Solution to puzzle 45: Area of regular 2ⁿ-gon

The regular octagon below will serve to represent a general 2ⁿ-gon.
Since the polygon has unit perimeter, each side is of length 2⁻ⁿ.

The area of the colored right triangle is ½ × 2⁻⁽ⁿ⁺¹⁾ × h.
The 2ⁿ-gon consists of 2ⁿ⁺¹ such triangles; therefore its area is ½h.

We also have tan x = 2⁻⁽ⁿ⁺¹⁾/h, from which h = 2⁻⁽ⁿ⁺¹⁾/tan x.
Further, x = /2ⁿ.
Hence the area of the 2ⁿ-gon is 2⁻⁽ⁿ⁺²⁾/tan(/2ⁿ).

The problem is therefore reduced to finding the tangent of /2ⁿ, which we can determine by repeated application of appropriate half angle formulae.

Although there is a half angle formula that gives tan (½a) directly in terms of tan a, this is not suitable for repeated application.  More fruitful is to recursively determine cos(/2ⁿ⁻¹), and thereby derive tan(/2ⁿ).

We shall use the following two identities.  Since our angles all lie in the first quadrant (0 to /2), we will always take the positive square root.

We begin with the standard result that cos(/4) = /2.
Then, using (1), cos(/8) = [(1 + /2)/2] = ½.
Similarly, cos(/16) = ½.
From the algebraic manipulation, it's clear that each application of (1) will generate one more nested radical sign, and an additional prefix of "2 +".  (This can be proved more formally using mathematical induction, if required.)

Therefore cos(/2ⁿ) = ½, where there are n−1 twos under the nested radical signs.

Substituting the expression for cos(/2ⁿ⁻¹) into (2), we have:

tan(/2ⁿ) = , where there are n−1 twos under the nested radical signs.

Therefore the area of the 2n-gon	= 2−(n+2)/tan(/2n)
	= , where there are n−1 twos under both sets of nested radical signs.

---

Remarks

As n increases, we should expect the area of the regular 2ⁿ-gon to approach that of a circle with unit perimeter, that is 1/(4).
The table below, which shows the approximate area, A_n, of the regular 2ⁿ-gon for various values of n, and of the circle, illustrates this.

2ⁿ-gon areas

n	2n	An
2	4	0.0625
3	8	0.0754442
4	16	0.0785522
5	32	0.0793216
6	64	0.0795135
7	128	0.0795615
8	256	0.0795735
9	512	0.0795765
10	1024	0.0795772
	Circle	0.0795775

As n tends to infinity, A_n tends to 1/(4).

Furthermore, since cos(/2ⁿ) tends to 1 as n tends to infinity, it must be the case that a_n = tends to 2 as n, the number of twos under the nested radical signs, tends to infinity.

An independent, rigorous, proof of this result begins by showing that the limit of sequence {a_n}, as n tends to infinity, exists.  To do this we show that {a_n} is monotonic increasing and bounded above.

Firstly, we prove by mathematical induction that a_n < 2, for all n.
The basis is straightforward: a₁ = < 2.
For the induction step, note that, if a_k < 2, then a_k+1 = (2 + a_k) < 2.
Therefore a_n < 2, for all n.

Now consider a_n+1 = (2 + a_n).
Then a_n+1² = 2 + a_n.
So (a_n+1 + 1)(a_n+1 − 1) = a_n + 1.
Since for all n, a_n < 2, a_n+1 − 1 < 1, and so a_n+1 + 1 > a_n + 1.
Therefore a_n is monotonic increasing; in fact, it is strictly monotonic increasing.

Putting these two results together, by the Monotonic Convergence Theorem, {a_n} has a limit.

Having proved that the limit exists, we can calculate its value, L, from the quadratic equation L² = 2 + L.
The only positive solution is L = 2, and therefore this is the limit of {a_n}.

Viète's formula for pi

Having devised and solved this puzzle, I realised that, in the limit, the solution affords a formula for .  Of course, such a result must already be known, and indeed a little searching on the web turned up the closely related Viète's formula: number 64 in this list of pi formulas.  This was the first ever exact formula for , and was developed in 1593.

Source: Original (but anticipated by Viète by 410 years!)

Back to top