Download MAT 3272: Selected solutions to problem set 1 Chapter 4, Exercises

MAT 3272: Selected solutions to problem set 1
Chapter 4, Exercises:
21. Please refer to figure 4.30 in the text, page 140. First suppose BC < BA. (Do not suppose BA is
the greatest side of 4ABC; it need not be.)
The fact that ∠A < ∠CDB (EAT) and ∠CDB ' ∠DCB (Proposition 3.10: base angles) implies
that ∠A < ∠CDB is not a consequence of “transitivity”. It is a consequence of Proposition
3.21(b). Transitivity refers to only one relation at a time.
It is important to justify (or at least know how to justify) the fact that ∠DCB < ∠ACB. Point
D is in the interior of ∠ACB by Proposition 3.7. (We established A ∗ D ∗ B when point D
−−→
was introduced, in accordance with the definition of < for segments.) Therefore, ray CD is
−−→
−−→
between CA and CB by definition, which implies ∠DCB < ∠ACB by definition (since ∠DCB
is congruent to itself, by CA5).
Next, to prove the converse implication, suppose that ∠A < ∠C. Use Proposition 3.10 and the
implication above to derive contradictions if BC = BA or BC > BA, respectively. Trichotomy
(Proposition 3.21(a)) implies that BC < BA.
22. Please refer to figures 4.31 and 4.32. First, given 4ABC and 4A0 B 0 C 0 , assume BA ' B 0 A0 ,
BC ' B 0 C 0 , and ∠B < ∠B 0 .
Let us understand exactly what it means to “reduce to the case that A = A0 , B = B 0 , and C
interior to ∠ABC, so that you must show AC < AC 0 .” By definition of < for angles, there is a
−−−→
−−−→
−−−→
ray B 0 C 00 between B 0 A0 and B 0 C 0 such that ∠A0 B 0 C 00 ' ∠ABC. By CA1, we may choose C 00
so B 0 C 00 ' BC.
The essential point now is that, by SAS, 4ABC ' 4A0 B 0 C 00 ; therefore, AC ' A0 C 00 . By
Proposition 3.13(c), it thus suffices to show A0 C 00 < A0 C. To simplify the notation, we may now
relabel A0 as A, B 0 as B, and C 00 as C and forget about the original 4ABC. (But note that it
would have been less efficient to copy angle B 0 at B, since then, unable to apply the definition of
<, we would have needed more steps and some results about betweenness for rays to prove that
−−−→
−−→
−−→
BC is between BA and BC 0 .)
−−→
By the crossbar theorem, ray BC intersects CA0 at a point D. We now divide into cases depending
−−→
on the location of D; by definition of ray BC , one of the following must hold: B ∗ D ∗ C, D = C,
or B ∗ C ∗ D. (Clearly D 6= B, since A, B, and C 0 are not collinear.) We outlined in class the
sequence of arguments needed in each case. In the first and last cases, Proposition 4.5 must finally
be applied to 4ACC 0 .
To prove the converse implication, assume BA ' B 0 A0 , BC ' B 0 C 0 , and AC < AC 0 . (We may
revert to the original picture with two separate triangles.) Use trichotomy for angles ∠B and B 0
(and segments AC and A0 C 0 ), and eliminate two of the possibilities by contradiction, using SAS
and the result of the first part of this proof.
27. Once again, please refer to the diagram in the text, Figure 4.34. Some of you used the SaccheriLegendre Theorem to do this one. Your arguments are correct, but you actually only need to use
Corollary 1 of angle measurement (Theorem 4.3.A). While the S-L Theorem is obviously equivalent
to Corollary 1 (make sure you understand why this fact is obvious), the proof of the S-L Theorem
1
from Corollary 1 requires a delicate argument involving replacement of a triangle with another
triangle having the same angle sum but an arbitrarily small angle.
We argue as follows: ∠BCD + ∠DBC < 180◦ (Cor. 1 to Theorem 4.3). Since (∠BCD)◦ = 90◦ , it
follows that (∠DBC)◦ < 90◦ , and therefore that ∠DBC < ∠BCD. By Theorem 4.5, DC < DB.
By a similar argument, ∠A < ∠C. By the Exterior Angle Theorem, ∠ABD > ∠C. Therefore, by
the transitivity of angle ordering (Proposition 3.21 (d); this really is transitivity of the < relation),
∠A < ∠ABD. By Theorem 4.5, DB < DA.
28. Given 4DAC, let B be a point between A and C. Let E be the foot of the perpendicular from
←−→
D to line AC . The desired result (that DB is less than one of the two sides DA or DC) follows
easily from the result of the previous problem. The only trick is to correctly and efficiently divide
the proof into cases. Following a nice idea of Jeff Melone, we can reduce to three possibilities:
Claim: E = B or A ∗ B ∗ E or E ∗ B ∗ C. In the first instance, DB is less than both DA and DC.
In the second instance DB < DA by the result of problem 27, and in the last instance DB < DC
for the same reason.
←−→
To prove the claim, we use the fact that E is on line AC , so there are five cases (by BA3): E = A,
E = C, E ∗ A ∗ C, A ∗ E ∗ C, or A ∗ C ∗ E. In the first case, it is clear just by substituting into
the given that E ∗ B ∗ C. Similarly, if E = C, then A ∗ B ∗ E. If E ∗ A ∗ C, then combining this
information with A ∗ B ∗ C and applying Proposition 3.3 yields E ∗ B ∗ C. A similar argument
works if A ∗ C ∗ E. In case A ∗ E ∗ C, Proposition 3.5 implies that either E = B (one of the
possibilities of the claim), A ∗ E ∗ B, or B ∗ E ∗ C. Applying Proposition 3.3 again completes the
proof.
29. Recall the definition of convex: a set (of points) S is convex if A ∈ S ∧ B ∈ S ⇒ AB ⊆ S. (I
think the definition is easiest to understand if the defining condition is written in formal symbols,
showing it as a unit and removing any chance of ambiguity. Note that, in order to render the
definition into readable English, the word “whenever” is typically used as a synonym for “if”, as
in the following paraphrase.) In other words, a set is convex if, whenever two points are in it,
every point on the segment joining those two points is in it.
So let A and B be points in the interior of a circle with center O and radius OP , and let C be a
point between A and B. (We obviously don’t have to check the endpoints of the segment, since
they are given to be in the set.) By definition, OA < OP and OB < OP . By the result of the
preceding problem, OC < OA or OC < OB. In either case, by transitivity, OC < OP , so C is in
the interior of the circle.
Do you find working with concepts like convexity to be a bit difficult? This is probably because
their definitions involve a double layer of implication: “a set is convex if, if..., then....” As the
logic we use acquires more layers, it becomes more difficult to sort out. I recommend taking small
steps, using defined terms and logical concepts to keep track of what you are doing: to prove a
set is convex, we must verify the definition; the condition for a set to be convex, according to the
definition, is that if two points are in the set, then the entire segment joining them is contained in
the set; so we must prove that, if two points A and B are in set S, then AB ⊆ S; so let A and B
be points S (assuming the hypothesis); to prove AB ⊆ S, we must show that every point of AB is
an element of S; that is, we must prove that, if a point C is an element of AB, then C belongs to
S as well; so let C be an element of AB. By the definition of segment AB, we know .... You must
end up by showing that C satisfies the defining condition of set S (in this case, that OC < OP ).
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