The maximum area of a pentagon inside a circle

I thought of this problem, which is nothing new, but searching over here and elsewhere I found nothing explicit or satisfactory. So I decided to try to solve it by my own, but I am not fully convinced my work is ok. So I want you to tell me where I am wrong and how to proceed.

I considered a priori a regular pentagon, hence the area is $$A = p\cdot a/2$$ where $$p = 5\ell$$ is the perimeter and $$\ell$$ is the side, and $$a$$ is the apothem. Now by drawing a regular pentagon, the inscribed circle and the circle inside which the pentagon is inscribed (which is $$x^2+y^2 = R^2$$), I can write

$$a = \sqrt{R^2 - \frac{\ell^2}{4}} = \frac{1}{2}\sqrt{4R^2 - \ell^2}$$

Hence the area of the pentagon is $$A = \frac{5}{4}\ell \sqrt{4R^2 - \ell^2}$$

At this point I thought of Lagrange multipliers:

$$L(R, \ell, \lambda) = \frac{5}{4}\ell\sqrt{4R^2 - \ell^2} - \lambda(x^2+y^2-R^2)$$

I thought to treat $$x, y$$ as dummy indexes since they don't tell me anything: what matters is the radius $$R$$ of the exterior circle. And indeed in doing this, setting $$\nabla L = 0$$ I obtain two equations. After arranging:

$$\begin{cases}4R^2 - 2\ell^2 = 0 \\\\ 10\ell R + 4\lambda R \sqrt{4R^2 - \ell^2} = 0 \end{cases}$$

The first one tells me $$\ell = \sqrt{2}R$$ and the second one only tells me $$\lambda = 5/2$$, which is irrelevant.

I indeed obtained the fact that given a random circle of radius $$R$$, the regular pentagon inscribed in it whose area is the maximum possible needs to have a side of $$\sqrt{2}R$$.

This seems plausible, but I am in doubt if this is correct at all.

EDIT

I would like to proceed without using angles, sine and cosine, if possible.

• But the side of a regular pentagon inscribed in a circle of radius $R$ is $$\ell=2R\sin36°={\sqrt{10-2\sqrt5}\over2}R$$ Commented Jul 19 at 9:23
• @Intelligentipauca A difference of $0.23\sim$, cool! Anyway, thank you for this, didn't thought to check. But at this point, what went wrong in my approach? I didn't want to use angles, sine and cosine.
– J.N.
Commented Jul 19 at 9:26
• What you found is the maximum of $f(\ell)={1\over4}\ell\sqrt{4R^2-\ell^2}$, where $\ell$ is the lenght of a chord ($0<\ell<2R$) and $f(\ell)$ is the area of the triangle. It's true you multiplied that by $5$, but that's irrelevant as far as the value of $\ell$ at maximum is concerned. Commented Jul 19 at 15:29
• @Intelligentipauca Ah, thank you! Really didn't think in this optics. Well, then the next question, hoping I don't stress you too much, is: how would we solve the max problem explicitly with Lagrange multipliers? If you believe I should write a new post, tell me!
– J.N.
Commented Jul 19 at 15:37
• Yes, I think you'd better ask a new question, exactly stating what you need. Commented Jul 19 at 15:41

Consider any two consecutive sides $$AB$$, $$BC$$ of a convex polygon inscribed in a given circle. Suppose to keep all vertices fixed apart $$B$$: the area of the polygon is maximum when $$B$$ is the midpoint of arc $$AC$$ (distance from $$B$$ to fixed base $$AC$$ is maximum).

But that implies $$AB=BC$$, hence the polygon with maximum area must have any two consecutive sides equal between them, i.e. it must be a regular polygon.

• You still have to prove that a maximum exists. Commented Jul 19 at 13:39
• @trula $(S^1)^5$ (or $(D^2)^5$ if you don't assume a priori that the vertices are on the circle but not outside) is compact and the area depends continuously on the five vertices. Commented Jul 19 at 19:21
• @trula: this answer shows clearly that any pentagon that is not regular must have a smaller area than the regular pentagon. Therefore, trivially, the regular pentagon has the maximum possible area. Commented Jul 19 at 23:46
• It is clear , that you can augment every natural number except 1 by taking its square, So it follows, that 1 is the largest number since its square is not larger. Commented Jul 20 at 14:00
• @IntelligentiPauca what software you use to make this geometry image? Commented Jul 20 at 15:45

Let's look at general (convex) $$n$$-sided polygons inscribed in a unit circle. Join the centre of the circle to each vertex of the polygon to form $$n$$ isosceles triangles. Say their apex angles are $$\theta_1,\theta_2,\ldots,\theta_n$$.

We know that $$\sum \theta_i = 2\pi$$. The area of the $$i\text{th}$$ triangle is $$\frac12 \sin \theta_i$$, so the area of the whole polygon is $$P=\frac12 \sum \sin \theta_i$$

So we want to maximise $$\sum \sin \theta_i$$ subject to $$\sum \theta_i = 2\pi$$, which we can indeed do with a Lagrange multiplier. Let $$F\left(\theta_1,\ldots,\theta_n,\lambda\right)=\sum \sin \theta_i+\lambda\left(\sum \theta_i - 2\pi\right)$$

Then $$\frac{\partial F}{\partial \theta_i}=\cos \theta_i+\lambda$$

Setting these to zero, $$\cos \theta_i=-\lambda$$ for every $$i$$. Without solving the system we can see that this implies all $$\theta_i$$ are equal; ie the maximum is achieved by a regular polygon.

• Thank you! Is there a way to proceed that doesn't involve angles, sine and cosine functions?
– J.N.
Commented Jul 19 at 9:26
• +1. But it suffices to apply the Jensen's inequality for the maximization of the function $F$.
– NN2
Commented Jul 19 at 10:05