1982 IMO Problems/Problem 2
A non-isosceles triangle has sides , , with the side lying opposite to the vertex . Let be the midpoint of the side , and let be the point where the inscribed circle of triangle touches the side . Denote by the reflection of the point in the interior angle bisector of the angle . Prove that the lines , and are concurrent.
Rename the vertices so that and . Let be the incenter of .
Claim 1: all lie on the incircle. Proof: By construction, is the reflection of with respect to the angle bisector since lies on the angle bisector of . So, so lies on the incircle. Similarly, and lie on the incircle.
Note that is the circumcenter of because of equal radii due to Claim 1.
Claim 2: Proof: In cyclic quadrilateral , . Let be the point where the angle bisector intersects . Due to the construction of , But, so Thus,Similarly, Therefore,
Hence, we use Claim 2 to get and we similarly get and
Claim 3: Let be the midpoint of Then, is collinear. Proof: We already derived that and Recall that a radius is perpendicular to the midpoint of a chord so Then, Thus,as desired.
Then, because a diameter is perpendicular to a chord at its midpoint and because is part of a diameter, by Claim 3. Hence, But, so . Similarly, we get and so with corresponding parallel sides. Thus, there exists a homothety that maps to and and intersect at the center of the homothety.
This solution was posted and copyrighted by Jzhang21. The original thread for this problem can be found here: 
|1982 IMO (Problems) • Resources|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All IMO Problems and Solutions|