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

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

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 36	Followed by Problem 38
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1957 AHSME Problems/Problem 37

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