Solución del problema 59

Enunciado del problema 59

En el triángulo

$$\begin{array}{ccc}{r}^2+r^2& =& {\left(\frac{a\sqrt{2}}{2}-r\right)}^2\\ r\sqrt{2}& =& \frac{a\sqrt{2}}{2}-r\\ r\left(1+\sqrt{2}\right)& =& \frac{a\sqrt{2}}{2}\\ r& =& \frac{a\sqrt{2}}{2\left(\sqrt{2}+1\right)}\\ r& =& \frac{a\sqrt{2}\left(\sqrt{2}-1\right)}{2}\\ r& =& a\frac{2-\sqrt{2}}{2}\end{array}$$