AT 2355 Introduction to Geometry – THIRD ASSIGNMENT -

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MAT2355 – I NTRODUCTION TO G EOMETRY – T HIRD A SSIGNMENT P ROFESSOR : H ADI S ALMASIAN Deadline: October 17 Name

Student Number

? ? ? Please print the assignment, write your solutions in the designated spaces, and staple the sheets together.

Problem Grade

1

2

3

4

5

6 Total out of 17

Q UESTION 1. In this question you should only write the final answers. Simplify your answers as much as possible. (a) [1 point] Compute the product of the following quaternion numbers: (−1 − i + 2k)(i − j) =

1 + i + 3j + k

(b) [1 point] Let w = 1 − i + 2k. Then ... √ √ 1+1+4= 6 |w| =

and w−1 =

1 1 1 6 + 6i − 3k

(c) [1 point] Let R : E3 → E3 be the rotation in E3 with axis in the direction of the vector π ~a = (−1, 2, −2) and angle θ = . If p ∈ E3 denotes the point (0, 1, 0) then ...

3 √

√

3 3 1 13 2 − − i + j + − R(p) = 3 9 18 6 9 k (c) When using the formula for the spatial rotation, do not forget to normalize the vector u! (d) [2 points] Let R : E2 → E2 be the reflection such that R(0, 0) = (1, 1) and R(0, −1) = (2, 1).

Then R is the reflection by the line y = mx + b where ... m=

−1

and b =

1

(d) The formula for a reflection in terms of complex numbers is R(z) = γ1 z + γ2 . From the constraints of the problem we obtain: R(0) = 1 + i ⇒ γ2 = 1 + i and R(−i) = 2 + i ⇒ iγ1 + (1 + i) = 2 + i ⇒ γ1 = −i ⇒ R(z) = −iz + 1 + i.

The line of a reflection is the set of fixed points, i.e., points z = x + iy which satisfy R(z) = z. We have: R(z) = z ⇒ −i(x − iy) + (1 + i) = x + iy ⇒ y(1 + i) = x(−1 − i) + (1 + i) ⇒ y = −x + 1 Grading Scheme: (a): -.5 pts for each incorrect coefficient. (b): .5 +.5 (c): -.5 pts for each incorrect coefficient. (d): 1+1.

Q UESTION 2. For each of the following isometries determine if it is orientation-preserving or orientation-reversing. You do not need to justify your answer.

2

(a) [1 point] The map f : C → C given by f (z) = iz − i + 2. (b) [1 point] The composition of five reflections in E3 .

orientation-reversing

orientation-reversing

Grading Scheme: 1 + 1 Solution of part (c): The normal vector in the direction of ~a is ~u = k~~aak , that is:

Thus

1 2 2 ~u = − i + j − k 3 3 3 √ √ π π 3 1 1 2 2 3 1 1 1 w = cos( ) + sin( )~u = + (− i + j − k) = − i+ j− k 6 6 2 2 3 3 3 2 6 3 3

Therefore √

! ! √ 3 1 3 1 1 1 1 1 R(p) = w jw = + i− j+ k j − i+ j− k 2 6 3 3 2 6 3 3 ! ! √ √ 3 1 1 1 1 1 3 1 + i− j+ k − − i+ j+ k = 2 6 3 3 3 3 2 6 ! ! √ √ 3 1 13 3 2 = − − i+ j+ − k 3 9 18 6 9 ∗

3

Q UESTION 3. [3 points] For each of the following statements, determine if the statement is true or false.

Circle T for the true statements and F for the false statements. For each correct answer you will receive 1 point. For each incorrect answer you will receive -1 points. If you do not circle any of the letters next to a statement, you will receive 0 points for that statement.

T F For every two line segments AB and A0 B 0 of equal length in E2 , there exists a rotation R : E2 → E2 such that R(A) = A0 and R(B) = B 0 .

T F If the triangles 4ABC and 4A0 B 0 C 0 in E2 are similar (that is, ∠A = ∠A0 , ∠B = ∠B 0 , and ∠C = ∠C 0 ), then

AB AC BC Area(4ABC) = 0 0 = 0 0 = 0 0 AB AC BC Area(4A0 B 0 C 0 )

T F If 4ABC and 4A0 B 0 C 0 are two triangles in E2 such that ∠A = ∠B 0 and ∠B = ∠C 0 then ∠C = ∠A0 .

Answers from top to bottom: • False: if AB and A0 B 0 are parallel, then there is no such rotation.

