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$ (Error compiling LaTeX. Unknown error_msg)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 $is o the diagonal$ 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 around the triangle$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)\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