Definition D1
For a circle of radius 1, let Lₙ, Uₙ be the perimeters of the inscribed and circumscribed regular n-gons divided by the diameter 2.
Proof.
(1)By A2 these values do not depend on the size of the circle, so taking the radius to be 1 loses no generality.A2
(2)Writing sₙ, tₙ for the side of the inscribed and circumscribed n-gon, Lₙ = n sₙ / 2 and Uₙ = n tₙ / 2.definition, A7
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Lemma L1
L₆ = 3 and U₆ = 2√3
Proof.
(1)The central angle of the inscribed hexagon is 60°, and the triangle formed by two radii is isosceles, so its other two angles are 60° as well. It is equilateral, hence s₆ = 1.definition, A7
(2)L₆ = 6 · 1 / 2 = 3(1), D1, A7
(3)For the circumscribed hexagon, the triangle joining the centre to half of one side has angles 30°, 60°, 90°. By A3 its sides are in the ratio 1 : √3 : 2, so t₆ / 2 = 1/√3.A3, A7
(4)U₆ = 6 · (2/√3) / 2 = 6/√3 = 2√3 = 3.464101…(3), D1, A7
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Lemma L2
U₂ₙ = 2 Lₙ Uₙ / (Lₙ + Uₙ) and L₂ₙ = √( Lₙ U₂ₙ )
Proof.
(1)Draw the ray bisecting the central angle. By A4 it divides a side of the circumscribed polygon in the stated ratio.A4
(2)Rearranging gives 2/U₂ₙ = 1/Uₙ + 1/Lₙ on the circumscribed side: U₂ₙ is the harmonic mean of Lₙ and Uₙ.(1), A4, A7
(3)Applying A3 to the right triangle formed by the point where the bisector meets the circle and the two original vertices gives L₂ₙ² = Lₙ U₂ₙ.(1), A3, A7
(4)Both sides are positive, so L₂ₙ = √( Lₙ U₂ₙ ).(3), A7, E
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Lemma L3
For every n, Lₙ < L₂ₙ < U₂ₙ < Uₙ
Proof.
(1)L₆ < U₆ holds (3 < 2√3). Assume Lₙ < Uₙ.L1, A7
(2)A harmonic mean lies between the two numbers: Lₙ < U₂ₙ < Uₙ.L2, (1), A7
(3)A geometric mean lies between them too: Lₙ < √( Lₙ U₂ₙ ) = L₂ₙ < U₂ₙ.L2, (2), A7
(4)Hence (Lₙ) increases, (Uₙ) decreases, and Lₙ < Uₙ at every stage.(2), (3)
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Lemma L4
Doubling n repeatedly, Uₙ − Lₙ → 0
Proof.
(1)Dividing U₂ₙ − L₂ₙ by Uₙ − Lₙ and simplifying, the ratio is less than 1/2 at every stage.L2, L3, A7
(2)The actual ratios are 0.2361, 0.2467, 0.2492, 0.2498 — close to 1/4. The gap falls from U₆ − L₆ = 0.464101… to U₉₆ − L₉₆ = 0.001682…, below 1/275 of its initial value after four doublings.(1), L1, L2, A7
(3)The gap tends to 0.(1), (2)
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Theorem T1
(Lₙ) and (Uₙ) converge to the same value. Call that common value π.
Proof.
(1)(Lₙ) increases and is bounded above by U₆.L3
(2)By A5, (Lₙ) converges.(1), A5
(3)Since Uₙ = Lₙ + (Uₙ − Lₙ) and the second term tends to 0, (Uₙ) converges to the same value.(2), L4, A7
(4)Write that common value as π.(2), (3), definition
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Theorem T2
For a circle of radius 1, (circumference) ÷ (diameter) = π
Proof.
(1)The perimeter of the inscribed n-gon is less than the circumference, and that of the circumscribed n-gon is greater.A1
(2)Dividing by the diameter, Lₙ ≤ (circumference)/(diameter) ≤ Uₙ for every n.(1), D1, A7
(3)Both sides tend to π, so by A6 (circumference)/(diameter) = π.(2), T1, A6
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Theorem T3
3.1410 < π < 3.1428
Proof.
(1)Start from L1 and apply L2 four times: n = 6 → 12 → 24 → 48 → 96.L1, L2
(2)L₉₆ = 3.141031…, U₉₆ = 3.142714…(1), A7
(3)By L3, (Lₙ) keeps increasing and (Uₙ) keeps decreasing, so their limit π lies between these two values.L3, T1
(4)Truncating the lower value down and the upper value up gives 3.1410 < π < 3.1428.(2), (3), A7
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