Kvant Math Problem 689

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Problem

Prove that it is impossible to form a rectangle from identical tiles shaped as isosceles trapezoids with bases 3 cm and 1 cm and height 1 cm.

S. Rukshin

Exploration

Each tile is an isosceles trapezoid with bases $3$ and $1$ and height $1$. Its area is

$$\frac{3+1}{2}\cdot 1=2.$$

The nonparallel sides have length

$$\sqrt{1^2+1^2}=\sqrt2,$$

because the difference of the bases is $2$, so each leg is displaced horizontally by $1$.

A rectangle tiled by these pieces has boundary consisting entirely of segments belonging to edges of tiles. Every boundary segment of a tile has length $1$, $3$, or $\sqrt2$.

The irrational length $\sqrt2$ suggests an invariant involving coordinates. Place a tile so that its bases are horizontal. Then every vertex has integer coordinates. After translating such tiles, all vertices remain lattice points. Rotations are possible, so this observation alone is insufficient.

A more promising idea is to color the plane periodically. Since the bases differ by $2$, the horizontal projection of a tile has width $3$ at one side and $1$ at the other. The area argument gives no contradiction.

Consider the boundary directions. In a standard position the sides have slopes $\pm1$. Thus every edge of every tile is parallel either to one of the bases or to a direction making angle $45^\circ$ with the bases. Since all tiles are congruent, every edge direction belongs to four directions separated by $45^\circ$.

A rectangle has only two boundary directions, perpendicular to each other. An invariant modulo $2$ involving the slanted edges may work. Let us assign weights to edges. The crucial observation is that each tile has exactly two slanted sides of length $\sqrt2$. Any tiling identifies interior edges in pairs, so only boundary contributions survive.

Take coordinates and color the plane by the function $f(x,y)=x+y \pmod 2$. Along horizontal or vertical unit segments the endpoints have opposite colors, while along a diagonal segment of slope $\pm1$ and length $\sqrt2$ the endpoints have the same color. This suggests a Dehn-type additive invariant.

A simpler route is to use a signed edge count. Associate to every edge parallel to the line $y=x$ the value $+1$, to every edge parallel to $y=-x$ the value $-1$, and to edges parallel to the bases the value $0$. For one trapezoid the two legs contribute $+1$ and $-1$, so the total is $0$. That invariant is too weak.

A stronger invariant uses lengths. Let an edge parallel to $y=x$ contribute its length, an edge parallel to $y=-x$ contribute minus its length, and horizontal or vertical edges contribute $0$. Every tile still has invariant $0$ because its two legs have equal length $\sqrt2$.

For a tiling, interior edges cancel. Hence the invariant of the rectangle must be $0$. But a rectangle has no slanted boundary edges at all, so its invariant is also $0$. No contradiction arises.

A different invariant is needed. Since the slanted edges have irrational length $\sqrt2$ while the bases have rational length $1$ or $3$, a Dehn invariant relative to the vector space basis ${1,\sqrt2}$ is natural. Let horizontal and vertical directions have weight $1$, and diagonal directions weight $\sqrt2$. For a tile, the total weighted edge contribution may be nonzero, whereas for a rectangle it may vanish.

The standard Dehn-style choice is to assign to a segment the product of its length and a direction label modulo rational relations. For our tile, the two bases contribute $3+1=4$ in one direction class, and the two legs contribute $2\sqrt2$ in another. Since $4$ and $2\sqrt2$ are linearly independent over $\mathbb Q$, cancellation in a rectangle seems impossible.

The likely invariant is the Dehn invariant of polygon dissections. A rectangle has only right angles. The trapezoid has angles $45^\circ,45^\circ,135^\circ,135^\circ$. Since $135^\circ\equiv -45^\circ$ modulo $\pi$, every angle contributes the same class $45^\circ$. Then the Dehn invariant is proportional to

$$(3+1+ \sqrt2+\sqrt2)\otimes 45^\circ =(4+2\sqrt2)\otimes45^\circ.$$

This is nonzero because $4+2\sqrt2\neq0$. A rectangle has Dehn invariant $0$ since $\pi/2$ represents $0$ in $\mathbb R/\pi\mathbb Q$. Additivity under dissection yields the contradiction.

