Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it Estonia

Triangle $ABC$ satisfies $AB = AC$ . Medians $AD$ and $BE$ intersect at $G$ . Let $P$ be the midpoint of the line segment $GE$ .

a. Prove that if $GP = GD$ then the quadrilateral $CEPD$ is cyclic.

b. Does it hold that if the quadrilateral $CEPD$ is cyclic then $GP = GD$ ?

Solution

Let $BD = DC = x$ and $GP = PE = y$ . Then $BG = 2 \cdot 2y = 4y$ , $BP = 4y + y = 5y$ and $BE = 4y + 2y = 6y$ . Thus $GP = GD$ if and only if $y^2 = BG^2 - BD^2 = 16y^2 - x^2$ or, equivalently, $x^2 = 15y^2$ . The quadrilateral $CEPD$ is cyclic if and only if $BP \cdot BE = BD \cdot BC$ , i.e., $5y \cdot 6y = x \cdot 2x$ . The latter simplifies to $x^2 = 15y^2$ , too. Hence $GP = GD$ if and only if the quadrilateral $CEPD$ is cyclic, which solves both parts of the problem.

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