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\title[Sources of Clavius' Book IV]
{The Eclectic Content and Sources of Book IV of Clavius' {\it Geometria Practica}}

\institute[College of the Holy Cross]
{
   John B. Little\\
   Department of Mathematics and CS\\
   College of the Holy Cross

}

\date[ ] 
{H.C. Faculty Seminar\\September 5, 2019}

\begin{document}

\begin{frame}
\maketitle
\end{frame}

\begin{frame}
{\bf Overview}
\tableofcontents
\end{frame}

\section{Introduction -- Clavius and {\it Geometria Practica}}

\begin{frame}
\begin{itemize}
\item I've always been interested in the history of mathematics 
(in addition to my nominal specialty in algebraic geometry/computational methods/coding theory, etc.)
\item To pursue historical work, I've been studying Latin and Greek with our 
excellent (and very welcoming) Classics department 
\item The subject for today relates to a Latin-to-English translation project I 
am currently pursuing -- working 
with the {\it Geometria Practica} of Christopher Clavius, S.J. (1538 - 1612, CE)
\end{itemize}
\end{frame}

\begin{frame}
{\bf Christophorus Clavius, S.J. 1538 - 1612}
\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics[height=2.5in, width=1.8in]{clavius.jpg}
\end{figure}
\column{.5\textwidth}
\begin{itemize}
\item {\it Euclidis Elementorum, Libri XV } (1574)
\item {\it Epitome artithmeticae practicae} (1583)
\item {\it Geometria practica} (1604)
\item {\it Algebra} (1608)
\item Commentary on the {\it Sphere} of Sacrobosco (astronomy), other
textbooks
\item Chief mathematician on Gregorian calendar reform
\end{itemize}
\end{columns}
\end{frame}

\begin{frame}
{\bf Context of Clavius' work}

\begin{itemize}
\item A long tradition of books with Islamic and Greek roots, starting in the late Medieval 
period (e.g. Leonardo Pisano -- ``Fibonacci'', 1170 - 1240 CE {\it De Practica Geometriae}) 
\item ``Orthogonal to'' the Platonic orientation of Euclid in some ways:
\begin{itemize} 
\item Discussed how to make and use physical instruments -- astrolabe, quadrants, 
proportional compasses, etc.
\item Intended to be \emph{practically useful} for land surveying, architecture, engineering, 
military science, etc.  
\item Used {\it numerical measures of lengths and angles}; mensuration formulas; numerical examples 
in addition to, or instead of, Euclidean proofs
\end{itemize}
\end{itemize}
\end{frame}

%\begin{frame}
%{\bf The context, cont.}

%\begin{itemize}
%\item Overtly ``practical'' problems: measuring heights of towers or mountains, 
%depths of valleys, etc. via trigonometry
%\item Mensuration -- area and volume formulas, often expressed in quasi-Archimedean terms: e.g.
%\emph{The area of a circle equals the area of the right triangle with sides about the right angle
%equal to the radius and the semi-circumference of the circle.}
%\item ``Geodesy'' -- problems about subdividing plane areas or solid volumes
%into parts in given ratios
%\end{itemize}
%\end{frame}

\begin{frame}
{\bf Book 4 of Clavius' {\it Geometria Practica}}
\begin{itemize}
\item Overall topic -- {\it measuring plane areas}
\item Rectangles, triangles, nonrectangular quadrilaterals, general polygons, then
\item Regular polygons, circles, and finally 
\item Sectors and segments of circles, and,
\item (as an ``appendix''), ellipses and segments of parabolas (after Archimedes, 
from {\it Conoids and Spheroids}, {\it Quadrature of the Parabola})
\end{itemize}
\end{frame}

\section{The Archimedean connection}

\begin{frame}
{\bf Some results are only quoted}

\begin{itemize}
\item (p. 202) ``It is pleasing to conclude this fourth book with two theorems most acutely
discovered and proved by Archimedes the Syracusan.''
\item The first (p. 202): If $ABCD$ is an ellipse with minor diameter $AC$ and major diameter $BD$, 
then let $HI$ be the mean proportional between $AC$ and $BD$.  The area of the ellipse is equal 
to the area of the circle with diameter $HI$.
\item The second (p. 203): If $ABC$ is a segment of a parabola with vertex $B$, and base $AC$, then the ratio of 
the area of the segment to the area of the triangle is $\frac{4}{3}$ (``sesquitertia'') 
\item But Clavius doesn't give the proofs (in effect, they are much more advanced than
most of the topics he includes)
\end{itemize}
\end{frame}

