Web Homework College Algebra

Chapter 1: Exercises

1.1 Algebra

1.1.1 Simplifying Expressions

  1. 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,

    images/rev1.png

    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?

  2. Arithmetic Series How do you add all the numbers from 1 to N? This is the arithmetic series images/rev2.png . 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 images/rev3.png 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.

  3. 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.
    1. As an example, make up a geometric figure to explain the binomial expansion:

      images/rev4.png

      The figure should be a square with sides of length a + b, and the square should be cut up into four subrectangles.

    2. You can also do the same thing in three dimensions. Construct a geometric figure corresponding to the identity

      images/rev5.png

  4. Let T be the set of all subsets of the set of real numbers. One has two binary operators: the union operator images/rev6.png and the intersection operator images/rev7.png 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..
    1. images/rev8.png is commutative.
    2. images/rev9.png is commutative.
    3. images/rev10.png is associative.
    4. images/rev11.png is associative.
    5. images/rev12.png distributes over images/rev13.png .
    6. images/rev14.png distributes over images/rev15.png .
    7. There is an identity element in T for images/rev16.png . What is it?
    8. There is an identity element in T for images/rev17.png . What is it?
    9. Every element of T has an inverse for images/rev18.png .
    10. Every element of T has an inverse for images/rev19.png .
  5. 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.
    1. Proposition 2: If x = y + z, then images/rev20.png .
    2. Proposition 3: If x = y + z, then images/rev21.png .
    3. Proposition 4: If x = y + z, then images/rev22.png .
    4. Proposition 5-6: images/rev23.png .
    5. Proposition 7: If x = y + z, then images/rev24.png .
    6. Proposition 8: If x = y + z, then images/rev25.png .
    7. Proposition 9-10: images/rev26.png .
  6. Make geometric figures which illustrate each of these results.

1.1.2 Solving Equations

  1. Given numbers a and b, the problem is to find all x and y such that

    images/rev27.png

    1. 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.
    2. Solve the problem by reducing it to the problem treated in the text where subtraction was substituted with addition.
  2. Given numbers a and b, find all the solutions of

    images/rev28.png

    (Warning: There is one value of b that requires special treatment.)

  3. A quadratic equation images/rev29.png has two solutions, one of which is twice as large as the other. What relation holds between a and b?
  4. A quadratic equation images/rev30.png has two roots images/rev31.png and images/rev32.png .
    1. Find a quadratic equation whose roots are images/rev33.png and images/rev34.png .
    2. Assuming that the roots images/rev35.png and images/rev36.png are both zero, find a quadratic equation whose roots are images/rev37.png and images/rev38.png .
    3. Assuming that the roots images/rev39.png and images/rev40.png are both zero, find a quadratic equation whose roots are images/rev41.png and images/rev42.png .
  5. The arithmetic mean of two numbers a and b is defined to be their average images/rev43.png . The geometric mean of two numbers a and b is defined to the square root of their product.
    1. For which values of a and b is the geometric mean not defined?
    2. Given the real number m and the non-negative number g, analyze the problem of determining the pairs (x, y) of real numbers whose arithmetic mean is m and geometric mean is g.
    3. 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.
  6. 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.
  7. 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.
  8. 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.

1.1.3 Solving Equations of One Variable

  1. Metrodorus compiled a collection of puzzles including one which gives detais of the life of Diophantus. He says, "... his boyhood lasted one-sixth of his life; he married after one-seventh more; his beard grew after one-twelfth 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.
  2. Solve the equation images/rev44.png for the real number x where a is a constant (Euclid, Elements Book II, Proposition 11).
  3. Consider the real number images/rev45.png where a, b, and c are rational numbers and images/rev46.png .
    1. Such a number satisfies a polynomial equation of the form images/rev47.png where M and N are rational numbers. Find M and N in terms of a, b, and c.
    2. Find necessary and sufficient conditions in order for x to satisfy a linear equation images/rev48.png with A and B both rational and A non-zero.
  4. Let images/rev49.png and images/rev50.png be roots of

