Appendix C — Appendix C: Solutions to More Exercises

C.1 Chapter 2: Star Polygons

C.1.1 Solutions to More Exercises

C.1.1.1 1. True or False.

Problem: The star polygon \(\{5/2\}\) is the same as \(\{5/3\}\). Explain your reasoning.

Solution: True. \(\{5/2\}\) and \(\{5/3\}\) trace the same pentagram edges in opposite directions, so they represent the same star polygon.

C.1.1.2 2. Short Answer.

Problem: For a star polygon \(\{n/k\}\) to be connected (a single piece), the values \(n\) and \(k\) must be relatively prime (share no common factors greater than 1). Determine whether each is connected or disconnected:

  1. \(\{8/2\}\)
  2. \(\{9/2\}\)
  3. \(\{10/4\}\)
  4. \(\{7/3\}\)

Solution: a. \(\{8/2\}\): disconnected because \(\gcd(8,2)=2\). b. \(\{9/2\}\): connected because \(\gcd(9,2)=1\). c. \(\{10/4\}\): disconnected because \(\gcd(10,4)=2\). d. \(\{7/3\}\): connected because \(\gcd(7,3)=1\).

C.1.1.3 3. Calculation.

Problem: What is the tip angle of the star polygon \(\{7/2\}\)? Show your work using the angle formula.

Solution: For \(\{7/2\}\),

\[ \theta=\frac{360^\circ(2)}{7}=\frac{720^\circ}{7} \]

\[ \alpha=180^\circ-\theta=180^\circ-\frac{720^\circ}{7}=\frac{540^\circ}{7}\approx77.14^\circ \]

So the tip angle is \(\frac{540^\circ}{7}\approx77.14^\circ\).

C.1.1.4 4. Multiple Choice.

Problem: Which correctly names the star polygon formed by connecting every 3rd point on a set of 8 equally-spaced points?

    1. \(\{3/8\}\)
    1. \(\{8/5\}\)
    1. \(\{8/3\}\)
    1. \(\{3/5\}\)

Solution: (c) \(\{8/3\}\).

C.1.1.5 5. Art & Fashion Connection.

Problem: The Star of David (\(\{6/2\}\)) and the five-pointed star (\(\{5/2\}\)) appear in fashion, jewelry, printed fabrics, and logos.

  • Identify the star polygon notation for a six-pointed star formed by two overlapping equilateral triangles. Is it connected or disconnected? How many triangles make it up?
  • Sketch or describe how to construct it starting from 6 equally spaced points on a circle.

Solution: The six-pointed star made from two equilateral triangles is \(\{6/2\}\). It is disconnected (since \(\gcd(6,2)=2\)) and is made of 2 triangles.

Construction idea: place 6 equally spaced points on a circle, connect every other point to form one equilateral triangle, then connect the three remaining points to form the inverted triangle.

C.2 Chapter 3: Sequences

C.2.1 Solutions to More Exercises

C.2.1.1 1. Identify the Sequence Type.

Problem: For each sequence, state whether it is arithmetic, geometric, or neither. If arithmetic, give \(d\). If geometric, give \(r\).

  1. \(3, 7, 11, 15, 19, \ldots\)
  2. \(2, 6, 18, 54, \ldots\)
  3. \(1, 4, 9, 16, 25, \ldots\)
  4. \(100, 50, 25, 12.5, \ldots\)

Solution: a. Arithmetic, \(d=4\). b. Geometric, \(r=3\). c. Neither. d. Geometric, \(r=\frac{1}{2}\).

C.2.1.2 2. Write a Formula.

Problem: The sequence \(5, 8, 11, 14, \ldots\) is arithmetic.

  • Write an explicit formula for the \(n\)th term.
  • Write a recursive formula.
  • Use your explicit formula to find the 20th term.

Solution: For \(5,8,11,14,\ldots\):

  • Explicit: \(a_n=5+3(n-1)=3n+2\)
  • Recursive: \(a_1=5\), \(a_n=a_{n-1}+3\) for \(n\ge2\)
  • 20th term: \(a_{20}=3(20)+2=62\)

C.2.1.3 3. Method of Successive Differences.

Problem: Use the Method of Successive Differences to find the next two terms of:

\(2, 5, 10, 17, 26, \ldots\)

Solution: Sequence: \(2,5,10,17,26,\ldots\)

First differences: \(3,5,7,9\). Second differences are constant: \(2,2,2\).

So next first differences are \(11,13\).

Next two terms:

  • \(26+11=37\)
  • \(37+13=50\)

Answer: \(37,50\).

C.2.1.4 4. Partial Sum.

Problem: Find the sum of the first 10 terms of the geometric sequence:

\(1, 2, 4, 8, \ldots\)

Solution: This is geometric with \(a_1=1\), \(r=2\), \(n=10\).

\[ S_{10}=a_1\frac{1-r^{10}}{1-r}=\frac{1-2^{10}}{1-2}=1023 \]

So the sum is \(1023\).

