Archimedes Uses a Lever to Prove Geometry Theorems

This post follows on from these two posts on Archimedes (287 BCE-212 BCE):

• The Rediscovery of The Archimedes Palimpsest and His Use of the Infinite

• André Koch Torres Assis And Ceno Pietro Magnaghi Write About Archimedes's Method

In the latter post, I wrote that André Koch Torres Assis and Ceno Pietro Magnaghi had published a small book entitled The Illustrated Method of Archimedes: Utilizing the Law of the Lever to Calculate Areas, Volumes and Centers of Gravity¹, in which they presented several mechanical proofs, as opposed to geometrical proofs, of Archimedes’s results in geometry, and that they had published a Portuguese translation and commentary of his Method².

This approach to proving mathematical theorems was also presented in Chapter 8 of Reviel Netz and William Noel’s book The Archimedes Codex: How a Medieval Prayer Book Is Revealing the True Genius of Antiquity’s Greatest Scientist³; that specific chapter was written by Netz.

In this post, I will focus on one specific proof found in Archimedes’s Method and which is discussed in both of the referenced books. It corresponds to Proposition 1 in Thomas Heath’s (1861-1940) book The Method of Archimedes Recently Discovered by Heiberg: A Supplement to the Works of Archimedes, 1897⁴.

What is interesting about Archimedes’s proof is that he will materialize geometric constructs, such as lines, triangles and parabolic segments. In other words, he will manipulate these abstract objects as if they were material. This is opposite to the way that we commonly understand mathematics, i.e., as a means to abstract away from the physical.

In particular, Archimedes will balance, using his famous law of the lever, geometrical objects of the same nature. For example, two lines AB and CD of equal lengths will exactly balance out each other on a lever whose fulcrum is the center of the bar; should AB be twice as long as CD, then they would balance each out other if line CD were placed twice as far from the fulcrum as is placed AB. Similarly, two-dimensional surfaces—in this case triangles and parabolic segments—can also be hung from a lever.

In the presentation below, all of the diagrams but one come from the book The Illustrated Method of Archimedes; the last comes from the book O Método de Arquimedes. These diagrams were graciously sent to me by Prof. Assis.

In the first diagram we see a parabola ρφγ with vertex φ, diameter φη and chord ργ perpendicular to the diameter, divided into two equal parts at η. Although not illustrated, the parabola obviously continues beyond ρ and γ.

Archimedes considered the general case of a parabolic segment αβγ with chord αγ inclined relative to the diameter φη. The point δ is chosen so that divides αγ into two equal segments, and the line βδ is drawn parallel to φη. [Figure 4.1, p.22]

Archimedes proceeds to demonstrate that the area of the parabolic segment αβγ is 4/3 of the area of the triangle αβγ. This result holds both in the specific case where βδ coincides with φη (case (a) below), i.e., when chord αγ is perpendicular to the diameter φη, and in the general case when βδ and φη do not coincide (case (b) below). [Figure 4.2, p.22]

The proof begins with the construction of the diagram below.

A line parallel to line δβ is drawn from point α, to meet at point ζ the tangent to the parabolic segment αβγ at point γ. Line δβ is extended to meet tangent γζ at point ε. Line γβ is extended to θ, crossing line αζ at point κ in such a way that γκ is equal to κθ.

An arbitrary point ξ is chosen along chord αδγ, and then a line ξoνμ is drawn parallel to δβε, crossing the parabolic segment at point o, line γβκθ at point ν, and tangent γεζ at point μ.

Finally, a line ητ is drawn parallel to line αζ, with center θ and of the same length as line ξo. [Figure 4.3, p.23]

Now we can begin the reasoning. First, since δ is at the midpoint of αδγ, and κ is at the midpoint of ακζ, it follows that the area of triangle αβγ is one quarter the area of triangle αζγ.

From a property proven by Archimedes in the Quadrature of the Parabola [Heath, p.16], we have μξ : oξ = γα : ξα = γκ : κν = θκ : κν.

So, since τη = oξ, we have μξ : τη = θκ : κν.

So now we are ready for the introduction of the lever, as illustrated below.

The last equation can be summarized by considering the line θγ as the bar of a lever whose fulcrum is at κ, and then balancing the lever by placing μξ on the bar at distance θκ, and τη on the bar at distance κν. [Figure 4.4, p.24]

Now it is important to remember that the point ξ was arbitrarily chosen along chord αδγ. This means that the same balancing effect applies for every point along chord αδγ. So what we want to do is to balance out on a lever the parabolic segment αβγ and the triangle αζγ. To do this, we need to know the center of gravity of the triangle, which is the point χ corresponding to the intersection of the bisectors of each of the angles α, ζ and γ. This point χ lies on line κγ such that κγ/κχ = 3. This gives us the following diagram:

Here is how Assis and Magnaghi summarize the diagram:

The horizontal lever remains in equilibrium about κ with the segment of parabola αβγ suspended by a weightless string at θ, while the triangle αζγ is suspended by a weightless string at χ such that κχ = κγ/3. [Figure 4.6, p.25]

In their Portuguese translation of the Method, Magnaghi and Assis actually build a physical version of this lever in equilibrium.

Here is a translation of their caption:

Lever in equilibrium with the parabolic segment and the triangle supported only by their centres of gravity. [Figura 10.8, p.80]

It follows that the area of triangle αζγ has three times the area of parabolic segment αβγ. Hence the area of the segment is 4/3 of the area of triangle αβγ, which is what we were trying to prove.

Archimedes also uses the law of the lever to prove results about three-dimensional geometrical objects. In particular, he proves:

That the volume of any sphere is four times that of the cone which has its base equal to the greatest circle of the sphere and its height equal to the radius of the sphere, and that the volume of the cylinder which has its base equal to the greatest circle of the sphere and its height equal to the diameter of the sphere is one and a half times that of the sphere. [p.27]

The illustrated presentation of this proof can be found in The Illustrated Method of Archimedes. Assis also recorded, for an online conference, a video of this proof, in Portuguese. It is available on YouTube at the following link:

https://www.youtube.com/watch?v=CzY6BOYirPw&t=16080s

The slides for the talk are available here:

http://www.ifi.unicamp.br/~assis/Arquimedes-24-02-2022.pdf

What is fascinating is that these mechanical proofs of geometry theorems seem to have been Archimedes’s original proofs. His geometrical proofs of these results came after his mechanical proofs. For Archimedes, it seems, mathematics, science and engineering was one complex whole.

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1. Andre Koch Torres Assis and Ceno Pietro Magnaghi. The Illustrated Method of Archimedes: Utilizing the Law of the Lever to Calculate Areas, Volumes and Centers of Gravity. Montreal: Apeiron, 2012.

2. Ceno Pietro Magnaghi and André K.T. Assis. O Método de Arquimedes: Análise e Tradução Comentada. Montreal: Apeiron, 2019.

3. Reviel Netz and William Noel. The Archimedes Codex: How a Medieval Prayer Book Is Revealing the True Genius of Antiquity’s Greatest Scientist. Philadelphia: Da Capo Press, 2007.

4. Thomas L. Heath. The Method of Archimedes Recently Discovered by Heiberg: A Supplement to the Works of Archimedes, 1897. Cambridge University Press, 1912.