1957 AHSME Problems/Problem 37
Problem
In right triangle 
, and 
;
is on 
. If 
, one-half the perimeter of rectangle 
, 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]](http://latex.artofproblemsolving.com/c/2/9/c29954e48d094adb9f8ad33c142a06c2b1cba5a6.png)
Solution
Because 
and 
, 
. Let 
. Then, because 
is a rectangle, 
and 
, and so 
. By AA similarity, 
. From this similarity, we can solve the following proportion for 
:
\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 
, we can now substitute for and find 
in terms of 
:
\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 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
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