Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 AIME, harder Prove it Italy

Problem:

Let there be given in the plane two circles $\gamma_{1}$ and $\gamma_{2}$ with centers $A$ and $B$ respectively, intersecting at two points $C$ and $D$ . Suppose that the circle passing through $A$ , $B$ and $C$ further intersects $\gamma_{1}$ and $\gamma_{2}$ at $E$ and $F$ respectively, and that the arc $E F$ not containing $C$ lies outside the two disks bounded by $\gamma_{1}$ and $\gamma_{2}$ . Prove that the arc $E F$ not containing $C$ is bisected by the line $C D$ .

Solution

Solution:

By the central angle theorem, $\widehat{C A D}=2 \widehat{C E D}$ . Since by symmetry $\widehat{C A B}=\widehat{D A B}$ , we have $\widehat{C A B}=\widehat{C E D}$ . Moreover, $\widehat{C A B}=\widehat{C E B}$ , because both subtend the arc $C B$ , and therefore $\widehat{C E D}=\widehat{C E B}$ . It follows that $E, D$ and $B$ are collinear (because $D$ and $B$ lie on the same side of the line $C E$ ). Since the arcs $C B$ and $B F$ are equal, we have $\widehat{C E B}=\widehat{B E F}$ , hence $D$ belongs to the bisector of $\widehat{C E F}$ . By an analogous argument, one shows that $D$ belongs to the bisector of $\widehat{C F E}$ , which means that $D$ is the incenter of triangle $C E F$ . Therefore the line $C D$ bisects

the angle $\text{E C F}$ , and hence the arc $E F$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.