# A pentagon that can measure the first 7 integer distances

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?

• 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? Jul 23 at 12:01
• Well if you have edge 1 then you can measure any integer distance... Jul 23 at 14:03
• Seeing just the title, I guessed that this was one of @DmitryKamenetsky's puzzles ;) Jul 24 at 8:40
• @Oliphaunt wow how did you do that? Jul 24 at 9:31
• For those interested, a hexagon can measure the first 9 distances. See if you can find it. Sep 10 at 11:34

yes.

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.

• Very nice work! Do you think it's possible to add distance 8? Jul 22 at 14:27
• @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. Jul 22 at 14:31
• Here's a degenerate pentagon that can count up to 9: i.stack.imgur.com/slrz7.png
– Bass
Jul 22 at 16:39
• The proof isn't quite convincing though: having three vertices on the same line doesn't necessarily make a pentagon degenerate.
– Bass
Jul 23 at 23:40
• @Bass You're right, that only holds for convex pentagons. Jul 24 at 0:17

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

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

• That's very nice Jul 23 at 7:25
• Sriram Sathyamoorthy comments: There are quite a few answers if we are not constrained by a convex pentagon. Sep 9 at 7:25
• 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. Sep 9 at 7:32

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.