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Difference between revisions of "2024 AMC 12B Problems/Problem 13"

(Created page with "Adding up the first and second statement, we get: h+k = 2x^2 + 2y^2 - 16x - 4y = 2(x^2 - 8x) + 2(y^2 - 2y) = 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1)...")
 
(→‎Solution 2 (Coordinate Geometry and AM-QM Inequality))
 
(21 intermediate revisions by 6 users not shown)
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Adding up the first and second statement, we get:
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==Problem 13==
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h+k = 2x^2 + 2y^2 - 16x - 4y
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There are real numbers <math>x,y,h</math> and <math>k</math> that satisfy the system of equations<cmath>x^2 + y^2 - 6x - 8y = h</cmath><cmath>x^2 + y^2 - 10x + 4y = k</cmath>What is the minimum possible value of <math>h+k</math>?
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    = 2(x^2 - 8x) + 2(y^2 - 2y)
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    = 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1)
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<math>
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    = 2(x - 4)^2 + 2(y - 1)^2 - 34
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\textbf{(A) }-54 \qquad
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All squared values must be greater or equal to 0. As we are aiming for the minimum value, we let the 2 squared terms be 0. This leads to (h+k)min = 0 + 0 - 34 = (C) -34
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\textbf{(B) }-46 \qquad
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\textbf{(C) }-34 \qquad
 +
\textbf{(D) }-16 \qquad
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\textbf{(E) }16 \qquad
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</math>
 +
 
 +
 
 +
==Solution 1 (Easy and Fast)==
 +
 
 +
Adding up the first and second equation, we get:
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
All squared values must be greater than or equal to <math>0</math>. As we are aiming for the minimum value, we set the two squared terms to be <math>0</math>.  
 +
 
 +
This leads to <math>\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}</math>
 +
 
 +
~mitsuihisashi14
 +
 
 +
==Solution 2 (Coordinate Geometry and AM-QM Inequality)==
 +
 
 +
[[Image:2024_amc_12B_P13_V2.PNG|thumb|center|500px|]]
 +
 
 +
<cmath>(x-3)^2 + (y-4)^2 = h + 25 </cmath>
 +
<cmath>(x-5)^2 + (y+2)^2 = k + 29 </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>
 +
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>
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<cmath>\sqrt{h+25} + \sqrt{k+29}  \geq  2\cdot\sqrt{10} </cmath>
 +
Note that they will be equal if and only if the circles are tangent,
 +
 
 +
Applying the AM-QM inequality (<math> 2(a^2 + b^2) \geq (a+b)^2</math>) in the steps below, we get 
 +
<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}
 +
\geq  \frac{\left(2\sqrt{10}\right)^2}{2} = 20.
 +
</cmath>
 +
 
 +
Therefore, <math>h + k \geq 20 - 54 = \boxed{C -34} </math>. 
 +
 
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 
 +
==Solution 3==
 +
[[Image: 2024_AMC_12B_P13.jpeg|thumb|center|600px|]]
 +
~Kathan
 +
 
 +
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=U0PqhU73yU0
 +
 
 +
==See also==
 +
{{AMC12 box|year=2024|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 20:30, 17 November 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-QM 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-QM 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

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

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