Bulgarian Mathematical Competitions

Solution:

a) It follows from the Pythagorean theorem that the altitude of $\triangle ABC$ through $C$ is equal to $\sqrt{1 - x^2}$. Then
$$r = \frac{S}{p} = \frac{x \sqrt{1 - x^2}}{1 + x} = x \sqrt{\frac{1 - x}{1 + x}}$$

b) We have to find the maximum of the function
$$f(x) = \frac{x^2 (1 - x)}{1 + x}$$
in the interval $(0, 1)$. Since
$$f'(x) = \frac{2x(1 - x - x^2)}{(x + 1)^2}$$
the function $f(x)$ increases in the interval $\left(0, \frac{\sqrt{5} - 1}{2}\right]$ and decreases in the interval $\left[\frac{\sqrt{5} - 1}{2}, 1\right)$. Therefore the maximum of $f(x)$ in $(0, 1)$ is attained for $x = \frac{\sqrt{5} - 1}{2}$ and is equal to
$$f\left(\frac{\sqrt{5} - 1}{2}\right) = \frac{5\sqrt{5} - 11}{2}.$$
Hence the maximum possible value of $r$ is $\sqrt{\frac{5\sqrt{5} - 11}{2}}$.