The single step most likely to hide an error is the verification that the rectangle indeed has zero Dehn invariant and that the tile has nonzero Dehn invariant in the chosen quotient.

Problem Understanding

We must prove that no finite collection of congruent isosceles trapezoids with bases $3$ cm and $1$ cm and height $1$ cm can tile a rectangle.

This is a Type B problem. The task is to prove impossibility.

The core difficulty is that elementary area and parity arguments do not rule out a rectangle. One needs an additive invariant preserved under dissections and capable of distinguishing the given trapezoid from every rectangle.

Proof Architecture

The proof uses the Dehn invariant.

First, define the Dehn invariant of a polygon as

$$D(P)=\sum l_i\otimes \alpha_i,$$

where $l_i$ are side lengths and $\alpha_i$ are the corresponding interior angles, considered in the tensor product over $\mathbb Q$ with angles taken modulo $\pi\mathbb Q$.

Second, use the additivity of the Dehn invariant under dissections. When two polygons are cut and reassembled, the invariants add because contributions from internal edges cancel.

Third, compute the Dehn invariant of a rectangle. Every interior angle is $\pi/2$, which is $0$ in $\mathbb R/\pi\mathbb Q$, hence the invariant is $0$.

Fourth, compute the Dehn invariant of the trapezoid. Its angles are $45^\circ$ and $135^\circ$, and $135^\circ\equiv -45^\circ\equiv45^\circ$ modulo $\pi\mathbb Q$. Its side lengths are $3$, $1$, $\sqrt2$, $\sqrt2$. Hence

$$D(T)=(4+2\sqrt2)\otimes \frac{\pi}{4}.$$

Fifth, prove that this element is nonzero. Since $\sqrt2\notin\mathbb Q$, the number $4+2\sqrt2$ is nonzero and not rationally forced to vanish in the tensor product.

The most delicate point is proving that $(4+2\sqrt2)\otimes \pi/4\neq0$.

Solution

Let $T$ be the given isosceles trapezoid.

We use the Dehn invariant. For a polygon $P$ with side lengths $l_1,\dots,l_n$ and interior angles $\alpha_1,\dots,\alpha_n$, define

$$D(P)=\sum_{i=1}^{n} l_i\otimes \alpha_i,$$

where lengths are regarded as elements of the $\mathbb Q$-vector space $\mathbb R$, angles are regarded as elements of the quotient space $\mathbb R/\pi\mathbb Q$, and the tensor product is taken over $\mathbb Q$.

A standard property of the Dehn invariant is its additivity under dissections: if a polygon is partitioned into finitely many polygons, then the Dehn invariant of the whole polygon equals the sum of the Dehn invariants of the pieces. Contributions coming from internal edges cancel in pairs, because each internal edge is counted twice with supplementary angles, whose classes in $\mathbb R/\pi\mathbb Q$ are negatives of one another.

Suppose that a rectangle $R$ can be tiled by $N$ copies of $T$. By additivity,

$$D(R)=N,D(T).$$

We compute both sides.

Every angle of a rectangle equals $\pi/2$. Since

$$\frac{\pi}{2}\in \pi\mathbb Q,$$

its class in $\mathbb R/\pi\mathbb Q$ is $0$. Hence every term in the definition of $D(R)$ is zero, and therefore

$$D(R)=0.$$

Now consider the trapezoid $T$. The difference of the bases is $2$, and the height is $1$, so each leg has length

