Solutions to Selected Odd-numbered Exercises

 

Section 1.1

 

1.      In the Egyptian approach, .  In the Babylonian approach, , which is closer to the actual value.  So, the Babylonians had a more accurate method for computing the area of a circle.

 

 

 

3.     

Using similar triangles, compute the volume of the small pyramid and the large pyramid as

Large pyramid =                    Small pyramid =

 

Large pyramid volume - small pyramid volume = 64 - 8 = 56.  The procedure computes the volume of the truncated pyramid.

 

5.      a)  Answers may vary, such as 6, 8, 10; 5, 12, 13; 10, 24, 26

 

b) , which leads to the triple 3, 4, 5.

 

c)   Since n is always an integer, k must always be irrational.  Consequently, no integer values of k are possible.

 

d)  Pythagorean triples take the form , where m and n are integers such that gcd(m,n) = 1.  If then we are done.  If 3 does not divide m and 3 does not divide n, then   m = 1(mod3) or m = 2(mod3) and n = 1(mod3) or n = 2(mod3).  If m = 1(mod3), then m2 = 1(mod3).  Likewise, if m = 2(mod3), then m2 = 1(mod3).  [Why?]   Similar argument applies to n.  Then m2 - n2 = 1 - 1 = 0(mod3).  So, m2 - n2  is divisible by 3.

 

 

7.      Assume that is rational.  Then , where p and q are integers.  Furthermore, we may assume that p and q have no common factors.  Squaring both sides yields   This forces p2 to be even, which forces p to be even.  If p is even then p2 is divisible by 4. This forces 2q2 to be divisible by 4.  Therefore, q2 must be even, forcing q to be even.  Since p and q are both even, then have a common factor, contradicting the assumption.  So cannot be rational.  It is irrational.

 

 

9.       is the total area within the figure.  The four rectangles have a total area of 4ab.  The square in the interior of the figure must then have an area of -4ab = (a-b)2.   Consequently, (a+b)2 > (a-b)2. 

 

 

11.  1+2+3+ ... = n(n+1)/2

  

 

 

 

 

 

 

 

 

 

 

 

 

 

Section 1.2

 

1.      Answers will vary.

 

3.      Answers will vary.

 

5.      a)  Model  

Points   A, B, C

Lines   A    B    A

            B    C    C

 

A1:  Note that there are exactly 3 points.

A2:  Note that every pair of points is represented as a column (e.g., as a line).

A3:  Note that

- lines AB and BC have B in common.

- lines BC and AC have C in common.

- lines AB and AC have A in common.

            A4:  Note that no line contains more than 2 points.

            So, the system is consistent

 

 

        b)  The following models may be used to demonstrate the independence of the axioms:

 

Model 1

A  D  D  D

B  A  B  C

C

Model 2

A  A

B  C

Model 3

A

B

C

Model 4

A  B  A  C

B  C  C

 

 

Section 1.3

1.      Answers will vary.

 

3.       a)

Assume there are two distinct midpoints, M and M'.   Since M and M' are midpoints of segment AB, d(AM) = d(MB)= .5d(AB).  Furthermore, d(AM) + d(MB) = d(AB) and d(AM) + d(MM') + d(M'B) = d(AB).  Since d(M'B) = .5 d(AB) = d(AM), d(AM) + d(MM') + d(M'B) = d(AB) + d(MM') = d(AB), so d(MM') = 0.  Consequently, M and M' are not distinct and there is only one midpoint.

 

 

b)  Use the same approach taken in part a), this time in terms of angle sums.

 

c)  Let a = f.  Since 180° = a + b = f + d, b = d.   A similar argument applies to angles complementary to congruent angles.

 

 

 

 

d)    Let line m intersect side BC at point P.  Then B and C are on opposite sides of line m.  Assume that m does not intersect sides AB or AC.  Then A and B must be on the same side of line m.  Likewise, A and C must be on the same side of line m.  Then, by transitivity, B and C must be on the same side of line m.  Consequently, line m must not intersect side BC, which is a contradiction.

