ganeshie8
  • ganeshie8
show that sum of the vectors drawn from center of a regular polygon to its vertices is 0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sdfgsdfgs
  • sdfgsdfgs
for regular polygon w even no. of vertices, it is easy to show the sum of vecto5rs from center to vertices is zero by symmetry... it can be proven by symmetry as well for odd no. of vertices but it is less obvious....hmmmm.....
dan815
  • dan815
all the vector sides of an n sided regular polygon centered around the origin can be rewritten was r*e^(2pi*k/n) k is an int from 1 to n and r is the length of each side by this definition we are producing vector from the fact that they are summing to 0
ganeshie8
  • ganeshie8
Consider an octagon for a concrete example with even number of vertices, |dw:1436576807411:dw| How do we use symmetry ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dan815
  • dan815
|dw:1436576906626:dw|
ganeshie8
  • ganeshie8
@dan815 do you mean r*e^(\(\color{red}{i}\)2pi*k/n) ?
dan815
  • dan815
oh yes sry
ganeshie8
  • ganeshie8
so based on that i think the problem translates to proving \[\large \sum\limits_{k=0}^{n-1} e^{i2\pi k/n} = 0\]
dan815
  • dan815
yeah
sdfgsdfgs
  • sdfgsdfgs
|dw:1436577086602:dw|
dan815
  • dan815
what if its odd careful there
ganeshie8
  • ganeshie8
Ahh nice, so we always have opposite vectors for polygons with even number of vertices
sdfgsdfgs
  • sdfgsdfgs
@dan815 it can still be proven by symmetry for odd no. but its less obvious...
dan815
  • dan815
adding vectors to vectors with the same angle of separation between them and the angles add to 360 has to equal 0 another way to say that complex expression in words
geerky42
  • geerky42
*
dan815
  • dan815
or also like the outer angles of all regular polygon = 360
dan815
  • dan815
their sum
dan815
  • dan815
|dw:1436577429209:dw|
sdfgsdfgs
  • sdfgsdfgs
|dw:1436577289096:dw|
ganeshie8
  • ganeshie8
Building on dan's method using complex numbers, the vectors can be viewed as solutions to the equation \[\large x^n = r^n\] Clearly the sum of roots of the polynomial \(x^n-r^n\) is \(0\) so i think we're done ?
sdfgsdfgs
  • sdfgsdfgs
@ganeshie8 @dan815 nicely done! :)
dan815
  • dan815
you know kai actually found that equation randomly to generate any n sided regular polygon
ganeshie8
  • ganeshie8
yeah earlier i almost forgot we could use complex numbers here, that equation is same as the equation in vectors because we can treat complex numbers literally as 2D vectors
dan815
  • dan815
right
ganeshie8
  • ganeshie8
ofcourse wid some extra algebra as we cant multiply vectors like we multiply complex numbers
ganeshie8
  • ganeshie8
wonder if there is an useful interpretation dot product/cross product of complex numbers
sdfgsdfgs
  • sdfgsdfgs
btw - just realized that proving the y-componet all vectors in a regular polygon w odd no. of vertices by symmetry will be much less OBVIOUS than I had initially thought! so good that u guys have proven it by using complex nos.! :)
ganeshie8
  • ganeshie8
for even no. of vertices the symmetry argument worked like a charm as we found pairs of opposite vectors !
sdfgsdfgs
  • sdfgsdfgs
by lining up 1 of the vertices in the y-direction, the x-component of all other vectors (except the 1 pointing in y) in an odd-no polygon will equal out as the vectors will be in pair by symmetry. but showing the y-component of the vectors will add to zero is much tricky than i though.

Looking for something else?

Not the answer you are looking for? Search for more explanations.