Difference between revisions of "Talk:2021 USAMO Problems/Problem 1"
(Some ideas on this problem) |
(Figured out a nice solution to Q1. Challenging question!) |
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− | + | We are given the acute triangle <math>ABC, 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>. | |
− | + | 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>. | |
− | + | </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 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>\angle A2CC_1 = \gamma<math> (given), and </math>\angle A'OC' = \angle A'AC' = \gamma<math> (angles on the same arc). But since </math>A'<math> is on </math>AA_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>. | ||
+ | |||
+ | Finally, since </math>\angle A_2AC = \frac{\pi}{2}<math>, </math>A_2C<math> is a diameter of </math>X_3<math>, and </math>\angle A_2OC = \frac{\pi}{2}<math>. Similarly, </math>\angle B_1OC = \angle C_1OA = \angle B_2OA = \frac{\pi}{2}<math>, so O is on </math>B_2C_1, B_1A_2$, and the three diagonals are concurrent. |
Revision as of 15:34, 11 September 2021
We are given the acute triangle AA_1B_2B, BB_1C_2C, CC_1A_2A
\angle AA_1B + \angle BB_1C + \angle CC_1A = \pi
\angle AA_1B = \alpha, \angle BB_1C = \beta, \angle CC_1A = \gamma$.
Construct circumcircles$ (Error compiling LaTeX. Unknown error_msg)X_1, X_2AA_1B_2B, BB_1C_2C
X_1, X_2
B
O
A_1B
X_1
C_2B
X_2
\angle A_1OB = \angle C_2OB = \frac{\pi}{2}
\angle A_1OC_2 = \pi
O
A_1C_2
\angle A_1BB_2 = \angle A_1OB_2 = \alpha
X1
\angle B_1BC_2 = \angle B_1OC_2 = \beta
\angle B_2OB_1 = \pi - (\alpha + \beta) = \gamma$.
Construct another circumcircle around the triangle$ (Error compiling LaTeX. Unknown error_msg)AOCAA_2
A'
CC_1
C'
A'=A_2, C'=C_2
\angle A2CC_1 = \gamma
\angle A'OC' = \angle A'AC' = \gamma
A'
AA_2
\angle A_2AC' = \angle A_2CC_1
C'
AC_1
C'
CC_1
C'=C1
A'=A_2$.
Finally, since$ (Error compiling LaTeX. Unknown error_msg)\angle A_2AC = \frac{\pi}{2}A_2C
X_3
\angle A_2OC = \frac{\pi}{2}
\angle B_1OC = \angle C_1OA = \angle B_2OA = \frac{\pi}{2}
B_2C_1, B_1A_2$, and the three diagonals are concurrent.