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

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==Problem==
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== Problem ==
 
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.
 
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.
  
==Solution==
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== Solution ==
 
{{solution}}
 
{{solution}}
  
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== See also ==
 
{{IMO box|year=2001|before=First question|num-a=2}}
 
{{IMO box|year=2001|before=First question|num-a=2}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 20:58, 21 April 2008

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

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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