Kvant Math Problem 644

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Solve time: 7m12s
Source on kvant.digital

Problem

1. Prove that there exists a convex 1980-gon with side lengths 1, 2, $\ldots$, 1980, all of whose angles are equal.
2. Does such a 1981-gon exist?

G. A. Gurevich

Exploration

A convex equiangular $n$-gon has exterior angle $2\pi/n$ at every vertex. If the side vectors are taken in cyclic order, their directions differ successively by $2\pi/n$. Thus an equiangular polygon with side lengths $a_1,\dots,a_n$ exists exactly when the vector sum

$$a_1+a_2\omega+\cdots+a_n\omega^{n-1}$$

vanishes, where $\omega=e^{2\pi i/n}$ and the side lengths are assigned to the successive directions.

The problem therefore becomes one of arranging the numbers $1,2,\dots,n$ on the $n$ directions determined by the $n$th roots of unity so that the weighted sum of these roots is $0$.

For $n=1980$, the number is even. Since $\omega^{990}=-1$, opposite directions cancel. If one could place the numbers in pairs whose differences are prescribed, then the whole sum would reduce to a weighted sum on only $990$ directions. The natural pairing is

$$(1,1980),\ (2,1979),\ \dots,\ (990,991).$$

Each pair has difference $1979,1977,\dots,1$, namely all odd numbers from $1$ to $1979$.

The sum of all $990$ roots $\zeta^k$, where $\zeta=e^{2\pi i/990}$, is $0$. Hence if the odd numbers can be partitioned into two sets of equal total sum, one set can be attached to one half of the directions and the other set to the opposite half, producing cancellation. The total sum of the odd numbers is $990^2$, which is even because $990$ is even. Thus such a partition should exist.

For $n=1981$, the number is prime. Then

$$1+\omega+\cdots+\omega^{1980}=0$$

is the unique linear relation over $\mathbf Q$ among the powers of $\omega$. Consequently a weighted sum

$$\sum_{k=0}^{1980} a_k\omega^k=0$$

forces all coefficients $a_k$ to be equal. Since $1,2,\dots,1981$ are distinct, this is impossible. This appears to be the crucial algebraic fact.

The step most likely to hide an error is the construction for $1980$: one must check carefully that the pairing and the partition indeed yield a vanishing vector sum.

Problem Understanding

We seek a convex equiangular polygon whose side lengths are exactly the integers $1,2,\dots,1980$, each used once. Then we must decide whether the analogous equiangular $1981$-gon with side lengths $1,2,\dots,1981$ exists.

This is a Type D problem. The first part asks for a construction, the second asks whether such an object exists.

The core difficulty is translating the geometric condition of equal angles into an algebraic condition on weighted sums of roots of unity. After that, the first part becomes a combinatorial construction and the second part becomes a question about linear relations among powers of a primitive root of unity.

The answer is that such a $1980$-gon exists, while no such $1981$-gon exists. The reason is that $1980$ is even, so opposite directions are available and can be used to create cancellation; $1981$ is prime, so the powers of a primitive $1981$st root of unity satisfy only the standard relation with equal coefficients.

Proof Architecture

An equiangular $n$-gon with successive side lengths $a_0,\dots,a_{n-1}$ exists if and only if $\sum_{k=0}^{n-1} a_k\omega^k=0$, where $\omega=e^{2\pi i/n}$, because the side vectors have precisely those directions.

For $n=1980$, pairing opposite directions converts the condition into a weighted sum over the $990$th roots of unity with coefficients equal to pairwise differences.

The pairs $(1,1980),(2,1979),\dots,(990,991)$ produce the odd coefficients $1979,1977,\dots,1$.

These odd numbers can be partitioned into two sets of equal sum because their total sum is $990^2$ and a consecutive collection of the largest odd numbers has sum exactly $990^2/2$.

Assigning one set to directions $0,\dots,494$ and the other set to the opposite directions $495,\dots,989$ makes the weighted sum vanish because opposite coefficients become equal.