$$\sqrt{1^2+1^2}=\sqrt2.$$

Thus the side lengths are

$$3,\quad 1,\quad \sqrt2,\quad \sqrt2.$$

Its interior angles are

$$\frac{\pi}{4},\quad \frac{\pi}{4},\quad \frac{3\pi}{4},\quad \frac{3\pi}{4}.$$

In the quotient space $\mathbb R/\pi\mathbb Q$,

$$\frac{3\pi}{4}-\left(-\frac{\pi}{4}\right)=\pi\in\pi\mathbb Q,$$

so

$$\frac{3\pi}{4}\equiv -\frac{\pi}{4}.$$

Also,

$$-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2}\in\pi\mathbb Q,$$

hence

$$-\frac{\pi}{4}\equiv \frac{\pi}{4}.$$

Therefore all four angles represent the same element of $\mathbb R/\pi\mathbb Q$, namely the class of $\pi/4$. Consequently,

$$D(T) =(3+1+\sqrt2+\sqrt2)\otimes\frac{\pi}{4} =(4+2\sqrt2)\otimes\frac{\pi}{4}.$$

We show that this tensor is nonzero. Since $\pi/4\notin\pi\mathbb Q$, its class in $\mathbb R/\pi\mathbb Q$ is nonzero. Choose a $\mathbb Q$-linear functional

$$\varphi:\mathbb R\to\mathbb Q$$

such that

$$\varphi(1)=1,\qquad \varphi(\sqrt2)=0.$$

Such a functional exists because ${1,\sqrt2}$ is linearly independent over $\mathbb Q$ and can be extended to a basis of $\mathbb R$ over $\mathbb Q$.

Applying $\varphi\otimes\mathrm{id}$ gives

$$(\varphi\otimes\mathrm{id})!\left((4+2\sqrt2)\otimes\frac{\pi}{4}\right) =4\otimes\frac{\pi}{4}.$$

Since $4\neq0$ in $\mathbb Q$ and the class of $\pi/4$ is nonzero, the element $4\otimes\pi/4$ is nonzero. Hence

$$D(T)\neq0.$$

Therefore

$$D(R)=0, \qquad N,D(T)\neq0.$$

This contradicts the equality $D(R)=N,D(T)$ obtained from the assumed tiling.

Hence no rectangle can be tiled by congruent copies of the given trapezoid.

This completes the proof.

∎

Verification of Key Steps

The computation of the leg length uses the geometry of an isosceles trapezoid. The bases differ by $3-1=2$. Symmetry splits this excess equally between the two sides, so each leg forms the hypotenuse of a right triangle with legs $1$ and $1$. Thus the leg length is $\sqrt2$.

The angle reduction requires two congruences. First,

$$\frac{3\pi}{4}\equiv-\frac{\pi}{4} \pmod{\pi\mathbb Q}$$

because their difference is $\pi$. Second,

$$-\frac{\pi}{4}\equiv\frac{\pi}{4} \pmod{\pi\mathbb Q}$$

because their difference is $-\pi/2$. Since $\pi/2\in\pi\mathbb Q$, all four angles indeed represent the same class.

The nonvanishing of $(4+2\sqrt2)\otimes\pi/4$ is the place where a careless argument may fail. The fact that both factors are nonzero does not by itself imply that a tensor is nonzero. The linear functional $\varphi$ isolates the rational component of $4+2\sqrt2$. After applying $\varphi\otimes\mathrm{id}$, the image is $4\otimes\pi/4$, which is nonzero because $\pi/4$ has nonzero class in $\mathbb R/\pi\mathbb Q$. Since a nonzero image cannot come from a zero tensor, the original tensor is nonzero.

Alternative Approaches

A proof can also be formulated as a special case of Dehn's theorem on equidecomposability. For polygons whose angles are rational multiples of $\pi$, one may define a simpler angle-length invariant tailored to the angle class $\pi/4$. The trapezoid contributes a nonzero amount proportional to $4+2\sqrt2$, while every rectangle contributes zero because all its angles are right angles. Additivity under tilings again yields a contradiction.

The Dehn invariant approach is preferable because it gives a complete and systematic explanation of why the trapezoid and a rectangle cannot belong to the same dissection class. The contradiction emerges directly from a standard additive invariant, with no need for ad hoc coloring or parity constructions.