# Geometry(2)

• Dec 3rd 2009, 04:14 PM
Drexel28
Geometry(2)
Problem(1): Let $P_n$ be a regular $n$-gon with points $Q_1,\cdots,Q_n$. Fix one point $Q_i\quad 1\le i\le n$ and define $D_k$ as the distance from $Q_k$ to $Q_i$. Evaluate $\prod_{\stackrel{k=1}{\scriptstyle{k}\ne i}}^{n}D_k$

Problem(2): What about $\sum_{k=1}^{n}D_k$?
• Dec 4th 2009, 04:09 AM
Unbeatable0
Spoiler:

First of all, I'm sorry for just describing without a drawing - I'm having problems with that.
Let the $d_k (k=0,...,n-2)$ be a line segment connecting a point on the regular $n$-gon to another, which is k vertices far (by symmetry all of those are congruent). That is, there are k vertices between the two connected vertices. By drawing the circumscribed circle of the regular polygon, it can be seen that the angle between two line segments ending in two adjacent vertices and starting at another vertex is $\frac{\pi}{n}$. Draw two consecutive $d_k$s: $d_k$ and $d_{k+1}$ for some k. The angle between them is, as said, $\frac{\pi}{n}$. Also, the angle between $d_{k+1}$ and the side of the polygon inside the triangle formed is $\frac{(k+1)\pi}{n}$ or $\pi - \frac{(k+1)\pi}{n}$ (this too can be seen by drawing the circumscribed circle and counting equal arcs). In either case, from the sinus law in the triangle formed (by $d_k$, $d_{k+1}$ and a side of the polygon) we get $d_k= \frac{\sin{\frac{(k+1)\pi}{n}}}{\sin{\frac{\pi}{n} }} a$.
Because the $D_i$s and $d_k$s are equal in some order, the answers are

$
\prod_{\stackrel{k=1}{\scriptstyle{k}\ne i}}^{n}D_k = \prod_{k=0}^{n-2} d_k = \prod_{k=0}^{n-2} \frac{\sin{\frac{(k+1)\pi}{n}}}{\sin{\frac{\pi}{n} }} a = \frac{a^{n-1}}{\sin^{n-1}{\frac{\pi}{n}}} \prod_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}}
$

Is there a way to proceed with this expression?

$
\sum_{k=1}^{n} D_k = \sum_{k=0}^{n-2} d_k = \sum_{k=0}^{n-2} \frac{\sin{\frac{(k+1)\pi}{n}}}{\sin{\frac{\pi}{n} }} a = \frac{a}{\sin{\frac{\pi}{n}}} \sum_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}} = \frac{a}{2 \sin^2{\frac{\pi}{2n}}}
$

Explanation to the last step:

First we derive a formula for $\sum_{k=0}^{n} \sin{k\theta}$:

$
\sum_{k=0}^{n} (\cos{\theta} + i\sin{\theta})^k = \frac{1 - [\cos((n+1)\theta)+i\sin((n+1)\theta)]}{1-(\cos{\theta}+i\sin{\theta})} = ... =
$

$
= \frac{\sin{\frac{(n+1)\theta}{2}} \cos{\frac{n\theta}{2}}}{\sin{\frac{\theta}{2}}} + \frac{\sin{\frac{(n+1)\theta}{2}} \sin{\frac{n\theta}{2}}}{\sin{\frac{\theta}{2}}} i
$

Therefore:

$
\sum_{k=0}^{n} \sin{k\theta} = \sum_{k=0}^{n} Im(\cos{k\theta} + i\sin{k\theta}) = \sum_{k=0}^{n} Im(\cos{\theta} + i\sin{\theta})^k =
$

$
= Im \sum_{k=0}^{n} (\cos{\theta} + i\sin{\theta})^k = \frac{\sin{\frac{(n+1)\theta}{2}} \sin{\frac{n\theta}{2}}}{\sin{\frac{\theta}{2}}}
$

And thus

$
\sum_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}} = \sum_{k=0}^{n-1} \sin{\frac{k\pi}{n}} = \frac{\sin{\frac{(n+1)\pi}{2n}} \sin{\frac{\pi}{2}}}{\sin{\frac{\pi}{2n}}} = \frac{\cos{\frac{\pi}{2n}}}{\sin{\frac{\pi}{2n}}}
$

Subtituting in $\frac{a}{\sin{\frac{\pi}{n}}} \sum_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}}$ and using $\sin{\frac{\pi}{n}} = 2\sin{\frac{\pi}{2n}}\cos{\frac{\pi}{2n}}$ yields the last expression for the sum of distances above.