 

 

e)   Extend AD from A to form line FD.  Extend AC to a point G and draw segment GB.  Consider DGBC.  By Pasch, ray AD must intersect either BG or BC.  Because point D is in the interior of ÐBAC, all points of the ray AD are in the interior of the angle.  Since no points of BG are in the interior of the angle, ray AD must intersect BC.

 

f)   Let Q be the midpoint of BC and R the midpoint of BA.  Draw segment AQ.  DABQ @ DACQ by side-side-side.  Then ÐABC @ ÐACB.  Draw segment CR.  DCRB @ DCRA by side-side-side.  Then ÐABC @ ÐBAC. 

 

 

Section 1.4

1.      a)  Let m ^ n at point A.  Construct a circle with center A and radius AB.  Find the midpoint of AB, point E.  Construct a circle with center A and radius AE.  Let point D be the intersection of that circle and line n.  Construct a line parallel to AB through D.  Label the intersection of that line and the larger circle point C.  Ray AC forms a 30° angle with line m, since the sine of ÐBAC = 0.5. To complete the trisection, bisect ÐCAD.

 

 

b)   Construct a perpendicular to line AB at C.  Bisect ÐDCB.  This completes the trisection. 

 

c)    Given  ÐCAB = 45°, construct an equilateral triangle DDEF and bisect ÐDEF, forming ÐFEG = 30°.  Copy ÐFEG at point A so that EF overlaps AB.  Bisect ÐGAB to complete the trisection.

  

 

 

3.      Starting with the demonstration of this problem seen on the first screen of the Modern Geometry CD-ROM, complete the task by finding the geometric mean of the two sides of the triangle (See Figure 1.4.14) created in that demonstration.

 

 

5.      DABC ~ DAED, so 1/b - a/x.  Then x = ab.

 

7.      DCBD ~ DDBA, so 1/x = x/a.  Then x2 = a, so x = .

 

9.      x3 - 2 = 0.  By the rational root theorem, there are no rational roots to this equation, nor would any composition of square roots yield the necessary result.

 

 

Section 2.1

1.      Answers will vary

a.       STEP   ACTION

1          Using a compass, mark points D and E on the sides of angle BAC such that DA=EA.

2          Draw a segment from D to E.

3          Using the length of segment DE as a radius, construct arcs centered at D and E and mark

            their point of intersection as point F.

4          Draw segments FD and FE.

5          Dray ray AF

 

 

b.      STEP   ACTION

1          Using the length of segment AB as a radius, construct arcs centered at A and B and mark

            their point of intersection as point C.

2          Using the procedure described in Proposition 9, bisect angle ACB.

3          The bisector of angle ACB also bisects segment AB.

 

 

c.       STEP   ACTION

1          Using point C as the center, construct arcs of equal radius on either side of C.  Mark the

points of intersection of these arcs with line AB as points D and E.

2          Using the procedure described in proposition 1, construct an equilateral triangle with side

            DE as the base.  Label the third vertex of this triangle as point F.

3          Draw segment FC.  This segment is perpendicular to AB at point C.

 

 

3.             STEP ACTION

1          Using the procedure described in Proposition 10, find the midpoint of segment AB.

            Label the midpoint F. 

2          Repeat the same procedure on segments BC and AC.

3          Draw segments AD, BE, CF

 

5.            STEP   ACTION

1          Using the procedure described in Proposition 12, draw a perpendicular line from point C

to the line containing segment AB

            2          Repeat this procedure to draw perpendicular lines from the other two vertices to the lines

                        containing the other two sides of the triangle.

 

 

Section 2.2

1.      a)  No

b)  Yes, the midpoint of the hypotenuse of a right triangle.

c)  The "center" of the triangle coincident with the incenter and centroid.

 

3.      Answers may vary, for instance:

 

5.      Equilateral triangles.  Since the incenter is always in the interior of a triangle and the circumcenter

Is in the interior only for acute triangles, only acute triangles need be considered.  Let P be the incenter of triangle ABC.  Then Ð1 = Ð2; Ð3 = Ð4; Ð5 = Ð6.   If P is also the circumcenter, then AP ^ BC and DADB @ DACD by angle-side-angle.  Consequently, AB = AC.  Using similar reasoning, AC = BC.  Then AC = AB = BC and the triangle is equilateral.

