## Problem I - 1

Given a right isosceles triangle $A B C$, with the right angle at $C$, and the legs of length 2. An arc of a circle $l$ with center $A$ divides the triangle into two parts of equal area, while the arc of a circle $m$ with center at $B$ is tangent to the arc $l$ at a point on the hypotenuse $A B$.

Find the area of the portion of the triangle not covered by the circular sectors corresponding to the two arcs.
Solution.

Let $r$ be the radius of the arc $l$. The area of the sector thus determined in the triangle is 1/8 of the area of the circle. Therefore,

$$
\frac{1}{8} \pi r^{2}=1 \Rightarrow r=2 \sqrt{\frac{2}{\pi}}
$$

The radius of the circle $m$ is

$$
r_{1}=|A B|-r=2 \sqrt{2}-2 \sqrt{\frac{2}{\pi}}=2 \sqrt{2}\left(1-\frac{1}{\sqrt{\pi}}\right)
$$

The area of the desired region is then

$$
S=1-\frac{1}{8} \pi \cdot 8\left(1-\frac{1}{\sqrt{\pi}}\right)^{2}=2 \sqrt{\pi}-\pi
$$