Does there exist a circle whose boundary contains 6 points whose 15 pairwise distances are distinct integers?
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1$\begingroup$ I think 4 points would already be hard/impossible, let alone 6 $\endgroup$– IvoCommented Jan 18, 2016 at 15:06
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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? $\endgroup$– Tim CouwelierCommented Jan 18, 2016 at 15:21
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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? $\endgroup$– KevinCommented Jan 18, 2016 at 15:39
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1$\begingroup$ I've added the number "15" to the problem statement. $\endgroup$– dshinCommented Jan 18, 2016 at 15:45
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1$\begingroup$ I found that 4 is possible :) check this . Stuff mentioned there could perhaps help with 6 too $\endgroup$– IvoCommented Jan 18, 2016 at 16:30
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3 Answers
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They do exist. And they are actually called Brahmagupta Hexagons. An example is this:
Which I took from this paper which has a lot more info on them
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$\begingroup$ Surprising! +1 for the reference $\endgroup$ Commented Jan 18, 2016 at 17:00
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$\begingroup$ 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 $\leq5$. $\endgroup$– dshinCommented Jan 18, 2016 at 17:06
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This puzzle of mine appears to have languished with no solution for some time. I don’t remember the solution I initially had in mind, but I thought of one recently that might be of interest, so answering my own puzzle here.
It suffices to show that for any $n$, there exists an $n$-gon inscribed in the unit circle with pairwise-distinct rational distances, since any such $n$-gon can be converted to one with integer distances by a simple dilation.
We prove this by induction, starting with two diametrically opposite points $P_1$ and $P_2$. Each subsequent point can be found by picking a primitive Pythagorean triple $(a, b, c)$, and picking a point $Q$ such that the right triangle $\triangle P_1P_2Q$ has side lengths $(2a/c, 2b/c, 2)$. The induced distances to the other existing points are guaranteed to be rational thanks to Ptolemy’s Theorem and the induction hypothesis. Now, some choices of Pythagorean triple might lead to distance collisions. But only finitely many choices will do so, and there are an infinite number of primitive Pythagorean triples, implying that a good choice of Pythagorean triple exists. This completes the proof.
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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
answered Sep 12, 2021 at 14:29