Difference between revisions of "Talk:2021 USAMO Problems/Problem 1"

Latest revision as of 13:14, 15 September 2021

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 $A'=A_2, C'=C_1$ . Note that $A'AOC$ is a cyclic quadrilateral in $X_3$ , so since $\angle A'AC = \frac{\pi}{2}$ , $A'C$ is a diameter of $X_3$ and $\angle A'OC = \angle A'C'C = \frac{\pi}{2}$ - so $A'ACC'$ is a rectangle.

Since $\angle A'AC = \frac{\pi}{2}$ , $A'C$ is a diameter of $X_3$ , and $\angle A'OC = \frac{\pi}{2}$ . Similarly, $\angle B_1OC = \angle C'OA = \angle B_2OA = \frac{\pi}{2}$ , so O is on $B_2C', B_1A'$ , and by opposite angles, $\angle A'OC' = \gamma = \angle A'CC'$ .

Finally, since $A'$ is on $AA_2$ and $C'$ is on $CC_2$ , 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 $C'=C_1$ . Similarly, $A'=A_2$ . So the three diagonals $A_1C_2, B_1A_2, C_1B_2$ intersect in $O$ .

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 Since <math>\angle A'AC = \frac{\pi}{2}</math>, <math>A'C</math> is a diameter of <math>X_3</math>, and <math>\angle A'OC = \frac{\pi}{2}</math>. Similarly, <math>\angle B_1OC = \angle C'OA = \angle B_2OA = \frac{\pi}{2}</math>, so O is on <math>B_2C', B_1A'</math>, and by opposite angles, <math>\angle A'OC' = \gamma = \angle A'CC'</math>.
-Finally, since <math>A'</math> is on <math>AA_2</math> and <math>C'</math> is on <math>CC_2</math>, that gives <math>\angle A_2AC' = \angle A_2CC_1</math> - meaning <math>C'</math> is on the line <math>AC_1</math>. But <math>C'</math> is also on the line <math>CC_1</math> - so <math>C'=C1</math>. Similarly, <math>A'=A_2</math>. So the three diagonals <math>A_1C_2, B_1A_2, C_1B_2</math> intersect in <math>O</math>.
+Finally, since <math>A'</math> is on <math>AA_2</math> and <math>C'</math> is on <math>CC_2</math>, that gives <math>\angle A_2AC' = \angle A_2CC_1</math> - meaning <math>C'</math> is on the line <math>AC_1</math>. But <math>C'</math> is also on the line <math>CC_1</math> - so <math>C'=C_1</math>. Similarly, <math>A'=A_2</math>. So the three diagonals <math>A_1C_2, B_1A_2, C_1B_2</math> intersect in <math>O</math>.
 [[File:USAMO_Q1.png|alt="USAMO Q1 graph"|800px|USAMO 2021 Q1 set-up]]

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Difference between revisions of "Talk:2021 USAMO Problems/Problem 1"

Latest revision as of 13:14, 15 September 2021