2005 IMO Shortlist Problems/G1
(Greece) In a triangle satisfying the incircle has centre and touches the sides and at and , respectively. Let and be the symmetric points of and with respect to . Prove that the quadrilateral is cyclic.
This was also Problem 2 of the second round of the 2006 Poland Math Olympiad and Problem 6 of the final round of the 2006 Costa Rica Math Olympiad.
Remark. The converse of this problem is also true, i.e., if is cyclic, then . This can be proven easily along the lines of the first solution.
Let intersect the circumcircle of again at ; let be the midpoint of . We note that is the perpendicular bisector of , as it passes through the midpoint of segment and the midpoint of the arc . We note that the condition is equivalent to the condition .
Since , and and are both right angles, triangles , are similar.
Let be the projection of onto the line . We note that and are both right angles, and , so triangles are similar. Thus triangles are similar. But we note that by measures of intercepted arcs, , so (in fact, this is a known result) and triangles are congruent. This means that , so is the perpendicular bisector of , and . By symmetry, , so all lie on a circle centered at .
Let the bisector of angle meet at , at , at , and the circumcircle of a second time at . Let be the midpoint of , and let the incircle of touch at . We note that is the perpendicular bisector of , since this passes both through the midpoint of arc and through the midpoint of . The condition is equivalent to the condition .
Now, we have , and , so triangles are similar. Thus we have . Considering the power of the point to the circumcircle of with respect to lines gives us
where and are the circumradius and inradius of .
Then, the Pythagorean theorem on triangle gives us
If is the foot of the altitude from to , then we have . Furthermore, triangles are homothetic about , so
It follows that . The extended law of sines gives us . By the angle bisector theorem, . Considering the power of the point with respect to the circumcircle of and lines gives us
Now, since is the reflection of about , the two triangles are congruent. Since are right triangles with a common acute angle, they are similar. Specifically,
Finally, applying the Pythagorean theorem to triangle , we have
By symmetry, , so all lie 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.