Additional Exercises for Chapter 3
About Parabolas, Ellipses, and Circles
21. A parabola is given. The point V is the intersection of the parabola with its axis and d is the
axis
x
2d
F
d
P
y
Q
V
directrix
distance from V to the focus F .
i. Why is the tangent of the parabola at V parallel to the directrix and hence perpendicular
to the axis? [Hint: Suppose the tangent at V where to slope downward. Since the tangent
“hugs” the parabola near the point V , there would be points on the parabola closer to
the directrix than V but farther from the focus than V . Can this be?]
ii. Show that the point Q specified in the figure is on the parabola.
iii. Let P be any point on the parabola, let x be the distance from P to the axis, and let y
be the distance from P to the tangent at V . Use one of the propositions of Section 3.1
1
of the text to show that y = 4d
x2 .
22. What follows is a proof of Proposition 3.1 that uses a basic property of light. The property
in question, formulated by the French mathematician Pierre de Fermat, asserts that a light
ray will “choose” a path of shortest distance. [See Chapter 9.] Consider a parabolic mirror.
Suppose that there is a light source at the focus F with light shining in all directions. Let a
Q
Q
axis
axis
P1
P1
P
P
P2
F
P2
directrix
F
directrix
ray emanate from F , reflect off the mirror, and proceed to a point Q. As to the precise path
of the ray from F to Q, consider the three possibilities shown in the diagram above (on the
left). Those striking the mirror at P1 , P2 , and P . In the last case, the ray - after striking the
mirror at P - continues to Q on a path that is parallel to the axis of the parabola. In the
diagram on the right the three segments from F to the parabola are replaced by segments
- all parallel to the axis - from the parabola to the directrix. Why do these two diagrams
taken together show that of all the possible paths from F to the mirror to Q, the shortest one
approaches Q on a path parallel to the axis? So by the shortest path principle, this is the
path that a ray will take from F to the mirror to Q. Since Q can be a point on any reflected
ray, it follows (as was already observed in Section 3.1) that the mirror reflects all rays coming
from F parallel to the axis of the parabola.
23. Consider an elliptical mirror. Suppose that there is a light source at one of the focal points
F . What do all the light rays that emanate from F and strike the mirror have in common?
24. The figure below shows an ellipse, its center C, and its major and minor axes. Let a and b
denote one half of the lengths of the major and minor axes respectively. Let P be any point
on the ellipse and let x and y be the respective distances from P to the minor and major axes.
2
2
Use one of the propositions in Section 3.1 to show that xa2 + yb2 = 1.
P
x
b
a
y
C
25. Suppose that a triangle ∆ABC sits inside a circle in such a way that the side AB is a diameter
of the circle and the vertex C is on the circle. Show that the angle at C is 90◦ . [Hint: Let
O be the center of the circle, and draw in the segment OC. Let α be the angle and A, β the
angle at B, and study the two triangles ∆AOC and ∆BOC.]
x
a
b
diameter
2
P
26. Consider a point P on the diameter of a circle and let a and b be the lengths of the two segments that P determines along
√ the diameter. Let x be the length of the segment from P up
to the circle. Show that x = ab. Do so twice. First by making use of Proposition 3.5, and
then again by using either the Pythagorean Theorem (three times) or similar triangles.
Circles and Tangent Lines
27. Consider a circle, a point P on it, and the tangent line to the circle at P . Take a point Q on
the circle different from P and let L be the line through Q and P . By making use of the figure
below show that the tangent to the circle at P is perpendicular to the diameter through P .
L
Q
ϕ
θ
diameter
P
tangent at
P
[Hint: Push Q to P and observe that L rotates towards the tangent. What is the angle at
Q of the moving triangle equal to throughout the process? What happens to ϕ? Therefore
what happens to θ?]
28. Consider a tangent line of a circle and draw a radius to the point of tangency. By Exercise
27, the angle between the tangent and the radius is 90◦ . Verify this fact again by using a
proposition from Section 3.1.