    images/rev51.png

    1. Express B and C in terms of images/rev52.png and images/rev53.png .
    2. The expression images/rev54.png appears in the quadratic formula (with A = 1). Express images/rev55.png in terms of images/rev56.png and images/rev57.png and explain why its being zero is equivalent to having both roots equal. The quantity images/rev58.png or more generally images/rev59.png is called the discriminant of the quadratic polynomial p(x).
  5. Let images/rev60.png and images/rev61.png be two solutions of the equation images/rev62.png where M and N are integers.
    1. Let images/rev63.png for images/rev64.png . Explain why this sequence satisfies the linear recurrence relation:

      images/rev65.png

    2. Show that images/rev66.png and images/rev67.png .
    3. Show that if one has two sequences, say images/rev68.png and images/rev69.png both of which satisfy the above linear recurrence relation, then so does the sequence images/rev70.png where c and d are constants.
    4. Show that images/rev71.png satisfies the linear recurrence relation:

      images/rev72.png

      and that images/rev73.png and images/rev74.png . In particular, it follows from the linear recurrence relation that images/rev75.png are all integers. In special case in which M = N = 1, the sequence images/rev76.png is called the sequence of Fibonacci numbers. The sequence looks like images/rev77.png and one has

      images/rev78.png

    5. Now suppose that we know that images/rev79.png . Explain why images/rev80.png for large n is approximately equal to images/rev81.png .
    6. Suppose we wanted to find some rational numbers which were good approximations of, say images/rev82.png . Then images/rev83.png is less than 1 in absolute value. Also, this number and images/rev84.png are roots of images/rev85.png . Use the last part of this exercise to find rational number approximations of images/rev86.png . Use these to get rational number approximations of images/rev87.png .

1.1.4 Solving Systems of Equations

  1. 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.
    1. Given values P and A for the perimeter and area of a rectangle, find formulas for the dimensions of the rectangle.
    2. 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.
    3. 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.
  2. A cylindrical can with top and bottom removed has area images/rev88.png and volume images/rev89.png where r is the radius of the can and h is its height.
    1. Given the area A and volume V, find formulas for the radius and height of the can.
    2. 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?
  3. Solve the general system of two equations in two unknowns: images/rev90.png and images/rev91.png where a, b, c, d, e, and f are real numbers where images/rev92.png . What happens if ad - bc = 0?
  4. The system of simultaneous equations

    images/rev93.png

    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 images/rev94.png .

  5. Parametrized Circle: The unit circle is the set of solutions of the equation images/rev95.png . 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

    images/rev96.png

    where f(t) and g(t) are functions. Such a description is called a parametrization. One can do this with trigonometric functions:

    images/rev97.png

    which describes all the solutions as t varies in the interval images/rev98.png . 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 re-arrange the equation in the form images/rev99.png . Assuming that x is not zero and y is not -1, we can re-arrange this as:

    images/rev100.png

    Let t be this common value, i.e. images/rev101.png .

    1. 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.
    2. We started off assuming that images/rev102.png and images/rev103.png . 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.
  6. 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:

    images/rev104.png

    where t is allowed to be any real number other than 0.

    1. Using the last problem as a model, try and work out a parametrization of the set of solutions of the equation of the hyperbola: images/rev105.png .
    2. Modify your solution to obtain a parametrization of the set of solutions of the hyperbola: images/rev106.png .
  7. 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

    images/rev107.png

    This is a problem known to the Babylonians.

    1. Using the last two problems as models, try to parametrize the points on the circle of radius 2: images/rev108.png using a parametrization in which the coefficients of the rational functions are integers.
    2. Show that if (x, y) lies on the unit circle, then the point (x + y, x - y) lies on the circle of radius 2.
    3. Show that if (x, y) lies on the circle of radius 2, then the point images/rev109.png lies on the unit circle.
    4. Can you use the last two parts to do the first part of this problem in another way?

