Download ASSIGNMENT 2 SOLUTIONS

Homework Exercises
Due January 23, 2002
ASSIGNMENT 2
SOLUTIONS
January 16, 2002
Exercise 1.15. 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.
To show that f is a reflection, we need to show that it fixes every point on the line P Q and
that f (R) 6= R for any point R 6∈ P Q.
Let R lie on the line P Q. 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
|P R| + |RQ| = |P Q|,
since R is on P Q. Since f is an isometry
|f (P )f (R)| + |f (R)f (Q)| = |f (P )f (Q)|
|P f (R)| + |f (R)Q| = |P Q|
Thus, f (R) lies on P Q. But |P R| = |f (P )f (R)| = |P f (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
line P Q.
If there is no point R 6∈ P Q so that f (R) 6= R, then f fixes every point in the plane, and f is
the identity. We are done.
Assume that there is a point R 6∈ P Q. Assume that f (X) = X for some other point X not in
P Q. If RX intersects P Q, then RX contains two points that are fixed by f , thus by our previous
work every point in the line RX is fixed, which makes f the identity. If RX does not intersect P Q,
then the two lines are parallel. Then using the fact that R must remain equidistant from P , Q,
and X fixes R. Thus, if f fixes three points, it must be the identity.
Exercise 1.17. Prove that if a line `1 6= ` is sent to itself under a reflection through `, then `1
and ` intersect at right angles.
Let Q be the point of intersection of ` ∩ `1 . Let A ∈ ` and B ∈ `1 . By our hypothesis B 0 , the
image of B under reflection in `, lies in `1 . Thus, ∠BQA is sent to ∠B 0 QA, one of its adjacent
angles. Thus, by definition, this is an right angle.
Exercise 1.19. 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 4ABC. Show that f = g. That is, show that
f (P ) = g(P ) for any point P .
Consider the isometry g −1 ◦ f . It fixes three points A, B, and C. Thus, it must be the identity.
That means that g −1 ◦ f (P ) = P for any point P . Thus, f (P ) = g(P ) for any point P .
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MATH 6118-090
Homework Exercises
Due January 23, 2002
Exercise 1.26 Prove the Star Trek lemma for an acute angle for which
the center O is outside the angle.
From Figure 1, we have three isosceles triangles: 4AOB, 4AOC and
4BOC. Then,
B
C
r
r
O
∠BOC = ∠AOB − ∠AOC
r
= (180 − 2∠OAB) − (180 − 2∠OAC)
A
= 2(∠OAC − ∠OAB)
= 2∠CAB
Figure 1: Exercise 1.26
A0 ,
Exercise 1.29 (Bow Tie Lemma) Let A,
B and C lie on a cir0
cle, and suppose ∠BAC and ∠BA C subtend the same arc. Show that
∠BAC ∼
= ∠BA0 C.
From the Star Trek Lemma:
1
∠BAC = ∠BOC = ∠BA0 C.
2
A'
B
A
A
D
O
O
C
B
Figure 2: Exercise 1.29
C
Figure 3: Exercise 1.29
Exercise 1.30 If |AB| = |AC| = |BC|, what is the angle at D?
From the previous problem since 4ABC is equilateral, it is equiangular, so ∠ADC = ∠ABC =
60◦ .
Exercise 1.31 If |AB = 12|, |BD| = 9, |BC| = 16 and |AC| = 20, then what is the length of the
diameter?
Since |AB|2 + |BC|2 = 122 + 162 = 144 + 256 = 400 = |AC|2 , the
angle at B is a right angle. Thus, AD is a diameter.
p
p
|AD| = |AB|2 + |BD|2 = 122 + 92 = 15.
A
Exercise 1.34 Suppose that two line intersect at P inside a circle and
meet the circle at A and A0 and at B and B 0 , as shown in Figure 4.
Let α and β be the measures of the arcs A0 B 0 and AB respectively.
Prove that
α+β
.
