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

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We are given the acute triangle <math>ABC</math>, rectangles <math>AA_1B_2B, BB_1C_2C, CC_1A_2A</math> such that <math>\angle AA_1B + \angle BB_1C + \angle CC_1A = \pi</math>. Let's call <math>\angle AA_1B = \alpha, \angle BB_1C = \beta, \angle CC_1A = \gamma</math>.
 
We are given the acute triangle <math>ABC</math>, rectangles <math>AA_1B_2B, BB_1C_2C, CC_1A_2A</math> such that <math>\angle AA_1B + \angle BB_1C + \angle CC_1A = \pi</math>. Let's call <math>\angle AA_1B = \alpha, \angle BB_1C = \beta, \angle CC_1A = \gamma</math>.
  
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Construct circumcircles <math>X_1, X_2</math> around the rectangles <math>AA_1B_2B, BB_1C_2C</math> respectively. <math>X_1, X_2</math> intersect at two points: <math>B</math> and a second point we will label <math>O</math>. Now <math>A_1B</math> is a diameter of <math>X_1</math>, and <math>C_2B</math> is a diameter of <math>X_2</math>, so <math>\angle A_1OB = \angle C_2OB = \frac{\pi}{2}</math>, and <math>\angle A_1OC_2 = \pi</math>, so <math>O</math> is o the diagonal <math>A_1C_2</math>.
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Construct circumcircles <math>X_1, X_2</math> around the rectangles <math>AA_1B_2B, BB_1C_2C</math> respectively. <math>X_1, X_2</math> intersect at two points: <math>B</math> and a second point we will label <math>O</math>. Now <math>A_1B</math> is a diameter of <math>X_1</math>, and <math>C_2B</math> is a diameter of <math>X_2</math>, so <math>\angle A_1OB = \angle C_2OB = \frac{\pi}{2}</math>, and <math>\angle A_1OC_2 = \pi</math>, so <math>O</math> is on the diagonal <math>A_1C_2</math>.
  
 
<math>\angle A_1BB_2 = \angle A_1OB_2 = \alpha</math> (angles standing on the same arc of the circle <math>X1</math>), and similarly, <math>\angle B_1BC_2 = \angle B_1OC_2 = \beta</math>. Therefore, <math>\angle B_2OB_1 = \pi - (\alpha + \beta) = \gamma</math>.
 
<math>\angle A_1BB_2 = \angle A_1OB_2 = \alpha</math> (angles standing on the same arc of the circle <math>X1</math>), and similarly, <math>\angle B_1BC_2 = \angle B_1OC_2 = \beta</math>. Therefore, <math>\angle B_2OB_1 = \pi - (\alpha + \beta) = \gamma</math>.

Revision as of 14:38, 11 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 $A'=A_2, C'=C_2$. 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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