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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.7 AIME Prove it Soviet Union

Problem:
A tangent to the inscribed circle of a triangle drawn parallel to one of the sides meets the other two sides at $X$ and $Y$ . What is the maximum length $XY$ , if the triangle has perimeter $p$ ?

Solution

Solution:
Let $BC$ be the side parallel to $XY$ , $h$ the length of the altitude from $A$ , and $r$ the radius of the incircle. Then $XY/BC = (h - 2r)/h$ . But $r p = h BC$ . So $XY = \frac{(p - 2BC) BC}{p} = \frac{p^2}{8} - \frac{2(BC - p/4)^2}{p}$ . So the maximum occurs when $BC = p/4$ and has value $p/8$ .

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