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Difference between revisions of "2021 USAMO Problems/Problem 1"
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Rectangles <math>BCC_1B_2,</math> <math>CAA_1C_2,</math> and <math>ABB_1A_2</math> are erected outside an acute triangle <math>ABC.</math> Suppose that<cmath>\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.</cmath>Prove that lines <math>B_1C_2,</math> <math>C_1A_2,</math> and <math>A_1B_2</math> are concurrent.
 
Rectangles <math>BCC_1B_2,</math> <math>CAA_1C_2,</math> and <math>ABB_1A_2</math> are erected outside an acute triangle <math>ABC.</math> Suppose that<cmath>\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.</cmath>Prove that lines <math>B_1C_2,</math> <math>C_1A_2,</math> and <math>A_1B_2</math> are concurrent.
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==Solution==
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Let <math>D</math> be the second point of intersection of the circles <math>AB_1B</math> and <math>AA_1C.</math> Then
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<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 AB_1B + \angle AA_1C  \implies \angle BDC + \angle BC_1C = 180^\circ \implies</cmath>
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<math>BDCC_1B_2</math> is cyclic with diameters <math>BC_1</math> and <math>CB_2 \implies \angle CDB_2 = 90^\circ.</math>

Revision as of 07:20, 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

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 $CB_2 \implies \angle CDB_2 = 90^\circ.$

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