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 $AD \parallel BC$ . Notice that $\angle CBD = \angle BCA$ , so $\angle ABD \ge 30^\circ$ . Hence $\angle AOD \ge 60^\circ$ . Let $Z$ be the midpoint of $AD$ and $Y$ the midpoint of $BC$ . Then $AZ \ge R/2$ , where $R$ is the radius of the circumcircle. But $AZ = YP$ (since $AZYP$ is a rectangle).

Now $O$ cannot coincide with $Y$ (otherwise $\angle A$ would be $90^\circ$ and the triangle would not be acute-angled). So $OP > YP \ge R/2$ . But $PC = YC - YP < R - YP \le R/2$ . So $OP > PC$ .

Hence $\angle COP < \angle OCP$ . Let $CE$ be a diameter of the circle, so that $\angle OCP = \angle ECB$ . But $\angle ECB = \angle EAB$ and $\angle EAB + \angle BAC = \angle EAC = 90^\circ$ , since $EC$ is a diameter. Hence $\angle COP + \angle BAC < 90^\circ$ .

Solution 2

Notice that because $\angle{PCO} = 90^\circ - \angle{A}$ , it suffices to prove that $\angle{POC} < \angle{PCO}$ , or equivalently $PC < PO.$

Suppose on the contrary that $PC > PO$ . By the triangle inequality, $2 PC = PC + PC > PC + PO > CO = R$ , where $R$ is the circumradius of $ABC$ . But the Law of Sines and basic trigonometry gives us that $PC = 2R \sin B \cos C$ , so we have $4 \sin B \cos C > 1$ . But we also have $4 \sin B \cos C \le 4 \sin B \cos (B + 30^\circ) = 2 (\sin (2B + 30^\circ) - \sin 30^\circ) \le 2 (1 - \frac{1}{2}) = 1$ because $\angle{C} \ge \angle{B} + 30^\circ$ , and so we have a contradiction. Hence $PC < PO$ and so $\angle{PCO} + \angle{A} < 90^\circ$ , as desired.

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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Problem

Solution

Solution 2

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