Difference between revisions of "2001 IMO Problems/Problem 1"
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== Solution == | == Solution == | ||
− | Take D on the circumcircle with AD parallel to BC. | + | Take <math>D</math> on the circumcircle with <math>AD \parallel to 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 = YX</math> (since <math>AZYX</math> is a rectangle). |
− | Now O cannot coincide with Y (otherwise angle A would be | + | 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>OX > YX \ge R/2</math>. But <math>XC = YC - YX < R - YX \le R/2</math>. So <math>OX > XC</math>. |
− | Hence angle COX < angle OCX. Let CE be a diameter of the circle, so that angle OCX = angle ECB. But angle ECB = angle EAB and angle EAB + angle BAC = angle EAC = | + | Hence <math>\angle COX < \angle OCX</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCX = \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 COX + \angle BAC < 90^\circ</math>. |
== See also == | == See also == |
Revision as of 18:39, 11 July 2012
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 .
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 |