\begin{frame}
{\bf Others are proved in detail}

\begin{itemize}
\item Earlier in Book 4, Clavius gives a reworked version of essentially all
of what survives of Archimedes' {\it Measurement of the Circle}
\item ``It will not be a digression, therefore, if I include this
truly most acute and precise book on the measurement of the circle, 
partly because it is very brief ... , 
partly so that the student, in order to understand something so useful and
so widely disseminated by other masters, should be encouraged to go to 
Archimedes himself, and finally mostly because the writings of Archimedes,
as a result of their brevity, are somewhat more obscure, and we hope to be
able to bring some light to them.  Nor do we doubt that these things will be welcome 
and enjoyable for a studious reader."
\end{itemize}
\end{frame}

\begin{frame}
{\bf A first result}

Clavius and Archimedes begin by showing this result:
\vskip 10pt
\begin{theorem}[{\it Measurement of the Circle, Prop. 1}] The area of any circle is equal to the [area of] a right triangle, 
one of whose sides about the right angle is equal to the semidiameter
of the circle, while the other is equal to the circumference of the same circle. 
\end{theorem}
\vskip 10pt
The proof is given in full by the classical ``method of exhaustion''
in the form perfected by Archimedes.
\end{frame}

\begin{frame}
{\bf Background -- The ``Archimedean Principle''}

\begin{itemize}
\item Clavius (and less explicitly, Archimedes) make use of the following.  I'll use
modern notation for simplicity:
\item Let $\varepsilon > 0$ be arbitrary.  {\it If $A_1 > A_2 > A_3 > \cdots$ is a sequence of areas, in which 
$A_{n+1} < \frac{1}{2} A_n$ for all $n$, then for some $n_0$, $A_{n_0} < \varepsilon$.}
\item An easy consequence of the standard Archimedean principle and the unboundedness of the sequence $2^n$.  
\end{itemize}
\end{frame}

\begin{frame}
{\bf An Archimedean proof ``by exhaustion''}

\begin{itemize}
\item First suppose the circle is larger than the stated triangle
by a certain positive magnitude (i.e. area -- Clavius calls this $z > 0$)
\item  A sequence of non-overlapping areas is removed from the interior 
of the circle. 
\item Specifically, the inscribed square is removed at the first step, then
 four triangles on the sides of the square inscribed in the circle, so a
regular octagon has been removed, then eight triangles on the sides
of the octagon, so the total removed figure is a regular 16-gon, etc.  
\end{itemize}
\end{frame}

\begin{frame}
{\bf What this looks like}
\begin{center}
\includegraphics<1>[width=2.5in]{ExhaustionFrame1.png}
\includegraphics<2>[width=2.5in]{ExhaustionFrame2.png}
\includegraphics<3>[width=2.5in]{ExhaustionFrame3.png}
\end{center}
\end{frame}

\begin{frame}
{\bf Exhaustion proof, cont.}
\begin{itemize}
\item The ``{\it Archimedean Principle}'' implies that eventually the 
remaining area will be less than z, since at each stage, more than half of the 
area left at the previous stage is removed (a nice trigonometric exercise!)
\item But this leads to a contradiction for $n_0$ sufficiently large:
The area of the circle is supposed to equal $({\rm area\ of\ triangle}) + z$, but
the area of the circle also equals:
$$({\rm area\ of\ polygon}) + ({\rm remaining\ area}) < ({\rm area\ of\ triangle}) + z$$
(since the perimeter of the polygon is smaller than the circumference of the circle
and the apothem is less than the radius) 
\item Hence the area of the circle cannot equal the area of the triangle $+ z$.
\end{itemize}
\end{frame}

\begin{frame}
{\bf Exhaustion proof, cont.}
\begin{itemize}
\item Now suppose that the circle is less than the stated triangle by 
a certain magnitude (again denoted $z$)
\item Now Archimedes starts from the circumscribed square about
the circle (area definitely greater than the triangle since the perimeter
of the square is greater than the circumference of the circle, and the 
apothem is the same as the radius).
\item Now begin removing areas from the square (first the circle), then
triangles with base tangent to the circle.  The remaining regions now
are collections of ``mixed triangles'' with one side an arc of the circle.
\end{itemize}
\end{frame}

\begin{frame}
{\bf What this looks like}
\begin{center}
\includegraphics<1>[width=2.5in]{ExhaustionFrame4.png}
\includegraphics<2>[width=2.5in]{ExhaustionFrame5.png}
\includegraphics<3>[width=2.5in]{ExhaustionFrame6.png}
\end{center}
\end{frame}


