Difference between revisions of "2022 AIME II Problems/Problem 12"

m (Solution)
(Undo revision 188789 by Jacob3abraham (talk))
(Tag: Undo)
Line 1: Line 1:
Vishal has 5 slavicks. How much people does he own?
+
==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>
  
 
==Problem==
 
==Problem==

Revision as of 13:14, 7 February 2023

Problem

Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that\[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\]Find the least possible value of $a+b.$

Problem

Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that\[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\]Find the least possible value of $a+b.$

Video Solution

https://youtu.be/4qiu7GGUGIg

~MathProblemSolvingSkills.com

Video Solution

https://youtu.be/dx5vgznIyt0

See Also

2022 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png