1990 IMO Problems/Problem 6
Contents
Problem
Prove that there exists a convex with the following two properties:
(a) All angles are equal.
(b) The lengths of the
sides are the numbers
in some order.
Solution 1
Let and
.
Then the problem is equivalent to that there exists a way to assign
such that
.
Note that
, then if we can find a sequence
such that
and
, the problem will be solved.
Note that
, then
can be written as a sum from two elements in sets
and
.
If we assign the elements in
in the way that
, then clearly
Similarly, we could assign elements in
in that way (
) to
.
Then we make
according to the previous steps. Let
(mod 5),
(mod 199), and
, then each
will be some
and
And we are done.
This solution was posted and copyrighted by YCHU. The original thread for this problem can be found here: [1]
Solution 2
Throughout this solution, denotes a primitive
th root of unity.
We first commit to placing and
on opposite sides,
and
on opposite sides, etc. Since
,
,
, etc., this means the desired conclusion is equivalent to
being true for some permutation
of
.
Define . Then notice that\begin{align*} z &= 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ \omega^5 z &= 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ \omega^{10} z &= 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ &\vdotswithin{=} \end{align*}and so summing yields the desired conclusion, as the left-hand side becomes
and the right-hand side is the desired expression.
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]
See Also
1990 IMO (Problems) • Resources | ||
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