Let's call a set of points good if every pair of points in the set are an integer distance apart. Let's call a circle small if its radius is less than 7.

A good set lies on the boundary of a small circle. How many points can the set contain?

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    $\begingroup$ Without the requirement that the circle is small, it is possible to construct a good set of arbitrary size such that all its points lie on the same circle. $\endgroup$
    – f''
    Jan 31, 2016 at 18:49

1 Answer 1


This is not a proper answer! I'll guess that the number is


But this is just a guess. (EDIT: now with ugly proof)

First off,

any three-point subset of a good set is an integer-sided triangle, and the circumradius of each such triangle must be the same. The formula by the way is $$r=\frac{abc}{\sqrt{(a+b+c)(a+b-c)(c+a-b)(b+c-a)}}$$ So to start we are looking for different triples of integers which have give the same value of r when you put them in this formula (r doesn't need to be an integer).

I cheated a bit here:

I clicked on OP's profile, and see a previous problem, and a comment about a certain hexagon whose side lengths are 5 or less

I see by trial-and-error that

the equilateral triangle (7,7,7) matches (3,5,7) and (3,7,8) and (5,7,8), r = $\frac{7}{\sqrt{3}}$ which is just over 4. You can make a skewed hexagon out of this by taking two equilateral triangles of side length 7 to make a Star-of-David. Now rotate one triangle until the side lengths of the outer hexagon are 3-5-3-5-3-5, and this will make the diagonals 8, so everything is an integer and this is a "good" set. I think this is OP's hexagon.

I also see by trial-and-error that

the next equilateral triangle that can match anything else is (13,13,13), and r = $\frac{13}{\sqrt{3}}$ which is just past OP's critical value of 7. This is a puzzle and not just a math question. It could turn out that in fact there is a heptagon solution containing no equilateral triangle. But probably OP's hexagon is involved, and OP chose 7 for a reason.

Again, this is just a guess

based on assumptions about OP :-)

not a proof.

ADDED: Well since nobody is answering ... I found all the

different triples of integers up to 13 which have the same circumradius. There are a bunch of pairs of triples, and then three other higher-order coincidences: first, the one I found already; second, (4,4,7), (4,6,8), (4,8,8), (6,7,8), $r = \frac{16}{\sqrt{15}}$, just a bit bigger than the previous r; and, (3,8,10), (3,12,12), (8,8,12), and (8,10,12), $r=\frac{16}{\sqrt{7}}$, just over 6. Trial-and-error trying to fit these around a circle shows that these both make pentagons. The first has sides 4-4-6-4-6, one of the diagonals has length 7 and the other 4 have length 8. The other has sides 3-8-8-8-8, two diagonals have length 10, three have length 12. Both "good" sets.

To finish, we need to rule out the

pairs. { EDIT: Example: (2,3,4) and (2,4,4) are the smallest pair, $r=\frac{8}{\sqrt{15}}$ which is just over 2. This makes a quadrilateral with sides length 2-3-2-4 and both diagonals length 4. BTW we also know it fits on a circle by Ptolemy's theorem, since $2 \cdot 2 + 3 \cdot 4 = 4 \cdot 4$. } Well, a subset of a good set is a good set, so we just need to show that any heptagon would need more than two different (ie. not congruent) kinds of triangle. Uhhhh. Well, it really looks like it ought to be true, and I want it to be true, so I'll just call that "obvious" ;-)

So my guess was right. But maybe OP had a more elegant proof in mind, and didn't intend for me to hax0r it with my 1337 programming skillz (by that I mean, I made a spreadsheet.. hey don't judge)

  • 1
    $\begingroup$ Or the OP was someone like me, and had no idea what to expect. $\endgroup$ Feb 2, 2016 at 17:12

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