Download MA 3330 Practice Final Answers in red Name April 24, 2009 1. True

MA 3330 Practice Final Answers in red
Name
April 24, 2009
1.
True or False.
a.
Euclid’s The Elements, according to Howard, is the optimal approach to the study of geometry. False
b.
Euclid’s The Elements, according to Howard, is important primarily as an inspiration for the development
of axiom systems and the development of non-Euclidean geometry. True
c.
Euclid’s The Elements, according to Howard, is really a guide to doing straightedge and compass constructions, as opposed to intending to be an axiom system for plane geometry. True
d.
Descartes, according to Howard, admired the Greeks, and thought Euclid did things the best way
possible. False
e.
Hilbert, according to Howard, fixed up Euclid’s geometry, so that now, Euclid’s and Hilbert’s approach
to geometry is the best way to study geometry. False
f.
Euclid is a contemporary of Descartes (i.e. they lived at about the same time). False
g.
We’re not sure if Hilbert was an actual person, and it may be that a group of ancient Greeks simply
published under this assumed name. False
h.
For Euclid, only right triangles have hypotenuses. False
i.
According to Euclid’s definition, two lines in a plane are parallel, if they do not intersect. True
j.
According to Euclid’s definition, if two lines are not equidistant, then they are not parallel. False
2.
In the Figure below, let AB = 1, AC = x, and AD = y. Show how you would construct the point E so
that AE = xy. Draw the segment BD, then draw a segment from C parallel to BD. The point E is the
intersection between the parallel segment and the line AD.
1
MA 3330 Practice Final Answers in red
E
D
2
F
A
C
B
3.
In Euclid’s proof of the Pythagorean theorem, he shows that in the Figure above, the following areas
are equal
1
1
2
3 1
(1)
rectangleCD = 4ABD = 4ACE = rectangleAF.
2
2
Each equality is established, because the two figures share a common base and height. For each equality,
give me a segment that has the common base and a segment that has the common height. (For example, in
Equality 1, the rectangle and triangle both have bases equal to segment AD.)
a.
Equality 1. The square and triangle both have base AD and height AC. Therefore, the area of the
square is double the area of the triangle.
b.
Equality 2. Actually, the two triangles are congruent. Since AD = AC, ∠DAB = ∠CAE, and
AB = AE, the two triangles are congruent by SAS.
c.
Equality 3. The rectangle and triangle both have base AE and height EF . Therefore, the area of the
rectangle must be double the area of the triangle.
MA 3330 Practice Final Answers in red
3
A
D
4.
c
b
b0
a0
C
a
B
Another proposition is given below.
In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than
the squares on the sides containing the obtuse angle by twice the rectangle contained by one
of the sides about the obtuse angle, namely that on which the perpendicular falls, and the
straight line cut off outside by the perpendicular towards the obtuse angle.
a.
What modern theorem is this describing? This is the Law of Cosines.
b.
In terms of the letters, a, b, c, A, B, and/or C, how long is “the straight line cut off outside by the
perpendicular towards the obtuse angle”? a0 = b cos(180◦ − C) = −b cos(−C) = −b cos(C).
c.
Euclid is basically saying that in the Figure above,
(2)
c2 = a2 + b2 + 2a0 a.
Prove this using the Pythagorean theorem.
By the Pythagorean theorem, b2 = (b0 )2 + (a0 )2 , and also c2 = (b0 )2 + (a + a0 )2 .
Therefore, (b0 )2 = b2 − (a0 )2 and (b0 )2 = c2 − a2 − 2aa0 − (a0 )2 .
This gives us the equation b2 − (a0 )2 = c2 − a2 − 2aa0 − (a0 )2 .
Solving for c2 gives us a2 + b2 + 2aa0 = c2 .
5.
Suppose we have some triangle 4ABC with B = 30◦, a = 5, and b = 7. Find all possible values for c.
You may round your answer(s) to four decimal places. The equation is 72 = 52 + c2 − 2(5)c cos(30◦ ). This
simplifies to c2 − 10 cos(30◦ ) · c − 24 = 0. Therefore,
p
10 cos(30◦ ) ± 102 cos2 (30◦) − 4(1)(−24)
(3)
c=
= 10.8685, −2.2082,
2(1)
but we only want the positive solution.
