Question about symmetry

Question

What is the area of the largest trapezoid that can be inscribed in a semi-circle with radius $r=1$?

Here in @Hagen von Eitzen answer He mentioned: "By symmetry alone the largest hexagon that can be inscribed in a circle is regular hexagon."

What is the meaning of symmetry here? and how to prove the statement that largest n-gon inscribed in a circle is regular n-gon. I never saw this method before.So I am looking for some references or similar geometry problems like this (for example I think the problem that asks for largest rectangle that can be inscribed in a quarter circle has something to do with symmetry) but I couldn't find anything can you send some links for that? Thank you.

Answer 1

Let's ignore the symmetry argument and start as a minimization/maximization problem. Your n-gon inscribed in a circle of radius $1$ has the angles at the center $2\alpha_i$. Then the length of the side is $2\sin\alpha_i$ and the length of the perpendicular to the side is $\cos\alpha_i$. Then the area of the n_gon is $$A=\sum_{i=1}^n\sin\alpha_i\cos\alpha_i,$$ subject to the constraint $$\sum_{i=1}^n\alpha_i=\pi$$ We rewrite $$A=\frac12 \sum_{i=1}^n\sin(2\alpha_i)$$ Using Lagrange multiplier method:$$\frac{d}{d\alpha_i}\left(\frac12 \sum_{i=1}^n\sin(2\alpha_i)-\lambda(\sum_{i=1}^n\alpha_i-\pi)\right)=0$$ This results in $$\cos(2\alpha_i)=\lambda$$ So all $\alpha_i$ are the same, therefore you have a regular n-gon.

Answer 2

I suppose, theoretically, Hagen von Eitzen didn't prove that the largest inscribed $n$-gon is regular, but that for every irregular $n$-gon the is a larger $n$-gon. Thus if there is a maximal $n$-gon it must be the regular one. It is, I suppose, hypothetically possible that no $n$-gon is maximal. Hagen von Eitzen gave no argument that there is maximal $n$-gon.

The argument that for every irregular $n$-gon there is a larger $n$-gon is like this.

If an $n$-gon is irregular not all the sides are equal and there will be two that unequal and there will be a pair of adjacent unequal sides. Let this pair of unequal sides be $AB$ between vertices $A,B$ and $BC$ between vertices $A,C$.

These three vertices form a triangle with base $AC$ and the area of this triangle will be $\frac 12 AB\cdot h$ where $h$ is the height of the perpendicular from $B$ to $AC$. Vertices $A$ and $C$ are connected by a circular arch and $B$ is a point on that arch.

If we choose a different point other than $B$ on the arch, say $B'$ then the height of the perpendicular from $B'$ to $AC$ may be a different value and the triangle $\triangle AB'C$ will have a different area than $\triangle ABC$.

The point on the arch that has the greatest perpendicular value is the point that is on the perpendicular bisector of $AC$, let's call it $D$. As $D$ is on the perpendicular bisector $AD=DC$ and as $AB\ne BC$ we know $B \ne D$ and that the height of the perpendicular from $B$ to $AC$ is less than the height of the perpendicular for $D$ to $AC$. And therefore the area of $\triangle ABC$ is less than the area of triangle $\triangle ADC$.

ANd so the area of the $n$-gon with vertices $A,B,C$ (and $n-3$ other vertices) will be smaller than the $n$-gon with the same $n-1$ verices but only $B$ replaced with $D$.

So our first inscribe $n$-gon is smaller than another existing inscribed $n$-gon.

Answer 3

Let's consider a hexagon inscribed in circle. Let's divide the area of hexagon into 6 triangles - with sides as two radii and a side of hexagon. Note that each SIDE of the hexagon has identical contribution to the area. So, to maximise area the best you can do is keep all of them equal.