1.1.5 Inequalities

  1. Describe the following sets:
    1. The set of all real numbers with images/rev110.png .
    2. The set of all real numbers with images/rev111.png .
    3. The set of all real numbers with images/rev112.png .
    4. The set of all real numbers with images/rev113.png .
    5. The set of all real numbers with images/rev114.png .
    6. The set of all real numbers with images/rev115.png .
    7. The set of all real numbers with images/rev116.png .
    8. The set of all real numbers with images/rev117.png .
    9. The set of all real numbers with images/rev118.png .
  2. Describe the following sets:
    1. The set of all real numbers with images/rev119.png .
    2. The set of all real numbers with images/rev120.png .
    3. The set of all real numbers with images/rev121.png .
    4. The set of all real numbers with images/rev122.png .
    5. The set of all real numbers with images/rev123.png .
    6. The set of all real numbers with images/rev124.png .
    7. The set of all real numbers with images/rev125.png .
    8. The set of all real numbers with images/rev126.png .
    9. The set of all real numbers with images/rev127.png .
  3. The triangle inequality says: images/rev128.png . Determine when equality holds.
  4. Prove that images/rev129.png for all real numbers a and b.
  5. Show that images/rev130.png if images/rev131.png and images/rev132.png . Is the result true if one drops the 1 out of the inequality?
  6. Consider the set T of all subsets of the set of real numbers. As mentioned earlier, there are two binary operators images/rev133.png and images/rev134.png . One also has a binary relation: images/rev135.png . Determine the truth or falsity of each of the following statements. If true, indicate why; otherwise give a counter-example.
    1. images/rev136.png satisfies trichotomy.
    2. images/rev137.png is transitive.
    3. images/rev138.png is reflexive, i.e. images/rev139.png 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 images/rev140.png ? What about images/rev141.png ?

1.2 Geometry

1.2.1 The Pythagorean Theorem

  1. Show that the sum of the interior angles of a quadrilateral is 360 degrees (or images/rev142.png radians).
  2. A regular n-gon is a polygon with n equal length sides. Prove that the sum of the interior angles of a regular n-gon is 180(n - 2) degrees (or images/rev143.png radians).
  3. Let's try an alternate proof of the Pythagorean Theorem.
    1. Given a right triangle images/rev144.png 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 images/rev145.png .
    2. 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.
    3. Prove that the five pieces together make a square with a side of length c.
    4. Use the diagram to prove the Pythagorean Theorem.
  4. Let images/rev146.png be a triangle with obtuse angle C (i.e. angle C is greater than images/rev147.png ). 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

    images/rev148.png

    This is essentially the Law of Cosines, as it appears in Euclid's Elements Book II, Proposition 12.

  5. What would be the result corresponding to the last exercise in case the angle C is acute (i.e. less than images/rev149.png radians), rather than obtuse? State the result and modify your proof appropriately. The result would be Proposition 13 of Euclid's Elements Book II.
  6. The areas of polygons are typically calculated by breaking the figures up into rectangles and triangles. Use this method to handle the following:
    1. 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.
    2. 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.
    3. 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?
  7. Let L be a line and images/rev150.png and images/rev151.png 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 images/rev152.png to P and the distance from P to images/rev153.png is smallest possible? (What is the best way to get a drink of water from a straight river while traveling from images/rev154.png to images/rev155.png ?)

    This looks hard until you get the idea of reflecting the point images/rev156.png across the line. To be precise, this means draw a line through images/rev157.png 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 images/rev158.png and Q on opposite sides of the line. Now, the distance from any point R on the line L to images/rev159.png 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 images/rev160.png to P makes the same angle with L as the the line from images/rev161.png 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.

  8. Let L be a line and images/rev162.png and images/rev163.png 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 images/rev164.png to P and the distance from images/rev165.png to P is largest possible.

    Again, choose a point Q which is the reflection of images/rev166.png 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 images/rev167.png to P makes the same angle with L as the the line from images/rev168.png to P. This is another way of determining P.

