# The area is greater than that of the original polygon - Edit

In the book Methods of mathematical physics - Volume $1$ by Courant-Hilbert page $166$, it is indicated :

Among all polygons which are not self-intersection and have an even number $2n$ sides and a given perimeter $2l$ find the one with the greatest area. The desired polygon $\Pi(A_1, ...., A_{2n})$ is the regular $2n$-gon.To prove this we first convince ourselves that $\Pi$ is convex. If $\Pi$ were not convex, a straight line lying entirely outside $\Pi$ could be drawn through two non-adjacent vertices, say $A_k$ and $A_l$; we could then reflect the polygonal sequence $A_k, A_{k+1}, \dots, A_{l-1}, A_l$ in this straight line and obtain a polygon with the same perimeter and with an area greater than that of the original polygon by twice the area of the triangle $A_1 A_2 A_3$.

I think this citation wants to transform a concave $2n$-gon into a convex $2n$-gon having the same perimeter, but a different area. Could anyone be able to explain to me the bold part of that citation?

Edit :

The transformation $r$ on that example do not preserve the dimension of $A$, and it is to highlight that shape is concave. I know it is not the case here, but why $A$ couldn't be the one with the greater area among all $6$-gon?

• What is the full question? – Batominovski Aug 6 '16 at 19:54
• @Batominovski I'll edit my question, and you could see the whole question. – CHSLD Aug 6 '16 at 19:59
• I guess the proof of the isoperimetric inequality through Steiner symmetrization method. This is quite a trivial question: the OP just needs to draw a sketch ;) – Jack D'Aurizio Aug 6 '16 at 19:59
• @JackD'Aurizio This is the thing I've done so far. I know we want to maximize the area, and to do that we need at least a convex polygon. However, the bold part is not quite understandable so far (why necessarily with the triangle $A_1 A_2 A_3$). – CHSLD Aug 6 '16 at 20:03
• The point is just that the area gets strictly increased by the reflection process. The amount of this increase is at least twice the area of the smallest triangle we may form with three vertices from $\{A_1,\ldots,A_n\}$. – Jack D'Aurizio Aug 6 '16 at 20:11

The situation below could have been meant:

(Large image versions here and here)

To prove this we first convince ourselves that Π is convex. If Π were not convex, a straight line lying entirely outside Π could be drawn through two non-adjacent vertices,

See the line $m$.

say Ak and Al;

Above we have vertices $A$ and $C$.

we could then reflect the polygonal sequence Ak,Ak+1,…,Al−1,Al in this straight line

Above vertex $B$ is reflected into $B'$ along $m$.

and obtain a polygon with the same perimeter and with an area greater than that of the original polygon by twice the area of the triangle A1A2A3.

Above we have twice the area of triangle $ABC$.

Note the perimeters of the left and right polygon are the same, the area increased.