1976 AHSME Problems/Problem 24

Problem

In the adjoining figure, circle $K$ has diameter $AB$ ; circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$ ; and circle $M$ tangent to circle $K$ , to circle $L$ and $AB$ . The ratio of the area of circle $K$ to the area of circle $M$ is $[asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(150); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); label("$A$", A, W); label("$K$", K, S); label("$B$", B, E); label("$L$", L); label("$M$", M); [/asy]$ $\textbf{(A) }12\qquad \textbf{(B) }14\qquad \textbf{(C) }16\qquad \textbf{(D) }18\qquad \textbf{(E) }\text{not an integer}$

Solution

Let $R$ and $r$ be the radius of $\odot K$ and the radius of $\odot M,$ respectively. It follows that the radius of $\odot L$ is $\frac{R}{2}.$

Suppose $P$ is the foot of the perpendicular from $M$ to $\overline{KL}.$ We construct the auxiliary lines, as shown below: $[asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(200); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25),I=(2*sqrt(2)/3,1/3),E=(sqrt(2)/3,1/3),P=(0,0.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); draw(L--K,red); draw(L--M,red); draw(K--I,red); draw(P--M,red); label("$A$", A, (-5/4,0)); label("$K$", K, (0,-5/4)); label("$B$", B, (5/4,0)); label("$L$", L, (0,5/4)); label("$M$", M, (0,5/4)); label("$P$", P, (-5/4,0)); dot(K,linewidth(4)); dot(L,linewidth(4)); dot(M,linewidth(4)); dot(I,linewidth(4)); dot(E,linewidth(4)); dot(P,linewidth(4)); [/asy]$ In right $\triangle KPM,$ we have $KP=r$ and $KM=R-r.$ By the Pythagorean Theorem, we get $PM^2=(R-r)^2-r^2.$

In right $\triangle LPM,$ we have $LP=\frac{R}{2}-r$ and $LM=\frac{R}{2}+r.$ By the Pythagorean Theorem, we get $PM^2=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2.$

We equate the expressions for $PM^2,$ then simplify: $\begin{align*} (R-r)^2-r^2&=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2 \\ \left(R^2-2Rr+r^2\right)-r^2&=\left(\frac{R^2}{4}+Rr+r^2\right)-\left(\frac{R^2}{4}-Rr+r^2\right) \\ R^2-2Rr&=2Rr \\ R^2&=4Rr \\ R&=4r. \end{align*}$ Therefore, the ratio of the area of $\odot K$ to the area of $\odot M$ is $\frac{\pi R^2}{\pi r^2}=\left(\frac{R}{r}\right)^2=\boxed{\textbf{(C) }16}.$

~MRENTHUSIASM

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1976 AHSME Problems/Problem 24

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