2. The chord $AB$ divides a circle with radius $r$ into two arcs in the ratio $1:2$. In the larger segment, a square is inscribed such that one of its sides lies on this chord. Express the length of the side of the square in terms of the radius $r$.
Solution. Just as the chord $AB$ divides the circle into two arcs in the ratio $1:2$, so the corresponding central angles are in the ratio $1:2$, i.e., one is $120^{\circ}$, and the other is $240^{\circ}$. Let $S$ be the center of the circle. Let $C$ be a point on the smaller arc such that $SC \perp AB$, $SC \cap AB=\{P\}$, and $SC \cap DE=\{Q\}$. Then we have $\measuredangle ASB=120^{\circ}$, and $\Varangle CSB=60^{\circ}$. From the fact that triangle $CSB$ is equilateral ($\overline{SB}=\overline{SC}=r$, and the angle between them is $60^{\circ}$), it follows that $BP$ is the altitude in this triangle and divides the side $SC$ in half, i.e., $\overline{SP}=\frac{r}{2}$, and therefore $\overline{SQ}=a-\frac{r}{2}$, where $a$ is the side of the square $DEFG$.

From the right triangle $\triangle SDQ$ using the Pythagorean theorem, we have:

$$
r^{2}=\left(\frac{a}{2}\right)^{2}+\left(a-\frac{r}{2}\right)^{2} \quad \text{ i.e. } 5a^{2}-4ar-3r^{2}=0
$$

Since $a>0$, we get $a=\frac{4r+\sqrt{16r^{2}+60r^{2}}}{10}$, i.e., $a=\frac{2+\sqrt{19}}{5}r$

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