Difference between revisions of "1992 AHSME Problems/Problem 11"
(→Solution) |
(→Solution (Similarity)) |
||
Line 18: | Line 18: | ||
== Solution (Similarity) == | == 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. | + | 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> | ||
Revision as of 12:00, 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,
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.