Two squares $A B C D$ and $K L M N$ are positioned such that the vertices $B, C, K$ and $N$ lie on the same line, and the four remaining vertices are located on opposite sides of $B C$ and lie on the same circle. It is known that the side of one of the squares is 1 unit longer than the side of the other. Find the distance from the center of the circle to the line $B C$.
Let $x$ be the side of the smaller square $KLMN$, then the side of the square $ABCD$ is $x+1$. Let the line passing through the center $O$ of the given circle, perpendicular to $BC$, intersect $BC$ at point $F$, and $OF = a$.

First method. Let the line $OF$ intersect $LM$ at point $Q$, $P$ - the projection of point $O$ on $DC$, $R$ - the radius of the circle.

By the Pythagorean theorem, $(a+x)^{2} + \frac{1}{4} x^{2} = OQ^{2} + QM^{2} = R^{2} = OP^{2} + DP^{2} = \frac{1}{4}(x+1)^{2} + (x+1-a)^{2}$.

From this, $(a+x)^{2} - (x+1-a)^{2} = \frac{1}{4}((x+1)^{2} - x^{2})$, or $2a - 1 = \frac{1}{4}$, hence $a = \frac{5}{8}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-34.jpg?height=563&width=583&top_left_y=-1&top_left_x=743)

Second method. Notice that the segment $MN$ lies on a chord of length $2(x+a)$. Let the segment $BC$ lie on a chord of length $2b$. Then $b^{2} = MF \cdot FA = \frac{5}{4} x(x+1)$.

On the other hand, calculating the power of point $N$ in two ways, we get $(b - \frac{x}{2})(b + \frac{x}{2}) = x(x + 2a)$, which means $2ax = b^{2} - \frac{5}{4} x^{2} = \frac{5x}{4}$, hence $a = \frac{5}{8}$.

## Answer

$\frac{5}{8}$.