 

7.      The centroid is always between the orthocenter and circumcenter.  The distance from the orthocenter to the centroid is twice the distance from the circumcenter to the centroid.

 

 

Section 2.3

1.      Answers will vary.

 

3.      Answers will vary.

 

5.      a = r + x ® x = a - r;     b = r + y ® y = b - r;  c = x + y

a2 + b2 = c2;  (r + x)2 + (r + y)2 = (x + y)2 ® (r2 + 2rx + x2) + (r2 + 2ry + y2) = x2 + 2xy + y2 ®

2 r2 + 2rx + 2ry = 2xy ® r2 + rx + ry = xy ® r2 + r(x + y) = xy.  Then r2 + rc = (a-r)(b-r) ®

r2 + rc = ab -ar -br + r2, so rc = ab - ar - br ® rc + ra + rb = ab ®

 

7.            a = 180° - 2(Ð1 + Ð2) ® Ð1 + Ð2 = (180° -  a)/2

b = 180° -  (Ð1 + Ð2) ® 180° -  (180° -  a)/2  = 90 - a/2

 

Section 3.1

1.      DCAF @ DDBF by side-angle-side.  So CF @ DF and Ð1 @  Ð2.  Then DCFE @ DDFE by side-side-side.  Then Ð1 @  Ð4  Ð5 @  Ð6.  Since Ð5 +  Ð6 = 180°, Ð5 =  Ð6 = 90°.  Then Ð1 +  Ð3 = Ð2 +  Ð4 ® ÐAFE and ÐBFE are right angles so EF ^ AB.

 

3.      Logical equivalence is established by proving each statement from the other. 

a)

Playfair:  Through a point not on a given line exactly one parallel may be drawn to the given line.

 

Given: Line k is parallel to line l through point P.

 

Proof:  Construct rays PA and PB through point A and B on l.  Label the angles formed as shown.  Ð1 + Ð2 + Ð3 = 180°.  Since k is parallel to l, we know that Ð1 = Ð4 and Ð3 = Ð5.  Then Ð4  + Ð5 + Ð2 = 180°. 

The sum of the angles of any triangle is equal to 2 right angles.

 

 

Given: Ð1 + Ð2 + Ð3 = 180°. 

 

Proof: Copy Ð4 at point P as shown. Label a point E on the side of Ð4 as shown, to the left of P.  Since Ð4 = Ð1, line l is parallel to AB.  Draw a line through EP, forming Ðx.  Then Ð4 + Ð3 + Ðx = 180°.  (We must now show that this is the same parallel obtained if Ð2 is copied at point P to form Ðx.)  Since Ð1 = Ð4 and Ð4 + Ð2 + Ð3 = 180°, Ð2 = Ðx.  Thus, the line through P drawn parallel to AB because Ð1 = Ð4 is the same line drawn through P parallel to AB because Ð2 = Ðx.  So there is only one line parallel to AB through P.

 

     b) - d) are demonstrated using the same general approach.

 

 

Section 3.2

1.       

 

3.      Answers will vary.

 

5.      C = 2psinh(R) = 2psinh(2) = 22.79;  C = 2psinh(R) = 2psinh(4) = 171.47

 

 

7.        If the center of the circle is O and the distance from O to A is r,

 

 

Section 3.3

1.         Defect = 180° - 153° = 27°

 

 

 

3.         cosh(c) = cosh(a)*cosh(b) = cosh(2)*cosh(3) =  37.88 ® c = 4.327

 

 

5.      Answers will vary.

 

7.      The legs of a Saccheri quadrilateral are perpendicular to the base.  The base is a common perpendicular between the sides.  Assume that a second common perpendicular existed.  It would complete a rectangle with the base and sides, a figure that is not possible in this space.  So the segment joining the midpoints of the sides cannot be perpendicular to the sides.

