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Themed maths · 14 March

Pi Day maths for IB Maths SL and HL

14 March is Pi Day (3/14). This ready-to-teach lesson estimates π three ways: random darts, Buffon’s needle and infinite series. A 5-minute starter, a 35-minute main activity, an HL extension and full worked answers.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Geometric probability and relative frequency; The binomial distribution; Definite integrals (Buffon’s needle); Series and convergence; Trigonometry in regular polygons
Equipment
The starter is non-calculator. A GDC is needed for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A11 minMonte Carlo π
Main: task B12 minBuffon’s needle
Main: task C12 minSeries for π
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC.

  1. A point is chosen at random inside a square of side 2. Find the probability that it lies inside the circle of radius 1 drawn in the square.
  2. Using π ≈ 22/7, find the circumference of a circle of diameter 7.
  3. Work out 1 − 1/3 + 1/5 − 1/7 as a single fraction.
  4. Work out ∫0π sin θ dθ.

Main activity (35 minutes)

Task A: Monte Carlo π (11 min)

Darts land at random points (x, y) with 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. A dart is a ‘hit’ if x2 + y2 ≤ 1.

  1. Explain why the probability of a hit is π/4.
  2. A class throws 400 darts (on a computer) and gets 314 hits. Estimate π.
  3. The number of hits in 1000 darts is X ~ B(1000, π/4). Find E(X) to the nearest whole number and the standard deviation of X to 3 significant figures.

Task B: Buffon’s needle (12 min)

A needle of length l is dropped on a floor with parallel lines d apart (l ≤ d). The probability that it crosses a line is 2l/(πd).

  1. When l = d, find the probability that the needle crosses a line, to 3 significant figures.
  2. A class drops 500 needles with l = d and 318 cross a line. Estimate π to 3 significant figures.
  3. If the needle makes angle θ with the lines, it crosses with probability (l sin θ)/d, and θ is equally likely to be anywhere from 0 to π. Show that the overall probability is (1/π)∫0π (l sin θ)/d dθ = 2l/(πd).

Task C: Series for π (12 min)

Two infinite series for π: Leibniz, π/4 = 1 − 1/3 + 1/5 − 1/7 + …, and Nilakantha, π = 3 + 4/(2 × 3 × 4) − 4/(4 × 5 × 6) + 4/(6 × 7 × 8) − …

  1. Use the first 5 terms of the Leibniz series to estimate π, to 3 significant figures.
  2. Use the first 4 terms of the Nilakantha series to estimate π, to 4 decimal places.
  3. Which series converges faster? How many terms of the Leibniz series are needed before 4 × (partial sum) is within 0.01 of π? (Use your GDC’s sequence or sum function.)

Extension (10 minutes)

Extension 1 is for everyone; extension 2 is HL.

  1. A regular 12-sided polygon is drawn inside a circle of diameter 1. Its perimeter is 12 sin 15°. Find it to 4 decimal places and compare with π.
  2. (HL) Use the substitution x = tan θ to show that ∫01 4/(1 + x2) dx = π.

For teachers

Teacher notes and full worked answers
  • Pi Day is 14 March because 3/14, written month first, matches π = 3.14…
  • Run task A or B live: a spreadsheet with RAND() or a GDC random-number program, or real cocktail sticks on a floor of floorboards.
  • Task C (c) shows why the Leibniz series is beautiful but slow: discuss what ‘converges’ means in practice.

Starter

  1. π/4
    • Circle area ÷ square area = π ÷ 4
  2. 22
    • 22/7 × 7 = 22
  3. 76/105
    • (105 − 35 + 21 − 15)/105 = 76/105
  4. 2
    • [−cos θ]0π = 1 + 1 = 2

Task A: Monte Carlo π

  1. π/4
    • The hits fill a quarter of a circle of radius 1, area π/4; the square has area 1.
  2. 3.14
    • 314/400 ≈ π/4, so π ≈ 4 × 314/400 = 3.14
  3. 785; 13.0
    • E(X) = 1000 × π/4 = 785.39…
    • SD = √(1000 × π/4 × (1 − π/4)) = √168.5… = 12.98…

Task B: Buffon’s needle

  1. 2/π ≈ 0.637
    • 2l/(πl) = 2/π = 0.6366…
  2. 3.14
    • 318/500 ≈ 2/π, so π ≈ 2 × 500/318 = 3.144…
  3. 2l/(πd)
    • Average (l sin θ)/d over 0 ≤ θ ≤ π: (1/π) × (l/d) × ∫0π sin θ dθ.
    • ∫0π sin θ dθ = 2, so the probability is 2l/(πd).

Task C: Series for π

  1. 3.34
    • 1 − 1/3 + 1/5 − 1/7 + 1/9 = 263/315
    • π ≈ 4 × 263/315 = 3.339…
  2. 3.1452
    • 3 + 1/6 − 1/30 + 1/84 = 3.14523…
  3. Nilakantha; 100 terms
    • The error of the Leibniz estimate after n terms is close to 1/n.
    • After 99 terms the error is 0.0101…; after 100 terms it is 0.00999…, so 100 terms.

Extension

  1. 3.1058
    • Each side subtends 30° at the centre, so it is 2 × ½ × sin 15° = sin 15°.
    • 12 sin 15° = 3.10582…, a little less than π.
  2. π
    • dx = sec2 θ dθ and 1 + tan2 θ = sec2 θ, so the integral is ∫ 4 dθ.
    • Limits 0 and π/4: 4 × π/4 = π.

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

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