2022 AIME II Problems/Problem 12

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)

2022 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2022 AIME II Problems/Problem 12

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