Chapter 1: Exercises
 Your intuitive grasp of the basic rules of arithmetic can be enhanced by
some simple geometric operations.
Since whole number operations are basically counting operations, think
in terms of counting. For example, a number can be a set of n objects, e.g.
you might represent 3 as * * *, a sequence of three asterisks. Similarly
2 would be * *. To do addition, you simply put the objects together and
count the result. For example,
is what we mean by 3 + 5. To do multiplication of m by n, simply make n
rows of m objects; when counted, there are mn objects in all.
Now that you have the operations, you can see if various properties
are true. For example, addition is commutative because, if we put out
3 objects and then add 2 more to them, we can draw out 2 objects from this
stack and then draw out another 3 and there will be none left. For commutativity
of multiplication, you can lay out 2 rows of 3 objects and then rotate
the array ninety degrees to obtain 3 rows of 2 objects.
Describe similar operations for associativity and the distributive
law. Can you do the same for identity elements and inverses? How do you
describe subtraction and division?
 Arithmetic Series How do you add all the numbers from 1 to N?
This is the arithmetic series
.
For example the sum of the integers from 1 to 100 is 100(101)/2 = 5050.
But why is this true? One sometimes explains it by noting that if you
add
you always get 101. There
are 100 such additions and doing them all you have done the sum of all
integers from 1 to 100 twice.
Make up a simple geometric interpretation of the same proof using the
ideas of the last problem.
 One can also make up similar geometric representations for a variety
of algebraic results involving real numbers. The idea here is to use
line segments of specified lengths. Multiplication can still be accomplished
by making a rectangle whose sides have lengths equal to the factors; the
area of the rectangle is precisely equal to the product of the two lengths.

As an example, make up a geometric figure to explain the binomial
expansion:
The figure should be a square with sides of length a + b, and the square
should be cut up into four subrectangles.
 You can also do the same thing
in three dimensions. Construct a geometric figure corresponding to the
identity
 Let T be the set of all subsets of the set of real numbers. One has
two binary operators: the union operator
and the intersection
operator
defined on T. Which of the following are true?
If it is true, explain why. If it is false, give an example of where the
property fails to be true..

is commutative.

is commutative.

is associative.

is associative.

distributes over
.

distributes over
.
 There is an identity element in T for
. What is it?
 There is an identity element in T for
. What is it?
 Every element of T has an inverse for
.
 Every element of T has an inverse for
.
 The following are from
Book II of Euclid's Elements; for the
sake of simplicity, they are stated in modern algebraic notation.
Prove the following results valid for any elements x, y, and z of a field F.
Your proof should be a sequence of steps with justification for each step.
 Proposition 2: If x = y + z, then
.
 Proposition 3: If x = y + z, then
.
 Proposition 4: If x = y + z, then
.
 Proposition 56:
.
 Proposition 7: If x = y + z, then
.
 Proposition 8: If x = y + z, then
.
 Proposition 910:
.
 Make geometric figures which illustrate each of these results.
 Given numbers a and b, the problem is to find all x and y such that
 Use the method of the text to solve this problem. You might suppose
that your solution takes the form x = a/2 + t where t is a number to be
determined.
 Solve the problem by reducing it to the problem treated in the text
where subtraction was substituted with addition.
 Given numbers a and b, find all the solutions of
(Warning: There is one value of b that requires special treatment.)
 A quadratic equation
has two solutions, one of which
is twice as large as the other. What relation holds between a and b?
 A quadratic equation
has two roots
and
.
 Find a quadratic equation whose roots are
and
.
 Assuming that the roots
and
are both zero, find a
quadratic equation whose roots are
and
.
 Assuming that the roots
and
are both zero, find a
quadratic equation whose roots are
and
.
 The arithmetic mean of two numbers a and b is defined to be their
average
. The geometric mean of two numbers a and b
is defined to the square root of their product.
 For which values of a and b is the geometric mean not defined?
 Given the real number m and the nonnegative number g, analyze the problem
of determining the pairs (x, y) of real numbers whose arithmetic mean is m
and geometric mean is g.
 Show that the square of the arithmetic mean of two numbers cannot
be smaller than the the square of the geometric mean of the same two numbers.
 Analyze the problem of determining the numbers x and y whose sum is a
fixed number s and whose quotient is a fixed number q.
 Analyze the problem of determining the numbers x and y whose sum is a
fixed number s and whose difference is a fixed number d.
 Analyze the problem of determining the numbers x and y whose product
is a fixed number p and whose quotient is a fixed number q.
 Metrodorus compiled a collection of puzzles including one which
gives detais of the life of Diophantus. He says, "... his boyhood
lasted onesixth of his life; he married after oneseventh more; his
beard grew after onetwelfth more, and his son was born five years
later; the son lived to half his father's age (i.e. the son's lifetime was half
that of his father), and the father died four
years after his son." Use this information to determine how long Diophantus
lived, at what age he married, etc.
 Solve the equation
for the real number x where a is
a constant (Euclid, Elements Book II,
Proposition 11).
 Consider the real number
where a, b, and c
are rational numbers and
.
 Such a number satisfies a
polynomial equation of the form
where M and N are
rational numbers. Find M and N in terms of a, b, and c.
 Find necessary and sufficient conditions in order for x to satisfy
a linear equation
with A and B both rational and A nonzero.
 Let
and
be roots of
 Express B and C in terms of
and
.
 The expression
appears in the quadratic formula
(with A = 1). Express
in terms of
and
and explain why its being zero is equivalent to having both roots equal.
The quantity
or more generally
is called the
discriminant of the quadratic polynomial p(x).
 Let
and
be two solutions of the equation
where M and N are integers.
