1st Eötvös 1894/3 solution

1st Eötvös 1894

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.

1st Eötvös 1894