MA 3330 Practice Final Answers in red
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6.
Classify the following as true in neutral geometry or true only in Euclidean (assuming that what I’ve
told you about Gauss, Lobachevski, and Bolyai is true).
a.
∠4 + ∠6 < 180◦ =⇒ l 6k m. Only Euclidean
b.
∠2 = ∠6 =⇒ l k m. Neutral
c.
l 6k m =⇒ ∠5 6= ∠4. Neutral
d.
l k m =⇒ ∠1 = ∠8. Only Euclidean
7.
For these problems, refer to the figure above.
a.
Explain why 4AEC ∼
= 4F EB. CE = EB by construction, AE = EF by construction, and ∠4 = ∠40
because they’re vertical angles. The two triangle are congruent by SAS.
b.P Prove that
4ABF .
P
4ABC =
P
4ABF .
P
4ABC = ∠1a + ∠1b + ∠2 + ∠3 = ∠5 + ∠1b + ∠2 + ∠30 =
8.
In the Figure for Problem 7, show how you would construct two triangles which have the same angle
sum as 4ABC and at least one of the triangles has an angle that is less than or equal to one-fourth of ∠A.
Draw two separate pictures below, clearly indicating how the construction was done. In other words, they
should look something like the Figure from Problem 7. We need to split either ∠1b or ∠5 “in half.” To split
∠1b, we would find the midpoint of BF , call it G maybe, and then draw a segment from A through the G to
a point H so that AG = GH. Then add side HB. To split ∠5, you would draw a segment from F through
the midpoint of AB.
MA 3330 Practice Final Answers in red
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9a.
In the Poincaré disk model of the hyperbolic plane shown above, two of the ‘lines” are not really lines
in the model. Cross out these two “lines,” and explain why they’re not really lines. There is a straight line
and a circle that do not meet the boundary at right angles.
9b.
In the half-plane model of the hyperbolic plane shown above, two of the “lines” are not really lines in
the model. Cross out these two “lines,” and explain why they’re not really lines. There is a straight line
and a circle that do not meet the x-axis at right angles.
G
H
F
A
C
E
B
I
D
MA 3330 Practice Final Answers in red
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10a.
A quadruply asymptotic quadrilateral is shown in the picture above and to the right. Assuming what you
know about triply asymptotic triangles, find the area of the quadrilateral. Clearly explain your reasoning.
You can add a diagonal to split the quadrilateral into two triply asymptotic triangles, which each have area
π.
b.
In the picture on the left, there is a triangle 4ABC. Let’s say that the interior angles at vertices A,
B, and C are α, β, and γ, respectively. You may assume the area formulas for singly, doubly, and triply
asymptotic triangles. Prove the area formula for 4ABC. Quadrilateral HIDE has area 4π, and is covered
by triangles 4HCI, 4EAD, and 4IBD, but 4ABC is covered twice. Therefore,
(4)
2π = (π − γ) + (π − α) + (π − β) − 4ABC.
Solving for 4ABC gives you
(5)
4ABC = π − α − β − γ.
11a.
On the unit sphere, find the area of a 45◦-45◦ -135◦-triangle. In radians, this is a
π
π
angle sum, therefore, is 5π
4 = π + 4 . Therefore, the area must be 4 .
π π 3π
4 - 4 - 4 -triangle.
The
b.
Consider a regular quadrilateral on the unit sphere with area π4 . What is the measure of the angles?
(Regular means that the sides and angles are all equal, so this thing is kind of like a square.) If you take
any quadrilateral on the sphere, and add a diagonal, you’ll get two triangles. It follows that the angle sum
of a quadrilateral must be 2π + A, where A is the area of the quadrilateral. In this case, A = π4 , so the
. Since this is a regular quadrilateral, each of the four angles must be the same, so each must
angle sum is 9π
4
9π
measure 16 .
12*.
