IMO Shortlist 1992 Problem #7
by KingSmasher3, Mar 31, 2013, 11:00 PM
Two circles 
and 
are externally tangent to each other at a point 
, and both of these circles are tangent to a third circle 
which encloses the two circles and .
The common tangent to the two circles and at the point meets the circle at a point 
. One common tangent to the circles and which doesn't pass through meets the circle at the points 
and 
such that the points and lie on the same side of the line 
.
Prove that the point is the incenter of triangle 
.
_______
We invert about point
Line 
is invariant, with 
somewhere on it. 
and 
become lines on either side of line with 
From the given tangent circles, we know that 
and 
map to two intersecting circles both tangent to 
and 
These two circles intersect at 
and 
The inversion of contains 
and the inversion of contains 
Clearly, these two circles are congruent. Let the inversion of intersect line at another point 
besides By symmetry, 
Therefore, 
Similarly, also lies on the angle bisector of 
so is the incenter as desired.
If we let
intersect at another point 
besides 
then we could also finish by noting that since 
by symmetry and and are equidistant from 
must be the orthocenter of 
Through inversion, this means that is the incenter of 
and 
are externally tangent to each other at a point 
, and both of these circles are tangent to a third circle 
which encloses the two circles and .The common tangent to the two circles
and at the point meets the circle at a point 
. One common tangent to the circles and which doesn't pass through meets the circle at the points 
and 
such that the points and lie on the same side of the line 
.Prove that the point
is the incenter of triangle 
._______
We invert about point
Line 
is invariant, with 
somewhere on it. 
and 
become lines on either side of line with 
From the given tangent circles, we know that 
and 
map to two intersecting circles both tangent to 
and 
These two circles intersect at 
and 
The inversion of contains 
and the inversion of contains 
Clearly, these two circles are congruent. Let the inversion of intersect line at another point 
besides By symmetry, 
Therefore, 
Similarly, also lies on the angle bisector of 
so is the incenter as desired.If we let
intersect at another point 
besides 
then we could also finish by noting that since 
by symmetry and and are equidistant from 
must be the orthocenter of 
Through inversion, this means that is the incenter of 
be the midpoint of arc 
Let 
It follows that 
Then 
so it lies on their radical axis.This post has been edited 3 times. Last edited by KingSmasher3, Apr 27, 2013, 10:07 PM
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