Solutions to PS#3
MATH 355 –Spring 2009
p.73 #6. Let abe a line and let Bbe a point o¤ a. Prove that 'Ba'Ba'Ba'Bis a
re‡ection in some line parallel to a:
Proof: By Theorem 113b, ('Ba'B)a('Ba'B) = 'B(a)a'B(a):Let b='B(a) ; then
akbsince 'Bis a dilatation by Theorem 50:Furthermore, bab=b(a)=m;where m=b(a):
Thus mis the re‡ection of ain some line bparallel to a: Conclude that mis parallel to a:
p.75 #9. Use conjugation and Exercises 10 and 11 in Section 3.2 to prove that: A re‡ection reverses
orientation.
Proof: Let `be a line and let Pbe any point on `: Then m=P;O (`)is a line through the origin since
O=P;O (P):Note that `=O;P (m)and by Theorem 113b we have `=O;P(m)=O ;P m1
O;P =
O;P mP;O :Now by Exercise 10, P;O preserves orientation; mreverses orientation by Exercise 11;
and O;P preserves orientation by Exercise 10. Therefore `reverses orientation.
p.75 #10. Use Exercise 9 to prove Proposition 109: Even isometries preserve orientation; odd isometries
reverse orientation.
Proof: An even isometry is a composition of two re‡ections, each of which reverse orientation by Ex-
ercise 9. But reversing orientation twice preserves orientation. An odd isometry is a single re‡ection or a
composition of three re‡ections. A single re‡ection reverses orientation by Exercise 9. A composition of
three isometries reverses orientation three times, and consequently reverses the given orientation.
p. 76 #4. Given a translation and a non-identity rotation rotation C;;prove there is a point Bsuch
that C;=B;:
Proof: Let mbe the line through Cperpendicular to the direction of translation; let `and nbe the
unique lines such that =m`and C;=nm:Then C;=nmm`=n`with `km:
Since C;6=; =20and 1
2=2180:Thus nis a transversal for parallels `and m: Let B=`\n;then
the corresponding angles from `to nand from mto nmeasure 1
2. Therefore B ;=n`=C;.
p.80 #3e. Let ;and denote the respective interior angles of 4ABC: Let a; b and cdenote
the sides of 4ABC opposite from A; B and C; respectively. Show that C;2 B ;2 A;2 =2but
A;2 B;2 C;2 =:
Proof: By Theorem 124, =abcis a glide-re‡ection. Thus C;2 B;2 A;2 = (ab)
(ca)(bc) = (abc)2=2=6=by Theorem 123c. On the other hand, A;2 B ;2
C;2 = (bc)(ca)(ab) = b(cc)(aa)b=bb=:
p.80 #5. Let Aand Bbe distinct points and let cbe a glide re‡ection with axis c. Prove that A;B c=
cA;B if and only if A;B (c) = c:
Proof: First note that A;B c=cA;B if and only if c=A;B c1
A;B:By Theorem 126,
A;B c1
A;B =c0is a glide re‡ection with axis c0=A;B (c):But c=c0implies c=c0so that
A;B (c) = c: Conversely, if A;B (c) = c, we may substitute and write c=A;B (c):But again by Theorem
126 we have A;B(c)=A;B c1
A;B:Therefore c=A;B c1
A;B and equivalently, A;B c=cA;B :
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