Difference between revisions of "2024 AMC 12B Problems/Problem 13"
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==Solution 1 (Easy and Fast)== | ==Solution 1 (Easy and Fast)== | ||
− | Adding up the first and second | + | Adding up the first and second equation, we get: |
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | h + k &= 2x^2 + 2y^2 - 16x - 4y \ | ||
+ | &= 2(x^2 - 8x) + 2(y^2 - 2y) \ | ||
+ | &= 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 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | All squared values must be greater than or equal to <math>0</math>. As we are aiming for the minimum value, we set the two squared terms to be <math>0</math>. | ||
− | = | + | This leads to <math>\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}</math> |
− | + | ~mitsuihisashi14 | |
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− | + | ==Solution 2 (Coordinate Geometry and AM-QM Inequality)== | |
− | + | [[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> | ||
<cmath>(x-5)^2 + (y+2)^2 = k + 29 </cmath> | <cmath>(x-5)^2 + (y+2)^2 = k + 29 </cmath> | ||
− | + | Distance between 2 circle centers is <cmath>(O_{1}O_{2})^2 = (5-3)^2 + (4 - (-2)) ^2 = 40 </cmath> | |
− | <cmath>\sqrt{h+25} + \sqrt{k+29} | + | the 2 circle must intersect given there exists one or more pair of (x,y), connecting <math>O_{1}O_{2}</math> and any one of the 2 circle intersection point we get a triangle with 3 sides ( radius (<math>O_{1}</math>) , radius (<math>O_{2}</math>) , <math>O_{1}O_{2}</math>) , then |
+ | <cmath> radius (O_{1}) + radius (O_{2}) \geq O_{1}O_{2} </cmath> | ||
+ | <cmath>\sqrt{h+25} + \sqrt{k+29} \geq 2*\sqrt{10} </cmath> | ||
+ | the equal sign will be reached when 2 circles are external tangent to each other, | ||
+ | |||
+ | 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} | ||
− | + | \geq \frac{\left(2\sqrt{10}\right)^2}{2} = 20. | |
</cmath> | </cmath> | ||
− | + | ||
+ | Therefore, h + k <math> \geq 20 - 54 = \boxed{C -34} </math>. | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Solution 3== | ||
+ | [[Image: 2024_AMC_12B_P13.jpeg|thumb|center|600px|]] | ||
+ | ~Kathan | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2024|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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.