 Let
for
. Explain why
this sequence satisfies the linear recurrence relation:
 Show that
and
.
 Show that if one has two sequences, say
and
both of which
satisfy the above linear recurrence relation, then so does the sequence
where c and d are constants.
 Show that
satisfies the linear recurrence relation:
and that
and
. In particular, it follows from the
linear recurrence relation that
are all integers.
In special case in which M = N = 1, the sequence
is called the
sequence of Fibonacci numbers. The sequence looks like
and one has
 Now suppose that we know that
. Explain why
for large n is approximately equal to
.
 Suppose we wanted to find some rational numbers which were good
approximations of, say
. Then
is
less than 1 in absolute value. Also, this number and
are roots of
. Use the last part of this exercise to find
rational number approximations of
. Use these to get rational
number approximations of
.
 Given the length L and width W of a rectangle, the perimeter P is
given by P = 2L + 2W and the area A is A = LW.
 Given values P and A for the perimeter and area of a rectangle,
find formulas for the dimensions of the rectangle.
 In the first part of the problem, you probably solved a quadratic
equation. You know that quadratics have zero, one, or two solutions.
Interpret in terms of the rectangle what it means to have zero, one, or
two solutions.
 For a given perimeter, the rectangle with this perimeter and largest
area is a square. Explain why your solution to the first part of the problem
proves this assertion.
 A cylindrical can with top and bottom removed has area
and volume
where r is the radius of the can and h is its
height.
 Given the area A and volume V, find formulas for the radius and height
of the can.
 For a given volume, how do you make a can using arbitrarily large amounts
of material? For a given volume, how do you make a can using arbitrarily
small amounts of material?
 Solve the general system of two equations in two unknowns:
and
where a, b, c, d, e, and f are real numbers
where
. What happens if ad  bc = 0?
 The system of simultaneous equations
can be interpreted as finding the points of intersection of a line and
a hyperbola. Explain why solving this system of equations is equivalent
to solving the quadratic equation
.
 Parametrized Circle: The unit circle is the set of solutions of
the equation
. There are infinitely many such solutions and
so one cannot just make a list of the solutions. Instead, we would like to
express the solutions in parametric form, like
where f(t) and g(t) are functions. Such a description is called a
parametrization. One can do this with trigonometric
functions:
which describes all the solutions as t varies in the interval
.
But, we would like instead to obtain a a parametrization in which the
functions f(t) and g(t) are rational functions, i.e. quotients of polynomials.
To find the parametrization, let's assume that we have a solution (x, y)
and rearrange the equation in the
form
. Assuming that x is not zero and y is not 1,
we can rearrange this as:
Let t be this common value, i.e.
.
 Treating this as a system of simultaneous equations in the variables
x and y, solve for x and y as functions of t. Your answer will be a parametrization of the
unit circle. Make sure that you verify that every solution described by
your parametrization in fact is a solution.
 We started off assuming that
and
. How many
solutions were omitted by this assumption? How many solutions are, in fact,
not covered by the parametrization? A full description of the set of solutions
would have to include not only the parametrization but also the list of
these exceptional solutions.
 Parametrized Hyperbola There are three very commonly used
hyperbolas. One of them is the set of solutions of the equation xy = 1.
This is particularly easy to very parametrize:
where t is allowed to be any real number other than 0.
 Using the last problem as a model, try and work out a parametrization of
the set of solutions of the equation of the hyperbola:
.
 Modify your solution to obtain a parametrization of the
set of solutions of the hyperbola:
.
 Bisecting a Trapezoid Given a trapezoid with base and top
parallel, let x and y be the lengths of the base and top. Using a line
parallel to the base, divide the trapezoid into two smaller trapezoids
of equal area. Let w be the length of the this line segment. Show that
This is a problem known to the Babylonians.
 Using the last two problems as models, try to parametrize the points
on the circle of radius 2:
using a parametrization in which
the coefficients of the rational functions are integers.
 Show that if (x, y) lies on the unit circle, then the point (x + y, x  y)
lies on the circle of radius 2.
 Show that if (x, y) lies on the circle of radius 2, then the
point
lies on the unit circle.
 Can you use the last two parts to do the first part of this problem in
another way?
 Describe the following sets:
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 Describe the following sets:
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The set of all real numbers with
.
 The triangle inequality says:
.
Determine when equality holds.
 Prove that
for all real numbers a and b.
 Show that
if
and
. Is the result true if
one drops the 1 out of the inequality?
 Consider the set T of all subsets of the set of real numbers. As
mentioned earlier, there are two binary operators
and
.
One also has a binary relation:
. Determine the truth or
falsity of each of the following statements. If true, indicate why; otherwise
give a counterexample.

satisfies trichotomy.

is transitive.

is reflexive, i.e.
for all x in T.
If you were to define a set of positive elements of T, what would the set be?
Is the set of positive elements closed under
? What about
?
 Show that the sum of the interior angles of a quadrilateral is 360
degrees (or
radians).
 A regular ngon is a polygon with n equal length sides. Prove that
the sum of the interior angles of a regular ngon is 180(n  2) degrees
(or
radians).
 Let's try an alternate proof of the Pythagorean Theorem.
 Given a right triangle
where the sides opposite angles
A, B, and C respectively are of length a, b, and c. Assume that the vertices
have been labeled so that
.
 Start with a square of side b  a. Make four copies of the original
triangle, placing each on a side of the square with the right angle at the
vertex and the side b along the side of the square.
 Prove that the five pieces together make a square with a side of length c.
 Use the diagram to prove the Pythagorean Theorem.