In our proof that the Euler characteristic of a disk is χ = 1, we started with an arbitrary triangulation
of a disk, and we proceeded to remove faces so that after removing each face, we still had a disk. There were
two kinds of faces that we removed. Describe each kind and the effect on the Euler characteristic. In one
case, the triangle shared two edges with the rest of the disk. Removing such a triangle, therefore, removes
one face and one edge. Reducing f and e by one leaves χ unchanged. In the other case, only one edge is
shared with the rest of the disk. Removing a triangle like this removes one face, two edges, and one vertex.
Therefore, f goes down by one, e goes down by two, and v goes down by one. The net change in χ is zero.
A.
You’re going to prove Descartes’ theorem from Euler’s theorem. We have a polyhedral sphere, and
WLOG, we will assume that all of the faces are triangles.
a.
In terms of f, the number of faces, what is the sum of all of the angles from all of the faces in radians?
π · f.
b.
Therefore, in terms of v and f, what is the sum of all of the external angles? 2πv − πf.
c.
Express e in terms of f. e =
3f
2
d.
Using Euler’s theorem, show that the sum of the external angles must sum to 4π. From Euler’s theorem,
2 = f − e + v = f − 3f
+ v = − f2 + v. Multiplying this by 2π gives 4π = 2πv − πf, and this is the sum of
2
the external angles from Part a.
B.
There are five Platonic solids. In our terminology, these are the five regular polyhedral spheres (faces
are regular, faces are congruent to each other, and the same number of faces around each vertex).
a.
In radians, determine the measure of each angle in a regular n-gon. We can add a diagonal to a
quadrilateral to split it into two triangles. We add two diagonals to a pentagon to form three triangles. In
general, we add n − 3 diagonals to an n-gon to split it into n − 2 triangles. It follows that the angle sum is
(n − 2)π. Dividing this by n gives us the measures of the individual angles, so each angle of a regular n-gon
must measure (n−2)π
= π − 2π
n
n .
MA 3330 Practice Final Answers in red
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b.
Triangular faces: It doesn’t make a whole lot of sense to have only two triangles around one vertex,
but you can have three or more. What is the maximum number of faces around a vertex that makes sense?
Why? Figure out how many vertices a regular polyhedral sphere must have for three faces around a vertex,
four faces around a vertex, etc. The external angles must sum to 4π, and they must all be the same, so at
the very least, they must all be positive. Six equilateral triangles around a vertex gives an external angle
of zero, and any more than that makes the external angle negative. We can only have 3, 4, or 5 triangles
around each vertex. Case I (3 faces): External angle is π, so four vertices will add up to 4π. Case II (4
faces): External angle is 2π
, so there must be six vertices. Case III (5 faces): External angle is π3 , so there
3
must be twelve vertices. For all three cases, you can use Euler’s formula to find the number of faces. Case
I is a tetrahedron, Case II is an octahedron, and Case III is a icosahedron. You can look these up on the
Internet to see what they look like.
c.
Square faces: Same as Part b. Two faces around a vertex doesn’t make sense, and four or more faces
won’t give you positive external angles. Therefore, three faces around a vertex is the only possibility. In this
case, the external angle is π2 , and there must be eight vertices. This is a cube.
d.
Pentagonal faces: Same as part b. The angles on a regular pentagon measure 3π
5 . Two faces around
a vertex doesn’t make sense, and four or more faces give negative external angles. Three faces makes each
external angle measure π5 , so there would have to be 20 vertices. This is the dodecahedron.
e.
Hexagonal faces: Same as part b. The angles on a regular hexagon measure 2π
3 . Two faces doesn’t
make sense, and three faces gives a zero external angle. Four or more gives negative external angles. This
just won’t work.
f.
Does it make sense to have faces that have more sides than six? Why? For n > 6, the angles of a regular
n-gon measure more than 2π
3 , so putting three of these faces around a vertex will give a negative external
angle. Two faces around a vertex still makes no sense, so we can’t have faces with more sides than five.
Note: We have shown that all but five combinations are impossible. We can actually construct a polyhedral
sphere that satisfies these five combinations, so we must have found all of the possibilities. These five shapes
are called the Platonic solids.