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Bulgarian Mathematical Olympiad

Geometry Difficulty 6.3 National Olympiad Prove it Bulgaria

Problem:

Let $ABC$ be a triangle with $\angle BAC = 30^\circ$ and $\angle ABC = 45^\circ$ . Consider all pairs of points $X$ and $Y$ such that $X$ and $Y$ lie on the rays $AC^{\rightarrow}$ and $BC^{\rightarrow}$ , respectively, and $OX = BY$ , where $O$ is the circumcenter of $\triangle ABC$ . Prove that the perpendicular bisectors of the segments $XY$ pass through a fixed point.

Solution

Solution:

We shall prove that the perpendicular bisector of $XY$ passes through the point $C'$ which is symmetric to the point $C$ with respect to $AB$ . Denote by $R$ the circumradius of $\triangle ABC$ . The Sine theorem for $\triangle ABC$ gives $AC = R \sqrt{2}$ and $BC = R$ . Suppose that $C$ lies between $A$ and $X$ . Then $OX = BY$ , $C'A = CA$ , $C'B = CB$ , $\angle C'BY = 90^\circ$ and the Cosine theorem for

$\triangle C'AX$ gives
$\begin{aligned} C'X^2 & = C'A^2 + AX^2 - C'A \cdot AX = AC^2 + AX^2 - AC \cdot AX \\ & = AC^2 + XA \cdot XC = 2R^2 + (OX^2 - R^2) = R^2 + BY^2 \\ & = C'B^2 + BY^2 = C'Y^2 \end{aligned}$
Therefore $C'$ is a point on the perpendicular bisector of $XY$ . In the case when $X$ lies between $A$ and $C$ the proof is similar.

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