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 ${\displaystyle \pi /2,}$ at ${\displaystyle A}$, ${\displaystyle B}$, ${\displaystyle C}$, ${\displaystyle D}$, and ${\displaystyle E.}$

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

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

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

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

Gauss introduced the notation

$${\displaystyle (\alpha ,\beta ,\gamma ,\delta ,\varepsilon )=(\tan ^{2}TP,\tan ^{2}PQ,\tan ^{2}QR,\tan ^{2}RS,\tan ^{2}ST).}$$

The following identities hold, allowing the determination of any three of the above quantities from the two remaining ones:[2]

$${\displaystyle {\begin{aligned}1+\alpha &=\gamma \delta &1+\beta &=\delta \varepsilon &1+\gamma &=\alpha \varepsilon \\1+\delta &=\alpha \beta &1+\varepsilon &=\beta \gamma .\end{aligned}}}$$

Gauss proved the following "beautiful equality" (schöne Gleichung):[2]

$${\displaystyle {\begin{aligned}\alpha \beta \gamma \delta \varepsilon &=\;3+\alpha +\beta +\gamma +\delta +\varepsilon \\&=\;{\sqrt {(1+\alpha )(1+\beta )(1+\gamma )(1+\delta )(1+\varepsilon )}}.\end{aligned}}}$$

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

Proof of the first part of the equality:

${\displaystyle {\begin{aligned}\alpha \beta \gamma \delta \varepsilon &=\alpha \beta \gamma \left({\frac {1+\alpha }{\gamma }}\right)\left({\frac {1+\gamma }{\alpha }}\right)=\beta (1+\alpha )(1+\gamma )\\&=\beta +\alpha \beta +\beta \gamma +\alpha \beta \gamma =\beta +(1+\delta )+(1+\varepsilon )+\alpha (1+\varepsilon )\\&=2+\alpha +\beta +\delta +\varepsilon +1+\gamma \\&=3+\alpha +\beta +\gamma +\delta +\varepsilon \end{aligned}}}$

Proof of the second part of the equality:

${\displaystyle {\begin{aligned}\alpha \beta \gamma \delta \varepsilon &={\sqrt {\alpha ^{2}\beta ^{2}\gamma ^{2}\delta ^{2}\varepsilon ^{2}}}\\&={\sqrt {\gamma \delta \cdot \delta \varepsilon \cdot \varepsilon \alpha \cdot \alpha \beta \cdot \beta \gamma }}\\&={\sqrt {(1+\alpha )(1+\beta )(1+\gamma )(1+\delta )(1+\varepsilon )}}\end{aligned}}}$

From Gauss comes also the formula[2]

$${\displaystyle (1+i{\sqrt {^{^{\!}}\alpha }})(1+i{\sqrt {\beta }})(1+i{\sqrt {^{^{\!}}\gamma }})(1+i{\sqrt {\delta }})(1+i{\sqrt {^{^{\!}}\varepsilon }})=\alpha \beta \gamma \delta \varepsilon e^{iA_{PQRST}},}$$

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

The image of spherical pentagon ${\displaystyle PQRST}$ 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 ${\displaystyle P'Q'R'S'T'}$ unambiguously determine a conic section; in this case — an ellipse. Gauss showed that the altitudes of pentagram ${\displaystyle P'Q'R'S'T'}$ (lines passing through vertices and perpendicular to opposite sides) cross in one point ${\displaystyle O'}$, 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 ${\displaystyle O'}$, then the coordinates of vertices ${\displaystyle P'Q'R'S'T'}$: ${\displaystyle (x_{1},y_{1}),\ldots ,}$ ${\displaystyle (x_{5},y_{5})}$ satisfy the equalities ${\displaystyle x_{1}x_{4}+y_{1}y_{4}=}$ ${\displaystyle x_{2}x_{5}+y_{2}y_{5}=}$ ${\displaystyle x_{3}x_{1}+y_{3}y_{1}=}$ ${\displaystyle x_{4}x_{2}+y_{4}y_{2}=}$ ${\displaystyle x_{5}x_{3}+y_{5}y_{3}=-\rho ^{2}}$, where ${\displaystyle \rho }$ is the length of the radius of the sphere.[3]

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