 Let
be a triangle with obtuse angle C (i.e. angle C is
greater than
). Drop a line from A perpendicular to the side CB;
let D be the point of intersection of this line and CB. Let a, b, and c
be the lengths of the side of the triangle opposite angles A, B, and C
respectively and let e = DC. Then
This is essentially the Law of Cosines, as it appears in Euclid's Elements
Book II,
Proposition 12.
 What would be the result corresponding to the last exercise in case
the angle C is acute (i.e. less than
radians), rather than obtuse?
State the result and modify your proof appropriately. The result would be
Proposition 13 of Euclid's Elements Book II.
 The areas of polygons are typically calculated by breaking the figures
up into rectangles and triangles. Use this method to handle the following:
 A parallelogram is a quadrilateral in which opposite sides
are parallel. Show that the area of a parallelogram is the product of
its height and length of its base.
 A trapezoid is a quadrilateral in which the base and top
are parallel. Explain why the area of a trapezoid is the product of its
height and the average of the lengths of its base and top.
 A triangle might be thought of as a degenerate trapezoid, one where
the top has been shrunk down to a point. Does the formula for the area
of a trapezoid apply to triangles?
 Let L be a line and
and
be two distinct points on the
same side of the line. We would like to find the point P on the line L
such that the sum of the distance from
to P and the distance from
P to
is smallest possible? (What is the best way to get a drink of
water from a straight river while traveling from
to
?)
This looks hard until you get the idea of reflecting the point
across the line. To be precise, this means draw a line through
perpendicular to the line. Let Q be the point on this new line which is
the same distance to the line as P_2 is but with
and Q on opposite
sides of the line. Now, the distance from any point R on the line L
to
is the same as the distance to Q. Make a sketch and see if
you now know how to find point P.
Suppose you know the optimal point P. Show that the line
to P makes the
same angle with L as the the line from
to P. This is another way of
determining P. These results as well as those of the next exercise
are due to the Greek mathematician
Heron.
 Let L be a line and
and
be two points on opposite sides
of the line L. Now, we would like to find a point P on the line L such
that the absolute value of the difference of the distance from
to P
and the distance from
to P is largest possible.
Again, choose a point Q which is the reflection of
across the
line L. Sketch in a number of possible solutions until you see how to
locate the point P.
Suppose you know the optimal point P. Show that the line
to P makes the
same angle with L as the the line from
to P. This is another way of
determining P.
 Parabolas: Fix a point F = (0, p) and a line y = p.
 The distance from a point P = (x, y) to a line is the smallest of
the distances from (x, y) to any point on the line. In our case, the
line is horizontal: y = p. For a given P = (x, y), which point on the
line looks like it would be closest to P. Use the distance formula to
show that your answer is correct. You can do this by comparing the distance
between P and an arbitrary point (u, p) on the line to the distance from
your P to the point you think is closest.
 A parabola with focus F = (0, p) and directrix y = p is defined to
be the set of points P = (x, y) such that the distance from P to F is
equal to the (shortest) distance from P to the directrix y = p.
Use the distance formula to write down an equation for the parabola. Simply
the expression by getting rid of the square root signs and verify that the
equation reduces down to
.
 Find the focus and the directrix for the parabola
.
 Ellipses: Fix two points
and
where
c > 0 and let a > c. Set
. The ellipse with
foci
and
and semimajor axis a is defined to be the locus
of points P = (x, y) such that the sum of the distances from P to each of
the foci is equal to 2a. Express this condition as an equation using the
distance formula. Simplify the expression by eliminating the square root
signs (through squaring) to see that the equation of the ellipse
simplifies to
 Hyperbolas: Fix two points
and
where
c > 0 and let a satisfy 0 < a < c. Set
. The hyperbola with
foci
and
and semitraverse axis a is defined to be the locus
of points P = (x, y) such that the difference of the distances from P to each of
the foci is equal to 2a. Express this condition as an equation using the
distance formula. Simplify the expression by eliminating the square root
signs (through squaring) to see that the equation of the ellipse
simplifies to
 The equation of the ellipse looks similar to the equation for
the unit circle, if we think of the variables as being x/a and y/b. What is
the relation between the set of solutions
and
? If (u, v) is a solution of the
first equation, then (au, bv) is the solution of the second equation;
and if (r, s) is a solution of the second equation, then (r/a, s/b) is a
solution of the first equation. We can describe this by saying that
the graph of the ellipse is obtained by expanding the graph of the unit
circle by a factor of a horizontally and by a factor b vertically. Alternatively,
we can say that the unit circle is obtained from the graph of the ellipse by
contracting it by a factor of a horizontally and a factor of b vertically.
In the section on solving systems of equations, we obtained a
parametrization of the unit circle in terms of a variable t. Use it to find a parametrization of
the ellipse. Now, draw the parametrization and label several points to
get a good idea of which points on the ellipse correspond to which values
of the parameter t. For example, which point corresponds to t = 0, 1, 1, etc.?
As t grows without bound, how does the corresponding point change?
 Repeat the last exercise for hyperbolas instead of ellipses.
 This problem will study the family of lines which pass through a
single point. To make the discussion concrete, start with the fixed point
(1, 2) and let P be the set of all lines which pass through the fixed point
(1, 2). Clearly P is an infinite set. Let's start by taking a sample
of lines in P.
 For each equation below, determine whether or not the corresponding
line is in P.
 y = 3x  2
 y = 3x  5
 6x  2y = 10
 y = 2
 How many lines in P have slope 1/2? Find the equations of all these
lines and sketch their graphs.
 How many lines in P have slope 2? Find the equations of all these
lines and sketch their graphs.
 Find all the vertical lines in P and write down their equations.
 Find all the horizontal lines in P and write down their equations.
