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1982 AHSME Problems/Problem 14

Revision as of 14:00, 22 August 2015 by LOTRFan123 (talk | contribs) (Created page with "Solution: Since <math>GP</math> is 15, <math>AP</math> is 75, and <math>\angle{AGP}=90</math>, <math>AG=15\sqrt{24}</math>. Now drop an altitude from <math>N</math> to <mat...")
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Solution:

Since $GP$ is 15, $AP$ is 75, and $\angle{AGP}=90$, $AG=15\sqrt{24}$.

Now drop an altitude from $N$ to $AG$ at point $H$. $AN=45$, and since $\triangle{AGP}$ is similar to $\triangle{AHN}$. $NH=9$. $NE=NF=15$ so by Pythagorean Theorem, $EH=HF=12$. Thus $EF=\boxed{24}$. $\boxed{C}$

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