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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 10 · IDU — Aesthetics

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.4
2.6
3.5
4.No (order 1)
5.Equilateral triangle, square, or regular hexagon
6.120°120°
7.Yes, order 2
8.B, M, X
9.ϕ≈1.6180\phi \approx 1.6180
10.34, 55
Silver
11.Interior angle 108° does not divide 360°
12.5 lines; rotational order 5
13.2
14.13/8=1.62513/8 = 1.625
15.24 cm
16.9
17.Yes — ratio 8:5 = 1.6 ≈ ϕ\phi
18.(a) 60°60° (b) 6
19.1800°1800°
20.Translation only (group p1 in frieze notation)
Gold
21.8 (group D4D_4)
22.ϕ−1≈0.618\phi - 1 \approx 0.618; 1/ϕ≈0.6181/\phi \approx 0.618
23.(a) 55/34≈1.617655/34 \approx 1.6176 (b) ϕ≈1.6180\phi \approx 1.6180 — very close
24.3
25.4.8.8 (square, octagon, octagon)
26.10π10\pi
27.12
28.6 (cyclic group C6C_6)
29.25.89 cm²
30.Rotational order 2; no reflective symmetry (chirality)
Platinum
31.19.09 m
32.3×135=405°>360°3 \times 135 = 405° > 360°; no way to fit
33.x=3x = 3
34.About 55.0 (actual F10=55F_{10} = 55)
35.x=1±52x = \dfrac{1 \pm \sqrt{5}}{2}
36.1/161/16 (one cell of 16)
37.4 types: translation, rotation, reflection, glide reflection
38.They tile the plane without periodic repetition (aperiodic)
39.Perimeter ×4/3\times 4/3 per iteration → infinity; area enclosed converges
40.6 lines, rotational order 6, group D6D_6 (12 elements)

Pack B — Answers

Bronze
1.3
2.8
3.7
4.Yes (order 2)
5.Equilateral triangle and square (besides hexagon)
6.108°108°
7.Yes, order 2
8.D, H, T
9.ϕ−1≈0.6180\phi - 1 \approx 0.6180
10.34, 55, 89
Silver
11.Interior angle 128.57°128.57° does not divide 360°
12.6 lines; rotational order 6
13.2
14.21/13≈1.61521/13 \approx 1.615
15.42 cm
16.16
17.Yes — ratio 13:8 = 1.625 ≈ ϕ\phi
18.(a) 45°45° (b) 8
19.3240°3240°
20.Translation + horizontal reflection
Gold
21.12 (group D6D_6)
22.ϕ2≈2.618=ϕ+1\phi^2 \approx 2.618 = \phi + 1
23.(a) 89/55≈1.618289/55 \approx 1.6182 (b) very close to ϕ\phi
24.7
25.3.6.3.6 (alternating triangles and hexagons)
26.33π2\dfrac{33\pi}{2}
27.48
28.5 (C5C_5)
29.58.25 cm²
30.Rotational order 2; no reflection (chirality)
Platinum
31.5.82 m
32.Heptagon interior ≈128.57°\approx 128.57°; doesn't divide 360°
33.x=5x = 5
34.About 144.0 (actual F12=144F_{12} = 144)
35.x=1±52x = \dfrac{1 \pm \sqrt{5}}{2}
36.1/251/25
37.Translation, rotation, reflection, glide reflection
38.No translational symmetry — pattern never exactly repeats
39.Area ×3/4\times 3/4 per iteration → 0
40.10 lines, order 10, group D10D_{10} (20 elements)

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) a+ba=ab\frac{a+b}{a} = \frac{a}{b} (b) ϕ=1+52\phi = \frac{1+\sqrt{5}}{2} (c) Longer side ≈ 9.71 cm
Full working
(a) Let the golden rectangle have shorter side aa and longer side a+ba + b (where bb is the remainder after removing the square of side aa). The proportion is:

a+ba=ab\frac{a+b}{a} = \frac{a}{b}


(b) Let ϕ=ab\phi = \frac{a}{b}. Then a+ba=1+ba=1+1ϕ\frac{a+b}{a} = 1 + \frac{b}{a} = 1 + \frac{1}{\phi}. Setting this equal to ϕ\phi:

ϕ=1+1ϕ⇒ϕ2=ϕ+1⇒ϕ2−ϕ−1=0.\phi = 1 + \frac{1}{\phi} \Rightarrow \phi^2 = \phi + 1 \Rightarrow \phi^2 - \phi - 1 = 0.


