, 2015

Geometry Difficulty 6.3 National olympiad Prove it Romania

Let $ABC$ be a triangle, let $A_1$ , $B_1$ , $C_1$ be the antipodes of the vertices $A$ , $B$ , $C$ , respectively, in the circle $ABC$ , and let $X$ be a point in the plane $ABC$ , collinear with no two vertices of the triangle $ABC$ . The line through $B$ , perpendicular to the line $XB$ , and the line through $C$ , perpendicular to the line $XC$ , meet at $A_2$ ; the points $B_2$ and $C_2$ are defined similarly. Show that the lines $A_1A_2$ , $B_1B_2$ and $C_1C_2$ are concurrent.

Solution

Let the lines $B_2C_2$ , $C_2A_2$ , $A_2B_2$ meet the circle $ABC$ again at $A_3$ , $B_3$ , $C_3$ , respectively. The angle $AA_3A_1$ is a right angle, so $A_1A_3$ is perpendicular to $B_2C_2$ and hence parallel to $XA$ , and the parallel lines $XA$ and $A_1A_3$ are equidistant from the circumcentre $O$ of the triangle $ABC$ . Hence $A_1A_3$ passes through the reflection $X'$ of $X$ across $O$ .

Similarly, the lines $B_1B_3$ and $C_1C_3$ pass through $X'$ , and therefore $X'A_1 \cdot X'A_3 = X'B_1 \cdot X'B_3 = X'C_1 \cdot X'C_3$ , a quantity denoted $r^2$ .

It follows that $A_1$ , $B_1$ , $C_1$ are the poles of the lines $B_2C_2$ , $C_2A_2$ , $A_2B_2$ , respectively, relative to the circle of radius $r$ centered at $X'$ , so the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are in perspective, by the well-known lemma below.

Lemma. If $XYZ$ is a triangle, and $X'$ , $Y'$ , $Z'$ are the poles of the lines $YZ$ , $ZX$ , $XY$ , respectively, relative to a conic $\gamma$ , then the triangles $XYZ$ and $X'Y'Z'$ are in perspective.

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