Difference between revisions of "2001 IMO Problems/Problem 1"
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== Solution == | == Solution == | ||
− | Take <math>D</math> on the circumcircle with <math>AD \parallel BC</math>. Notice that <math>\angle CBD = \angle BCA</math>, so <math>\angle ABD \ge 30^\circ</math>. Hence <math>\angle AOD \ge 60^\circ</math>. Let <math>Z</math> be the midpoint of <math>AD</math> and <math>Y</math> the midpoint of <math>BC</math>. Then <math>AZ \ge R/2</math>, where <math>R</math> is the radius of the circumcircle. But <math>AZ = | + | Take <math>D</math> on the circumcircle with <math>AD \parallel BC</math>. Notice that <math>\angle CBD = \angle BCA</math>, so <math>\angle ABD \ge 30^\circ</math>. Hence <math>\angle AOD \ge 60^\circ</math>. Let <math>Z</math> be the midpoint of <math>AD</math> and <math>Y</math> the midpoint of <math>BC</math>. Then <math>AZ \ge R/2</math>, where <math>R</math> is the radius of the circumcircle. But <math>AZ = YP</math> (since <math>AZYP</math> is a rectangle). |
− | Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math> | + | Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math>OP > YP \ge R/2</math>. But <math>PC = YC - YP < R - YP \le R/2</math>. So <math>OP > PC</math>. |
− | Hence <math>\angle | + | Hence <math>\angle COP < \angle OCP</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCP = \angle ECB</math>. But <math>\angle ECB = \angle EAB</math> and <math>\angle EAB + \angle BAC = \angle EAC = 90^\circ</math>, since <math>EC</math> is a diameter. Hence <math>\angle COP + \angle BAC < 90^\circ</math>. |
==Solution 2== | ==Solution 2== |
Latest revision as of 16:24, 21 July 2018
Contents
[hide]Problem
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 .
Solution
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 .
Solution 2
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.
See also
2001 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |