2008 Mock ARML 1 Problems/Problem 6

Problem

Square $ABCD$ has side length $2$ . $M$ is the midpoint of $CD$ , and $N$ is the midpoint of $BC$ . $P$ is on $MN$ such that $N$ is between $M$ and $P$ , and $m\angle MAN = m\angle NAP$ . Compute the length of $AP$ .

Solution

$[asy] pointpen=black;pathpen=black+linewidth(0.7); pair A=(0,2), D=(0,0), C=(2,0), B=(2,2), M=(C+D)/2, N=(B+C)/2, P=8/3*(N-M)+M; D(MP("A",A,(0,1))--MP("B",B,(0,1))--MP("C",C)--MP("D",D)--cycle); D(D(MP("M",M))--D(MP("N",N,E))--D(MP("P",P,E))); D(M--A--N--A--P); D(anglemark(M,A,P)); MP("2",(A+D)/2,W); [/asy]$

By the Pythagorean Theorem, $MA = NA = \sqrt{5}$ and $MN = \sqrt{2}$ . Let $\theta = \angle MAN = \angle NAP$ . By the Law of Cosines on $\triangle MAN$ , $\left(\sqrt{2}\right)^2 = \left(\sqrt{5}\right)^2 + \left(\sqrt{5}\right)^2 - 2 \cdot \left(\sqrt{5}\right) \cdot \left(\sqrt{5}\right) \cos \theta \Longrightarrow \cos \theta = \frac{4}{5}.$ The Law of Cosines on $\triangle NAP$ yields

$NP^2 = AP^2 + \left(\sqrt{5}\right)^2 - 2 \cdot AP \cdot \left(\sqrt{5}\right) \cos \theta = AP^2 - \frac{8}{\sqrt{5}}AP + 5.$

The Angle Bisector Theorem on $\triangle MAP$ yields $\frac{AP}{NP} = \frac{AM}{MN} = \sqrt{\frac{5}{2}} \Longrightarrow NP = \sqrt{\frac{2}{5}}AP.$ Substituting, $\begin{align*} 0 &= 3AP^2 - 8\sqrt{5}AP + 25\\ AP &= \frac{8\sqrt{5} \pm \sqrt{20}}{6} = \boxed{\frac{5\sqrt{5}}{3}}, \sqrt{5}. \end{align*}$

2008 Mock ARML 1 (Problems, Source)
Preceded by Problem 5	Followed by Problem 7
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2008 Mock ARML 1 Problems/Problem 6

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