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Thread: Geometry(2)

  1. December 3rd 2009, 04:14 PM #1
    Drexel28
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    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?
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  2. December 4th 2009, 04:09 AM #2
    Unbeatable0
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    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_ks: 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_is and d_ks are equal in some order, the answers are

    <br /> \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}}<br />

    Is there a way to proceed with this expression?


    <br /> \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}}}<br />

    Explanation to the last step:

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

    <br /> \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})} = ... =<br />

    <br /> = \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<br />


    Therefore:

    <br /> \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 =<br />

    <br /> = 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}}}<br />


    And thus

    <br /> \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}}}<br />

    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.

    Last edited by Unbeatable0; December 4th 2009 at 09:00 AM.
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  3. December 4th 2009, 10:00 PM #3
    Drexel28
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    Quote Originally Posted by Unbeatable0 View Post
    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_ks: 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_is and d_ks are equal in some order, the answers are

    <br /> \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}}<br />

    Is there a way to proceed with this expression?


    <br /> \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}}}<br />

    Explanation to the last step:

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

    <br /> \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})} = ... =<br />

    <br /> = \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<br />


    Therefore:

    <br /> \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 =<br />

    <br /> = 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}}}<br />


    And thus

    <br /> \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}}}<br />

    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?
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  4. December 4th 2009, 10:09 PM #4
    simplependulum
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    Suppose it has a unit radius ( circumscribed circle)


    (1) : n

    (2) : 2\cot(\frac{\pi}{2n})
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  5. December 5th 2009, 04:27 AM #5
    Drexel28
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    Quote Originally Posted by simplependulum View Post
    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?
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  6. December 5th 2009, 05:04 AM #6
    simplependulum
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    Quote Originally Posted by Drexel28 View Post
    Can you give proof? Did you do an alternate proof to Unbeatable0's, or did you just apply his/her formulae?

    Oh , i just complete his answer to your questions :


    for (1)


    <br /> \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}}<br />

    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}))
    Last edited by simplependulum; December 5th 2009 at 05:21 AM.
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  7. December 5th 2009, 06:35 AM #7
    Unbeatable0
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    I have to say, simplependulum - your solution is very inspiring!

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

    <br /> \sum_{k=1}^{n} D_k = \frac{a}{2 \sin^2{\frac{\pi}{2n}}} = 2R\cot{\frac{\pi}{2n}}<br />

    <br /> \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}<br />

    Where R is the radius of the circumscribed circle.

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


    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?
    Last edited by Unbeatable0; December 5th 2009 at 09:14 AM.
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  8. December 7th 2009, 09:00 PM #8
    Drexel28
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    Quote Originally Posted by Unbeatable0 View Post
    I have to say, simplependulum - your solution is very inspiring!

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

    <br /> \sum_{k=1}^{n} D_k = \frac{a}{2 \sin^2{\frac{\pi}{2n}}} = 2R\cot{\frac{\pi}{2n}}<br />

    <br /> \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}<br />

    Where R is the radius of the circumscribed circle.

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


    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 nth 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.
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  9. December 27th 2009, 12:28 AM #9
    Unbeatable0
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    Quote Originally Posted by Drexel28 View Post
    For the product one (in the unit circle) you can think of the points Q_1,\cdots,Q_n as being the nth 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).
    Thanks in advance

    Too bad, I couldn't send you a private message because my post count is not above 15.
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