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Junior Balkan Mathematics Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it JBMO

Problem:
Let $ABC$ be an acute angled triangle, let $O$ be its circumcentre, and let $D, E, F$ be points on the sides $BC, AC, AB$ , respectively. The circle $(c_1)$ of radius $FA$ , centred at $F$ , crosses the segment $(OA)$ at $A'$ , and the circumcircle $(c)$ of the triangle $ABC$ again at $K$ . Similarly, the circle $(c_2)$ of radius $DB$ , centred at $D$ , crosses the segment $(OB)$ at $B'$ , and the circle $(c)$ again at $L$ . Finally, the circle $(c_3)$ of radius $EC$ , centred at $E$ , crosses the segment $(OC)$ at $C'$ , and the circle $(c)$ again at $M$ . Prove that the quadrilaterals $BKFA'$ , $CLDB'$ , and $AMEC'$ are all cyclic, and their circumcircles share a common point.

Solution

Solution:
We will prove that the quadrilateral $BKFA'$ is cyclic and its circumcircle passes through the center $O$ of the circle $(c)$ .

The triangle $AFK$ is isosceles, so $m(\widehat{KFB}) = 2 m(\widehat{KAB}) = m(\widehat{KOB})$ . It follows that the quadrilateral $BKFO$ is cyclic.

The triangles $OFK$ and $OFA$ are congruent (S.S.S.), hence $m(\widehat{OKF}) = m(\widehat{OAF})$ . The triangle $FAA'$ is isosceles, so $m\left(\widehat{FA'A}\right) = m(\widehat{OAF})$ . Therefore $m\left(\widehat{FA'A}\right) = m(\widehat{OKF})$ , so the quadrilateral $OKFA'$ is cyclic.

(1) and (2) prove the initial claim.

Along the same lines, we can prove that the points $C, D, L, B', O$ and $A, M, E, C', O$ are concyclic, respectively, so their circumcircles also pass through $O$ .

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