2014 IMO Problems/Problem 4
Points and lie on side of acute-angled so that and . Points and lie on lines and , respectively, such that is the midpoint of , and is the midpoint of . Prove that lines and intersect on the circumcircle of .
We are trying to prove that the intersection of and , call it point , is on the circumcircle of triangle . In other words, we are trying to prove . Let the intersection of and be point , and the intersection of and be point . Let us assume . Note: This is circular reasoning. If , then should be equal to and . We can quickly prove that the triangles , , and are similar, so . We also see that . Also because angles and and are equal, the triangles and , and must be two pairs of similar triangles. Therefore we must prove angles and and are equal. We have angles . We also have , . Because the triangles and are similar, we have , so triangles and are similar. So the angles and and are equal and we are done.
Let be the midpoint of . Easy angle chasing gives . Because is the midpoint of , the cotangent rule applied on triangle gives us Hence, by the cotangent rule on , we have Because the period of cotangent is , but angles are less than , we have
Similarly, we have Hence, if and intersect at , then by the Angle Sum in a Triangle Theorem. Hence, is cyclic, which is equivalent to the desired result.
--Suli 23:27, 7 February 2015 (EST)
Let be the midpoint of . By AA Similarity, triangles and are similar, so and . Similarly, , and so triangle is isosceles. Thus, , and so . Dividing both sides by 2, we have , or But we also have , so triangles and are similar by similarity. In particular, . Similarly, , so . In addition, angle sum in triangle gives . Therefore, if we let lines and intersect at , by Angle Sum in quadrilateral concave , and so convex , which is enough to prove that is cyclic. This completes the proof.
--Suli 10:38, 8 February 2015 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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