MATH 243   Winter 2008 
                            Geometry II: Transformation Geometry 
                            Solutions to Final Examination Practice Problems 
                            Department of Mathematical and Statistical Sciences 
                            University of Alberta 
              Question 1. Show that Thomsen’s relation is still true if each reflection σx is replaced by a halfturn σX,
              that is, show that if A, B, and C are three distinct points in the plane, then
                            σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ =ι.
                             C A B C A B A B C A B C B A C B A B A C B A
              Explain how the proof must be changed.
              Solution: Note that if A, B, and C are three distinct points in the plane and D is a fourth point such
              that ABCD is a parallelogram, then
                                                    −−→  −−→    −−→ −−→   →−
                                      σDσCσBσA=τ2CDτ2AB =τ2(CD+AB) =τ0 =ι,
              so that
                                                    σCσBσA =σD,
              and
                                                 (σ σ σ )2 =σ2 =ι.
                                                   C B A       D
              Thus, each product σC σB σA is an involution, and therefore
                            σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ
                             C A B C A B A B C A B C B A C B A B A C B A
                             =σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ
                                C A B C A B A B C A B C B A C B A C C B A C B A
                             =(σ σ σ )2(σ σ σ )2(σ σ σ )2(σ σ σ )2
                                 C A B     A B C     B A C     C B A
                             =ι,
              which is Thomsen’s relation.
              Question 2. Given a line b and a point A, show that the following conditions are equivalent:
                (a) A ∈ b,
               (b) σAσb = σbσA,
                (c) σb(A) = A,
               (d) σA(b) = b,
                (e) σbσA (or σAσb) is an involution,
                (f) σbσA is a reflection in the line through A perpendicular to b.
               Solution:
               (a) =⇒ (b). If A ∈ b, then
                                                     σbσAσb = σσb(A) = σA
               since σb(A) = A if A ∈ b. Therefore,
                                                         σAσb = σbσA.
               (b) =⇒ (c). Suppose that σ σ = σ σ , then
                                         A b    b A
                                                         σ σ σ =σ
                                                          b A b    A
               and
                                                        σbσAσb = σσb(A)
               imply that σb(A) = A.
               (c) =⇒ (d). Suppose that σb(A) = A, since the only fixed points of σb are on b, then A ∈ b. Now let P
               be an arbitrary point on b, then σA(P) is on the line joining A and P, so that σA(P) ∈ b also, therefore
               σA(b) ⊆ b. However, σA(b) is a line, and so σA(b) = b.
               (d) =⇒ (e). Suppose that σA(b) = b, then
                                                     σ σ σ =σ       =σ ,
                                                      A b A     σA(b)   b
               and so
                                                σ σ σ σ =σ σ σ σ =σ2 =ι,
                                                 A b A b     b A b A     b
               that is, σAσb and σbσA are involutions.
               (e) =⇒ (f). Suppose that σAσb is an involution, and let a be the line through A perpendicular to b.
               Since σAσb is an involution, then
                                                   σAσbσAσb(A) = ι(A) = A,
               so that
                                                    σ σ σ (A) = σ (A) = A,
                                                     b A b       A
               and the halfturn σ σ σ = σ    has A as a fixed point, and therefore A = σ (A), which implies that A ∈ b.
                               b A b     σb(A)                                     b
               Now let P ∈ a be arbitrary and let P′ = σ (P), since b is perpendicular to a, then P′ ∈ a and σ (P′) = P,
                                                     b                                               A
               so that σAσb(P) = P for all points P ∈ a, and σAσb is a reflection in a. The inverse σbσA is also a reflection
               in the line a.
               (f) =⇒ (a). Suppose that σ σ  is a reflection in the line a through A perpendicular to b. Let b′ = σ (b),
                                         b A                                                              A
               then the line b′ is parallel to b, and since A ∈ a, then
                                                     A=σbσA(A)=σb(A),
               that is, A is a fixed point of σb, and so A ∈ b.
              Question 3. If A 6= C, show that the following conditions are equivalent:
                (a) B is the midpoint of AC,
                (b) σCσB = σBσA,
                (c) σB(A) = C, and
                (d) σBσAσB = σC.
              Solution:
              (a) =⇒ (b). Suppose that B is the midpoint of AC, then
                                                    −−→  −→     −−→
                                                   2AB=AC=2BC,
              so that
                                              σ σ =τ −−→ =τ −−→ = σ σ .
                                               C B    2BC    2AB    B A
              (b) =⇒ (c). Suppose that σCσB = σBσA, then
                                                     σ σ σ =σ
                                                      B C B     A
              and
                                                   σBσCσB =σσB(C)
              so that σ (C) = A, and σ (A) = σ2 (C) = C.
                      B             B       B
              (c) =⇒ (d). Suppose that σB(A) = C, then
                                                 σ σ σ =σ       =σ .
                                                  B A B    σB(A)   C
              (d) =⇒ (a). Suppose that σBσAσB = σC, then
                                                     σ σ =σ σ ,
                                                      B A    C B
              so that
                                                     τ −−→ = τ −−→,
                                                      2AB    2BC
                     −−→  −−→
              so that AB = BC.
              Therefore                          −→ −−→ −−→       −−→
                                                 AC=AB+BC=2AB
              and                                −→ −−→ −−→       −−→
                                                 AC=AB+BC=2BC,
              so that B is the midpoint of the line segment AC.
                 Question 4. Using halfturns, prove that the diagonals of a parallelogram bisect each other.
                 Hint: Show that if N is the midpoint of the diagonal AC of parallelogram ABCD so that σAσN = σNσC,
                 then N is the midpoint of BD also, that is, that σ σ  =σ σ .
                                                                  D N      N B
                 Solution: Since ABCD is a parallelogram, then
                                                             σDσCσBσA =ι,
                 that is,
                                                              σBσA =σCσD.
                 If N is the midpoint of the diagonal AC, then
                                                             σAσN =σNσC,
                 and so
                                                     σ σ σ =σ σ σ =σ σ σ ,
                                                      B A N       B N C      C N B
                 where the last equality follows from the fact that σBσNσC is an involution.
                 Therefore,
                                                           σCσDσN =σCσNσB,
                 so that
                                                             σDσN =σNσB,
                 and N is also the midpoint of the diagonal BD.
                 Question 5. If the isometry α is an involution, show that for any point P in the plane, the midpoint of the
                 line segment joining P and α(P) is a fixed point of α.
                 Solution: If P is a fixed point of the involutive isometry, then we are done.
                 If P is not a fixed point of α, let M be the midpoint of the line segment from P to α(P), then
                                                          d(M,P)=d(M,α(P)),
                 and since α is an isometry which is an involution, we have
                                 d(α(M),α(P)) = d(M,P) =d(M,α(P)) =d(α(M),α2(P)) = d(α(M),P).
                 Therefore, α(M) is on the perpendicular bisector of the segment from P to α(P).
                 Now, α maps the line joining P and α(P) onto itself, so that α(M) is on the line segment joining P and
                 α(P), and
                                                       d(P,α(M)) =d(α(P),α(M)),
                 so that α(M) = M, and therefore M is a fixed point of α.