[ Inscribed Circles in a Segment ]

In a rectangular sector $A O B$, a chord $A B$ is drawn, and a square is inscribed in the resulting segment. Find the ratio of the side of the square to the radius of the circle that touches the chord $A B$, the arc $A B$, and the side of the square perpendicular to the chord $A B$.

#
Express the radius of the sector in terms of the side of the square.

## Solution

Let $a$ be the side of the square $M N P Q$ (points $P$ and $Q$ lie on the chord $A B$), $R$ be the radius of the sector, $F$ be the center of the circle $\omega$, $r$ be its radius, $E$ and $H$ be the points of tangency of $\omega$ with the side $N P$ of the square and the arc $A B$, $T$ be the midpoint of $M N$, $K$ be the midpoint of the chord $A B$, and $D$ be the intersection point of the line $F E$ with the segment $O T$.

By the Pythagorean theorem, $O M^{2}=O T^{2}+M T^{2}$, or $R^{2}=\left(\frac{R}{\sqrt{2}}+a\right)^{2}+\frac{a^{2}}{4}$, which simplifies to $2 R^{2}-4 a R \sqrt{2}-5 a^{2}=0$.

From this, we get $R=\frac{5 a \sqrt{2}}{2}$.

$O F^{2}=O D^{2}+D F^{2}$, or $(R-r)^{2}=\left(\frac{R}{\sqrt{2}}+r\right)^{2}+\left(r+\frac{a}{2}\right)^{2}$. Substituting $R=\frac{5 a \sqrt{2}}{2}$, after simplification, we obtain the equation $6 a^{2}-a r(5 \sqrt{2}+6)-r^{2}=0$, from which we find $a / r$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-52.jpg?height=603&width=806&top_left_y=1236&top_left_x=635)

Answer

$\frac{6+5 \sqrt{2}+(3+\sqrt{2}) \sqrt{10}}{12}=\frac{(\sqrt{5}+\sqrt{2})(3+\sqrt{5})}{6 \sqrt{2}}$