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Let $n$ be a positive integer, prove that it is possible to put $n$ points on a circle so that the distances among them are all integers.

For $n \leq 3$ this is trivial. I have shown it for $n=4$ by considering a rectangle. I don't know how to do it for larger numbers though.

Batominovski
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mtheorylord
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  • How do you do it with a rhombus? Vertices of a rhombus lie on a circle only if the rhombus is a square, and then the diagonals divided by the sides is not a rational number. Maybe you meant a rectangle? – Oscar Lanzi Oct 31 '19 at 00:42
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    yes, my bad, rectangle – mtheorylord Oct 31 '19 at 00:47

2 Answers2

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This is easier than it looks. All you need to do is use Pythagorean triples in the appropriate way.

Draw a circle with a diameter of five units. Select any point on the circle and generate other points from it by drawing chords with length three units joined end to end. All chords being congruent, they intercept congruent arcs and the arcs add together. Therefore the distance between any two points will have the form

$|5\sin(k\theta/2)|$

where $k$ is a whole number and $\theta$ is the minor arc intercepted by a chord. The multiple angle identities then guarantee that $\sin(k\theta/2)$ will be rational given the rational values $\sin(\theta/2)=3/5$ and $\cos(\theta/2)=4/5$. So all point-to-point distances are rational and they may be converted to integers by scaling appropriately. Since the value of $\theta$ as constructed here is not a rational number times $2\pi$, an unlimited number of distinct points may be identified, so there is no limit on $n$.

Oscar Lanzi
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  • Alright, I drew your picture as an answer. It appears the final pair of points also have rational distance with an odd number of points, I'm not sure how well this works with even number of edges. I suppose in that case you don't go for exact $n,$ solve the problem for a larger, but odd, number of points. – Will Jagy Oct 31 '19 at 19:46
  • You get rational distances for all n. With two chords for n=3 you get $5\sin(2θ/2)$ when $\sin(θ/2)=3/5$ and $\cos(θ/2)=4/5$, thus $24/5$. Compare with another Pythagorean triple, $7−24−25$. – Oscar Lanzi Oct 31 '19 at 20:04
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This is with radius 5 and edge length 6. It appears that the final edge (not drawn in) also has rational length since it is vertical and both endpoints have rational coordinates

enter image description here

Will Jagy
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  • this is what I am doing (except my circle has half the radius of yours). The distance between pairs of points is rational because it's related to trig functions that are themselves rational ($\sin(\theta/2)=3/5, \cos(\theta/2)=4/5$) through polynomial functions derived from multiple angle formulas. – Oscar Lanzi Oct 31 '19 at 19:49