1.2.2 Analytic Geometry

  1. Parabolas: Fix a point F = (0, p) and a line y = -p.
    1. 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.
    2. 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 images/rev169.png .
    3. Find the focus and the directrix for the parabola images/rev170.png .
  2. Ellipses: Fix two points images/rev171.png and images/rev172.png where c > 0 and let a > c. Set images/rev173.png . The ellipse with foci images/rev174.png and images/rev175.png and semi-major 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

    images/rev176.png

  3. Hyperbolas: Fix two points images/rev177.png and images/rev178.png where c > 0 and let a satisfy 0 < a < c. Set images/rev179.png . The hyperbola with foci images/rev180.png and images/rev181.png and semi-traverse 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

    images/rev182.png

  4. 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 images/rev183.png and images/rev184.png ? 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?

  5. Repeat the last exercise for hyperbolas instead of ellipses.

1.2.3 Lines, Circles, and Parabolas

  1. 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.
    1. For each equation below, determine whether or not the corresponding line is in P.
      1. y = 3x - 2
      2. y = 3x - 5
      3. 6x - 2y = 10
      4. y = 2
    2. How many lines in P have slope 1/2? Find the equations of all these lines and sketch their graphs.
    3. How many lines in P have slope -2? Find the equations of all these lines and sketch their graphs.
    4. Find all the vertical lines in P and write down their equations.
    5. Find all the horizontal lines in P and write down their equations.
    6. Find the equations of all the lines in P which intersect the x-axis at a 45 degree angle. Did you find two of these?
    7. Find the equations of all the lines in P which have y-intercept 5.
    8. Find the equations of all the lines in P which have x-intercept 5.
    9. How many lines in P have y-intercept 2001?
  2. Let P be the set of all lines which pass through the point (1, 2). For an arbitrary non-vertical line L in P, let m be its slope and b be its y-intercept
    1. Find an expression for m in terms of b.
    2. Find an expression for b in terms of m.
    3. 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?
    4. Try to find an expression for x-intercept 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?
    5. We restricted our first two expressions to the case where the line L was non-vertical. Why was this done?
  3. 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 non-vertical line L in P, let m and b denote the slope and y-intercept of L.
    1. Find an expression for m in terms of b. Your expression will involve the fixed quantities c and d.
    2. Find an expression for b in terms of m.
    3. 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.
  4. 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 x-coordinates and the average of the y-coordinates, i.e. the midpoint should be images/rev185.png .
    1. 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.
    2. Show that the distances from M to each of the two original points are equal. This verifies that M is the midpoint.
  5. Repeat the previous exercise for two arbitrary points (a, b) and (c, d).
  6. 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?
  7. Just as points are determineed by their x-coordinate and y-coordinate pair, non-vertical lines are determined by their slope and y-intercept 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 y-intercepts. (Note: midline is NOT a standard mathematical term.)
    1. Let P be the set of all lines through the point (1, 2). Show that if images/rev186.png and images/rev187.png are two lines in P, then so is the midline between the two lines.
    2. A bisector an angle images/rev188.png is a line BD where images/rev189.png is equal to images/rev190.png . How many bisectors are there for a given angle? Is the midline of AB and BC a bisector of images/rev191.png ? (You should try to give a convincing argument for your answer.)
  8. 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.
  9. For two distinct points A and B, show that the set of points equidistant to A and B is a line perpendicular to AB.
  10. 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?

    1. Repeat the same for the points A = (-2, -1) and (-2, 4).
    2. Repeat the same for the points A = (-2, -1) and (3, 4).
    3. Repeat the same for the points A = (a, b) and B = (c, d).
    4. Check your last result for exceptional cases and fix your solution.
  11. Let A and B be two distinct points and n be a positive integer. Find n - 1 points images/rev192.png all on the line AB such that the distance from images/rev193.png to images/rev194.png is images/rev195.png for images/rev196.png .
  12. Let A and B be two distinct points and let r be a real number with images/rev197.png ,
    1. Find all points P on the line AB such that |AP| = r|AB|.
    2. Find all points on the line AB such that |AP| = r|AB| and |PB| = (1-r)|AB|.
  13. Let D, E, and F be the midpoints of the sides of the triangle images/rev198.png where D, E, and F are on the sides opposite the angles A, B, and C respectively.
    1. Use analytic geometry to find a point P on all three of the lines AD, BE, and CF.
    2. Is it true that images/rev199.png ? If so, what is the common numerical value of these fractions?
  14. Three points images/rev200.png for i = 1, 2, and 3 are said to be collinear if they all lie on the same line.
    1. Show that if images/rev201.png , images/rev202.png , and images/rev203.png are distinct, then the three points are collinear if and only if

