2001 IMO Problems/Problem 1

Problem

Consider an acute triangle $\triangle ABC$ . Let $P$ be the foot of the altitude of triangle $\triangle ABC$ issuing from the vertex $A$ , and let $O$ be the circumcenter of triangle $\triangle ABC$ . Assume that $\angle C \geq \angle B+30^{\circ}$ . Prove that $\angle A+\angle COP < 90^{\circ}$ .

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.

2001 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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2001 IMO Problems/Problem 1

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