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Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $ABC$ be an obtuse triangle with circumcenter $O$ such that $\angle ABC = 15^{\circ}$ and $\angle BAC > 90^{\circ}$ . Suppose that $AO$ meets $BC$ at $D$ , and that $OD^{2} + OC \cdot DC = OC^{2}$ . Find $\angle C$ .

Solution

Solution:

Let the radius of the circumcircle of $\triangle ABC$ be $r$ .

$\begin{gathered} OD^{2} + OC \cdot DC = OC^{2} \\ OC \cdot DC = OC^{2} - OD^{2} \\ OC \cdot DC = (OC + OD)(OC - OD) \\ OC \cdot DC = (r + OD)(r - OD) \end{gathered}$

By the power of the point at $D$ ,

$\begin{gathered} OC \cdot DC = BD \cdot DC \\ r = BD \end{gathered}$

Then, $\triangle OBD$ , $\triangle OAB$ , and $\triangle AOC$ are isosceles triangles. Let $\angle DOB = \alpha$ . $\angle BAO = 90 - \frac{\alpha}{2}$ . In $\triangle ABD$ , $15 + 90 - \frac{\alpha}{2} = \alpha$ . This means that $\alpha = 70$ .

Furthermore, $\angle ACB$ intercepts minor arc $AB$ , thus $\angle ACB = \frac{\angle AOB}{2} = \frac{70}{2} = 35$ .

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