2004 IMO Shortlist Problems/G3
(South Korea) Let be the circumcenter of an acute-angled triangle with . The line meets the side at . The circumcenters of the triangles and are and , respectively. Extend the sides and beyond , and choose, on the respective extensions points and such that and . Prove that the quadrilateral is a rectangle if and only if .
(This was also Problem 2, Day 3 of the 2005 Moldova TST and Problem 5 of the 2005 Taiwan 2nd TST final exam.)
Lemma. In any triangle with circumcenter , the altitude from is the reflection of over the angle bisector of .
Proof. This is well-known, but we prove it anyway. Let meet sides at , and let be the foot of the altitude from . Let us denote , , , and let us use the notation for the angles of triangle . By virtue of inscribed arcs in the circumcircle of , we know , , , so , and again by inscribed arcs, . The lemma follows. ∎
We first note that is the reflection of over the exterior angle bisector of . It follows that line is the altitude from in triangle , i.e., . Since both and line on the perpendicular bisector of , it follows that and are always parallel.
We extend and to meet a point . Since , is a convex quadrilateral. In particular, if we use the notation , , , then , , , so . It follows that line makes an angle of with . Now, if is the midpoint of and is the midpoint of , we note that and are perpendicular to . Hence . But if is a rectangle, then , so and . Thus the condition is necessary for to be a rectangle.
We now prove that it is sufficient. From the previous paragraph, we know that if , then is a parallelogram. It is sufficient to show that if is the intersection of line with and is the intersection of line and , then , since is perpendicular to and . Indeed, since the cosine of the angle between lines and is , it is sufficient to show that if is the projection of onto , then . Let be the projection of onto . Since are congruent, . On the other hand, since is the midpoint of , is the midpoint of the projection of onto , namely, , so , as desired. Thus is a rectangle if and only if , as desired.
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