IMO 1963 Problem 3
by djmathman, Feb 10, 2013, 11:48 PM
Darn I got a bit lazy after I proved the odd integer 
case for this problem, since I had some stuff to do afterwards and the cases are only marginally different.
case for this problem, since I had some stuff to do afterwards and the cases are only marginally different.Problem wrote:
In an -gon 
, all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
![\[a_{1}\geq a_{2}\geq \dots \geq a_{n}. \]](//latex.artofproblemsolving.com/5/7/7/577a7e8985158e8f42b6869d5381accd1fabac76.png)
Prove that 
-gon 
, all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation![\[a_{1}\geq a_{2}\geq \dots \geq a_{n}. \]](http://latex.artofproblemsolving.com/5/7/7/577a7e8985158e8f42b6869d5381accd1fabac76.png)
Prove that 
being the origin and 
lying on the 
-
.
we obtain![\begin{align*}\Im\left(\sum_{k=1}^{n}b_k\right)&=\sum_{k=1}^{n}\Im (b_k)=\sum_{k=1}^{n}a_k\sin\left(\dfrac{2(k-1)\pi}{n}\right)\\&=\sin 0 + \sum_{k=2}^{(n-1)/2}\left[a_k\sin\left(\dfrac{2(k-1)\pi}{n}\right)-a_{n-k+2}\sin\left(\dfrac{2(k-1)\pi}{n}\right)\right]\\&=\sum_{k=2}^{(n-1)/2}\sin\left(\dfrac{2(k-1)\pi}{n}\right)(a_k-a_{n-k+2}).\end{align*}](http://latex.artofproblemsolving.com/9/3/7/9372491976b080c2323a1e5e316b5a6ce8eadcb9.png)
Since 
never reaches 
,
and 
for all 
is a closed planar shape, so we must have 
.
for all 
.
is located on the origin and 
By using a similar procedure to the one above, we obtain that![\[\Im\left(\sum_{k=1}^{n}b_k\right)=\sum_{k=1}^{(n-1)/2-1}\sin\left(\dfrac{2k\pi}{n}\right)(a_k-a_{n-k}),\]](http://latex.artofproblemsolving.com/2/4/9/249c9b899da574fc3f03e4bc96191b763b99820e.png)
so 
for all 
.
at least once with all these equalities, we must have 
.
as we did at first, we see that 
,This post has been edited 3 times. Last edited by djmathman, Jan 2, 2018, 4:10 AM
Reason: latex
Reason: latex
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