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Problem 1131

AIME late

Geometry Difficulty 5.0 Prove it NMO Selection Tests for JBMO · Romania

Let $ABC$ be an acute triangle with $AB < AC$ , let $G$ be its centroid and $D$ the foot of the altitude from $A$ . The line $DG$ meets the small arc $BC$ of the circumcircle of triangle $ABC$ at point $E$ . Prove that the line $AB$ is tangent to the circumcircle of triangle $BDE$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $F$ be the point in which the parallel through $A$ to $BC$ intersects again the circumcircle of triangle $ABC$ . We prove that the points $D$ , $G$ and $F$ are collinear. $AFCB$ is a cyclic trapezoid, hence a cyclic one. If $T$ is the orthogonal projection of point $F$ onto $BC$ , then $AFTD$ is a rectangle. It is easy to prove that triangles $ABD$ and $FCT$ are equal; it follows that $BD = CT$ , i.e. $M$ is the midpoint of the line segment $DT$ . It follows that $\frac{DM}{FA} = \frac{DM}{DT} = \frac{1}{2} = \frac{GM}{GA}$ and, since $\angle GMD = \angle GAF$ , triangles $GMD$ and $GAF$ are similar. We deduce that $\angle DGM = \angle FGA$ , i.e. points $D$ , $G$ , $F$ are collinear.

Then $\angle BED = \angle BEF = \angle BCF = \angle ABC = \angle ABD$ , which shows that the line $AB$ is tangent to the circumcircle of triangle $BDE$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.