Difference between revisions of "2024 AMC 12B Problems/Problem 13"
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+ | ==Problem 13== | ||
+ | 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>? | ||
+ | |||
+ | <math> | ||
+ | \textbf{(A) }-54 \qquad | ||
+ | \textbf{(B) }-46 \qquad | ||
+ | \textbf{(C) }-34 \qquad | ||
+ | \textbf{(D) }-16 \qquad | ||
+ | \textbf{(E) }16 \qquad | ||
+ | </math> | ||
+ | |||
+ | |||
==Solution 1 (Easy and Fast)== | ==Solution 1 (Easy and Fast)== | ||
Revision as of 02:26, 14 November 2024
Contents
[hide]Problem 13
There are real numbers and that satisfy the system of equationsWhat is the minimum possible value of ?
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)
distance between 2 circle centers is min( h + k ) = .
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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