Download CHAPTER 2. REVIEW: GEOMETRY Part 1: Angles. Angles can be

CHAPTER 2. REVIEW: GEOMETRY
Part 1: Angles.
Angles can be measured in degree or in radian. For example, the right angle in degree
is 90o and in radian is π/2. The “full angle” in degree is 360o and in radian is 2π.
As you know, each angle of an equilateral triangle is 60o , which is π/3 in radian. In
everyday life, we use degree, but in mathematics, natural sciences and engineering, we
use radian. So we have to pay more attention to angle measurement in radian.
In the Cartesian plane, draw a circle C of radius 1 centered at the origin O. (The
equation describing this circle is x2 + y 2 = 1.) We call C the unit circle. Let A be the
intersection of the unit circle and the positive part of the x-axis. In Cartesian coordinates,
A = (1, 0). Take a point B ≡ (a, b) on the unit circle. Then the angle ∠AOB in radian
is defined to be the length of the circular arc between A and B.
Figure 1.
The “full angle” corresponds to the whole circle, which has length 2π. Hence the “full
angle” is 2π. A right angle is a quarter of the “full angle” and hence is
2π/4 ≡ π/2 in radian.
Since 360o in radian is 2π, we have the following conversion between degree and radian
2π
360o
× angle in radian, angle in radian =
× angle in degree.
2π
360o
For example, π/15 converted to degree is (π/15) × (360o /2π) = 12o and 270o converted
angle in degree =
to radian is 270o × 2π/360o = 3π/2.
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Review Exercise 1. Match angles in degree in List A with angles in radian in List B.
Then plot the points on the unit circle to represent these angles:
List A: 180o , 60o , 15o , 45o , 135o , 120o , 240o , 150o , 40o , 300o , 315o .
List B: π/3, π/4, 2π/3, 7π/4, π, 2π/9, 5π/3, π/12, 3π/4, 4π/3, 5π/6.
We say that two lines ℓ1 and ℓ2 in the Euclidean plane are perpendicular, and we write
ℓ1 ⊥ ℓ2 , if these two lines intersect at a right angle. (Similarly, we say that two vectors
−→
−→
u ≡ OP and v ≡ OQ are perpendicular and we write u ⊥ v if ∠P OQ is a right angle.)
Figure 2.
We say that a line ℓ and a plane π in the Euclidean 3D–space are perpendicular
and we write ℓ ⊥ π if they intersect at some point A, and if, for each point P on π, the
line AP is perpendicular to ℓ: ℓ ⊥ AP .
Figure 3.
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A basic fact about angles is this: the sum of internal angles of any triangle is equal to
180o , or in radian, is equal to π. Thus, if ∆ABC is an arbitrary triangle, then
∠CAB + ∠ABC + ∠BCA = π.
(∗)
Figure 4.
Why? One explanation is that, as indicated in the following figure, you can glide ∆ABC
−−→
along BC in the direction of BC to form a new triangle ∆ECD. Then you can insert
a triangle between them by turning ∆ABC up side down to get another triangle ∆CEA
to fill the gap between them.
Figure 5.
Now we have ∠BCA + ∠ACE + ∠ECD = π. Also, ∠ACE = ∠CAB and ∠ECD =
∠ABC. So (∗) above is clear.
Exercise 2. What is the sum of all internal angles of a quadrilateral (i.e. a polygon with
4 sides)? What about a pentagon (a polygon with 5 sides)? What about an n-gon (a
polygon with n sides) in general?
Exercise 3. What is the size of an internal angle of a regular n-gon? (By a regular n–gon
we mean a polygon of n sides with equal length such that all of its vertices lie in a circle.
See the following figure.) Check your answer for n = 3.
3
Figure 6.
Exercise 4. (This is a hard problem.) Explain why, among all regular polygons, the
triangle, the square and the hexagon are the only possible shapes for making tiles? [Hint:
First prove that if your bathroom floor can be filled with tiles in the shape of a regular
n–gon, then 2π must be an integral multiple of (n − 2)π/n.]
Another interesting fact about angles is this: if AB is a diameter of a circle C and if
P is any point on C, then ∠AP B is necessarily a right angle:
Figure 7.
