1957 AHSME Problems/Problem 37
Problem
In right triangle , and ; is on . If , one-half the perimeter of rectangle , then:
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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