Points, circumference and circles

Shape and space, angle and circle properties, ...
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Alhazen
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Points, circumference and circles

Post by Alhazen » Tue Feb 14, 2012 4:59 pm

Hi,

What is the maximal number of points you can place in the circumference of circle such as the distance between any 2 of them is always integer?
Can you find the minimal diameter d(k) such as for k points all the distances between any 2 of them are all integers?

drwhat
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Re: Points, circumference and circles

Post by drwhat » Thu Feb 23, 2012 3:16 am

When you say distance do you mean the chord length between 2 points, or the arc length along the circumference?

Alhazen
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Re: Points, circumference and circles

Post by Alhazen » Thu Feb 23, 2012 3:41 am

drwhat wrote:When you say distance do you mean the chord length between 2 points, or the arc length along the circumference?
The chord not the arc length

drwhat
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Re: Points, circumference and circles

Post by drwhat » Thu Feb 23, 2012 4:31 am

Posted a proof that it could only be 2 points, but I was assuming angle subtended by the points and center had to be rational. Which it does not.
Given angle 106.26..and 73.73.. (the angles of a 345 triangle) you can place 4 points around the circumfernce all of which will be 6 , 8 or 10 units from each other.

thundre
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Re: Points, circumference and circles

Post by thundre » Fri Feb 24, 2012 6:52 pm

4 is easy. You can construct a rectangle from any Pythagorean triple. The 4 corners of a rectangle are always on a circle.

6 points is not very hard either. For example, in a circle of diameter 25, inscribe four (15,20,25) triangles, all using the same diameter as the hypotenuse. The four right-angle vertices form a 7x24 rectangle. Distances are {7,15,20,24,25}.
Image

Alhazen
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Re: Points, circumference and circles

Post by Alhazen » Sat Feb 25, 2012 2:27 am

thundre wrote:4 is easy. You can construct a rectangle from any Pythagorean triple. The 4 corners of a rectangle are always on a circle.

6 points is not very hard either. For example, in a circle of diameter 25, inscribe four (15,20,25) triangles, all using the same diameter as the hypotenuse. The four right-angle vertices form a 7x24 rectangle. Distances are {7,15,20,24,25}.
Thank you for your answer.
I did not find the 6 points (4 was really easy).
I think that we can find more points using big radius.

thundre
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Re: Points, circumference and circles

Post by thundre » Tue Feb 28, 2012 11:10 am

Alhazen wrote:Can you find the minimal diameter d(k) such as for k points all the distances between any 2 of them are all integers?
d(2) = 1 (trivially)

Distances given below are from the first point, so the first one is always 0, the distance to itself.

d(3) = 1.1547005383792515 = sqrt(4/3), distances = 0 1 1
d(4) = 4.131182235954578 = sqrt(256/15), distances = 0 2 4 4
d(6) = 8.082903768654761 = sqrt(196/3), distances = 0 3 7 8 7 5
d(7) = 33.04945788763662 = sqrt(16384/15), distances = 0 10 24 32 33 28 16
d(9) = 56.58032638058333 = sqrt(9604/3), distances = 0 16 35 49 55 56 49 39 21
d(12) = 105.07774899251189 = sqrt(33124/3), distances = 0 11 49 65 91 96 105 104 91 85 56 39
Image

Alhazen
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Re: Points, circumference and circles

Post by Alhazen » Tue Feb 28, 2012 1:54 pm

thundre wrote:
Alhazen wrote:Can you find the minimal diameter d(k) such as for k points all the distances between any 2 of them are all integers?
d(2) = 1 (trivially)

Distances given below are from the first point, so the first one is always 0, the distance to itself.

d(3) = 1.1547005383792515 = sqrt(4/3), distances = 0 1 1
d(4) = 4.131182235954578 = sqrt(256/15), distances = 0 2 4 4
d(6) = 8.082903768654761 = sqrt(196/3), distances = 0 3 7 8 7 5
d(7) = 33.04945788763662 = sqrt(16384/15), distances = 0 10 24 32 33 28 16
d(9) = 56.58032638058333 = sqrt(9604/3), distances = 0 16 35 49 55 56 49 39 21
d(12) = 105.07774899251189 = sqrt(33124/3), distances = 0 11 49 65 91 96 105 104 91 85 56 39
Thanx a lot for the results. I will try to check in Sloane Encyplopedia (OEIS) if there is some sequence nearing d(i).
Sure that there is a room to build some formula d(n)=something

thundre
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Re: Points, circumference and circles

Post by thundre » Tue Feb 28, 2012 2:38 pm

Alhazen wrote:Thanx a lot for the results. I will try to check in Sloane Encyplopedia (OEIS) if there is some sequence nearing d(i).
Sure that there is a room to build some formula d(n)=something
OEIS is for integers. I don't think it's the right place to put a list of square roots of rationals. There are also several "missing" numbers on the list, which are equal to their successors.

The formula I derived takes a triangle and gives the diameter of the circumcircle. So, for two chords with a common endpoint, define the distance between the other two endpoints (without violating the triangle inequality), and there's your triangle.

d2 = 4*a2*b2*c2/(2*a2*b2 + 2*a2*c2 + 2*b2*c2 - a4 - b4 - c4)

The numerator in that expression is a perfect square, but is the numerator of the reduced fraction also, or might the GCF of the numerator and denominator be non-square?
Image

Alhazen
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Re: Points, circumference and circles

Post by Alhazen » Wed Feb 29, 2012 3:37 pm

Hi,

Thank you for your comments. I know that OEIS is dedicated to integers. I said a sequence nearing yours (sequence nearing d(i)).
I tried to understand your formula without success because you did not define a,b,c.
I was surprised that you could put 12 points on the circumference of a circle such as any 2 of them are integers (66 connections or chords if you want).
Can you please send a picture or at least the 66 integers values (some are surely repeated).

1000 thanks!

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kevinsogo
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Re: Points, circumference and circles

Post by kevinsogo » Thu Mar 01, 2012 12:02 pm

Alhazen wrote:I tried to understand your formula without success because you did not define a,b,c.
The formula that thundre posted is the formula for the diameter of the circumcircle of a triangle. The sides of the triangle are a, b, c and the diameter is d.
This link shows a derivation of the equivalent formula r = abc/4A. Here r is the radius of the circumcircle and A is the area of the triangle. Substitute r = d/2 and A = [Heron's formula] and you'll arrive at the first formula.

For the current problem:
To calculate the distance between two points in the circle (a chord): substitute the values a, b (chord distances from first point), and d (diameter) and then solve for c.

Grey
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Re: Points, circumference and circles

Post by Grey » Fri Apr 13, 2012 9:11 pm

Sorry, i did not understand clearly that, are you telling about distance between two places? If you are telling about it then you can measure distance between two places using a distance calculator.

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