@Alfred_F The proof I know uses a Hamel basis. If we are given such a basis, the proof is "constructive", but for the basis itself the Axiom of Choice is needed.

@JAMET Actually the fixed point is the symmetric w.r.t. O  of the projection:

solve({x/a+y/b=1, a*x-b*y=0},{x,y});

        {x = a*b^2/(a^2 + b^2), y = b*a^2/(a^2 + b^2)}

@dharr Unit triangles are not enough.
Actually I forgot to mention that the 8 points are also not enough, but 10 are.

For example:  

Here monochromatic unit triangles do not exist.
If we remove the upper vertex to obtain a 9 point configuration, there are no monochromatic triangles!

@dharr Very nice proof. It should be converted into an answer!

However some corrections are needed. The triangles you find are only those with length=1, and the color of a monochromatic triangle could be 0 or 1.

@Christopher2222 

https://math.stackexchange.com/questions/3618369/show-that-there-are-3-vertices-of-the-same-color-forming-an-isosceles-triangle 

https://www.quora.com/The-points-of-the-plane-are-colored-in-two-colors-red-and-blue-show-that-there-are-three-points-of-the-same-color-vertices-of-an-equilateral-triangle-How-do-I-solve

I'll stop here!

@EugeneKalentev  The "source" is wrong, the integral is obviously +oo  for a>0. 

@one man Assuming that a plex basis G can be computed in a reasonable amount of time and memory (I do not have a very powerful computer) then one of the polynomials in G is univariate (in x4 in our situation), so that all the solutions are easy to compute.

@Alfred_F 

1. Accepting the existence of a maximal area polygon, the fact that it is cyclic follows easily from the elementary case n=4.

2. The polygon being cyclic, any 3 distinct vertices A[i], A[j], A[k] will do.

@MaPal93 Forget abaut Maple, think mathematically:

 d/dX[i] sum(X[i]^2, i= 1..3)  =  d/dX[i] ( X[1]^2 + X[2]^2 + X[3]^2 ) 

= d/dX[i]  X[1]^2 + d/dX[i]  X[2]^2 + d/dX[i]  X[3]^2 = ?

The last sum should be 2X[i]  if i is in {1,2,3} and 0 otherwise, but it is much better to avoid such expressions in Maple; for Sum d/dX[i] is applied to the summand.

@MaPal93 

DXI = diff(OBJ, X[i]);
has two problems:
--   you want :=  instead of  = 

--   diff(OBJ, X[i])  cannot work well (actually it makes no sense) when X[i] appears in sum( ..., i=1..n)

@MaPal93 

factor(expand(value(DB_wrong-DB_correct)));

So,  DB_wrong <> DB_correct;  DB_wrong is correct! 

You have changed the code with something unexplained!

@MaPal93 X__i and X__r are just symbols, they do not depend on the variables i and r.
So, for example sum(X__i, i=1..3)  simplifies to 3*X__i  and appears as  3 X_i

Use only subscripted names X[i]; avoid  X__i which is used only for typesetting purposes.

@Alfred_F The computations are easy (by hand) and are not Maple related. 
The perimaters can be arbitrarily large. By replacing a portion of a ray by an isometric copy of the graph of x -> b*x*sin(1/x), x in [0,a], the perimeters will be infinite.