Kvant Math Problem 1191
Let $S_A$ denote the central symmetry with center $A$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m15s
Source on kvant.digital
Problem
Let $A_1$, $A_2$, $A_3$, $\ldots$ be a sequence of points in the plane. Starting from some point $T_0$, construct a sequence $T_1$, $T_2$, $T_3$, $\ldots$, where $T_n$ is the point symmetric to $T_{n-1}$ with respect to $A_n$ ($n=1$, 2, 3, $\ldots$).
What necessary and sufficient conditions must the sequence $A_n$ satisfy so that the sequence $T_n$ is periodic for every choice of the point $T_0$ (that is, $T_{n+p}=T_n$ for some $p$ and for all $n$)?
I. F. Akulich
Exploration
Let $S_A$ denote the central symmetry with center $A$. In vector form,
$$S_A(X)=2A-X.$$
The recurrence is
$$T_n=S_{A_n}(T_{n-1}).$$
The first task is to understand compositions of central symmetries. For two centers $A,B$,
$$S_B\circ S_A(X) =2B-(2A-X) =X+2(B-A),$$
so the composition of two central symmetries is a translation.
For three centers,
$$S_C\circ S_B\circ S_A(X) =2C-\bigl(X+2(B-A)\bigr) =-X+2(A-B+C),$$
which is again a central symmetry.
Hence every finite composition is either a translation or a central symmetry, according to whether the number of factors is even or odd.
Let
$$F_n=S_{A_n}\circ\cdots\circ S_{A_1}.$$
Then $T_n=F_n(T_0)$. The condition of the problem is that for every $T_0$ there exists a positive integer $p$ such that
$$T_{n+p}=T_n \qquad\text{for all }n.$$
Since
$$T_{n+p}=F_{n+p}(T_0) =(F_{n+p}F_n^{-1})(T_n),$$
it is natural to study
$$G_n:=F_{n+p}F_n^{-1}.$$
The condition says that $G_n$ fixes every point of the orbit ${T_n}$, and this must hold for every initial point.
A better approach is to write $F_n$ explicitly. Induction gives
$$F_n(X)=(-1)^nX+2\sum_{k=1}^{n}(-1)^{,n-k}A_k.$$
Let
$$B_n=\sum_{k=1}^{n}(-1)^{,n-k}A_k.$$
Then
$$T_n=(-1)^nT_0+2B_n.$$
Now periodicity with period $p$ means
$$(-1)^{n+p}T_0+2B_{n+p} = (-1)^nT_0+2B_n.$$
Since this must hold for every $T_0$, the coefficient of $T_0$ must vanish:
$$(-1)^p=1.$$
Thus $p$ must be even.
For even $p$ the condition becomes
$$B_{n+p}=B_n \qquad\text{for all }n.$$
The recurrence
$$B_n=A_n-B_{n-1}$$
gives
$$A_n=B_n+B_{n-1}.$$
Therefore, if $B_n$ is periodic with period $p$, then $A_n$ is periodic with the same period:
$$A_{n+p} = B_{n+p}+B_{n+p-1} = B_n+B_{n-1} = A_n.$$
Conversely, if $A_n$ is periodic with an even period $p$, then
$$B_{n+p} = \sum_{k=1}^{n+p}(-1)^{,n+p-k}A_k.$$
Subtracting $B_n$ and grouping the last $p$ terms,
$$B_{n+p}-B_n = \sum_{j=1}^{p}(-1)^{,p-j}A_{n+j}.$$
Because $p$ is even and $A_{n+j}=A_j$,
$$B_{n+p}-B_n = \sum_{j=1}^{p}(-1)^jA_j,$$
a constant vector independent of $n$.
Hence $B_n$ is periodic iff this constant is zero. Thus the condition is
$$\sum_{j=1}^{p}(-1)^jA_j=0.$$
Equivalently,
$$A_2+A_4+\cdots+A_p = A_1+A_3+\cdots+A_{p-1}.$$
This seems both necessary and sufficient. The most delicate point is proving necessity of the vanishing alternating sum.
Problem Understanding
We are given a sequence of points $A_1,A_2,\ldots$ in the plane. Starting from an arbitrary point $T_0$, we obtain $T_n$ by reflecting $T_{n-1}$ through $A_n$.
