Pentagramma mirificum

On a sphere, both the angles and the sides of a triangle (arcs of great circles) are measured as angles.

There are five right angles, each measuring $\backslash pi/2,$ at $A$, $B$, $C$, $D$, and $E.$

There are ten arcs, each measuring $\backslash pi/2:$ $PC$, $PE$, $QD$, $QA$, $RE$, $RB$, $SA$, $SC$, $TB$, and $TD.$

In the spherical pentagon $PQRST$, every vertex is the pole of the opposite side. For instance, point $P$ is the pole of equator $RS$, point $Q$ — the pole of equator $ST$, etc.

At each vertex of pentagon $PQRST$, the external angle is equal in measure to the opposite side. For instance, $\backslash angle\; APT\; =\; \backslash angle\; BPQ\; =\; RS,\backslash ;\; \backslash angle\; BQP\; =\; \backslash angle\; CQR\; =\; ST,$ etc.

Napier's circles of spherical triangles $APT$, $BQP$, $CRQ$, $DSR$, and $ETS$ are rotations of one another.

Gauss introduced the notation

: $$(\backslash alpha,\; \backslash beta,\; \backslash gamma,\; \backslash delta,\; \backslash varepsilon)\; =\; (\backslash tan^2\; TP,\; \backslash tan^2\; PQ,\; \backslash tan^2\; QR,\; \backslash tan^2\; RS,\; \backslash tan^2\; ST).$$

The following identities hold, allowing the determination of any three of the above quantities from the two remaining ones:{{cite journal |url=http://matwbn.icm.edu.pl/ksiazki/aa/aa18/aa18132.pdf |title=Frieze patterns |journal=Acta Arithmetica |first=H. S. M. |last=Coxeter |author-link=Harold Scott MacDonald Coxeter |volume=18 |pages=297–310 |year=1971 |doi=10.4064/aa-18-1-297-310|doi-access=free }}

: $$\backslash begin\{align\}$$

1 + \alpha &= \gamma\delta &1 + \beta &= \delta\varepsilon &1 + \gamma &=\alpha \varepsilon \\

1 + \delta &= \alpha\beta &1 + \varepsilon &= \beta\gamma.

\end{align}

Gauss proved the following "beautiful equality" (schöne Gleichung):

: $$\backslash begin\{align\}$$

\alpha\beta\gamma\delta\varepsilon &=\; 3 + \alpha + \beta + \gamma + \delta + \varepsilon \\

&=\; \sqrt{(1+\alpha)(1+\beta)(1+\gamma)(1+\delta)(1+\varepsilon)}.

\end{align}

It is satisfied, for instance, by numbers $(\backslash alpha,\; \backslash beta,\; \backslash gamma,\; \backslash delta,\; \backslash varepsilon)\; =\; (9,\; 2/3,\; 2,\; 5,\; 1/3)$, whose product $\backslash alpha\backslash beta\backslash gamma\backslash delta\backslash varepsilon$ is equal to $20$.

Proof of the first part of the equality:

: $\backslash begin\{align\}$

\alpha\beta\gamma\delta\varepsilon

&= \alpha\beta\gamma\left(\frac{1+\alpha}{\gamma}\right)\left(\frac{1+\gamma}{\alpha}\right)

= \beta(1+\alpha)(1+\gamma)\\

&= \beta + \alpha\beta + \beta\gamma + \alpha\beta\gamma = \beta + (1 + \delta) + (1 + \varepsilon) + \alpha(1 + \varepsilon) \\

&= 2 + \alpha + \beta + \delta + \varepsilon + 1 + \gamma \\

& = 3 + \alpha + \beta + \gamma + \delta + \varepsilon

\end{align}

Proof of the second part of the equality:

:$\backslash begin\{align\}$

\alpha\beta\gamma\delta\varepsilon &= \sqrt{\alpha^2\beta^2\gamma^2\delta^2\varepsilon^2} \\

&= \sqrt{\gamma\delta \cdot \delta\varepsilon \cdot \varepsilon\alpha \cdot \alpha\beta \cdot \beta\gamma}\\

&= \sqrt{(1+\alpha)(1+\beta)(1+\gamma)(1+\delta)(1+\varepsilon)}

\end{align}

From Gauss comes also the formula

$$(1+i\backslash sqrt\{^\{^\{\backslash !\}\}\backslash alpha\})(1+i\backslash sqrt\{\backslash beta\})(1+i\backslash sqrt\{^\{^\{\backslash !\}\}\backslash gamma\})(1+i\backslash sqrt\{\backslash delta\})(1+i\backslash sqrt\{^\{^\{\backslash !\}\}\backslash varepsilon\})\; =\; \backslash alpha\backslash beta\backslash gamma\backslash delta\backslash varepsilon\; e^\{iA\_\{PQRST\}\},$$

where $A\_\{PQRST\}\; =\; 2\backslash pi\; -\; (|\backslash overset\{\backslash frown\}\{PQ\}|\; +\; |\backslash overset\{\backslash frown\}\{QR\}|\; +\; |\backslash overset\{\backslash frown\}\{RS\}|\; +\; |\backslash overset\{\backslash frown\}\{ST\}|\; +\; |\backslash overset\{\backslash frown\}\{TP\}|)$ is the area of pentagon $PQRST$.

The image of spherical pentagon $PQRST$ in the gnomonic projection (a projection from the centre of the sphere) onto any plane tangent to the sphere is a rectilinear pentagon. Its five vertices $P\text{'}Q\text{'}R\text{'}S\text{'}T\text{'}$ unambiguously determine a conic section; in this case — an ellipse. Gauss showed that the altitudes of pentagram $P\text{'}Q\text{'}R\text{'}S\text{'}T\text{'}$ (lines passing through vertices and perpendicular to opposite sides) cross in one point $O\text{'}$, which is the image of the point of tangency of the plane to sphere.

Arthur Cayley observed that, if we set the origin of a Cartesian coordinate system in point $O\text{'}$, then the coordinates of vertices $P\text{'}Q\text{'}R\text{'}S\text{'}T\text{'}$: $(x\_1,\; y\_1),\backslash ldots,$ $(x\_5,\; y\_5)$ satisfy the equalities $x\_1\; x\_4\; +\; y\_1\; y\_4\; =$ $x\_2\; x\_5\; +\; y\_2\; y\_5\; =$ $x\_3\; x\_1\; +\; y\_3\; y\_1\; =$ $x\_4\; x\_2\; +\; y\_4\; y\_2\; =$ $x\_5\; x\_3\; +\; y\_5\; y\_3\; =\; -\backslash rho^2$, where $\backslash rho$ is the length of the radius of the sphere.{{cite journal |url=https://archive.org/stream/collectedmathema07cayluoft#page/36/mode/2up |title=On Gauss's pentagramma mirificum |journal=The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science |first=Arthur |last=Cayley |author-link=Arthur Cayley |volume=42 |issue=280 |pages=311–312 |year=1871 |doi=10.1080/14786447108640572}}

{{reflist}}

• {{commons category-inline|Pentagramma mirificum}}
• {{YouTube|id=wQKMAx6jxGw|title=Pentagramma mirificum}}

Category:Spherical trigonometry

Category:Types of polygons