∠AP B =
2
2
B
D
C
MATH 6118-090
Homework Exercises
Due January 23, 2002
α
and
From the Star Trek Lemma, we have that ∠B 0 AA0 =
2
β
∠AB 0 B = . Now,
2
∠AB 0 B + ∠B 0 BA + ∠BAB 0 = 180◦
α
β
+ ∠P BA + ( + ∠BAP ) = 180◦
2
2
α+β
= 180◦ − (∠P BA + ∠BAP )
2
α+β
= ∠AP B
2
A
O
B'
P
B
A'
Figure 4: Exercise 1.34
Exercise 1.35 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 5, ∠AOB = β and ∠A0 OB 0 = γ.) Show
α=
β−γ
.
2
β
and
From the Star Trek Lemma, we have that ∠AA0 B =
2
γ
∠A0 BP = . Now,
2
A
∠AP B + ∠P AB + ∠ABP = 180◦
α + ∠P AB + (∠ABA0 + ∠A0 BP ) = 180◦
γ
α + ∠P AB + (180 − (∠P AB + ∠AA0 B) + = 180◦
2
α = ∠AA0 B −
O
A'
B
B'
β−γ
γ
=
2
2
Figure 5: Exercise 1.35
Exercise 1.36 Prove that the opposite angles in a convex quadrilateral inscribed in a circle sum to
180◦ . Conversely, prove that if the opposite angles in a convex quadrilateral sum to 180 ◦ , then the
quadrilateral can be inscribed in a circle. Such a quadrilateral is called a cyclic quadrilateral.
The opposite angles of a quadrilateral inscribed in a circle subtend complimentary arcs. The
sum of the central angles then is 360◦ . The sum of the angles will be half of this, or 180◦ .
Any three points lie on a circle. So choose A, C, and D and let Γ be the circle that circumscribes
4ACD. Let BD intersect the circle in a point X. Now, ∠DXC subtends the same arc as does
∠DAC. We have
∠CAX = ∠CDX = α
∠DAC = ∠DXC = β
∠ACX = ∠ADX = γ
∠ACD = ∠ACD = δ
We know that ∠ADC + ∠ABC = 180 = ∠ADC + ∠AXC, so ∠ABC = ∠AXC. Then in 4ABC
we have that ∠BAC > ∠XAC and ∠BCA > ∠XCA. However,
∠ABC + ∠BCA + ∠CAB > ∠AXC + ∠XCA + ∠CAX = α + β + γ + δ = 180 ◦ .
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MATH 6118-090
Homework Exercises
Due January 23, 2002
Thus, there is a triangle with angle sum more than 180◦ . This means that X = B, and we are
done.
Exercise 1.41 Suppose 4ABC ∼ 4A0 B 0 C 0 . Show that
0
0
0
|4A B C | =
µ
|A0 B 0 |
|AB|
¶2
|4ABC|.
If the two triangles are similar, then by dropping a perpendicular from B to AC and from B 0
to A0 C 0 , we will create similar triangles so that
|BD|
|B 0 D0 |
= 0 0 = k,
|AB|
|A B |
where k is the ratio of similarity. Another way to convey this is |BD| = k|B 0 D0 |.
Thus,
1
|4ABC| = |BD| · |AC|
2
1
= (k|B 0 D0 |)(k|A0 C 0 |)
2
¶
µ
|AB| 2
= k 2 |4A0 B 0 C 0 | =
|4ABC|
|A0 B 0 |
Exercise 1.43 The Angle Bisector Theorem In an arbitrary triangle 4ABC, let the interior
angle bisector at A intersect the side BC at D. Show that
|BD|
|AB|
=
.
|DC|
|AC|
Construct the line through B parallel to AD. Let E be the point of intersection with AC.
Then, ∠BEA ∼
= ∠DAC, we
= ∠DAC. Now, ∠EAB + ∠BAD + ∠DAC = 180◦ . Since ∠BAD ∼
have ∠EAB = 180◦ − 2∠DAC. But, ∠BEA + ∠EAB + ∠ABE = 180◦ . Applying the previous
equation gives us that ∠BEA = ∠ABE and 4ABE is isosceles. Thus, AE ∼
= AB.