      images/rev204.png

    2. Find a necessary and sufficient condition for the three points to be collinear in case one has images/rev205.png , but images/rev206.png is distinct from both images/rev207.png and images/rev208.png .
    3. Find a necessary and sufficient condition for the three points to be collinear in case one all three images/rev209.png are equal.
  15. 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 images/rev210.png .

    Let A, B, and C be distinct points.

    1. Find the equation of the line L of all points (x, y) equidistant from A and B.
    2. Find the equation of the line M of all points (x, y) equidistant from B and C.
    3. 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 images/rev211.png .
    4. Still assuming that the lines L and M intersect, find a formula for the radius of the circle which circumscribes images/rev212.png . Find the equation of the circle.
    5. Show that the lines L and M intersect if and only if A, B, and C are not collinear.
  16. Let A, B, and C be three non-collinear points. Let L, M, and N be the lines perpendicular to and through the midpoint of one of the sides of the triangle images/rev213.png . Show that there is a common point of intersection of all three lines.
  17. 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.
    1. Let two distinct points be images/rev214.png for i = 1, 2. If images/rev215.png , then the line is vertical and its equation is images/rev216.png .
    2. If images/rev217.png , then the line has a slope m and y-intercept 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.
  18. Now suppose one has three distinct points images/rev218.png 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:

    images/rev219.png

    which can be written as images/rev220.png where A = -2a, B = - 2b, and images/rev221.png 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 images/rev222.png , images/rev223.png , and images/rev224.png are all different, show that this system of equations has a unique solution precisely when the three points are non-collinear.

  19. 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 images/rev225.png is the equation of a circle with center (a, b) and radius r. Let images/rev226.png be a point on this circle. Consider the line with equation

    images/rev227.png

    1. Show that the line intersects the circle at the point images/rev228.png .
    2. 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.
    3. Show that the line is perpendicular to the radius line from the center O = (a, b) to the point images/rev229.png .
  20. Finding the points of intersection of a circle and a parabola can be non-trivial. In fact, Descartes showed that this is equivalent to solving an arbitrary quartic (fourth degree) equation. This is how it is done:
    1. Let images/rev230.png be an arbitrary quartic equation. Show that if you substitute x = y - a/4, then one gets an equation of the form images/rev231.png . So, in order to solve an arbitrary quartic, it is enough to solve a quartic with no cubic term.
    2. Consider the intersection of the parabola images/rev232.png with the circle images/rev233.png . Show that if you could find the points of intersection of an arbitrary circle and the parabola images/rev234.png , then you would be able to solve an arbitrary fourth degree equation.
  21. Let images/rev235.png 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.
  22. Let images/rev236.png 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.
  23. Let images/rev237.png be the parabola with focus (0, p) and directrix y = -p. If images/rev238.png is any point on the parabola. Show that the line through P with y-intercept images/rev239.png intersects the parabola in exactly one point. This line is called the tangent line to the parabola at the point images/rev240.png . Find the slope of this tangent line.
  24. Given an angle images/rev241.png and a point D, we say that AD bisects the angle images/rev242.png if the ray AD lies inside the angle and images/rev243.png .

    Consider a triangle images/rev244.png . Let D be on AC such that BD bisects angle images/rev245.png , and let E be on AB such that CE bisets angle images/rev246.png . 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.

