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

$[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("$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 $\boxed{C}$ .

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1982 AHSME Problems/Problem 14

Problem 14

Solution