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Estonian Math Competitions

Geometry Difficulty 5.9 AIME, harder Prove it Estonia

Let $O$ be the circumcentre of the acute triangle $ABC$ . Let $c_1$ and $c_2$ be the circumcircles of triangles $ABO$ and $ACO$ . Let $P$ and $Q$ be points on $c_1$ and $c_2$ respectively, such that $OP$ is a diameter of $c_1$ and $OQ$ is a diameter of $c_2$ . Let $T$ be the intersection of the tangent to $c_1$ at $P$ and the tangent to $c_2$ at $Q$ . Let $D$ be the second intersection of the line $AC$ and the circle $c_1$ . Prove that the points $D, O$ and $T$ are collinear.

Solution

Since $\angle OAP = \angle OAQ = 90^\circ$ , the points $P$ , $A$ and $Q$ are collinear. Since $\angle OPT = \angle OQT = 90^\circ$ , $OPTQ$ is cyclic. Since $OA = OB$ , the diameter $OP$ of $c_1$ is perpendicular to the chord $AB$ . Therefore $PT$ and $AB$ are parallel.

Now $\angle TOQ = \angle TPQ = \angle TPA = \angle BAP = \angle BOP = 90^\circ - \angle ABO$ . On the other hand, equality of inscribed angles subtending the arc $AO$ of circle $c_1$ gives $\angle CDO = \angle ABO$ (figures 20 and 21 show two possible situations). Therefore $\angle DOQ = 90^\circ - \angle CDO = 90^\circ - \angle ABO$ .

In summary, $\angle TOQ = \angle DOQ$ , whence $D$ , $O$ and $T$ are collinear.

Fig. 20

Fig. 21

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