1. MATH 3581 College Geometry Spring 2011 University of Memphis Mathematical Sciences Dwiggins Homework Assignment # 6 Rigid Motions b D ℓ # 16 # 17 # 18 # 19 Part I. # 16. Translation of the given triangle along the given vector b. # 17. 120º rotation (counter- clockwise) of the given triangle about the point D. # 18. Reflection of the given triangle about the line ℓ. # 19. A glide reflection, first reflecting about the line ℓ, and then translating along a vector which is parallel to ℓ and has the same length as b. # 20–23 is a similar set of exercises, but it would be messy to try to show them all in the same figure, so I’ll omit these from the solutions. 120º # 24. (a) In the illustration at right, rectangle R1 is rotated 180º about point O in order to obtain rectangle R2, congruent to R1. R2 is also obtained from R1 by translating along the vector b. (b) As illustrated at right, it is possible to have congruent squares S1 and S2 without being able to obtain S2 by translating S1. (The congruence requires a rotation followed by a translation or reflection.) 180º R1R2 O b (c) Two circles  (O 1, r 1 ) and  (O 2, r 2 ) are congruent if r 1 = r 2 but O 1  O 2. S1S2

2. # 25. (a) As illustrated at right, reflection about the altitude through the apex of an isosceles triangle may be used to prove the two medians to the congruent sides are also congruent. (Note: this is actually the method Euclid used to prove the base angles in an isosceles triangle are congruent.) (b) When two parallel lines are cut by a transversal, we can use translation along the transversal (indicated at right by the vector b) in order to prove corresponding angles must be congruent. (c) If we reflect about either of the diagonals in a rhombus we obtain a congruent figure. Since reflection involves perpendicular bisectors, this may be used to give a proof that the diagonals in a rhombus are perpendicular. (Other answers are also possible.) (d) If a parallelogram is not a rhombus then we can’t use the same method of proof as in part (c). However, we can still use 180º rotation about the center of the parallelogram in order to prove each diagonal divides the parallelogram into two congruent triangles. (e) Suppose two circles intersect at two points A and B, and let ℓ denote the line through the centers of the two circles. Each circle is symmetric about this line of reflection, which means A and B must be the reflected images of each other. But this then means ℓ must be the perpendicular bisector of the chord joining A and B, as required. 1 b 2 180º A ℓ B

3. X X O (2,0) (  3,1) ℓ ℓ x y Illustration for # 9. Illustration for # 13.Illustration for # 14.

4. x y P Q Q P Ox y P Q Q P O1O1 x y P P