# 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, An, of the regular 2n-gon for various values of n, and of the circle, illustrates this.

2n-gon areas
n2nAn
240.0625
380.0754442
4160.0785522
5320.0793216
6640.0795135
71280.0795615
82560.0795735
95120.0795765
1010240.0795772
Circle0.0795775

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

Furthermore, since cos(/2n) tends to 1 as n tends to infinity, it must be the case that an = 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 {an}, as n tends to infinity, exists.  To do this we show that {an} is monotonic increasing and bounded above.

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

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

Putting these two results together, by the Monotonic Convergence Theorem, {an} 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 {an}.

### 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!)