The areas of the triangles in Archimedes’ construction: Take the parabola y=x^2, and cut off the segment between x=−1 and x=1. The first triangle has vertices at the ends and midpoint of the parabola. Between each of its lower edges and the parabola we insert a new triangle whose third vertex is also on the parabola, halfway (in x value) between the other two. Then we repeat the process with the lower edges of the new triangles. This creates successive “generations” of triangles, with each triangle in generation n+1 having two vertices from a triangle in generation n and its third vertex also on the parabola halfway (in x value) between them. The area of generation n + 1 is 1/4 of the area of generation n, so we can find the area by summing a geometric series: area of parabolic segment(1+1/4+1/4^2+1/4^3+...)x(area of the first triangle)=(4/3)x(1)=4/3
Given that the triangle with vertices (which lie on the parabola $y = x^2$) $(a,a^2), (\frac{a+b}{2}, (\frac{a+b}{2})^2), (b, b^2)$ has area $(\frac{b-a}{2})^3$, deduce that each triangle in generation $n$ has area $2^{3-3n}$and hence that the total area of generation $n$ is $2^{2-2n}$.