Difference between revisions of "2024 AMC 12B Problems/Problem 13"
(→Solution 2 (Coordinate Geometry and HM-GM)) |
(→Solution 2 (Coordinate Geometry and QM-GM Inequality)) |
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~mitsuihisashi14 | ~mitsuihisashi14 | ||
− | ==Solution 2 (Coordinate Geometry and QM | + | ==Solution 2 (Coordinate Geometry and AM-QM Inequality)== |
− | [[Image: | + | [[Image:2024_amc_12B_P13_V2.PNG|thumb|center|500px|]] |
<cmath>(x-3)^2 + (y-4)^2 = h + 25 </cmath> | <cmath>(x-3)^2 + (y-4)^2 = h + 25 </cmath> | ||
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the equal sign will be reached when 2 circles are external tangent to each other, | the equal sign will be reached when 2 circles are external tangent to each other, | ||
− | Apply QM | + | Apply AM-QM inequality <math> 2(a^2 + b^2) \geq (a+b)^2</math> in step below, we get |
<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} | 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} |
Revision as of 15:16, 15 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 equation, we get: All squared values must be greater than or equal to . As we are aiming for the minimum value, we set the two squared terms to be .
This leads to
~mitsuihisashi14
Solution 2 (Coordinate Geometry and AM-QM Inequality)
Distance between 2 circle centers is the 2 circle must intersect given there exists one or more pair of (x,y), connecting and any one of the 2 circle intersection point we get a triangle with 3 sides ( radius () , radius () , ) , then the equal sign will be reached when 2 circles are external tangent to each other,
Apply AM-QM inequality in step below, we get
Therefore, h + k .
Solution 3
~Kathan
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.