Difference between revisions of "1992 AHSME Problems/Problem 11"
(Created page with "== 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)); d...") |
(→Solution) |
||
Line 16: | Line 16: | ||
\text{(E) } 26</math> | \text{(E) } 26</math> | ||
− | == Solution == | + | == Solution (Similarity) == |
+ | |||
+ | 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. | ||
+ | |||
+ | |||
<math>\fbox{B}</math> | <math>\fbox{B}</math> | ||
Revision as of 10:52, 24 November 2016
Problem
The ratio of the radii of two concentric circles is . If is a diameter of the larger circle, is a chord of the larger circle that is tangent to the smaller circle, and , then the radius of the larger circle is
Solution (Similarity)
We are given that is tangent to the smaller circle. Using that, we know where the circle intersects , it creates a right triangle. We can also point out that since is the diameter of the bigger circle and triangle is inscribed the semi-circle, that angle is a right angle. Therefore, we have similar triangles.
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.