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Difference between revisions of "1976 AHSME Problems/Problem 24"

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== Problem ==
 
== Problem ==
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In the adjoining figure, circle <math>\mathit{K}</math> has diameter <math>\mathit{AB}</math>; circle <math>\mathit{L}</math> is tangent to circle <math>\mathit{K}</math>  
+
In the adjoining figure, circle <math>K</math> has diameter <math>AB</math>; circle <math>L</math> is tangent to circle <math>K</math> and to <math>AB</math> at the center of circle <math>K</math>; and circle <math>M</math> tangent to circle <math>K</math>, to circle <math>L</math> and <math>AB</math>. The ratio of the area of circle <math>K</math> to the area of circle <math>M</math> is
−
and to <math>\mathit{AB}</math> at the center of circle <math>\mathit{K}</math>; and circle <math>\mathit{M}</math> tangent to circle <math>\mathit{K}</math>,  
 
−
to circle <math>\mathit{L}</math> and <math>\mathit{AB}</math>. The ratio of the area of circle <math>\mathit{K}</math> to the area of circle <math>\mathit{M}</math> is
 
 
<asy>
 
<asy>
 
/* Made by Klaus-Anton, Edited by MRENTHUSIASM */
 
/* Made by Klaus-Anton, Edited by MRENTHUSIASM */
Line 19: Line 17:
 
\textbf{(C) }16\qquad
 
\textbf{(C) }16\qquad
 
\textbf{(D) }18\qquad  
 
\textbf{(D) }18\qquad  
−
\textbf{(E) }\text{not an integer} </math>  
+
\textbf{(E) }\text{not an integer}</math>
  
 
== Solution ==
 
== Solution ==
 +
Let <math>R</math> and <math>r</math> be the radius of <math>\odot K</math> and the radius of <math>\odot M,</math> respectively. It follows that the radius of <math>\odot L</math> is <math>\frac{R}{2}.</math>
 +
 +
Suppose <math>P</math> is the foot of the perpendicular from <math>M</math> to <math>\overline{KL}.</math> 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 <math>\triangle KPM,</math> we have <math>KP=r</math> and <math>KM=R-r.</math> By the Pythagorean Theorem, we get <math>PM^2=(R-r)^2-r^2.</math>
 +
 +
In right <math>\triangle LPM,</math> we have <math>LP=\frac{R}{2}-r</math> and <math>LM=\frac{R}{2}+r.</math> By the Pythagorean Theorem, we get <math>PM^2=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2.</math>
 +
 +
We equate the expressions for <math>PM^2,</math> then simplify:
 +
<cmath>\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*}</cmath>
 +
Therefore, the ratio of the area of <math>\odot K</math> to the area of <math>\odot M</math> is <math>\frac{\pi R^2}{\pi r^2}=\left(\frac{R}{r}\right)^2=\boxed{\textbf{(C) }16}.</math>
 +
 +
~MRENTHUSIASM
  
 
== See Also ==
 
== See Also ==
 
{{AHSME box|year=1976|num-b=23|num-a=25}}
 
{{AHSME box|year=1976|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:58, 6 September 2021

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

See Also

1976 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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