    1. Explain why images/rev247.png is congruent to images/rev248.png . Then repeat your argument to show that images/rev249.png is congruent to images/rev250.png .
    2. Why does it follow that MF, MH, and MG are all of the same length?
    3. Now explain why images/rev251.png is congruent to images/rev252.png .
    4. Why does it follow that AM bisects the angle images/rev253.png ?
    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.
  25. Distance from point to line: Let L be a line and P be a point not on the line.
    1. If L is a vertical line, then its equation is of the form x = a. If P has coordinates images/rev254.png . 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.
    2. Modify the last part to handle horizontal lines.
    3. Now suppose that the line L has equation y = mx + b where images/rev255.png . Show that the shortest distance from images/rev256.png to the line L is

      images/rev257.png

  26. Elliptical Refection: Let E be an ellipse with foci images/rev258.png and images/rev259.png . 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 images/rev260.png 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 images/rev261.png . Why is the distance from R to images/rev262.png smaller than the distance the sum of the distances from R to Q and from Q to images/rev263.png ? 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 images/rev264.png 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 images/rev265.png is tangent to the ellipse images/rev266.png at the point images/rev267.png . (You might start by trying to show the result in case a = b = 1. Then see if you can reduce to this case.)

  27. Hyperbolic Reflection: Repeat the last exercise with a hyperbola.
  28. Parabolic Reflection:Let images/rev268.png be a point on the parabola images/rev269.png . 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 images/rev270.png intersects the parabola in precisely one point. We say that this line is tangent to the parabola at P.

    The vertical line images/rev271.png 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.

1.2.4 Trigonometry

  1. 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 x-axis 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 images/rev272.png you arrive at a point on the unit circle which is the reflection across the x-axis of the point you would have arrived at by sweeping out the angle images/rev273.png . Use this property to show that the cosine function is even and the sine function is odd, i.e.

    images/rev274.png

    for all angles images/rev275.png .

  2. Complementary Angles: Starting from the positive y-axis and sweeping out an angle images/rev276.png in a clockwise direction brings you to the point on the circle corresponding to the angle images/rev277.png as measured from the x-axis in a counter-clockwise direction. Notice that the two points on the circle, one corresponding to the angle images/rev278.png and the other corresponding to the angle images/rev279.png 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 x-coordinate with the y-coordinate.

    The point on the unit circle corresponding to the angle images/rev280.png is images/rev281.png . What are the coordinates of the point corresponding to the angle images/rev282.png ? Show that this implies that

    images/rev283.png

    These identities are useful to convert properties of the sine function into properties of the cosine function and vice versa.

  3. Special Triangles: The values of the trigonometric functions for some very special angles angles can be calculated from simple properties of some special triangles.
    1. 45-45-90 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 images/rev284.png radians.
    2. 30-60-90 Degree Triangle Draw such a triangle and label the angles A, B, and C where the angles are images/rev285.png , images/rev286.png , and images/rev287.png 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 images/rev288.png is equilateral. Use this information to read off the the values of the trig functions for the angles images/rev289.png and images/rev290.png .
    3. The 30-60-90 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 images/rev291.png 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 30-60-90 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 images/rev292.png 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 45-45-90 degree triangle. On the other hand, if you make the length of DC exactly equal to a, then you have the 30-60-90 degree triangle.

      36-72-72 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 images/rev293.png would be isosceles.

      1. Explain why the triangle is isosceles.
      2. Explain why triangle images/rev294.png is similar to images/rev295.png .
      3. 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 images/rev296.png .
      4. Find the value of a, being careful to explain which root of the equation is the value of a.
      5. 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.
    4. 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 n-gon 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 n-gon.
      1. Show that a square inscribed in the circle has sides of length images/rev297.png .
      2. Show that the length of the sides of a regular hexagon inscribed in the circle is exactly 1.
      3. Show that the length of the sides of the a regular decagon (a 10-gon) is precisely the length a you calculated for the 36-72-72 degree triangle.
      4. Use the Pythagorean Theorem to show how to construct segments of length images/rev298.png and images/rev299.png . Use this to show that you can construct segments of length the sides of the regular quadrilateral, hexagon and decagon.
      5. Explain why this tells you that you can construct regular pentagons and equilateral triangles. Do you see how to construct regular n-gons where n has the form images/rev300.png , images/rev301.png , and images/rev302.png for all images/rev303.png ? (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.)
    5. Angle Trisection: Start with our triangle images/rev304.png 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 images/rev305.png . So, the diagram allows one to trisect the angle images/rev306.png .