• Dec 4th 2009, 10:00 PM
Drexel28
Quote:

Originally Posted by Unbeatable0
Spoiler:

First of all, I'm sorry for just describing without a drawing - I'm having problems with that.
Let the $d_k (k=0,...,n-2)$ be a line segment connecting a point on the regular $n$-gon to another, which is k vertices far (by symmetry all of those are congruent). That is, there are k vertices between the two connected vertices. By drawing the circumscribed circle of the regular polygon, it can be seen that the angle between two line segments ending in two adjacent vertices and starting at another vertex is $\frac{\pi}{n}$. Draw two consecutive $d_k$s: $d_k$ and $d_{k+1}$ for some k. The angle between them is, as said, $\frac{\pi}{n}$. Also, the angle between $d_{k+1}$ and the side of the polygon inside the triangle formed is $\frac{(k+1)\pi}{n}$ or $\pi - \frac{(k+1)\pi}{n}$ (this too can be seen by drawing the circumscribed circle and counting equal arcs). In either case, from the sinus law in the triangle formed (by $d_k$, $d_{k+1}$ and a side of the polygon) we get $d_k= \frac{\sin{\frac{(k+1)\pi}{n}}}{\sin{\frac{\pi}{n} }} a$.
Because the $D_i$s and $d_k$s are equal in some order, the answers are

$
\prod_{\stackrel{k=1}{\scriptstyle{k}\ne i}}^{n}D_k = \prod_{k=0}^{n-2} d_k = \prod_{k=0}^{n-2} \frac{\sin{\frac{(k+1)\pi}{n}}}{\sin{\frac{\pi}{n} }} a = \frac{a^{n-1}}{\sin^{n-1}{\frac{\pi}{n}}} \prod_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}}
$

Is there a way to proceed with this expression?

$
\sum_{k=1}^{n} D_k = \sum_{k=0}^{n-2} d_k = \sum_{k=0}^{n-2} \frac{\sin{\frac{(k+1)\pi}{n}}}{\sin{\frac{\pi}{n} }} a = \frac{a}{\sin{\frac{\pi}{n}}} \sum_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}} = \frac{a}{2 \sin^2{\frac{\pi}{2n}}}
$

Explanation to the last step:

First we derive a formula for $\sum_{k=0}^{n} \sin{k\theta}$:

$
\sum_{k=0}^{n} (\cos{\theta} + i\sin{\theta})^k = \frac{1 - [\cos((n+1)\theta)+i\sin((n+1)\theta)]}{1-(\cos{\theta}+i\sin{\theta})} = ... =
$

$
= \frac{\sin{\frac{(n+1)\theta}{2}} \cos{\frac{n\theta}{2}}}{\sin{\frac{\theta}{2}}} + \frac{\sin{\frac{(n+1)\theta}{2}} \sin{\frac{n\theta}{2}}}{\sin{\frac{\theta}{2}}} i
$

Therefore:

$
\sum_{k=0}^{n} \sin{k\theta} = \sum_{k=0}^{n} Im(\cos{k\theta} + i\sin{k\theta}) = \sum_{k=0}^{n} Im(\cos{\theta} + i\sin{\theta})^k =
$

$
= Im \sum_{k=0}^{n} (\cos{\theta} + i\sin{\theta})^k = \frac{\sin{\frac{(n+1)\theta}{2}} \sin{\frac{n\theta}{2}}}{\sin{\frac{\theta}{2}}}
$

And thus

$
\sum_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}} = \sum_{k=0}^{n-1} \sin{\frac{k\pi}{n}} = \frac{\sin{\frac{(n+1)\pi}{2n}} \sin{\frac{\pi}{2}}}{\sin{\frac{\pi}{2n}}} = \frac{\cos{\frac{\pi}{2n}}}{\sin{\frac{\pi}{2n}}}
$

Subtituting in $\frac{a}{\sin{\frac{\pi}{n}}} \sum_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}}$ and using $\sin{\frac{\pi}{n}} = 2\sin{\frac{\pi}{2n}}\cos{\frac{\pi}{2n}}$ yields the last expression for the sum of distances above.

