HMMT February, 2022

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with centroid $G$ , and let $E$ and $F$ be points on side $BC$ such that $BE = EF = FC$ . Points $X$ and $Y$ lie on lines $AB$ and $AC$ , respectively, so that $X$ , $Y$ , and $G$ are not collinear. If the line through $E$ parallel to $XG$ and the line through $F$ parallel to $YG$ intersect at $P \neq G$ , prove that $GP$ passes through the midpoint of $XY$ .

Solution

Solution:

Let $CG$ intersect $AB$ at $N$ . Then $N$ is the midpoint of $AB$ and it is known that $\frac{CG}{AB} = 2 = \frac{CE}{EB}$ , so $EG \parallel AB$ . Moreover, since $FE = EB$ , we have $[EFG] = [EXG]$ . Similarly, $[EFG] = [FYG]$ . Now we have $[PXG] = [EXG] = [EFG] = [FYG] = [PYG]$ , so $PG$ bisects $XY$ , as desired.

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