IGCSE Math Revision Start revising
Themed maths · 23 November

Fibonacci Day maths activities for IGCSE and GCSE

23 November is Fibonacci Day: written month first, 11/23 reads 1, 1, 2, 3. This ready-to-teach lesson has a 5-minute starter, a 35-minute main activity on Fibonacci-type sequences, the golden ratio and the Fibonacci spiral, an extension and full worked answers.

Level
IGCSE and GCSE (Higher and Extended)
Time
40 minutes, plus a 10-minute extension
Topics
Fibonacci-type sequences; Forming and solving equations; The golden ratio and the quadratic formula; Area of squares and rectangles; Arc length
Equipment
The starter is non-calculator. A calculator is allowed in the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minFibonacci-type sequences
Main: task B12 minThe golden ratio
Main: task C11 minFibonacci squares and the spiral
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator. In a Fibonacci-type sequence each term is the sum of the two terms before it.

  1. Write down the next three terms: 1, 1, 2, 3, 5, 8, …
  2. Write down the next two terms of the Fibonacci-type sequence 3, 4, 7, 11, …
  3. Expand and simplify (1 + √5)2.
  4. Write 8 : 5 as a decimal, in the form n : 1.

Main activity (35 minutes)

Task A: Fibonacci-type sequences (12 min)

Each term is the sum of the two terms before it.

  1. The first two terms are 2 and 5. Find the 6th term.
  2. The first two terms are a and b. Show that the 5th term is 2a + 3b. If the 5th term is 41 and a = 4, find b.
  3. A Fibonacci-type sequence has 3rd term 10 and 6th term 42. Find the first two terms.

Task B: The golden ratio (12 min)

Divide each Fibonacci number by the one before it.

  1. Work out 13 ÷ 8, 21 ÷ 13 and 34 ÷ 21, each to 3 decimal places. What do you notice?
  2. The ratios get closer to the golden ratio φ, the positive solution of x2 = x + 1. Use the quadratic formula to find φ to 3 decimal places.
  3. A golden rectangle has length φ times its width. A golden rectangle is 20 cm wide. Find its length to 1 decimal place.

Task C: Fibonacci squares and the spiral (11 min)

Squares with sides 1, 1, 2, 3, 5 and 8 cm fit together to make a rectangle.

  1. What are the length and width of the rectangle?
  2. Show that the total area of the squares equals the area of the rectangle.
  3. A quarter circle is drawn in each square, with radius equal to the side. Find the total length of the spiral, to 3 significant figures.

Extension (10 minutes)

For fast finishers, or as homework.

  1. φ = (1 + √5)/2. Show that φ2 = φ + 1 by working out both sides.
  2. Every third Fibonacci number is even. How many of the first 60 Fibonacci numbers are even? Explain why the pattern works.

For teachers

Teacher notes and full worked answers
  • Fibonacci Day is 23 November because 11/23, written month first, reads 1, 1, 2, 3.
  • Task A (c) is simultaneous equations in disguise: let students find the general terms first.
  • Task C works well on squared paper with compasses: the finished spirals make a good display.

Starter

  1. 13, 21, 34
    • 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34
  2. 18, 29
    • 7 + 11 = 18, 11 + 18 = 29
  3. 6 + 2√5
    • 1 + 2√5 + 5 = 6 + 2√5
  4. 1.6 : 1
    • 8 ÷ 5 = 1.6

Task A: Fibonacci-type sequences

  1. 31
    • 2, 5, 7, 12, 19, 31
  2. b = 11
    • a, b, a + b, a + 2b, 2a + 3b
    • 8 + 3b = 41, so 3b = 33 and b = 11.
  3. 4 and 6
    • The terms are a, b, a + b, a + 2b, 2a + 3b, 3a + 5b.
    • a + b = 10 and 3a + 5b = 42.
    • Substitute b = 10 − a: 3a + 50 − 5a = 42, so a = 4 and b = 6.

Task B: The golden ratio

  1. 1.625, 1.615, 1.619: they get closer to about 1.618
    • 13 ÷ 8 = 1.625
    • 21 ÷ 13 = 1.6153…
    • 34 ÷ 21 = 1.6190…
    • The answers go up and down, getting closer to 1.618…
  2. 1.618
    • x2 − x − 1 = 0, so x = (1 ± √5)/2.
    • The positive solution is (1 + √5)/2 = 1.618 (3 d.p.).
  3. 32.4 cm
    • 20 × 1.6180… = 32.36…
    • 32.4 cm (1 d.p.)

Task C: Fibonacci squares and the spiral

  1. 13 cm by 8 cm
    • The last square is 8 by 8 and sits beside a 5 by 8 rectangle made from the others, so the rectangle is 8 by (8 + 5) = 8 by 13.
  2. 104 cm2
    • 1 + 1 + 4 + 9 + 25 + 64 = 104
    • 8 × 13 = 104
  3. 31.4 cm
    • Each quarter circle is ¼ × 2πr = πr/2.
    • Total = π/2 × (1 + 1 + 2 + 3 + 5 + 8) = π/2 × 20 = 10π = 31.41…
    • 31.4 cm (3 s.f.)

Extension

  1. Both sides equal (3 + √5)/2
    • φ2 = (1 + √5)2/4 = (6 + 2√5)/4 = (3 + √5)/2.
    • φ + 1 = (1 + √5)/2 + 2/2 = (3 + √5)/2.
  2. 20
    • odd + odd = even, odd + even = odd, even + odd = odd, so the pattern odd, odd, even repeats.
    • 60 ÷ 3 = 20 even numbers.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

Practise the topics

e Day maths · Valentine’s Day maths · Women in maths maths · All themed maths

More for lessons: Weekly starters · Worksheet builder · Competition maths