\begin{frame}
{\bf Exhaustion proof, concluded}
\begin{itemize}
\item Again, more than half the remaining area is removed at each step, 
so eventually the remaining area becomes less than $z$.
\item This also leads to a contradiction, since both 
$$({\rm area\ of\ polygon}) > ({\rm area\ of\ triangle})$$
and
$$({\rm area\ of\ polygon}) < ({\rm area\ of\ circle})+ z = ({\rm area\ of\ triangle})$$
\item Hence since the triangle is neither greater nor less than
the circle, it can only {\it equal the circle}.  QED
\end{itemize}
\end{frame}


\begin{frame}
{\bf An amusing sidelight}

In a ``scholium'' to this proposition, Clavius takes the French Protestant 
humanist Joseph Justus Scaliger (1540-1609) to task.  Scaliger was an eminent classicist and
literary scholar who also fancied himself a brilliant mathematician.  His mathematical
{\it magnum opus} was grandly titled {\it Cyclometrica}, 
and published in 1594. (The 1st edition is lavishly printed with statements of theorems in both Latin and 
ancient Greek!)  Scaliger claimed this proof by Archimedes was incorrect (what he said was
essentially, amazingly enough, knowing $a > b$ and $a < b$ are both false does not show 
$a = b$!)  He went on to ``prove'' that 
a circle of radius 1 was equal in area to a square of side $\frac{9}{5}\sqrt{3}$, i.e. $\pi \doteq 3.11769$.
This ``implied'' the circle was squared with straightedge and compass!  
\end{frame}



\begin{frame}
{\bf Scaliger and Clavius}

Scaliger
had a number of other ``run-ins'' with Jesuits on various subjects; Clavius' satisfaction
at being able to point out his gross errors is palpable in this passage. 
\vskip 10pt
\href{https://www.e-rara.ch/zuz/content/pageview/14258610}{\beamergotobutton{Link}}
\vskip 10pt
In public Scaliger said something gracious(?) to the effect
that he would rather be criticized by Clavius than praised by other men, but he also 
privately lampooned Clavius for his appetite for food and drink (``c'est un gros ventre d'Aleman'').
\end{frame}

\begin{frame}
{\bf Continuing with Archimedes {\it Measurement of the Circle}}
\begin{itemize}
\item The remainder of Archimedes' work and Clavius' treatment of it is
devoted to a painstaking numerical computation of approximations to $\pi$, the 
ratio between the circumference and the diameter of a circle.
\item By considering repeatedly bisecting angles in circum- and inscribed squares
until he reaches regular 96-gons, Clavius, following Archimedes gives these estimates: 
 $\frac{223}{71} = 3\frac{10}{71} < \pi < \frac{22}{7} = 3\frac{1}{7}$.
\item This sort of numerical calculation involved was revolutionary in Greek geometry.
Its usefulness was recognized immediately and this work of Archimedes was justly
famous.  
\end{itemize}
\end{frame}

\begin{frame}
{\bf More accurate approximations}

Somewhat later, on page 198, Clavius gives the following, quite amazing figures for a circle:
\vskip 10pt
Diameter:  100000000000000000000.
\vskip 10pt
Circumference, overestimate: 314159265358979323847.
\vskip 10pt
Circumference, underestimate: 314159265358979323846.
\end{frame}

\begin{frame}
{\bf $\pi$ to 20 decimal places in 1604(!)}

\begin{itemize}
\item These approximate values were computed by the Dutch calculator
Ludolph van Ceulen (1540 - 1610) (who also had run-ins with Scaliger), and Clavius gives 
him credit, mentioning also his own younger colleague, Christopher Grienberger, S.J. (1561-1638).
\item `` ... while the difference
between the true area and the computed area may be very small for small
circles [using the Archimedean ratios], yet in larger circles it may not be negligible.  
If one should desire more accurate areas of circles, one may take the closer 
approximations [above].  ''
\end{itemize}
\end{frame}

