%\title{Lesson on Area and Gauss' %Theorem\footnote{Revision of {\it Post-Euclidean %Geometry: Class Notes and Workbook}, UpClose Printing \& Copies, %Champaign, IL 1995} %} %\author{George Francis\footnote{(C) 2004, Prof. George K. Francis, Mathematics %Department, University of Illinois.}} Last update 21apr15.## Non-EuclideanArea and Gauss' Theorem (Part 1)

\textit{(C) 2004, 2015 George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{Introduction} Informally, everybody knows what area is. The area of a bounded plane region is a positive number. (A region is bounded if it fits inside a sufficiently large circle.) The area of two non-overlapping regions is the sum of the areas of each. As the very name of the subject says, the earliest task of geometry was to measure the area of real-estate for the ancient farming civilizations . We all agree that the area of a rectangle is the product of its base times its height. And some still rememember the area of a trapezoid from high school. As we have seen, there are no rectangles in non-Euclidean geometry to measure. However, as we shall see in this lesson, non-Euclidean the area of non-Euclidean triangles is proportional to their angle defect. Non-Euclidean ancients would have needed only a protractor to measure areas of agricultural plots instead of a complicated surveying science. But, as we have already seen, the price of this convenience is the absence of similarity and hence impossibility of scaling shapes to make the larger or smaller. In what follows we will refer to another plane geometry where the postulates of Euclidean geometry do not hold. The geometry of the surface of a sphere, or \textit{spherical geometry} were well known to Euclid. And imagining the differences between Euclidean areas and spherical areas is relatively easy and instructive. So we (again) adopt the synonym \textit{hyperbolic geometry} for non-Euclidean geometry properly, along with spherical geometry. \section{Areas by Dissection } From just a few rules, practically everything we need to know about areas can be deduced by elementary methods of \textit{dissection} (cutting and pasting), and when that fails, by the integral calculus. But to actually assign a number to an area, we need a unit of measurement, microns, inches, meters, light-years. Similar figures have proportional areas. The proportion is the square of the proportion of their linear measures: double the side of a square, and it's area quadruples. This figure reminds you how to deduce the area of a triangle by dissection from the area of a rectangle. The Greeks, who preferred beauty over practicality, also knew how to relate the area $ab$ of a rectangle to that of a square of side $s$. Recall their \textit{geometric mean}, $s = \sqrt{ab}$ was constructible. Indeed, geometers like Newton still referred to finding the area under a curve as \textit{squaring the figure}. (Finding the length of a curve, which Newton also used his calculus for, was called \textig{rectifying the curve} since, to the Greeks, it meant constructing a segment of that length. this lesson. \section{On the Sphere } On the sphere, triangles have an angle sum greater than $\pi$. Experimenting with a dynamical geometry construction package, such as such as GEX2.0, or with some spherical trigonometry leads you to discover the \textin{angular excess theorem} for spherical geometry. One example can be done in your head. Imagine walking west on the equator and making a right turn, heading due north. At the north pole make another right turn, heading south. At the equator you will have made a rectangular triangle. Compared to a Euclidean triangle, it has one too many right angles. It has angular excess of $\pi/2$. \section{On the Hyperbolic Plane} Recall the construction that showed that the Euclidean \textit{Exterior Angle Theorem} (EXAT) fails to be true in the Klein model of $\mathbb{H}^2$. The amount by which it fails to be true is also the angular defect, because the exterior angle opposite two interior angles $\alpha, \beta$ is $c=\pi-\gamma$. Hence \[c - (\alpha + \beta) = \pi -(\alpha+\beta+\gamma) = \mbox{defect} \] This too may be determined experimentally with Hvidsten's GEX2.0 or Joel Castellano's \textit{NonEuclid}. The smaller the triangle, the smaller the angular defect. And, as you moved the vertices closer to the boundary unit circle of the Poincar'e disk, the defect approached zero. The method of dissection alone suffices to determine that angular defect, like area, is \textit{additive}, in the sense that the angular defect of of a triangle composed of the sum of two adjacent triangle is the sum of their angular defects. \textbf{Theorem:} The angle defect is additive. \textbf{Proof:} We rewrite the angle defect of the sum of two triangles formed by a Cevian of a third triangle. \begin{eqnarray*} \mbox{Defect of both}&=&\pi - (\alpha + (\gamma + \delta) + \eta ) \\ &=& (\pi - \alpha -\beta - \gamma) + \beta - \delta - \eta \\ &=& \mbox{left} + (\pi - \epsilon) -\delta - \eta \\ &=& \mbox{left} + (\pi - (\epsilon + \delta + \eta ) \\ &=& \mbox{left defect } + \mbox{ right defect} \end{eqnarray*} \section{Ideal Triangles} Every triangle is a trilateral in that is is formed by three non-concurrent lines. But the converse is not true, even in the Euclidean plane. There, a trilateral could consists of two parallel rays attached to a line segment. That would be a \textit{bi-angle} because it has only two proper angles. In $\mathbb{H}^2$ trilaterals come in four interesting kinds, look at the Klein model. With three positive angles, it's an ordinary triangle. With two angles, but with two lines \textit{asymptotically parallel} (in the model the lines meet on the boundary), we'll call it a \textit{biangle}. But now we can have \textit{monangles}, with but one positive angle. The \textit{nulangles}, which have no positive angles, are of special interest, and these are often called by their historical name \textit{ideal triangles}. (Only the first and last terms are universal in geometry, the others I made up.) What is