Difference between revisions of "1976 AHSME Problems/Problem 24"
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== Solution == | == Solution == | ||
− | Let <math>R</math> and <math>r</math> be the | + | 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> | <asy> | ||
/* Made by Klaus-Anton, Edited by MRENTHUSIASM */ | /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ | ||
size(200); | size(200); | ||
− | pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25), | + | 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(K,1)^^A--B); | ||
draw(circle(L,0.5)^^circle(M,.25)); | draw(circle(L,0.5)^^circle(M,.25)); | ||
− | label("$A$", A, | + | draw(L--K,red); |
− | label("$K$", K, | + | draw(L--M,red); |
− | label("$B$", B, | + | 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("$L$", L, (0,5/4)); | ||
label("$M$", M, (0,5/4)); | label("$M$", M, (0,5/4)); | ||
+ | label("$P$", P, (-5/4,0)); | ||
dot(K,linewidth(4)); | dot(K,linewidth(4)); | ||
dot(L,linewidth(4)); | dot(L,linewidth(4)); | ||
dot(M,linewidth(4)); | dot(M,linewidth(4)); | ||
− | dot( | + | dot(I,linewidth(4)); |
+ | dot(E,linewidth(4)); | ||
+ | dot(P,linewidth(4)); | ||
</asy> | </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 07:58, 6 September 2021
Problem
In the adjoining figure, circle has diameter
; circle
is tangent to circle
and to
at the center of circle
; and circle
tangent to circle
, to circle
and
. The ratio of the area of circle
to the area of circle
is
Solution
Let and
be the radius of
and the radius of
respectively. It follows that the radius of
is
Suppose is the foot of the perpendicular from
to
We construct the auxiliary lines, as shown below:
In right
we have
and
By the Pythagorean Theorem, we get
In right we have
and
By the Pythagorean Theorem, we get
We equate the expressions for then simplify:
Therefore, the ratio of the area of
to the area of
is
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.