C.2.1.5 5. Pop Culture Connection.

Problem: Suppose the number of songs performed each night follows an arithmetic sequence: first night 22 songs, each subsequent night adds 2 songs.

  • Write an explicit formula for songs on night \(n\).
  • How many songs on night 15?
  • If the total across all nights is 500 songs, how many nights did the tour run?

Solution: Given \(a_1=22\), \(d=2\):

  • Explicit formula: \(a_n=22+2(n-1)=2n+20\)
  • Night 15: \(a_{15}=2(15)+20=50\) songs

Total after \(n\) nights:

\[ S_n=\frac{n}{2}\big(2\cdot22+(n-1)2\big)=n(n+21) \]

Set \(n(n+21)=500\):

\[ n^2+21n-500=0 \]

This has no whole-number solution, so under this model there is no exact integer number of nights that gives total 500 songs.

C.3 Chapter 4: Figurate Numbers, Fibonacci, and Sums

C.3.1 Solutions to More Exercises

C.3.1.1 1. Figurate Numbers.

Problem: Find the 8th triangular number, the 5th square number, and the 4th pentagonal number. Show the formula used for each.

Solution: - 8th triangular number:

\[ T_8=\frac{8(9)}{2}=36 \]

  • 5th square number:

\[ S_5=5^2=25 \]

  • 4th pentagonal number:

\[ P_4=\frac{4(3\cdot4-1)}{2}=\frac{4(11)}{2}=22 \]

C.3.1.2 2. Fibonacci Sequence.

Problem: The Fibonacci sequence starts \(1,1,2,3,5,8,\ldots\)

  • Write a recursive formula for \(F_n\).
  • List the first 12 terms.
  • Compute \(F_{12}/F_{11}\). What constant does this ratio approach?

Solution: Recursive formula:

\[ F_1=1,\;F_2=1,\;F_n=F_{n-1}+F_{n-2}\;(n\ge3) \]

First 12 terms:

\(1,1,2,3,5,8,13,21,34,55,89,144\)

Ratio:

\[ \frac{F_{12}}{F_{11}}=\frac{144}{89}\approx1.61798 \]

This approaches the golden ratio \(\varphi\approx1.61803\).

C.3.1.3 3. Gauss’s Method.

Problem: Use Gauss’s Method to find the sum of integers from 1 to 200. Show your reasoning.

Solution: \[ 1+2+\cdots+200=\frac{200(201)}{2}=20100 \]

So the sum is \(20100\).

C.3.1.4 4. Sigma Notation.

Problem: Evaluate:

\[ \sum_{k=1}^{6} (2k-1) \]

What do you notice about the result?

Solution: \[ \sum_{k=1}^{6}(2k-1)=1+3+5+7+9+11=36 \]

Observation: the sum of the first \(n\) odd numbers is \(n^2\); here \(6^2=36\).

C.3.1.5 5. Fashion & Art Connection.

Problem: The 10th and 11th Fibonacci numbers are 55 and 89.

  • If a designer wants a golden-ratio rectangle with shorter side 55 cm, approximately how long is the longer side?
  • A triangular scarf display has 6 rows (1 on top, then 2, 3, and so on). How many scarves total?

Solution: Using the golden-ratio approximation for consecutive Fibonacci numbers:

\[ \text{long side}\approx55\cdot1.618\approx88.99\text{ cm}\approx89\text{ cm} \]

For 6 rows of a triangular scarf display:

\[ T_6=\frac{6(7)}{2}=21 \]

So the display has 21 scarves.

C.4 Chapter 5: Polygons and Polyhedra

C.4.1 Solutions to More Exercises

C.4.1.1 1. Vocabulary Matching.

Problem: Match each term with the correct description:

    1. Convex polygon (1) All sides and angles are equal
    1. Regular polygon (2) No interior angle greater than 180°
    1. Concave polygon (3) At least one interior angle greater than 180°

Solution: A-2, B-1, C-3.

C.4.1.2 2. Interior Angle Sum.

Problem: Find the sum of the interior angles for each polygon:

  1. Hexagon
  2. Decagon (10 sides)
  3. 15-gon

Solution: Use \((n-2)180^\circ\).

  1. Hexagon (\(n=6\)):

\[ (6-2)180^\circ=720^\circ \]

  1. Decagon (\(n=10\)):

\[ (10-2)180^\circ=1440^\circ \]

  1. 15-gon (\(n=15\)):

\[ (15-2)180^\circ=2340^\circ \]

C.4.1.3 3. Find the Missing Angle.

Problem: A pentagon has four interior angles measuring \(90^\circ\), \(105^\circ\), \(118^\circ\), and \(97^\circ\). Find the fifth angle.

