1981 IMO Problems/Problem 5

Revision as of 23:36, 12 December 2018 by Jzhang21 (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.

Invalid username
Login to AoPS