• Source: Pentagramma mirificum

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.




Geometric properties


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


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:








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):








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




(
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


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.




References






External links


Media related to Pentagramma mirificum at Wikimedia Commons
Pentagramma mirificum on YouTube

Kata Kunci Pencarian:

• John Napier
• Pentagramma mirificum
• Star polygon
• Carl Friedrich Gauss
• Spherical trigonometry
• Compound of five octahedra
• Mirifici Logarithmorum Canonis Descriptio
• List of trigonometric identities
• Vadim Schechtman