1957 AMC 12 Problem 37

Attempt Problem 37 of the 1957 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1957 AMC 12 solutions, or check the answer key.

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37.

In right triangle ABC,ABC, BC=5,BC=5, AC=12,AC=12, and AM=x;AM=x; MN⊥AC,MN\perp AC, NP⊥BC;NP\perp BC; NN is on AB‾.\overline{AB}. If y=MN+NP,y=MN+NP, one-half the perimeter of rectangle MCPN,MCPN, then:
y=12(5+12)y=\dfrac12(5+12)
y=5x12+125y=\dfrac{5x}{12}+\dfrac{12}{5}
y=144−7x12y=\dfrac{144-7x}{12}
y=12y=12
y=5x12+6y=\dfrac{5x}{12}+6
Answer: C
Concepts:similarityrectanglelinear equation
Small hint:
Triangles AMNAMN and ACBACB are similar
Big hint:
Use MNx=512\frac{MN}{x}=\frac{5}{12} and NP=MC=12−xNP=MC=12-x
Solution:
By similarity, MN=512AM=5x12. MN=\frac5{12}AM=\frac{5x}{12}. Also NP=MC=AC−AMNP=MC=AC-AM =12−x.=12-x. Hence y=MN+NP=5x12+12−x=144−7x12. \begin{aligned} y=MN+NP &=\frac{5x}{12}+12-x\\ &=\frac{144-7x}{12}. \end{aligned}
Therefore, the correct answer is C.
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