Pentagramma mirificum

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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, π/2B, π/2c, π/2A, 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

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, at A, B, C, D, and E.

There are ten arcs, each measuring π/2: PC, PE, QD, QA, RE, RB, SA, SC, TB, and TD.

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

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

Napier's circles of spherical triangles APT, BQP, CRQ, DSR, and ETS are rotations of one another.

Gauss's formulas

Gauss introduced the notation

(α,β,γ,δ,ε)=(tan2TP,tan2PQ,tan2QR,tan2RS,tan2ST).

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

1+α=γδ1+β=δε1+γ=αε1+δ=αβ1+ε=βγ.

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

αβγδε=3+α+β+γ+δ+ε=(1+α)(1+β)(1+γ)(1+δ)(1+ε).

It is satisfied, for instance, by numbers (α,β,γ,δ,ε)=(9,2/3,2,5,1/3), whose product αβγδε is equal to 20.

Proof of the first part of the equality:

αβγδε=αβγ(1+αγ)(1+γα)=β(1+α)(1+γ)=β+αβ+βγ+αβγ=β+(1+δ)+(1+ε)+α(1+ε)=2+α+β+δ+ε+1+γ=3+α+β+γ+δ+ε

Proof of the second part of the equality:

αβγδε=α2β2γ2δ2ε2=γδδεεααββγ=(1+α)(1+β)(1+γ)(1+δ)(1+ε)

From Gauss comes also the formula[2]

(1+iα)(1+iβ)(1+iγ)(1+iδ)(1+iε)=αβγδεeiAPQRST, where APQRST=2π(|PQ|+|QR|+|RS|+|ST|+|TP|) is the area of pentagon PQRST.

Gnomonic projection

The image of spherical pentagon 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 PQRST unambiguously determine a conic section; in this case — an ellipse. Gauss showed that the altitudes of pentagram PQRST (lines passing through vertices and perpendicular to opposite sides) cross in one point 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, then the coordinates of vertices PQRST: (x1,y1),, (x5,y5) satisfy the equalities x1x4+y1y4= x2x5+y2y5= x3x1+y3y1= x4x2+y4y2= x5x3+y5y3=ρ2, where ρ is the length of the radius of the sphere.[3]

References

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