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Themed maths · 14 March

Pi Day maths for A Level

14 March is Pi Day (3/14). This ready-to-teach lesson estimates π with random darts, Buffon’s needle and polygons, and asks how good each estimate is. A 5-minute starter, a 35-minute main activity, an extension and full worked answers.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
The binomial distribution; Integration (Buffon’s needle); Small-angle approximations and radians; Integration by substitution (Year 13)
Equipment
The starter is non-calculator. A calculator is needed for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minMonte Carlo π and the binomial distribution
Main: task B11 minBuffon’s needle
Main: task C12 minArchimedes’ polygons
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. A point is chosen at random in a square of side 2. Find the probability that it is inside the circle of radius 1 in the square.
  2. Work out ∫0π sin x dx.
  3. Work out 1 − 1/3 + 1/5 − 1/7 as a single fraction.
  4. Using sin θ ≈ θ for small θ (in radians), estimate 12 sin(π/12).

Main activity (35 minutes)

Task A: Monte Carlo π and the binomial distribution (12 min)

A computer throws 1000 random darts at a unit square; each lands inside the quarter circle x2 + y2 ≤ 1 with probability π/4. Let X be the number of hits, so X ~ B(1000, π/4).

  1. Find E(X), to the nearest whole number.
  2. One run gives 791 hits. Find the estimate of π.
  3. Find the standard deviation of X, to 3 significant figures, and so the standard deviation of the estimate 4X/1000.
  4. Use your calculator’s binomial distribution function to find P(X ≥ 800), to 3 significant figures. What estimate of π does 800 hits give?

Task B: Buffon’s needle (11 min)

A needle of length l is dropped on a floor with parallel lines d apart, where l ≤ d. At angle θ to the lines it crosses one with probability (l sin θ)/d, and θ is equally likely to be anywhere from 0 to π.

  1. Show that the probability of crossing is (1/π)∫0π (l sin θ)/d dθ = 2l/(πd).
  2. Cocktail sticks 4 cm long are dropped on lines 5 cm apart. Find the probability that a stick crosses a line, to 3 significant figures.
  3. 1000 sticks are dropped and 512 cross a line. Estimate π.

Task C: Archimedes’ polygons (12 min)

A regular n-sided polygon is drawn inside a circle of radius 1. Half its perimeter is n sin(π/n), which approaches π as n grows.

  1. Find n sin(π/n) for n = 6, 12 and 96, to 4 decimal places.
  2. Using sin θ ≈ θ − θ3/6, show that n sin(π/n) ≈ π − π3/(6n2). Find the smallest n for which this error, π3/(6n2), is less than 0.001.
  3. Use the first 5 terms of π/4 = 1 − 1/3 + 1/5 − 1/7 + … to estimate π, to 3 significant figures. Which method is better?

Extension (10 minutes)

Year 13 stretch.

  1. Use the substitution x = tan θ to show that ∫01 4/(1 + x2) dx = π.
  2. Use the trapezium rule with 4 strips to estimate ∫01 4/(1 + x2) dx, to 4 decimal places.

For teachers

Teacher notes and full worked answers
  • Pi Day is 14 March because 3/14, written month first, matches π = 3.14…
  • Task A (c) explains why random methods are slow: halving the error needs four times as many darts.
  • Run task B with real cocktail sticks and a lined floor or sheet of paper; pool the class results.

Starter

  1. π/4
    • π ÷ 4
  2. 2
    • [−cos x]0π = 2
  3. 76/105
    • (105 − 35 + 21 − 15)/105
  4. π
    • 12 × π/12 = π

Task A: Monte Carlo π and the binomial distribution

  1. 785
    • 1000 × π/4 = 785.39…
  2. 3.164
    • 4 × 791/1000 = 3.164
  3. 13.0; 0.0519
    • √(1000 × π/4 × (1 − π/4)) = 12.98…
    • The estimate is 4X/1000, so its SD is 0.004 × 12.98… = 0.0519…
  4. 0.138; 3.2
    • P(X ≥ 800) = 1 − P(X ≤ 799) = 0.1383…
    • 800 hits give 4 × 0.8 = 3.2.

Task B: Buffon’s needle

  1. 2l/(πd)
    • (1/π)(l/d)∫0π sin θ dθ = (1/π)(l/d) × 2
  2. 0.509
    • 2 × 4/(5π) = 8/(5π) = 0.5092…
  3. 3.125
    • 512/1000 ≈ 8/(5π), so π ≈ 8000/(5 × 512) = 3.125

Task C: Archimedes’ polygons

  1. 3, 3.1058, 3.1410
    • 6 sin(π/6) = 3
    • 12 sin(π/12) = 3.10582…
    • 96 sin(π/96) = 3.14103…
  2. n = 72
    • n(π/n − π3/(6n3)) = π − π3/(6n2).
    • π3/(6n2) < 0.001 gives n2 > 5167.6…, so n > 71.88…, and n = 72.
  3. 3.34: the polygons are much better
    • 4 × (1 − 1/3 + 1/5 − 1/7 + 1/9) = 4 × 263/315 = 3.339…

Extension

  1. π
    • dx = sec2 θ dθ and 1 + tan2 θ = sec2 θ, so the integrand becomes 4.
    • θ goes from 0 to π/4: 4 × π/4 = π.
  2. 3.1312
    • h = 0.25; the values at 0, 0.25, 0.5, 0.75, 1 are 4, 3.7647…, 3.2, 2.56, 2.
    • 0.125 × (4 + 2 + 2(3.7647… + 3.2 + 2.56)) = 3.13117…

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

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