1answer of the first one =2na2
a=radius of circumscribing circle
Part 1: On a circle which circumscribes a n-sided polygon A1,A2,A3,.... An, a point M is taken. Prove that the sum of the squares of the distances from this point to all the vertices of the polygos is a number(independent of the position of the point M) What is this equal to?
Part 2: Prove that the sum of the squrares of the distances from an arbitrary point M, taken in the plane of a regular n-sided polygon A1,A2,A3,.... An, to all the vertices of the polygon, depends only upon the distance l of the point M from the center O of the polygon and find its value? in terms of R(radius of the circle circumscribing the n sided polygon.)
This is a problem that has come somewhere earlier (Will tell the source after the solution :)
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16 Answers
1do i use the eqn of circle. |z-z0|=r, where z0 is the centre and r is the radius
62take z0 as zero.. it wont matter.. (will it?)
u can actually remove r aas well.. and later reintroduce it.. but that could be a bit confusing..
1say I take the pt M to be x,y then I take the other pts to be x1 y1 x2 y2 and so on. then I take the sum of the distances as
d=√(x-x1)2+(y-y1)2and so on upto n then I will get on solving the underroot term
√x2+y2+x12+y12+2xx1+2yy1=√2(r+xx1+yy1) am i on the right track
62how about using something more..
we know that xi,y i lie on a circle
so there is something of the form reiθ
1i proved it as follows =
let the random point be (acosθ,asinθ)
the points of the polygon can be written as of the form = (acos(2Ï€K/n),asin(2Ï€K/n))
where k varies from 1 to n
then the required sum is
Σa2((cosθ-cos(2πK/n))2+(sinθ-sin(2πK/n))2)
=2a2n -(a2)(2cosθΣcos(2πK/n) - 2sinθΣsin(2πK/n))
but from
(cosθ+isinθ)n=cosnθ+isinnθ
we have to find the sum of the roots of cosnθ=1
in the above expression we find that the coefficient of cosn-1θ =0
and we can express all the other terms as function of cosθ
hence sum of roots =coefficient of cos^(n-1)θ/coeff. of cos^nθ = 0
hence Σcos(2πK/n) = 0
similarly we can prove that Σsin(2πK/n)=0
hence we get the required sum as 2na2
by the above results it can be proven similarly the question below
its answer will come out to be =
n(l2+R2)
l=dist. of point form centre
R=circumscribing radius of polygon
62Awesome dude :)
good work :)
I really wish u all the best.. these are not very easy problems.. good that u are ammunitioned enuf to handle these :)
I hope u are not one of the guys who gives up chem bcos they are good in maths and physics..
Good luck dear :)
9it dint seem as tough as u project it to be wen u use concept of vectors
ans is
Σ ( Z - ai )2 = Z2 + ai2 - 2Zai
Σ ai =0 (closed polgon)
ai2=z2=R2 as all lie on circle
hence ans is 2nR2
similarly when Z lies at dist l from O
ans is n( l2 + R2)
62Ur first expansion...
Σ ( Z - ai )2 = Z2 + ai2 - 2Zai
will not be true!!
it will be -Zai - Zai +Z2+ai2
wont it?
9im using vector algebra and not complex algebra
i think that expansion is correct if i write in detailed form as
|Z - ai|2 = |Z|2 + |ai|2 + 2Z.ai where Z annd ai are vectors
62Sorry dude.. u are perfectly correct :)
I was too much into complex numbers.. U have a better solution than I had in my mind :)
Well done :)
62It is from a USSR math olympiad practice question! (But i found it very relevant to Complex.) (I did not want to use the word Olympiad bcos it scares away so many ppl)
U did a Better job to see vectors here :)