23
$\begingroup$

A pentagon can be used to measure 10 distances - one distance between each pair of its 5 vertices. Can you find a pentagon that can measure every integer distance from 1 to 7, inclusive?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Can't you use it to measure a distance using more than one pair of vertices? Like if the neighbouring edges are 1 and 2 then you can roll the pentagon on its side and measure 3? $\endgroup$
    – Vilx-
    Commented Jul 23, 2022 at 12:01
  • 3
    $\begingroup$ Well if you have edge 1 then you can measure any integer distance... $\endgroup$
    – Dmitry Kamenetsky
    Commented Jul 23, 2022 at 14:03
  • 1
    $\begingroup$ Seeing just the title, I guessed that this was one of @DmitryKamenetsky's puzzles ;) $\endgroup$
    – Oliphaunt
    Commented Jul 24, 2022 at 8:40
  • $\begingroup$ @Oliphaunt wow how did you do that? $\endgroup$
    – Dmitry Kamenetsky
    Commented Jul 24, 2022 at 9:31
  • 1
    $\begingroup$ For those interested, a hexagon can measure the first 9 distances. See if you can find it. $\endgroup$
    – Dmitry Kamenetsky
    Commented Sep 10, 2022 at 11:34

3 Answers 3

Reset to default
19
$\begingroup$

The answer is

yes.

pentagon

This can be proven by

computing the length of the diagonal determined by the 2-3-4 and 4-5-6 triangles as $$\sqrt{\frac{983+45\sqrt{105}}{32}}\approx6.71778$$ using the Cayley-Menger determinant, which is strictly between 6 and 8, thereby satisfying the triangle inequality for the sides of lengths 1 and 7.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Very nice work! Do you think it's possible to add distance 8? $\endgroup$
    – Dmitry Kamenetsky
    Commented Jul 22, 2022 at 14:27
  • 3
    $\begingroup$ @DmitryKamenetsky No because the segment of length 1 together with the 3 vertices not on that segment form 3 triangles with 3 distinct pairs of segments. Each pair can have at most 1 integral length by the triangle inequality (unless we permit degenerate pentagons), so we can have at most 10 - 3 = 7 integral lengths. $\endgroup$
    – noedne
    Commented Jul 22, 2022 at 14:31
  • 25
    $\begingroup$ Here's a degenerate pentagon that can count up to 9: i.sstatic.net/slrz7.png $\endgroup$
    – Bass
    Commented Jul 22, 2022 at 16:39
  • 3
    $\begingroup$ The proof isn't quite convincing though: having three vertices on the same line doesn't necessarily make a pentagon degenerate. $\endgroup$
    – Bass
    Commented Jul 23, 2022 at 23:40
  • 2
    $\begingroup$ @Bass You're right, that only holds for convex pentagons. $\endgroup$
    – noedne
    Commented Jul 24, 2022 at 0:17
29
$\begingroup$

A possibly more elegant solution for 1..7 if we don't insist on a convex pentagon.

enter image description here

Note: some angles appear to be right angles but none are.

Improve this answer
$\endgroup$
3
  • $\begingroup$ That's very nice $\endgroup$
    – Dmitry Kamenetsky
    Commented Jul 23, 2022 at 7:25
  • $\begingroup$ Sriram Sathyamoorthy comments: There are quite a few answers if we are not constrained by a convex pentagon. $\endgroup$
    – Florian F
    Commented Sep 9, 2022 at 7:25
  • 1
    $\begingroup$ That is true. For instance the figure above can be seen as a quadrilateral split along the diagonals, with one sector removed. Any other sector could be removed instead. I chose this one for purely esthetic reasons. $\endgroup$
    – Florian F
    Commented Sep 9, 2022 at 7:32
18
$\begingroup$

I first drew the lines 7, 6 and 5, then connected the ends with 4 and 3, then calculated the length of the top line, which turned out to be 2.583. $1 + 2 > 2.583$, so 1 and 2 can fit on top.

enter link description here

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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