See this link http://www.mapleprimes.com/questions/136257-Symbolic-Operations-With-Arbitrary-Matrices

Two drawbacks:

1) The perimeter of a triangle is equal to 2p, not p.

2) If S>0, then all the vertices of the triangle are the different points.

Fixed lines:

if type(2*p, integer) and type(S, posint)

then L:=[op(L), [[0, 0, 0], [x1, y1, z1], [x2, y2, z2]]]: fi:

From a geometrical point of view is actually found  only one triangle with sides 3, 4, 5. To find other such triangles  not looking for them in space but on the plane, because search range can be substantially increased.

It can be shown that there is no a triangle in space, for which coordinates of all the vertices are integers, with the angles 2*Pi/3 and Pi /4. So write the code for searching such triangles does not make sense!

1) If you want to [0,0,0] always been in the first position, then replace the braces on the brackets.

2) It's an obvious Maple bug.

Your main mistake: you look for only in the first octant in which there are no such triangles, as if two vectors out of the origin and lie entirely in the first octant, the angle between them does not exceed 90 degrees.

 Write code as follows:

N:=5:

L:=[]:

for x1 from -N to N do

for y1 from x1 to N do

for z1 from y1 to N do

for x2 from -N to N do

for y2 from -N to N do

for z2 from -N to N do

a:=x1^2+y1^2+z1^2:

b:=x2^2+y2^2+z2^2:

c:=a+b-2*sqrt(a*b)*cos(2*Pi/3):

if type(a, positive) and type(b, positive) and c=(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 then L:=[op(L), {[0, 0, 0], [x1, y1, z1], [x2, y2, z2]}]: fi:

od: od: od: od: od: od:

nops(L);

L;

Code in Maple 13:

line1:=20-0.05*x:

line2:=2+0.0002*x^2:

A:=plot([line1,line2], x=0..450, color=[red, blue], thickness=3):

B:=plots[implicitplot]((y-line1)*(y-line2),x=0..450, y=-5..50, coloring=[yellow, white], filledregions=true, numpoints=50000):

plots[display](A, B);

v:=Vector([5, -1, 6, 3]):
op(1, v);

4

Trivial examples:  Pythagorean triangles [[0, 0, 0], [2ab, 0, 0], [0, 0, a^2-b^2]] , where a and b - any unequal positive integers.

Nontrivial examples can be found using a simple code:

N:=10:

L:=[]:

for x1 to N do

for y1 from x1 to N do

for z1 from y1 to N do

for x2 to N do

for y2 to N do

for z2 to N do

a:=sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2):

if type(a, positive) and type(sqrt(x1^2+y1^2+z1^2)+ sqrt(x2^2+y2^2+z2^2)+a, integer) then L:=[op(L), {[0, 0, 0], [x1, y1, z1], [x2, y2, z2]}]: fi:

od: od: od: od: od: od:

nops(L);

L;

Yes, of course! See help for the  addcoords  command.

Your code can be significantly improved:

1) Since always  x2=x1+x3 and y2=y1+y3 and z2=z1+z3 , then variables x2, y2, z2  should not be written as a variables in a loop. This will speed up program execution.

2) You can exclude the  different trivial variants. If the points (x1, y1, z1) and (x3, y3, z3) lie in a coordinate plane, eg, xOy, then it will always be

(x1, y1, z1) = (x, y, 0) , (x3, y3, z3) = (-y, x, 0)  or  (x1, y1, z1) = (x, y, 0) , (x3, y3, z3) = (y, -x, 0)

(x and y - any numbers)

3) You can also foreclose variants, which are obtained from one another by simple transformations, such as reflections in the coordinate planes and axes.

4) Lists  [[0,0,0], [x1, y1, z1], [x2, y2, z2], [x3, y3, z3]]  and [[0,0,0], [x3, y3, z3], [x2, y2, z2], [x1, y1, z1]] give the same square, so from two lists, you can leave one list.

To implement  3) and 4) can be used  ListTools[Categorize]  command.

Write your 5 variables   (a,b,c,d,e)  in the list.

