Pentagramma mirificum

By Wikipedia Contributors

Sample configurations of pentagramma mirificum
Relations between angles and sides of pentagramma mirificum

Pentagramma mirificum (Latin for miraculous pentagram) is a star polygon on a sphere, composed of five great circle arcs, whose all internal angles are right angles. This shape was described by John Napier in his 1614 book Mirifici logarithmorum canonis descriptio (Description of the wonderful rule of logarithms) along with rules that link the values of trigonometric functions of five parts of a right spherical triangle (two angles and three sides). The properties of pentagramma mirificum were studied, among others, by Carl Friedrich Gauss.[1]

Geometric properties

On a sphere, both the angles and the sides of a triangle (arcs of great circles) are measured as angles. Angles A {\displaystyle A} A, B {\displaystyle B} B, C {\displaystyle C} C, D {\displaystyle D} D, and E {\displaystyle E} E are right angles. Arcs P C {\displaystyle PC} PC, P E {\displaystyle PE} {\displaystyle PE}, Q D {\displaystyle QD} {\displaystyle QD}, Q A {\displaystyle QA} {\displaystyle QA}, R E {\displaystyle RE} RE, R B {\displaystyle RB} RB, S A {\displaystyle SA} {\displaystyle SA}, S C {\displaystyle SC} {\displaystyle SC}, T B {\displaystyle TB} {\displaystyle TB}, and T D {\displaystyle TD} TD are equal to π / 2 {\displaystyle \pi /2} \pi /2. In spherical pentagon P Q R S T {\displaystyle PQRST} {\displaystyle PQRST}, every vertex is the pole of the opposite side. For instance, point P {\displaystyle P} P is the pole of equator R S {\displaystyle RS} RS, point Q {\displaystyle Q} Q — the pole of equator S T {\displaystyle ST} ST, etc.[2]

Gauss’s formulas

Gauss introduced the notation

( α , β , γ , δ , ε ) = ( tan 2 T P , tan 2 P Q , tan 2 Q R , tan 2 R S , tan 2 S T ) . {\displaystyle (\alpha ,\beta ,\gamma ,\delta ,\varepsilon )=(\tan ^{2}TP,\tan ^{2}PQ,\tan ^{2}QR,\tan ^{2}RS,\tan ^{2}ST).}
{\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 to determine any three of the above quantities from the two remaining ones:[3]

1 + α = γ δ 1 + β = δ ε 1 + γ = α ε 1 + δ = α β 1 + ε = β γ . {\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}}}
{\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):[3]

α β γ δ ε = 3 + α + β + γ + δ + ε = ( 1 + α ) ( 1 + β ) ( 1 + γ ) ( 1 + δ ) ( 1 + ε ) . {\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}}}
{\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 ( α , β , γ , δ , ε ) = ( 9 , 2 / 3 , 2 , 5 , 1 / 3 ) {\displaystyle (\alpha ,\beta ,\gamma ,\delta ,\varepsilon )=(9,2/3,2,5,1/3)} {\displaystyle (\alpha ,\beta ,\gamma ,\delta ,\varepsilon )=(9,2/3,2,5,1/3)}, whose product α β γ δ ε {\displaystyle \alpha \beta \gamma \delta \varepsilon } {\displaystyle \alpha \beta \gamma \delta \varepsilon } is equal to 20 {\displaystyle 20} 20.

Proof of the first part of the equality:

α β γ δ ε = α β γ ( 1 + α γ ) ( 1 + γ α ) = β ( 1 + α ) ( 1 + γ ) = β + α β + β γ + α β γ = β + ( 1 + δ ) + ( 1 + ε ) + α ( 1 + ε ) = 2 + α + β + δ + ε + 1 + γ = 3 + α + β + γ + δ + ε {\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}}} {\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:

α β γ δ ε = α 2 β 2 γ 2 δ 2 ε 2 = γ δ δ ε ε α α β β γ = ( 1 + α ) ( 1 + β ) ( 1 + γ ) ( 1 + δ ) ( 1 + ε ) {\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}}} {\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[3]