Why is that so? Well, that AB is a diameter means that the center, say O, of the circle C
is on AB. So |OA| = |OB| = |OP | = the radius of C. (Here, |OA| stands for the length
of the line segment connecting O and A; in other words, |OA| is the distance between O
and A.) Now OP cut the triangle ∆ABP into two: ∆AOP and ∆BOP . Let us take a
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look of one of them: ∆AOP . Since the sides OA and OP has the same length, the angles
facing them, namely ∠P AO and ∠AP O, are equal – call their common size α. For the
same reason, we can write ∠OP B = ∠OBP ≡ β:
Figure 8.
From the above figure we see that the sum of internal angles of ∆AP B is 2α+2β. Thus,
from the fact that the sum of internal angles of any triangle is π, we get 2α + 2β = π and
consequently α + β = π/2. Another look of the figure tells us that ∠AP B = α + β = π/2.
Hence ∠AP B is a right angle.
The following set of three related exercises are challenging. Please don’t be upset if
you find them too hard.
Exercise 5. With the above notation, verify that, if Q is any point inside the circle, then
∠AQB > π/2 and, if R is any point outside the circle, then ∠ARB < π/2.
Exercise 6. Prove that if C is an arc joining the end points of a line segment AB having
the property that ∠AP B is a right angle for any point P on C, then C is necessarily a
semicircle with AB as a diagonal.
Exercise 7. (Dido’s problem) Prove that among all planar arcs with given length L
and with its end points on a straight line ℓ, a semicircle (together with ℓ) enclose the
maximal area.
The subject of studying the relations between sides and angles of triangles or other
planar rectilinear figures is called trigonometry. The two basic trigonmetric functions,
sine and cosine, can be described via a right triangle ∆ABC, as given in Figure 9
5
below. (We use the word “function” here because the value of sine (or cosine) depends on
the given angle, which may be considered as a variable.) Here we have ∠ACB = 90o so
that AB is the hypotenuse, with length c. The lengths of sides AC aad BC are denoted
by b and c respectively, and the angles ∠CAB and ∠CAB are simply denoted by
α and β. Then
sin α =
opposite side
a
=
c
hypertenuse
cos α =
b
neighboring side
=
.
c
hypertenuse
Figure 9.
The following relation between sine and cosine is basic:
sin2 α + cos2 α = 1
Most people simply read this as “sine squared plus cose squared is one”. To verify this,
we notice that, from the Pythagoras theorem (see the begining of next part), we have
a2 + b2 = c2 , which gives
2
2
sin α + cos α =
³ a ´2
c
µ ¶
b
a2 b 2
a2 + b 2
c2
1= 2 + 2 =
=
= 1.
+
c
c
c
c2
c2
The other trigonometric functions are tangent, cotangent, secant and cosecant, given by
tan α =
sin α
a
= ,
cos α
b
cot α =
cos α
b
1
= =
,
sin α
a
tanα
sec α =
1
,
cos α
csc α =
1
.
sin α
Typical applications of trigonometry in geographic survey goes back to ancient histry of
civilizations. We give two “baby versions” of such, in the following example and exercise.
Example. Adele raises her head about 70o in order to see the top of a spruce about 23
meters from her. What is the height of the spruce? To answer this question, we draw a
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right triangle ∆ABC, where point A is where Adele is, point B is near the bottom of
the tree and point C is at the top of it.
We are asked to find the height h ≡ |BC|, given d ≡ |AB| = 23 and α ≡ ∠A = 70o .
Now h/d ≡ |BC|/|AB| = tan α and, by using a calculator or a trigonometry table, we
find that tan 70o is approximately 3. Thus h = d tan 70o = 23 × 3 ≈ 69. Adding Adele’s
height, we estimate that the spruce is roughly 70 meter tall.
Execise 8. Both Adele and Bernado stands in front of a spruce, keeping a distance of 42
meters from each other, with Adele standing hehind. In order to see the top of tree, Adele
has to raise her head about 30o and Bernado has to raise his head about 45o . Estimate
the height of the tree. (Note: Bernado is unable to find his distance to the tree because
√
3. Ignore the
there is a ditch in front of him.) Take 1.7 as the approximate value of
heights of Adele and Bernado in your estimation.
Figure 10.
