Difference between revisions of "1981 IMO Problems/Problem 5"
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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]]. | 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 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. | ||
| − | + | ==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. | 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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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. | 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 == {{IMO box|year=1981|num-b=4|num-a=6}} | ||
Latest revision as of 22:09, 29 January 2021
Contents
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 1
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.
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 | ||
