\section*{Problem 7}

A tangent to the inscribed circle of a triangle drawn parallel to one of the sides meets the other two sides at \(\mathrm{X}\) and \(\mathrm{Y}\). What is the maximum length \(\mathrm{XY}\), if the triangle has perimeter \(\mathrm{p}\) ?


\section*{Solution}

Let \(\mathrm{BC}\) be the side parallel to \(\mathrm{XY}, \mathrm{h}\) the length of the altitude from \(\mathrm{A}\), and \(\mathrm{r}\) the radius of the in-circle. Then \(X Y / B C=(h-2 r) / h\). But r.p \(=\) h.BC. So \(X Y=(p-2 B C) B C / p=\left(p^{2} / 8-2(B C-\right.\) \(\left.\mathrm{p} / 4)^{2}\right) / \mathrm{p}\). So the maximum occurs when \(\mathrm{BC}=\mathrm{p} / 4\) and has value \(\mathrm{p} / 8\).

