My question: Let $n$ points $A_1, A_2,\ldots,A_n$ lie on given circle then show that $\operatorname{Area}(A_1A_2\cdots A_n)$ maximum when $A_1A_2\cdots A_n$ is an $n$-regular polygon.
Asked
Active
Viewed 1,616 times
3 Answers
3
My answer: Show that if the polygon is not regular, then you can get a larger area by moving a sigle vertex along the circle.
Arthur
- 199,419
-
Thank you dear @Arthur, But, show that area of $(A_1A_2...A_n)$ maximum when $A_1A_2...A_n$ is regular polygon? – Oai Thanh Đào Sep 08 '16 at 16:33
-
@OaiThanhĐào This answer shows that for any irregular polygon there is always a larger polygon inscribed in the same circle. But no polygon can be larger than the maximal polygon. Hence the maximal polygon can only be the regular one. In fact this is the same idea as the answer you accepted, except that this answer does not contain incorrect statements. – David K Sep 08 '16 at 18:19
-
@DavidK It could be argued that one also needs to show that a maximal polygon does in fact exist. Depending on context and assumptions that may or may not be entirely obvious (for example, there is no non-degenerate polygon of minimal area). – dxiv Sep 08 '16 at 18:28
-
@dxiv Good point, although for the minimum it is easy to establish that if the set of cyclical polygons is allowed to include degenerate polygons then those degenerate polygons are minimal (provided that our definition of "area" doesn't allow negative values). – David K Sep 08 '16 at 18:35
3
Let's call the center of the circle $O$.
First, it's clear that none of the angles $\angle A_1OA_2$, $\angle A_2OA_3$,…,$\angle A_nOA_1$ should be greater than $\pi$. If they were, the polygon would be smaller then half of circle $O$. Obviously, this is not maximal.
Let's call $m\angle A_1OA_2=\theta_1$, $m\angle A_2OA_3=\theta_2$,…, $m\angle A_1OA_n=\theta_n$.
The area of the polygon is: $$\frac{r^2}{2}\sum_{i=1}^n \sin\left(\theta_i\right)$$ Note that $\sin(x)$ is concave on $[0,\pi]$. From Jensen's inequality: $$\sin\left(\frac{1}{n}\sum_{i=1}^n \theta_i\right)\geq \frac{1}{n}\sum_{i=1}^n \sin\left(\theta_i\right)$$ We know that $\displaystyle \sum_{i=1}^n \theta_i=2\pi$ so we can deduce the following: $$n\sin\left(\frac{2\pi}{n}\right)\geq \sum_{i=1}^n \sin\left(\theta_i\right)$$
This means that: $$\frac{r^2}{2}\sum_{i=1}^n \sin\left(\theta_i\right)\leq \frac{nr^2}{2}\sin\left(\frac{2\pi}{n}\right)$$
This suggests that the area of the polygon is maximized when $\theta_1, \theta_2, \ldots, \theta_n = \displaystyle\frac{2\pi}{n}$.
Hrhm
- 3,303
2
This answer is different from my previous answer, and maybe even better.
If our polygon is not regular, we can find $A_{i-1}$, $A_i$, and $A_{i+1}$, such that $A_i$ is not the midpoint of the arc between $A_{i-1}$ and $A_{i+1}$. Notice that when we move $A_i$ to the midpoint of this arc, the area of the polygon becomes bigger, since the area of $\triangle A_{i-1} A_i A_{i+1}$ increases. Ergo, the maximal polygon is the polygon for which $A_i$ is the midpoint of arc $A_{i-1}A_{i+1}$ for all $i$.
Hrhm
- 3,303
-
Haha, that was the argument I wanted to give before getting lost into the details. I goofed. – zarathustra Sep 08 '16 at 19:01
-
Ergo, the maximal polygon is ...IMHO this needs more elaborating (just like @Arthur's hint). Moving one vertex at a time to the middle position does indeed increase the area at each step, but it does not (generally) lead to the regular polygon in a finite number of steps. Depending on how the next vertex to be moved is chosen, the process may not even converge to the regular polygon. To complete the proof, one would need to show that there is a choice of steps that converges to the regular polygon, and that choice yields the largest area among all possible such sequences. – dxiv Sep 08 '16 at 22:11 -
@dxiv: why? The proof need not be constructive, it just shows that if the polygon is not regular the area is not maximal. So assuming there is indeed a polygon with maximal area, it needs to be regular. Right? – zarathustra Sep 09 '16 at 07:32
-
@zarathustra
assuming there is indeed a polygon with maximal areaThat's precisely where I see the difficulty along this line of argumentation. You need to prove rather than assume that there is such a maximal-area polygon, and that it's unique. Don't get me wrong, I believe that the idea is intuitive and fundamentally sound, but formalizing it into an actual proof requires more than anergo. – dxiv Sep 09 '16 at 07:42 -
@zarathustra--If every inscribed n-gon is either regular or irregular, and the regular n-gon has greater area than any irregular n-gon, and the area of the regular n-gon is unique for a given circle and given n, then doesn't it follow that there is an n-gon with maximum area, and it is the regular n-gon? – Edward Porcella Apr 22 '17 at 15:51
I wonder that: The journals AMM and Crux also proposed problem is the same way above? But why I can not posted my problem at here in the same way???
– Oai Thanh Đào Sep 13 '16 at 03:42