Now, since BE is parallel to AD, we have that 4BEC ∼ 4DAC.
Thus,
E
|EC|
|AC|
|EA| + |AC|
|AC|
|EA|
|AC|
|AB|
|AC|
4
|BC|
|DC|
|BD| + |DC|
=
|DC|
|BD|
=
|DC|
|BD|
=
|DC|
A
=
B
D
C
Figure 6: Exercise 1.43
MATH 6118-090
Homework Exercises
Due January 23, 2002
Exercise 1.44 The pentagon in Figure 7 is regular and each side has length one. Show that
√
1+ 5
|AF |
=
.
|F D|
2
From the fact that the pentagon is regular, we have that ∠AED =
108◦ . Now, 4AED is isosceles, since the sides are equal. Thus, the base
angles are equal. Since the sum is 180◦ , we get that ∠EDA = ∠EAD =
36◦ . From the Law of Sines, we find that
1
AD
=
◦
sin 108
sin 36◦
sin 108◦
.
AD =
sin 36◦
B
A
C
F
E
D
Now, we can also show that ∠CED = 36◦ in a similar manner. Thus,
4F ED is isosceles with angles 36◦ − 36◦ − 108◦ . It follows by the Law of
Figure 7: Exercise 1.44
Sines that
1
FD
=
◦
sin 36
sin 108◦
sin 36◦
1
FD =
=
sin 108◦
AD
Now, it is easy to show that ∠AEF = ∠AF E = 72◦ . Thus, 4AEF is isosceles, which makes
|AF | = 1. Now,
|AD| = |AF | + |F D|
1
|AD| = 1 +
|AD|
|AD|2 − |AD| − 1 = 0
Using the Quadratic Formula, we get that
|AD| =
Since 1 −
√
5 < 0, we must have
1±
√
1+4
.
2
√
1+ 5
.
|AD| =
2
Thus,
|AF |
1
=
|F D|
|AD| − 1|
1
√
=
−1+ 5
2
√
1+ 5
=
2
5
MATH 6118-090
Homework Exercises
Due January 23, 2002
Exercise 1.53 (The Tangential Version of Power of the Point) In Figure 8, suppose P R is
tangent to the circle at R. Show that
|P Q||P Q0 | = |P R|2 .
Let ∠OQQ0 = ∠OQ0 Q = α, ∠OQR = ∠QRO = β, and ∠Q0 RO =
∠RQ0 O = γ. These are equal since they are base angles of isosceles triangles. We intend to show that 4P QR ∼ 4P RQ0 . Clearly,
∠QP R = ∠RP Q0 , since it is the same angle.
Now,
R
r
O
r
r
P
Q
∠QOQ0 + ∠QOR = ∠ROQ0
(180 − 2α) + (180 − 2β) = 180 − 2γ
180 + 2γ = 2(α + β)
Q'
Figure 8: Exercise 1.53
90 + γ = α + β
Then, ∠P QR = 180 − (α + β) = 180 − (90 + γ) = 90 − γ = ∠P RQ 0 . Thus, the third angles will
be equal. Therefore, as we wanted, 4P QR ∼ 4P RQ0 . Thus,
|P Q|
|P R|
=
|P R|
|P Q0 |
|P Q| · |P Q0 | = |P R|2
Exercise 1.54 Use the tangential version of the power of the point to come up with yet another
proof of the Pythagorean Theorem.
Draw a circle centered at B with radius BC. This will make AC tangent to the circle at C.
Let Q denote the point where AB intersects the circle and let Q0 be the other point of intersection.
The above theorem gives us that
|AC|2 = |AQ| · |AQ0 |,
Now, |AC| = b, |AQ| = c − r = c − a, and |AQ0 | = c + r = c + a. Thus we get
b2 = (c − a)(c + a)
a2 + b2 = c 2 .