 Find the equations of all the lines in P which intersect the xaxis
at a 45 degree angle. Did you find two of these?
 Find the equations of all the lines in P which have yintercept 5.
 Find the equations of all the lines in P which have xintercept 5.
 How many lines in P have yintercept 2001?
 Let P be the set of all lines which pass through the point (1, 2).
For an arbitrary nonvertical line L in P, let m be its slope and b be its yintercept
 Find an expression for m in terms of b.
 Find an expression for b in terms of m.
 Use your expressions to check as many of the answers to the last question
as you can. Which parts couldn't be checked with your expressions?
 Try to find an expression for xintercept in terms of m. For which
values of m does your expression fail to work? How do you explain this
in terms of the geometry of the situation?
 We restricted our first two expressions to the case where the line L
was nonvertical. Why was this done?
 Let (c, d) be an arbitrary point. Let P be the set of all lines which
pass through the point (c, d). For a given nonvertical line L in P,
let m and b denote the slope and yintercept of L.
 Find an expression for m in terms of b. Your expression will involve
the fixed quantities c and d.
 Find an expression for b in terms of m.
 Carefully examine your justification for the answers of the previous
two parts. For which points (c, d) does each expression fail? Explain
geometrically, why one would expect these cases to be exceptional. Fix
your answers to handle the exceptional cases.
 Choose two distinct points, say (2, 1) and (3, 4). We want to
find the midpoint of the line segment from one point to the other.
By the midpoint, we mean the point which is on the line through the two points and
is equidistant from each of the points. You might guess (or maybe even
remember from previous math courses) that you can find the midpoint by
taking the average of the xcoordinates and the average of the ycoordinates,
i.e. the midpoint should be
.
 Find the equation of the line which passes through the two original
points. Use the equation to show that M lies on the line between the two
points.
 Show that the distances from M to each of the two original points are
equal. This verifies that M is the midpoint.
 Repeat the previous exercise for two arbitrary points (a, b) and
(c, d).
 Carefully check your solution to the last exercise to see if there
are any exceptional cases where your argument does not work. What are
the exceptional cases and how can you handle those exceptions?
 Just as points are determineed by their xcoordinate and ycoordinate
pair, nonvertical lines are determined by their slope and yintercept pair.
Since we defined a midpoint by taking the average of the coordinates of the
points, one might define a midline by taking the average of the
slopes and yintercepts. (Note: midline is NOT a standard mathematical term.)
 Let P be the set of all lines through the point (1, 2). Show that
if
and
are two lines in P, then so is the midline between the
two lines.
 A bisector an angle
is a line
BD where
is equal to
. How many bisectors are there
for a given angle? Is the midline of AB and BC a bisector of
?
(You should try to give a convincing argument for your answer.)
 Given two points A = (a, b) and B = (c, d), determine the set of all
points P = (x, y) which are equidistant to A and B. For this, you should
find an equation which whose solutions are all the points in this set
and show that it works.
 For two distinct points A and B, show that the set of points
equidistant to A and B is a line perpendicular to AB.
 The midpoint divides a line sebment AB into two equal pieces. Suppose
we wanted to divide it into more equal pieces. How could you do it?
Let A and B be two distinct points. Then the line is horizontal. What
are the points C and D on the line AB such that the distances from
A to C, C to D, and D to B are all equal?
 Repeat the same for the points A = (2, 1) and (2, 4).
 Repeat the same for the points A = (2, 1) and (3, 4).
 Repeat the same for the points A = (a, b) and B = (c, d).
 Check your last result for exceptional cases and fix your solution.
 Let A and B be two distinct points and n be a positive integer. Find
n  1 points
all on the line AB such that
the distance from
to
is
for
.
 Let A and B be two distinct points and let r be a real number with
,
 Find all points P on the line AB such that AP = rAB.
 Find all points on the line AB such that AP = rAB
and PB = (1r)AB.
 Let D, E, and F be the midpoints of the sides of the triangle
where D, E, and F are on the sides opposite the angles A, B, and C
respectively.
 Use analytic geometry to find a point P on all three of the
lines AD, BE, and CF.
 Is it true that
?
If so, what is the common numerical value of these fractions?
 Three points
for i = 1, 2, and 3 are said to be
collinear if they all lie on the same line.
 Show that if
,
, and
are distinct, then the three
points are collinear if and only if
 Find a necessary and sufficient condition for the three points to
be collinear in case one has
, but
is distinct from both
and
.
 Find a necessary and sufficient condition for the three points to
be collinear in case one all three
are equal.
 Two distinct points determine the line which passes through both
points. Similarly, we will show that usually three distinct points
determine circle; the circle which passes through 3 given points A, B, and C
is said to circumscribe the triangle
.
Let A, B, and C be distinct points.
 Find the equation of the line L of all points (x, y) equidistant from
A and B.
 Find the equation of the line M of all points (x, y) equidistant from
B and C.
 Assuming that the lines L and M intersect, find their point of
intersection. Show that this point P is the center of the circle which
circumscribes
.
 Still assuming that the lines L and M intersect, find a formula for the radius of the circle
which circumscribes
. Find the equation of the circle.
 Show that the lines L and M intersect if and only if A, B, and C are
not collinear.
 Let A, B, and C be three noncollinear points. Let L, M, and N be the
lines perpendicular to and through the midpoint of one of the sides of
the triangle
. Show that there is a common point of intersection
of all three lines.
 Finding the equation of a line which passes through two points can be
done with the two point form of the equation of the line. Here is a
different approach.
 Let two distinct points be
for i = 1, 2. If
,
then the line is vertical and its equation is
.
