*Supplementum Geometriae,*1593 as described by Robin Hartshorne)

*approaches to trisecting a general angle. I was first struck by how Burgi's approach seemed to be very algebraic, while Viete, father of algebra, uses a ancient geometric approach with the neusis, (sort of a marked straight edge approach) .*

In addition, even though I own Smith's 'Sourcebook', and had probably read the Pitiscus/Burgi article several times, I had never worked through his actual math. Doing so left me somewhat confused, confounded, and perhaps enlightened.

Links to both documents on the internet are below.

First, I will explain Viete's use, somewhat amplified to make it clearer to younger readers, and a bit about why he might have chosen this method. Then I will do somewhat the same with Burgi's method as presented by Pitiscus.

Viete explains clearly that he will extend the classical compass and straightedge approach by allowing, "from any point to any two straight lines, to draw a straight line cutting off between them any segment fixed in advance."

**The Neusis Approach of Viete (and Archimedes)**

Students who may hear about Viete's introduction of letters as variables in mathematical operations may not realize he was an excellent geometer. In fact he chose to follow a homogeneity of dimension restriction that kept him from addressing simple quadratics like x^2 -2x = 3 since it involved adding the area of a square to the length of a line. His ability to adjust such problems to become problems of proportions, so that x^2 + Zx = B^2 where the S and B were considered as known values (or pronumerals) allowed him to effect the same results. But in this case, he was seeking a solution to cubics, the so called "casus irreducibilis", and trying to avoid the use of Cardano's imaginaries (complex numbers) which were still unexplained geometrically. Caspar Wessel's paper on the geometric meaning of the complex numbers would be printed within a few years (1597) but even then, it would make hardly a ripple in the world of math outside his home country. Wessel was as much unknown as Cardano was known.

The relation of trisecting to the cubic equation \(x^3 - 3x = 2a)\) was well known, and Viete knew that with a suitable change of variables, it could be used to solve any cubic with all real roots.The method he would use dates back to the early Greeks, and was known and used by Archimedes.

I have tried to illustrate the "neusis" approach with static pictures from a very dynamic program, but teachers who are interested in making this clearer to their students can easily construct an interactive model for their students.

We begin with an isosceles triangle with a base angle at the origin and the angle to be trisected and the vertex on a circle centered at the origin. The marked ruler is represented her by the ray EG, and for our purpose has G fixed on the circle passing through the top vertex of the triangle. In use, the mathematician would lay his straightedge along one of the two congruent sides of the triangle, and mark EG so that it was the length of a side. Then holding E on the x-axis, and G on the circle, he would move the points along the respective curves until the straight edge passed through the vertex angle at C . The resulting at \(\angle \) AEG is the trisector.

For ease in seeing a simple proof, I add one drawing with a second isosceles triangle shown:

There are actually three isosceles triangles in the figure, the two shaded ones, \(\Delta \)ACH , and \(\Delta \)AGE and the one between them \(\Delta \)CAG. If we label the \( \angle \) CAH to be trisected as y

**,**then it's supplementary angle, \( \angle \) CAE is 180 - y. Now if we call the equal angles \( \angle \)AEG and \( \angle \)GAE each x, then the exterior angle \( \angle \)AGC to this isosceles triangle must be 2x, as must it's base angle partner \( \angle \)ACG. This leaves \( \angle \) CAG as 180 - 4x. Adding back the value x at \( \angle \)GAE means that \( \angle \)CAE is equal to 180 - 3x, but we already know it is equal to 180-y, so y=3x as we sought.

**Burgi Using Brahmagupta's theorem and double false position.**

Writing at the same time, Pituscus describes Jost Burgi's approach with a much more algebraic or arithmetical approach. They will create a complex construction involving Brahmagupta's theorem for inscribed quadrilaterals, then make an two educated guess of the trisecting angle's chord, then use the errors and a linear approximation technique called the rule of double false position to find the chord of the trisected angle. The idea of double false position as used by Piticus/Burgi is almost certainly worth showing to students for the historical value alone. Fibonacci described double false position in his well know “Liber Abaci” , using it's Arabic name, and says the method is the one “by which the solutions of nearly all problems are found." I would think a simple illustration to students is essential before exposing them to the approach used by Burgi and is well within the grasp of any student who has worked with linear equations.

**Double False Position**

Here is an example using a linear equation for which the method is exact. For non-linear functions, like the trigonometric functions, it serves as a method of approximation.

Suppose we take a simple expression like 5x-3 = 7 and have no idea how to solve it. We might begin with a guess of x=1. Trying that out we get 5(1)-3 = 2, an error that is five too low, or -5. We decide to guess larger, and try x=4. This time 5(4)-3 gives us 17, which is 10 above our target. The method of double false position was to take the cross products of each trial value, and the opposite error, and find their difference, then divide the result by the difference in the errors. If we label the x values x

_{1}and x

_{2}and the errors e

_{1}and e

_{2}, then the equation can be written as \( \frac{x_1 * e_2 - x_2 * e_1}{e_2 - e_1}\). If we apply this to the trials above we get \( \frac{1 * 10 - 4 * (-5)}{10 - (-5)}\), and to none of our surprise, it tells us the correct value of x was 2.

