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Greece — Selection Examination

Geometry Difficulty 5.3 AIME, harder Prove it Greece

Let $ABC$ be an acute angled triangle inscribed in a circle $c(O, R)$ and $F$ a point on the side $AB$ such that $AF < \frac{AB}{2}$ . The circle $c_1(F, FA)$ intersects the line $OA$ at point $A'$ and the circle $(c)$ at $K$ . Prove that the quadrilateral $BKFA'$ is inscribed in a circle passing through $O$ .

Solution

The triangle $AFK$ is isosceles and hence $\hat{F}_1 = 2\hat{A}_1$ . The angle $\hat{A}_1$ is inscribed in the circle $(c)$ and $\hat{O}_1 = 2\hat{A}_1 = \hat{F}_1$ , and hence the quadrilateral $BKFO$ is cyclic.

Next we will prove that the quadrilateral $OBKA'$ is cyclic. In fact, if $S$ be the counter point of $A$ in the circle $(c_1)$ . Then the triangle $AKS$ is right angled at $K$ , and hence $\hat{S}_1 = 90^\circ - \hat{A}_1$ . The angles $\hat{S}_1$ and $\hat{A}'_1$ are equal (inscribed in the circle $(c_1)$ and they correspond to the same arch $KA$ ). Hence: $\hat{A}'_1 = 90^\circ - \hat{A}_1$ (1).

From the isosceles triangle $OKB$ we have: $\hat{B}_1 = 90^\circ - \frac{\hat{O}_1}{2} = 90^\circ - \hat{A}_1$ (2)

From (1), (2) we have: $\hat{A}'_1 = \hat{B}_1$ . Hence the quadrilateral $OBKA'$ is cyclic.

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