Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1982 AHSME Problems/Problem 14

Revision as of 13:00, 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...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $EF=\boxed{24}$. $\boxed{C}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_14&oldid=71729"