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

(→‎Solution)
(I change axis to semi axes and and prove that a+b=23 exist)
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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>.
 
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>.
  
−
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>.
+
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>.
  
 
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>.
 
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>.
  
−
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>.
+
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 23: Line 23:
  
 
Hence, <math>2 a + 2 b = 26 + 20 = 46</math>.
 
Hence, <math>2 a + 2 b = 26 + 20 = 46</math>.
 +
 +
Therefore, <math>a + b >= 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, 12 \right)</math> has the equation <math>3x-12=4y</math>.
 +
These lines intersect at the point <math>\left(14, 15/2 \right)</math>.
 +
This point satisfies both equations for <math>a = 16, b = 7</math>.
 +
Hence, <math>a + b = 23</math> is possible.
  
 
Therefore, <math>a + b = \boxed{\textbf{(023) }}.</math>
 
Therefore, <math>a + b = \boxed{\textbf{(023) }}.</math>

Revision as of 11:27, 17 May 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 $P = \left( x , y \right)$.

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 semi-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 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 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, $a + b >= 23$.

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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