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Ukrainian National Mathematical Olympiad

Geometry Difficulty 6.0 AIME, harder Prove it Ukraine

Let $D$ be the point on the arc $AC$ of the circumcircle of the triangle $ABC$ ( $AB < BC$ ), that doesn't contain point $B$ . Let $X$ and $X'$ be any two points on the side $AC$ such that $\angle ABX = \angle CBX'$ . Show that regardless of the choice of point $X$ , the circumcircle of $\triangle DXX'$ passes through a fixed point different from $D$ .

Solution

Let $W$ be the circumcircle of $\triangle ABC$ , let $w$ be the circumcircle of $\triangle BXX'$ , and let $v$ be the circumcircle of $\triangle DXX'$ . Let $TB$ be a tangent line to the circle $W$ , where $T$ belongs to the line $AC$ . Then $TB^2 = TA \cdot TC$ , as well as (Fig. 12)
$\angle TBX = \angle TBA + \angle ABX = \angle ACB + \angle CBX' = \angle BX'X.$
Thus, the line $TB$ is also tangent to the circle $w$ , thus $TB^2 = TX \cdot TC$ . Let $w \cap TD = F \neq D$ . Then $TX \cdot TC = TF \cdot TD = TB^2$ . Thus, the points $X, X', D, F$ lie on the circle $v$ (due to the property of inscribed quadrilateral). Since $T, B$ and $D$ are fixed points, then so is the point $W \cap TD = F$ .

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