 

9.      cosh(1)*cosh(BC) = cosh(3) ® BC = 2.56

 

11.  Answers will vary, but resemble the proof of Case 1.

 

13.  Begin with any two non-intersecting lines l and m.  since there is only one common perpendicular between l and m, forming two right angles, no second common perpendicular  is possible to complete the rectangle.

 

15.  Since the shortest distance between two non-intersecting lines is along a common perpendicular, and since there is only one common perpendicular between such lines, all other separations are different.  Consequently, no all separations are the same.

 

17.  Since the sum of the interior angles is less than 360° and the sum of the three given angles is 270°, the fourth angle must be acute.

 

19.  Area DABD + Area DADC = Area DABC.  Since area = defect, Defect DABD + Defect DADC = Defect DABC.

 

 

 

21.    Area DABC > Area DDEF so Defect DABC > Defect DDEF

 

 

 

Section 3.4

1.      By Theorem 3.4.1, if ÐH = ÐI, ÐC = ÐD, and if ÐA = ÐB, the triangles are congruent.  Since all lines are orthogonal to the bounding circle, the angles formed at omega points A and B both have measure zero.  Consequently, the triangles are congruent.

 

 

3.      DADB @ DADC by Theorem 3.4.3.  Then ÐBDA @ ÐCDA, so AD ^ BC.

 

 

 

5.      Ð1 + Ð2 + Ð3 = Ð4 + Ð5 + Ð6 is true only in Euclidean space.

 

Section 4.1

1.          

a.      

 

b.     

 

c.      

 

d.     

 

e.      

 

3.          

a.      

 

b.     

 

 

c.      

 

d.     

 

5.        Answers will vary.

 

7.        To prove this theorem, simply restate Example 4.1.1 with suitable discussion.

 

9.        Let .  Eliminating x using linear combination, one obtains .  Eliminating y using the same approach yields .  Now expand to obtain .  Regrouping terms yields .  Dividing each term by  and expressing the result in matrix form yields

.  So, the matrix procedure yields the correct results as determined by the method of linear combination.

 

11.  Consider two lines .  From analytic geometry, any line perpendicular to the line          y = mx+b has slope –1/m.  Since the slope of line u is given by , the slope of any line perpendicular to u must have slope .  Therefore, the slope of line v, given by , must equal .  If =, then . 

 

 

Section 4.2

1.           

a.      

 

b.     

 

 

c.      

 

d.     

 

 

e.      

f.       

 

g.      

 

 

h.      

 

i.        

 

 

j.       

 

k.     

 

 

l.        

 

3.      ku’ = uT-1

 

5.     

 

7.      Answers will vary.

 

 

Section 4.3

1.          

 

3.       

 

5.        Geometric thinking: Under rotations, only the turn center is invariant.  So the only invariant point for this transformation is the origin.  Since rotations do not have invariant lines, there are none for this transformation.

Matrix methods:  Consequently, the origin is invariant under this transformation.  Since ,

Letting k=1, this leads to [0 0 u3], which isn’t a line, so there are no invariant lines.

 

 

7.       

 

 

9.        Given two points  and , the distance between them is given by . 

A rotation about the origin is given by the transformation matrix.  Under this transformation, points  and  become .  When the distance formula is applied to these points, one obtains .

 

 

11.     By exercise #9, distance is preserved under rotation, which forces angles to be preserved as well.  This may be verified using the same approach used in exercise #9 applied to two points and their images under translation.  .  Comparing distances yields .  So, distance is preserved under translation.

 

 

Section 4.4

1.                     

   

 

 

3.       

 

 

5.        Using matrix methods:  so all points  are invariant, i.e., all point on the y-axis.  To investigate invariant lines, the line equation ku’ = uR-1 is used.  In this case,

, so

If k = 1, u1 = -u1 so u1 = 0, with u2 and u3 unconstrained.  This leads to the line [0 u2 u3], the set of horizontal lines in the plane.  If k = -1, then u2 = 0 = u3 , leading to the line [u1 0 0], the y-axis.

