Spherical Triangle

8/11/2019 Spherical Triangle

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Spherical Triangles and Girard’s Theorem

Abhijit Champanerkar

College of Staten Island, CUNY

MTH 329, Spring 2013

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Spherical geometry

Let S 2 denote the unit sphere in R3 i.e. the set of all unit vectors

i.e. the set {(x , y , z ) ∈ R3| x 2 + y 2 + z 2 = 1 }.

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Spherical geometry

Let S 2 denote the unit sphere in R3 i.e. the set of all unit vectors

i.e. the set {(x , y , z ) ∈ R3| x 2 + y 2 + z 2 = 1 }.

A great circle in S 2 is a circle which divides the sphere in half. In

other words, a great circle is the interesection of S 2 with a planepassing through the origin.

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Great circles are straight lines

Great circles play the role of straight lines in spherical geometry.

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Given two distinct points on S 2, there is a great circle passingthrough them obtained by the intersection of S 2 with the planepassing through the origin and the two given points.

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Given two distinct points on S 2, there is a great circle passingthrough them obtained by the intersection of S 2 with the planepassing through the origin and the two given points.

You can similarly verify the other three Euclid’s posulates forgeometry.

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Diangles

Any two distinct great circles inter-sect in two points which are nega-tives of each other.

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Diangles


The angle between two great circles at an intersection point is theangle between their respective planes.

A region bounded by two great circles is called a diangle.

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Diangles


The angle between two great circles at an intersection point is theangle between their respective planes.

A region bounded by two great circles is called a diangle.

The angle at both the vertices are equal. Both diangles boundedby two great circles are congruent to each other.

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Area of a diangle

Proposition

Let θ be the angle of a diangle. Then the area of the diangle is 2θ.

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Area of a diangle

Proposition


Proof: The area of the diangle is proportional to its angle. Sincethe area of the sphere, which is a diangle of angle 2π, is 4π, the

area of the diangle is 2θ.

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Area of a diangle

Proposition




Alternatively, one can compute this area directly as the area of asurface of revolution of the curve z =

1 − y 2 by an angle θ. This

area is given by the integral 1−1

θz

1 + (z )2 dy .

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Area of a diangle

Proposition




Alternatively, one can compute this area directly as the area of asurface of revolution of the curve z =

1 − y 2 by an angle θ. This

area is given by the integral 1−1

θz

1 + (z )2 dy .

If the radius of the sphere is r then the area of the diangle is 2θr 2.

This is very similar to the formula for the length of an arc of theunit circle which subtends an angle θ is θ.

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Spherical polygons

A spherical polygon is a polygon on S 2 whose sides are parts of great circles.

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Gi d’ Th A f h i l i l



Girard’s Theorem: Area of a spherical triangle

Girard’s Theorem

The area of a spherical triangle with angles α, β and γ is α+β +γ −π.

Gi d’ Th A f h i l i l

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Girard’s Theorem


Proof:

Gi d’ Th A f h i l t i l

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Girard’s Theorem


Proof:

Gi d’ Th A f h i l t i l

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Girard s Theorem: Area of a spherical triangle

Girard’s Theorem


Proof:


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Girard’s Theorem


Proof:


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Girard’s Theorem


Proof:

Area of a spherical triangle

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B

A

C

F

E

D

ABC as shown above is formed by the intersection of three greatcircles.

Vertices A and D are antipodal to each other and hence have the

same angle. Similarly for vertices B , E and C , F . Hence thetriangles ABC and DEF are antipodal (opposite) triangles andhave the same area.

Assume angles at vertices A, B and C to be α, β and γ respectively.


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B

A

C

F

E

D

ABC R AD R BE R CF

Let R AD , R BE and R CF denote pairs of diangles as shown. ThenABC and DEF each gets counted in every diangle.


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B

A

C

F

E

D

ABC R AD R BE R CF


R AD ∪ R BE ∪ R CF = S 2, Area(ABC ) = Area(DEF ) = X .


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B

A

C

F

E

D

ABC R AD R BE R CF


R AD ∪ R BE ∪ R CF = S 2, Area(ABC ) = Area(DEF ) = X .

Area(S 2) = Area(R AD ) + Area(R BE ) + Area(R CF ) − 4X

4π = 4α + 4β + 4γ − 4X

X = α + β + γ − π

Area of a spherical polygon

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Corollary

Let R be a spherical polygon with n vertices and n sides with interiorangles α1, . . . , αn. Then Area(R ) = α1 + . . . + αn − (n − 2)π.


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Corollary

Let R be a spherical polygon with n vertices and n sides with interiorangles α1, . . . , αn. Then Area(R ) = α1 + . . . + αn − (n − 2)π.

Proof: Any polygon with n sides for n ≥ 4 can be divided inton − 2 triangles.

The result follows as the angles of these triangles add up to theinterior angles of the polygon.

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Spherical Triangle

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