Download MATH 6118-090 Non-Euclidean Geometry Exercise Set 2: Solutions

MATH 6118-090
Non-Euclidean Geometry
Exercise Set 2: Solutions
1. Suppose f is an isometry and suppose there exist two distinct points P and Q such that
f ( P) = P and f (Q) = Q . Show that f is either the identity or a reflection.
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To show that f is a reflection, we need to show that it fixes every point on the line PQ and
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that f ( R ) π R for any point R œ PQ .
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Let R lie on the line PQ . Then, we may assume that R lies between P and Q. If not, then
it is simply a matter of relabeling the points in the argument below. We know that
PR + RQ = PQ ,
since R is on PQ. Since f is an isometry
f ( P) f ( R) + f ( R) f (Q) = f ( P) f (Q)
Pf ( R ) + f ( R )Q = PQ
Thus, f(R) lies on PQ. But PR = f ( P) f ( R) = Pf ( R ) , so f(R) and R are equidistant from P.
Likewise, R and f(R) are equidistant from Q. Thus, f ( R) = R and f fixes every point on the
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line PQ .
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If there is no point R œ PQ so that f ( R ) π R , then f fixes every point in the plane, and f
is the identity. We are done.
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Assume that there is a point R œ PQ so that f ( R) π R . We want to show that f moves
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every point not on PQ . Assume that f ( X ) = X for some other point X not on PQ and
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consider the line RX . If RX intersects PQ , then RX contains two points that are fixed by
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f, thus by our previous work every point in the line RX is fixed so f ( R) = R , contradicting
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our hypothesis. If RX does not intersect PQ , then the two lines are parallel. Then using the
fact that R must remain equidistant from P, Q, and X we can show that again f ( R ) = R . This
contradiction gives us our result.
2. Prove that if a line A1 π A is sent to itself under a reflection through A , then A1 and A
intersect at right angles.
Let Q be the point of intersection A ∩ A1 = Q . Let A ŒA and B ŒA1 . By our hypothesis B', the
image of B under reflection in A , lies in A1 . Thus, –BQA is sent to –B ¢QA , one of its
adjacent angles. Thus, by definition, this is a right angle.
3. Suppose that f and g are two isometries such that f ( A) = g ( A) and f ( B) = g ( B) , and
f (C ) = g (C ) for some nondegenerate triangle DABC . Show that f = g . That is, show that
f ( P) = g ( P) for any point P.
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Consider the isometry ϕ = g −1 D f . It fixes three points A, B, and C. Thus by Problem 1, it
must be the identity. That means that
ϕ ( P) = P
g −1 D f ( P) = P
f ( P) = g ( P)
for any point P.
4. Prove the Star Trek lemma for an acute angle for which the center O is outside the angle.
From Figure 1 below, we have three isosceles triangles: DAOB , DAOC and DBOC . Then,
–BOC = –AOB - –AOC
(
) (
= 180D - 2–OAB - 180D - 2–OAC
= 2 (–OAC - –OAB )
)
= 2–CAB
B
C
r
r
O
r
A
Figure 2
5. (Bow Tie Lemma) Let A, A', B and C lie on a circle, and suppose –BAC and –BA¢ C
subtend the same arc. Show that –BAC @ –BA¢ C .
From the Star Trek Lemma:
–BAC =
1
–BOC = –BA¢ C
2
A'
A
O
C
B
Figure 3
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6. In Figure 4, if AB = AC = BC , what is the angle at D?
B
A
D
O
C
Figure 4
Since DABC is equilateral, it is equiangular, so
–ADC = –ABC = 60D .
7. Suppose that two lines intersect at P inside a circle and meet the circle at A and A' and at B
A¢ B ¢ and p
AB
and B', as shown in Figure 5. Let α and β be the measures of the arcs q
respectively. Prove that
α +β
∠APB =
.
2
A
O
B'
P
B
A'
Figure 5
From the Star Trek Lemma, we have that ∠B′AA′ =
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α
2
and ∠AB′B =
β
2
. Now,
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∠AB′B + ∠B′BA + ∠BAB′ = 180D
β
⎛α
⎞
+ ∠PBA + ⎜ + ∠BAP ⎟ = 180D
2
⎝2
⎠
α +β
= 180D − ( ∠PBA + ∠BAP )
2
α +β
= ∠APB
2
A second proof follows from noting that ∠APB is an exterior angle to ∆ABP′ . Thus,
∠APB = ∠PB′A + ∠B′AP
=
=
β
2
+
α
2
α +β
2
8. Suppose an angle α is defined by two rays which intersect a circle at four points. Suppose
the angular measure of the outside arc it subtends is β and the angular measure of the inside
arc it subtends is γ .(So in Figure 6, ∠AOB = β and ∠A′OB′ = γ .) Show
α=
β −γ
2
A
O
A'
B
B'
Figure 6
From the Star Trek Lemma, we have that ∠AA′B =
β
and ∠A′BP =
γ
2
2
D
∠APB + ∠PAB + ∠ABP = 180
. Now,
α + ∠PAB + ( ∠ABA′ + ∠A′BP ) = 180D
γ⎞
⎛
α + ∠PAB + ⎜180D − (∠PAB + ∠AA′B) + ⎟ = 180D
2
⎝
⎠
α = ∠AA′B −
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γ
2
=
β −γ
2
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