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

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Vishal has 5 slavicks. How much people does he own?
 
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==Problem==
 
==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>
 
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==
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Denote <math>P = \left( x , y \right)</math>.
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Because <math>\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1</math>, <math>P</math>  is on an ellipse whose center is <math>\left( 0 , 0 \right)</math> and foci are <math>\left( - 4 , 0 \right)</math> and <math>\left( 4 , 0 \right)</math>.
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Hence, the sum of distance from <math>P</math> to <math>\left( - 4 , 0 \right)</math> and <math>\left( 4 , 0 \right)</math> is equal to twice the major axis of this ellipse, <math>2a</math>.
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Because <math>\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1</math>, <math>P</math>  is on an ellipse whose center is <math>\left( 20 , 11 \right)</math> and foci are <math>\left( 20 , 10 \right)</math> and <math>\left( 20 , 12 \right)</math>.
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Hence, the sum of distance from <math>P</math> to <math>\left( 20 , 10 \right)</math> and <math>\left( 20 , 12 \right)</math> is equal to twice the major axis of this ellipse, <math>2b</math>.
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Therefore, <math>2a + 2b</math> is the sum of the distance from <math>P</math> to four foci of these two ellipses.
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To make this minimized, <math>P</math> is the intersection point of the line that passes through <math>\left( - 4 , 0 \right)</math> and <math>\left( 20 , 10 \right)</math>, and the line that passes through <math>\left( 4 , 0 \right)</math> and <math>\left( 20 , 12 \right)</math>.
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The distance between <math>\left( - 4 , 0 \right)</math> and <math>\left( 20 , 10 \right)</math> is <math>\sqrt{\left( 20 + 4 \right)^2 + \left( 10 - 0 \right)^2} = 26</math>.
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The distance between <math>\left( 4 , 0 \right)</math> and <math>\left( 20 , 12 \right)</math> is <math>\sqrt{\left( 20 - 4 \right)^2 + \left( 12 - 0 \right)^2} = 20</math>.
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Hence, <math>2 a + 2 b = 26 + 20 = 46</math>.
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Therefore, <math>a + b = \boxed{\textbf{(023) }}.</math>
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~Steven Chen (www.professorchenedu.com)
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/4qiu7GGUGIg
 
https://youtu.be/4qiu7GGUGIg
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~MathProblemSolvingSkills.com
 
~MathProblemSolvingSkills.com
  
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==Video Solution==
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https://youtu.be/dx5vgznIyt0
 
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2022|n=II|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:21, 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 $a+b.$

Solution

Denote $P = \left( x , y \right)$.

Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$, $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$.

Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice the major axis of this ellipse, $2a$.

Because $\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1$, $P$ is on an ellipse whose center is $\left( 20 , 11 \right)$ and foci are $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$.

Hence, the sum of distance from $P$ to $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ is equal to twice the major axis of this ellipse, $2b$.

Therefore, $2a + 2b$ is the sum of the distance from $P$ to four foci of these two ellipses.

To make this minimized, $P$ is the intersection point of the line that passes through $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$, and the line that passes through $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$.

The distance between $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ is $\sqrt{\left( 20 + 4 \right)^2 + \left( 10 - 0 \right)^2} = 26$.

The distance between $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$ is $\sqrt{\left( 20 - 4 \right)^2 + \left( 12 - 0 \right)^2} = 20$.

Hence, $2 a + 2 b = 26 + 20 = 46$.

Therefore, $a + b = \boxed{\textbf{(023) }}.$

~Steven Chen (www.professorchenedu.com)

Video Solution

https://youtu.be/4qiu7GGUGIg

~MathProblemSolvingSkills.com


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

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