# 1957 AHSME Problems/Problem 37

## Problem

In right triangle $ABC, BC = 5, AC = 12$, and $AM = x; \overline{MN} \perp \overline{AC}, \overline{NP} \perp \overline{BC}$; $N$ is on $AB$. If $y = MN + NP$, one-half the perimeter of rectangle $MCPN$, then:

$[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair M = (1,0); pair C = (2,0); pair P = (2,0.5); pair B = (2,1); pair Q = (1,0.5); draw(A--B--C--cycle); draw(M--Q--P); label("A",A,SW); label("M",M,S); label("C",C,SE); label("P",P,E); label("B",B,NE); label("N",Q,NW);[/asy]$

$\textbf{(A)}\ y = \frac {1}{2}(5 + 12) \qquad \textbf{(B)}\ y = \frac {5x}{12} + \frac {12}{5}\qquad \textbf{(C)}\ y =\frac{144-7x}{12}\qquad \\ \textbf{(D)}\ y = 12\qquad \qquad\quad\,\, \textbf{(E)}\ y = \frac {5x}{12} + 6$

## Solution

Because $AC=12$ and $AM=x$, $MC=12-x$. Let $MN=z$. Then, because $MNPC$ is a rectangle, $NP=12-x$ and $PC=z$, and so $BP=5-z$. By AA similarity, $\triangle AMN \sim \triangle NPB$. From this similarity, we can solve the following proportion for $z$: \begin{align*} \frac{BP}{PN} &= \frac{NM}{AM} \\ \frac{5-z}{12-x} &= \frac z x \\ 5x-xz &= 12z-xz \\ 5x &= 12z \\ z &= \frac{5x}{12} \end{align*} Because $y=MN+NP=z+12-x$, we can now substitute for $z$ and find $y$ in terms of $x$: \begin{align*} y &= z+12-x \\ &= \frac{5x}{12}-x+12 \\ &= \frac{-7x}{12}+\frac{144}{12} \\ &= \frac{144-7x}{12} \end{align*} Thus, our answer is $\boxed{\textbf{(C) }\frac{144-7x}{12}}$.