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

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== Solution ==
 
== Solution ==
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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).
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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).
  
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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.
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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>.
  
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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.
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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 $\triangle ABC$. 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 $AD \parallel to BC$. 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 $\angle COX + \angle BAC < 90^\circ$.

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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