Math 5-335 Fall 2003
Hints and
supplementary exercises for the October 16 homework
Supplementary exercise #1 Consider the same triangle ABC as in problem 30 of §2.5.
Supplementary exercise #2 Consider the same triangle ABC as in problem 30 of §2.5.
Hint for §2.5, problem 30 In the first part of the
problem, we're supposed to plot the line consisting of points whose points
have barycentric coordinates (r,s,t) where
r has a constant value. We can draw such a line
by finding two points on it. The two easiest such points to find
are the ones where the line r = (constant)
intersects the lines
Hint for §3.12, problem 64
(Re-written retroactively)
Here's a diagram of the setup:
Now, the text says that the only other loose end is about showing that
A and B are on the same side of
k. Actually, in the application of
Proposition 33 of Chapter 1 [about addition of angular measure],
one needs to check that B is in
the interior of angle
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and
. This was done in
Supplementary exercise #1. You don't need to use the formula here; it's
good enough just to solve on a case-by-case basis, noting that the lines
and
are characterized
by the equations
t = 0 and s = 0
respectively. Lines of the form s = (constant)
and t = (constant) are handled similarly.
and further
that A and D are on opposite sides of
while
B and E are on opposite sides of
. Under this setup, we need to
prove that
and
are opposite rays, so that
we can apply Proposition 34 of Chapter 1 [about complementary angles].
We can reformulate the question about opposite rays as being about
whether D - C and E - C are
negative scalar multiples of each other. Since we're given that
k is parallel to
, it is possible to
compare either D - C or E - C
with B - A, determining
in each case whether the vector in question is a scalar multiple of
B - A, with a positive coefficient or with a
negative coefficient.
ACD. Showing
that A and B are on the same side of
k is part of this. But it also involves
showing that B and D are on the
same side of
.
What was proved about oppositeness of rays in the other part of the
problem can be very helpful here.