Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Estonia

Let $\omega$ be a circle with center $O$ and diameter $AB$ . A circle with center $B$ intersects $\omega$ at $C$ and $AB$ at $D$ . The line $CD$ intersects $\omega$ at the point $E$ ( $E \neq C$ ). The intersection of lines $OE$ and $BC$ is $F$ .

a. Prove that the triangle $OBF$ is isosceles.

b. Find the ratio $\frac{FB}{BD}$ , given that $D$ is the midpoint of $OB$ .

Solution

(a) Let $\angle BCD = \alpha$ , then the equal radii $BC = BD$ give us $\angle BDC = \alpha$ (Fig. 2) and $\angle CBD = 180^\circ - 2\alpha$ . Then $\angle OBF = \angle CBD = 180^\circ - 2\alpha$ . But on the other hand $\angle BOE = 2\angle BCE = 2\angle BCD = 2\alpha$ and therefore $\angle BOF = 180^\circ - 2\alpha$ . Since $\angle OBF = \angle BOF$ , the triangle $OBF$ is isosceles.

(b) Due to equal radii the triangle $CBO$ is isosceles (Fig. 3). Furthermore, it is similar to $OBF$ as they share the angle at $B$ . We have $OB = 2BD = 2BC$ , so by the similarity also $FB = 2OB$ . Therefore $FB = 4BD$ and $\frac{FB}{BD} = 4$ .

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