ϕ=1+52≈1.618\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618.

(c) Shorter side a=6a = 6 cm. Longer side a+b=aϕ=6×1.618≈9.71a + b = a\phi = 6 \times 1.618 \approx 9.71 cm. So b≈3.71b \approx 3.71 cm. Check: remaining rectangle is 6×3.716 \times 3.71 cm; ratio 6/3.71≈1.618=ϕ6/3.71 \approx 1.618 = \phi ✓.
2Problem 2
Answer
(a) No — inconsistent; 4-fold rotation forces 4 reflection lines (or 0) (b) Adjust to 4 reflections (c) Still 4-fold, but now D4D_4
Full working
(a) **No.** A rotational symmetry of order nn forces either 0 reflection lines (cyclic group CnC_n) or nn reflection lines (dihedral group DnD_n). You can't have 4-fold rotation and exactly 2 reflections.

**Why?** If you have any reflection RR in axis ℓ\ell and a rotation ρ\rho of order 4, then the conjugates ρRρ−1,ρ2Rρ−2,ρ3Rρ−3\rho R \rho^{-1}, \rho^2 R \rho^{-2}, \rho^3 R \rho^{-3} are also reflections in 3 other axes obtained by rotating ℓ\ell by 90°, 180°, 270°. So you must have 4 reflection lines, not 2.

(b) A design with D4D_4 symmetry: a square with two diagonal "X" marks. 4 rotations + 4 reflection lines.

(c) Adding 2 reflection lines (total 4) to a design with 4-fold rotation gives the full dihedral group D4D_4. The rotational order is still 4 — but the symmetry group is now larger (8 elements: C4C_4 has 4, D4D_4 has 8).
3Problem 3
Answer
(a) 13/8=1.62513/8 = 1.625, very close to ϕ\phi (b) 34/21≈1.61934/21 ≈ 1.619; 55/34≈1.61855/34 ≈ 1.618 (c) Optimal packing — most efficient seed placement
Full working
(a) 13/8=1.62513/8 = 1.625. ϕ≈1.618\phi \approx 1.618. Very close.

(b) 34/21=1.6190‾34/21 = 1.619\overline{0}. 55/34=1.6176…55/34 = 1.6176\ldots. Both extremely close to ϕ\phi.

(c) **Hypothesis:** plants arrange leaves/seeds to maximise sunlight or space per unit. Mathematical analysis (Vogel's formula, 1979) shows that the angle that gives the most efficient seed packing without overlap is 360°ϕ2≈137.5°\frac{360°}{\phi^2} \approx 137.5°, the "golden angle." When seeds are placed at this angle, the resulting spirals naturally have counts that are consecutive Fibonacci numbers — because Fibonacci ratios approximate ϕ\phi as the values grow.

In short: **the golden angle is the most efficient irrational angle for biological packing**, and Fibonacci spirals are a consequence of this.
4Problem 4
Answer
(a) Triangle (60°), square (90°), hexagon (120°) (b) e.g., 3.6.3.6 (triangle-hexagon) (c) Interior angle 108° doesn't divide 360°
Full working
(a) A regular polygon tessellates alone iff its interior angle divides 360° (so the angles meeting at a vertex sum exactly to 360°):

- Equilateral triangle: 60° × 6 = 360° ✓
- Square: 90° × 4 = 360° ✓
- Regular hexagon: 120° × 3 = 360° ✓

Other regular polygons have angles that don't divide 360° (e.g. pentagon 108°, heptagon ≈128.57°), so they don't tessellate alone.

(b) **Semi-regular tessellations** — there are 8 of these. Examples:
- **3.3.4.3.4**: two triangles, square, triangle, square at each vertex. 60+60+90+60+90=360°60+60+90+60+90 = 360° ✓.
- **3.6.3.6**: triangle-hexagon alternating. 60+120+60+120=360°60+120+60+120 = 360° ✓.
- **4.8.8**: square + two octagons at each vertex. 90+135+135=360°90+135+135 = 360° ✓.

(c) Pentagons have interior angle 108°108°. 360°/108°=3.3‾360°/108° = 3.\overline{3}, not a whole number. Three pentagons leave a 36°36° gap; four would overlap. So they can't tile alone. (Note: *irregular* pentagons can — there are 15 known types of convex pentagon that tessellate.)
5Problem 5
Answer
(a) x=3x = 3 (b) y=−12x+154y = -\frac{1}{2}x + \frac{15}{4} (c) (3,2.25)(3, 2.25)
Full working
(a) Midpoint of ABAB: (3,0)(3, 0). ABAB is along the xx-axis (gradient 0). Perpendicular bisector is vertical: x=3x = 3.

