1. Draw a figure for each case of Table 2.3.

Ans:

3.Using congruent triangles, draw a specific example showing that the product of a reflection and a rotation could be a glide reflection.

Ans:

5. Using congruent triangles, draw a specific example showing that the product of two glide reflections could be a translation.

Ans:

7. Does the set of all glide reflections form a subgroup of all the motions of a plane?

Ans: No. Solution: This set is not closed (see exercise 5 above), so it cannot be a subgroup.

9. Does the set of all translations and all reflections form a subgroup of the group of all motions of the plane?

Ans: No.

11. In part 3 of the proof outline of Leonardo's theorem, show that the number of rotations and the number of reflections are equal.

Solution: Let n be the number of rotations and m be the number of reflections. For each reflection, the product of this reflection and one of the rotations is an odd isometry in the group, so nm. But the m relections multiplied on the right by one particular relection in the group gives m distinct even isometries in the group, so mn. Thus, m = n.

13. Name two properties of sets of points that are not invariant under the group of motions.

Ans: position and slope of a segment.

15. Prove that the image of a segment not passing through the center of rotation is a congruent segment under a rotation.

Solution:

Since OA and OA' are congruent, OB and OB' are congruent, and angles AOB and A'OB' are congruent (both are equal to the angle of rotation minus the angle AOB') the two triangles are congruent by SAS. Thus, the corresponding sides AB and A'B' are congruent.

17. Prove that the product of two halfturns is a translation.

Solution: If the two half-turns have the same center, then the product is the identity which is a translation. If the half-turns have different centers then the translation is in the direction of the line joining the centers. In the diagram below, line AA'' is parallel to and twice the measure of OO', since OO' joins the midpoints of the sides of triangle AA'A'' and line BB'' is also parallel to and twice the measure of OO', since OO' joins the midpoints of the sides of triangle BB'B''. Therefore, ABB''A'' is a parallelogram, and AB and A''B'' are parallel and of the same length. Thus, there is a translation by a vector in the direction of OO' and of magnitude twice that of the line segment OO' which maps AB to A''B''.

19. Show analytically that a translation has no invariant points.

Solution: If (x, y) is an invariant point of a translation in the direction of (a, b), then we would have that:

21. Use Theorem 2.4 to prove that if three noncollinear points of a plane remain invariant under a motion, the motion is the identity.

Solution: The identity transformation is one of the motions that causes three noncollinear points to remain invariant. By Theorem 2.4, there is only one such transformation; therefore, it must be the identity.

23. Prove analytically that the angle between two intersecting lines is an invariant under the set of all translations.

Solution: It is sufficient to prove this for two lines that meet at the origin. Let these lines be y = m₁x and y = m₂x with m₁m₂. Consider the translation given by x' = x + a and y' = y + b. The tangent of the angle between the two lines is (m₂ - m₁)/(1 + m₁m₂). After the translation the lines are y' - b = m₁(x' - a) and y' - b = m₂(x' - a) respectively. But the tangent of the angle between these two image lines is still (m₂ - m₁)/(1 + m₁m₂). So, this angle is an invariant of the set of all translations.

25. Prove analytically that the angle between two intersecting lines is an invariant for a reflection about the y-axis..

Solution: Let the two lines have equations y = m₁x + b₁ and y = m₂x + b₂, with m₁m₂. Then under reflection about the y-axis these lines become y = -m₁x + b₁ and y = -m₂x + b₂. The tangent of the angle between the original lines is (m₂ - m₁)/(1 + m₁m₂), while the tangent of the angle between their images is (-m₁ -(- m₂))/(1 +(- m₁)(-m₂)). So, the angles are equal (but their orientation has been reversed).