Problem 12.1. Let $A B C$ be an isosceles triangle such that $A C=B C=1$ and $A B=2 x, x>0$.

a) Express the inradius $r$ of $\triangle A B C$ as a function of $x$.

b) Find the maximum possible value of $r$.

Oleg Mushkarov
12.1. a) It follows from the Pythagorean theorem that the altitude of $\triangle A B C$ 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 $f^{\prime}(x)=\frac{2 x\left(1-x-x^{2}\right)}{(x+1)^{2}}$ 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 $\sqrt{\frac{5 \sqrt{5}-11}{2}}$