( 1 + i α ) ( 1 + i β ) ( 1 + i γ ) ( 1 + i δ ) ( 1 + i ε ) = α β γ δ ε e i S , {\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^{iS},}
{\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^{iS},}
where S = 2 π ( | P Q | + | Q R | + | R S | + | S T | + | T P | ) {\displaystyle S=2\pi -(|{\overset {\frown }{PQ}}|+|{\overset {\frown }{QR}}|+|{\overset {\frown }{RS}}|+|{\overset {\frown }{ST}}|+|{\overset {\frown }{TP}}|)} {\displaystyle S=2\pi -(|{\overset {\frown }{PQ}}|+|{\overset {\frown }{QR}}|+|{\overset {\frown }{RS}}|+|{\overset {\frown }{ST}}|+|{\overset {\frown }{TP}}|)} is the area of pentagon P Q R S T {\displaystyle PQRST} {\displaystyle PQRST}.

Gnomonic projection

The image of spherical pentagon P Q R S T {\displaystyle PQRST} {\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 P Q R S T {\displaystyle P'Q'R'S'T'} {\displaystyle P'Q'R'S'T'} unambiguously determine a conic section; in this case — an ellipse. Gauss showed that the altitudes of pentagram P Q R S T {\displaystyle P'Q'R'S'T'} {\displaystyle P'Q'R'S'T'} (lines passing through vertices and perpendicular to opposite sides) cross in one point O {\displaystyle O'} O', which is the image of the point of tangency of the plane to sphere.[4]

Arthur Cayley observed that, if we set the origin of a Cartesian coordinate system in point O {\displaystyle O'} O', then the coordinates of vertices P Q R S T {\displaystyle P'Q'R'S'T'} {\displaystyle P'Q'R'S'T'}: ( x 1 , y 1 ) , {\displaystyle (x_{1},y_{1}),\ldots } {\displaystyle (x_{1},y_{1}),\ldots } ( x 5 , y 5 ) {\displaystyle (x_{5},y_{5})} {\displaystyle (x_{5},y_{5})} satisfy the equalities x 1 x 4 + y 1 y 4 = {\displaystyle x_{1}x_{4}+y_{1}y_{4}=} {\displaystyle x_{1}x_{4}+y_{1}y_{4}=} x 2 x 5 + y 2 y 5 = {\displaystyle x_{2}x_{5}+y_{2}y_{5}=} {\displaystyle x_{2}x_{5}+y_{2}y_{5}=} x 3 x 1 + y 3 y 1 = {\displaystyle x_{3}x_{1}+y_{3}y_{1}=} {\displaystyle x_{3}x_{1}+y_{3}y_{1}=} x 4 x 2 + y 4 y 2 = {\displaystyle x_{4}x_{2}+y_{4}y_{2}=} {\displaystyle x_{4}x_{2}+y_{4}y_{2}=} x 5 x 3 + y 5 y 3 = ρ 2 {\displaystyle x_{5}x_{3}+y_{5}y_{3}=-\rho ^{2}} {\displaystyle x_{5}x_{3}+y_{5}y_{3}=-\rho ^{2}}, where ρ {\displaystyle \rho } \rho is the length of the radius of the sphere.[5]

References

  1. ^ Carl Friedrich Gauss (1866). "Pentagramma mirificum". In Königliche Gesellschaft der Wissenschaften zu Göttingen. Carl Friedrich Gauss Werke: Band III. Analysis. pp. 481–490.
  2. ^ Bruce Director (2005-10-07). "From Plato's Theaetetus to Gauss's Pentagramma Mirificum: A Fight for Truth" (PDF). Executive Intelligence Review. 32 (39): 40–49.
  3. ^ a b c H.S.M. Coxeter (1971). "Frieze patterns" (PDF). Acta Arithmetica. 18: 297–310.
  4. ^ Bruce Director. "On the 375th Anniversary of Kepler's Passing". Retrieved 2018-12-25.
  5. ^ Professor Cayley F.R.S. (1871). "On Gauss's pentagramma mirificum". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. 42 (280): 311–312. doi:10.1080/14786447108640572.