1991 USAMO Problems/Problem 5
Contents
Problem
Let be an arbitrary point on side of a given triangle and let be the interior point where intersects the external common tangent to the incircles of triangles and . As assumes all positions between and , prove that the point traces the arc of a circle.
Solution 1
Let the incircle of and the incircle of touch line at points , respectively; let these circles touch at , , respectively; and let them touch their common external tangent containing at , respectively, as shown in the diagram below.
We note that On the other hand, since and are tangents from the same point to a common circle, , and similarly , so On the other hand, the segments and evidently have the same length, and , so . Thus If we let be the semiperimeter of triangle , then , and , so Similarly, so that Thus lies on the arc of the circle with center and radius intercepted by segments and . If we choose an arbitrary point on this arc and let be the intersection of lines and , then becomes point in the diagram, so every point on this arc is in the locus of .
Solution 2
Define the same points as in the first solution. First extend to intersect at a point ; without loss of generality let lie in between and . Then the incircle of is also the incircle of , while the incircle of is the -excircle of . It follows that ; denote this equality by .
Now remark that Hence is a constant equal to , and so lies on the circle with center and radius .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1991 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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