Regional Competition

Geometry Difficulty 5.9 AIME, harder Prove it Austria

Let $ABC$ be a triangle with $\angle AC > \angle AB$ and circumcenter $O$ . The tangents to the circumcircle at $A$ and $B$ intersect at $T$ . The perpendicular bisector of the side $BC$ intersects $AC$ at $S$ .

a. Prove that the points $A$ , $B$ , $O$ , $S$ and $T$ lie on a common circle.

b. Prove that the line $ST$ is parallel to the side $BC$ .

Solution

Since $\angle AT$ and $\angle BT$ are perpendicular to $\angle AO$ and $\angle BO$ , the points $A$ , $B$ , $T$ and $O$ lie on a circle $k_1$ by Thales' theorem. By the central angle theorem we have $\angle AOB = 2\gamma$ . Since $BCS$ is an isosceles triangle, we find $\angle BCS = \angle CBS = \gamma$ . Now $\angle ASB = 2\gamma$ , because an exterior angle of a triangle equals the sum of the other two interior angles. Thus
$\angle ASB = 2\gamma = \angle AOB.$
and by the inscribed angle theorem we find that the points $A$ , $B$ , $S$ and $O$ lie on a circle $k_2$ . Since the circles $k_1$ and $k_2$ have the three points $A$ , $B$ and $O$ in common, we have $k_1 = k_2$ and the points $A$ , $B$ , $O$ , $S$ and $T$ lie on a circle.

Finally we have $\angle TSB = \angle TOB = \gamma = \angle SBC$ , from which it follows that $ST$ is parallel to $BC$ .

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