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(D) Let the square be $ABCD$ with side length $1$. Let $O$ be the center of the circle $\Gamma$ and $r$ be its radius.Let $M$ be the midpoint of $BC$.

Vedclass · Accepted Answer

$\frac{5}{8}$

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(D) Let the square be $ABCD$ with side length $1$. Let $O$ be the center of the circle $\Gamma$ and $r$ be its radius.Let $M$ be the midpoint of $BC$. Since $BC$ is a chord of the circle,the perpendicular from the center $O$ to $BC$ passes through $M$.Thus,$OM \perp BC$. Since $BC$ is vertical and parallel to $AD$,$OM$ is horizontal.Let $N$ be the point of tangency on $AD$. Then $ON \perp AD$. Since $AD$ is vertical,$ON$ is horizontal.Since $AD$ and $BC$ are parallel and separated by a distance of $1$,the total horizontal distance between $AD$ and $BC$ is $1$.Let $O$ be at a distance $r$ from $N$ (along the horizontal line). The distance from $O$ to $M$ is $1-r$.In the right-angled triangle $	riangle OMC$,we have $OC = r$,$CM = \frac{1}{2} BC = \frac{1}{2}$,and $OM = 1-r$.Using the Pythagorean theorem: $OC^2 = OM^2 + CM^2$.$r^2 = (1-r)^2 + (\frac{1}{2})^2$.$r^2 = 1 - 2r + r^2 + \frac{1}{4}$.$2r = 1 + \frac{1}{4} = \frac{5}{4}$.$r = \frac{5}{8}$.

Let $ABCD$ be a square of side length $1$,and $\Gamma$ be a circle passing through $B$ and $C$,and touching $AD$. The radius of $\Gamma$ is

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