Difference between revisions of "2021 USAMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
+ | [[File:2021 USAMO 1.png|400px|right]] | ||
+ | Let <math>D</math> be the second point of intersection of the circles <math>AB_1B</math> and <math>AA_1C.</math> Then: | ||
+ | <cmath>\begin{align*} | ||
+ | \angle ADB &= 180^\circ – \angle AB_1B,&\angle ADC &= 180^\circ – \angle AA_1C\\ | ||
+ | \angle BDC &= 360^\circ – \angle ADB – \angle ADC\\ | ||
+ | &= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C)\\ | ||
+ | &= \angle AB_1B + \angle AA_1C\\ | ||
+ | \angle BDC + \angle BC_1C &= 180^\circ | ||
+ | \end{align*}</cmath> | ||
+ | Therefore, <math>BDCC_1B_2</math> is cyclic with diameters <math>BC_1</math> and <math>CB_2</math>, and thus <math>\angle CDB_2 = 90^\circ.</math> | ||
+ | Similarly, <math>\angle CDA_1 = 90^\circ</math>, meaning points <math>A_1</math>, <math>D</math>, and <math>B_2</math> are collinear. | ||
− | + | Similarly, the points <math>A_2, D, C_1</math> and <math>C_2, D, B_1</math> are collinear. | |
− | + | ||
− | + | (After USAMO 2021 Solution Notes – Evan Chen) | |
− | + | ||
− | + | '''vladimir.shelomovskii@gmail.com, vvsss''' | |
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 12:32, 25 December 2023
Rectangles and are erected outside an acute triangle Suppose thatProve that lines and are concurrent.
Solution
Let be the second point of intersection of the circles and Then: Therefore, is cyclic with diameters and , and thus Similarly, , meaning points , , and are collinear.
Similarly, the points and are collinear.
(After USAMO 2021 Solution Notes – Evan Chen)
vladimir.shelomovskii@gmail.com, vvsss
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.