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University of Toronto Mississauga
Department of Mathematics
Instructor: Ester Dalvit
TA: Jennifer Vaughan
Classical geometries - MAT402
Some solutions for the midterm practice
Exercise 6. Show that any three points not on a line lie on a unique circle (hint: the centre must be
equidistant to any pair of the points). Give a Euclidean construction to find it.
Use the hint. Let P = (p1 , p2 ), Q = (q1 , q2 ), R = (r1 , r2 ) be three points not on a line. The perpendicular
bisector (axis) of segment P Q is the set of points X = (x1 , x2 ) such that d(P, X) = d(Q, X).
Similarly for the perpendicular bisector of segment QR.
Squaring both equations (distance is always non negative) we get the equations of two lines which are not
parallel since P, Q, R are not on a line. Hence they intersect in a single point X. Then the three points lie
on the circle with center X and radius d(P, X).
The Euclidean construction is the same: draw the perpendicular bisectors and intersect them.
Exercise 8. Spherical polygons are intersections of semispheres. What is the area of a spherical polygon?
Give a formula and justify it.
If the polygon has n angles αi (1 ≤ i ≤ n), then it can be cut into n − 2 triangles whose total angle sum
is the angle sum of the polygon. Since by Girard’s theorem the area of each triangle is the spherical excess
times the radius of the sphere, εR2 , then the area of the polygon will be
!
n
X
αi − (n − 2)π R2
i=1
Exercise 9. Do the following facts hold in Euclidean plane geometry and in spherical geometry? Why?
• Given two points, there is a unique line passing through them.
Holds in Euclidean geometry (it is a straightforward consequence of the axioms: there is line through
the 2 points and uniqueness comes from the fifth postulate).
Does not hold in spherical geometry: take two antipodal points.
• Similar triangles are congruent.
Does not hold in Euclidean geometry.
Holds in spherical geometry: it is a consequence of Girard’s theorem.
• Given a segment of length a, there is a segment with length 2a.
Holds in Euclidean geometry (compare with Euclid’s second postulate).
Does not hold in spherical geometry: on a sphere of radius R lines have length 2πR so length of
segments is also bounded.
• Given three points A, B, C, there is a unique line passing through A and perpendicular to BC.
Holds in Euclidean geometry: use analytic geometry to prove it; you should also know a way to
construct it.
Does not hold in spherical geometry: if B, C are on a line and A is a pole of that line all lines through
A are perpendicular to BC.
• There is a unique triangle with angles measuring 55◦ , 82◦ , 43◦ .
Does not hold in Euclidean geometry: the sum is 180◦ .
Does not hold in spherical geometry: the sum is greater than 180◦ .
• The area of a triangle can not exceed a certain quantity (if so, which quantity?).
Does not hold in Euclidean geometry since we can extend segments and area of a right angled triangle
is proportional to the product of the legs.
Holds in spherical geometry. The area is less than the area of a semisphere, since by definition the
triangle is a intersection of semispheres.
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Exercise 10. Show that the set of rotations about a fixed point in the Euclidean plane is a group under
composition. Is it a commutative group?
Rotations about a fixed point C can be obtained by first translating of a vector −C, then rotating about O
and then translating back of a vector C. In formulas RC,θ = TC ◦ Rθ ◦ T−C .
Composition of two such rotations of angles θ, ψ is then RC,ψ ◦ RC,θ = (TC ◦ Rψ ◦ T−C ) ◦ (TC ◦ Rθ ◦ T−C ) =
TC ◦ Rψ ◦ Rθ ◦ T−C . That is, we only need to show that rotations about O form a group, i.e. it is closed
under composition, the identity isometry can be expressed as a rotation about O (of angle 0), for each Rθ
there is an inverse, the composition is associative.
This group is commutative because
cos θ − sin θ
cos ψ − sin ψ
cos θ cos ψ − sin θ sin ψ − cos θ sin ψ − sin θ cos ψ
=
sin θ
cos θ
sin ψ
cos ψ
sin θ cos ψ + cos θ sin ψ − sin θ sin ψ + cos θ cos ψ
Which is (using trigonometric identities)
cos(θ + ψ) − sin(θ + ψ)
sin(θ + ψ) cos(θ + ψ)
which is a rotation Rθ+ψ about O of angle θ + ψ. Then writing the product with inverse order of factors
will give a rotation about O of angle ψ + θ which is clearly the same as Rθ+ψ .
If you are disappointed because I revealed much of the answer in class, you can show that the set of orientation
preserving isometries of R2 is a group and check if it is commutative.
Exercise 12. What are the possible types of isometries obtained from the composition of a reflection and
a rotation? Justify your answer.
A rotation is a composition of two reflections, so composing a reflection and a rotation we get the composition
of 3 reflections. This is orientation reversing thus it can be a reflection or a glide reflection.
Example to obtain a reflection (without the glide part): reflection S{x=0} and rotation S{y=0} ◦ S{x=0} .
Composition is S{y=0} ◦ S{x=0} ◦ S{x=0} = S{y=0} .
Yet in general it will have no fixed points so it will be a glide reflection.
Exercise 13. How many reflections are necessary and sufficient to describe rotations in R2 ? Justify your
answer.
Two reflections in incident lines. Reason: by the three reflections theorem any isometry can be written as the
composition of 1, 2 or 3 reflections. Rotations are orientation preserving, so can be obtained only composing
an even number of reflections.