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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.
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==Problem==
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Rectangles <math>BCC_{1}B_{2}</math>, <math>CAA_{1}C_{2}</math>, and <math>ABB_{1}A_{2}</math> are erected outside an acute triangle <math>ABC</math>. Suppose that <cmath>\angle BC_{1}C + \angle CA_{1}A + \angle AB_{1}B = 180^{\circ}.</cmath> Prove that lines <math>B_{1}C_{2}</math>, <math>C_{1}A_{2}</math>, and <math>A_{1}B_{2}</math> are concurrent.
  
 
==Solution==
 
==Solution==
 
[[File:2021 USAMO 1.png|400px|right]]
 
[[File:2021 USAMO 1.png|400px|right]]
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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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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>\begin{align*}
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<cmath>\angle BDC = 360^\circ – \angle ADB – \angle ADC =</cmath>
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\angle ADB &= 180^\circ – \angle AB_1B,&\angle ADC &= 180^\circ – \angle AA_1C\\
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<cmath>= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C) =</cmath>
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\angle BDC &= 360^\circ – \angle ADB – \angle ADC\\
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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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&= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C)\\
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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>
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&= \angle AB_1B + \angle AA_1C\\
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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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\angle BDC + \angle BC_1C &= 180^\circ
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\end{align*}</cmath>
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Therefore, <math>BDCC_1B_2</math> is cyclic with diameters <math>BC_1</math> and <math>CB_2</math>, and thus <math>\angle CDB_2 = 90^\circ.</math>
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Similarly, <math>\angle CDA_1 = 90^\circ</math>, meaning points <math>A_1</math>, <math>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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Similarly, the points <math>A_2, D, C_1</math> and <math>C_2, D, B_1</math> are collinear.
 
   
 
   
 
(After USAMO 2021 Solution Notes – Evan Chen)
 
(After USAMO 2021 Solution Notes – Evan Chen)
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
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 +
==Video Solution==
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https://youtube.com/watch?v=6e_IGnpQGEg
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==See also==
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{{USAMO newbox|year=2021|before=First Problem|num-a=2}}
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{{USAJMO newbox|year=2021|num-b=1|num-a=3}}
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[[Category:Olympiad Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:21, 1 September 2025

Problem

Rectangles $BCC_{1}B_{2}$, $CAA_{1}C_{2}$, and $ABB_{1}A_{2}$ are erected outside an acute triangle $ABC$. Suppose that \[\angle BC_{1}C + \angle CA_{1}A + \angle AB_{1}B = 180^{\circ}.\] Prove that lines $B_{1}C_{2}$, $C_{1}A_{2}$, and $A_{1}B_{2}$ are concurrent.

Solution

2021 USAMO 1.png

Let $D$ be the second point of intersection of the circles $AB_1B$ and $AA_1C.$ Then: \begin{align*} \angle ADB &= 180^\circ – \angle AB_1B,&\angle ADC &= 180^\circ – \angle AA_1C\\ \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\\ \angle BDC + \angle BC_1C &= 180^\circ \end{align*} Therefore, $BDCC_1B_2$ is cyclic with diameters $BC_1$ and $CB_2$, and thus $\angle CDB_2 = 90^\circ.$ Similarly, $\angle CDA_1 = 90^\circ$, meaning points $A_1$, $D$, and $B_2$ are collinear.

Similarly, the 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

Video Solution

https://youtube.com/watch?v=6e_IGnpQGEg

See also

2021 USAMO (Problems • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
2021 USAJMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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