Review Exercise 9. Verify that 1 + tan2 α = sec2 α. (Hint: Use sin2 α + cos2 α = 1.)
√
Review Exercise 10. Verify that sin 45o = cos 45o = 2/2. (Hint: Draw a right
triangle with a 45o angle.)
Review Exercise 11. Verify that sin 30o = cos 60o = 1/2 and sin 60o = cos 30o =
(Hint: Draw an equilateral triangle and divide it into two smaller right triangles.)
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√
3/2.
Part 2: Areas, Volumes and Dimensions.
The most basic formula for areas is that the area A of a rectangle is of sides a and b
is equal to ab: A = ab.
Figure 11.
For example, the area of a rectangle with sides of lengths 2cm and 3cm is 2 × 3 = 6cm2 . A
square is a rectangle with four equal sides. The area of a square with side a is a2 . A rather
slick argument using areas as follows gives a proof of the celebrated Pythagoras theorem.
Take any right triangle with sides a, b, c, where c is (the length of) the hypotenuse. We
can make four copies of this triangle and fit them along the sides (of length a + b) of a
square S as indicated in the following figure:
Figure 12.
The part of the square S not covered by these four copies of the right triangle forms
a smaller square T with sides equal to c. Notice that these four copies of the right
triangle can be combined into two rectangles, each with area ab. Now I ask you to finish
the argument:
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Exercise 12. Compute the area of S in two ways: one gives (a + b)2 and the other gives
c2 + 2ab. This will tells us (a + b)2 = c2 + 2ab. Derive a2 + b2 = c2 from the last identity.
The area formula for the rectangle can be extended to parallelograms: the area of a
parallelogram with base b and height h is hb; (see the Figure 13 below).
Figure 13.
Why? Because we can cut out a triangle along the dotted line in Figure 14 and paste it to
the other side of the parallelogram to form a rectangle with sides h and b. Since cutting
and pasting do not alter the area, the original parallelogram has the area of the rectangle
obtained in this way and hence its area is A = hb.
Figure 14.
Exercise 13. By a rhombus (a diamond shaped parallelogram) we mean a parallelogram
with four equal sides. What is the area of a rhombus with sides equal to 2cm and one of
its internal angle equal to 30o ?
What is the area of a triangle with base b and height h? The answer is
1
hb.
2
To see
this, we take a copy of the given triangle, turn it up side down and paste it on top of the
original triangle to obtain a parallelogram with base b and height h; (see Figure 15). The
area of the triangle is a half of the area hb of the parallelogram.
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Figure 15.
Example. In the movie Lord of the Ring, a castle is built against a cliff (for the
obvious defense reason). Suppose that the castle is in rectangular shape and its wall is
4 km long. (An areal view of the castle is shown in the following figure.) What is the
maximal area that can be enclosed by the wall of castle and the cliff?
Figure 16.
Let x be the “depth” of the castle and y be its “width”. Then 2x + y = 4 (km). The area
enclosed is
A = xy = x(4 − 2x) = −2x2 + 4x = −2(x2 − 2x).
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Now we use an important basic technique in algebra called “completing squares” to deal
with x2 − 2x. Certainly the last expression cannot be written as a square of something
neat. However, something close to it can be, namely x2 − 2x + 1. Indeed, x2 − 2x + 1 =
(x − 1)2 . This suggests that you can give 1 to x2 − 2x to make a square, and then take
away 1 to get “the account balanced”:
x2 − 2x = x2 − 2x + 1 − 1 = (x2 − 2x + 1) − 1 = (x − 1)2 − 1.
(At this point you probably don’t see why we do this “give, then take away” manoeuvre.
In the next step you will see.) Substitute this back to A to obtain:
A = −2(x2 − 2x) = −2((x − 1)2 − 1) = 2 − 2(x − 1)2 .
Since (x − 1)2 ≥ 0 for all x (the square of a real number cannot be negative), we have
A ≡ 2 − 2(x − 1)2 ≤ 2 for all x. Now it is easy to see that A attains its maximum 2 at
x = 1. Thus the maximal area enclosed by the castle is 2 km2 when its depth is x = 1
km and its width is 4 − 2x = 2 km. (We will see more about completing squares in the
future.)
