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@Zeineb Maple can help to compute it, but only if one uses the right theorem.
 

restart;
alpha := x->floor(x):
f:=x->x^2+1:
int(f(x)*diff(alpha(x),x), x=0..3);  # wrong
#                               0
f(3)*alpha(3)-f(0)*alpha(0) - int(alpha(x)*diff(f(x),x), x=0..3); # correct
#                               17

As I have mentioned, Int_[0..3] alpha d alpha   does not exisist.

If P[1], ..., P[4] are your points, consider a fifth point P[5] of unknown coordinates, say (a,b).
Then take the conic (via its quadratic equation) thru P[1], ..., P[5] and impose the condition to be a parabola.
In general there will two parabolas, but it is also possible to have 0 (!). E.g. a parabola cannot pass thru the vertices of a square.

Student:-Basics:-LongDivision(A,B,x, output=printf);

                4/3*x^2+80/3*x+1552/3
              -----------------------
3*x^2-60*x+36 ) 4*x^4
                4*x^4-80*x^3+48*x^2
                -------------------
                     80*x^3-48*x^2
                     80*x^3-1600*x^2+960*x
                     ---------------------
                           1552*x^2-960*x
                           1552*x^2-31040*x+18624
                           ----------------------
                                 30080*x-18624
 

You must export in a directory with write permission. By default the current directory is Maple's directory (without such permission).
So, you must change the directory, e.g. use "c:/temp/Test.jpg", (if you have such a directory),
or, execute first
currentdir(kernelopts(homedir));

Your second equation Y, equivalent to Y = 0,  reduces to a = 0, and does not depend on m. So, you have nothing to eliminate.

For higher-order functionals, use variables to represent derivatives. E.g.
L1:=eval(L, [diff(x(t),t,t)=diff(x1(t),t), diff(theta(t),t,t)=diff(theta1(t),t)]):
L2:=L1 + A*(diff(x(t),t) - x1(t))^2 + B*(diff(theta(t),t) - theta1(t))^2:
EulerLagrange(L2, t, [x(t), theta(t), x1(t), theta1(t)]);
 

Such an assignment is not possible due to the standard evaluation in any CAS: psi(t) would be evalaluated to psi(t)(t)(t)... (i.e. a nonsense).
It is possible to use macro (for input only)  or alias (for both input and output). You must be aware that in both cases psi(a) will be evaluated to psi(t)(a).

# x,y,z = the unknowns, top to bottom
# h1,h2,h3  = halves of the heights 
# a,b,c,d,e = the bases of the triangles
restart
eq:=
[260=a*h2, 247=b*h2, 287=d*h3, 209=e*h3,
x=a*h1, y=c*h2, z=(b+c)*h3,
h1/(h1+h2)=a/(b+c), h1/(h1+h2+h3)=a/(d+e)]:
S:=eliminate(eq,{h1,h2,h3,a,b,c,d,e}):
elim:=S[1]:
XY:=S[-1]:
XY1:=[isolve(XY[1])]:
xy:=select( u -> (eval(x,u)::posint and eval(y,u)::posint), XY1):
for v in xy do
  zz:={solve( eval(XY[2],v) )};
  for z1 in zz do  if z1::posint then print(v[],'z'=z1, eval(elim,[v[],'z'=z1])) fi od;
od:

So, two solutions found.
I don't see any relation with the star (100). Maybe I miss something.

It often works.

restart;
r := x*(diff(theta(t), t))^2+y*(diff(varphi(t), t))^2+z*(diff(theta(t), t$2))+w*diff(varphi(t), t$2):
g := (4*(f+T))*(diff(theta(t), t))^2+u*(diff(varphi(t), t))^2+(f+9)*(diff(theta(t), t$2))+4*s*diff(varphi(t), t$2):

eval(r-g, [theta=(t->t^3), varphi=(t->t^7)]): # or similar
solve([coeffs(%, t)], [x,y,z,w]);
simplify(eval(r-g, %[1]));   #check
#          [[x = 4*f+4*T, y = u, z = f+9, w = 4*s]]
#                               0

A brute-force search shows that the problem has no solution with at a least one tuple (xi,yi,zi) from the set {-N,...,N}, N = 40.
It shoud not be too hard to prove this in general (mathematically).

It's only a display decision when interface(typesetting=extended) .
For interface(typesetting=standard)  ==>  a^0.5

`@W`  is a name (symbol)  enclosed in back quotes. It could be used as a generic procedure, or maybe a procedure `@W` is defined somewhere. Such a name can be used in 2022 too!