 

Thinking geometrically, the reflecting line x = 0 will be pointwise invariant.  Lines perpendicular to x = 0 will map onto themselves, crossing the y-axis in the process.  Each of these lines are horizontal.

 

7.        Answers will vary.

 

 

9.       

 

 

 

11.     .  Let .  Then .  But  .  Since the absolute value function “removes” the negative sign, the areas will be the same.

 

 

 

13.        

 

 

Section 4.5

1.        Using t, u, and v as translation vectors, we may write t+u = v.

 

3.        With reflecting lines j and k parallel, we may show

 

 

 

5.          

a.        A reflection in the line y = x is given by the matrix .

 

b.       A reflection in the line y = x+1 is given by the matrix

 

 

c.        A reflection in the line y = x+1 is given by the matrix

 

d.       A reflection in the line y = x-1 is given by the matrix

 

 

e.        A reflection in the line y = x-2 is given by the matrix

 

 

 

7.            So the results are not the same.

 

 

9.        The sequence of transformations is 

a.        T1: Translate the reflecting line (and the plane in which it is embedded) down 4 units to that the y-intercept becomes the origin

b.       R1: Rotate the reflecting line through an angle q = arctan(5) = 1.3734 radians to make it coincide with the x-axis

c.        F: Reflect the plane in the x-axis

d.       R2: Rotate the reflecting line through an angle -q = -arctan(5) = -1.3734 radians

e.        T2: Translate the reflecting line up 4 units to its original position

 

Which is equivalent to T2*R2*F*R1*T1*X =

 

 

11.     Let l =  The steps in the procedure are:

a.        T1: Translate P to the origin

b.       R1: Rotate the line 90 ° clockwise

c.        T2: Translate P back to its original position

 

Taking into consideration the manner in which the line equation ku’ = uT-1 multiples lines by inverse transformations, this sequence may be written as

 l*T2*R1-1*T1 =

 

 

13.        

a.          Invariant lines

                                                   i.      [u1 0 u3]

                                                 ii.      [0 u2 0] and [u1 0 u3]

                                                iii.      [0 2u3 u3] and [u1 0 u3]

                                               iv.      [0 -2u3 u3] and [u1 0 u3]

b.       Eigen vectors

                                                   i.      [u1 0 u3]

                                                 ii.      [0 u2 0] and [u1 0 u3]

                                                iii.      [0 2 u3 u3] and [u1 0 u3]

                                               iv.      [0 -2u3 u3] and [u1 0 u3]

c.        The same, because eigenvectors always lie on invariant lines.

 

15.  Answers will vary.

 

 

Section 4.6

1.         

 

3.        

 

5.        

a.      

 

b.     

 

 

c.      

 

 

 

7.      Answers will vary.  For instance,

 

 

 

9.     

 

 

 

11. 

 

 

 

Section 5.1

1.        

a.        

 

b.       

 

 

3.         

n

r = # rows

b = #boxes

A = area of 1box

Cumulative Area

0

1

1

1

1 = 1

1

4 = 3 +1

16 = 1+3+5+7 = 42

1/9

16(1/9) = 1.7777

2

13 = 32 + 3 + 1

169 = 1+3+5+…+ 23+25 = 132

(1/9)2

169(1/9)2 = 2.0864

3

40 =33 + 32 + 3 + 1

1600 = 402

(1/9)3

1600(1/9)3 = 2.1948

 

5.        

n

#segments

segment length

curve length

0

2

0.577350269

1.154700538

1

4

0.333333333

1.333333333

2

8

0.19245009

1.539600718

3

16

0.111111111

1.777777778

4

32

0.06415003

2.052800957

5

64

0.037037037

2.37037037

 

 

Section 5.2

1.        

a.      

 

 

 

 

b.        

