Difference between revisions of "2022 AIME II Problems/Problem 12"
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| + | ==Problem== | ||
| + | Let <math>a, b, x,</math> and <math>y</math> be real numbers with <math>a>4</math> and <math>b>1</math> such that<cmath>\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.</cmath>Find the least possible value of <math>a+b.</math> | ||
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| + | ==Solution== | ||
| + | |||
| + | ==See Also== | ||
| + | {{AIME box|year=2022|n=II|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Revision as of 07:27, 18 February 2022
Problem
Let 
and 
be real numbers with 
and 
such that![\[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\]](http://latex.artofproblemsolving.com/c/5/e/c5e3bf454b8347b6d4c3d8441be9334a5b2ea1c3.png)
Find the least possible value of 
Solution
See Also
| 2022 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
