Difference between revisions of "1992 AHSME Problems/Problem 11"
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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. Let's label the center of the smaller circle (which is also the center of the larger circle) as <math>D</math>. Let's also label the point where the smaller circle intersects <math>BC</math> as <math>E</math>. So <math>ABC</math> is similar to <math>DEC</math>. Since <math>DE</math> is the radius of the smaller circle, call the length <math>x</math> and since <math>DC</math> is the radius of the bigger circle, call that length <math>3x</math>. The diameter, <math>AC</math> is <math>6x</math>. So, | 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. Let's label the center of the smaller circle (which is also the center of the larger circle) as <math>D</math>. Let's also label the point where the smaller circle intersects <math>BC</math> as <math>E</math>. So <math>ABC</math> is similar to <math>DEC</math>. Since <math>DE</math> is the radius of the smaller circle, call the length <math>x</math> and since <math>DC</math> is the radius of the bigger circle, call that length <math>3x</math>. The diameter, <math>AC</math> is <math>6x</math>. So, | ||
− | <math>\frac{AB}{AC} = \frac {DE}{DC} \Rightarrow \frac{12}{6x} = \frac{x}{3x}</math> | + | <math>\frac{AB}{AC} = \frac {DE}{DC} \Rightarrow \frac{12}{6x} = \frac{x}{3x} \Rightarrow 36x = 6x^2 \Rightarrow 6 = x</math> |
+ | |||
+ | But they are asking for the larger circle radius, so | ||
+ | |||
+ | <math>3x= \fbox{18}</math> | ||
Latest revision as of 11:02, 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. Let's label the center of the smaller circle (which is also the center of the larger circle) as . Let's also label the point where the smaller circle intersects as . So is similar to . Since is the radius of the smaller circle, call the length and since is the radius of the bigger circle, call that length . The diameter, is . So,
But they are asking for the larger circle radius, so
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 |
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