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

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== Solution ==
 
== Solution ==
{{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).
 
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.
 
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.  
+
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.
  
 
== See also ==
 
== See also ==

Revision as of 01:51, 8 February 2009

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.

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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