      To trisect any angle, we need only start with the angle and label it images/rev307.png 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.

  4. Product Identities: Prove the identities

    images/rev308.png

    These were used, before the advent of logarithms, as a means for calculating products of numbers. For example, to use the first one:

    1. 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.
    2. Using tables, find x and y such that images/rev309.png and images/rev310.png .
    3. Calculate u = x + y, and v = x - y.
    4. Using tables, find images/rev311.png and images/rev312.png .
    5. 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.
  5. Addition Theorems: Let images/rev313.png and images/rev314.png be any angles.
    1. The addition theorem for cosines is

      images/rev315.png

      One also has a subtraction result:

      images/rev316.png

      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.)

    2. Let the points images/rev317.png , images/rev318.png , images/rev319.png , images/rev320.png be the points on the unit circle corresponding to the angles 0, images/rev321.png , images/rev322.png , and images/rev323.png (measured counter-clockwise from the x-axis). The chord images/rev324.png is the same length as the chord images/rev325.png . 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.
    3. Use the distance formula to express images/rev326.png .
    4. Simplify the expressions using the Pythagorean Theorem to show that the second form of the law of cosines is true.
    5. The addition theorem for the sine function is:

      images/rev327.png

      As with the law of cosines, one has a subtraction form that looks like:

      images/rev328.png

      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.

    6. 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

      images/rev329.png

      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.

  6. Law of Cosines Let images/rev330.png be an arbitrary triangle with sides a, b, and c opposite angles A, B, and C respectively. The Law of Cosines says

    images/rev331.png

    In case the angle C is a right angle ( images/rev332.png radians), the result is just the Pythagorean Theorem.

    1. 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 images/rev333.png and images/rev334.png . 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.
    2. 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 images/rev335.png in terms of other squares and work with the expression until it becomes the Law of Cosines.
  7. Law of Sines Let images/rev336.png be an arbitrary triangle with sides a, b, and c opposite angles A, B, and C respectively. The Law of Sines says

    images/rev337.png

    1. To prove the last equality, start with a triangle images/rev338.png and drop a line perpendicular from angle A to a point D on the side BC. Again we have two right triangles images/rev339.png and images/rev340.png . Label all the sides of the triangles and express the images/rev341.png and images/rev342.png as quotients of various of the sides. Use these to complete the proof.
    2. Once you have proved the second equality, what needs to be done to prove the first one?
  8. Heron's Theorem: Consider the triangle images/rev343.png with sides of length a, b, and c opposite angles A, B, C respectively. Let s = (a + b + c)/2 be the semi-perimeter of the triangle. Then the area of the triangle is images/rev344.png .
    1. 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 images/rev345.png .
    2. Consider the system of equations: images/rev346.png and images/rev347.png . Solve this system to get the half angle formulas: images/rev348.png and images/rev349.png .
    3. Substitute the expression for images/rev350.png obtained by the Law of Cosines in the formulas of the last part and show that they can be simplified to: images/rev351.png and images/rev352.png .
    4. Use the double angle formula: images/rev353.png together with the results of the last part to conclude that images/rev354.png . This completes the proof of Heron's Theorem.
  9. 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 images/rev355.png 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

    images/rev356.png

    From the triangle images/rev357.png , one sees by the Pythagorean Theorem that images/rev358.png . Furthermore, the area of the triangle is clearly images/rev359.png . Combining all these results show that

    images/rev360.png

    Show that this expression is equivalent to the one in by Heron's Formula.

