Difference between revisions of "1991 USAMO Problems/Problem 5"
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356427#356427 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356427#356427 Discussion on AoPS/MathLinks] | ||
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=175552#175552 Further Discussion] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=175552#175552 Further Discussion] | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 20:53, 3 July 2013
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 • 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.