https://www.youtube.com/watch?v=TyZCM76Sx4gOP stole from this, there are infinite answers in the interval (20, 32)
The problem of dividing a square (3:17) has infinitely many solutions, because there can be infinitely many “branch point” positions that give the correct solution depending on the length of the side of the square. The length of the side of the square is a free variable in this problem.
The video shows only one particular solution, but does not show any condition that leads to this solution.
Let's denote:
a) the known area of the polygons as A1=16, A2=20, A3=32;
b) the area sought as Ax,
c) the length of the side of the square as a.
Now, if we assume the field proportions from the figure, i.e., assume A2<Ax<A3 (i.e., 20<Ax<32), then any solution for Ax in the interval (20;32) is correct when 2*Sqrt(22)<a<10 and each solution corresponds to one value of a.
The proof of the above statement uses the equality of the areas of the triangles, which are built in the lower half of the square (in this case, one needs to assume that the length of the base of the triangles is known).
For example, Ax=24 is the correct solution for a=2*Sqrt(23).
On the other hand, for Ax=28, we have a=2*Sqrt(24).
In the general case, we have 0<Ax<Sqrt(a). So, any number in the interval (0;Sqrt(a)) is a valid solution, provided that the “branch point” can lie anywhere in the square.
This is the complete solution to the problem of dividing the square.
The video shows a particulatr solution with implicite assumption that a=2*Sqrt(24).
