k), where k = sin²A. Maximising k(1 - 2k) is equivalent to maximising 2k(1 - 2k) which is achieved by taking 2k = 1 - 2k, so we want k = 1/4 and hence sin A = 1/2, so A = B = 30^o and C = 120^o. The value then of sin A sin B cos C is indeed -1/8, so as predicted we have sin A sin B cos C >= -1/8 with equality iff A = B = 30^o and C = 120^o. Since we are told that we do have equality, the angles are fixed. 

Problem 4

Each edge of a connected graph with n points is colored red, blue or green. Each point has exactly three edges, one red, one blue and one green. Show that n must be even and that such a colored graph is possible for any even n > 2. X is a subset of 1 < k < n points. In order to isolate X from the other points (so that there is no edge between a point in X and a point not in X) it is necessary and sufficient to delete R red edges, B blue edges and G green edges. Show that R, B, G are all even or all odd.

Solution

The number of edges is 3n/2. This must be integral, so n must be even. Take a regular 2m-gon. Make the sides alternately red and blue and draw the main diagonals (connecting each point to the opposite point) green. Then each vertex has three edges, one of each color.

Suppose X has m vertices. There must be R red, B blue, and G green edges with just one end a point of X. Suppose there are r red, b blue and g green edges with both ends at a point of X. We have R + 2r = m, B + 2b = m, G + 2g = m. Hence R, B and G have the same parity.