>>11012797Consider a curve y = f(x). We are interested in knowing how big the square under the curve is, presupposing that the square is contained on its left by x = a and both sides must be equal in size. The diagonal of the square has to touch the curve, it shall not cut it.
Eq 1 -> y = x-a, Eq 2 -> y = f(x)
Obviously x-a = f(x).
Suppose now a = -1 and f(x) = (1-x^2)^0.5.
We get from Eq1 and Eq 2 -> x+1 = (1-x^2)^0.5 or -(1-x^2)^0.5+x+1 = 0. Now, I don't care about solving it but it's obvious that it contains two roots because doubling it results in a polynome of the form ax^2+bx^1.5+cx+dx^0.5+e = 0, which must have 4 different roots as becomes clear by substituting x = y^2. The roots of the equation -(1-x^2)^0.5+x+1 are in fact -1+(sqrt(2)-1)*cos(45°) and 1-(sqrt(2)-1)*cos(45°).
The point of this whole post is showing how any such problem can be reduced to polynomic equations whose roots and solutions are hardly non-trivial.