29. We are in Islamic North Africa in the later Middle Ages. From a vantage point M on the slope
of a mountain 1.5 miles (or 7920 feet) above sea level a clear view is had of the Mediterranean
Sea. From this site, a geometer from the court of the Emir measures the angle that a plumb
M
H
r
C
3
line makes with his line of sight to the sea at the horizon. He first measures this angle to be
◦
◦
◦
88 16 , then 88 12 , but finally concludes that 88 13 is most accurate. He uses this measurement
to estimate the radius of the Earth. How does he do it and what estimate does he achieve?
What estimates did his first two measurements given him? How susceptible is this method to
error? [Just in case you’re wondering ... this is fiction. However, Islamic scholars did make
estimates of the size of the Earth in precisely this way. They perfected the astrolabe of the
Greeks before the tenth century and used it to measure such angles with the accuracy that
this exercise calls for.]
30. In her Parade Magazine column of June 15th, 2003, Marilyn Vos Savant answers her reader’s
question “If a person were standing at sea level looking out over the ocean, how far away
would the horizon be?” with “The horizon− in this case, where the sky meets the sea−isn’t
so far away: only about 2.5 miles.” Assess the accuracy of Marilyn’s reply.
President Garfield (and Pythagoras)
31. A trapezoid is a quadrilateral with two parallel sides. Let a and b be the lengths of the two
parallel sides of a trapezoid and let h be the distance between them. Show that the area of
the trapezoid is 12 (a + b)h by making use of the figure below.
b
D
h
h
A
C
a
B
32. Let ∆ABC be a right triangle with sides a, b and c and right angle at C as shown below.
Extend the side CA to a straight line and attach a copy ∆ADE of the given triangle to the line
as indicated. Why is the angle ∠EAB equal to 90◦ ? Compute the area of the quadrilateral
D
b
a
E
c
A
c
b
a
C
4
B
CBED in two ways to show that a2 + b2 = c2 . [This proof of the Pythagorean Theorem was
devised by James Garfield before he became President of the United States. Garfield was a
college professor, a major general in the Civil War, and a Republican senator from Ohio (in
that order) before he became President in 1881. He was assassinated in the same year. His
proof appeared in the New England Journal of Education in 1876.]
More about Circles and Areas
33. Depicted below are an equilateral triangle and a square. In each case the sides have length 2.
All circular arcs all have radius 1 and are centered at the appropriate vertex. Find the areas
of the star shaped regions at the center.
34. The figure below shows a right triangle as well as three semicircular arcs. The diameter of
each semicircle is a side of the triangle. Each of the two shaded regions that this configuration
determines is a lune.
i. Show that the sum of the areas of the two lunes is equal to the area of the right triangle.
ii. For what configuration is the sum of the areas of the two lunes as large as possible
and what is this largest sum equal to? For what configuration is the sum of the areas
smallest?
35. The Book of Lemmas of Archimedes is a collection of 13 propositions about geometry. It
came down to us only in transcription from the Arabic. One of the problems considers the
upper half of a circle with diameter AB. See the figure below. The points C and D divide
the diameter in such a way that AC = DB. Two semicircles with diameters AB and DB are
placed above the segment AB and another semicircle with diameter CD is placed below the
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segment AB. Archimedes calls the geometrical shape with boundary determined by the four
semicircles a salinon or salt cellar. Let O be the center of the segment AB and let P Q be the
perpendicular to AB through O from the one semicircle to the other. Show that the area of
the salinon is equal to the area of the circle with diameter P Q. [Don’t ask me why this figure
P
A
a
O
b
C
D
a
B
Q
is called a salinon, but consult http://www.cut-the-knot.org/proofs/Lemma.shtml after you
have solved the problem.]
36. Consider a circle of radius r and a diameter of the circle. Place any square into the upper
semicircle so that the base is on the diameter and one corner is on the circle. Let a be the
length of the side of this square. The square is shown on the left in the figure below. Now
a
b
place a second square to the right of the first one as shown in the figure and let b be the length
of its side. Describe how the placement of the second square is possible by considering the
first figure below. Show that the sum of the areas a2 + b2 of the two squares is equal to r2 .
