2000 USAMO Problems/Problem 5
Problem
Let be a triangle and let be a circle in its plane passing through and Suppose there exist circles such that for is externally tangent to and passes through and where for all . Prove that
Solution
Solution 1
Let the circumcenter of be , and let the center of be . and are externally tangent at the point , so are collinear.
is the intersection of the perpendicular bisectors of , and each of the centers lie on the perpendicular bisector of the side of the triangle that determines . It follows from that .
Since , and the perpendicular bisector of are fixed, the angle determines the position of (since lies on the perpendicular bisector). Let ; then, and together imply that .
Now (due to collinearility). Hence, we have the recursion , and so . Thus, .
implies that , and circles and are the same circle since they have the same center and go through the same two points.
Solution 2
Using the collinearity of certain points and the fact that is isosceles, we quickly deduce that From ASA Congruence we deduce that and are congruent triangles, and so , that is .
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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