2004 IMO Problems/Problem 1
Let be an acute-angled triangle with . The circle with diameter intersects the sides and at and respectively. Denote by the midpoint of the side . The bisectors of the angles and intersect at . Prove that the circumcircles of the triangles and have a common point lying on the side .
Let , , and . Call the circle with diameter and the circumcircle of .
Our ultimate goal is to show that . To show why this solves the problem, assume this statement holds true. Call the intersection point of the circumcircle of with side . Then, , and . Since , , implying also lies on the circumcircle of , thereby solving the problem.
We now prove that . Note that and are radii of , so is isosceles. The bisector of is thus the perpendicular bisector of . Since lies on the bisector of , . Angle computations yield that from and from .
The bisector of hits at the midpoint of the arc not containing . This point must lie on the perpendicular bisector of segment , which is the bisector of . It follows that is indeed the midpoint of arc , so , , , , are concyclic. Since and subtend the same arc , = . With being the bisector of , we have We know that . so we have . Since , and , we have
The problem is solved.
We have . Noting that , we then have which is indeed the measure of . This implies that lies on the bisector of , and from this, it is clear that must lie on the interior of segment . Not proving that had to lie in the interior of was a reason that a large portion of students who submitted a solution received a 1-point deduction.