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Difference between revisions of "2001 IMO Problems/Problem 1"

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Notice that because <math>\angle{PCO} = 90^\circ - \angle{A}</math>, it suffices to prove that <math>\angle{POC} < \angle{PCO}</math>, or equivalently <math>PC < PO.</math>
 
Notice that because <math>\angle{PCO} = 90^\circ - \angle{A}</math>, it suffices to prove that <math>\angle{POC} < \angle{PCO}</math>, or equivalently <math>PC < PO.</math>
  
−
Suppose on the contrary that <math>PC > PO</math>. By the triangle inequality, <math>2 PC = PC + PC > PC + PO = CO = R</math>, where <math>R</math> is the circumradius of <math>ABC</math>. But the Law of Sines and basic trigonometry gives us that <math>PC = 2R \sin B \cos C</math>, so we have <math>4 \sin B \cos C \ge 1</math>. But we also have <math>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</math> because <math>\angle{C} \ge \angle{B} + 30^\circ</math>, and so we have a contradiction. Hence <math>PC < PO</math> and so <math>\angle{PCO} + \angle{A} < 90^\circ</math>, as desired.
+
Suppose on the contrary that <math>PC > PO</math>. By the triangle inequality, <math>2 PC = PC + PC > PC + PO = CO = R</math>, where <math>R</math> is the circumradius of <math>ABC</math>. But the Law of Sines and basic trigonometry gives us that <math>PC = 2R \sin B \cos C</math>, so we have <math>4 \sin B \cos C > 1</math>. But we also have <math>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</math> because <math>\angle{C} \ge \angle{B} + 30^\circ</math>, and so we have a contradiction. Hence <math>PC < PO</math> and so <math>\angle{PCO} + \angle{A} < 90^\circ</math>, as desired.
  
 
== See also ==
 
== See also ==

Revision as of 22:38, 16 May 2015

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 $AZ = YX$ (since $AZYX$ 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 $OX > YX \ge R/2$. But $XC = YC - YX < R - YX \le R/2$. 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 $\angle EAB + \angle BAC = \angle EAC = 90^\circ$, since $EC$ is a diameter. Hence $\angle COX + \angle BAC < 90^\circ$.

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.

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
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