# 1982 AHSME Problems/Problem 14

## Problem 14

In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB, BC$, and $CD$ are diameters of circle $O, N$, and $P$, respectively. Circles $O, N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length $[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1)), F=intersectionpoints(A--G, Circle(N,1)); draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("A", A, W); label("B", B, SE); label("C", C, NE); label("D", D, dir(0)); label("P", P, S); label("N", N, S); label("O", O, S); label("E", E, dir(120)); label("F", F, NE); label("G", G, dir(100));[/asy]$

## Solution

Since $GP$ is 15, $AP$ is 75, and $\angle{AGP}=90$, $AG=30\sqrt{6}$.

Now drop a perpendicular 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 the Pythagorean Theorem, $EH=HF=12$. Thus $EF=\boxed{24.}$ Answer is then $\boxed{C}$.