Download HOMEWORK ASSIGNMENT #6 SOLUTIONS

MATH 3181 001
SPRING 1999
HOMEWORK ASSIGNMENT #6
SOLUTIONS
1. (Exercises 9.1.22 — 9.1.27, page 369) It can be proved that each of the statements in
Exercise 22–27 is equivalent to Euclid’s fifth postulate. Rewrite each sentence, using the
negation of the conclusion, to give a statement assume to be true in non-Euclidean geometry.
9.1.22: If a straight line intersects one of two parallel lines, then it will always intersect the
other.
If a straight line intersects one of two parallel lines, then it is not necessary
that it will intersect the other.
9.1.23: Straight lines parallel to the same straight line are always parallel to one another.
There exist two intersecting straight lines both of which are parallel to the
same line..
9.1.24: There exists one triangle for which the sum of the measures of the angles is π
radians.
For every triangle the sum of the measures of the angles is less than π
radians.
9.1.25: There exists a pair of similar but non-congruent triangles.
Every pair of similar triangles is congruent.
9.1.26: There exists a pair of straight lines the same distance apart at every point.
For every pair of straight lines, the set of points equidistant from each
other contains at most 2 points.
9.1.27: It is always possible to pass a circle through three non-collinear points.
There exist three non-collinear points through which no circle passes.
2. (Exercise 9.2.4, page 374) Without using an axiom of parallelism, prove that if a transversal
of two lines makes the alternate angles congruent, then the two given lines do not intersect.
Let m and n be the two lines cut by the
A
Q
transversal, k. Let A and B be the points of
n
intersection of k and n and k and m,
respectively. We are told that the angle at A and
M
the angle at B are congruent. Let M be the
midpoint of AB. Let P be the foot of M on m
m
B
P
and let Q be the point of intersection of n and
uuuuv
the ray PM . Then we have than AM ≅ BM ,
RMBP ≅ RMAQ (given), and RBMP ≅ RAMQ by vertical angles, since P, M, and Q are
©David Royster
Hyperbolic Geometry
FOR CLASSROOM USE ONLY
MATH 3181 001
HOMEWORK ASSIGNMENT #6 SOLUTIONS
SPRING 1999
collinear. Thus, ∆BMP ≅ ∆AMQ , and it follows that RBPM ≅ RAQM . This makes
RAQM a right angle. Hence, m and n have a common perpendicular. By our previous
theorem, they must be non-intersecting.
3. Let ¨ABDC be a Saccheri quadrilateral, so D
that RA and RB are right angles and
AD ≅ BC . Prove that RC ≅ RD . See Figure
1.
C
Since RA and RB are right angles, AD ≅ BC ,
and AB ≅ AB , we have that ∆ADB ≅ ∆BCA .
Thus, AC ≅ BD . Then, since CD ≅ CD , it
A
follows that ∆ADC ≅ ∆BCD . Then, it follows
that RBCD ≅ RADC .
B
4. This problem has five parts. We first construct the Saccheri quadrilateral associated with a
given triangle, and then apply this construction to triangles in the hyperbolic plane.
(a) Given ∆ABC , let I, J, and K be the midpoints of
BC, CA, and AB, respectively.
Drop B
K
A
perpendiculars AD, BE, and CF from the vertices
sur
to IJ . Prove that AD ≅ CF ≅ BE , and, hence
that ¨EDAB is a Saccheri quadrilateral. Why
F
should this quadrilateral have the same area as
E
I
J
D
the triangle? See Figure 2.
We know that BI ≅ CI , RBIE ≅ RCIF , and
RBEI ≅ RCFI . Thus, ∆BIE ≅ ∆CIF by AAS.
This means that BE ≅ CF . Similarly, AD ≅ CF
C
using an identical argument. By transitivity it
follows BE ≅ AD . Therefore, by definition, ¨EDAB is a Saccheri quadrilateral.
sur
IJ . What does this
(b) Prove that the perpendicular
bisector
of
AB
is
also
perpendicular
to
suur
sur
say about the lines IJ and AB .
From a theorem in class we know that the line perpendicular to the summit of Saccheri
quadrilateral at its midpoint
isralso perpendicular to the base of the Saccheri quadrilateral.
suu
sur
Thus, we have that IJ and AB are hyperparallel.
(c) Prove that IJ ≅ 12 ED . Conclude that in hyperbolic geometry IJ < 12 AB .
