1992 AIME Problems/Problem 9
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 .
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.
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 .
From above, and . Adding these equations yields . Thus, , and .
We can use from Solution 1 to find that and .
This implies that so
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, .
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 .
|1992 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|