ROSS Problem 1

1 (a)

An equilibrium 2-wheel is a quadrilateral with vertex numbers that satisfy the fact that the sum of any two adjacent numbers is equal to the sum of the two opposite numbers. Let the vertices be $A,B,C,D$ and the equilibrium condition be:

$$\{\begin{array}{l}A+B=C+D\\ B+C=D+A\end{array}$$

The solution is $A=C$ and $B=D$, i.e., the vertices are arranged as $A,B,A,B$. Thus, all balanced 2-wheels are of the alternating permutation form $A,B,A,B$, where $A,B$ are arbitrary real numbers.
All balanced 2-wheels are of alternating permutation form $A,B,A,B$, where $A$ and $B$ are arbitrary real numbers.

1 (b)

The primal n-wheel is defined as each number equal to the opposite number. Let the vertices be $a_1,a_2,\dots ,a_{2n}$ satisfying $a_i=a_{i+n}$. For any two adjacent numbers $a_i+a_{i+1}$, the two opposite numbers are $a_{i+n}+a_{i+n+1}$. Since $a_i=a_{i+n}$ and $a_{i+1}=a_{i+n+1}$, it follows:

$$a_i+a_{i+1}=a_{i+n}+a_{i+n+1}.$$

Thus, the primal n-wheel necessarily balances.

1 (c)

Case 1: When n is even

For each $i$, the equilibrium condition can be expanded as:

$$a_i+a_{i+1}=a_{i+n}+a_{i+n+1}.$$

There are a total of $2n$ equations, but due to circularity, the number of independent equations is $n$. The system of equations is:

$$\{\begin{array}{l}{a}_1+a_2=a_{1+n}+a_{2+n}\\ a_2+a_3=a_{2+n}+a_{3+n}\\ \dots \\ a_{n-1}+a_n=a_{2n-1}+a_{2n}\\ a_n+a_{n+1}=a_{2n}+a_1\end{array}$$

Put the 1st, 3rd, 5th… .n-1 equations and add the 2nd, 4th, 6th … .n equations are added to get

$$\{\begin{array}{l}{a}_1+a_2+a_3+a_4+\dots +a_{n-1}+a_n=a_{1+n}+a_{2+n}+a_{3+n}+a_{4+n}+\dots +a_{2n-1}+a_{2n}\\ a_2+a_3+a_4+a_5+\dots +a_n+a_{n+1}=a_{2+n}+a_{3+n}+a_{4+n}+a_{5+n}+\dots +a_{2n}+a_1\end{array}$$

Subtract this to get

$$\begin{array}{rl}{a}_{n+1}-a_1& =a_1-a_{n-1}\\ a_1& =a_{n-1}\end{array}$$

From this we are left with, for all i

$$a_i+a_{i+1}=a_{i+n}+a_{i+n+1}$$

So for even n there is no non-primal equilibrium $n$-wheel

Case 2: When n is odd

There are a total of $2n$ equations, but due to circularity, the number of independent equations is $n$. The system of equations is:

$$\{\begin{array}{l}{a}_1+a_2=a_{1+n}+a_{2+n}\\ a_2+a_3=a_{2+n}+a_{3+n}\\ \dots \\ a_{n-1}+a_n=a_{2n-1}+a_{2n}\\ a_n+a_{n+1}=a_{2n}+a_1\end{array}$$

When n is odd, we need to show that the given system of equations still has a solution in the case $a_1\ne a_{n+1}$.

Divide the variables into the first n variables $A=[a_1,a_2,\dots ,a_n]$ and the last n variables $B=[a_{n+1},a_{n+2},\dots ,a_{2n}]$. Each equation can be expressed as:
For $i=1,2,\dots ,n-1$:

$$A_i+A_{i+1}=B_i+B_{i+1}$$

When $i=n$:
$A_n+B_1=B_n+A_1$
Suppose there exists a constant $c$ such that $A_i=B_i+c(-1{)}^i$. Substituting into the first $n-1$ equations, the left and right sides are equal, so these equations are satisfied.
For the last equation, substituting $A_i=B_i+c(-1{)}^i$ gives:

$$\begin{array}{rl}{A}_n+B_1& =B_n+A_1\\ [B_n+c(-1{)}^n]+B_1& =B_n+[B_1+c(-1{)}^1]\end{array}$$

When n is odd, $(-1{)}^n=-1$, so the equation becomes:
$B_n-c+B_1=B_n+B_1-c$
i.e., the equation holds.
When $c\ne 0$, since $A_1=B_1-c\ne B_1$ i.e., $a_1\ne a_{n+1}$, such a solution exists.
Therefore

$$\boxed{\text{ exists a nonprimitive equilibrium if and only if }n\text{ is odd }n\text{-wheel}}$$

1 (d)

For all odd numbers $n$, envision a construction that fills the series $[1,2n,2,2n-1,3,2n-2,\dots ,n,n+1]$ into the positions $[1,2n,2n-1,2,3,2n-2,\dots ,n,n+1]$ respectively.
That is, for all $i$ (where $1\le i\le n-1$):

• If $i$ is odd, position $i$ is filled with the number $i$;
• If $i$ is even, the position $2n-i+1$ is filled with the number $i$.
  For all $i$ (where $n\le i\le 2n$):
• If $i$ is even, position $i$ is filled with the number $i$;
• If $i$ is odd, the position $2n-i+1$ is filled with the number $i$.
  Thus for any position $j$, if $j$ is even

$$\begin{array}{rl}& (a_j+a_{j+1})-(a_{j+n}+a_{j+n+1})\\ =& (2n-j+1+j+1)-(2n-(j+n)+1+j+n+1)\\ =& 0\end{array}$$

If $j$ is an odd number of pages, we have $a_i+a_{i+1}=a_{i+n}+a_{i+n+1}$.
So

$$\boxed{\text{ when and only when }n\text{ is odd there is a classical balanced }n\text{-wheel}}$$

1 (e)

There is a special case. Every pair of neighboring numbers produces a sum that is different except for the faces. This case has a specific placement.
For all odd numbers $n$, imagine a construction in which the series $[1,2,3,\dots ,n]$ is filled in positions $[1,3,5,\dots ,2n-1]$ respectively; and the series $[n+1,n+2,n+3,\dots ,2n]$ is filled in positions $[n,n+2,n+4,\dots ,n-2]$ respectively. This way the sum of each pair of numbers is completely different.