Solution: For a pentagon, interior sum is:

\[ (5-2)180^\circ=540^\circ \]

Given four angles total:

\[ 90^\circ+105^\circ+118^\circ+97^\circ=410^\circ \]

Missing angle:

\[ 540^\circ-410^\circ=130^\circ \]

C.4.1.4 4. Triangle Classification.

Problem: Classify each triangle by sides and by angles.

  • Triangle 1: sides 5, 5, 8; angles \(75^\circ\), \(75^\circ\), \(30^\circ\)
  • Triangle 2: sides 3, 4, 5; angles \(90^\circ\), \(53^\circ\), \(37^\circ\)
  • Triangle 3: sides 6, 6, 6; angles \(60^\circ\), \(60^\circ\), \(60^\circ\)

Solution: - Triangle 1 (5,5,8; 75,75,30): isosceles, acute. - Triangle 2 (3,4,5; 90,53,37): scalene, right. - Triangle 3 (6,6,6; 60,60,60): equilateral, acute.

C.4.1.5 5. Fashion & Design Connection.

Problem: - What is the measure of each interior angle of a regular octagon? Show work. - A custom pendant is cut as a regular hexagon. If each side is 1.2 cm, what is the sum of interior angles? Is it convex or concave?

Solution: Regular octagon interior angle:

\[ \frac{(8-2)180^\circ}{8}=135^\circ \]

Regular hexagon interior angle sum:

\[ (6-2)180^\circ=720^\circ \]

A regular hexagon is convex.

C.5 Chapter 6: Tessellations and Fractals

C.5.1 Solutions to More Exercises

C.5.1.1 1. True or False.

Problem: Every regular polygon can tessellate the plane on its own. If false, give a counterexample.

Solution: False. Not every regular polygon tessellates by itself; for example, a regular pentagon does not tessellate the plane.

C.5.1.2 2. Angle Check.

Problem: Determine whether each regular polygon can tessellate the plane by checking if its interior angle evenly divides \(360^\circ\).

  1. Equilateral triangle
  2. Regular pentagon
  3. Regular hexagon
  4. Regular octagon

Solution: Check whether interior angle divides \(360^\circ\).

  1. Equilateral triangle: interior angle \(60^\circ\), and \(360/60=6\) (integer), so yes.
  2. Regular pentagon: interior angle \(108^\circ\), and \(360/108\) is not an integer, so no.
  3. Regular hexagon: interior angle \(120^\circ\), and \(360/120=3\), so yes.
  4. Regular octagon: interior angle \(135^\circ\), and \(360/135\) is not an integer, so no.

C.5.1.3 3. Semi-Regular Tessellation.

Problem: A vertex in a tiling is surrounded by a regular triangle, a square, a regular hexagon, and another square (in some order). Do these form a valid tiling at that vertex? Show the angle sum.

Solution: Angles at the vertex:

\[ 60^\circ+90^\circ+120^\circ+90^\circ=360^\circ \]

Yes, this is a valid vertex configuration.

C.5.1.4 4. Fractals.

Problem: The Sierpinski triangle is formed by repeatedly removing the middle triangle from each remaining triangle.

  • After Step 0 there is 1 filled triangle, after Step 1 there are 3, after Step 2 there are 9. Write a formula for the number of filled triangles after Step \(n\).
  • Is this arithmetic or geometric?

Solution: The count triples each step: \(1,3,9,\ldots\).

If Step 0 is \(n=0\), then:

\[ N_n=3^n \]

This is a geometric sequence with common ratio \(3\).

C.5.1.5 5. Art Connection.

Problem: - In your own words, what property must a shape have to tessellate the plane? - If starting from a square with interior angle \(90^\circ\), how many squares meet at each vertex in a standard square tessellation? Verify using angle sum. - Describe or sketch how you might modify one side of an equilateral triangle to create an Escher-like interlocking shape that still tessellates.

Solution: - A shape tessellates if repeated copies cover the plane with no gaps and no overlaps. - In a square tessellation, 4 squares meet at each vertex because \(4\cdot90^\circ=360^\circ\). - One way to modify an equilateral triangle and keep tessellation: cut a small curve/notch from one side and translate that exact piece to the adjacent side so matching edges still interlock under translations/rotations.

C.6 Chapter 7: Symmetry, Isometries, and Frieze Patterns

C.6.1 Solutions to More Exercises

C.6.1.1 1. Lines of Symmetry.

Problem: How many lines of reflection symmetry does each shape have?

  1. Equilateral triangle
  2. Square
  3. Regular hexagon
  4. Circle

Solution: a. Equilateral triangle: 3 b. Square: 4 c. Regular hexagon: 6 d. Circle: infinitely many

C.6.1.2 2. Rotational Symmetry.

