1991 USAMO Problems/Problem 5
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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
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 . 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
| 1991 USAMO (Problems) | ||
| Preceded by Problem 4 | 1 • 2 • 3 • 4 • 5 | Followed by Last Question |
| All USAMO Problems and Solutions | ||