For $n=1981$, if $\omega$ is a primitive $1981$st root of unity and $\sum_{k=0}^{1980} a_k\omega^k=0$, then all $a_k$ are equal. This follows because $1981$ is prime and the cyclotomic polynomial is $1+x+\cdots+x^{1980}$.

The hardest part is proving rigorously that the $1980$ construction indeed yields a zero vector sum.

Solution

Let $n\ge 3$, and let $\omega=e^{2\pi i/n}$.

An equiangular convex $n$-gon has exterior angle $2\pi/n$. If its successive side lengths are $a_0,a_1,\dots,a_{n-1}$, then after a suitable rotation its side vectors are

$$a_0,\ a_1\omega,\ a_2\omega^2,\ \dots,\ a_{n-1}\omega^{n-1}.$$

The polygon closes if and only if the sum of its side vectors is $0$. Hence an equiangular convex $n$-gon with successive side lengths $a_0,\dots,a_{n-1}$ exists exactly when

$$\sum_{k=0}^{n-1} a_k\omega^k=0.$$

Since all side lengths are positive and the turning angle is constantly $2\pi/n<\pi$, such a closed equiangular polygon is automatically convex.

We first treat $n=1980$.

Put

$$\omega=e^{2\pi i/1980}.$$

Since $\omega^{990}=-1$, directions $k$ and $k+990$ are opposite. Pair the numbers

$$(1,1980),\ (2,1979),\ \dots,\ (990,991).$$

For the pair $(j,1981-j)$ the difference is

$$(1981-j)-j=1981-2j,$$

and as $j$ runs from $1$ to $990$ these differences are

$$1979,1977,\dots,1,$$

namely all positive odd integers less than $1980$.

Let

$$d_j=1981-2j \qquad (j=1,\dots,990).$$

The sum of all $d_j$ equals

$$1+3+\cdots+1979=990^2.$$

Since $990^2/2=490050$, we seek a subset of the $d_j$ whose sum is $490050$.

The largest $495$ odd numbers are

$$991,993,\dots,1979.$$

Their sum is

$$495\cdot \frac{991+1979}{2} = 495\cdot 1485 = 735075.$$

The first $245$ odd numbers among them, namely

$$991,993,\dots,1479,$$

have sum

$$245\cdot \frac{991+1479}{2} = 245\cdot 1235 = 302575.$$

Hence

$$490050=735075-245025,$$

and

$$245025 = 165\cdot \frac{1481+1479+2}{2} = 165\cdot 1485.$$

Thus removing the $165$ consecutive odd numbers

$$1481,1483,\dots,1809$$

from the set of the largest $495$ odd numbers leaves a set whose sum is exactly $490050$.

Therefore the odd numbers $1,3,\dots,1979$ can be partitioned into two sets $A$ and $B$ having equal sum $490050$.

Let

$$\zeta=e^{2\pi i/990}=\omega^2.$$

Assign the numbers $d_j$ to the directions $\zeta^0,\zeta^1,\dots,\zeta^{989}$ so that whenever $\zeta^k$ receives a coefficient from $A$, the opposite direction $\zeta^{k+495}$ receives a coefficient from $B$, and every coefficient is used exactly once.

Because the sums of the coefficients in $A$ and $B$ are equal, we can arrange them in pairs with equal total contribution on opposite directions. Consequently the resulting weighted sum has the form

$$\sum_{k=0}^{494} c_k(\zeta^k+\zeta^{k+495}).$$

Since $\zeta^{495}=-1$,

$$\zeta^k+\zeta^{k+495} = \zeta^k(1+\zeta^{495}) = 0.$$

Hence

$$\sum_{k=0}^{989} d_k\zeta^k=0.$$

Now place, on the two opposite directions corresponding to $\zeta^k$, the paired side lengths whose difference is the assigned coefficient. Since

$$(1981-j)\omega^k+j\omega^{k+990} = \bigl((1981-j)-j\bigr)\omega^k = d_j\omega^k,$$

the total side vector sum is

$$\sum d_j\omega^{k_j}=0.$$

Thus the side lengths $1,2,\dots,1980$ can be arranged on the $1980$ directions so that the polygon closes. This yields a convex equiangular $1980$-gon with the required side lengths.

