Difference between revisions of "1992 AHSME Problems/Problem 11"

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\text{(E) } 26</math>
 
\text{(E) } 26</math>
  
== Solution ==
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== Solution (Similarity) ==
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We are given that <math>BC</math> is tangent to the smaller circle. Using that, we know where the circle intersects <math>BC</math>, it creates a right triangle. We can also point out that since <math>AC</math> is the diameter of the bigger circle and triangle <math>ABC</math> is inscribed the semi-circle, that angle <math>B</math> is a right angle. Therefore, we have <math>2</math> similar triangles.
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<math>\fbox{B}</math>
 
<math>\fbox{B}</math>
  

Revision as of 11:52, 24 November 2016

Problem

[asy] draw(circle((0,0),18),black+linewidth(.75)); draw(circle((0,0),6),black+linewidth(.75)); draw((-18,0)--(18,0)--(-14,8*sqrt(2))--cycle,black+linewidth(.75)); dot((-18,0));dot((18,0));dot((-14,8*sqrt(2))); MP("A",(-18,0),W);MP("C",(18,0),E);MP("B",(-14,8*sqrt(2)),W); [/asy]

The ratio of the radii of two concentric circles is $1:3$. If $\overline{AC}$ is a diameter of the larger circle, $\overline{BC}$ is a chord of the larger circle that is tangent to the smaller circle, and $AB=12$, then the radius of the larger circle is

$\text{(A) } 13\quad \text{(B) } 18\quad \text{(C) } 21\quad \text{(D) } 24\quad \text{(E) } 26$

Solution (Similarity)

We are given that $BC$ is tangent to the smaller circle. Using that, we know where the circle intersects $BC$, it creates a right triangle. We can also point out that since $AC$ is the diameter of the bigger circle and triangle $ABC$ is inscribed the semi-circle, that angle $B$ is a right angle. Therefore, we have $2$ similar triangles.


$\fbox{B}$

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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