 

 

Section 5.3

1.      Consider two arbitrary points   Let the distance between these points be given by .  Since reversing the order of the terms in each difference has no effect on the result,   Since every pair of points has real, finite coordinates, the differences in the x-coordinates is real and finite, as it the difference in the y-coordinates.  Consequently, the computed distances are real and finite.  If a = b, then the difference terms for both x- and y-coordinates are zero, hence the distance is zero.  The final point to be proven is that   Assume  for some set of points a,b,c.  Then , which forces c to be closer to a than b.  But it must also be true that , which forces c to be closer to b than a.  Both of these statements cannot be true, so the assumption is false.  Consequently, the triangle inequality holds in

 

3.      Answers will vary, but the result will not be a strange attractor.  In particular, it will not be bounded.

 

 

5.      Change Map 3 to

 

 

7.        

a.       The fixed point is .    A sequence of iterates is given in the table below

 

 

Seed Point

Fixed Point

 

 

 

x-value

1

-2

 

 

 

y-value

1

2

 

 

 

 

 

 

 

 

 

Iterate

Old x-value

New x-value

Old y-value

New y-value

Distance from fixed point

1

1

-0.5

1

1.5

1.58113883

2

-0.5

-1.25

1.5

1.75

0.790569415

3

-1.25

-1.625

1.75

1.875

0.395284708

4

-1.625

-1.8125

1.875

1.9375

0.197642354

5

-1.8125

-1.90625

1.9375

1.96875

0.098821177

6

-1.90625

-1.953125

1.96875

1.984375

0.049410588

7

-1.953125

-1.9765625

1.984375

1.9921875

0.024705294

8

-1.9765625

-1.9882813

1.9921875

1.99609375

0.012352647

9

-1.98828125

-1.9941406

1.99609375

1.998046875

0.006176324

10

-1.99414063

-1.9970703

1.99804688

1.999023438

0.003088162

 

 

 

b.      The fixed point is .  A sequence of iterates is given in the table below:

 

 

Seed Point

Fixed Point

 

 

 

x-value

1

-0.608365

 

 

 

y-value

1

1.8718923

 

 

 

 

 

 

 

 

 

Iterate

Old x-value

New x-value

Old y-value

New y-value

Distance from fixed point

1

1

-0.06

1

2.05

0.576564416

2

-0.06

-0.545

2.05

2.0372

0.177036038

3

-0.545

-0.638672

2.0372

1.922834

0.059275383

4

-0.638672

-0.6276712

1.922834

1.87612892

0.019765623

5

-0.62767124

-0.6133278

1.87612892

1.868383977

0.006077614

6

-0.61332777

-0.6084454

1.86838398

1.869979089

0.0019149

7

-0.60844539

-0.6078836

1.86997909

1.87145296

0.00065171

8

-0.60788364

-0.6081545

1.87145296

1.871906414

0.000210975

9

-0.6081545

-0.6083266

1.87190641

1.871943877

6.43213E-05

10

-0.60832657

-0.6083707

1.87194388

1.871912542

2.10354E-05

 

 

Section 5.4

1.        

a.       f(x) = x(1-x) ® x = x(1-x) ® x = 0 ® (0, 0) is a fixed point.

b.      f(x) = 2x(1-x) ® x = 2x(1-x) ® both (0, 0) and (.5, .5) are fixed points.

c.       f(x) = 3x(1-x) ® x = 3x(1-x) ® both (2/3, 2/3) and (0, 0) are fixed points.

 

 

3.      Using the applet at URL 5.4.2, the following web diagrams may be plotted.

a.                      

 

 

b.         

 

c.          

 

d.        

 

Section 5.5

1.      .

 

3.        

a.         

C Real

C Imag

Iterate

Current R

Current I

New r

New I

0.5

-0.1

1

0

0

0.5

-0.1

 

 

2

0.5

-0.1

0.74

-0.2

 

 

3

0.74

-0.2

1.0076

-0.396

 

 

4

1.0076

-0.396

1.35844176

-0.8980192

 

 

5

1.35844176

-0.8980192

1.53892553

-2.5398136

 

 

6

1.53892553

-2.5398136

-3.5823612

-7.9171679

 

 

7

-3.5823612

-7.9171679

-49.348236

56.6243093

 

 

8

-49.348236

56.6243093

-770.56403

-5588.7195

 

 

9

-770.56403

-5588.7195

-30640017

8612932.37

 

 

10

-30640017

8612932.37

8.6463E+14

-5.278E+14

 

b.        