Exercise 15. Find the real number r > 1 having the following property: if R is a
rectangular board with the ratio of two sides equal to r, as shown in the following figure,
then, by sawing off a square, the remaining part is a rectangle with the ratio of two sides
again equal to r.
Figure 17.
The volume V of a rectangular box with sides a, b and c is abc: V = abc.
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Figure 18.
If each side of this rectangular box is doubled, its volume becomes 2a × 2b × 2c = 8abc,
which is enlarged by 8 times; (see Figure 17 below). But doubling the sides of a rectangle,
the area is enlarged only by 4 times.
Figure 19.
In general, if a body is inflated k times, then its diameter is enlarged by k times, its
surface area is enlarged by k 2 times, its volume is enlarged by k 3 times. In general, a
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d–dimensional measurement of an object is enlarged by k d if the objected in inflated by
k times.
Examples. Roughly speaking, the strength of an animal is proportional to the area
of a cross section of its limps and its weight is proportional to the volume of its body.
A grasshopper can jump higher than 10 times of its body length. By if a grasshopper
were 102 times bigger, say about the size of an elephant, then it is (102 )2 = 104 times
stronger but (102 )3 = 106 heavier, which is 100 times more than 104 . So, an “elephantine
grasshopper” cannot jump. This also explains why an elephant cannot jump, even though
it has powerful legs – so powerful that it could crush your chair if it steps on it. Elephants
are strong animals. But an elephant would collapse if another elephant stands on its
back. An ant is small and weak. However it can lift up an object more than 10 times
of its weight. Next, notice that, if a camel is as small as a rat, then it cannot cross a
desert. Indeed, suppose a camel shrinks to a size 100 times smaller. Then its perspiration
(through the skin, of course) is 104 times less, but the water it can carry is reduced by
106 times. A “thought experiment” once considered by Laplace is something like this.
Suppose that you wake up in one morning and God plays a joke to make everything in the
Universe ten times larger. Would you notice the change? Many people would think No,
because relative to each other things are not changed. This answer is wrong! According
to the above analysis, a person in such a situation would be 100 times stronger. His body
mass would be enlarged by 1000 times, and so would be the mass of the earth. Hence he
would become 1,000,000 times heavier! If that person was lucky enough not to be crushed
to death under his own weight, for sure he could not get out of his bed, even his bed
had already been crushed. For the similar reason King Kong would find that overcoming
his own weight in order to get up in the morning was impossible (but we must excuse
movie makers who are not scientists, just as many movie goers). Huge animals like whales
live in oceans in order to move. Scientists believe that huge dinosaurs had their bodies
submerged in lakes to counter their weights.
So far we have only considered the low dimensions of 1, 2 and 3. For applications,
we often need to consider higher dimensions. Indeed, if a function or relation involves n
variables, say x1 , x2 , . . . , xn , we should think of these variables combined to form a point
(or a vector), denoted by x ≡ (x1 , x2 , . . . , xn ), varying in an n–dimensional space.
Certainly n here may be greater than 3. For example, when linear programming (a branch
of mathematics) is applied to a practical problem in industry or business, the number of
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variables involved is often about several hundreds and, as a consequence, we have to
work with spaces having huge dimensions. If the state of a physical system is completely
determined by n (independent) variables, then we call n the degree or freedom of the
system. We may say that degree of freedom is just another word in physics or engineering
for dimension.
Example. What is the degree of freedom of an oxygen molecule O2 ? To answer this
question, Let us take the midpoint C of the oxygen atoms (representing the center of mass
of the molecule). The location of C in the space needs three coordinates. The distance of
one of the oxygen to C is fixed and we can think of this oxygen lies in a sphere centered
at C. Since a sphere is a two dimensional object, to specify the “spatial orientation” of
the molecule, two numbers are needed (in science, these two numbers are usually taken to
be the Euler angles). All together we need 5 parameters to describe the configuration
of the oxygen molecule. So the degree of freedom for an oxygen is 5.
Exercise 15. (Hard problem) Why is the degree of freedom for a satellite is 6? What
is the degree of freedom of the water molecule H2 O?
Weird objects called fractals with non-integer dimensions were noticed in mathematics
about 150 years ago. More recently scientists find out that their occurence is actually
quite common in nature. The amazing subtlety in their structure fascinates many mathematicians, scientists, as well as artists. Some engineers have applied theory of fractals
to image compression in communication theory.
