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

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<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>.
  
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Construct another circumcircle <math>X_3</math> around the triangle <math>AOC</math>, which intersects <math>AA_2</math> in <math>A'</math>, and <math>CC_1</math> in <math>C'</math>. We will prove that <math>A'=A_2, C'=C_2</math>. Note that <math>A'AOC</math> is a cyclic quadrilateral in <math>X_3</math>, so 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 = \angle A'C'C = \frac{\pi}{2}</math> - so <math>A'ACC'</math> is a rectangle.
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Construct another circumcircle <math>X_3</math> around the triangle <math>AOC</math>, which intersects <math>AA_2</math> in <math>A'</math>, and <math>CC_1</math> in <math>C'</math>. We will prove that <math>A'=A_2, C'=C_1</math>. Note that <math>A'AOC</math> is a cyclic quadrilateral in <math>X_3</math>, so 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 = \angle A'C'C = \frac{\pi}{2}</math> - so <math>A'ACC'</math> is a rectangle.
  
 
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>.
 
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>.

Revision as of 13:12, 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 $C'=C1$. Similarly, $A'=A_2$. So the three diagonals $A_1C_2, B_1A_2, C_1B_2$ intersect in $O$.

"USAMO Q1 graph"

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