 If
, then the line has a slope m and yintercept b.
Substituting the two points into y = mx + b gives a system of two equations
in two unknowns. If you can solve this system, for m and b, one will know
the equation of the line.
Show that the above method works and that in the case ii, the system of
equations always has a unique solution, and that unique solution gives
the same answer as you would have gotten using the two point form of the
equation of the line.
 Now suppose one has three distinct points
for
i = 1, 2, and 3. Let's work analogously with the method of the last problem.
The general form of the equation of a circle is:
which can be written as
where A = 2a, B =  2b, and
are to be determined. Substituting the three
points in to this equation again gives a system of 3 equations in 3 unknowns
A, B, and C. In the case where
,
, and
are all different,
show that this system of equations has a unique solution precisely when
the three points are noncollinear.
 A circle and a straight line should intersect in 0, 1, or 2 points.
Furthermore, it is always easy to determine the points of intersection
because solving the system of equations reduces down to solving
a quadratic equation. If
they intersect in exactly one point, then we say that the line is tangent
to the circle. We know that
is the equation of
a circle with center (a, b) and radius r. Let
be a
point on this circle. Consider the line with equation
 Show that the line intersects the circle at the point
.
 Show that there are no other points of intersection of the circle and
the line. So, the line is tangent to the circle at the point.
 Show that the line is perpendicular to the radius line from the
center O = (a, b) to the point
.
 Finding the points of intersection of a circle and a parabola can
be nontrivial. In fact,
Descartes showed that this is equivalent to
solving an arbitrary quartic (fourth degree) equation. This is how it
is done:
 Let
be an arbitrary quartic equation.
Show that if you substitute x = y  a/4, then one gets an equation of
the form
. So, in order to solve an arbitrary
quartic, it is enough to solve a quartic with no cubic term.
 Consider the intersection of the parabola
with the
circle
. Show that if you could find
the points of intersection of an arbitrary circle and the parabola
,
then you would be able to solve an arbitrary fourth degree equation.
 Let
be a triangle. From each vertex of the triangle,
construct a line through the vertex and perpendicular to the opposite side.
Let L, M, and N be these points. Determine whether or not the three lines
L, M, and N intersect in a common point P.
 Let
be a triangle where A lies on a circle and BC is
a diameter of the same circle. Show that the angle A of the triangle is
a right angle.
 Let
be the parabola with focus (0, p) and
directrix y = p. If
is any point on the parabola. Show that
the line through P with yintercept
intersects the parabola in
exactly one point. This line is called the tangent line to the
parabola at the point
. Find the slope of this tangent line.
 Given an angle
and a point D, we say that AD bisects
the angle
if the ray AD lies inside the angle and
.
Consider a triangle
. Let D be on AC such that BD bisects
angle
, and let E be on AB such that CE bisets angle
.
Let M be the point of intersection of the lines BD and CE.
Finally, let F, G, and H be on BC, AC, and AB respectively such that
MF, MG, and MH are each perpendicular to BC, AC, and AB respectively.
 Explain why
is congruent to
. Then repeat
your argument to show that
is congruent to
.
 Why does it follow that MF, MH, and MG are all of the same length?
 Now explain why
is congruent to
.
 Why does it follow that AM bisects the angle
?
You have shown that the three angles are bisected by lines which meet in
a common point M. Further, the perpendiculars from this common point M
to each of the three sides are equal in length. This means that we can
draw a circle with center M and radius equal to this common length. The
three sides of the triangle are tangent to this circle; the circle is said
to be inscribed in the triangle.
 Distance from point to line: Let L be a line and P be a point
not on the line.
 If L is a vertical line, then its equation is of the form x = a. If
P has coordinates
. The point on the line closest to P is the
one which you reach by proceeding from P to the line along a line perpendicular
to L. Why? Find a formula for the distance from P to the line.
 Modify the last part to handle horizontal lines.
 Now suppose that the line L has equation y = mx + b where
.
Show that the shortest distance from
to the line L is
 Elliptical Refection: Let E be an ellipse with foci
and
. Lines L intersect the
ellipse in either 0, 1, or 2 points. If the line intersects in exactly
one point, then we way that the line is tangent to the ellipse. We
would like to characterize the lines which are tangent to the ellipse E.
Draw the ellipse E and a line L tangent to the ellipse at a point P on
the ellipse. The sum of the distances from the foci to any point on the
ellipse is always the same value. Now, draw a line from
to any point
on the line L. It intersects L at a point Q and the ellipse R at some
point R. Draw the line from Q back to
. Why is the distance from
R to
smaller than the distance the sum of the distances from R to Q
and from Q to
? You should now have an intuitive understanding of
why P is the point on the line such that the sum of the distances from P
to the foci is least possible.
By Heron's result from the exercises of the last section, it follows
that the lines from the foci to P make the same angle with the line L.
This tells us how we construct a tangent line to any point P on the ellipse.
To do so, start by bisecting the angle
with a line PD. Then
make L perpendicular to PD and through P. Show that this is the desired
tangent line.
Show that the line
is tangent to
the ellipse
at the point
.
(You might start by trying to show the result in case a = b = 1. Then see
if you can reduce to this case.)
 Hyperbolic Reflection: Repeat the last exercise with a hyperbola.
 Parabolic Reflection:Let
be a point on the parabola
.
This parabola is the locus of all points which are equidistant from the
focus F = (0,p) and the line y = p.
Show that the line
intersects the parabola in precisely
one point. We say that this line is tangent to the parabola at P.
The vertical line
intersects the line y = p at some point Q.
Let R be a point other than Q on the same line such that its distance from P is the same distance
as the distance from P to Q. Let M be the midpoint of the line segment
FR. Calculate the coordinates of the these points and the slope of
the line PM, showing that PM is perpendicular to the tangent line.
