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ganeshie8

  • one year ago

A triangle is formed by picking \(3\) points at random from the vertices of a regular \(2n+1\)-gon. Find the probability for it to be a right triangle.

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  1. anonymous
    • one year ago
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    There doesn't seem to be any such right triangle for an equilateral triangle. In that case, the probability should be zero. And it seems to me that it should be zero for the other 2n+1 -gons too unless I'm not understanding something.

  2. Empty
    • one year ago
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    Yeah I like that answer @adxpoi I'm going to sleep so maybe this will be pointed out to either be wrong or be correct when I check this tomorrow morning.

  3. amilapsn
    • one year ago
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    n\(\in \mathbb{Z}^+\) right?

  4. ganeshie8
    • one year ago
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    Yes \(n\) is positive integer Indeed it is easy to guess that the probability is \(0\) for any \(2n+1\)-gon as the equilateral triangle case gives \(0\)... However I feel this is only a conjecture, requires proof..

  5. ganeshie8
    • one year ago
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    \(n=1\) case is an equilateral triangle, the probability is \(0\) what about other cases

  6. ganeshie8
    • one year ago
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    how do you know the probability is constant, it could be some weird function of \(n\) right ?

  7. anonymous
    • one year ago
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    Consider the circumcircle of the regular \(2n+1\) gon . If three vertices form a right triangle, then the side opposite to the right angle must be a diameter. But in a regular \(2n+1\)- gon , if we draw the diameter through a vertex , it bisect the opposite side and the other end of the diameter is not a vertex of the polygon. So the scenario described above is impossible. Now to see why this happens. Take any vertex, A and draw the diameter(say, AB) through it. Since it's a regular polygon there must be equal number of vertices on each side of the diameter, say k. So, we have accounted for 2k+1 points (including A) . If B is also a vertex then we would have 2k+2 vertices in all for the polygon, impossible since it's a \(2n+1\) gon. [I'll post a more rigorous argument later]

  8. amilapsn
    • one year ago
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    I got a result that any internal angle of any triangle formed using three vertices of a regular (2n+1)-gon = \(\Huge\frac{\pi}{\frac{2n+1}{k}}\) where \(k\leq 2n-1\). We can prove that it cannot be equal to a right angle.

  9. ganeshie8
    • one year ago
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    I think that proof works pretty nicely @adxpoi ! In short, if an angle is \(90\) degrees, then its arc measure is \(180^{\circ}\) from inscribed angle theorem. Since \(k*\dfrac{360}{2n+1}=180\) has no solutions for all pairs of positive integers \((k,n)\), there cannot be any right triangles...

  10. ganeshie8
    • one year ago
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    @amilapsn exactly! I think we can use that equation to show that there cannot be any equilateral triangles in a \(2n+1\)-gon too

  11. amilapsn
    • one year ago
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    @ganeshie8 except for n=1

  12. ganeshie8
    • one year ago
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    oops! right :)

  13. amilapsn
    • one year ago
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    :D

  14. anonymous
    • one year ago
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    @ganeshie8 , ah well, that is one neat way to say it :)

  15. ganeshie8
    • one year ago
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    omg! i take back my earlier statement about equilateral triangles, it is super wrong... equilateral triangles are possible for all \(2n+1\)-gons such that \(n\equiv 1\pmod{3}\). Because then we can split the polygon into \(3\) equal segments : |dw:1439719371233:dw|

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