A proof that the Area of a Parabolic Segment
by Todd Martin Miller, Ph.D. (Chem.)
April 17, 2014
This proof is based on the translation and work by Reviel Netz and William Noel in “The Archimedes Codex”, 2007, published by Da Capo Press (a member of the Perseus Books Group) in Philadelphia, PA, found to begin on page 150.
A parabola is “a plane curve formed by the intersection of a right circular cone with a plane parallel to a generator of the cone; the set of points in a plane that are equidistant from a fixed line and a fixed point in the same plane or in a parallel plane.” (Random House Webster's Unabridged Dictionary 1999). Many describe a parabola as the result of a cone sliced perpendicular to the axis of symmetry and at an angle, the planar object produced is a parabola. In this proof the parabolic segment is bounded on the bottom by a line segment to produce a closed figure parabolic segment.
Inside this closed figured parabolic segment, when lines are drawn between the end points of the base line, or directx line, and the farthest point from it: a triangle is created. The points of the triangle are labeled A, B, and C. A and C are endpoints of the base, directx line, and B is in the middle of the parabolic curve. The authors of “The Archimedes Codex” refer to the middle of the base, (directx line,) as point D and it is part of a line BD that is perpendicular to AC. Extending the line BD an equal distance beyond the parabola to a point E will establish a point on a tangent line from the end of the parabolic segment, at the base, directx line, beginning at C (or A). Doubling the length of that line (CE) produces a point Z, and the line CZ is a tangent segment to the parabola with E as its midpoint. A perpendicular line to A at A joining A to Z creates a larger triangle, the midpoint of which is found at a point K that intersects with the point found when the side of the triangle ABC, BC, is extended the same length as the line BC (the side of the triangle) towards the line AZ.
This establishes the picture on page 151 of “The Archimedes Codex” reproduced here using the GNU Image Manipulation Program (GIMP).
There is a linear relationship between distances (Axs produced by moving line segment MX perpendicularly along AC) and the area of the parabola that is not obvious. A directx is a line constructed by using lines from the focus of the parabola to points on the parabolic curve, and for each line making an angle so as the resulting lines are parallel and end in equal length on the line the focus point lies on. The result is a directx, a line upon which the length from an end to the point is the same as the radius of the line from the focus to the point on the parabolic curve perpendicular to the line called the directx. The operation in progress would look like spokes from point D radiating out to each point on the curve and each having a vertical line from that point to the base directx line.
The ratio between AC and AX is the same as the ratio between KN and KC. The points D and B are both in the middle of their segments, AC and KC, respectively, and the same is true of N and X for KC and AC, and also true of MX and OX since AX on AC is exactly proportional to OX and MX through the directx that is able to be over laid upon AC.
Archimedes' proof relies on a proof of “The Law of Balances” (ibidem, page 148) in which the ratio of the lengths of the lever's ends from the fulcrum are inversely proportional to the weights on each end at balance.
To use the Law of Balances, KC is extended as KT. SH, a line segment equal in length to OX is placed on the end of it to show how OXs will be transposed to T. An infinite number of OXs will be generated as an infinite number of them are determined as MX sweeps across the inside of the triangle and through the parabola.
The transposed OX segment is labeled SH, and the center of its gravity is at T. This is shown for one position of MX. At C, SH has a length zero and AX to AC has a ratio of 1 to 1. As MX moves from C towards A, parallel to AZ, the parabola is traced by segments of OX and finishes with its tip ( from C) at T as shown below
We know the triangle is 1/3rd of the way past K towards C at its center of gravity from another of Archimedes' proofs translated beginning on page 142 (ibidem) as the center of gravity is always located 1/3rd of the way from the midpoint of a side to the opposite vertex. ,The parabola is 3 times as long away from K since the distance from K to T equals the distance from C to K and C corresponds to the point T where the parabola is balanced. The distance C to K is 3 times the distance from K to the center of gravity of the triangle. That is the ratio of their weights. The parabola balances the large rectangle ACZ in a 3 to 1 ratio and the weight of the triangle is three times the weight of the parabola, the weight of the parabola is 1/3rd of the triangle.
The reason the center of gravity of the parabola is not used is the ratio of AX to AC and a problem with the diagrams. To balance the parabola, begin at C, where OX is zero, XC is zero CX and points M, X, and C are coincidental instead of from A towards C. Beginning at the right of the figure, the two figures, the one that will be CAZ and the other that will be the parabola after MX is moved to AX equal zero area and they will balance in the beginning since a zero area parabola, cut by MX and a zero area triangle, cut by MX have zero area. If I started at A, using X to determine the distance, the parabola would hang over the lever and it would not look balanced. Since they balance in the beginning, the farther to the left I go inside the triangle using MX, the more weight is added to each side to keep the areas in balance.
Another change that makes this proof clearer is moving the segment that begins at the fulcrum K and ends 1/3rd of the way towards C , the lever's on the side of the large triangle, outside of the triangle as shown below.
I did this proof to understand Archimedes' method for the area of a parabola to be determined. I found it difficult, and had to know about the directx line to understand it. I hope those who helped me will be glad to see this finalized on this website, and for those who missed parts of it while it was being done, are able to understand it themselves as they may have had less time than I did to work on it.
I can only do this problem with the entire human race cooperating. I believe the integration in Calculus is only possible the same way, and the integration in politics is the same integration in Calculus. Nobody can work on difficult problems in their minds while others are fighting, and without it some human beings could make this work impossible.