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Difference between revisions of "1957 AHSME Problems/Problem 37"

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== Solution ==
 
== Solution ==
−
<math>\boxed{\textbf{(C) }\frac{144-7x}{12}}</math>.
+
Because <math>AC=12</math> and <math>AM=x</math>, <math>MC=12-x</math>. Let <math>MN=z</math>. Then, because <math>MNPC</math> is a rectangle, <math>NP=12-x</math> and <math>PC=z</math>, and so <math>BP=5-z</math>. By [[AA similarity]], <math>\triangle AMN \sim \triangle NPB</math>. From this similarity, we can solve the following proportion for <math>z</math>:
 +
\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 <math>y=MN+NP=z+12-x</math>, we can now substitute for <math>z</math> and find <math>y</math> in terms of <math>x</math>:
 +
\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 <math>\boxed{\textbf{(C) }\frac{144-7x}{12}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 16:09, 26 July 2024

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 (Problems • Answer Key • Resources)
Preceded by
Problem 36 		Followed by
Problem 38
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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