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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]] [[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]].
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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]].
  
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== Solution ==
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== Solution 1==
  
 
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.
 
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.
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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.
 
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.
  
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{{alternate solutions}}
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==Solution 2==
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Suppose 3 congruent circles with centres P,Q,R lie inside ABC and are such that the circle with centre P touches AB & AC and the circle with centre Q touches CA & BC.an R with remaining 2.
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Since O lies in all 3 circles, PO=QO=RO. Therefore, O is circumcentre of PQR. let O' be circumcentre of ABC.
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Since BC is tangent to the circles with centers Q & R, the lengths of perpendiculars from Q & R, the lengths are equal. therefore,  QR//BC,RP//CA,PQ//AB.
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Again, since AB and AC both touch circle with centre P. Therefore P is equidistant from AB & AC. Therefore P lies on the internal bisector of angle A. Similarly Q & R lie internal bisectors of angle B and angle C respectively. Therefore, AP,BQ,CR produced meet at incenter I. Since, QR//BC,RP//CA,PQ//AB, it follows that I is also incentre of PQR, I being the centre of homothety. By the property of enlargements, O and O' must be co-linear with I , the centre of enlargement.
  
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{{IMO box|num-b=4|num-a=6|year=1981}}
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== See Also == {{IMO box|year=1981|num-b=4|num-a=6}}
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[[Category:Olympiad Geometry Problems]]
 

Latest revision as of 22:09, 29 January 2021

Problem

Three congruent circles have a common point $O$ 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 $O$ are collinear.

Solution 1

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

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

Solution 2

Suppose 3 congruent circles with centres P,Q,R lie inside ABC and are such that the circle with centre P touches AB & AC and the circle with centre Q touches CA & BC.an R with remaining 2.

Since O lies in all 3 circles, PO=QO=RO. Therefore, O is circumcentre of PQR. let O' be circumcentre of ABC.

Since BC is tangent to the circles with centers Q & R, the lengths of perpendiculars from Q & R, the lengths are equal. therefore, QR//BC,RP//CA,PQ//AB.

Again, since AB and AC both touch circle with centre P. Therefore P is equidistant from AB & AC. Therefore P lies on the internal bisector of angle A. Similarly Q & R lie internal bisectors of angle B and angle C respectively. Therefore, AP,BQ,CR produced meet at incenter I. Since, QR//BC,RP//CA,PQ//AB, it follows that I is also incentre of PQR, I being the centre of homothety. By the property of enlargements, O and O' must be co-linear with I , the centre of enlargement.

See Also

1981 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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