\section{Areas of triangles}

\begin{frame}
{\bf Something of a surprise}

\begin{itemize}
\item Clavius \emph{begins} his discussion of areas of triangles by discussing the 
following ``rule:''
\item  \emph{Add all the sides
together in one sum; each of the sides are subtracted from half of this sum, 
so that three differences between the sides and [the semiperimeter] are
obtained; finally, these three differences and the semiperimeter are multiplied
together.  The square root of the number produced will be the area of the 
triangle which is sought.}
\item Today known as {\it Heron's formula}:  If $a,b,c$ are
the sides of a triangle and $s = \frac{a + b + c}{2}$, then the
area is
$$A = \sqrt{s(s - a)(s - b)(s - c)}$$
\item In addition to the ``rule'' and numerical examples, Clavius does
give a complete, Euclidean-style proof.
\end{itemize}
\end{frame}

\begin{frame}
{\bf A perplexing fact}
\begin{itemize}
\item Clavius doesn't attribute this to anyone (unlike many other things such as 
the treatment of Archimedes described above).
\item Heron, sometimes written as Hero, is Heron of Alexandria (ca 10 - 70 CE(?)),
a fascinating figure in Greek engineering, science, and mathematics.
\item He, even more so than Archimedes, was a sort of ``anti-Euclidean''
figure interested in mechanical inventions and applications of geometry.  
\item The formula appears in Heron's {\it Metrica}, but this work was
considered lost \emph{until 1896}, when R. Sch\"one found a manuscript copy
in Istanbul.
\end{itemize}
\end{frame}

\begin{frame}
{\bf Tracking down Clavius' source}
\begin{itemize}
\item Clavius' proof itself is equivalent mathematically to the proof in Heron's
{\it Metrica}, but there are differences (mostly filling in steps and giving 
direct arguments rather than quoting results about the inscribed circle
of the triangle and cyclic quadrilaterals).  
\item The actual text is quite a bit 
 closer (i.e. extremely close; virtually word for word) to the proof of the same result in Leonardo 
of Pisa's (i.e. Fibonacci's)  {\it De Practica Geometriae} (!)
\end{itemize}
\end{frame}

\begin{frame}
{\bf Moving back in time}
\begin{itemize}
\item M.  Clagett (in {\it Archimedes in the Middle Ages}, v I, p. 224) says that Fibonacci's
      source was the Latin "Verba filiorum Moysi filii ... " or "Liber trium 
      fratrum de geometria" by Gerard of Cremona (1114 - 1187 CE)
\item From Arabic book on mensuration
      by the three "Banu Musa" brothers in 9th century Baghdad (sons of Musa ibn Shakir) -- also includes
      Archimedes' results from {\it Measurement of the Circle!}
\item The Banu Musa were translators at the ``House of Wisdom'' in Baghdad engaged
in translating Greek mathematical works into Arabic (and other scientific pursuits!)
\item   Gerard's version was the first proof of Heron's formula in Latin
\end{itemize}
\end{frame} 
          
\begin{frame}
{\bf Questions!}  
\begin{itemize}
\item Clagett says that the 
      proof of Heron's formula in Fibonacci is essentially copied directly 
      from the translation of the book by the Banu Musa.  
\item (I haven't been able to verify this directly yet, 
      but I'm willing to believe him!)
\item The Banu Musa probably knew Heron's {\it Metrica}.
\item But interestingly, they wrote that they thought the result 
       ultimately came from Archimedes (though, of course, no text of his about
       this has survived, if there ever was one). 
\end{itemize}
\end{frame}

\begin{frame}
{\bf More questions!}
\begin{itemize}
\item  Interestingly enough, there are some features of the proof that are perplexingly 
        foreign to standard Greek geometry (like Euclid, or Apollonius, Archimedes?) 
\item Especially: multiplying four lengths 
        together and interpreting that as a geometric magnitude (the square of an area).  
\item Some of Clavius' argument, in fact, also tries to deal with this by saying that 
        the proof is working with numbers representing the lengths, rather than with the 
        lengths themselves, etc. (but that's a later point of view!)   
\end{itemize}
\end{frame}

\begin{frame}
{\bf Conclusions}
\begin{itemize}
\item So the jury is still out, so to speak, about how 
        Archimedes or some other pre-Heron Greek might have thought about the proof, 
        although there have been a number of attempts at reconstructions.  
\item My guess: Clavius either had Fibonacci or Gerard of Cremona's book at hand and 
followed that.  
\item But why didn't he attribute this to either of them??
\end{itemize}
\end{frame}


\begin{frame}
{\bf References}
\begin{enumerate}
\item[]\hskip-20pt [1] Barnabas Hughes, ed.  {\it Fibonacci's De Practica Geometrie}, Springer, NY, 2008.
\item[]\hskip-20pt [2] Christophorus Clavius, S.J., {\it Geometria practica} (1606 edition), online at {\tt http://www.e-rara.ch}
\item[]\hskip-20pt [3] Marshall Clagett, {\it Archimedes in the Middle Ages}, vol I.  University of Wisconsin
Press, Madison, 1964.  
\end{enumerate}
\end{frame}

\end{document}










