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12-27-2018
07:30 AM

12-27-2018
07:30 AM

(This is only a puzzle to fun! )

Tokoro.

All distances of the points to be Integer

- Make a circle with its center (0,0) and radius r.
- Make five points on the circle with all distances of the points to be integer.
- The puzzle is to find the minimum of r.
- Next, the same puzzle except the number of points are 10.
- The answer of 4 points attached.

Solved! Go to Solution.

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1 ACCEPTED SOLUTION

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01-03-2019
11:46 PM

12 REPLIES 12

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12-28-2018
03:16 AM

12-28-2018
03:16 AM

Re: All distances of the points to be Integer

Hi,

I'll play (The definition does not need the radius be an integer?)

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12-28-2018
03:55 AM

12-28-2018
03:55 AM

Re: All distances of the points to be Integer

You are not free on the radius. Because the diagonals are twice the radius. Hence radius must be an integer times 1/2.

Success!

Luc

Success!

Luc

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12-28-2018
04:45 AM

12-28-2018
04:45 AM

Re: All distances of the points to be Integer

@LucMeekes wrote:

You are not free on the radius. Because the diagonals are twice the radius. Hence radius must be an integer times 1/2.

Which diagonal of which pentagram must be a diameter of the circle? There __may__ be a solution where one diagonal or maybe even one side of the pentagon is a diameter but sure not necessarily.

Even with four points its not demanded to be a regular quadrilateral or to have any symmetries. Its just the solution shown which happens to be at least a rectangle. And I see no proof that 5/2 is the minimum for the radius in case of a quadrilateral.

I have no clue how to proof that the solution for 4 points shown really is the minimum (wrt the radius) and that not maybe an irregular quadrilateral would not lead to a smaller radius.

Nor have I any idea how to use Mathcad to help finding any solution with 5 points, let alone the minimum one.

Its clear that it suffices to demand for all distances to be rational as we then can blow up the figure as necessary to make them integer, but I don't think that knowing this is much of help.

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12-28-2018
04:53 AM

12-28-2018
04:53 AM

Re: All distances of the points to be Integer

@terryhendicott wrote:

Hi,

I'll play (The definition does not need the radius be an integer?)

Correct, but I guess you overlooked that ALL distances of any two points must be integer. This includes the diagonals which in case of your pentagon are not integer.

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12-28-2018
05:08 AM

12-28-2018
05:08 AM

Re: All distances of the points to be Integer

Thanks Werner

thought it was too easy

Cheers

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12-28-2018
05:35 AM

12-28-2018
05:35 AM

Re: All distances of the points to be Integer

I stand corrected.

Luc

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12-28-2018
11:45 PM

12-28-2018
11:45 PM

Re: All distances of the points to be Integer

Thanks for your time to fun! Your answer is correct for 3 points.

By the way, connect all n points on the circle is good programing exercise for the students.

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01-03-2019
03:10 PM

01-03-2019
03:10 PM

Re: All distances of the points to be Integer

Hi,

Have an answer for 5 pointed problem.

Sides are 4,6,4,6,4 in a 4.131182 radius circle.

Diagonals are 8,8,8,7,8 all integers.

Calculation sheet enclosed.

It is based on solution of intersection of two circles one centred at origin and radius R, the other centred at R,0 with radius equal to side length.

Starting with next side, place a circle of radius the length of the side and centre at last location obtain the intersection point.

Repeat this for each side.

This is repeated around the circle to get a closing angle that should equal 2 pi.

Manually ratio the closing angle to 2 pi to get the next starting radius. Just do this till maximum accuracy is obtained

Cheers Terry

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01-03-2019
06:12 PM

01-03-2019
06:12 PM

Re: All distances of the points to be Integer

To be exact, that radius is 16/sqrt(15).

Luc

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