Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.4 National Olympiad Prove it Ireland

Consider three points $A$ , $B$ , $C$ on a circle $\Gamma$ , with $\angle BAC > 90^\circ$ . Let $d$ denote the line tangent to $\Gamma$ at $A$ . Points $M$ and $N$ are chosen on $d$ such that $\angle MBA = \angle ABC$ and $\angle NCA = \angle ACB$ . The circumcircles of triangles $ABM$ and $ACN$ intersect at $A$ and $T$ . Prove the following:

a. $TA$ is the angle bisector of $\angle BTC$ .

b. The circumcircles of triangles $TMN$ and $TBC$ are tangent at $T$ .

Solution

a. As $BTAM$ is cyclic, we have $\angle BTM = \angle BAM$ . By the Alternate Segment Theorem, $\angle BAM = \angle ACB$ . Since $CTAN$ is cyclic, $\angle ACB = \angle ACN = \angle ATN$ , hence $\angle BTM = \angle ATN$ . Similarly $\angle ATM = \angle CTN$ . Adding up we get $\angle BTA = \angle CTA$ , i.e. $TA$ is the angle bisector of $\angle BTC$ .

b. Let $B'$ be the intersection point of the lines $BA$ and $TN$ , and $C'$ the intersection point of the lines $CA$ and $TM$ . From (a) we have $\angle BTC' = \angle BTM = \angle BCA = \angle BCC'$ and $\angle CTB' = \angle CTN = \angle CBA = \angle CBB'$ , hence $C'$ and $B'$ lie on $(BTC)$ .

Hence also $\angle B'C'T = \angle B'BT$ , and $\angle B'BT = \angle ABT = \angle AMT = \angle NMT$ as $AMBT$ is cyclic. Thus $\angle B'C'T = \angle NMT$ and so $B'C' \parallel MN$ . Thus $\triangle TB'C'$ is homothetic to $\triangle TNM$ with centre $T$ , hence their circumcircles $(TMN)$ and $(TB'C') = (TBC)$ are tangent at $T$ .

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