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Difference between revisions of "1981 IMO Problems/Problem 5"

 
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== Problem ==
 
== Problem ==
  
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Three congruent circles have a common point <math> \displaystyle O </math> and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point <math>\displaystyle O </math> are collinear.
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Three [[congruent]] [[circle]]s have a common point <math> \displaystyle O </math> and lie inside a given [[triangle]]. Each circle touches a pair of sides of the triangle. Prove that the [[incenter]] and the [[circumcenter]] of the triangle and the point <math>\displaystyle O </math> are [[collinear]].
  
 
== Solution ==
 
== Solution ==
  
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Let the triangle have vertices <math>\displaystyle A,B,C</math>, and sides <math>\displaystyle a,b,c</math>, respectively, and let the centers of the circles inscribed in the angles <math>\displaystyle A,B,C</math> be denoted <math>\displaystyle O_A, O_B, O_C </math>, respectively.
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Let the triangle have vertices <math>\displaystyle A,B,C</math>, and sides <math>\displaystyle a,b,c</math>, respectively, and let the centers of the circles inscribed in the [[angle]]s <math>\displaystyle A,B,C</math> be denoted <math>\displaystyle O_A, O_B, O_C </math>, respectively.
  
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The triangles <math> \displaystyle O_A O_B O_C </math> and <math> \displaystyle ABC </math> are [[homothetic]], as their corresponding sides are parallel.  Furthermore, since <math>\displaystyle O_A</math> lies on the bisector of angle <math>\displaystyle A</math> and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles.  Since <math>\displaystyle O</math> is clearly the circumcenter of <math>\displaystyle O_A O_B O_C </math>, <math>\displaystyle O</math> is collinear with the incenter and circumcenter of <math>\displaystyle ABC</math>, as desired.
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The triangles <math> \displaystyle O_A O_B O_C </math> and <math> \displaystyle ABC </math> are [[homothetic]], as their corresponding sides are [[parallel]].  Furthermore, since <math>\displaystyle O_A</math> lies on the [[angle bisector | bisector]] of angle <math>\displaystyle A</math> and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles.  Since <math>\displaystyle O</math> is clearly the circumcenter of <math>\displaystyle O_A O_B O_C </math>, <math>\displaystyle O</math> is collinear with the incenter and circumcenter of <math>\displaystyle ABC</math>, as desired.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 16:18, 29 October 2006

Problem

Three congruent circles have a common point $\displaystyle O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point $\displaystyle O$ are collinear.

Solution

Let the triangle have vertices $\displaystyle A,B,C$, and sides $\displaystyle a,b,c$, respectively, and let the centers of the circles inscribed in the angles $\displaystyle A,B,C$ be denoted $\displaystyle O_A, O_B, O_C$, respectively.

The triangles $\displaystyle O_A O_B O_C$ and $\displaystyle ABC$ are homothetic, as their corresponding sides are parallel. Furthermore, since $\displaystyle O_A$ lies on the bisector of angle $\displaystyle A$ and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles. Since $\displaystyle O$ is clearly the circumcenter of $\displaystyle O_A O_B O_C$, $\displaystyle O$ is collinear with the incenter and circumcenter of $\displaystyle ABC$, as desired.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

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