AustriaMO

Geometry Difficulty 5.3 AIME, harder Prove it Austria

Points $A$ , $B$ and $C$ lie on a line in this order. For every circle $k$ passing through $B$ and $C$ , let $D$ be one of the common points of $k$ and the bisector of $BC$ . Furthermore, let $E$ be the second common point of the line $AD$ and $k$ .

*Prove that the ratio $BE : CE$ is constant for all circles $k$ .*

G. Baron, Vienna

Solution

Let $F$ be the diametrically opposite point to $D$ on $k$ . Since $F$ is a point on the bisector of $BC$ , we have $FB = FC$ , and therefore $\angle BEF = \angle CEF$ .

We see that $EF$ is the internal bisector of $\angle BEC$ , and since $ED \perp EF$ , $ED = AD$ is the external bisector. If we name $BC \cap EF = H$ , we see that $A$ and $H$ are harmonic with respect to $B$ and $C$ , since they are the points of intersection of perpendicular bisectors $EA$ and $EF$ with $BC$ . Since $A$ is independent of the choice of $k$ , we see that $H$ must be as well, and since $BC : CE = BH : CH$ , the ratio is independent of the choice of $k$ , as claimed. □

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