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

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m (→‎Solution)
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==Solution==
 
==Solution==
  
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Denote <math>P = \left( x , y \right)</math>.
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Denote <cmath>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.</cmath><math>\xi_1</math> is 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>. <math>\xi_2</math> is 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>.  
  
−
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>.
+
Since <math>P</math>  is on <math>\xi_1</math>, 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  semi-major axis of this ellipse, <math>2a</math>.
  
−
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 semi-major axis of this ellipse, <math>2a</math>.
+
Since <math>P</math> is on <math>\xi_2</math>, 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 semi-major axis of this ellipse, <math>2b</math>.
−
 
 
−
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>.
 
  
 
[[File:AIME-II-2022-12.png|400px|right]]
 
[[File:AIME-II-2022-12.png|400px|right]]
−
 
−
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 semi-major axis of this ellipse, <math>2b</math>.
 
  
 
Therefore, <math>2a + 2b</math> is the sum of the distance from <math>P</math> to four foci of these two ellipses.
 
Therefore, <math>2a + 2b</math> is the sum of the distance from <math>P</math> to four foci of these two ellipses.
Line 24: Line 20:
 
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>.
 
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>.
  
−
Hence, <math>2 a + 2 b = 26 + 20 = 46</math>. Therefore, <math>a + b \ge 23</math>.
+
Hence, <math>2 a + 2 b \ge 26 + 20 = 46</math>, i.e. <math>a+b\ge 23</math>.
  
 
The straight line connecting the points <math>\left(–4, 0 \right)</math> and <math>\left(20, 10 \right)</math> has the equation <math>5x+20=12y</math>.
 
The straight line connecting the points <math>\left(–4, 0 \right)</math> and <math>\left(20, 10 \right)</math> has the equation <math>5x+20=12y</math>.

Revision as of 15:07, 25 December 2022

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, $2b$.

AIME-II-2022-12.png

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)

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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