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

(Solution 3)
(Solution 3)
Line 49: Line 49:
 
[[Image: 2024_AMC_12B_P13.jpeg|thumb|center|600px|]]
 
[[Image: 2024_AMC_12B_P13.jpeg|thumb|center|600px|]]
 
~Kathan
 
~Kathan
 +
 +
==See also==
 +
{{AMC12 box|year=2024|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Revision as of 14:20, 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 and HM-GM)

2024 amc 12B P13.PNG

\[(x-3)^2 + (y-4)^2 = h + 25\] \[(x-5)^2 + (y+2)^2 = k + 29\] 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

Solution 3

2024 AMC 12B P13.jpeg

~Kathan

See also

2024 AMC 12B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png