Problem 165

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Problem 165: (2020 AMC 12b, 10)

In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$ $\textbf{(A) } \frac{\sqrt5}{12} \qquad$ $ \textbf{(B) } \frac{\sqrt5}{10} \qquad$ $ \textbf{(C) } \frac{\sqrt5}{9} \qquad$ $ \textbf{(D) } \frac{\sqrt5}{8} \qquad$ $ \textbf{(E) } \frac{2\sqrt5}{15}$

Solution

Exercise 165: Solution

In square $ABCD$ with side length $2$, the inscribed circle $\omega$ intersects $\overline{CD}$ at $M$. Let $E$ be on $\overline{AB}$ with $AE=\dfrac14$, and let $\overline{EM}$ meet $\omega$ again at $P$. What is $AP$?

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