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Difference between revisions of "2021 USAMO Problems/Problem 1"
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==Solution==
 
==Solution==
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[[File:2021 USAMO 1.png|400px|right]]
 
Let <math>D</math> be the second point of intersection of the circles <math>AB_1B</math> and <math>AA_1C.</math> Then
 
Let <math>D</math> be the second point of intersection of the circles <math>AB_1B</math> and <math>AA_1C.</math> Then
 
<cmath>\angle ADB = 180^\circ – \angle AB_1B,\angle ADC = 180^\circ – \angle AA_1C \implies</cmath>
 
<cmath>\angle ADB = 180^\circ – \angle AB_1B,\angle ADC = 180^\circ – \angle AA_1C \implies</cmath>
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<cmath>\angle BDC = 360^\circ – \angle ADB – \angle ADC = 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C) =</cmath>
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<cmath>\angle BDC = 360^\circ – \angle ADB – \angle ADC =</cmath>
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<cmath>= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C) =</cmath>
 
<cmath>=\angle AB_1B + \angle AA_1C  \implies \angle BDC + \angle BC_1C = 180^\circ \implies</cmath>
 
<cmath>=\angle AB_1B + \angle AA_1C  \implies \angle BDC + \angle BC_1C = 180^\circ \implies</cmath>
 
<math>BDCC_1B_2</math> is cyclic with diameters <math>BC_1</math> and <math>CB_2 \implies \angle CDB_2 = 90^\circ.</math>
 
<math>BDCC_1B_2</math> is cyclic with diameters <math>BC_1</math> and <math>CB_2 \implies \angle CDB_2 = 90^\circ.</math>
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Similarly, <math>\angle CDA_1 = 90^\circ \implies</math> points <math>A_1, D,</math> and <math>B_2</math> are collinear.
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Similarly, triples of points <math>A_2, D, C_1</math> and <math>C_2, D, B_1</math> are collinear.
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(After USAMO 2021 Solution Notes – Evan Chen)
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'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 06:27, 15 September 2022

Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.

Solution

2021 USAMO 1.png

Let $D$ be the second point of intersection of the circles $AB_1B$ and $AA_1C.$ Then \[\angle ADB = 180^\circ – \angle AB_1B,\angle ADC = 180^\circ – \angle AA_1C \implies\] \[\angle BDC = 360^\circ – \angle ADB – \angle ADC =\] \[= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C) =\] \[=\angle AB_1B + \angle AA_1C \implies \angle BDC + \angle BC_1C = 180^\circ \implies\] $BDCC_1B_2$ is cyclic with diameters $BC_1$ and $CB_2 \implies \angle CDB_2 = 90^\circ.$ Similarly, $\angle CDA_1 = 90^\circ \implies$ points $A_1, D,$ and $B_2$ are collinear.

Similarly, triples of points $A_2, D, C_1$ and $C_2, D, B_1$ are collinear.

(After USAMO 2021 Solution Notes – Evan Chen)

vladimir.shelomovskii@gmail.com, vvsss

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