On Cesàro triangles and spherical polygons

Abstract

In Donnay’s and Van Brummelen’s monographs on spherical trigonometry, the Cesàro method is revitalized to derive various results on spherical triangles. Using Cesàro’s triangles, we derive in this paper some further results on spherical triangles that are not given in these books. Among these are results concerning the relation of sides of a spherical triangle and their opposite angles, Lexell’s theorem, and a result concerning the area of spherical triangles with one variable side-length. These results are applied to derive theorems of isoperimetric type for spherical polygons and the corresponding extremal properties of regular spherical polygons.

Appendix

Let M be the midpoint of the circular arc , and let X be a moving point on the sub-arc of $\gamma$. Put $x=B_1X,\,y(x)=B_2X$ and $f(x)=x+y(x)$. First, we show that

$$\begin{aligned} f(x)\text { is monotonically increasing in }0<x<B_1M. \end{aligned}$$

(2)

Fig. 8

$B_1P=B_1Y=x+\varepsilon$

For a small $\varepsilon >0$, let P be the point on the extension of the geodesic segment $B_1X$ beyond X such that $XP=\varepsilon$, and Y be the point on such that $B_1Y=x+\varepsilon$; see Fig. 8, where is a circular arc with center $B_1$. Put

Then, and . Hence we have

$$\begin{aligned} \beta -\alpha >\angle MB_2X. \end{aligned}$$

(3)

Choose a small $\varepsilon _0$ so that holds for $0<\varepsilon <\varepsilon _0$. Then, from (3), we have $\angle YXB_2>\angle YXP$. Put

$$\begin{aligned} \delta =y(x)-y(x+\varepsilon )=XB_2-YB_2. \end{aligned}$$

If $\delta \le 0$, then $f(x+\varepsilon )=x+\varepsilon +y(x)-\delta >f(x)$. So, we assume $\delta >0$. This implies $\beta <\pi /2$, and hence $\angle YXB_2<\pi /2$.

Fig. 9

A neighborhood of X in a large scale

Now, consulting Fig. 9 (in which shows a circular arc with center $B_2$ and $\varGamma$ shows the circle with center X and spherical radius XY), we can see that

$$\begin{aligned} \varepsilon =XP>XP', \text { and } \delta =XQ<XQ'. \end{aligned}$$

Since $\angle YXY_2=2\angle YXB_2>2\angle YXP=\angle YXY_1$, we have $YY_2>YY_1$ by Theorem 5. This implies $XP'>XQ'$. Therefore, for every $0<\varepsilon <\varepsilon _0$,

$$\begin{aligned} f(x+\varepsilon )-f(x)=\varepsilon -\delta>XP'-XQ'>0. \end{aligned}$$

Thus f(x) is monotonically increasing in $0<x<B_1M$.

Proof of Lemma 1

Let D be the point on the sub-arc such that $B_2D=B_1C$. Then E lies on the sub-arc of . Since M is also the midpoint of the arc , we may consider the case that E lies on the sub-arc . In this case, $B_1E+B_2E>B_1C+B_2C$ follows from (2). $\square$