Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it Italy

Problem:

Let $ABC$ be a triangle with centroid $G$ . Let $D \neq A$ be a point on the line $AG$ such that $AG = GD$ , and let $E \neq B$ be a point on the line $GB$ such that $GB = GE$ . Finally, let $M$ be the midpoint of $AB$ . Prove that the quadrilateral $BMC D$ is inscribable in a circle if and only if $BA = BE$ .

Solution

Solution:

Let $N$ be the midpoint of $BC$ . $N$ is also the midpoint of $GD$ , since $GD = AG$ by hypothesis, and $GN = \frac{1}{2} AG$ by the well-known properties of the centroid. Then $BDCG$ is a parallelogram, since its diagonals bisect each other, and $BDCM$ is a trapezoid, with bases $BD$ and $CM$ . A trapezoid is inscribable in a circle if and only if it is isosceles. Since $BDCG$ is a parallelogram, we have $DC = BG$ ; therefore $BDCM$ is inscribable if and only if $BM = BG$ .

By hypothesis $M$ is the midpoint of $AB$ and $G$ is the midpoint of $BE$ , so $BM = BG$ if and only if $BA = BE$ ; this concludes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.