Gothman E.G.

A chord of a circle is at a distance $h$ from the center. In each of the segments bounded by the chord, a square is inscribed such that two adjacent vertices of the square lie on the arc, and the other two lie on the chord. What is the difference in the lengths of the sides of the squares?
Find the relationships between the sides of the squares, the radius of the circle, and the given value $h$.

## Solution

Let $O$ be the center of the circle, $P Q$ the given chord, $H$ its midpoint ($O H=h$), $A B C D$ and $K L M N$ the given squares (Fig.1), $E$ and $F$ the midpoints of segments $M N$ and $B C$. Let the sides of these squares be denoted by $a$ and $b$ respectively, and the radius of the circle by $r$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7b30f89ee421e63350f8g-11.jpg?height=389&width=412&top_left_y=589&top_left_x=823)

From the right triangles $O E N$ and $O F B$, we find that

\[
\begin{gathered}
b+h=H E+O H=E O=\sqrt{O N^{2}-N E^{2}}=\sqrt{r^{2}-\frac{b^{2}}{4}} \\
a-h=H F-O H=F O=\sqrt{O B^{2}-F B^{2}}=\sqrt{r^{2}-\frac{a^{2}}{4}}
\end{gathered}
\]

Therefore,

\[
(b+h)^{2}=r^{2}-\frac{b^{2}}{4},(a-h)^{2}=r^{2}-\frac{a^{2}}{4}
\]

Subtracting these equations term by term, we get

\[
(b+h)^{2}-(a-h)^{2}=\frac{a^{2}-b^{2}}{4}, 2 h-(a-b)=\frac{a-b}{4}
\]

From this, we find that $a-b=\frac{8}{5} h$.

## Answer

$\frac{8}{5} h$.