# 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 $$P(x,y), \qquad \xi_1 : \frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1, \qquad \xi_2 : \frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1.$$ $\xi_1$ is an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$. $\xi_2$ is an ellipse whose center is $\left( 20 , 11 \right)$ and foci are $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$.

Since $P$ is on $\xi_1$, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice the semi-major axis of this ellipse, $2a$.

Since $P$ is on $\xi_2$, the sum of distance from $P$ to $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ is equal to twice the semi-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 minimize this, $P$ must be 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, $2 a + 2 b \ge 26 + 20 = 46$, i.e. $a+b\ge 23$.

The straight line connecting the points $\left(–4, 0 \right)$ and $\left(20, 10 \right)$ has the equation $5x+20=12y$. The straight line connecting the points $\left(4, 0 \right)$ and $\left(20, 12 \right)$ has the equation $3x-12=4y$. These lines intersect at the point $\left(14, 15/2 \right)$. This point satisfies both equations for $a = 16, b = 7$. Hence, $a + b = 23$ is possible.

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

~Steven Chen (www.professorchenedu.com)

## Video Solution

~MathProblemSolvingSkills.com

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 