IMO 1968 SL 14

Origin: BUL

Problem

A line in the plane of a triangle $ABC$ intersects the sides $AB$ and $AC$ respectively at points $X$ and $Y$ such that $BX = CY$. Find the locus of the center of the circumcircle of triangle $XAY$.

Solution

Let $O$, $D$, $E$ be the circumcenter of $\triangle ABC$ and the midpoints of $AB$ and $AC$, and given arbitrary $X \in AB$ and $Y \in AC$ such that $BX = CY$, let $O_1$, $D_1$, $E_1$ be the circumcenter of $\triangle AXY$ and the midpoints of $AX$ and $AY$, respectively. Since $AD = AB/2$ and $AD_1 = AX/2$, it follows that $DD_1 = BX/2$ and similarly $EE_1 = CY/2$. Hence $O_1$ is at the same distance $BX/2 = CY/2$ from the lines $OD$ and $OE$ and lies on the half-line bisector $l$ of $\angle DOE$.

If we let $X$, $Y$ vary along the segments $AB$ and $AC$, we obtain that the locus of $O_1$ is the segment $OP$, where $P \in l$ is a point at distance $\min(AB, AC)/2$ from $OD$ and $OE$.