From the triangles in part (a) we have that EI ≅ FI and FJ ≅ JD . Then,
EI + IF + FJ + JD = ED
2 IF + 2 FJ = ED
2 IJ = ED
1
So, IJ ≅ 2 ED . Since the summit of a Saccheri quadrilateral is longer than its base, We
know that AB > ED, so IJ < 12 AB .
©David Royster
Hyperbolic Geometry
FOR CLASSROOM USE ONLY
MATH 3181 001
HOMEWORK ASSIGNMENT #6 SOLUTIONS
SPRING 1999
(d) Suppose, in this part, that RC is a right angle. Prove that the Pythagorean Theorem
doe NOT hold in hyperbolic geometry. (If it were valid, what would applying to the
triangles ∆ABC and ∆JIC imply?)
Assume that the Pythagorean Theorem does hold, then in ∆ABC we have
AC 2 + BC 2 = AB 2 , and in ∆JIC we have, IC 2 + JC 2 = IJ 2 . Since I is the midpoint of
AC and J is the midpoint of BC, the first equation gives us
AB 2 = (2 IC ) 2 + (2 JC ) 2
= 4( IC 2 + JC 2 )
AB 2 = 4 IJ 2
Thus, from the last equation we get that 2IJ ≅ AB , which we just showed to be
impossible. Thus, the Pythagorean Theorem does not hold in hyperbolic geometry.
(e) In this part, assume that ∆ABC is an isosceles triangle with AC ≅ BC . Prove that K, F,
and C are collinear but F is not the midpoint of CK.
Since AC ≅ BC and I and J are the midpoints, it follows
that IC ≅ JC . F is the foot of C on ED, so by
Hypotenuse-Leg, we have that RICF ≅ RJCF . Let X
uuur
denote the point at which the ray CF intersects AB. It
must intersect AB by the Crossbar Theorem. Then by
SAS, ∆BCX ≅ ∆ACX . Hence X is the midpoint of AB,
and X = K . Thus, K, F, and C are collinear.
K
B
A
F
E
I
J
D
If F is the midpoint of CK, then KF ≅ FC ≅ BE ≅ AD
suur
C
sur
and the lines AB and IJ have three equidistant points.
This contradicts the theorem that we proved in class. Thus, F cannot be the midpoint of
CK.
5. Let ∆ABC be any triangle, and let L, M, and N be the midpoints of BC, AB, and AC,
respectively.
A
(a) Prove that ∆AMN is not
similar to ∆ABC . (HINT:
Otherwise the defect of
¨MBCN would be 0.)
N
D
M
Assume that ∆AMN is similar
to ∆ABC . Then we have that
RAMN ≅ RABC and
RANM ≅ RACB . Since
RAMN + RBMN = 180° and
RANM + RCNM = 180° , it
B
L
C
follows that
RCBM + RBMN + RMNC + RNCB = 360° . Thus, ¨MBCN has defect 0, contradicting
the fact that no quadrilateral in hyperbolic geometry has defect 0. Thus the two triangles
cannot be similar.
©David Royster
Hyperbolic Geometry
FOR CLASSROOM USE ONLY
MATH 3181 001
HOMEWORK ASSIGNMENT #6 SOLUTIONS
SPRING 1999
(b) Prove that MN is not congruent to BL by assuming the contrary and deducing that
∆ABC has defect 0. (HINT: Choose D such that N is between M and D and ND ≅ MN .)
Assume that MN ≅ BL . Choose D such that N is between M and D and ND ≅ MN .
Since N is the midpoint of AC and MNA
DNC , by vertical angles, it follows that
∆AMN ≅ ∆CDN . Therefore,
≅
≅
RAMN ≅ RCDN and RMAN ≅ RDCN .
Now, we have that MD ≅ BC , CD ≅ MB , and CM ≅ CM . Thus, by SSS,
∆MBC ≅ ∆CDM and RB = RMBC ≅ RCDM ≅ RAMN . Likewise, RBMC ≅ MCD .
Now,
RBMC + RCMD + RDMA = 180°
RBMC + RCMD + RB = 180°
RBMC + RBCM + RB = 180°
RMCD + RBCM + RB = 180°
RBCA + RNCD + RB = 180°
RA + RB + RC = 180°
Thus, the defect of ∆ABC is 0. This contradiction proves that MN is not congruent to
BL.
©David Royster
Hyperbolic Geometry
FOR CLASSROOM USE ONLY