Explain why this means that the angle between the tangent line and the
vertical is equal to the angle between the tangent line and the line from
P to the focus F.
 Even and Odd Properties The radian measure of an angle is
equal to the length of the arc of the unit circle swept out as you move
through the angle from the positive xaxis to the radius line corresponding
to the angle. If you sweep out the angle in a counterclockwise motion, it
is a positive value; otherwise, it is negative.
If you sweep out the angle
you arrive at a point on the
unit circle which is the reflection across the xaxis of the point you
would have arrived at by sweeping out the angle
. Use this property
to show that the cosine function is even and the sine function is odd, i.e.
for all angles
.
 Complementary Angles: Starting from the positive yaxis and sweeping out an angle
in a clockwise direction brings you to the point on the circle corresponding
to the angle
as measured from the xaxis in a counterclockwise
direction. Notice that the two points on the circle, one corresponding to
the angle
and the other corresponding to the angle
are reflections of each other across the 45 degree line y = x. In particular,
the coordinates of one point can be obtained from the coordinates of the
other point by swapping the xcoordinate with the ycoordinate.
The point on the unit circle corresponding to the angle
is
. What are the coordinates of the point
corresponding to the angle
? Show that this implies that
These identities are useful to convert properties of the sine function into
properties of the cosine function and vice versa.
 Special Triangles: The values of the trigonometric functions
for some very special angles angles can be calculated from simple properties
of some special triangles.
 454590 Degree Triangle Since two of the angles are equal, their
opposite sides equal lengths. (Triangles with two sides of equal length are
called isosceles.) Make the length of one of these sides equal to 1
unit. Use the Pythagorean Theorem to calculate the length of the
longest side (called the hypotenuse). Now use the triangle to
write down the values of the trig functions at
radians.
 306090 Degree Triangle Draw such a triangle and label the
angles A, B, and C where the angles are
,
, and
respectively. Let D be the midpoint of the hypotenuse AB.
Draw a line segment from D to C as well as line segments from D perpendicular
to the sides AC and CB. Show that all four of the inner triangles are
all congruent right triangles similar to the original triangle. In particular,
this means that the triangle
is equilateral. Use this
information to read off the the values of the trig functions for the
angles
and
.
 The 306090 degree triangle has one of its angles equal to twice
another of its angles. One can make lots of triangles of this type.
To do so, start with a line L and make an isosceles triangle
with the base BD on the line L. Let a be the common length of the CB and CD.
Now mark off a point A on L so that the distance from A to D is exactly a
and such that D lies between A and B. Now connect the line segment AC.
Your drawing should look somewhat like your 306090 degree triangle with
the additional line segment CD added in  of course, angle C will not in
general be a right angle. Use the result that the sum of the angles of
any triangle is
in order to show that angle B is exactly
twice as large as angle A.
The angle B can be made to be any size between
0 and 90 degrees, and as it changes, angle A ranges from 0 to 45 degrees.
In the limiting case where B is ninety degrees, the two segments CB and CD
coalesce and you are left with a 454590 degree triangle. On the other
hand, if you make the length of DC exactly equal to a, then you have
the 306090 degree triangle.
367272 Degree Triangle: Make a triangle as in the last part of
the exercise with the angle B equal to exactly 72 degrees. Then angle A
will be exactly 36 degrees. We chose the value of 72 degrees so that our triangle
would be isosceles.
 Explain why the triangle is isosceles.
 Explain why triangle
is similar to
.
 Let's fix our unit of measure so that the sides AC and AB are of length
exactly one. Then the segment DB has length exactly 1  a. Using similar
triangles, show that
.
 Find the value of a, being careful to explain which root of the equation
is the value of a.
 Add additional line segments into your triangle in order to be
able to read off the values of the trig functions for 72 degrees and
18 degrees. Use your results to calculate the values of the trig functions
for 36 degrees.
 You can use the above triangles to construct regular polygons inscribed
in circles. Start with a circle of radius 1. Suppose that you knew that
the side of a regular ngon inscribed in the circle was precisely L units
long and that you could construct such a line segment. Then, you could
simply mark off points on the circle, where the distance between successive
points is always L. By connecting the points, you would have your regular
ngon.
 Show that a square inscribed in the circle has sides of length
.
 Show that the length of the sides of a regular hexagon inscribed in
the circle is exactly 1.
 Show that the length of the sides of the a regular decagon (a 10gon)
is precisely the length a you calculated for the 367272 degree triangle.
 Use the Pythagorean Theorem to show how to construct segments of
length
and
. Use this to show that you can construct
segments of length the sides of the regular quadrilateral, hexagon and decagon.
 Explain why this tells you that you can construct regular pentagons
and equilateral triangles. Do you see how to construct regular ngons
where n has the form
,
, and
for all
? (If so, you know how to construct all the regular
polygons whose construction was known prior to 1798, the date when
Gauss discovered that you could construct a regular 17 sided polygon.)
 Angle Trisection: Start with our triangle
and point
D on AB with the lengths of AD, DB, and DC being equal to, say r. Extend
the AC beyond C to a point E such that CE also has length r. Complete
the diagram by adding the segment BE. Show that the angle A is exactly
one third the measure of the angle
. So, the diagram allows
one to trisect the angle
.
To trisect any angle, we need only start with the angle and label it
where BC and CE are chosen equal in length. If we can construct
the rest of the diagram, then we will have a construction for the trisection
of any angle. Draw a circle of radius r with center C. Then our diagram
will have to have D on this circle because CD is also of length r. Mark
off a distance r on the straightedge by putting two marks on it. Lay the
straightedge on the paper and slide it along in such a manner that one mark is always on
the line determined by C and E and the other mark is always on the circle.
