Difference between revisions of "2021 USAJMO Problems/Problem 2"
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+ | ==Problem== | ||
+ | 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. | ||
+ | ==Solution== | ||
+ | [[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. | ||
+ | |||
+ | Let the second intersection of <math>(BCC_1B_2)</math> and <math>(CAA_1C_2)</math> be <math>K</math>. Then | ||
+ | <cmath>\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*}</cmath> | ||
+ | which implies that <math>AA_1C_2CK</math> is cyclic as desired. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ~Leonard_my_dude | ||
+ | |||
+ | == See Also == | ||
+ | {{USAJMO newbox|year=2021|num-b=1|num-a=3}} | ||
+ | |||
+ | [[Category:Olympiad Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 15:02, 15 April 2021
Problem
Rectangles and are erected outside an acute triangle Suppose thatProve that lines and are concurrent.
Solution
We first claim that the three circles and are 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
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.