1957 AHSME Problems/Problem 37

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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 $\boxed{\textbf{(C) }\frac{144-7x}{12}}$.

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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