Normat 60:3, 133–134 (2012) 133
Lexell’s theorem
Ulf Persson
Matematiska Institutionen
Chalmers Tekniska Högskola och
Göteborgs Universitet
ulfp@chalmers.se
Introduction
Given a line segment AB in the plane, and consider all triangles with AB as a base
and with fixed area. What is the locus of the opposite vertex? This has an obvious
answer, as all those triangles must have the same height h. The answer is hence
the line parallel to that given by the base and hence w ith a fixed distance from it.
Lexell’s theorem states that in the case of a sphere, we should replace the line
with a (small) circle. I.e. the intersection with the sphere of a plane, not necessarily
passing through the center of the sphere (the latter case corresponding to a great
circle, i.e. a geodesic on the sphere).
At first one may be a bit puzzled. Think of a circle in the plane and a segment AB
outside it. For some points C on the circle the triangle ABC is bound to intersect
the circle, meaning that part of its circumference is contained in the interior of
the triangle. For any point C
Õ
on such an arc the triangle ABC
Õ
is clearly properly
contained in ABC and hence has strictly smaller area. However, by choosing points
A
Õ
,B
Õ
such that AA
Õ
,BB
Õ
are tangent to the circle, they determine an arc, points
C on which, determine triangles ABC which only intersect the circle in C.The
same argument holds for any convex area in the plane. However, the plane picture
is misleading, on a sphere this does not necessarily happen.
An exercise in Absolute Geometry
Consider the triangle ABC and let A
1
,B
1
be the midpoints on the lines AC and AB
respectively
1
. Let L be the line through them. Furthermore let A
p
,B
p
,C
p
be the
projections of A, B, C on L. Now the triangles AA
1
A
p
and A
1
,C,C
p
are congruent.
(Two angles equal and the hypothenuse). The same argument works as well on
BB
1
B
p
and B
1
CC
p
. From this we conclude that the area of the quadrilateral is
equal to that of the triangle ABC, and that the distances of A and B to the line
L are equal, and equal as well as to the distance of C to L. Note that we do not
1
The following argument is referred to in ’On some classical constructions extended to hyper-
bolic geometry’ by A.V.Akopyan, arxiv.org/pdf/1105.2153.pdf
134 Ulf Persson Normat 3/2012
use anywhere that the geometry is Euclidean, but it works as well in the spherical
as hyperbolic case. We also note that the distances involved depend on the area of
the triangle and are in fact uniquely determined by that. Thus if we fix the are a of
the triangle, the locus of C will form an equidistant curve to L. In the Euclidean
case this simply means that the locus is a line parallel to L and hence the line AB,
which as noted above could have been seen right away.
A
A
1
A
p
B
B
1
B
p
C
C
p
L
A
A
p
A
*
B
B
p
B
*
C
C
p
The Spherical Case
On the sphere the equidistant curves to a line L are given by the (small) circles
cut out by planes parallel to the plane cutting out L. (Think of L as the equator,
and the circles as the latitudes). This proves Lexell’s theorem. By considering the
anti-podal points A
ú
,B
ú
of A, B respectively, we can think of those curves as the
pencil of circles through their base points A
ú
,B
ú
Not so Fast!
Can this be true? What happens to the area of a triangle ABA
ú
? Could this take
any value? Or if C varies close to A
ú
any small change of C leads to a great
change in area. But if A and A
ú
are antipodal, there is an entire pencil of lines
(great circles) joining the two points, and for each choice we get a triangle with its
specific area. This also s hows the extreme sensitivity to area, when the point C is
chosen close to one of the anti-podal points of the end points of the base.
Bounds on areas of spherical triangles
Given any three points on a sphere, it determines in a sense two triangles, each
being the complement of the other. Those have the same area only if the points lie
on a great circle, and then it is given by 2fi (half of the area of the sphere). The
great circle is the largest circle through two points. The smallest is the one, having
the segment as the diameter. A nice exercise for the reader would be to compute
the area corresponding to the minimal circle.
