Difference between revisions of "1981 IMO Problems/Problem 5"
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== Problem == | == Problem == | ||
| − | Three [[congruent]] [[circle]]s have a common point <math> | + | Three [[congruent]] [[circle]]s have a common point <math>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>O </math> are [[collinear]]. |
== Solution == | == Solution == | ||
| − | Let the triangle have vertices <math> | + | Let the triangle have vertices <math>A,B,C</math>, and sides <math>a,b,c</math>, respectively, and let the centers of the circles inscribed in the [[angle]]s <math>A,B,C</math> be denoted <math>O_A, O_B, O_C </math>, respectively. |
| − | The triangles <math> | + | The triangles <math>O_A O_B O_C </math> and <math>ABC </math> are [[homothetic]], as their corresponding sides are [[parallel]]. Furthermore, since <math>O_A</math> lies on the [[angle bisector | bisector]] of angle <math>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>O</math> is clearly the circumcenter of <math>O_A O_B O_C </math>, <math>O</math> is collinear with the incenter and circumcenter of <math>ABC</math>, as desired. |
{{alternate solutions}} | {{alternate solutions}} | ||
| − | == | + | {{IMO box|num-b=4|num-a=6|year=1981}} |
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 19:46, 25 October 2007
Problem
Three congruent circles have a common point 
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 are collinear.
Solution
Let the triangle have vertices 
, and sides 
, respectively, and let the centers of the circles inscribed in the angles be denoted 
, respectively.
The triangles 
and 
are homothetic, as their corresponding sides are parallel. Furthermore, since 
lies on the bisector of angle 
and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles. Since is clearly the circumcenter of , is collinear with the incenter and circumcenter of , as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
| 1981 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