(b) Midpoint of ACAC: (1.5,3)(1.5, 3). Gradient of ACAC: 6−03−0=2\frac{6 - 0}{3 - 0} = 2. Perpendicular gradient: −12-\frac{1}{2}. Equation through (1.5,3)(1.5, 3): y−3=−12(x−1.5)y - 3 = -\frac{1}{2}(x - 1.5), i.e. y=−12x+3.75y = -\frac{1}{2}x + 3.75.

(c) Set x=3x = 3 in (b): y=−1.5+3.75=2.25y = -1.5 + 3.75 = 2.25. So intersection at (3,2.25)(3, 2.25).

This is the **circumcentre** of △ABC\triangle ABC — equidistant from all three vertices. It is the meeting point of the Voronoi cell boundaries.

(d) **Sketch**: from (3,2.25)(3, 2.25), draw rays along each perpendicular bisector. The plane is divided into 3 "wedge-like" cells (unbounded for an outer-edge configuration), one for each seed point.
6Problem 6
Answer
Open creative — see exemplar
Full working
**Example design.** Take a square grid. Inside each square, draw a smaller square rotated by 45° (a diamond), then a 4-pointed star inscribed in the diamond.

**Symmetries identified:**
1. **Rotation**: each square cell has 4-fold rotational symmetry (rotations by 0°, 90°, 180°, 270°).
2. **Reflection**: 4 lines per cell (2 along sides, 2 along diagonals).
3. **Translation**: by the lattice vectors (cell width horizontally and vertically).
4. **Glide reflection**: along the diagonals.

This pattern belongs to wallpaper group **p4m** — one of the 17 wallpaper groups.

**Student variations:**
- Pinwheel pattern: 4-fold rotation, no reflections (group p4).
- Brick pattern: translation only (p1).
- Honeycomb with motifs: 6-fold rotation (p6 or p6m).

**Submission note for Criterion D:** describe the design choices, justify aesthetically, and explain mathematically using the language of symmetry groups.
7Problem 7
Answer
(a) ≈ 932 cm² (b) ≈ 2098 cm² (c) k2k^2 where kk is linear scale factor
Full working
(a) Shorter side 24 cm, longer side 24×ϕ≈24×1.618=38.8324 \times \phi \approx 24 \times 1.618 = 38.83 cm. Area ≈24×38.83≈932\approx 24 \times 38.83 \approx 932 cm².

(b) Linear scale factor 1.5 → new sides 3636 and ≈58.25\approx 58.25 cm. Area ≈36×58.25≈2097\approx 36 \times 58.25 \approx 2097 cm². Or use scale-factor: 932×1.52=932×2.25≈2097932 \times 1.5^2 = 932 \times 2.25 \approx 2097 cm².

(c) For linear scale factor kk: area scale factor is k2k^2. (For volume in 3D: k3k^3.)
8Problem 8
Answer
(a) 3 (b) Rotations and translations (c) Wallpaper group p3
Full working
(a) Three distinct orientations — typically the lizard appears in three rotational positions, 120° apart.

(b) Each orientation is obtained from the others by **rotation by 120°** about specific centres. The pattern as a whole is also invariant under **translation** by the lattice vectors. There are no reflection axes (lizards are chiral — the design has handedness).

(c) The wallpaper group is **p3** — 3-fold rotational symmetry, no reflection. One of the 17 wallpaper groups.

(Escher famously studied wallpaper symmetries and used most of the 17 groups in his work — a beautiful intersection of mathematics, art, and crystallography.)
9Problem 9
Answer
(a) Exponentially growing along a "Fibonacci spiral" direction (b) Gradients approach ϕ\phi (c) (13,21),(21,34)(13, 21), (21, 34)
Full working
(a) Plotting: the points "race away" from the origin along a curving path. Each point's coordinates are roughly ϕ\phi times the previous — this is exponential growth, like the Fibonacci spiral.

(b) Gradients between consecutive points:
- (1,1)→(1,2)(1,1)→(1,2): undefined (vertical).
- (1,2)→(2,3)(1,2)→(2,3): (3−2)/(2−1)=1(3-2)/(2-1) = 1.
- (2,3)→(3,5)(2,3)→(3,5): (5−3)/(3−2)=2(5-3)/(3-2) = 2.
- (3,5)→(5,8)(3,5)→(5,8): (8−5)/(5−3)=1.5(8-5)/(5-3) = 1.5.
- (5,8)→(8,13)(5,8)→(8,13): (13−8)/(8−5)=5/3≈1.667(13-8)/(8-5) = 5/3 ≈ 1.667.

