56 CHAPTER 2 STRATEGIES FOR INVESTIGATING PROBLEMS
This procedure works for all three colors. One color must be used at l n/3 J vertices
at most (since otherwise, each color is used on more than l n/3 J vertices, which would
add up to more than n vertices). Choose that color, and we have shown that at most
l n/3 J guards are needed.
Thus, g( n) :::; l n/3 J. To see that g( n) = l n/3 J, we need only produce an example,
for each n, that requires l n/3 J guards. That is easy to do; just adapt the construction
used for the 9-gon above. If l n/3 J = r, just make r "spikes," etc. _
The next example (used for training the 1996 USA team for the IMO) is rather
contrived. At first glance it appears to be an ugly algebra problem. But it is actually
something else, crudely disguised (at least to those who remember trigonometry well).
Example 2.4.4 Find a value of x satisfying the equation5(VI=X+v'I+x) =6x+8�.
Solution: Notice the constants 5,6, 8 and the VI -x^2 and VI ±x terms. It all
looks like 3-4-5 triangles (maybe 6-8- 10 triangles) and trig. Recall the basic formulas:
sin^2 e +cos^2 e = 1,
. !!.-_
J
I-cose
SIn
2- 2'
!!.-_
J
I +cos e
cos
2- 2 ·
Make the trig substitution x = cos e. We choose cosine rather than sine, because
v'I ± cos e is involved in the half-angle formulas, but v'I ± sin e is not. This substitu
tion looks pretty good, since we immediately get v'I - x2 = sin e, and also
r,:;J
1 ± cos e
v'I ± x = VI ± cos e = v 2
2.
Thus the original equation becomes