As to why Burgi would have used this approach, when he might have used the much simpler neusis method of Archimedes and Viete, I think it was a matter of focus. Burgi was an astronomer and precision clock maker who had just before this writing, made an astronomical clock which was based on the Copernican system, and along the way created a table of sines (chord lengths) for astronomical use, his

*Canon Sinuum*which seems not to have been published

*.*Burgi had independently invented logarithms, almost certainly prior to Napier, as an aide to his computations. I believe that Burgi, and Pitiscus use of Burgi in his Trigonometria, was for the purpose of computing chord lengths for Sine Tables. To obtain more accurate chord lengths, astronomers increased the radius of their calculating circles to seven or more digits. In the time before decimals were common (although Viete supposedly, "

*... wrote decimal fractions with the fractional part printed in smaller type than the integral and separated from the latter by a vertical line."*) the use of a large radius allows the use of a chord to be expressed with greater accuracy and in integer values. When Pitiscus uses 5176381 for the chord length of a 30

^{o}arc, he is giving what we would now write as .5176381 for 2 sin(15

^{o}).

**Burghi's Use of Brahmagupta's Thm**

Burgi realized that if you start with the chord of an angle (in this case, the chord of AB in \( \angle \)ADB) and double the angle by adding another chord of the same length, (BC, thus making \( \angle \) ADB congruent to \( \angle \)BDC, and the chord of \( \angle \) ADC would be AC). Then with one more congruent chord (CD) we have the chord of triple the arc of AB, and thus an angle AD is the chord for an angle three times \( \angle \) ABD (you can pick the point P for this APD anywhere on the arc AD).

With this groundwork, Burgi recognized that the four points form an inscribed quadrilateral, and so Brahmagupta's theorem would apply. That is, in this case, AC * BD = AB*CD + AD * BC. But since AB = BC = CD, and AC = BD we can rewrite that as \( AC^2 = AB^2 + AB * AD \). And by a little algebra we arrive at \( \frac{AC^2 - AB^2}{AB} = AD \)

Now to trisect the angle, we will be able to construct the chord AD from the given angle, and we want to find the length of the chord that would subtend 1/3 of that angle, that is, the length of AB.

**The Mystery**

As this point, Pitiscus says to, take a third of the chord, add something to it (the chords of the three equal angles are simply longer than the chord of the triple angle, so we must have something a little larger), and use the method of problem 3 (Brahmagupta's thm) to compute the given chord. He knows this will have some error, and so he will repeat it again to get a second error and use the method of double false position to find a close approximation. But...... I can see no way he knew AC, or AC

^{2}. If we are building a table of chords, and we knew AC or AC

^{2}and hence the chord of 2/3 of the angle, we would have bisected this angle to begin with rather than trisecting the original angle since such a method was known at least as early as Hipparchus . And Burgi was an expert in prosthaphaeresis, the use of combining addition and subtraction of products to sums and is often credited with inventing at least two of them. How could we... how could HE carry out this operation?

**Doubt and Suspicion**(or guesses and wild conjectures)

Immediately after the explanation of this problem, he gives "another method by algebra" and described the relation which in modern language would be x^3-3x =2a. Written in terms of chord lengths this would be crd3δ = 3crdδ−crd

^{3}δ which might have been known as early as Ptolemy, and which Burgi and Pitiscus both well knew.

For the example of a 30

^{o}inscribed angle, we know that the chord length is about .5176381 (or 2 sin(15

^{o}) . Burgi's approach is to guess that the chord of 1/3 the arc is a little more than 1/3 this number, and he chooses .173 as his first guess. Letting this be x in the expression \( f(x) = 3x - x^3 \) which I now suspect is what he did, we arrive at .5138223 (Burgi / Pitiscus use this number without decimal points, as explained before) .

Since he intends to use the method of double false position, he picks a second guess (.174) and gets a second estimate for his triple chord, .5167320

He arrives at an estimate using the double false position method of .1743114.

You might compare this to the actual value of 2 sin (5

^{o}) to compare. The error depends on how close the original guesses were, since the chord lengths are not linear.

**Testing Alternatives**

What if we try to follow the approach using Brahmagupta's formula to get the values? Beginning with .173. We have to find a way to tease out a guess for AC from this. Since \( \Delta \) ABC is isosceles, with Arcs AB and BC both being 10

^{o}arcs, the inscribed angles \( \angle \) BAC and \( \angle \) BCA are both 5,sup>o

leaving 170 degrees for \( \angle \) ABC. Using my very modern calculator I arrive at a length for AC

^{2}of .1206146401. When this and .173 are used in the Brahmagupta formula, we get .5241944515; not miles away from the .5138... result he found, but obviously not the approach he used.

I tried a few other things as well and all were as far, or farther away from Pitiscus' value than this, and none seem to get me very close to his result other than the cubic. I would love to hear some other approaches to how he might have done this other than the cubic.

And my special thanks to my unnamed correspondent for his note, sending me on several interesting days of exploration and study. Now on to write about the other ways of trisecting an angle.

Robin Hartshorne Paper on Viete https://math.berkeley.edu/~robin/Viete/construction.html

D E Smith "Piticus on Bergi's trisection of an Arc \" https://books.google.com/books?id=awAfO7Ff_z0C&q=Pitiscus+trisection#v=snippet&q=Pitiscus%20trisection&f=false

## 1 comment:

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