Problem: State the smallest angle of rotation that maps each figure onto itself:

  1. A regular pentagon
  2. The letter S
  3. A regular octagon

Solution: a. Regular pentagon: \(72^\circ\) b. Letter S: \(180^\circ\) c. Regular octagon: \(45^\circ\)

C.6.1.3 3. Isometry Identification.

Problem: For each transformation, identify the isometry (reflection, rotation, translation, or glide reflection):

  1. Moving a shape 5 units to the right without turning or flipping.
  2. Flipping a shape over the \(x\)-axis.
  3. Reflecting a shape over a line, then translating it along that same line.
  4. Rotating a shape \(90^\circ\) around a fixed point.

Solution: a. Translation b. Reflection c. Glide reflection d. Rotation

C.6.1.4 4. Frieze Patterns.

Problem: Identify the frieze pattern type (hop, step, sidle, jump, spinning hop, spinning sidle, or spinning jump):

  1. The pattern only has translational symmetry.
  2. The pattern has a vertical line of reflection and translational symmetry, but no other symmetries.
  3. The pattern has a glide reflection and translational symmetry only.

Solution: a. Hop (translation only) b. Jump (vertical reflection plus translation) c. Step (glide reflection plus translation)

C.6.1.5 5. Fashion Connection.

Problem: - A scarf border repeats a feather motif that is reflected vertically and can also be flipped upside-down. Which frieze pattern type is this? - Does the LV logo itself have lines of reflection symmetry? Does it have rotational symmetry? Explain. - Name one capital letter with both a horizontal and a vertical line of symmetry.

Solution: - A border motif with vertical reflection and upside-down symmetry is commonly classified as spinning jump. - The LV monogram itself is generally treated as having no reflection symmetry and no nontrivial rotational symmetry. - One capital letter with both horizontal and vertical symmetry: H.

C.7 Chapter 8: Conic Sections

C.7.1 Solutions to More Exercises

C.7.1.1 1. Matching.

Problem: Match each conic section to how it is formed by a plane cutting a cone:

    1. Circle (1) Plane cuts both nappes of the cone
    1. Ellipse (2) Plane is parallel to the base of the cone
    1. Parabola (3) Plane cuts at an angle but only one nappe
    1. Hyperbola (4) Plane is parallel to one slant side of the cone

Solution: A-2, B-3, C-4, D-1.

C.7.1.2 2. Volume Calculations.

Problem: Round to the nearest hundredth where needed (\(\pi \approx 3.14159\)).

  1. Find the volume of a cylinder with radius 4 cm and height 10 cm.
  2. Find the volume of a cone with radius 3 cm and height 9 cm.
  3. Find the volume of a sphere with radius 5 cm.

Solution: Use \(\pi\approx3.14159\).

  1. Cylinder (\(r=4\), \(h=10\)):

\[ V=\pi r^2h=\pi(4)^2(10)=160\pi\approx502.65\text{ cm}^3 \]

  1. Cone (\(r=3\), \(h=9\)):

\[ V=\frac13\pi r^2h=\frac13\pi(3)^2(9)=27\pi\approx84.82\text{ cm}^3 \]

  1. Sphere (\(r=5\)):

\[ V=\frac43\pi r^3=\frac43\pi(125)=\frac{500\pi}{3}\approx523.60\text{ cm}^3 \]

C.7.1.3 3. Surface Area.

Problem: A soup can (cylinder) has radius 3.5 cm and height 11 cm. Find the total surface area and show all steps.

Solution: For a closed cylinder:

\[ SA=2\pi r^2+2\pi rh \]

With \(r=3.5\), \(h=11\):

\[ SA=2\pi(3.5)^2+2\pi(3.5)(11)=24.5\pi+77\pi=101.5\pi\approx318.87\text{ cm}^2 \]

C.7.1.4 4. Planetary Orbits.

Problem: - True or False: A perfect circle is a special case of an ellipse. - Which planet in our solar system has the most circular orbit, and which has the most elongated (eccentric) elliptical orbit?

Solution: - True. A circle is a special case of an ellipse. - Most circular planetary orbit: Venus (very small eccentricity). - Most elongated planetary orbit: Mercury (largest eccentricity among the 8 planets).

C.7.1.5 5. Pop Culture & Design Connection.

Problem: - Why is a parabolic shape used for satellite dishes and car headlights? Describe in terms of the reflective property of a parabola. - A decorative bowl is a hemisphere with inner radius 12 cm. What volume of liquid can it hold?

Solution: - Parabolic reflectors focus parallel incoming rays to the focus (dishes), and if the source is at the focus, reflected rays leave parallel (headlights).

Hemisphere volume with \(r=12\) cm:

\[ V_{\text{hemisphere}}=\frac12\left(\frac43\pi r^3\right)=\frac23\pi r^3 \]

\[ V=\frac23\pi(12)^3=1152\pi\approx3619.11\text{ cm}^3 \]