Stop when the straightedge also goes through the point B. Draw the line ADB
where A is the point where the first mark is on the line CE and D is the
at the point on the circle where the second mark is on the straightedge.
One has a diagram of exactly the right type and so we have trisected the
angle.
The problem of
trisecting an arbitrary angle is one of the most
famous problems of antiquity. The above construction is due to
Archimedes
and does require marking the straightedge, which is not a legal operation
in doing constructions with straightedge and compass. So, it is not
really a construction of the trisection in the ancient sense. It was
shown by
Wantzel in 1837 that the construction is not possible.
 Product Identities: Prove the identities
These were used, before the advent of logarithms, as a means for
calculating products of numbers. For example, to use the first one:
 Suppose you want to multiply a and b. By multiplying the two
by appropriate powers of 10, gets numbers c and d where c and d lie between
1 and 1. One can assume that c is the larger of the two.
 Using tables, find x and y such that
and
.
 Calculate u = x + y, and v = x  y.
 Using tables, find
and
.
 Calculate the average z of r and s. Then divide z by the powers of
10 used in the first step to get the product of a and b.
 Addition Theorems: Let
and
be any angles.
 The addition theorem for cosines is
One also has a subtraction result:
Use the results of the previous exercises to show that the second formula
implies the first. (It is also true that the first implies the second.)
 Let the points
,
,
,
be the points on the
unit circle corresponding to the angles 0,
,
, and
(measured counterclockwise from the xaxis). The chord
is the
same length as the chord
. Draw a sketch including all the points
and chords. Your sketch should label the points with their coordinates
expressed in terms of sines and cosines.
 Use the distance formula to express
.
 Simplify the expressions using the Pythagorean Theorem to show that
the second form of the law of cosines is true.
 The addition theorem for the sine function is:
As with the law of cosines, one has a subtraction form that looks like:
Show that either form implies the other.
Use the result of the last exercise to show that the addition theorem
for sine function is true.
 The main addition theorems are the ones for the sine function and the
cosine function. However, there is one for the tangent function looks like
What is the subtraction form of this addition theorem? Show that this
addition theorem is a consequence of the addition theorems for sine and cosine.
 Law of Cosines Let
be an arbitrary triangle with
sides a, b, and c opposite angles A, B, and C respectively. The Law of Cosines
says
In case the angle C is a right angle (
radians), the result is just
the Pythagorean Theorem.
 Dropping a perpendicular from the vertex A to the side CB, one obtains
a point D on the line CB.
One has two right triangles
and
. Draw a diagram
labeling all the points and all the sides of the triangles. Express the
cosine of the angle C in terms of a quotient of the lengths of two of the
sides.
 Use the Pythagorean Theorem
on the two triangles to get expressions relating the squares of the lengths
of various of the sides. Use these to express
in terms of other
squares and work with the expression until it becomes the Law of Cosines.
 Law of Sines Let
be an arbitrary triangle with
sides a, b, and c opposite angles A, B, and C respectively. The Law of Sines
says
 To prove the last equality, start with a triangle
and drop
a line perpendicular from angle A to a point D on the side BC. Again
we have two right triangles
and
. Label all the
sides of the triangles and express the
and
as quotients
of various of the sides. Use these to complete the proof.
 Once you have proved the second equality, what needs to be done to prove the
first one?

Heron's Theorem: Consider the triangle
with
sides of length a, b, and c opposite angles A, B, C respectively. Let
s = (a + b + c)/2 be the semiperimeter of the triangle. Then the
area of the triangle is
.
 Recall that the bisectors of the three angles intersect in a common
point M, which is the center of the inscribed circle. The radius of the
inscribed circle is the common distance from M to each of the three sides
of the triangle. Show that the triangle has area
.
 Consider the system of equations:
and
. Solve this system to get
the half angle formulas:
and
.
 Substitute the expression for
obtained by the Law of Cosines in
the formulas of the last part and show that they can be simplified to:
and
.
 Use the double angle formula:
together
with the results of the last part to conclude that
. This completes the proof of Heron's Theorem.
 The nice thing about the above proof is that it gave me the excuse to
use most of the geometry we have just studied. The original proof was
also quite circuitous and hard to understand. To understand things
better, let's look at the proof from a different point of view.
Start with a triangle
where angle C is chosen to be one
of the acute angles of the triangle. We let a, b, and c be the lengths of
the sides opposite angles A, B, and C respectively. Drop a line from the
point A perpendicular to the side BC, letting D be the point of intersection
of the line and BC. The length h of AD is the height of the triangle and
let d = CD. Does this look familiar, it is the picture you used to
prove Proposition 13 of Book II of Euclid's Elements. That result was
essentially the Law of Cosines, but it the cosine function did not appear
in it. Instead, one had
From the triangle
, one sees by the Pythagorean Theorem
that
. Furthermore, the area of the triangle is
clearly
. Combining all these results show that
Show that this expression is equivalent to the one in by Heron's Formula.
 What is arclength? The length of a polygonal line should be
the sum of the lengths of the line segments which make it up. If one wants to
define the length of a curved line, then things are much less clear. The
definition which will be used is: Take the sequence of points on the curve,
connect consecutive points with straight lines, take the length of this
polygonal line. Then the length of the curve is the least upper bound of
the lengths of all such polygonal lines.
The rest of this exercise is to illustrate how easy it is to go astray
with lengths of curves. Suppose we want to find the length of the unit circle
. Here is one approach:
 Start by circumscribing the square with the square whose sides are at
and
. The perimeter of this square is the first
approximation.
