2001 IMO Problems/Problem 1
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 D on the circumcircle with AD parallel to BC. Angle CBD = angle BCA, so angle ABD >= 30o. Hence angle AOD >= 60o. Let Z be the midpoint of AD and Y the midpoint of BC. Then AZ >= R/2, where R is the radius of the circumcircle. But AZ = YX (since AZYX is a rectangle).
Now O cannot coincide with Y (otherwise angle A would be 90o and the triangle would not be acute-angled). So OX > YX >= R/2. But XC = YC - YX < R - YX <= R/2. So OX > XC.
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 = 90o, since EC is a diameter. Hence angle COX + angle BAC < 90o.
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 |