2024 AMC 12B Problems/Problem 13

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

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2024 AMC 12B Problems/Problem 13

Contents

Problem 13

Solution 1 (Easy and Fast)

Solution 2 (Coordinate Geometry)

See also