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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.