@Rouben Rostamian  Thanks for finding this example!  I tested the  Into_4_Equal_Areas  procedure on this triangle and it found both solutions. To get symbolic solutions I set in the code Digits=15 and then set in identify command to evalf[8] , because unfortunately fsolve doesn't work very well, when calculating with 15  digits it returns  0.166666631418626  instead  1/6=0.166666666666667 .

L:=[[[0,0],[1,0]],[[1,0],[1/2,8/9]],[[1/2,8/9],[0,0]]];
Sol:=Into_4_Equal_Areas(L);
identify(evalf[8](Sol));
plots:-display(<Pic(L,Sol[1]) | Pic(L,Sol[2])>);

@Alfred_F  Thank you for your interest! This is a very important question about the number of essential solutions for a given flat region (solutions are called essential if they do not transform into one another during movements). It seems that for regular polygons, whose number of sides is divisible by 4, there are infinitely many essential solutions. I do not know of any other example where the number of essential solutions would be greater than 1.

@Rouben Rostamian  Thank you for your interest and high rating. The procedure will work correctly if there is a solution in which the cutting lines intersect the boundary of the domain at exactly 4 points. If the domain is "strongly non-convex", then this condition is not satisfied. For example, in the example below, a solution exists (it is easy to find without using the procedure), but the procedure does not work here, because the cutting lines intersect the boundary at more than 4 points.

The latest result for Maple <=2021 is also not quite correct, part of the parabola lies above the straight line (plotted in Maple 2018.2):

plots:-display(plots:-implicitplot(y=x^2, x=-1..1, y=0..(x+1)/2),plot((x+1)/2, x=-1..1), color=[red,green]);

Edit. Here is another example of incorrect plotting (Maple 2018.2):

restart;
plots:-display(plots:-implicitplot(y=-4*x^2+1, x=-1..1, y=0..(x+1)/2),plot((x+1)/2, x=-1..1), color=[red,green], scaling=constrained, size=[500,300]);

There are no errors in Maple 2018.2. The solution is expressed in terms of integrals of Kummer functions. Of course, we can solve this equation numerically if we specify the initial conditions.

Download dsolve_factor_dec_24_2024_new.mw

This problem can be solved with no calculations, if you make some additional constructions (dashed lines in the figure below). The inscribed angle FAC is 45 degrees. But the angle FEC subtends the same arc.

@Alfred_F  Good question! Obviously it won't change at all (a  is the length of the side of the square).

restart;
M:=<a/2,0>: C:=<a,0>: MB:=y=2*(x-a/2):
Circle:=x^2+y^2=a^2:
solve({MB,Circle},{x,y}):
E:=eval(<x,y>,%[2]):
EM:=M-E: EC:=C-E:
Angle_CEM=simplify(arccos(EM.EC/sqrt(EM.EM)/sqrt(EC.EC))) assuming real;

restart; # Clear all variables 
expr := (x^3 - 2*x^2 + x) / (x^2 - 1); 
simplify(expr);

@MaPal93 If both parameters are not simultaneously 0, then the polynomial  x  will always be positive for any  n>1  (not necessarily an integer). This is easily verified (see the code below), since  4*n^3*u^2 - 3*u^2>0  and  2*n^2*u*v-2*n*u*v>0  for any  n>1 :

Download x.mw

@Alfred_F  In fact, the full code for drawing these 64 hexagons, as well as the second drawing, is in my answer. I did the actual check for the presence of the necessary equilateral triangles in each of the options manually, simply running my eyes over these options. Of course, this can also be automated, but then the code would be unnecessarily complicated. I provide the code for the last drawing (in edit):

P1:=display(P[28],pointplot([0,sqrt(3)],symbol=solidcircle,symbolsize=40,color=red)):
P2:=display(P[28],pointplot([0,sqrt(3)],symbol=solidcircle,symbolsize=40,color=black)):
display(<P1 | P2>);

@Ronan  It seems that the problem is in using  DocumentTools:-Tabulate command. If you use only  DataFrame , then everything is OK:

restart;
QQFProj := proc(q12::algebraic, q23::algebraic,
                q34::algebraic, q14::algebraic,prnt::boolean:=true)
  description "Projective quadruple quad formula and intermediate 13 and 24 quads. Useful for cyclic quadrilaterals";
  local qqf,q13,q24, sub1,sub2,sub3, R,values,DF,lens,varp;
  uses   DocumentTools;
  sub1:= (q12 + q23 + q34 + q14);
  sub2:=-4*(q12*q23*q34);
  sub3:=64*q12*q23;
  qqf:=(sub1+sub2)^2-sub3;
  q13:=(q12-q23)^2;
  q24:=varp*(q23-q34)^2;
 varp:=alpha;
  if prnt then
  
   values:=<qqf,q13,q24>;
   DF:=DataFrame(<values>, columns=[`"Values Equations"`],rows=[`#1  QQF`,`#2  Q13`,`#3  Q24`(varp)]);
   lens := [4 +8* max(op(length~(RowLabels(DF)))),4+ min(max( 10*(length~(values))),1000)];#op(length~(ColumnLabels(DF)0)
print(DF);
#   Tabulate(DF,width=add(lens),widthmode = pixels,weights = lens);
  return qqf,q13,q24
  end if;
  return qqf,q13,q24
end proc:
 q12:=1/2:q23:=9/10:q34:=25/26:q41:=9/130:#Cyclic quadrilateral
q12:=sqrt(17+a)/2:q23:=(r^2+t^2)^2/10:q34:=((a+b+c)^4/26):q41:=sqrt(17+b)/130:
Q:=QQFProj(q12,q23,q34,q41,true):

@Carl Love  The widest range for the parameter  m  is easy to find. It is determined by the slopes  m  of two tangents to the circle c2 passing through  J .

Download latex_of_laplace_nov_21_2024_Maple_2018.2.mw

@EugeneKalentev  Yes, I accidentally missed  x  when I was typing the code in Maple. The correct answer was given by vv . 

@Blanc But you omitted thу  assuming  option in your solution. Just copy the code from my answer as text and paste it into your worksheet.