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| − | ==Problem==
| + | #redirect [[2021 USAMO Problems/Problem 1]] |
| − | 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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| − | [[Image:Leonard my dude.png|frame|none|###px|]] | |
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| − | We first claim that the three circles <math>(BCC_1B_2),</math> <math>(CAA_1C_2),</math> and <math>(ABB_1A_2)</math> share a common intersection.
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| − | Let the second intersection of <math>(BCC_1B_2)</math> and <math>(ABB_1A_2)</math> be <math>X</math>. Then
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| − | <cmath>\begin{align*}
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| − | \angle AXC &= 360^\circ - \angle BXA - \angle CXB \\
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| − | &= 360^\circ - (180^\circ - \angle AB_1B + 180^\circ - \angle BC_1C) \\&
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| − | = 180^\circ - \angle CA_1A,
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| − | \end{align*}</cmath>
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| − | which implies that <math>AA_1C_2CX</math> is cyclic as desired.
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| − | Now we show that <math>X</math> is the intersection of <math>B_1C_2,</math> <math>C_1A_2,</math> and <math>A_1B_2.</math> Note that <math>\angle C_1XB = \angle BXA_2 = 90^\circ,</math> so <math>A_2, X, C_1</math> are collinear. Similarly, <math>B_1, X, C_2</math> and <math>A_1, X, B_2</math> are collinear, so the three lines concur and we are done.
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| − | ~Leonard_my_dude
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| − | Edited by pear333
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| − | == See Also ==
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| − | {{USAJMO newbox|year=2021|num-b=1|num-a=3}}
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| − | [[Category:Olympiad Geometry Problems]]
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| − | {{MAA Notice}}
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