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Let $A_1,A_2,...,A_n$ be the vertice of an regular polygon inscribed in a unit circle, $P$ is a point on the circle.

What is the maximum and the minimum values of $\sum_{k=1}^n|PA_k|$?

My idea : Maybe we can define $\omega=e^{\frac{2\pi i}n}$, $A_k=\omega^k$ and $P=e^{i\theta}$, then

$$\sum_{k=1}^n|PA_k|=\sum_{k=1}^n|e^{i\theta}-e^{\frac{2k\pi i}n}|=\sum_{k=1}^ne^{i(-\arg (P-A_k))}(P-A_k)=...$$

In this way, perhaps we can write the final result in the form of trigonometric functions, so that it is convenient to find its maximum and minimum values. But I failed to expand and simplify it, can anyone help me?

Starlight237
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  • Because the distance function is symmetric it is enough to consider the case when $P$ moves between $A_1$ and $A_2$. Further on this arch the function is again symmetric with respect to "the middle point", say $A_{1,2}$. So, max and min can be achieved at $A_1,A_2$ (which are the same) or at the middle point. Now one has to compute these values and decide which one is bigger. – Salcio Jul 16 '22 at 14:19
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    @Sacio How to prove that it can take an extreme value on the midpoint? It's not an obvious fact, because we don't know anything about the properties of this function. – Starlight237 Jul 16 '22 at 14:25
  • Right, it is symmetric with respect to the middle point, one has to show that it is monotonic on the part which may not be obvious ... . – Salcio Jul 16 '22 at 14:28

2 Answers2

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Using simple trigonometry, we have

$$S(\theta)=\sum_{k=1}^n|PA_k|=2\sum_{k=0}^{n-1}\sin{\frac{\theta+\frac{2k\pi}{n}}{2}}, \text{where } 0\leq\theta\leq\frac{\pi}{n}.$$

Then

$$\frac{\text{d}S}{\text{d}\theta}=\sum_{k=0}^{n-1}\cos{\frac{\theta+\frac{2k\pi}{n}}{2}}$$

Based on symmetry in the unit circle,

$$\theta=\dfrac{\pi}{n}\implies \frac{\text{d}S}{\text{d}\theta}=\sum_{k=0}^{n-1}\cos{\left((1+2k)\frac{\pi}{2n}\right)}=0$$

As $\theta$ decreases from $\dfrac{\pi}{n}$ to $0$, every term in the series $\dfrac{\text{d}S}{\text{d}\theta}$ increases. From this it follows that $S_\text{min}=S(0)$ and $S_\text{max}=S\left(\dfrac{\pi}{n}\right).$

We know that the mean distance between two points on the unit circle is $\dfrac{4}{\pi}$. From this it follows that as $n\to\infty$, both $S_\text{min}$ and $S_\text{max}$ approach $\dfrac{4n}{\pi}$.

Dan
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EDIT1:

enter image description here

If centered at origin, the vector sum of all polar symmetric vectors or forces $AP$ ( inclined at same $\pi/k$ to each next/consecutive vector) vanishes. So,

$$=\sum_{k=1}^n|OA_k|+\sum_{k=1}^n| A_k P|== n|OA| +0 $$

where $\vec {OA}$ is the constant eccentric vector. Since polygon vectors sum up to zero (no matter radius length), the summation to $A$ or $P$ is same. The former constant eccentric vector is summed $n$ times.

Narasimham
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  • In the last expression, $k.|OP|$, what does $k$ mean? $k$ is the index of summation in the previously written series; $k$ has no meaning without the series. – Dan Jul 16 '22 at 22:12
  • Thanks, that should be $n$, number of vertices of the polygon. – Narasimham Jul 17 '22 at 00:58
  • Then $\sum_{k=1}^n|PA_k|=n|OP|=n$, which is independent of the location of $P$. But $\sum_{k=1}^n|PA_k|$ depends on the location of $P$ (this is easy to see when $n=2$). – Dan Jul 17 '22 at 01:04
  • Please see graphic edit. – Narasimham Jul 17 '22 at 16:55