vv

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@dharr 

for a=b=c=1 there is the valid solution x=y=z=k, for any k<>0.

@Alfred_F There is no chance to use GroupTheory. We would need RingTheory which Maple is missing.
But the idea was to show that Maple can be used to find the result and its proof, in case we don't know it.

@mmcdara Nice and much faster execution.
Note that my intention was to minimize: time(think) + time(code) + time(execute).

@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.

 

Int(1/( (a^2+x^m)^n * x^p ), x=0..infinity);

Int(1/((a^2+x^m)^n*x^p), x = 0 .. infinity)

(1)

ans:=simplify(value(%)) assuming a>0;

a^((-2*m*n-2*p+2)/m)*GAMMA((-p+1)/m)*GAMMA((m*n+p-1)/m)/(m*GAMMA(n))

(2)

I'll stop here!

J:=Int(1/( (a^2+x^2)^(3/2) * x^(1/2) ), x=0..infinity);

Int(1/((a^2+x^2)^(3/2)*x^(1/2)), x = 0 .. infinity)

(1)

'J'=simplify(convert(value(J),GAMMA)) assuming a>0

J = (1/2)*Pi^(3/2)/(a^(5/2)*GAMMA(3/4)^2)

(2)

evalf('J'=rhs(%));

J = 1.854074678/a^(5/2)

(3)

 

@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)

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