2001 IMO Problems/Problem 1
Consider an acute triangle . Let be the foot of the altitude of triangle issuing from the vertex , and let be the circumcenter of triangle . Assume that . Prove that .
Take on the circumcircle with . Notice that , so . Hence . Let be the midpoint of and the midpoint of . Then , where is the radius of the circumcircle. But (since is a rectangle).
Now cannot coincide with (otherwise would be and the triangle would not be acute-angled). So . But . So .
Hence . Let be a diameter of the circle, so that . But and , since is a diameter. Hence .
Notice that because , it suffices to prove that , or equivalently
Suppose on the contrary that . By the triangle inequality, , where is the circumradius of . But the Law of Sines and basic trigonometry gives us that , so we have . But we also have because , and so we have a contradiction. Hence and so , as desired.
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