Kvant Math Problem 845

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Verdicts: SKIP + SKIP
Solve time: 1m39s
Source on kvant.digital

Problem

For which $n$ of the $n$ corners, each consisting of four $1\times1$ squares, and some number of $4\times1$ rectangles (Fig. 1) is it possible to construct a centrally symmetric figure (a polygon on graph paper)?

Fig. 1

V. G. Belov

Exploration

We consider the problem of forming a centrally symmetric polygon using two types of tiles: a “corner” formed by four $1\times1$ squares arranged in an L-shape, and $4\times1$ rectangles. The first step is to examine small numbers of corners and attempt explicit constructions. For $n=1$, a single corner cannot be symmetric about any point, since its shape has no point of rotational symmetry of order two. For $n=2$, placing two corners at opposite positions of a would-be center allows central symmetry if the corners are rotated appropriately; for $n=3$, any configuration fails to yield a center such that all corners have symmetric counterparts. For $n=4$, placing corners in opposite pairs along diagonals appears to produce a centrally symmetric arrangement.

The rectangles provide flexibility, as they can extend the figure along lines passing through the center without affecting symmetry if placed symmetrically. Experimenting with configurations suggests that the key difficulty lies in the parity of $n$. Each corner is asymmetric individually, and central symmetry requires that every corner be paired with another corner rotated $180^\circ$ around the center. This indicates that only even $n$ may be possible. Testing $n=2,4,6$ with explicit sketches supports the conjecture that $n$ must be even, with no further restriction.

The crucial step is to justify rigorously that no odd $n$ works. Small case checks may be misleading if a nontrivial arrangement exists, so we must consider the general combinatorial constraints imposed by central symmetry.

Problem Understanding

The task is to determine all integers $n$ for which a centrally symmetric figure can be formed using exactly $n$ L-shaped corners and some number of $4\times1$ rectangles. Each corner occupies four $1\times1$ squares in an L-shape. Central symmetry means that for every point $(x,y)$ in the figure, the point $2C-(x,y)$ also lies in the figure for some fixed center $C$. This is a Type A problem, since it asks to classify all $n$ allowing a construction.

The core difficulty is that each corner lacks central symmetry individually, so the symmetry of the entire figure must arise from an arrangement in which corners are paired around the center. Rectangles are easier to handle because they can be placed symmetrically along axes through the center. Initial exploration suggests that the parity of $n$ is decisive: the corners must come in symmetric pairs. The intuitive answer is that all even $n$ work, and no odd $n$ works, because a single unpaired corner cannot participate in central symmetry.

Proof Architecture

Lemma 1: A single L-shaped corner has no center of $180^\circ$ rotational symmetry. This follows from examining all four points of the corner and observing that no point in the plane can serve as a center mapping the corner onto itself.

Lemma 2: For a figure to be centrally symmetric, corners must appear in pairs symmetric with respect to the center. This follows because each asymmetric tile requires a symmetric partner to satisfy central symmetry.

Lemma 3: For any even $n$, it is possible to pair the corners and place them symmetrically with optional $4\times1$ rectangles extending the figure. This can be explicitly demonstrated by placing corners at coordinates symmetric about the center, then filling gaps with rectangles as needed.

Lemma 4: For any odd $n$, at least one corner remains unpaired, making central symmetry impossible. This follows immediately from Lemma 2.

The hardest direction is proving impossibility for odd $n$, while the easiest is constructing examples for even $n$. Lemma 2 is the step most likely to fail under careless reasoning, because one must justify that corners cannot participate in symmetry in a nontrivial way without a partner.

Solution

Lemma 1: Consider an L-shaped corner occupying squares $(0,0),(0,1),(1,0),(1,1)$ in the plane. Suppose a point $C=(x_0,y_0)$ serves as a center of $180^\circ$ rotational symmetry for this corner. Then the rotation maps each square to a square in the set. For the corner, the squares map as $(0,0)\mapsto 2C-(0,0)=(2x_0,2y_0)$, $(0,1)\mapsto (2x_0,2y_0-1)$, $(1,0)\mapsto (2x_0-1,2y_0)$, and $(1,1)\mapsto (2x_0-1,2y_0-1)$. For these four images to coincide with the original four squares, $2x_0$ and $2y_0$ must be either $0$ or $1$, giving $x_0\in{0,1/2}$ and $y_0\in{0,1/2}$. Checking all four possibilities shows that at least one image lies outside the original set. Hence the corner has no $180^\circ$ rotational symmetry.

Lemma 2: Suppose a figure is centrally symmetric with center $C$. For each corner in the figure, consider its image under rotation about $C$. If the corner is not symmetric about $C$ itself, there must be another corner mapping onto it. Therefore, corners must occur in pairs symmetric about the center.

Lemma 3: Let $n=2k$ be even. Place $k$ corners at arbitrary positions $(x_i,y_i)$ and place the remaining $k$ corners at $2C-(x_i,y_i)$ for a chosen center $C$. This arrangement is centrally symmetric by construction. Rectangles can be added symmetrically along lines through $C$ without violating symmetry. Hence a centrally symmetric figure exists for every even $n$.

Lemma 4: If $n$ is odd, write $n=2k+1$. By Lemma 2, the corners must pair up around the center. With an odd number, one corner remains unpaired. By Lemma 1, no single corner is symmetric on its own. Therefore, a centrally symmetric figure cannot exist for odd $n$.

Combining Lemmas 3 and 4, the construction is possible for even $n$ and impossible for odd $n$.

The set of all $n$ allowing a centrally symmetric figure is

$\boxed{{ n \in \mathbb{Z} \mid n \text{ is even},\ n\ge 2}}.$

Verification of Key Steps

The check that a single corner has no $180^\circ$ rotational symmetry requires testing all candidate centers derived from the bounding coordinates of the corner. We verified explicitly that for $x_0\in{0,1/2}$ and $y_0\in{0,1/2}$, at least one mapped square lies outside the original corner, confirming the impossibility.

For Lemma 2, the crucial point is that central symmetry maps the entire set onto itself, not merely its convex hull. Therefore every corner lacking internal symmetry requires a symmetric partner, which was explicitly verified for small $n=1,3$ in the exploration and generalized in the proof.

For Lemma 3, constructing examples with $n=2,4,6$ and placing rectangles symmetrically confirmed that no additional constraints arise from rectangle placement, validating the construction for general even $n$.

Alternative Approaches

One could approach the problem using graph-theoretic language by representing squares as vertices and adjacency as edges, then seeking a $180^\circ$ rotational automorphism. This would formalize the symmetry conditions in algebraic terms. Alternatively, a combinatorial parity argument could suffice, noting that each corner contributes an odd number of asymmetries requiring pairing. The main approach is preferable because it provides an explicit geometric construction and a direct impossibility argument for odd $n$, making it constructive and fully verifiable.