IMO 1990 · Problem 6

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 \(1^2, 2^2, 3^2, \ldots, 1990^2\) in some order.

Official text of the problem. Official problem paper (PDF)

Solution

Identify the plane with the complex numbers. We will construct \(1990\) vectors having the required lengths, equally spaced directions, and sum zero, and then prove that they form a convex polygon.

Put \[ \alpha=e^{2\pi \mathrm{i}/5},\qquad \beta=e^{2\pi \mathrm{i}/199}. \] Thus \[ \sum_{p=0}^{4}\alpha^p=0, \qquad \sum_{q=0}^{198}\beta^q=0. \] For each pair \(0\le p\le4\), \(0\le q\le198\), let \[ r_{pq}=199p+q+1,\qquad u_{pq}=\alpha^p\beta^q, \] and take the two vectors \[ r_{pq}^{\,2}u_{pq}, \qquad -(1991-r_{pq})^2u_{pq}. \] The numbers \(r_{pq}\) run through \(1,\ldots,995\) exactly once, and their complements \(1991-r_{pq}\) run through \(996,\ldots,1990\). Consequently, these vectors have precisely the required lengths.

Their sum is zero. Indeed, \[ r_{pq}^{\,2}-(1991-r_{pq})^2 =1991(398p+2q-1989), \] so their total sum equals \[ \begin{aligned} &1991\sum_{p=0}^{4}\sum_{q=0}^{198} (398p+2q-1989)\alpha^p\beta^q\\ &\quad=1991\left[ \left(\sum_{p=0}^{4}(398p-1989)\alpha^p\right) \left(\sum_{q=0}^{198}\beta^q\right) + 2\left(\sum_{p=0}^{4}\alpha^p\right) \left(\sum_{q=0}^{198}q\beta^q\right) \right] =0. \end{aligned} \]

Next, the \(995\) directions \(u_{pq}\) are exactly the \(995\)-th roots of unity. They are all such roots, and they are distinct: if \[ \alpha^p\beta^q=\alpha^{p'}\beta^{q'}, \] then \(\alpha^{p-p'}=\beta^{q'-q}\) is both a fifth and a \(199\)-th root of unity. Since \(5\) and \(199\) are coprime, it equals \(1\), giving \(p=p'\) and \(q=q'\).

Because \(995\) is odd, these roots and their negatives together are exactly the \(1990\)-th roots of unity. Hence we can arrange our vectors in increasing angular order as \[ v_k=\ell_k e^{2\pi\mathrm{i}k/1990}, \qquad 0\le k<1990, \] where every \(\ell_k>0\). Define \[ P_0=0,\qquad P_{k+1}=P_k+v_k. \] The zero vector sum gives \(P_{1990}=P_0\).

We now verify convexity, including that this closed chain has no repeated vertices or crossings. Fix any side \(P_kP_{k+1}\), interpreting indices cyclically. The signed perpendicular displacement of \(P_{k+r}\) from its oriented line is \[ h_r=\sum_{t=0}^{r-1}\ell_{k+t} \sin\left(\frac{2\pi t}{1990}\right). \] For \(2\le r\le995\), this is strictly positive, since the terms with \(1\le t\le994\) are positive. For \(996\le r\le1989\), closure gives \[ h_r=-\sum_{t=r}^{1989}\ell_{k+t} \sin\left(\frac{2\pi t}{1990}\right)>0, \] since every sine in this last sum is negative.

Thus every vertex other than the endpoints of each side lies strictly to the left of that side’s oriented line. In particular, all vertices are distinct, and each side is an edge of their convex hull. The chain therefore forms a convex \(1990\)-gon.

Finally, consecutive side directions differ by \(2\pi/1990\), so every interior angle equals \[ \pi-\frac{2\pi}{1990}. \] Its side lengths are precisely \(1^2,2^2,\ldots,1990^2\), as required. \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).