Area(4ABC) AB 2 • False. The correct statement is Area(4A = ···.

0B0C 0) = A0 B 0 0 • True: this follows from ∠C = π − ∠A − ∠B = π − ∠B − ∠C 0 = ∠A0 .

Grading Scheme: See the question statement. The minimum grade will be zero (i.e., if their total grade is negative, they receive zero.

Q UESTION 4. [2 points] A pentagon is a polygon with five sides (possibly with different lengths). Prove that for every pentagon ABCDE we have ∠A + ∠B + ∠C + ∠D + ∠E = 3π.

D E C A

B

4

Connect A to C and D by line segments. Clearly ∠A + ∠B + ∠C + ∠D + ∠E is equal to the sum of the angles of the three triangles 4ABC, 4ACD, and 4ADE. Since the sum of angles of any triangle is π, it follows that: ∠A + ∠B + ∠C + ∠D + ∠E = π + π + π = 3π.

Grading Scheme: 1 pt for the idea of dividing into three triangles 1 pt for using the fact that the sum of angles of a triangle is π.

Remark. Any other valid solution receives for credit. Students who prove it for a regular pentagon do not receive any credit.

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Q UESTION 5. (a) [3 points] Using the method of Cartan–Dieudonné Theorem, write the isometry f : C → C , f (z) = −iz + i − 1 as a product of three reflections. Determine the equations of the fixed lines of the three reflections. You should give the equations of these fixed lines in the form y = mx + b. (b) [1 point] Can f be written as a composition of fewer reflections? Justify your answer. (a) According to the method explained in class, we have f = g ◦ R`0 where R`0 (z) = z is the reflection by the line y = 0 and g(z) = −iz + i − 1 = γ1 z + γ2 where γ1 = −i and γ2 = i − 1.

Since γ1 6= 1, by the formulas given in class we have g = R`2 ◦ R`1 where i − 1 −i − 1 R`1 (z) = z + − = z + 2i 1+i 1−i γ2 which is the reflection by the line `1 with complex equation z = t + 1−γ = t + i where t ∈ R. 1 Therefore the equation of `1 is y = 1. For R`2 we have R`2 (z) = −iz +

i − 1 (−i − 1)(−i) − = −iz + i + 1 1+i 1+i π

√ )t + i where t ∈ R. which is a reflection by the line`2 with complex equation z = e−i 4 t + i = ( 1−i 2 To find the equation of this line in the form y = mx + b, we set z = x + iy, and we obtain ( x = √12 t 1−i x + iy = ( √ )t + i ⇒ ⇒ y = 1 − x. y = 1 − √12 t 2

Conclusion. We can write f = R`2 ◦ R`1 ◦ R`0 where • `0 is the line y = 0, • `1 is the line y = 1, • `2 is the line y = −x + 1. (b) f cannot be written as a composition of fewer than three reflections. To see this, note that: • f is not a reflection: proof by contradiction. Suppose f is a reflection. Then it should have a fixed line. The points of the fixed line of the reflection should satisfy f (z) = z. If z = x + iy where x, y ∈ R then f (z) = z ⇒ −i(x − iy) + i − 1 = x + iy ⇒ y(1 + i) = x(−1 − i) + i − 1 ⇒ y = −x + i which is impossible because both x and y are real numbers.

• f is not a composition of two reflections because f is orientation-reversing but the product of any two reflections is orientation-preserving.

Grading Scheme: (a) 1 pt for the correct derivation of the equation of each of the lines `0 , `1 , and `2 . (Depending on the amount of work achieved, the grade can be 0, .5, or 1.) (b) .5 for ruling out (with a rigorous justification) each of the two cases.

6

Q UESTION 6. [1 point] You are standing at a point A between two mirrors m1 and m2 . You are looking at the reflection in m2 of your image in m1 (that is, point B in the picture). The mirrors suddenly start a counterclockwise rotation about their common axis (point O), but the angle between them does not change. What happens to the distance between you (point A) and your image (point B)? Does it increase? Decrease? Or does it remain fixed? Justify your answer.

B• m2 •A m1

O •

B is obtained from A by two reflections. The composition of the two reflections is a rotation by 2θ, where θ is the angle between the mirrors. As the mirrors rotate, this rotation does not change (because θ is fixed), and therefore the position of your image (point B) does not move either.

Therefore the distance between A and B does not change.

Grading Scheme: 0 or 1.