  10. 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 images/rev361.png . Here is one approach:

    1. Start by circumscribing the square with the square whose sides are at images/rev362.png and images/rev363.png . The perimeter of this square is the first approximation.
    2. 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 images/rev364.png 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.
    3. For the third step, add in vertices at images/rev365.png , images/rev366.png , images/rev367.png , etc. Now one has doubled the number of stair steps and we have a polygonal line which is closer to the circle.
    4. Repeat the process with images/rev368.png , images/rev369.png , images/rev370.png , etc. for the fourth step. Again the number of stair steps doubles.
    5. 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?

  11. Circumference of a circle: Again take the unit circle. Inscribe a polygon, for example a regular n-gon, 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.
    1. Explain why the perimeter of the circumscribed polygon is larger than the perimeter of the inscribed polygon.
    2. 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.)
  12. Calculation of images/rev371.png The circumference C of a circle of radius r can be calculated by images/rev372.png . The real number images/rev373.png is an irrational number whose value we propose to compute. One simple approach is to use the relation between images/rev374.png and the circumference of a circle of radius 1. The idea is to inscribe regular n-sided 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.
    1. The perimeter images/rev375.png of a regular n-sided polygon inscribed in the unit circle is simply n times the length images/rev376.png of one of its sides. So, to calculate images/rev377.png , it is enough to figure out how to calculate images/rev378.png . Finding a formula for images/rev379.png is not so easy to do. On the other hand, notice that it is enough to get a formula for images/rev380.png for some arbitrarily large values n. So, the first step is to note that it is enough to calculate images/rev381.png for n of the form images/rev382.png where images/rev383.png . To get the process started, calculate images/rev384.png . For this inscribe a square inside a circle and use the Pythagorean Theorem to show that images/rev385.png .
    2. The next step is to calculate images/rev386.png assuming that we can calculate images/rev387.png . To do so, start with a regular n-sided 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 images/rev388.png 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 images/rev389.png and images/rev390.png are congruent; so images/rev391.png . Now, explain why images/rev392.png .

      Now, let's try and compute images/rev393.png . Using the same formula gives

      images/rev394.png

      So, if we are to relate images/rev395.png and images/rev396.png , 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

      images/rev397.png

    3. The last equation says that images/rev398.png satisfies a fourth degree equation. Solve this fourth degree equation to obtain

      images/rev399.png

      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 images/rev400.png , there were two roots -- but only one is the correct one, and the reason uses an estimate on the size of images/rev401.png .

    4. You have already shown that images/rev402.png . Substitute this into the equation of the last part to show that

      images/rev403.png

      Now use this expression to calculate

      images/rev404.png

      Do one more step to show that

      images/rev405.png

      The form of all these results is

      images/rev406.png

      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 images/rev407.png sides:

      images/rev408.png

    5. For particular values of images/rev409.png , one can calculate images/rev410.png , assuming that one knows how to calculate square roots. These values then should approach images/rev411.png . Use the formula to calculate the images/rev412.png 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 images/rev413.png sides to the polygon and an approximate value for images/rev414.png of 3.1415877 which rounds to 3.14159.
    6. From the figure of a polygon inscribed in the circle, it is intuitively obvious that the estimates for images/rev415.png will all be underestimates. This seems to be validated in the numerical estimates which are clearly increasing and less than images/rev416.png . Use properties of the sine function to show that the approximations to images/rev417.png are in fact increasing as k increases.
    7. One could equally well have circumscribed the circle with regular n-sided polygons. Let images/rev418.png be length of the side of the n-sided regular polygon which circumscribes the unit circle. Show that images/rev419.png . But then images/rev420.png where images/rev421.png .
    8. Show that images/rev422.png . This says that we can find images/rev423.png by taking the square root of the average of images/rev424.png and 1.
    9. Use the circumscribed polygons with images/rev425.png sides in order to estimate images/rev426.png . 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 images/rev427.png . (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 images/rev428.png are indeed decreasing. By using BOTH estimates, we not only get an estimate for images/rev429.png , but also an estimate for how large the error can be. The estimate for images/rev430.png should be the average of the upper and the lower estimate, and the largest error is simply half the difference between the two estimates.

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Revised: June 28, 2001
All contents © copyright 2001 K. K. Kubota. All rights reserved