MATH3061 Geometry and Topology – Semester 2, 2022
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MATH3061 Geometry and Topology – Semester 2, 2022 – Geometry Assignment
Question 1. Let ↵ : E ! E be defined by ↵(x,y) = (6 − 2x,−y).
(a) Prove that ↵ is a transformation.
(b) Find all fixed points of ↵ .
(c) Is ↵ an involution, isometry, affine transformation?
Question 2. Let Q = (2, 1) and v = [ ]6(2) .
(a) Find the Cartesian equation of the line ` through Q in direction v.
(b) Find the Cartesian equation of the line m such that pQ,−⇡/2 = σ` σm .
(c) What is the isometry Tv ◦ pQ,−⇡/2 ◦ T1 ?
Question 3. Denote by a, b the lines with the Cartesian equations y = 0, x = 0 respectively,
and i = [ ]0(1) , j = [ ]1(0) .
(a) Find the functions f(x,y) and g(x,y) such that γa,i(x,y) = (f(x,y),g(x,y)). (b) If γa,i(P) = Q, show that the midpoint of the segment PQ belongs to a.
(c) What is the isometry γa,iγb,j?
Question 4. Let A, B, C be the vertices of an equilateral triangle in the plane.
(a) Describe all isometries which map the set {A,B,C} to itself. (Your answer should consist
of a list specifying the reflections, rotations, translations and/or glide-reflections with the requested property.)
(b) Is the set G of all those isometries a group?
(c) Find all subsets of G which are groups.
Question 1. (a) We need to prove that ↵ is a bijection, i.e. that for each point (u,v) 2 E , there is a unique point (x,y) 2 E, such that ↵(x,y) = (u,v). We have:
u
The last system gives those unique values for x and y.
(b) ↵(x,y) = (x,y) , 6 − 2x = x, −y = y , x = 2, y = 0. Since the only fixed
point is (2, 0), the fixed-point set is {(2, 0)}.
(c) If ↵ is an involution, then ↵ ◦ ↵ = L. We calculate:
(↵ ◦ ↵)(x,y) = ↵(6 − 2x,−y) = (6 − 2(6 − 2x), −(−y)) = (4x − 6,y).
We see that it is not the identity map: for example, (↵ ◦ ↵)(0, 0) = (−6, 0) (0, 0).
Let A = (0, 0), B = (1, 0). We have d(A,B) = 1 and d(↵(A), ↵(B)) = d((6, 0), (4, 0)) = 2 d(A,B).
Let ` be a line with the Cartesian equation ax + by + c = 0, (a,b) (0, 0). As calculated in
(a), for ↵(x,y) = (u,v) we have: x = 3 − 2 , y = −v, thus the points (u,v) belongs to ↵(`) if and only if it satisfies the equation: a ⇣3 − ⌘ − bv + c = 0, i.e. au +2bv − 6a − c = 0, which
is a line again.
We conclude that ↵ is neither an isometry nor an involution, but it is an affine transforma- tion.
Question 2. (a) A vector orthogonal to v is u = [ 1] = [ 2], u · v = 0. Point X = (x,y)
−−!
is: 3x − y − 5 = 0.
(b) Line m contains Q and \(m,`) = − 4 . If u1 is a vector orthogonal to m, we then have \(u1 , u) = − , i.e R⇡/4u is collinear with u1 :
] u = ][ 1] =^2 [ ]1(2) ,
so we can set u1 = [ ]1(2) . Now, the line m will be the set of all points X(x,y) satisfying
−−!
(c) Note that Tv (Q) is a fixed point of that isometry, and that the isometry is even, as a
composition of even isometries. Now, notice that translations do not change direction veors
Question 3. (a) We have: γa,i = σaTi , and σa (x,y) = (x,−y), Ti (x,y) = (x + 1,y). Thus, γa,i(x,y) = (x +1, −y), i.e. f(x,y) = x +1, g(x,y) = −y.
(b) Denote P1 = σa (P), P2 = Ti (P). Then PP1 QP2 is a rectangle and a is one of its axes of symmetry. Thus, the midpoint of the diagonal PQ belongs to a.
(c) We have: γa,i = Ti σa , γb,j = Tj σb , Tj = σa σ` , where ` is the line with Cartesian equation
y = − , and Ti = σm σb , where m is the line with Cartesian equation x = . We calculate:
γa,iγb,j = Ti σaTj σb = Ti σa σa σ` σb = Ti σ` σb = σm σb σ` σb = σm σb σb σ` = σm σ` . We used here that since σa , σb are involutions and σ` σb = ⌘`\b = σb σ` . Since m and ` are perpendicular lines, the isometry σm σ` is a half-turn about the intersection point of those two lines.
We conclude that γa,iγb,j = ⌘M , where M = ✓ , − ◆ .
Question 4. (a) Let a, b, c be the bisectors of the angles of the triangle ABC, and S their
(b) Yes, since L 2 G; the composition of two transformations which map any set into itself will also fix that set; and, if a transformation maps a set onto itself, then its inverse will also
map it onto itself.
(c) The subgroups of G are {L, }, {L, σa}, {L, σb}, {L, σc}, {L, pS,2⇡/3 , ps,−2⇡/3}, G .
2022-12-12