 

C Real

C Imag

Iterate

Current R

Current I

New r

New I

1

0

1

0

0

1

0

 

 

2

1

0

2

0

 

 

3

2

0

5

0

 

 

4

5

0

26

0

 

 

5

26

0

677

0

 

 

6

677

0

458330

0

 

 

7

458330

0

2.1007E+11

0

 

 

8

2.1007E+11

0

4.4128E+22

0

 

 

9

4.4128E+22

0

1.9473E+45

0

 

 

10

1.9473E+45

0

3.7919E+90

0

 

c.         

C Real

C Imag

Iterate

Current R

Current I

New r

New I

0

1

1

0

0

0

1

 

 

2

0

1

-1

1

 

 

3

-1

1

0

-1

 

 

4

0

-1

-1

1

 

 

5

-1

1

0

-1

 

 

6

0

-1

-1

1

 

 

7

-1

1

0

-1

 

 

8

0

-1

-1

1

 

 

9

-1

1

0

-1

 

 

10

0

-1

-1

1

 

 

 

d.        

C Real

C Imag

Iterate

Current R

Current I

New r

New I

1.414214

1.414214

1

0

0

1.41421356

1.41421356

 

 

2

1.41421356

1.41421356

1.41421356

5.41421356

 

 

3

1.41421356

5.41421356

-25.899495

16.7279221

 

 

4

-25.899495

16.7279221

392.374675

-865.07525

 

 

5

392.374675

-865.07525

-594395.89

-678865.83

 

 

6

-594395.89

-678865.83

-1.076E+11

8.0703E+11

 

 

7

-1.076E+11

8.0703E+11

-6.397E+23

-1.736E+23

 

 

8

-6.397E+23

-1.736E+23

3.7912E+47

2.2211E+47

 

 

9

3.7912E+47

2.2211E+47

9.4399E+94

1.6841E+95

 

 

10

9.4399E+94

1.6841E+95

-1.95E+190

3.18E+190

 

e.         

C Real

C Imag

Iterate

Current R

Current I

New r

New I

2

1

1

0

0

2

1

 

 

2

2

1

5

5

 

 

3

5

5

2

51

 

 

4

2

51

-2595

205

 

 

5

-2595

205

6692002

-1063949

 

 

6

6692002

-1063949

4.3651E+13

-1.424E+13

 

 

7

4.3651E+13

-1.424E+13

1.7026E+27

-1.243E+27

 

 

8

1.7026E+27

-1.243E+27

1.3535E+54

-4.233E+54

 

 

9

1.3535E+54

-4.233E+54

-1.61E+109

-1.15E+109

 

 

10

-1.61E+109

-1.15E+109

1.275E+218

3.687E+218

 

Section 6.1

 

1.         

a.          

 

3.      Answers will vary.

 

 

Section 6.2

1.         

 

3.         

 

5.        

a.           It is not a code word.

 

b.      It is not a code word.

 

 

c.       It is a code word.

 

 

 

7.        

A complete quadrangle consists of 4 points and the 6 lines they determine.  In this order 2 finite projective plane, every combination of 4 vertices is examined to see if the necessary lines exist.  If not, the combination of points is crossed through.  Five complete quadrangles remain.

ABCD       ADEG     CDFG

ABCE        BCDE     DEFG

ABCF        BCDF     ACDE

ABCG       BCDG

ABDE       BCEF

ABDF       BCEG

ABDG      BDFG

ACDF       CDEF

ACDG      CDEG

 

 By duality, five complete quadrilaterals exist.

 

 

Section 6.3

1.          

Since  By vertical angles, Then   Consequently,   Similarly,

  Then .  So,

 

 

3.      If concurrent lines k and l separate lines m and n harmonically, the rays form a harmonic set.

 

 

 

 

5.     

 

 

Section 6.4

1.      When the points are collinear.

 

3.       .

 

Section 6.5

1.         

a.      

 

b.        

 

3.        

 

 

 

 

 

5.           

 

 

Section 6.6

 

 

Section 6.7

1.