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 so that Extend to meet at and note that bisects let it meet at Using the angle bisector theorem, we let for some
Then thus which we can rearrange, expand and cancel to get hence . This gives us a final answer of
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, .
Note: this solution shows that the length of is irrelevant as long as there still exists a circle as described in the problem.
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|