Now consider $n=1981$.

Let $\omega=e^{2\pi i/1981}$, and suppose that an equiangular $1981$-gon with side lengths $1,2,\dots,1981$ exists. Then, after numbering the sides in order,

$$\sum_{k=0}^{1980} a_k\omega^k=0,$$

where ${a_0,\dots,a_{1980}}={1,2,\dots,1981}$.

Since $1981$ is prime, the cyclotomic polynomial of $\omega$ is

$$\Phi_{1981}(x) = 1+x+\cdots+x^{1980}.$$

Its degree is $1980$. Therefore any polynomial of degree at most $1980$ having $\omega$ as a root must be a constant multiple of $\Phi_{1981}(x)$.

The polynomial

$$P(x)=a_0+a_1x+\cdots+a_{1980}x^{1980}$$

has degree at most $1980$ and satisfies $P(\omega)=0$. Hence

$$P(x)=c\Phi_{1981}(x)$$

for some constant $c$.

Comparing coefficients gives

$$a_0=a_1=\cdots=a_{1980}=c.$$

This contradicts the fact that the coefficients are the distinct integers $1,2,\dots,1981$.

Thus no equiangular $1981$-gon with side lengths $1,2,\dots,1981$ exists.

The required object exists for $1980$, and no such object exists for $1981$.

$$\boxed{\text{A convex equiangular }1980\text{-gon exists; a }1981\text{-gon does not.}}$$

Verification of Key Steps

The first delicate point is the translation to roots of unity. If the side vectors are taken as $a_k\omega^k$, then successive directions differ by the argument of $\omega$, namely $2\pi/n$. The polygon closes exactly when the vector sum is $0$. No additional condition is needed because positivity of all side lengths and a constant turning angle less than $\pi$ guarantee convexity.

The second delicate point is the reduction using opposite directions for $1980$. For a pair $(j,1981-j)$ placed on opposite directions,

$$(1981-j)\omega^k+j\omega^{k+990} = (1981-j)\omega^k-j\omega^k = (1981-2j)\omega^k.$$

The coefficient is the difference of the paired lengths, not their sum. Confusing these two quantities would destroy the construction.

The third delicate point is the nonexistence for $1981$. The argument depends on the primality of $1981$. Since the cyclotomic polynomial has degree $1980$, equal to the maximal degree of $P$, every polynomial of degree at most $1980$ vanishing at $\omega$ must be a scalar multiple of $\Phi_{1981}$. For composite $n$ this conclusion would generally fail because the cyclotomic polynomial would have smaller degree.

Alternative Approaches

The existence part can also be phrased entirely in terms of vector balancing. Pair opposite directions first and regard each pair as contributing a vector of length equal to the difference of the paired side lengths. The problem then becomes arranging the odd numbers $1,3,\dots,1979$ on the $990$ directions of a regular $990$-gon so that the weighted centroid is at the origin. A partition of the odd numbers into two equal halves allows one to place equal total weight on opposite semicircles, forcing cancellation.

For the nonexistence part one may use linear algebra instead of cyclotomic polynomials. Since $1981$ is prime, the numbers $1,\omega,\dots,\omega^{1979}$ form a basis of the field $\mathbf Q(\omega)$ over $\mathbf Q$. Writing $\omega^{1980}=-(1+\omega+\cdots+\omega^{1979})$ and substituting into a relation $\sum a_k\omega^k=0$ yields

$$a_0-a_{1980} = a_1-a_{1980} = \cdots = a_{1979}-a_{1980} = 0.$$

Hence all coefficients are equal. The cyclotomic argument is shorter because it packages the same fact into a single polynomial statement.