Successive gradients 1,2,1.5,1.667,…1, 2, 1.5, 1.667, \ldots **oscillate around ϕ≈1.618\phi \approx 1.618**.

(c) Next pair: (13,21)(13, 21) and (21,34)(21, 34).
10Problem 10
Answer
(a) 3/4 (b) 9/16 (c) (3/4)n(3/4)^n (d) Tends to zero (yet infinitely many pieces remain)
Full working
(a) Removing the central triangle (1/4 of the area) leaves 3/43/4.

(b) Each remaining small triangle (there are 3) has 1/4 removed. So fraction remaining: 34×34=916\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}.

(c) Iteration nn: (34)n\left(\frac{3}{4}\right)^n.

(d) As n→∞n \to \infty: (34)n→0\left(\frac{3}{4}\right)^n \to 0. The area tends to **zero**, yet the Sierpinski triangle contains infinitely many disconnected points — a fractal with "fractional dimension" log⁡23≈1.585\log_2 3 \approx 1.585 (more than a line but less than a plane).
11Problem 11
Answer
See exemplar outline
Full working
**(a) Research question.**
"How is the golden ratio ϕ\phi, the Fibonacci sequence, and symmetry used by artists to create aesthetic appeal? Are these mathematical features perceived as 'beautiful' across cultures?"

**(b) Three examples.**
1. **Leonardo da Vinci's Mona Lisa**: composition allegedly based on golden rectangles. Investigate the actual proportions; verify or refute the claim.
2. **The Parthenon (architecture)**: golden rectangle in façade dimensions.
3. **Islamic geometric patterns** (e.g., Alhambra tilings): use of symmetry groups (wallpaper groups).

**(c) Mathematical principles.**
- For Mona Lisa: measure key features (face, body) and compute ratios. Compare to ϕ\phi.
- For Parthenon: measure column-spacing-to-height ratios.
- For Islamic patterns: identify symmetry group, count reflection lines, classify by the 17 wallpaper groups.

**(d) Presentation.**
- Photos with measurements overlaid.
- Tables showing ratios computed vs ϕ\phi.
- Diagrams illustrating symmetry types.
- Discussion: which artistic choices align with mathematics, which are myth?

**Criterion D angle.** Critique whether the maths is descriptive (artists notice ϕ\phi post-hoc) or prescriptive (artists deliberately use ϕ\phi). Compare across cultures.
12Problem 12
Answer
Open creative — see exemplar response
Full working
**Exemplar poster outline.**

**Visual elements:**
- Central motif: a **golden rectangle** (sides 21 cm × 13 cm, ratio close to ϕ\phi).
- Inside: a **Fibonacci spiral** drawn through quarter-circles in nested squares of sides 1, 1, 2, 3, 5, 8, 13.
- Frame around the rectangle: a **tessellation of equilateral triangles and squares** (3.3.4.3.4 semi-regular tiling).
- Colour scheme: monochrome blues from light to dark, following a Fibonacci spacing (1, 1, 2, 3, 5 shades of blue intensity).
- A **6-fold symmetric snowflake** in the top-right corner, suggesting dihedral symmetry D6D_6.

**Justification (200 words):**

This poster demonstrates three core unit concepts: (1) the golden ratio, (2) Fibonacci numbers and the related spiral, and (3) symmetry in tessellation. The central golden rectangle uses sides 21 and 13 — consecutive Fibonacci numbers whose ratio (1.615) approximates ϕ\phi. Drawing the spiral inside reinforces the link between Fibonacci and ϕ\phi. The framing tessellation uses the semi-regular 3.3.4.3.4 pattern; the interior angle sum (two triangles + two squares = 360°) gives a valid tiling. The snowflake illustrates dihedral symmetry D6D_6 — 6 rotational and 6 reflective symmetries forming a 12-element group. The colour scheme follows Fibonacci spacing because perceptually that ordering feels balanced — a known result from Gestalt psychology. The overall composition is asymmetric (deliberate, to feel dynamic) but each individual element carries strong internal symmetry. The poster therefore visually communicates the contrast between local mathematical order and global creative balance — a common feature of art that uses mathematics.

**Notes for Criterion D evaluation:** demonstrate creativity, justify each mathematical choice, and link to aesthetics.