304 CHAPTER 8 GEOMETRY FOR AMERICANS
how to construct an equilateral triangle, right? (If not, do Problems 8.2.28 and
8.3.3 1 !)
- Construct a line perpendicular to this rotated segment that pases through its
endpoint. This will be f4 , and its intersection with f 2 is the point B. •
That was fun and rather easy, but we won't always be so lucky. Sometimes we
will need to combine more than one rotation. What happens when we compose two
rotations? First, a simple lemma.
Fact 8.5.6 Let f be a line and let f' be its image under rotation by e about the center
C. Then the angle between f' and f is also e.
This is not at all surprising. We are asserting that LX PY = e in the picture below.
This follows from easy angle chasing. Do it!
,
, ,
, e ,
4-..' , ,
, ,
·c
This lemma should allow you to easily prove the following composition facts.
Fact 8.5.7 The composition of a rotation by a with a rotation by f3 (possibly about a
different center) is
- a rotation by a + f3, if a + f3 =I- 360 °.
- a translation (not the identity), if a + f3 = 360 ° and the centers are different.
- the identity transfonnation, if a + f3 = 360 ° and the centers are the same.
Fact 8.5.8 When the composition of two rotations is another rotation, here is a way to
find the center. Let RA,a denote rotation by a about A. Likewise, define the rotation
RB,{3. Draw rays AX, BY so that BAX = a /2 and Y BA = f3 /2. The intersection of AX
and BY is the center of the rotation RB,{3 0 RA,a. (Remember that composition To S
means "do S, then T .")
Fact 8.5.9 The composition of a rotation and a translation (taken in either order) is
a rotation by the same angle, but with a different center. (Finding the center is an
exercise.)
We conclude our discussion of rotations with a special case of a wonderful classic
problem.
Example 8.5.10 Let XYZ be a triangle. Erect exterior isosceles triangles on the sides
XY, YZ, XZ, with vertex points A, B, C, respectively, such that LA = 60 °,LB= 120 °,LC=