Difference between revisions of "1992 AIME Problems/Problem 9"
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Let <math>7a = AP</math>, then <math>AB = 7a + 5a = 12a</math>. <math>AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}</math>. Thus, <math>m + n = 164</math>. | Let <math>7a = AP</math>, then <math>AB = 7a + 5a = 12a</math>. <math>AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}</math>. Thus, <math>m + n = 164</math>. | ||
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+ | Note: this solution shows that the length of <math>CD</math> is irrelevant as long as there still exists a circle as described in the problem. | ||
==Solution 4== | ==Solution 4== |
Revision as of 20:56, 6 November 2017
Problem
Trapezoid has sides , , , and , with parallel to . A circle with center on is drawn tangent to and . Given that , where and are relatively prime positive integers, find .
Solution 1
Let be the base of the trapezoid and consider angles and . Let and let equal the height of the trapezoid. Let equal the radius of the circle.
Then
Let be the distance along from to where the perp from meets .
Then and so . We can substitute this into to find that and .
Remark: One can come up with the equations in without directly resorting to trig. From similar triangles, and . This implies that , so .
Solution 2
From above, and . Adding these equations yields . Thus, , and .
We can use from Solution 1 to find that and .
This implies that so
Solution 3
Extend and to meet at a point . Since and are parallel, . If is further extended to a point and is extended to a point such that is tangent to circle , we discover that circle is the incircle of triangle . Then line is the angle bisector of . By homothety, is the intersection of the angle bisector of with . By the angle bisector theorem,
Let , then . . Thus, .
Note: this solution shows that the length of is irrelevant as long as there still exists a circle as described in the problem.
Solution 4
The area of the trapezoid is , where is the height of the trapezoid.
Draw lines and . We can now find the area of the trapezoid as the sum of the areas of the three triangles , , and .
(where is the radius of the tangent circle.)
From Solution 1 above,
Substituting , we find , hence the answer is .
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.