IMO 2000 SL N5

Origin: BUL | Category: Number Theory

Problem

Prove that there exist infinitely many positive integers n such that p = nr, where p and r are respectively the semiperimeter and the inradius of a triangle with integer side lengths.

Solution

It is known that the area of the triangle is S = pr = p2/n and S =

p(p −a)(p −b)(p −c). It follows that p3 = n2(p −a)(p −b)(p −c), which by putting x = p −a, y = p −b, and z = p −c transforms into (x + y + z)3 = n2xyz. (1) We will be done if we show that (1) has a solution in positive integers for infinitely many natural numbers n. Let us assume that z = k(x + y) for an integer k > 0. Then (1) becomes (k + 1)3(x + y)2 = kn2xy. Further, by setting n = 3(k + 1) this equation reduces to (k + 1)(x + y)2 = 9kxy. (2) Set t = x/y. Then (2) has solutions in positive integers if and only if (k + 1)(t + 1)2 = 9kt has a rational solution, i.e., if and only if its discriminant D = k(5k −4) is a perfect square. Setting k = u2, we are led to show that 5u2 −4 = v2 has infinitely many integer solutions. But this is a classic Pell-type equation, whose solution is every Fibonacci number u = F2i+1. This completes the proof.