Definition D1
On the circle with center O and radius r, take the points dividing the circle into n equal parts (n ≥ 3). Let Iₙ be the regular n-gon with those points as vertices, and Cₙ the regular n-gon formed by the tangents drawn at those points.
Proof.
(1)Let sₙ be the length of one side of Iₙ, pₙ = n sₙ its perimeter, and aₙ the distance from O to one side.convention
(2)Let qₙ be the perimeter of Cₙ.convention
(3)The disk D is the set of all points at distance at most r from O, and the area of a figure X is written [X].convention
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Lemma L1
Iₙ ⊆ D ⊆ Cₙ, and therefore [Iₙ] ≤ K ≤ [Cₙ]
Proof.
(1)The vertices of Iₙ lie on the circle, hence belong to D.D1
(2)D is convex, so the polygon Iₙ formed by those points is contained in D.(1), A8
(3)Each side of Cₙ lies on a tangent to the circle, and the disk lies on one side of that tangent.D1, A9
(4)Cₙ is the intersection of the half-planes determined by those n tangents, so D ⊆ Cₙ(3), D1
(5)Iₙ ⊆ D ⊆ Cₙ(2), (4)
(6)[Iₙ] ≤ K ≤ [Cₙ](5), A5, A4
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Lemma L2
[Iₙ] = ½ aₙ pₙ and [Cₙ] = ½ r qₙ
Proof.
(1)The n triangles formed by O and adjacent pairs of vertices of Iₙ do not overlap and fill Iₙ.D1
(2)Each triangle has base sₙ and height aₙ, the distance from O to that side, so its area is ½ sₙ aₙ.D1, A2
(3)[Iₙ] = n × ½ sₙ aₙ = ½ aₙ (n sₙ) = ½ aₙ pₙ(1), (2), A3, A14, D1
(4)Each side of Cₙ is perpendicular to the radius at its point of tangency, so the distance from O to that side is r.D1, A7
(5)The same decomposition applied to Cₙ gives [Cₙ] = ½ r qₙ(4), (1), (2), (3), A2, A3, A14
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Lemma L3
As n → ∞, aₙ → r
Proof.
(1)Let M be the foot of the perpendicular from O to a side VW of Iₙ. Since OV = OW = r, M is the midpoint of VW.D1, A10
(2)Triangle OMV has a right angle at M and hypotenuse OV = r, so aₙ² + (sₙ/2)² = r²(1), D1, A6
(3)sₙ = pₙ × (1/n), and pₙ → 2πr, 1/n → 0, so sₙ → 0D1, A1, A15, A11, A14
(4)(sₙ/2)² → 0(3), A11
(5)aₙ² = r² − (sₙ/2)² → r²(2), (4), A11, A14
(6)Since aₙ ≥ 0, aₙ → r(5), A12
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Theorem T
The area of the disk of radius r is K = πr²
Proof.
(1)[Iₙ] = ½ aₙ pₙ → ½ × r × 2πr = πr²L2, L3, A1, A11, A14
(2)[Cₙ] = ½ r qₙ → ½ × r × 2πr = πr²L2, A1, A11, A14
(3)For every n, [Iₙ] ≤ K ≤ [Cₙ]L1
(4)K = πr²(1), (2), (3), A13, E
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