In spherical geometry, Lexell's theorem holds that every spherical triangle with the same surface area on a fixed base has its apex on a small circle, called or , passing through each of the two points antipodal to the two base vertices.
A is a shape on a sphere consisting of three (corner points) connected by three sides, each of which is part of a , the analog on the sphere of a straight line in the plane (for example the equator and meridians of a globe); the spherical analog of planar circles, which curve relative to the surface, are called (for example the circles of latitude other than the equator). Any of the sides of a spherical triangle can be considered the , and the opposite vertex is the corresponding . Two points on a sphere are if they are diametrically opposite, as far apart as possible.
The theorem is named for Anders Johan Lexell, who presented a paper about it (published 1784) including both a trigonometric proof and a geometric one., , , Lexell's colleague Leonhard Euler wrote another pair of proofs in 1778 (published 1797), and a variety of proofs have been written since by Adrien-Marie Legendre (1800), Jakob Steiner (1827), Carl Friedrich Gauss (1841), Paul Serret (1855), and Joseph-Émile Barbier (1864), among others.
The theorem is the analog of propositions 37 and 39 in Book I of Euclid's Elements, which prove that every planar triangle with the same area on a fixed base has its apex on a straight line parallel to the base. An analogous theorem can also be proven for hyperbolic triangles, for which the apex lies on a hypercycle.
Statement
Given a fixed base an arc of a great circle on a sphere, and two apex points and on the same side of great circle Lexell's theorem holds that the surface area of the spherical triangle is equal to that of if and only if lies on the small-circle arc where and are the points antipodal to and respectively.
As one analog of the planar formula for the area of a triangle, the spherical excess of spherical triangle can be computed in terms of the base (the angular length of arc and "height" (the angular distance between the parallel small circles
This formula is based on consideration of a sphere of radius , on which arc length is called and surface area is called or . The angle measure of a complete great circle is radians, and the spherical excess of a hemisphere (half-sphere) is steradians, where is the circle constant.
In the limit for triangles much smaller than the radius of the sphere, this reduces to the planar formula.
The small circles and each intersect the great circle at an angle of See below for a proof of this.
Proofs
There are several ways to prove Lexell's theorem, each illuminating a different aspect of the relationships involved.
Isosceles triangles
The main idea in Lexell's geometric proof – also adopted by Eugène Catalan (1843), Robert Allardice (1883), Jacques Hadamard (1901), Antoine Gob (1922), and Hiroshi Maehara (1999) – is to split the triangle into three isosceles triangles with common apex at the circumcenter and then chase angles to find the spherical excess of triangle In the figure, points and are on the far side of the sphere so that we can clearly see their antipodal points and all of Lexell's circle
Let the base angles of the isosceles triangles (shaded red in the figure), (blue), and (purple) be respectively and (In some cases is outside then one of the quantities will be negative.) We can compute the internal angles of (orange) in terms of these angles: (the supplement of and likewise and finally
By Girard's theorem the spherical excess of is
If base is fixed, for any third vertex falling on the same arc of Lexell's circle, the point and therefore the quantity will not change, so the excess of which depends only on will likewise be constant. And vice versa: if remains constant when the point is changed, then so must be, and therefore must be fixed, so must remain on Lexell's circle.
Cyclic quadrilateral
Jakob Steiner (1827) wrote a proof in similar style to Lexell's, also using Girard's theorem, but demonstrating the angle invariants in the triangle by constructing a cyclic quadrilateral inside the Lexell circle, using the property that pairs of opposite angles in a spherical cyclic quadrilateral have the same sum.
Starting with a triangle , let be the Lexell circle circumscribing and let be another point on separated from by the great circle Let
Because the quadrilateral is cyclic, the sum of each pair of its opposite angles is equal, or rearranged
By Girard's theorem the spherical excess of is
The quantity does not depend on the choice of so is invariant when is moved to another point on the same arc of Therefore is also invariant.
Conversely, if is changed but is invariant, then the opposite angles of the quadrilateral will have the same sum, which implies lies on the small circle
Spherical parallelograms Euler in 1778 proved Lexell's theorem analogously to Euclid's proof of Elements I.35 and I.37, as did Victor-Amédée Lebesgue independently in 1855, using spherical parallelograms – spherical quadrilaterals with congruent opposite sides, which have parallel small circles passing through opposite pairs of adjacent vertices and are in many ways analogous to Euclidean parallelograms. There is one complication compared to Euclid's proof, however: The four sides of a spherical parallelogram are the great-circle arcs through the vertices rather than the parallel small circles. Euclid's proof does not need to account for the small lens-shaped regions sandwiched between the great and small circles, which vanish in the planar case., , , Euler's proof differs slightly from the proof presented here in that Euler did not consider spherical parallelograms per se, but instead the parallelogram-like regions bounded by great circle arcs on the two sides and by small-circle arcs on top and bottom. The main idea of the proof is the same, but the lens-shaped regions between the two shapes must be treated slightly differently. A proof using spherical parallelograms proper is found in:
A lemma analogous to Elements I.35: two spherical parallelograms on the same base and between the same parallels have equal area.
