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Classical geometries - MAT402
Instructor: Ester Dalvit
HW 3 - Solutions to selected exercises
Ex 1. See http://www.math.toronto.edu/dalvit/courses/mat402/sphere.pdf
Ex 2. The same construction as in Euclidean geometry works.
Ex 3. Consider the three great circles defined as intersections of the coordinate planes (xy-plane, xz-plane,
yz-plane) with the sphere. Each of the eight regions defined in this way is a 3-right-angled triangle since
coordinate planes are perpendicular. Then the eight triangles are congruent by congruence criterion AAA
for spherical triangles. So the area of each of them is 1/8 the area of the sphere, i.e. πR/2 if the sphere has
radius R.
Ex 4. Two solutions:
(1) Three squares meet in each vertex, the total angle around a vertex is 2π, the angles of the squares are
all congruent, so their measure is 2π/3.
(2) Since the squares are congruent, they have the same area, so the area A of each of them is 1/6 the area
of the sphere. Each square has 4 conguent angles α, so it can be cut into two triangles whose total spherical
eccess s = 4α − 2π is related to the area A of the square by a straightforward consequence of Girard’s
theorem: A = s R (R is the radius of the sphere). [More generally this is true for any quadrilateral and
similarly for any polygon with more sides].
Then we have two expressions for the same area:
4πR
= s R
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so 4α − 2π = s = 2/3π so α = 2π/3.
Ex 5. The statement of Thales’ theorem can not be considered in spherical geometry as it is, because there
exist no parallel lines on the sphere.
Pythagora’s theorem does not hold in spherical geometry. A counterexample is the 3-right-angled triangle,
which is also equilateral with sides of length l = πR/2: Pythagora’s theorem would imply that l2 = 2l2 .
Remark, not requested in the homework: there is a theorem relating the lengths a, b of legs and the length
c of the hypotenuse of a spherical triangle. It says that cos(c/R) = cos(a/R) cos(b/R) where R is the radius
of the sphere. The proof uses elementary trigonometry and scalar products.
Ex 6. The Euclidean radius of the yellow circle in the figure is r0 , its spherical radius is α. If the sphere has
radius R, then r0 = R sin α.
Note: a spherical circle has two centers and two radii (take the antipodal point to A), we consider the radius
that is shorter/narrower.