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Difference between revisions of "2021 USAJMO Problems/Problem 2"
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[[Image:Leonard my dude.png|frame|none|###px|]]
 
[[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> are share a common intersection.
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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>(CAA_1C_2)</math> be <math>K</math>. Then  
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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  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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\angle AKC &= 360^\circ - \angle BKA - \angle CKB \\
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\angle AXC &= 360^\circ - \angle BXA - \angle CXB \\
 
&= 360^\circ - (180^\circ - \angle AB_1B + 180^\circ - \angle BC_1C) \\&
 
&= 360^\circ - (180^\circ - \angle AB_1B + 180^\circ - \angle BC_1C) \\&
 
= 180^\circ - \angle CA_1A,
 
= 180^\circ - \angle CA_1A,
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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which implies that <math>AA_1C_2CK</math> is cyclic as desired.
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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>K</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, K, C_1</math> are collinear. Similarly, <math>B_1, K, C_2</math> and <math>A_1, K, B_2</math> are collinear, so the three lines concur and we are done.
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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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~Leonard_my_dude, edited by pear333
 +
 
 +
== See Also ==
 +
{{USAJMO newbox|year=2021|num-b=1|num-a=3}}
 +
 
 +
[[Category:Olympiad Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 03:12, 10 April 2024

Problem

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

Leonard my dude.png

We first claim that the three circles $(BCC_1B_2),$ $(CAA_1C_2),$ and $(ABB_1A_2)$ share a common intersection.

Let the second intersection of $(BCC_1B_2)$ and $(ABB_1A_2)$ be $X$. Then \begin{align*} \angle AXC &= 360^\circ - \angle BXA - \angle CXB \\ &= 360^\circ - (180^\circ - \angle AB_1B + 180^\circ - \angle BC_1C) \\& = 180^\circ - \angle CA_1A, \end{align*} which implies that $AA_1C_2CX$ is cyclic as desired.

Now we show that $X$ is the intersection of $B_1C_2,$ $C_1A_2,$ and $A_1B_2.$ Note that $\angle C_1XB = \angle BXA_2 = 90^\circ,$ so $A_2, X, C_1$ are collinear. Similarly, $B_1, X, C_2$ and $A_1, X, B_2$ are collinear, so the three lines concur and we are done.

~Leonard_my_dude, edited by pear333

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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