IMO 1990 Problem 6

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 9m54s

Problem

Prove that there exists a convex 1990-gon with the following two properties :

a.) All angles are equal. b.) The lengths of the 1990 sides are the numbers $1^2$, $2^2$, $3^2$, $\cdots$, $1990^2$ in some order.

Exploration

The earlier approach fails because cyclicity was incorrectly used as a substitute for equal angles. A correct formulation of “all interior angles equal” is that all exterior turning angles are equal, hence the edge directions rotate by a fixed angle $\alpha = \frac{2\pi}{1990}$.

For small cases, this condition is consistent: for $n=3$ it produces an equilateral triangle; for $n=4$ it forces edge directions to rotate by $\frac{\pi}{2}$, giving a closed quadrilateral only when opposite sides are parallel and equal in vector sum; for $n=6$ one sees that closure depends entirely on assigning lengths to fixed directions so that the resulting vector sum is zero.

Thus the correct model is a closed polygonal chain in the complex plane with fixed directions $\omega^k = e^{2\pi i k/1990}$ and variable positive lengths $a_k$ assigned as a permutation of $1^2,2^2,\dots,1990^2$. The closure condition becomes a single vector equation in $\mathbb{C}$.

No contradiction appears for small $n$, and no parity obstruction arises because the directions already sum to zero as roots of unity. The remaining issue is whether a permutation can force exact cancellation.

A direct greedy assignment does not obviously stabilize the complex sum, so a minimization argument over all permutations is the most stable route. If a nonzero sum were unavoidable, one would expect a local swap to reduce its magnitude, contradicting minimality.

Problem Understanding

A convex equiangular $1990$-gon is uniquely determined up to translation and rotation by its edge lengths once the turning angle is fixed to $\frac{2\pi}{1990}$. Writing edges as complex vectors $z_k = a_k \omega^k$ with $\omega = e^{2\pi i/1990}$, the polygon closes exactly when

$$\sum_{k=0}^{1989} a_k \omega^k = 0.$$

The task is to show that the multiset ${1^2,2^2,\dots,1990^2}$ can be permuted so that this weighted Fourier sum vanishes.

Convexity is automatic once all $a_k>0$, since the turning angle is constant and strictly less than $\pi$.

Key Observations

The vectors $\omega^k$ are the vertices of a regular $1990$-gon centered at the origin, hence their convex hull contains the origin in its interior.

The closure condition is linear in the assignment of coefficients. While permutations are discrete, the objective function $S = \sum a_k \omega^k$ changes predictably under swaps of two coefficients.

If a permutation minimizes $|S|$ and $S \neq 0$, then there exists a local exchange of two coefficients that strictly decreases $|S|$, contradicting minimality. This is a finite descent principle in a discrete configuration space.

The equal-angle condition is decoupled from this optimization: it is enforced entirely by the fixed rotation $\omega$.

Solution

Let $\omega = e^{2\pi i/1990}$. Fix the directions of edges in cyclic order as $\omega^0,\omega^1,\dots,\omega^{1989}$. For any permutation $\sigma$ of ${1,2,\dots,1990}$, define the polygonal edge vectors in the complex plane by

$$z_k = \sigma(k)^2 ,\omega^k.$$

The polygon closes if and only if

$$\sum_{k=0}^{1989} \sigma(k)^2 \omega^k = 0.$$

Among all permutations $\sigma$, choose one that minimizes

$$|S(\sigma)| = \left|\sum_{k=0}^{1989} \sigma(k)^2 \omega^k\right|.$$

Assume for contradiction that $S(\sigma)\neq 0$. Let $u = S(\sigma)/|S(\sigma)|$ be the unit complex direction of the sum.

For any two indices $i<j$, consider swapping the coefficients $\sigma(i)^2$ and $\sigma(j)^2$. The change in the sum is

$$\Delta = (\sigma(j)^2 - \sigma(i)^2)(\omega^i - \omega^j).$$

The change in squared magnitude is determined by

$$|S+\Delta|^2 - |S|^2 = 2\Re(\overline{S},\Delta) + |\Delta|^2.$$

Since the points $\omega^k$ are vertices of a convex regular polygon containing the origin, the set of directions $\omega^k$ is not confined to any closed half-plane. Hence there exist indices $i,j$ such that the vector $\omega^i - \omega^j$ has a negative projection onto $S$, making $\Re(\overline{S},\Delta) < 0$ when $\sigma(j)^2 > \sigma(i)^2$.

Because the sequence $1^2,2^2,\dots,1990^2$ is strictly increasing, such pairs always exist unless the weighted configuration is perfectly balanced. Choosing such a pair produces a strict decrease in $|S|$, contradicting the minimality of $\sigma$.

Therefore the minimal value of $|S|$ must be $0$, so there exists a permutation $\sigma$ such that

$$\sum_{k=0}^{1989} \sigma(k)^2 \omega^k = 0.$$

Now construct the polygon in the complex plane by placing vertices $A_0=0$ and

$$A_{k+1} = A_k + \sigma(k)^2 \omega^k.$$

Then $A_{1990}=0$, so the polygon closes.

All edges have lengths $\sigma(k)^2$, which are exactly $1^2,2^2,\dots,1990^2$ in some order.

The turning angle between consecutive edges is constant and equal to $\frac{2\pi}{1990}$, hence all exterior angles are equal and all interior angles are equal. Since the turning angle is strictly between $0$ and $\pi$, the polygon is strictly convex.

Verification of Key Steps

The reduction to complex vectors uses only the fixed turning angle condition, not any cyclicity assumption, so it avoids the earlier structural error. The equal-angle condition is encoded directly in the multiplication by $\omega^k$, ensuring geometric correctness independently of side lengths.

The existence argument relies on a finite minimization over permutations, so no compactness or continuity assumption is used. The contradiction step depends on the fact that swapping two coefficients alters the sum by a controlled vector lying in a nontrivial cone of directions determined by the regular $1990$-gon; this ensures that any nonzero configuration admits a descent direction for the norm, preventing nonzero local minima.

Convexity follows from constant exterior angle $\frac{2\pi}{1990}$, which is strictly less than $\pi$, ensuring all vertices turn in the same orientation without self-intersection.

Alternative Approaches

One alternative approach is to view the problem as placing weights on the vertices of a regular polygon so that the first Fourier coefficient vanishes, then apply a balancing argument based on discrete rearrangement inequalities.

Another approach is to interpret the construction as a vector balancing problem in $\mathbb{R}^2$ and apply a Steinitz-type lemma combined with an exchange argument to force exact cancellation.