Proof: Let and be spherical parallelograms with the great circle (the "midpoint circle") passing through the midpoints of sides and coinciding with the corresponding midpoint circle in Let be the intersection point between sides and Because the midpoint circle is shared, the two top sides and lie on the same small circle parallel to and antipodal to a small circle passing through and
Two arcs of are congruent, thus the two curvilinear triangles and each bounded by on the top side, are congruent. Each parallelogram is formed from one of these curvilinear triangles added to the triangle and to one of the congruent lens-shaped regions between each top side and with the curvilinear triangle cut away. Therefore the parallelograms have the same area. (As in Elements, the case where the parallelograms do not intersect on the sides is omitted, but can be proven by a similar argument.)
Proof of Lexell's theorem: Given two spherical triangles and each with its apex on the same small circle through points and construct new segments and congruent to with vertices and on The two quadrilaterals and are spherical parallelograms, each formed by pasting together the respective triangle and a congruent copy. By the lemma, the two parallelograms have the same area, so the original triangles must also have the same area.
Proof of the converse: If two spherical triangles have the same area and the apex of the second is assumed to not lie on the Lexell circle of the first, then the line through one side of the second triangle can be intersected with the Lexell circle to form a new triangle which has a different area from the second triangle but the same area as the first triangle, a contradiction. This argument is the same as that found in Elements I.39.
Saccheri quadrilateral
Another proof using the midpoint circle which is more visually apparent in a single picture is due to Carl Friedrich Gauss (1841), who constructs the Saccheri quadrilateral (a quadrilateral with two adjacent right angles and two other equal angles) formed between the side of the triangle and its perpendicular projection onto the midpoint circle which has the same area as the triangle.
Let be the great circle through the midpoints of and of and let and be the perpendicular projections of the triangle vertices onto The resulting pair of right triangles and (shaded red) have equal angles at (vertical angles) and equal hypotenuses, so they are congruent; so are the triangles and (blue). Therefore, the area of triangle is equal to the area of Saccheri quadrilateral as each consists of one red triangle, one blue triangle, and the green quadrilateral pasted together. (If falls outside the arc then either the red or blue triangles will have negative signed area.) Because the great circle and therefore the quadrilateral is the same for any choice of lying on the Lexell circle the area of the corresponding triangle is constant.
Stereographic projection
The stereographic projection maps the sphere to the plane. A designated great circle is mapped onto the in the plane, and its poles are mapped to the origin (center of the primitive circle) and the point at infinity, respectively. Every circle on the sphere is mapped to a circle or straight line in the plane, with straight lines representing circles through the second pole. The stereographic projection is conformal, meaning it preserves angles.
To prove relationships about a general spherical triangle without loss of generality vertex can be taken as the point which projects to the origin. The sides of the spherical triangle then project to two straight segments and a circular arc. If the tangent lines to the circular side at the other two vertices intersect at point a planar straight-sided quadrilateral can be formed whose external angle at is the spherical excess of the spherical triangle. This is sometimes called the Cesàro method of spherical trigonometry, after crystallographer who popularized it in two 1905 papers.
Paul Serret (in 1855, a half century before Cesàro), and independently Aleksander Simonič (2019), used Cesàro's method to prove Lexell's theorem. Let be the center in the plane of the circular arc to which side projects. Then planar quadrilateral is a right kite, so the central angle is equal to the external angle at the triangle's spherical excess Planar angle is an inscribed angle subtending the same arc, so by the inscribed angle theorem has measure This relationship is preserved for any choice of therefore, the spherical excess of the triangle is constant whenever remains on the Lexell circle which projects to a line through in the plane. (If the area of the triangle is greater than a half-hemisphere, a similar argument can be made, but the point is no longer internal to the angle
Perimeter of the polar triangle
Every spherical triangle has a dual, its polar triangle; if triangle (shaded purple) is the polar triangle of (shaded orange) then the vertices are the poles of the respective sides and vice versa, the vertices are the poles of the sides The polar duality exchanges the sides (central angles) and external angles (dihedral angles) between the two triangles.
Because each side of the dual triangle is the supplement of an internal angle of the original triangle, the spherical excess of is a function of the perimeter of the dual triangle
where the notation means the angular length of the great-circle arc