Difference between revisions of "2021 USAJMO Problems/Problem 2"
m (→Solution) |
m (→Solution) |
||
| Line 17: | Line 17: | ||
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. | 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. | ||
| − | ~Leonard_my_dude | + | ~Leonard_my_dude, edited by pear333 |
| − | |||
== See Also == | == See Also == | ||
Latest revision as of 03:12, 10 April 2024
Problem
Rectangles 
and 
are erected outside an acute triangle 
Suppose that![\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/7/3378f0c4409d14bc0afddf2616992c13f624be3b.png)
Prove that lines 
and 
are concurrent.
Solution
We first claim that the three circles 
and 
share a common intersection.
Let the second intersection of 
and be 
. Then
which implies that 
is cyclic as desired.
Now we show that is the intersection of and 
Note that 
so 
are collinear. Similarly, 
and 
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. 
