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October 22

Scalene equilateral polygons

(For the purpose of this question, "scalene" means [for a polygon with any number of sides] having no symmetry.)

A scalene equilateral triangle is obviously not possible; every line from an angle to the center of the side opposite the angle is a line of symmetry. A scalene equilateral quadrilateral is also not possible; any equilateral quadrilateral (rhombus) has at lest 2 lines of symmetry; they are the diagonals. But is a scalene equilateral pentagon possible?? Georgia guy (talk) 16:18, 22 October 2023 (UTC)[reply]

Let the vertices be

(0,0),(5,0),(5,5),(1,8),(x,y),

where

x={\frac {13+8{\sqrt {91}}}{26}}\approx 3.4352,y={\frac {104-{\sqrt {91}}}{26}}\approx 3.6331.

This pentagon is not convex, but can be made into a convex one by mirroring

(x,y)

with respect to the line through

(0,0)

and

(1,8).

Conceivably, there are Diophantine solutions in which all five vertices have integer coordinates. --Lambiam 17:28, 22 October 2023 (UTC)[reply]

Lambiam, can you show me an image of the convex one you referred to in the phrase "can be made into a convex one"?? Georgia guy (talk) 17:42, 22 October 2023 (UTC)[reply]

Lookie here. --Lambiam 18:55, 22 October 2023 (UTC)[reply]

(I think the mirrored point would just be

\left({\frac {13-8{\sqrt {91}}}{26}},{\frac {104+{\sqrt {91}}}{26}}\right)

if that helps.) 2603:8001:4542:28FB:5D48:1153:A783:5728 (talk) 20:29, 22 October 2023 (UTC) (Please send talk messages here instead)[reply]

That is correct. --Lambiam 22:39, 22 October 2023 (UTC)[reply]

I think I have a proof that there are no Diophantine solutions. In fact no Diophantine solutions exist for any odd number of vertices whether you allow symmetry of not. Define the parity of a point (a, b) to the parity of a+b. Suppose there is a Diophantine solution, and let the side be √n where n is minimal. Wlog say (0, 0) is a vertex and let (x, y) be an adjacent vertex with x and y integers. Then x²+y²=n, which implies (x,y) has even parity if n is even and odd parity if n is odd. If n is odd then this argument can be repeated to show that all pairs of adjacent vertices have opposite parity. But this is impossible if the number of vertices is odd, so n must be even. The argument now shows that all vertices have the same parity, and if we start at (0,0) then all the points are even. But the set of even points forms a sublattice of the integer lattice, and if we rotate the figure by 45° and shrink by a factor of √2 we get a similar figure with all integer coordinates. In this new figure the side is √(n/2), contradicting that n is minimal.

It's straightforward to show this implies there are no solutions with rational coordinates. The parity argument fails with a triangular lattice and it's fairly easy to construct asymmetric pentagons with vertices on this lattice. RDBury (talk) 15:27, 23 October 2023 (UTC)[reply]

This is IMO a convincing proof. The rotate-and-shrink operation can be described succinctly in a formula by

(m,n)\mapsto ({\tfrac {1}{2}}(m+n),{\tfrac {1}{2}}(n-m)).

Following Courant, I should have used the terminology that "all five vertices are lattice points".^[1] --Lambiam 09:17, 24 October 2023 (UTC)[reply]