Difference between revisions of "2021 USAJMO Problems/Problem 2"

m (See Also)
m (Solution)
Line 5: Line 5:
 
[[Image:Leonard my dude.png|frame|none|###px|]]
 
[[Image:Leonard my dude.png|frame|none|###px|]]
  
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.
+
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.
  
 
Let the second intersection of <math>(BCC_1B_2)</math> and <math>(CAA_1C_2)</math> be <math>K</math>. Then  
 
Let the second intersection of <math>(BCC_1B_2)</math> and <math>(CAA_1C_2)</math> be <math>K</math>. Then  

Revision as of 22:00, 18 April 2021

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 $(CAA_1C_2)$ be $K$. Then \begin{align*} \angle AKC &= 360^\circ - \angle BKA - \angle CKB \\ &= 360^\circ - (180^\circ - \angle AB_1B + 180^\circ - \angle BC_1C) \\& = 180^\circ - \angle CA_1A, \end{align*} which implies that $AA_1C_2CK$ is cyclic as desired.

Now we show that $K$ 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 $A_2, K, C_1$ are collinear. Similarly, $B_1, K, C_2$ and $A_1, K, B_2$ are collinear, so the three lines concur and we are done.

~Leonard_my_dude

See Also

2021 USAJMO (ProblemsResources)
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

Invalid username
Login to AoPS