We must determine exactly which sequences $(A_n)$ have the property that, for every initial point $T_0$, the resulting sequence $(T_n)$ is periodic.
This is a Type A problem. We must classify all sequences $A_n$ satisfying the stated condition.
The answer is that there must exist an even positive integer $p$ such that
$$A_{n+p}=A_n \qquad (n\ge1),$$
and
$$A_1+A_3+\cdots+A_{p-1} = A_2+A_4+\cdots+A_p.$$
The intuition is that after an even number of reflections the dependence on the initial point survives unchanged, so periodicity for every $T_0$ forces the cumulative effect over one period to be the identity transformation. The latter condition becomes exactly the vanishing of the alternating sum of one period.
Proof Architecture
Define $F_n=S_{A_n}\circ\cdots\circ S_{A_1}$ and prove that
$$F_n(X)=(-1)^nX+2B_n,$$
where
$$B_n=\sum_{k=1}^{n}(-1)^{,n-k}A_k.$$
This follows by induction on $n$.
Show that
$$T_n=(-1)^nT_0+2B_n.$$
This is obtained by substituting $X=T_0$ into the formula for $F_n$.
Assume periodicity for every $T_0$ and derive that the period $p$ must be even and that $B_{n+p}=B_n$ for all $n$.
This follows from comparing $T_{n+p}$ and $T_n$ and using the arbitrariness of $T_0$.
Prove that $B_{n+p}=B_n$ implies $A_{n+p}=A_n$.
Use the recurrence $B_n=A_n-B_{n-1}$.
Assume $A_n$ is periodic with even period $p$ and compute $B_{n+p}-B_n$.
The difference turns out to be the constant alternating sum of one period.
Show that $B_{n+p}=B_n$ is equivalent to
$$\sum_{j=1}^{p}(-1)^jA_j=0.$$
This yields the complete classification.
The hardest direction is proving necessity, namely extracting from the periodicity of every orbit the periodicity of $A_n$ together with the vanishing alternating sum condition.
Solution
Let $S_A$ denote the central symmetry with center $A$:
$$S_A(X)=2A-X.$$
Define
$$F_n=S_{A_n}\circ S_{A_{n-1}}\circ\cdots\circ S_{A_1}.$$
Since $T_n=F_n(T_0)$, the problem reduces to understanding the transformations $F_n$.
We claim that
$$F_n(X)=(-1)^nX+2B_n,$$
where
$$B_n=\sum_{k=1}^{n}(-1)^{,n-k}A_k.$$
For $n=1$,
$$F_1(X)=2A_1-X=-X+2A_1,$$
which agrees with the formula.
Assume the formula holds for some $n$. Then
$$\begin{aligned} F_{n+1}(X) &=S_{A_{n+1}}(F_n(X))\ &=2A_{n+1}-\Bigl((-1)^nX+2B_n\Bigr)\ &=(-1)^{n+1}X+2(A_{n+1}-B_n). \end{aligned}$$
On the other hand,
$$\begin{aligned} B_{n+1} &=\sum_{k=1}^{n+1}(-1)^{,n+1-k}A_k\ &=A_{n+1}-\sum_{k=1}^{n}(-1)^{,n-k}A_k\ &=A_{n+1}-B_n. \end{aligned}$$
Hence
$$F_{n+1}(X)=(-1)^{n+1}X+2B_{n+1}.$$
The induction is complete.
Therefore
$$T_n=(-1)^nT_0+2B_n.$$
Assume now that for every choice of $T_0$ the sequence $(T_n)$ is periodic. Let $p$ be a period. Then
$$T_{n+p}=T_n \qquad (n\ge0).$$
Substituting the formula for $T_n$ gives
$$(-1)^{n+p}T_0+2B_{n+p} = (-1)^nT_0+2B_n.$$
Since this equality holds for every point $T_0$, the coefficients of $T_0$ must coincide:
$$(-1)^p=1.$$
Thus $p$ is even. The preceding equality then reduces to
$$B_{n+p}=B_n \qquad (n\ge0).$$
From
$$B_n=A_n-B_{n-1},$$
we obtain
$$A_n=B_n+B_{n-1}.$$
Consequently,
$$A_{n+p} = B_{n+p}+B_{n+p-1} = B_n+B_{n-1} = A_n.$$
Hence $(A_n)$ is periodic with period $p$.
