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> | |
| − | = 2x^2 + 2y^2 - 16x - 4y | + | \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) + 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> | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | This leads to (h+k) | ||
~mitsuihisashi14 | ~mitsuihisashi14 | ||
| − | ==Solution 2 (Coordinate Geometry and | + | ==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> | ||
<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> | + | The 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 circles must intersect given there exists one or more pairs of (x,y), connecting <math>O_{1}O_{2}</math> and any pair of the 2 circle intersection points gives us a triangle with 3 sides, then |
| + | <cmath> radius (O_{1}) + radius (O_{2}) \geq O_{1}O_{2} </cmath> | ||
| + | <cmath>\sqrt{h+25} + \sqrt{k+29} \geq 2\cdot\sqrt{10} </cmath> | ||
| + | Note that they will be equal if and only if the circles are tangent, | ||
| + | |||
| + | Applying the AM-QM inequality (<math> 2(a^2 + b^2) \geq (a+b)^2</math>) in the steps 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, <math>h + k \geq 20 - 54 = \boxed{C -34} </math>. | ||
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[[Image: 2024_AMC_12B_P13.jpeg|thumb|center|600px|]] | [[Image: 2024_AMC_12B_P13.jpeg|thumb|center|600px|]] | ||
~Kathan | ~Kathan | ||
| + | |||
| + | ==Video Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=U0PqhU73yU0 | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC12 box|year=2024|ab=B|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:30, 17 November 2024
Contents
Problem 13
There are real numbers 
and 
that satisfy the system of equations![\[x^2 + y^2 - 6x - 8y = h\]](http://latex.artofproblemsolving.com/6/e/1/6e1b822d6a797e597019f3e2dec998e71d787585.png)
![\[x^2 + y^2 - 10x + 4y = k\]](http://latex.artofproblemsolving.com/8/1/e/81e93689500b6956b5b22642888c380e29c2a60e.png)
What 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)
![\[(x-3)^2 + (y-4)^2 = h + 25\]](http://latex.artofproblemsolving.com/9/a/e/9ae17badb0db86f1f929a000c426857df7dd376a.png)
![\[(x-5)^2 + (y+2)^2 = k + 29\]](http://latex.artofproblemsolving.com/3/3/5/3359dd10b43d4594313208fe47bbaa76e2b1767a.png)
The distance between 2 circle centers is ![\[(O_{1}O_{2})^2 = (5-3)^2 + (4 - (-2)) ^2 = 40\]](http://latex.artofproblemsolving.com/8/3/5/83572b102bca95476f4268f1a7111291af8f7e69.png)
The 2 circles must intersect given there exists one or more pairs of (x,y), connecting 
and any pair of the 2 circle intersection points gives us a triangle with 3 sides, then
![\[radius (O_{1}) + radius (O_{2}) \geq O_{1}O_{2}\]](http://latex.artofproblemsolving.com/e/9/d/e9d61686d20a6ae035b157840c1ef6c2f4777cfb.png)
![\[\sqrt{h+25} + \sqrt{k+29} \geq 2\cdot\sqrt{10}\]](http://latex.artofproblemsolving.com/b/f/8/bf88d198b54e21e2bc39578f250c0b90d16ba34d.png)
Note that they will be equal if and only if the circles are tangent,
Applying the AM-QM inequality (
) in the steps below, we get
![\[h + k + 54 = (h + 25) + (k + 29) =\sqrt{(h + 25)}^2 + \sqrt{(k + 29)}^2 \geq \frac{\left(\sqrt{h + 25} + \sqrt{k + 29}\right)^2}{2} \geq \frac{\left(2\sqrt{10}\right)^2}{2} = 20.\]](http://latex.artofproblemsolving.com/3/2/c/32ca303b4f306d3bb7501a1ce7113a518393722d.png)
Therefore, 
.
Solution 3
~Kathan
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=U0PqhU73yU0
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. 
