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, 2013

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let points $A$ and $B$ be on circle $\omega$ centered at $O$ . Suppose that $\omega_{A}$ and $\omega_{B}$ are circles not containing $O$ which are internally tangent to $\omega$ at $A$ and $B$ , respectively. Let $\omega_{A}$ and $\omega_{B}$ intersect at $C$ and $D$ such that $D$ is inside triangle $A B C$ . Suppose that line $B C$ meets $\omega$ again at $E$ and let line $E A$ intersect $\omega_{A}$ at $F$ . If $F C \perp C D$ , prove that $O, C$ , and $D$ are collinear.

Solution

Solution:

Let $H = C A \cap \omega$ , and $G = B H \cap \omega_{B}$ . There are homotheties centered at $A$ and $B$ taking $\omega_{A} \rightarrow \omega$ and $\omega_{B} \rightarrow \omega$ that take $A: F \mapsto E$ , $A: C \mapsto H$ , $B: C \mapsto E$ and $B: G \mapsto H$ . In particular $C F \parallel E H \parallel C G$ , so $C, F, G$ are collinear, lying on a line perpendicular to $D C$ .

Because of the right angles at $C$ , $D F$ and $D G$ are diameters of $\omega_{A}$ , $\omega_{B}$ , respectively. Also, we have that the ratio of the sizes of $\omega_{A}$ and $\omega_{B}$ , under the two homotheties above, is $C F / E H \cdot E H / C G = C F / C G$ . Therefore, $D F / D G = C F / C G$ , but then $\triangle D C F$ and $\triangle D C G$ are both right triangles which share one side and have hypotenuse and other side in proportion; it is obvious now that the two circles $\omega_{A}$ and $\omega_{B}$ are congruent.

Therefore, $O$ has the same distance to $A$ and $B$ , and so the same distances to the centers of the two circles as well. As a result, $O$ lies on the radical axis $C D$ as desired.

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