Difference between revisions of "2021 USAJMO Problems/Problem 2"
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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. | 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> | + | Let the second intersection of <math>(BCC_1B_2)</math> and <math>(CAA_1C_2)</math> be <math>X</math>. Then |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \angle | + | \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> | ||
− | which implies that <math> | + | which implies that <math>AA_1C_2CX</math> is cyclic as desired. |
− | Now we show that <math> | + | 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 |
Revision as of 20:03, 19 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 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.