Sample configurations of pentagramma mirificum Relations between the angles and sides of five right triangles adjacent to the inner pentagon. Their Napier’s circles contain circular shifts of parts ( a , {\displaystyle (a,} π / 2 − B , {\displaystyle \pi /2-B,} π / 2 − c , {\displaystyle \pi /2-c,} π / 2 − A , {\displaystyle \pi /2-A,} b ) {\displaystyle b)} Pentagramma mirificum (Latin for "miraculous pentagram") is a star polygon on a sphere , composed of five great circle arcs , all of whose internal angles are right angles . This shape was described by John Napier in his 1614 book Mirifici Logarithmorum Canonis Descriptio (Description of the Admirable Table 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 [ edit ] 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 π / 2 , {\displaystyle \pi /2,} at A {\displaystyle A} , B {\displaystyle B} , C {\displaystyle C} , D {\displaystyle D} , and E . {\displaystyle E.}
There are ten arcs, each measuring π / 2 : {\displaystyle \pi /2:} P C {\displaystyle PC} , P E {\displaystyle PE} , Q D {\displaystyle QD} , Q A {\displaystyle QA} , R E {\displaystyle RE} , R B {\displaystyle RB} , S A {\displaystyle SA} , S C {\displaystyle SC} , T B {\displaystyle TB} , and T D . {\displaystyle TD.}
In the spherical pentagon P Q R S T {\displaystyle PQRST} , every vertex is the pole of the opposite side. For instance, point P {\displaystyle P} is the pole of equator R S {\displaystyle RS} , point Q {\displaystyle Q} — the pole of equator S T {\displaystyle ST} , etc.
At each vertex of pentagon P Q R S T {\displaystyle PQRST} , the external angle is equal in measure to the opposite side. For instance, ∠ A P T = ∠ B P Q = R S , ∠ B Q P = ∠ C Q R = S T , {\displaystyle \angle APT=\angle BPQ=RS,\;\angle BQP=\angle CQR=ST,} etc.
Napier's circles of spherical triangles A P T {\displaystyle APT} , B Q P {\displaystyle BQP} , C R Q {\displaystyle CRQ} , D S R {\displaystyle DSR} , and E T S {\displaystyle ETS} are rotations of one another.
Gauss's formulas [ edit ] 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).} The following identities hold, allowing the determination of any three of the above quantities from the two remaining ones:[2]
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}}} Gauss proved the following "beautiful equality" (schöne Gleichung ):[2]
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}}} 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)} , whose product α β γ δ ε {\displaystyle \alpha \beta \gamma \delta \varepsilon } is equal to 20 {\displaystyle 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}}} 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}}} From Gauss comes also the formula[2]
( 1 + i α ) ( 1 + i β ) ( 1 + i γ ) ( 1 + i δ ) ( 1 + i ε ) = α β γ δ ε e i A P Q R S T , {\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
A P Q R S T = 2 π − ( | P Q ⌢ | + | Q R ⌢ | + | R S ⌢ | + | S T ⌢ | + | T P ⌢ | ) {\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
P Q R S T {\displaystyle PQRST} .
Gnomonic projection [ edit ] The image of spherical pentagon P Q R S T {\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'} 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'} (lines passing through vertices and perpendicular to opposite sides) cross in one point O ′ {\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 O ′ {\displaystyle O'} , then the coordinates of vertices P ′ Q ′ R ′ S ′ T ′ {\displaystyle P'Q'R'S'T'} : ( x 1 , y 1 ) , … , {\displaystyle (x_{1},y_{1}),\ldots ,} ( 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}=} 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}=} 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}} , where ρ {\displaystyle \rho } is the length of the radius of the sphere.[3]
References [ edit ] External links [ edit ]