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Twelfth IMAR MATHEMATICAL COMPETITION, 2014

Geometry Difficulty 5.1 AIME, harder Prove it Romania

Let $ABC$ be a triangle and let $M$ be the midpoint of the side $BC$ . The circle of radius $MA$ centred at $M$ meets the lines $AB$ and $AC$ again at $B'$ and $C'$ , respectively, and the tangents to this circle at $B'$ and $C'$ meet at $D$ . Show that the perpendicular bisector of the segment $BC$ bisects the segment $AD$ .

Solution

Let $A'$ be the antipodal of $A$ in the circle $AB'C'$ , and let $A''$ be the point where this circle meets again the line through $A$ parallel to $BC$ (the points $A$ and $A''$ may coincide). Since $M$ is the midpoint of the side $BC$ , the lines $AA''$ , $AB'$ , $AA'$ , $AC'$ form a harmonic pencil. Consequently, so do the lines $XA''$ , $XB'$ , $XA'$ , $XC'$ for any point $X$ on the circle $AB'C'$ .

Now let $X = B'$ and $X = C'$ to infer that the pencils $B'A'', B'D$ , $B'A'$ , $B'C'$ and $C'A'', C'B'$ , $C'A'$ , $C'D$ are both harmonic. Since the two pencils share the line $B'C'$ , the points $A'', A'$ , $D$ lie on a line which is clearly perpendicular to $BC$ and the conclusion follows.

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