Answers to Exercises
1. 180o = π, 60o = π/3, 15o = π/12, 45o = π/4, 135o = 3π/4, 120o = 2π/3, 240o =
4π/3, 150o = 5π/6, 40o = 2π/9, 300o = 5π/3, 315o = 7π/4.
2. Joining any pair of opposite vertices of a quadrilateral, we can cut it into two
triangles. Hence the sum total of its internal angle is 2 × π = 2π. For a pentagon, we
can cut it into three triangles and hence the sum of its internal angles is 3 × π = 3π. In
14
general, we can cut an n-gon into n − 2 triangles and hence the sum of its internal angles
is (n − 2)π.
3. A regular n-gon has n angles of the same size. By the previous exercise, we known
that their total is (n − 2)π and hence each of them is (n − 2)π/n. For example, for an
equilateral triangle, n = 3 and hence (n − 2)π/n = (3 − 2)π/3 = π/3, that is, 60o . (For
a square, n = 4 and hence (n − 2)π/n = (4 − 2)π/4 = π/2, that is, 90o . For a regular
pentagon, n = 5 and hence (n − 2)π/n = (5 − 2)π/5 = 3π/5, or 216o .)
4. The internal angle of a tile with n sides is (n − 2)π/n, according to the previous
exercise. Now, in order to fit a number of tiles snugly at their corners, 2π must be divisible
by (n − 2)π/n, say
n−2
π,
or
2n = k(n − 2)
n
for some positive integer k. How do we find all possible values of n and k? Here is the
2π = k ×
trick: introduce a new letter m which stands for n−2: m = n−2. Notice that the minimal
number of sides of a polygon is 3, and hence n ≥ 3, which gives m = n − 2 ≥ 3 − 2 = 1.
Therefore m is a positive integer. Replace n by m + 2 in 2n = k(n − 2), we have
2(m + 2) = km, or 2m + 4 = km. Rearranging terms, we obtain (k − 2)m = 4. Now
we can list all possible ways of multiplying two positive integers to get 4 as the answer:
4 × 1 = 4, 2 × 2 = 4 and 1 × 4 = 4. So the only possible values of m are 1, 2 and 4.
Therefore the only possible values of n = m + 2 are 3, 4, 6, corresponding to the triangle,
the square and the hexagon, which are only the possible shapes of bathroom tiles.
5. Suppose that Q is inside the circle. Extend OQ to meet at a point P on the circle;
see Figure 18 below. Clearly ∠QAB < ∠P AB and ∠QBA < ∠P BA. Hence
∠AQB = π − ∠QAB − ∠QBA > π − ∠P AB − ∠P BA = ∠AP B = π/2.
The argument for ∠ARB < π/2 is similar.
Figure 20.
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6. Drawn a semicircle S using AB as a diameter, on the same side as the arc. Then
extend S to a full circle. We have to prove that, under the given condition, S is a
semicircle, that is, S = C. Suppose the contary that S and C are not the same. Then
either we have a point P on C which is either inside the circle or outside the circle.
Figure 21.
Our previous exercise tells us that either ∠AP B is > π/2, or < π/2, contradicting to
our assumption.
7. Let C be an arc with length L and with end points on ℓ such that the area enclosed
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by C and ℓ is maximal. According to the previous exercise, it is enough to show that,
if P is on C, then ∠AP B = π/2. Suppose the contrary that this is not true. Then we
can glide point B along ℓ to a new position, say B ′ , so that ∠AP ′ B ′ = π/2, where P ′ is
the new position of P . Now ∆AP B and ∆AP ′ B ′ have their “bases” of the same length
|AP | = |AP ′ |, but the “height” of ∆AP ′ B ′ is |P ′ B ′ | = |P B|, which is clearly greater than
the “height” of ∆AP B. As a consequence, the area of ∆AP ′ B ′ is greater than the area
of ∆AP B. We copy the part CA of the arc C between A and P rigidly to get an arc over
AP ′ , say CA′ , so that the lengths of CA and CA′ are the same and the area enclosed by CA
and AP is the same as the area enclosed by CA′ and ℓ. The same thing can be done to
get an arc CB′ ′ .