You have the same answer as I do for number one in the general case. What about when the polygon is inscribed in the unit circle?
• Dec 4th 2009, 10:09 PM
simplependulum
Suppose it has a unit radius ( circumscribed circle)

(1) : $n$

(2) : $2\cot(\frac{\pi}{2n})$
• Dec 5th 2009, 04:27 AM
Drexel28
Quote:

Originally Posted by simplependulum
Suppose it has a unit radius ( circumscribed circle)

(1) : $n$

(2) : $2\cot(\frac{\pi}{2n})$

Can you give proof? Did you do an alternate proof to Unbeatable0's, or did you just apply his/her formulae?
• Dec 5th 2009, 05:04 AM
simplependulum
Quote:

Originally Posted by Drexel28
Can you give proof? Did you do an alternate proof to Unbeatable0's, or did you just apply his/her formulae?

for (1)

$
\prod_{\stackrel{k=1}{\scriptstyle{k}\ne i}}^{n}D_k = \prod_{k=0}^{n-2} d_k = \prod_{k=0}^{n-2} \frac{\sin{\frac{(k+1)\pi}{n}}}{\sin{\frac{\pi}{n} }} a = \frac{a^{n-1}}{\sin^{n-1}{\frac{\pi}{n}}} \prod_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}}
$

note that

$\frac{a}{\sin(\frac{\pi}{n})}$ means the diameter of the polygon , as i set , its length is two .

that means

$2^{n-1} \prod_{k=0}^{n-2} \sin{\frac{(k+1)\pi}{n}} = 2^{n-1} \prod_{k=1}^{n-1} \sin{\frac{k\pi}{n}} = n$

Starting from

$\frac{ x^{2n} - 1}{x^2 - 1} = \prod_{k=1}^{n-1}( x^2 - 2\cos(\frac{k\pi}{n})x + 1 )$

By substituting $x = 1$ and $x = -1$
(takes limit )

$n = \prod_{k=1}^{n-1}( 4\sin^2(\frac{k\pi}{2n}))$

$n = \prod_{k=1}^{n-1}( 4 \cos^2(\frac{k\pi}{2n}))$

their porduct gives

$n^2 = \prod_{k=1}^{n-1}( 4 \cdot 4\sin^2(\frac{k\pi}{2n})\cos^2(\frac{k\pi}{2n})) = \prod_{k=1}^{n-1} (4\sin^2(\frac{k\pi}{n}))$

$n = 2^{n-1} \prod_{k=1}^{n-1} (\sin(\frac{k\pi}{n}))$
• Dec 5th 2009, 06:35 AM
Unbeatable0
I have to say, simplependulum - your solution is very inspiring! (Clapping)

So to sum it up, in the general case we have:

$
\sum_{k=1}^{n} D_k = \frac{a}{2 \sin^2{\frac{\pi}{2n}}} = 2R\cot{\frac{\pi}{2n}}
$

$
\prod_{\stackrel{k=1}{\scriptstyle{k}\ne i}}^{n} D_k = n \left( \frac{a}{2\sin{\frac{\pi}{n}}} \right)^{n-1} = n R^{n-1}
$

Where $R$ is the radius of the circumscribed circle.

Thank you Drexel28 for this interesting problem with surprisingly simple solution formulas. (Nod)

Edit:
After noticing that the last expression for the product of the distances does not include trigonometry, an interesting question came into my head: can you find this formula without the use of trigonometry?
• Dec 7th 2009, 09:00 PM
Drexel28
Quote:

Originally Posted by Unbeatable0
I have to say, simplependulum - your solution is very inspiring! (Clapping)

So to sum it up, in the general case we have:

$
\sum_{k=1}^{n} D_k = \frac{a}{2 \sin^2{\frac{\pi}{2n}}} = 2R\cot{\frac{\pi}{2n}}
$

$
\prod_{\stackrel{k=1}{\scriptstyle{k}\ne i}}^{n} D_k = n \left( \frac{a}{2\sin{\frac{\pi}{n}}} \right)^{n-1} = n R^{n-1}
$

Where $R$ is the radius of the circumscribed circle.

Thank you Drexel28 for this interesting problem with surprisingly simple solution formulas. (Nod)

Edit:
After noticing that the last expression for the product of the distances does not include trigonometry, an interesting question came into my head: can you find this formula without the use of trigonometry?

For the product one (in the unit circle) you can think of the points $Q_1,\cdots,Q_n$ as being the $n$th roots of unity $1,\cdots,e^{\frac{(n-1)\pi}{n}}$. That's how I did it. I'll post up a full solution later...if I remember. I did the general case similar to you.
• Dec 27th 2009, 12:28 AM
Unbeatable0
Quote:

Originally Posted by Drexel28
For the product one (in the unit circle) you can think of the points $Q_1,\cdots,Q_n$ as being the $n$th roots of unity $1,\cdots,e^{\frac{(n-1)\pi}{n}}$. That's how I did it. I'll post up a full solution later...if I remember. I did the general case similar to you.

I'm interested to see the proof you referred to.
Will you please post the proof? Or at least the idea of the proof (using roots of unity does not lead me too far).