Competition problem — Geometry — Maths Olympiad Prep

Library / /28 of 108

Geometry Difficulty 5.6 AIME, harder Prove it Mongolia

Let $\omega$ be the circumcircle of $ABC$ where $AB \neq AC$ , and let $M$ be the midpoint of side $BC$ . Tangent lines drawn at points $B$ and $C$ of circle $\omega$ intersect at point $T$ . The circumcircle of triangle $AMT$ intersects line $BC$ again at point $N$ . Let $S$ be the midpoint of $NT$ . Prove that $SA$ is tangent to the circle $\omega$ .
(Gerelkhuu Erdenetugs)

Solution

Since BTC is isosceles, TM is altitude.
$\angle CMT = \angle NMT = 90^\{\circ\}.$
Thus, S is the circumcenter of triangle AMT, making SA = ST and $\angle SAT = \angle STA$ . Considering AT as the A-symmedian of $ABC$ , we know $\angle BAM = \angle TAC$ .
$\begin{align*} \angle SAT &= \angle ATN = \angle AMN \\ &= \angle ABM + \angle BAM \\ &= \angle TAC + \angle ABM \end{align*}$
Since $\angle SAT = \angle TAC + \angle ABM$ , it follows that $\angle CAS = \angle BAM = \angle ABC$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.