Figure 22.
Putting CA′ and CB′ ′ together, we get an arc C ′ over AB ′ . Now C and C ′ have the
same length L, but the area enclosed by C ′ and ℓ is larger, because the area of the
triangle ∆AP ′ B ′ is larger, while the areas of the remaining pieces remain unchanged.
Thus the total area enclosed by C ′ and ℓ is larger. This certainly is a contradiction to
our assumption that the area enclosed by C and ℓ is maximal.
8. Let x = |A2 C|. Since ∆BA2 C is a right triangle with ∠BA2 C = 45o , we have
|BC| = |A2 C| = x. Notice that |A1 C| = |A1 A2 | + |A2 C| = 42 + x. So we have
|BC|
1
x
1
=
= tan ∠A1 = tan 30o = √ ≈
.
x + 42
|A1 C|
1.7
3
Thus 1.7x = x + 42, or 0.7x = 42, which gives x = 60. So the spruce is about 60
meters tall.
9. We have
2
1 + tan α = 1 +
µ
sin α
cos α
¶2
=1+
sin2 α
cos2 α + sin2 α
1
=
=
= sec2 α.
2
2
cos α
cos α
cos2 α
17
10. Let ∆ABC be a triangle with ∠A = 90o , ∠B = 45o , ∠C = 45o , and |AB| =
|AC| = 1. The Pythagoras theorem tells us that |BC|2 = |AB|2 + |AC|2 = 12 + 12 = 2.
√
Hence |BC| = 2. Therefore
√
√
1
2
|AC|
1
2
|AC|
o
o
=√ =
, cos 45 = cos ∠C =
=√ =
sin 45 = sin ∠B =
|BC|
2
|BC|
2
2
2
11. Let ∆ABC be an equilateral triangle. Then ∠A = ∠B = ∠C = 60o . Let us take
|AB| = |AC| = |BC| = 2. Let M be the midpoint of BC. Then |BM | = |CM | = 1.
The triangle ∆ABM
is a right triangle with angles ∠M = 90o , ∠B = 60o , and
∠A = 30o . The Pythagoras theorem tells us that |AM |2 = |AB|2 − |BM |2 = 22 − 12 = 3.
√
Hence |BC| = 3. Therefore
1
|BM |
= , cos 60o = cos ∠B =
|AB|
2
√
|AM |
3
o
=
, cos 30o = cos ∠A =
sin 60 = sin ∠B =
|AB|
2
sin 30o = sin ∠A =
|BM |
1
= ,
|AB|
2
√
|AM |
3
=
.
|AB|
2
12. A side of the square S has length a + b and hence the area of S is
(a + b)2 = a2 + 2ab + b2 .
On the other hand, S is divided into a square of sides of length c and four copies of the
right triangle, which can be assembled into two rectangles of sides a and b. The total
area of these pieces are c2 + 2ab. So the area of S is also equal to c2 + 2ab. Therefore
a2 + 2ab + b2 = c2 + 2ab.
Cancelling 2ab, we get a2 + b2 = c2 .
13. The height of the rhombus is h = a sin 30o = 2cm × sin 30o = 2 × 12 = 1cm. So the
area of the rhombus is A = ah = 2 × 1 = 2cm2 .
14. Let the sides of the rectangular board be a and b with a/b = r. Since r > 1, b is
the smaller side. Saw off a square of side b, we get a smaller rectangular board of sides b
and a − b. By assumption,
b
a
=
.
b
a−b
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Now we multiply both sides by (a − b)/b to obtain
aa−b
= 1,
b b
or
´
a ³a
− 1 = 1.
b b
Replace a/b by r to get r(r − 1) = 1, or r2 − r = 1. Thus we get a quadratic equation
r2 − r − 1 = 0. The solutions to this equation are
p
√
1 ± (−1)2 − 4 × 1 × (−1)
1± 5
=
.
r=
2
2
Since r is necessarily a positive number, our final answer is r = (1 +
√
5)/2.
15. We need three parameters to determine the center of the mass, two parameters
to determine the axis (of rotational symmetry) of the satellite, and one parameter for the
angle of spanning about this axis. Same answer for the water molecule.
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