
#1
Dec809, 07:34 AM

P: 57

In a math contest, the question goes somehow like this:
A lattice point is a point wherein the value of (x,y) is an integer. Determine the total number of lattice points in a circle which has a radius of 6 and the its center is at the origin. Any one knows the solution or shortcut for this? 



#2
Dec809, 10:00 AM

P: 14

you could draw a quadrant of a circle of radius 6 and check the number of points there and multiply that number by four, being careful not to double count points that lie on the axis,
as for a closed form, i would be surprised if one did not exist... Side note Wolfram indeed does have a interesting write up in regards to this problem. 



#3
Dec909, 06:51 AM

P: 57

Wolfram? Sorry I'm still new to the community.




#4
Dec909, 08:43 AM

P: 418

Lattice Points
I guess crd refers to: http://mathworld.wolfram.com/ which is a math resource, but I don't know the particular writeup he mentioned.
Anyway just work it out in cases. As crd suggested just count the points in the first quadrant (which we can take to include the positive xaxis, but not the positive yaxis because then we get simple rotational symmetry without doublecounting), and then use symmetry to deduce the total number. In that case the xcoordinate is 1,2,3,4,5 or 6. When the xcoordinate is x, then the ycoordinate must be less than or equal to [itex]\sqrt{6^2x^2}[/itex], so for any xcoordinate you want to count the integers in [tex][0,\sqrt{6^2x^2}][/tex] Try to see how far you can get, and if you get stuck at a particular step just ask for more help. 



#5
Dec909, 12:28 PM

P: 14




#6
Dec1109, 06:56 AM

P: 57

I'm still not familiar in self studying especially with those complex solutions.
Can someone write a general formula for me which I could use when I'm given the center of the circle and the length of the radius. That would be a very big help. Thanks. 



#7
Dec1109, 07:05 AM

P: 418

Gauss's circle problem asks for the number of lattice points within a circle of radius R [tex]N(R) = 1+4\lfloor R\rfloor + 4\sum_{i=1}^{\lfloor R\rfloor}\left\lfloor\sqrt{R^2i^2}\right\rfloor[/tex] Which is exactly what you would get if you split it into cases. 



#8
Dec1109, 05:12 PM

P: 14

I think thats the easiest way to make sense of formulas, once your hands already dirty in what you are working with, what other people have discovered falls into place just that much easier, other wise you are pushing around symbols that have no meaning to you. 


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