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Let $A B C D$ be a square of side length 1 , and $\Gamma$ a circle passing through $B$ and $C$, and touching $\mathrm{AD}$. The radius of $\Gamma$ is
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Verified Answer
The correct answer is:
$\frac{5}{8}$
$\mathrm{PC}=\mathrm{r}$
$\mathrm{PC}^{2}=\mathrm{r}^{2}$
$(\mathrm{r}-1)^{2}+\left(\frac{1}{2}-1\right)^{2}=r^{2}$
$1-2 \mathrm{r}+\frac{1}{4}=0$
$r=\frac{5}{8}$
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