5-sided star shaped polygon
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,}
A
{\displaystyle A}
B
{\displaystyle B}
C
{\displaystyle C}
D
{\displaystyle D}
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}
T
D
.
{\displaystyle TD.}
In the spherical pentagon
P
Q
R
S
T
{\displaystyle PQRST}
P
{\displaystyle P}
R
S
{\displaystyle RS}
Q
{\displaystyle Q}
S
T
{\displaystyle ST}
At each vertex of pentagon
P
Q
R
S
T
{\displaystyle PQRST}
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,}
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}
E
T
S
{\displaystyle ETS}
rotations of one another.
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)}
α
β
γ
δ
ε
{\displaystyle \alpha \beta \gamma \delta \varepsilon }
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}},}
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}}|)}
P
Q
R
S
T
{\displaystyle PQRST}
Gnomonic projection [ edit ] The image of spherical pentagon
P
Q
R
S
T
{\displaystyle PQRST}
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'}
O
′
{\displaystyle O'}
Arthur Cayley observed that, if we set the origin of a Cartesian coordinate system in point
O
′
{\displaystyle O'}
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})}
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}}
ρ
{\displaystyle \rho }
[ 3]
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