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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 1 (Easy and Fast))
 
(10 intermediate revisions by 2 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
 +
</math>
 +
 
 +
 
 +
==Solution 1 (Easy and Fast)==
 +
 
 +
Adding up the first and second statement, we get h+k with:
 +
 
 +
= 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)
 +
 
 +
= 2(x - 4)^2 + 2(y - 1)^2 - 34
 +
 
 +
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
 +
 
 +
~mitsuihisashi14
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 +
==Solution 2 (Coordinate Geometry)==
 +
<cmath>(x-3)^2 + (y-4)^2 = h + 25 </cmath>
 +
<cmath>(x-5)^2 + (y+2)^2 = k + 29 </cmath>
 +
distance between 2 circle centers is <cmath>d^2 = (5-3)^2 + (4 - (-2)) ^2 = 40 </cmath>
 +
<cmath>\sqrt{h+25} + \sqrt{k+29}  = 2*\sqrt{10} </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}
 +
=  \frac{\left(2\sqrt{10}\right)^2}{2} = 20.
 +
</cmath>
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min( h + k ) = <math>\boxed{C -34} </math>. 
 +
 
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 
 +
==See also==
 +
{{AMC12 box|year=2024|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 02:26, 14 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 statement, we get h+k with:

= 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)

= 2(x - 4)^2 + 2(y - 1)^2 - 34

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

~mitsuihisashi14

Solution 2 (Coordinate Geometry)

\[(x-3)^2 + (y-4)^2 = h + 25\] \[(x-5)^2 + (y+2)^2 = k + 29\] distance between 2 circle centers is \[d^2 = (5-3)^2 + (4 - (-2)) ^2 = 40\] \[\sqrt{h+25} + \sqrt{k+29} = 2*\sqrt{10}\] \[h + k + 54 = (h + 25) + (k + 29) =\sqrt{(h + 25)}^2 + \sqrt{(k + 29)}^2 \geq \frac{\left(\sqrt{h + 25} + \sqrt{k + 29}\right)^2}{2} = \frac{\left(2\sqrt{10}\right)^2}{2} = 20.\] min( h + k ) = $\boxed{C -34}$.


~luckuso

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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