1st Eötvös Competition Problems 1894

1.  Show that { (m, n): 17 divides 2m + 3n} = { (m, n): 17 divides 9m + 5n}.

2.  Given a circle C, and two points A, B inside it, construct a right-angled triangle PQR with vertices on C and hypoteneuse QR such that A lies on the side PQ and B lies on the side PR. For which A, B is this not possible?

3.  A triangle has sides length a, a + d, a + 2d and area S. Find its sides and angles in terms of d and S. Give numerical answers for d = 1, S = 6. 

Solutions

Problem 1

Show that { (m, n): 17 divides 2m + 3n} = { (m, n): 17 divides 9m + 5n}.

Solution

Put N = 2m+3n, M = 9m+5n. 17|N implies 17|(13N-17(m+2n)) = M. Similarly, 17|M implies 17|(4M-17(2m+n)) = N. 

Problem 2

Given a circle C, and two points A, B inside it, construct a right-angled triangle PQR with vertices on C and hypoteneuse QR such that A lies on the side PQ and B lies on the side PR. For which A, B is this not possible?

Solution

Since ∠APB = 90^o, P must lie on the circle C and on the circle diameter AB. So a necessary condition is that these circles should intersect. Conversely, if the circle do intersect, then it is clear that we can take a point of intersection as P. Then extend PA to meet the circle C again at Q and extend PB to meet the circle C again at R. 

Problem 3

A triangle has sides length a, a + d, a + 2d and area S. Find its sides and angles in terms of d and S. Give numerical answers for d = 1, S = 6.

Solution

It is slightly more convenient to use b = a+d. Then by Heron S² = (3b²/4)(b²/4 - d²). This is a quadratic in b². We need a positive solution, so b² = 2(d² + √(d⁴ + 4S²/3) ). That gives the sides a, b, c = b+d. Now sin A = 2S/bc, and sin B = 2S/ac. Since c is the longest side, A and B must be acute. Finally, C = 180^o - A - B.

For d = 1, S = 6, we find b = 4. Hence a = 3, c = 5, A = sin^-1(3/5). We recognize the triangle, so C = 90^o, B = 90^o - A.