Talk:2021 USAMO Problems/Problem 1

We are given the acute triangle $ABC$ , rectangles $AA_1B_2B, BB_1C_2C, CC_1A_2A$ such that $\angle AA_1B + \angle BB_1C + \angle CC_1A = \pi$ . Let's call $\angle AA_1B = \alpha, \angle BB_1C = \beta, \angle CC_1A = \gamma$ .

Construct circumcircles $X_1, X_2$ around the rectangles $AA_1B_2B, BB_1C_2C$ respectively. $X_1, X_2$ intersect at two points: $B$ and a second point we will label $O$ . Now $A_1B$ is a diameter of $X_1$ , and $C_2B$ is a diameter of $X_2$ , so $\angle A_1OB = \angle C_2OB = \frac{\pi}{2}$ , and $\angle A_1OC_2 = \pi$ , so $O$ is on the diagonal $A_1C_2$ .

$\angle A_1BB_2 = \angle A_1OB_2 = \alpha$ (angles standing on the same arc of the circle $X1$ ), and similarly, $\angle B_1BC_2 = \angle B_1OC_2 = \beta$ . Therefore, $\angle B_2OB_1 = \pi - (\alpha + \beta) = \gamma$ .

Construct another circumcircle $X_3$ around the triangle $AOC$ , which intersects $AA_2$ in $A'$ , and $CC_1$ in $C'$ . We will prove that $A'=A_2, C'=C_2$ . Note that $\angle A2CC_1 = \gamma$ (given), and $\angle A'OC' = \angle A'AC' = \gamma$ (angles on the same arc). But since $A'$ is on $AA_2$ , that gives $\angle A_2AC' = \angle A_2CC_1$ - meaning $C'$ is on the line $AC_1$ . But $C'$ is also on the line $CC_1$ - so $C'=C1$ . Similarly, $A'=A_2$ .

Finally, since $\angle A_2AC = \frac{\pi}{2}$ , $A_2C$ is a diameter of $X_3$ , and $\angle A_2OC = \frac{\pi}{2}$ . Similarly, $\angle B_1OC = \angle C_1OA = \angle B_2OA = \frac{\pi}{2}$ , so O is on $B_2C_1, B_1A_2$ , and the three diagonals are concurrent.

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Talk:2021 USAMO Problems/Problem 1