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Austrian Mathematical Olympiad

Geometry Difficulty 6.8 National Olympiad Prove it Austria

Let $\triangle ABC$ be an acute-angled triangle with $AC < AB$ and circumradius $R$ . Furthermore, let $D$ be the foot of the altitude from $A$ on $BC$ and let $T$ denote the point on the line $AD$ such that $AT = 2R$ holds with $D$ lying between $A$ and $T$ . Finally, let $S$ denote the mid-point of the arc $BC$ on the circumcircle that does not include $A$ .
Prove: $\angle AST = 90^\circ$ .

Solution

As usual, we denote the angles $\angle BAC$ , $\angle ABC$ and $\angle BCA$ by $\alpha$ , $\beta$ and $\gamma$ , respectively. The center of the circumcircle is denoted by $O$ , see Figure 1.

Figure 1: Problem 2
By assumption, we have $\beta < \gamma$ . Let $E$ be the point on the circumcircle of $ABC$ diametrically opposite to $A$ .
By the inscribed angle theorem, we have $\angle AOB = 2\gamma$ . By definition, $S$ is the intersection of the angular bisector of $\angle CAB$ and the circumcircle. We note that
$\angle EAS = \angle BAS - \angle BAO = \frac{\alpha}{2} - \frac{1}{2}(180^\circ - \angle AOB) = \frac{\alpha}{2} - \frac{1}{2}(180^\circ - 2\gamma) = \frac{\alpha}{2} + \gamma - 90^\circ$
Since we also have
$\angle SAT = \angle SAC - \angle DAC = \frac{\alpha}{2} - (90^\circ - \angle ACD) = \frac{\alpha}{2} + \gamma - 90^\circ$

it therefore follows that $\angle EAS = \angle TAS$ holds. Since we also have $AE = AT = 2R$ , triangles $ASE$ and $AST$ are congruent, and therefore $\angle AST = \angle ASE$ follows. Since $AE$ is a diameter of the circumcircle, we have $\angle ASE = 90^\circ$ , and the claim is proven.

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