 Clearly the approximation is rough as the corners stick out substantially.
So, at the second stage, cut out the four corners. The vertices to be added are
where the lines
intersect the unit circle. So, e.g. in the
first quadrant, add in a line segment from the new vertex both horizontally and
vertically. The picture should now look like a stair case with 2 steps.
Repeeat with the other new vertices. This is the second approximation.
 For the third step, add in vertices at
,
,
, etc.
Now one has doubled the number of stair steps and we have a polygonal line
which is closer to the circle.
 Repeat the process with
,
,
, etc. for the
fourth step. Again the number of stair steps doubles.
 Keep repeating this process. Soon, the polygonal arc appears to be
almost indistinguishable from the unit circle. The circumference of the
circle should be the limit of the lengths of the polygonal arcs.
Calculate the lengths of the polygonal arcs for each step. Why is this
not a good way to define the circumference of a circle?
 Circumference of a circle: Again take the unit circle. Inscribe
a polygon, for example a regular ngon, in the unit circle. Draw the rays
from the center of the circle toward each of the vertices on the polygon.
Extend these rays beyond the circle and draw lines tangent to the circle
and parallel to the sides of the inscribed polygon. One now has the
circle circumscribed.
 Explain why the perimeter of the circumscribed
polygon is larger than the perimeter of the inscribed polygon.
 Now refine the inscribed polygon by adding in additional vertices and
the lines connecting the adjacent vertices. Show that the perimeter of
the refined polygon is still smaller than the perimeter of the (unchanged)
circumscribed polygon. (Hint: Consider one of the sectors of the circle
formed by adjacent rays toward the vertices of the new inscribed polygon.
the two radius lines and the segment of the polygon make an isosceles
triangle and a segment of the circumscribed polygon cuts the rays at points
beyond the circle.)
 Calculation of
The circumference C of a circle of radius r can be calculated by
. The real number
is an irrational number
whose value we propose to compute. One simple approach is to use the relation between
and the circumference of a circle of radius 1. The idea is to inscribe
regular nsided polygons in the unit circle, compute their perimeters and
realize the circumference of the circle as the limit of the perimeters as
n grows without bound. Although the idea is simple enough to understand,
the mechanics of actually doing it will use many of the ideas of this
chapter.
 The perimeter
of a regular nsided polygon inscribed in the unit circle
is simply n times the length
of one of its sides. So, to calculate
,
it is enough to figure out how to calculate
. Finding a formula for
is not so easy to do. On the other hand, notice that it is enough to
get a formula for
for some arbitrarily large values n. So, the
first step is to note that it is enough to calculate
for n of the
form
where
. To get the process started,
calculate
. For this inscribe a square inside a circle and use the
Pythagorean Theorem to show that
.
 The next step is to calculate
assuming that we can calculate
. To do so, start with a regular nsided polygon inscribed in the
unit circle. Draw the n radius lines from the center of the circle to the
vertices of the polygon. Now, let's concentrate on one of the triangles
formed by two adjacent radius lines OA and OB and a side AB of the
polygon; O is the center of the circle. We know that OA = OB = 1
and so the triangle
is isosceles; its base angles A and B are
therefore equal. Drop a line OD from O perpendicular to AB; D can be
taken as the intersection of this line and AB. Then
and
are congruent; so
.
Now, explain why
.
Now, let's try and compute
. Using the same formula gives
So, if we are to relate
and
, then we need to find
a relation between the sine of an angle and the sine of twice that angle.
But, you already know such a relation from the addition theorem for
the sine function. Use this and the Pythagorean Theorem to show that
 The last equation says that
satisfies a fourth degree equation.
Solve this fourth degree equation to obtain
When you are done, go back over your work and check to be sure that you
can justify the particular square roots chosen at each step. For example,
when you solved for
, there were two roots  but only one is
the correct one, and the reason uses an estimate on the size of
.
 You have already shown that
. Substitute this into
the equation of the last part to show that
Now use this expression to calculate
Do one more step to show that
The form of all these results is
where there are precisely k  1 two's on the right side. So, one now has
for the perimeter of the regular inscribed polygon with
sides:
 For particular values of
, one can calculate
, assuming that one knows how to calculate square roots. These
values then should approach
. Use the formula to calculate the
for k = 2, 3, 4, and 5. Your final result should be approximately
3.1365, which rounds to 3.14. If you were to carry this out a bit further,
with k = 10, one has
sides to the polygon and an approximate
value for
of 3.1415877 which rounds to 3.14159.
 From the figure of a polygon inscribed in the circle, it is intuitively
obvious that the estimates for
will all be underestimates. This seems to
be validated in the numerical estimates which are clearly increasing and
less than
. Use properties of the sine function to show that the
approximations to
are in fact increasing as k increases.
 One could equally well have circumscribed the circle with regular
nsided polygons. Let
be length of the side of the nsided regular
polygon which circumscribes the unit circle. Show that
.
But then
where
.
 Show that
. This says that
we can find
by taking the square root of the average of
and 1.
 Use the circumscribed polygons with
sides in order to estimate
. By calculating the first examples, you will see that the estimates
appear to be decreasing as k increases and that the estimates are all greater
than
. (Note that this relationship is NOT obvious from the figure.)
Use properties of the trig functions to verify that the sequence of these
estimates for
are indeed decreasing. By using BOTH estimates, we
not only get an estimate for
, but also an estimate for how large the
error can be. The estimate for
should be the average of the upper and
the lower estimate, and the largest error is simply half the difference between
the two estimates.
Revised: June 28, 2001
All contents © copyright 2001 K. K. Kubota. All rights reserved
