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| | Vishal has 5 slavicks. How much people does he own? | | Vishal has 5 slavicks. How much people does he own? |
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| − | ==Solution== | + | ==Problem== |
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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>.
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| − | 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>.
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| − | 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>.
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| − | [[File:AIME-II-2022-12.png|400px|right]]
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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 minimize this, <math>P</math> must be 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 \ge 26 + 20 = 46</math>, i.e. <math>a+b\ge 23</math>.
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| − | 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>.
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| − | 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>.
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| − | These lines intersect at the point <math>\left(14, 15/2 \right)</math>.
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| − | This point satisfies both equations for <math>a = 16, b = 7</math>.
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| − | Hence, <math>a + b = 23</math> is possible.
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| − | Therefore, <math>a + b = \boxed{\textbf{023}}.</math>
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| − | ~Steven Chen (www.professorchenedu.com)
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| | + | 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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| | ==Video Solution== | | ==Video Solution== |
and 
be real numbers with 
and 
such that![\[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\]](http://latex.artofproblemsolving.com/c/5/e/c5e3bf454b8347b6d4c3d8441be9334a5b2ea1c3.png)
Find the least possible value of 