Now assume conversely that $(A_n)$ has period $p$, where $p$ is even. We compute
$$\begin{aligned} B_{n+p}-B_n &= \sum_{k=1}^{n+p}(-1)^{,n+p-k}A_k - \sum_{k=1}^{n}(-1)^{,n-k}A_k. \end{aligned}$$
Since $p$ is even, the terms with $k\le n$ cancel, and
$$\begin{aligned} B_{n+p}-B_n &= \sum_{j=1}^{p}(-1)^{,p-j}A_{n+j}. \end{aligned}$$
Using $A_{n+j}=A_j$,
$$\begin{aligned} B_{n+p}-B_n &= \sum_{j=1}^{p}(-1)^{,p-j}A_j. \end{aligned}$$
Because $p$ is even,
$$(-1)^{p-j}=(-1)^j,$$
so
$$B_{n+p}-B_n = \sum_{j=1}^{p}(-1)^jA_j.$$
This quantity is independent of $n$. Therefore
$$B_{n+p}=B_n \quad\Longleftrightarrow\quad \sum_{j=1}^{p}(-1)^jA_j=0.$$
If both conditions
$$A_{n+p}=A_n$$
and
$$\sum_{j=1}^{p}(-1)^jA_j=0$$
hold for an even $p$, then $B_{n+p}=B_n$. Hence
$$\begin{aligned} T_{n+p} &=(-1)^{n+p}T_0+2B_{n+p}\ &=(-1)^nT_0+2B_n\ &=T_n. \end{aligned}$$
Thus every sequence $(T_n)$ is periodic with period $p$.
We have proved both necessity and sufficiency.
The required sequences are exactly those for which there exists an even positive integer $p$ such that
$$A_{n+p}=A_n \qquad (n\ge1)$$
and
$$A_1+A_3+\cdots+A_{p-1} = A_2+A_4+\cdots+A_p.$$
Equivalently,
$$\sum_{j=1}^{p}(-1)^jA_j=0.$$
$$\boxed{\text{There exists an even }p\text{ such that }A_{n+p}=A_n\text{ and }\sum_{j=1}^{p}(-1)^jA_j=0.}$$
Verification of Key Steps
The first delicate step is extracting information from
$$(-1)^{n+p}T_0+2B_{n+p} = (-1)^nT_0+2B_n.$$
A careless argument might fix one particular $T_0$ and conclude only that the corresponding orbit is periodic. The hypothesis concerns every $T_0$. Taking two different initial points and subtracting the resulting identities shows that
$$(-1)^{n+p}=(-1)^n,$$
hence $p$ must be even. Without using the arbitrariness of $T_0$, this conclusion would not be justified.
The second delicate step is deriving periodicity of $A_n$ from periodicity of $B_n$. The relation
$$B_n=A_n-B_{n-1}$$
must first be rewritten as
$$A_n=B_n+B_{n-1}.$$
Only then can one compute
$$A_{n+p} = B_{n+p}+B_{n+p-1} = B_n+B_{n-1} = A_n.$$
The argument uses both $B_{n+p}=B_n$ and $B_{n+p-1}=B_{n-1}$.
The third delicate step is computing $B_{n+p}-B_n$. If one replaces $A_{n+j}$ by $A_j$ before checking that $p$ is even, the signs may be handled incorrectly. The equality
$$(-1)^{p-j}=(-1)^j$$
depends essentially on the parity of $p$. This is exactly where the condition that the period be even enters the final classification.
Alternative Approaches
Instead of introducing the quantities $B_n$, one may work directly with affine transformations. Every composition of an even number of central symmetries is a translation, and every composition of an odd number is a central symmetry. Let $F_p$ be the transformation obtained from one block of $p$ reflections. If every orbit is periodic, then $F_p$ must be the identity transformation. A nontrivial translation has no fixed points, and a nontrivial central symmetry has only one fixed point, so neither can produce periodicity for every initial point.
For even $p$, the transformation $F_p$ is a translation. Computing its translation vector yields
$$2\sum_{j=1}^{p}(-1)^jA_j.$$
Thus $F_p$ is the identity exactly when this alternating sum vanishes. Requiring the same block of reflections to repeat forces $A_n$ itself to be periodic. This approach is shorter conceptually, but the explicit formula for $T_n$ makes the necessity and sufficiency completely transparent.