Exercise 1.58 In Figure 9, the point C is on the diameter AB of a half circle. The two smaller
half circles have diameters AC and CB. The region bounded by the curved edges of the half circles
was called an arbelos or butcher’s knife by Archimedes. The perpendicular at C intersects the
larger circle at D. Prove that the circle with diameter CD has the same area as the arbelos.
Let |AB| = r and |AC| = p. Let |DC| = h. Then since D lies on the semicircle, the angle
∠ADB is a right angle and we have three right triangles. The Pythagorean theorem then gives us:
p2 + h 2 = x 2
(r − p)2 + h2 = y 2
x2 + y 2 = r 2
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MATH 6118-090
Homework Exercises
Due January 23, 2002
A
A
C
B
Figure 9: Exercise 1.58
where x = |AD| and y = |BD|. We need h in terms of r and p. Adding the first two together and
equating this in the third we have
r 2 = x2 + y 2
= (p2 + h2 ) + ((r − p)2 + h2 )
= 2h2 + 2p2 − 2pr + r 2
2h2 + 2p2 − 2pr = 0
p
h = p(p − r)
The area of the circle with diameter |CD| = h is πh2 /4. This gives
π
h2
π
= p(p − r)
4
4
The area of the arbelos is given by
1
2
Ã
π
³ r ´2
2
−
Ã
π
³ p ´2
2
+π
µ
(r − p)
2
¶2 !!
=
π
p(p − r).
4
Exercise 1.59 Show that the two inscribed circles in the arbelos in Figure 10 has the same radii.
Let C1 denote the outer circle of the arbelos, C2 , the left-hand inner circle, and C3 the righthand inner circle. Let C4 denote the circle that is tangent to C1 , C2 , and CD. Assume that the
center of circle Ci is Oi . Let Ci have radius ri . Thus, r1 = |AO1 | = |BO1 | and r2 = |AO2 | = |CO2 |.
Let |O1 O2 | = p. Note then that r1 = r2 + p and r2 + r3 = r1 .
We want to solve for the radius of C4 in terms of r1 , r2 , and r3 . This means that when we do
the same computation for the circle C5 , we must get the same answer, meaning that the two circles
have the same radius, which is what we want to show.
Let P be the foot of the perpendicular dropped from O3 to AB. Note that since the circle C4
is tangent to CD, then |P C| = r4 . Also, since C4 is tangent to C1 , then O4 lies a radius to C1
through O1 . Likewise, since C4 is tangent to C2 , then O4 lies a radius to C2 through O2 . The
triangles 4O4 O2 P and 4O4 O1 P are right triangles with common leg P O4 . Thus,
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MATH 6118-090
Homework Exercises
D
C3
C1
Due January 23, 2002
C4
O3
O4
C2
A
O2
O1
P
C
B
Figure 10: Exercise 1.59
|O4 O2 |2 − |O2 P |2 = |O4 O1 |2 − |O1 P |2
|O4 O2 |2 − |O4 O1 |2 = |O2 P |2 − |O1 P |2
(r2 + r4 )2 − (r1 − r4 )2 = (r2 − r4 )2 − (r2 − p − r4 )2
(r2 + r4 )2 − (r2 − r4 )2 = (r1 − r4 )2 − (r2 − p − r4 )2
4r2 r4 = (r2 + p − r4 )2 − (r2 − p − r4 )2
4r2 r4 = 4(r2 − r4 )p
r2 r4 + pr4 = r2 p
r2 (r1 − r2 )
r2 r3
r2 p
=
=
r4 =
r2 + p
r1
r1
Thus, the radius of this circle tangent to the outer circle, the inner circle and the perpendicular
segment depends only on the radii of the outer circle and the two inner circles. This means that
the computation for the radius of circle C5 will yield the same result.
