1963 IMO Problems/Problem 3
In an -gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
Prove that .
Let , , etc.
Plot the -gon on the cartesian plane such that is on the -axis and the entire shape is above the -axis. There are two cases: the number of sides is even, and the number of sides is odd:
In this case, the side with the topmost points will be . To obtain the -coordinate of this top side, we can multiply the lengths of the sides , , ... by the sine of the angle they make with the -axis:
We can obtain the -coordinate of the top side in a different way by multiplying the lengths of the sides , , ... by the sine of the angle they make with the -axis to get the -coordinate of the top side:
It must be true that . This implies that for all , and therefore .
This case is very similar to before. We will compute the -coordinate of the top point two ways:
It must be true that . Then, we get for all . Therefore, . It is trivial that is then equal to the other values, so . This completes the proof.
Define the vector to equal . Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length is parallel to . We then have that
But for all , so
for all . This shows that , with equality when . Therefore
There is equality only when for all . This implies that and , so we have that .
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