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Difference between revisions of "2024 AMC 12B Problems/Problem 13"
(→‎Solution 2 (Coordinate Geometry and AM-GM Inequality))
 
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~mitsuihisashi14
 
~mitsuihisashi14
  
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==Solution 2 (Coordinate Geometry and AM-QM Inequality)==
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==Solution 2 (Coordinate Geometry and AM-GM Inequality)==
  
 
[[Image:2024_amc_12B_P13_V2.PNG|thumb|center|500px|]]
 
[[Image:2024_amc_12B_P13_V2.PNG|thumb|center|500px|]]
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<cmath>(x-3)^2 + (y-4)^2 = h + 25 </cmath>
 
<cmath>(x-3)^2 + (y-4)^2 = h + 25 </cmath>
 
<cmath>(x-5)^2 + (y+2)^2 = k + 29 </cmath>
 
<cmath>(x-5)^2 + (y+2)^2 = k + 29 </cmath>
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Distance between 2 circle centers is <cmath>(O_{1}O_{2})^2 = (5-3)^2 + (4 - (-2)) ^2 = 40 </cmath>
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The distance between 2 circle centers is <cmath>(O_{1}O_{2})^2 = (5-3)^2 + (4 - (-2)) ^2 = 40 </cmath>
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the 2 circle must intersect given there exists one or more pair of (x,y), connecting <math>O_{1}O_{2}</math> and any one of the 2 circle intersection point we get a triangle with 3 sides ( radius (<math>O_{1}</math>) , radius (<math>O_{2}</math>) , <math>O_{1}O_{2}</math>) , then     
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The 2 circles must intersect given there exists one or more pairs of (x,y), connecting <math>O_{1}O_{2}</math> and any pair of the 2 circle intersection points gives us a triangle with 3 sides, then     
 
<cmath> radius (O_{1}) + radius (O_{2}) \geq O_{1}O_{2} </cmath>  
 
<cmath> radius (O_{1}) + radius (O_{2}) \geq O_{1}O_{2} </cmath>  
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<cmath>\sqrt{h+25} + \sqrt{k+29}  \geq  2*\sqrt{10} </cmath>
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<cmath>\sqrt{h+25} + \sqrt{k+29}  \geq  2\cdot\sqrt{10} </cmath>
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the equal sign will be reached when 2 circles are external tangent to each other,
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Note that they will be equal if and only if the circles are tangent,
  
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Apply AM-QM inequality <math> 2(a^2 + b^2) \geq (a+b)^2</math> in step below, we get   
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Applying the AM-GM inequality (<math> 2(a^2 + b^2) \geq (a+b)^2</math>) in the steps below, we get   
 
<cmath>
 
<cmath>
 
   h + k + 54 = (h + 25) + (k + 29) =\sqrt{(h + 25)}^2 + \sqrt{(k + 29)}^2 \geq \frac{\left(\sqrt{h + 25} + \sqrt{k + 29}\right)^2}{2}
 
   h + k + 54 = (h + 25) + (k + 29) =\sqrt{(h + 25)}^2 + \sqrt{(k + 29)}^2 \geq \frac{\left(\sqrt{h + 25} + \sqrt{k + 29}\right)^2}{2}
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</cmath>
 
</cmath>
  
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Therefore, h + k <math> \geq 20 - 54 = \boxed{C -34} </math>.   
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Therefore, <math>h + k \geq 20 - 54 = \boxed{C -34} </math>.   
  
  
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso],~[https://artofproblemsolving.com/wiki/index.php/User:ShortPeopleFartalot]
  
 
==Solution 3==
 
==Solution 3==

Latest revision as of 23:11, 25 December 2024

Problem 13

There are real numbers $x,y,h$ and $k$ that satisfy the system of equations\[x^2 + y^2 - 6x - 8y = h\]\[x^2 + y^2 - 10x + 4y = k\]What is the minimum possible value of $h+k$?

$\textbf{(A) }-54 \qquad \textbf{(B) }-46 \qquad \textbf{(C) }-34 \qquad \textbf{(D) }-16 \qquad \textbf{(E) }16 \qquad$


Solution 1 (Easy and Fast)

Adding up the first and second equation, we get: \begin{align*} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \\ &= 2(x - 4)^2 + 2(y - 1)^2 - 34 \end{align*} All squared values must be greater than or equal to $0$. As we are aiming for the minimum value, we set the two squared terms to be $0$.

This leads to $\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}$

~mitsuihisashi14

Solution 2 (Coordinate Geometry and AM-GM Inequality)

2024 amc 12B P13 V2.PNG

\[(x-3)^2 + (y-4)^2 = h + 25\] \[(x-5)^2 + (y+2)^2 = k + 29\] The distance between 2 circle centers is \[(O_{1}O_{2})^2 = (5-3)^2 + (4 - (-2)) ^2 = 40\] The 2 circles must intersect given there exists one or more pairs of (x,y), connecting $O_{1}O_{2}$ and any pair of the 2 circle intersection points gives us a triangle with 3 sides, then \[radius (O_{1}) + radius (O_{2}) \geq O_{1}O_{2}\] \[\sqrt{h+25} + \sqrt{k+29} \geq 2\cdot\sqrt{10}\] Note that they will be equal if and only if the circles are tangent,

Applying the AM-GM inequality ($2(a^2 + b^2) \geq (a+b)^2$) in the steps below, we get \[h + k + 54 = (h + 25) + (k + 29) =\sqrt{(h + 25)}^2 + \sqrt{(k + 29)}^2 \geq \frac{\left(\sqrt{h + 25} + \sqrt{k + 29}\right)^2}{2} \geq \frac{\left(2\sqrt{10}\right)^2}{2} = 20.\]

Therefore, $h + k \geq 20 - 54 = \boxed{C -34}$.


~luckuso,~[1]

Solution 3

2024 AMC 12B P13.jpeg

~Kathan

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=U0PqhU73yU0

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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