Track / Stage 5 / 15 of 400 #615 of 1964

Problem 615

AIME late

Geometry Difficulty 5.0 Prove it Brazilian Math Olympiad · Brazil

Let $P$ be a convex $2006$ -gon. The $1003$ diagonals connecting opposite vertices and the $1003$ lines connecting the midpoints of opposite sides are concurrent, that is, all $2006$ lines have a common point. Prove that the opposite sides of $P$ are parallel and congruent.

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 $A_1A_2\ldots A_{2006}$ be a polygon such that $A_1A_{1004}$ , $A_2A_{1005}$ , $\ldots$ , $A_{1003}A_{2006}$ and $M_1M_{1004}$ , $M_2M_{1005}$ , $\ldots$ , $M_{1003}M_{2006}$ are concurrent at $O$ , where $M_1, M_2, \ldots, M_{2006}$ are the midpoints of $A_1A_2, A_2A_3, \ldots, A_{2006}A_1$ , respectively.

Note that $A_1A_{1004}$ , $A_2A_{1005}$ , $M_1M_{1004}$ are concurrent, so $A_1A_2 \parallel A_{1004}A_{1005}$ . Similarly, we obtain that any opposite sides are parallel.

Now we have by applying the ratio theorem repetitively, we obtain
$\frac{OA_1}{OA_{1004}} = \frac{OA_2}{OA_{1005}} = \dots = \frac{OA_{1003}}{OA_{2006}} = \frac{OA_{1004}}{OA_1} = \frac{OA_{1005}}{OA_2} = \dots = \frac{OA_{2006}}{OA_{1003}}$
This implies that $O$ is the midpoint of every main diagonal. Therefore, all opposite sides are congruent and parallel.

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