1992 AIME Problems/Problem 9
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[hide]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
and
Let be the distance along from to where the perp from meets .
Then and so now substitute this into to get and .
you don;t have to use trig nor angles A and B ..From similar triangles,
h/r = 70/x and h/r = 50/ (92-x)
this implies that 70/x =50/(92-x) so x = 161/3
Solution 2
From above, and . Adding these equations yields . Thus, , and .
from solution 1 we get from 1 that h/r = 70/x and h/r = 50/ (92-x)
this implies that 70/x =50/(92-x) so x = 161/3
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 $\circle P$ (Error compiling LaTeX. Unknown error_msg), we discover that 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, .
See also
1992 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 10 | |
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