8
$\begingroup$

Does there exist a circle whose boundary contains 6 points whose 15 pairwise distances are distinct integers?

$\endgroup$
17
  • 1
    $\begingroup$ I think 4 points would already be hard/impossible, let alone 6 $\endgroup$ Jan 18 '16 at 15:06
  • 1
    $\begingroup$ As for the question: do you need integer distances between any two, or just integer lengths for any vertex on the hexagon they make within the circle? $\endgroup$ Jan 18 '16 at 15:21
  • 1
    $\begingroup$ I'm not entirely clear on what "pairwise" means. How many line segments would an illustration of the solution contain? 3, from connecting each point to exactly one other point? 6, from connecting each point to its neighbors on the boundary? 15, from connecting each point to each other point? $\endgroup$
    – Kevin
    Jan 18 '16 at 15:39
  • 1
    $\begingroup$ I've added the number "15" to the problem statement. $\endgroup$
    – dshin
    Jan 18 '16 at 15:45
  • 1
    $\begingroup$ I found that 4 is possible :) check this . Stuff mentioned there could perhaps help with 6 too $\endgroup$ Jan 18 '16 at 16:30
19
$\begingroup$

They do exist. And they are actually called Brahmagupta Hexagons. An example is this:

enter image description here

Which I took from this paper which has a lot more info on them

$\endgroup$
2
  • $\begingroup$ Surprising! +1 for the reference $\endgroup$ Jan 18 '16 at 17:00
  • $\begingroup$ Wow, never knew about these. The methodology in the paper seems similiar to the methodology I used to create the puzzle. However, my puzzle is different, as it contains the word "distinct". Without that word, we can actually construct a Brahmagupta Hexagon whose side lengths are $\leq5$. $\endgroup$
    – dshin
    Jan 18 '16 at 17:06
0
$\begingroup$

There exists a 6-gon whose 15 pairwise distances are all distinct integers, but not sure if its vertices lie on a circle.

sides 47, 663, 264, 169, 105, 1020, diagonals 700, 884, 975, 855, 952, 1001, 425, 520, 272, and area 196950. See https://oeis.org/A270558

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.