25th Eötvös Competition Problems 1918

1.  AC is the long diagonal of a parallelogram ABCD. The perpendiculars from C meet the lines AB and AD at P and Q respectively. Show that AC² = AB·AP + AD·AQ.

2.  Find three distinct positive integers a, b, c such that 1/a + 1/b + 1/c is an integer.

3.  The real quadratics ax² + 2bx + c and Ax² + 2Bx + C are non-negative for all real x. Show that aAx² + 2bBx + cC is also non-negative for all real x. 

Solutions

Problem 1

AC is the long diagonal of a parallelogram ABCD. The perpendiculars from C meet the lines AB and AD at P and Q respectively. Show that AC² = AB·AP + AD·AQ.

Solution

Let the foot of the perpendicular from D to AC be X. Then AXD is similar to AQC, so AQ/AC = AX/AD. Also CXD is similar to APC, so AP/AC = CX/CD. Hence AQ·AD/AC + AP·CD/AC = AX + CX = AC. 

Problem 2

Find three distinct positive integers a, b, c such that 1/a + 1/b + 1/c is an integer.

Solution

Unfortunately, it is well-known that 1/2 + 1/3 + 1/6 = 1.

Suppose we did not know it! wlog a < b < c, so 1/a + 1/b + 1/c ≤ 1/1 + 1/2 + 1/3 < 2. Hence we must have 1/a + 1/b + 1/c = 1. Since 1/3 + 1/4 + 1/5 < 1, we must have a = 1 or 2. We cannot have a = 1 for then 1/b + 1/c = 0. So a = 2. Hence 1/b + 1/c = 1/2. We have 1/4 + 1/5 < 1/2, so b = 3. Then c = 6. 

Problem 3

The real quadratics ax² + 2bx + c and Ax² + 2Bx + C are non-negative for all real x. Show that aAx² + 2bBx + cC is also non-negative for all real x.

Solution

ax² + 2bx + c is non-negative for all real x iff a ≥ 0 and b² -ac ≥ 0. So a, A ≥ 0 and hence aA ≥ 0, and b² ≥ ac and B² ≥ AC and hence (bB)² ≥ (aA)(cC), so aAx² + 2bBx + cC is non-negative for all real x.