Difference between revisions of "2024 AMC 12B Problems/Problem 13"
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distance between 2 circle centers is <cmath>d^2 = (5-3)^2 + (4 - (-2)) ^2 = 40 </cmath> | distance between 2 circle centers is <cmath>d^2 = (5-3)^2 + (4 - (-2)) ^2 = 40 </cmath> | ||
the 2 circle must interact, so | the 2 circle must interact, so | ||
− | <cmath>\sqrt{h+25} + \sqrt{k+29} | + | <cmath>\sqrt{h+25} + \sqrt{k+29} \geq 2*\sqrt{10} </cmath> |
the min value will be reached when 2 circles are external tangent to each other | the min value will be reached when 2 circles are external tangent to each other | ||
<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} | ||
− | + | \geq \frac{\left(2\sqrt{10}\right)^2}{2} = 20. | |
</cmath> | </cmath> | ||
+ | the above step uses HM-GM inequality <math> 2(a^2 + b^2) \geq (a+b)^2 </math> | ||
+ | |||
min( h + k ) = <math>\boxed{C -34} </math>. | min( h + k ) = <math>\boxed{C -34} </math>. | ||
Latest revision as of 20:50, 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 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 HM-GM)
distance between 2 circle centers is the 2 circle must interact, so the min value will be reached when 2 circles are external tangent to each other the above step uses HM-GM inequality
min( 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.