Example:

Case:= [ [1, 2, 3, 1.5, 6], [3, 3, 1, 3, 8], [5, 3, 3, 5, 3] ]:
Case[1], Case[2], Case[3];

       [1, 2, 3, 1.5, 6], [3, 3, 1, 3, 8], [5, 3, 3, 5, 3]

This problem can be solved similar to your problem of an equilateral triangle. Take advantage of the fact that if [[0,0,0], [x1, y1, z1], [x2, y2, z2], [x3, y3, z3]] are the vertices of a square, then x2=x1+x3, y2=y1+y3, z2=z1+z3. It is therefore sufficient to seek only two vertices [x1, y1, z1], [x3, y3, z3] .

Let us the solution to be non trivial if none of the vertices of other than [0, 0, 0] , does not lie in the coordinate plane, and the absolute values ​​of the coordinates of each vertex are unequal and integer numbers. The simplest nontrivial solution is  [[0, 0, 0], [6, 3, -2], [9, 1, 4], [3, -2, 6]] . Interestingly, the length of the side of the square is also an integer equal to 7.

If I understand your problem, it is simple. Should be in the loop to multiply all the elements of each row to the least common multiple of their denominators.

Code of the procedure:

DoIntegers:=proc(A)

local m, n, i, j, d, B, f;

uses LinearAlgebra;

m:=RowDimension(A);

n:=ColumnDimension(A);

for i to m do

d:=ilcm(seq(denom(A[i,j]), j=1..n));

for j to n do

B[i,j]:=A[i,j]*d;

od; od;

f:=(i,j)->B[i,j];

B:=Matrix(m, n, f);

end proc:

Example:

A:=Matrix([[1, 1/2, 1/3], [1/4, 1/5, 1/6], [7, 8, 9]]);

B=DoIntegers(A);

1) Your code is correct.

2) Since the triangle with the vertex at an arbitrary point can be obtained from the triangle with the vertex at the origin of the usual translation, there is no need to write new code.

3) Rewrote my previous code. Now the triangles, differing only in the numbering of the vertices are not distinct. In this code, the user himself determines the required number of triangles (parameter N). The search range is not fixed from the beginning, but gradually expands as new triangles. Therefore, if the number N is not too large, then the procedure is fast.

N_equil_triangles:=proc(N)

local x1, y1, z1, x2, y2, z2, k, n, a;

global L;

k:=0:

L:=[]:

for n while k<N do

for x1 from -n to n while k<N do

for y1 from -n to n while k<N do

for z1 from -n to n while k<N do

for x2 from -n to n while k<N do

for y2 from -n to n while k<N do

for z2 from -n to n while k<N do

a:=x1^2+y1^2+z1^2:

if a>0 and a=x2^2+y2^2+z2^2 and a=(x2-x1)^2+(y2-y1)^2+(z2-z1)^2 then

L:=[op(L),{[0,0,0], [x1,y1,z1], [x2,y2,z2]}]: k:=nops(L): fi:

od: od: od: od: od: od: od:

end proc:

Example:

t:=time():

N_equil_triangles(1000):

L[500..510];

Time=cat(convert(time()-t, symbol),`  sec`);

I'm not sure that all colors are identified correctly. Details can be improved:

R:=3: r:=1: h:=0.25:  # h specifies the width of the band

Colors:=["Red", "HotPink",  "Blue", "BlueViolet", "DeepSkyBlue", "LightBlue", "LightSteelBlue", "LimeGreen", "Lime", "Yellow", "Orange", "Gold"]:

E:=seq(seq(plottools[line]([(R+r*cos(11/3*t))*cos(t), (R+r*cos(11/3*t))*sin(t), r*sin(11/3*t)],[(R+r*cos(11/3*t))*cos(t+h), (R+r*cos(11/3*t))*sin(t+h), r*sin(11/3*t)], color=Colors[k+1], thickness=5), t=6*k/12*3.14..6*(k+1)/12*3.14, 0.01), k=0..11):

plots[display]( E, scaling=constrained);