P
a
Proceed as follows. Place the point O on the diameter by marking off a distance a from the
lower right vertex of the second square and draw in the two segments from O to the semicircle
√
as shown below. Show that both of these segments have length a2 + b2 . Conclude that
√
O must be the center of the circle and that a2 + b2 = r. [Hint: Suppose that a different
point is the center.] This problem with its solution was taken from the section Mathematical
Entertainments, edited by Alexander Shen, in The Mathematical Intelligencer, Springer6
b
a
O
a
Verlag, New York, Volume 23, Number 2, 2001.
37. Start with an equilateral triangle of radius r and vertices A, B and C. Draw three circular
arcs of radius r centered respectively at A, B, and C as shown in the figure. Compute the
C
r
r
r
A
B
area of the curved region that the three circular arcs determine. A curved region of this shape
forms the cross-section of the piston of the Wankel engine. This internal combustion engine
has a single cylinder in which this piston revolves. The engine is named after Felix Wankel,
its German inventor. The Mazda RX-8 has such engines.
About the Results of Archimedes
38. Explain why the figure below shows that
1
4
+
1
42
+
1
43
+ · · · = 31 . This is equivalent to Archi-
1
16
1
16
1
8
1
8
1
4
1
4
1
2
1
2
medes’s equality 1 + 14 +
1
42
+
1
43
+ · · · = 34 .
7
39. A parabola is cut with a straight line. The length of the cut is 5 and the area of the resulting
parabolic section is 16. What is the distance from the vertex of the parabolic section to the
cut?
40. The distance from the focus F of a parabola to its directrix is 3. The parabola is cut parallel
to the directrix at a distance of 7 units from the directrix. Determine the area of the parabolic
S'
S
F
directrix
section. [Hint: Refer to the figure, analyze the triangle SF S 0 , and find the length of the cut.]
41. The figure below shows a circle with center C and radius r that is cut by the segment SS 0 .
The segments CS and CS 0 determine the angle θ, and the segment CV bisects this angle.
Refer to Problem 3 on page 66 of the text for the fact that the area A of the circular section
SV S 0 is equal to 21 θr2 − r2 sin 2θ cos 2θ . Let B be the area of the triangle SV S 0 .
S
V
r
θ
2
C
θ
2
r
S'
i. Express first the area B and then the ratio
A
B
as functions of both r and θ.
ii. Compute the ratio A
in each of the following two cases: θ = π and θ = π2 . Conclude that
B
there is no fixed constant k such that the area A of the circular section is always equal
to k times the area B of the inscribed triangle. [Recall that Archimedes’s Theorem tells
us that there is such a fixed k for parabolic sections and that it is k = 43 .]
Roots of Polynomials
The next several problems continue the themes of Exercises 1C and Exercises 3F.
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42. Complete the square for the polynomial −2x2 + 7x − 5. What is the largest value of this
polynomial? What x provides this value? What are the roots of the polynomial?
43. Complete the square for the polynomial 3x2 −9x+8. What are the largest and smallest values
of the polynomial. What are the roots?
44. Consider
the generic quadratic
polynomial ax2 + bx + c. By the quadratic formula, x =
√
√
2
2
−b+ b −4ac
and x = −b− 2ab −4ac are both roots of the polynomial. Therefore both
2a
x +
b−
√
b2 − 4ac
b+
and x +
2a
√
b2 − 4ac
2a
divide ax2 + bx + c. Confirm this by computing the product (x +
and comparing what you get with ax2 + bx + c.
√
b− b2 −4ac
) · (x
2a
+
√
b+ b2 −4ac
)
2a
45. Factor the polynomial x3 + 6x2 − 9x − 14 as completely as possible. [Hint: x = −1 is a root.]
46. Factor x3 − 4x2 − 4x − 5 as completely as possible. [Hint: Check for roots.]
47. Factor x4 + x3 − x2 + x − 2 as completely as possible. [Hint: Look for roots between -2 and 2.]
The article Mirror, Mirror from the Science Section of the New York Times of August 30th, 2005,
will give you insight into the culture of large scale scientific research. It will also provide you with
an interesting example of a parabolic surface. Have a go at it.
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