Exercise 1.60 Let two circles Γ and Γ0 have distinct centers O and O 0 . Prove that the radical axis
of Γ and Γ0 is a line perpendicular to OO 0 .
Let me first so a calculation. Let 4ABC be a triangle, let M be the midpoint of AB and let
CH be the altitude with H ∈ AB. Label the sides BC, AC, and AB as a, b, and c respectively.
Then from the Pythagorean Theorem we have
a2 − b2 = |BH|2 − |AH|2
= (|BH| − |AH|)(|BH| + |AH|)
= 2c · M H.
Now, let P be a point on the radical axis of Γ and Γ0 with centers O1 and O2 and radii r1 and
r2 . Then, because P lies on the radical axis
|P O1 |2 − r12 = |P O2 |2 − r22
|r12 − r22 | = ||P O1 |2 − |P O2 |2 |
= 2|O1 O2 | · M H
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MATH 6118-090
Homework Exercises
Due January 23, 2002
where M is the midpoint of the segment O1 O2 and H is the foot of P in the line O1 O2 . Thus,
MH =
|r12 − r22 |
,
2O1 O2
is a constant. This means that H never moves any further from M . Thus, every point on the
radical axis has the same foot in O1 O2 , so the radical axis is perpendicular to O1 O2 .
Exercise 1.61 Suppose two circles intersect at A and B. Prove that the radical axis of the two
circles is the line AB.
Since 4O1 AO2 ∼
= 4O1 BO2 , A and B have equal powers with respect to the two circles. Since
the radical axis is a line and it must pass through A and B, it must be the line AB.
Exercise 1.62 Let the circle Γ intersect the circle Γ0 at A and B, and the circle Γ00 at C and D.
Let P be the point of intersection of AB and CD. Prove that P lies on the radical axis for Γ 0 and
Γ00 .
Since P lies on AB it lies on the radical axis of Γ. Since it lies on CD, it lies on the radical
axis of Γ0 . Both of these follow from the previous problem.
Exercise 1.63 The Radical Center. Let Γ1 , Γ2 , and Γ3 be three circles. Show that the three
radical axes of these circles intersect at a common point. This point is called the radical center
of the three circles.
Let the radical axis for Γ1 and Γ2 and the radical axis for Γ1 and Γ3 intersect in a point P 1 .
We want to show that P is on the radical axis for Γ2 and Γ3 .
Now,
|P O3 |2 − r32 = |P O1 |2 − r12 = |P O2 |2 − r22 .
The first equality is true because P lies on the radical axis for Γ1 and Γ3 and the second equality
is true because P lies on the radical axis for Γ1 and Γ2 . Thus, P lies on the radical axis for Γ2 and
Γ3 .
Exercise 1.64 Let P be the radical center of three circles Γ1 , Γ2 , and Γ3 . Suppose P is outside
circle Γ. Prove that P is also outside circles Γ0 and Γ00 .
Since |P O1 |2 − r12 > 0 because P lies outside Γ1 , then all three quantities are equal so |P O3 |2 −
2
r3 = |P O2 |2 − r22 = |P O1 |2 − r12 > 0 puts them all outside their respective circles.
Exercise 1.65 Let P be the radical center of three circles Γ1 , Γ2 , and Γ3 and suppose P is outside
circle Γ. Show that P is the center of a circle which
intersects Γ, Γ0 , and Γ00 perpendicularly.
p
Furthermore, show that the radius of this circle
p is Π(P ).
Take the circle centered at P with radius Π(P ). Then, note that
Π(P ) = |P Oi |2 − ri2 .
p
At the point at which the circle intersects circle Γi we have a triangle with sides Π(P ), ri , and
P Oi . Since ri2 + Π(P ) = |P Oi |2 , we have a right triangle, which means that the radius of the circle
Γi and our circle are perpendicular, and we are done.
1
Note that these could be parallel. Take three circles all centered on the same line, the middle circle equidistant
from the other two. Then the radical axes are all perpendicular to the line containing the centers, and hence parallel.
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MATH 6118-090