Difference between revisions of "1991 AHSME Problems/Problem 22"
(Created page with "== Problem == <asy> draw(circle((0,0),10),black+linewidth(.75)); MP(")",(0,0),S); </asy> Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overli...") |
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== Problem == | == Problem == | ||
<asy> | <asy> | ||
− | draw(circle((0,0), | + | draw(circle((0,6sqrt(2)),2sqrt(2)),black+linewidth(.75)); |
− | MP(")",(0,0),S); | + | draw(circle((0,3sqrt(2)),sqrt(2)),black+linewidth(.75)); |
+ | draw((-8/3,16sqrt(2)/3)--(-4/3,8sqrt(2)/3)--(0,0)--(4/3,8sqrt(2)/3)--(8/3,16sqrt(2)/3),dot); | ||
+ | MP("B",(-8/3,16*sqrt(2)/3),W);MP("B'",(8/3,16*sqrt(2)/3),E); | ||
+ | MP("A",(-4/3,8*sqrt(2)/3),W);MP("A'",(4/3,8*sqrt(2)/3),E); | ||
+ | MP("P",(0,0),S); | ||
</asy> | </asy> | ||
− | Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overline{PA'B'}</math> are common tangents with <math>A</math> and <math>A'</math> on the smaller circle <math>B</math> and <math>B'</math> on the larger circle. If <math>PA=AB=4</math>, then the | + | Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overline{PA'B'}</math> are common tangents with <math>A</math> and <math>A'</math> on the smaller circle <math>B</math> and <math>B'</math> on the larger circle. If <math>PA=AB=4</math>, then the area of the smaller circle is |
<math>\text{(A) } 1.44\pi\quad | <math>\text{(A) } 1.44\pi\quad | ||
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | Using the tangent-tangent theorem, <math>PA=AB=PA'=A'B'=4</math>. We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point <math>S</math> and the center of the larger circle be point <math>L</math>. If we let the radius of the larger circle be <math>x</math> and the radius of the smaller circle be <math>y</math>, we can see that, using similar triangle, <math>x=2y</math>. In addition, the total hypotenuse of the larger right triangles equals <math>2(x+y)</math> since half of it is <math>x+y</math>, so <math>y^2+4^2=(3y)^2</math>. If we simplify, we get <math>y^2+16=9y^2</math>, so <math>8y^2=16</math>, so <math>y=\sqrt2</math>. This means that the smaller circle has area <math>2\pi</math>, which is answer choice <math>\fbox{B}</math>. |
== See also == | == See also == |
Latest revision as of 18:09, 11 October 2016
Problem
Two circles are externally tangent. Lines and are common tangents with and on the smaller circle and on the larger circle. If , then the area of the smaller circle is
Solution
Using the tangent-tangent theorem, . We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point and the center of the larger circle be point . If we let the radius of the larger circle be and the radius of the smaller circle be , we can see that, using similar triangle, . In addition, the total hypotenuse of the larger right triangles equals since half of it is , so . If we simplify, we get , so , so . This means that the smaller circle has area , which is answer choice .
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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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