Proofs of Archimedes’ Parabolic Triangles Results

Proofs of Archimedes’ Parabolic Triangles Results


Proof: If r is the x-coordinate of the point C, there is a positive number h such that the x-coordinate of A is r – h and the x-coordinate of B is r + h C is the point of the parabola directly below the midpoint of the line segment AB. We need to prove that the area of the triangle ABC depends only on h and not on r. The coordinates of A, B and C are given by:

We will apply the following formula for the area of a triangle T with vertices at the points (a, b), (c, d), (e, f):

Apply this formula to the triangle ABC with the coordinates of A, B, and C given above and use the properties of determinants to obtain:

This shows that the area of ABC depends only on h and not on the location of the “shadow” of the interval AB on the x-axis. It also shows that the total area of the two new triangles, ACJ and CBK, is one-fourth of the area of ABC. For the “shadows” of
the triangles ACJ and CBK on the x-axis are half the length of the segment EF, so the total area of these two triangles is:

Archimedes continued this process by creating a new parabolic triangle on each of 4 line segments AJ, JC, CK and KB, with the total area of the new parabolic triangles being one-fourth of the total area of the triangles ACJ and CBK, or one-sixteenth of the area of the original parabolic triangle ABC. In this way, he “exhausted” the area of the parabolic section above the parabola and below the line segment AB. He concluded from this that the area of this parabolic section is 4/3 of the original parabolic triangle ABC. You are asked to verify this fact in Homework Problem 1 by using the formula for the sum of an infinite geometric series.