significant about two asymptotic lines is that they are parallel in that they do not meet in a hyperbolic point, but they do not have a common perpendicular. No parallel hyperbolic lines have two common perpendiculars, for that would form a forbidden rectangle. How different matters are in non-Euclidean geometry! Only triangles have finite area in the Euclidean plane. The region between two Euclidean parallels and a transversal has infinite area. (In on the sphere there are no parallels at all, and everything, even the entire spherical plane has finite area.) Not so, and that is what we shall show first, in the hyperbolic plane. We shall next prove the main lemma to Gauss's Theorem. The theorem states that all ideal triangles (nulangles) are congruent and have finite area $\pi$. This is appropriate because all three angles of a nulangle are zero, hence the angular defect is $\pi$. \section{Finite Area Lemma} Here we show that all four kinds of hyperbolic triangles we defined above have finite area. We begin with biangles where the construction and proof is the easiest to understand. And then use the method of dissection to dispose of the other three by using biangles to measure their area. \textbf{Theorem: } Biangles Have Finite Areas. For the proof of the \begin{itemize} \item [Step 1:] In the figure the biangle $\triangle ABC$ is given. We extend the side between its two positive angles $AB$ to a ray $AD$ where $D$ is on the unit circle. \item [Step 2:] Note that $D$ is not a hyperbolic point, and $\triangle$ is, in fact, a monangle with one side an entire hyperbolic line $\ell_{CD}$. We shall need the pole of secant $CD$ to drop hyperbolic perpendiculars. \item [Step 3:] There are, at this point only two finite points, $A,B$, to connect to this pole. And the perpendiculars to them cross the polar, $\ell_{CD}$, at the points $E,F$ respectively. \end{itemize} \subsection{Hyperbolic Reflections} At this point we pause the proof to explain that we need the hyperbolic isometries (congruence) of reflections across a line in the Klein model. We have not defined these yet. Here we shall merely make the plausible and continue. Recall that in the Euclidean plane, a reflection across a mirror line entails moving every point on one side of the mirror to a point on the other along a perpendicular and the same distance. In other words, the mirror is the perpendicular bisector between a point and its reflection in the mirror. If we use our familiar Poincare disk model, we can define such a reflection as follows. Suppose The mirror is the line $(PQ)$, use an isometry $h_{PQ}$ which takes it to the real-line by sending $P \mapsto 0$ and "endpoints" $P',Q'$ of $(PQ)$ to $-1, +1$ . We can use crossratios to define it implicitly thus: \[ CR(h_{PQ}(z), -1,0,1)=CR(z,P',P,Q')$ \] Note that, being an isometry, it also sends $Q \mapsto d_H(P,Q)$. Reflections across the real axis, $w\mapsto \bar{w}$ is clearly a reflection which is expressed by taking the conjugate. So \[\sigma_{PQ}(z) = h_{PQ}^{-1}(\bar{h_{PQ}(z)})\] is the general reflection across $(PQ)$. However, in the Klein Model there is a more constructive way of doing this and we shall have a look at it later. Here we only need to know that \textbf{K-reflection lemma:} Given a mirror line $m$ crossed by line $k$ with point $K$ on it, the reflection proceeds in 5 steps as shown in the figure so labelled. \subsection{Completion of the Proof of Biangle Theorem} \begin{itemize} \item [Step 4:] Draw mirror $m_1 = (AE)$ and reflect $B \mapsto B'$. \item [Step 5:] Since $m_1 \perp (CD) $ we have that \begin{eqnarray*} \sigma_1(CD) & = & (DC) \\ \sigma_1(A) & = & (A) \\ \sigma_1(CB)& = &(DB') \\ \sigma_1(PB)& = &(PB') \\ \sigma_1(\triangle_1)& = & \triangle_2\\ \triangle_1 & \cong & \triangle_2\\ \triangle_3 & \cong & \triangle_4\\ \end{eqnarray*} \item [Step 6:] Since $m_1 \perp (CD) $ we have that Note that 3 of these triangles are in two regions: the original biangle $\triangle ABC$ and the pentagonal \textit{pentahouse} $ABFF'B'$. But $\triangle_4$ is only in the pentahouse. \item [Step 6:] A second mirror $m_2=(PB')$, with reflection $\sigma_2$ lets us find a congruent cousin to $\triangle _{4'}$, which is only inside the biangle. \begin{eqnarray*} \triangle_3&=&\sigma_1(\triangle_4) \\ \triangle_{4'}&=&\sigma_2(\triangle_3) \\ \end{eqnarray*} \item [Step 7:] The strategy is now apparent. We have reflected $(CB)$ across $(DB')$ to obtain $(DB'')=m_2$, the next mirror. \item [Step 8:] Reflecting $(CB)$ across $m_2$ to $DB'''$ leads to four congruent triangles: \begin{eqnarray*} \triangle_6' \cong \triangle_5' \cong \triangle_5 \cong \triangle_6 \\ \end{eqnarray*} The first two are only in the bi-angle, the last two are only in pentahouse. \item [Step 8:] From now on it just repeats, except that there are longer and longer chains of reflections between the corresponding triangles in the bi-angle and the pentahouse. \end{itemize} We close in appealing to a second geometrical concept, \textit{continuity}. Recall that Euclid tacitly assumed this for the case of overlapping circles to actually intersect at points. Here we use it as follows: By continuity we fill up the infinite region with pairs of triangles which are congruent to their cousins in the finite pentagonal region. Thus the infinite region inside the biangle must have the same finite area as the pentahouse. And we have shown that biangles too have finite area. \subsection{Monangles and Nulangles also have Finite Area} Dissect the monangle into two adjacent biangles by a cevian from the vertex of the single positive angle. (A cevian is a line connecting the vertex of a triangle to the opposite side.) The area of the monangle is, thus the sum of the finite areas of the two biangles. And there are many more ways to dissect an ideal triangle (nulangle) to complete this proof. But there is one in particular that will interest us. \section{Preview} Next, we need to actually measure the area of triangles, beginning, odlly enough, with nulangles, then monangles. Finally actual triangles. \end{document}