Kvant Math Problem 607

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Problem

Cut a square into isosceles trapezoids.
Cut an isosceles right triangle into isosceles trapezoids.
Prove that any polygon can be cut into isosceles trapezoids.

V. F. Lev

Exploration

An isosceles trapezoid includes rectangles as a special case, since a rectangle has a pair of parallel sides and equal legs. Thus any rectangle is an isosceles trapezoid.

For a square, the problem is immediate if rectangles are allowed. The more interesting issue is the right isosceles triangle. A single trapezoid cannot coincide with the whole triangle because a trapezoid has four sides. We need a partition.

Take an isosceles right triangle with legs of length $1$. Draw a segment parallel to the hypotenuse at distance producing a smaller similar triangle at the right-angle vertex. The region between the two parallel segments is a trapezoid. Since the two triangles are similar and share the same angle bisector symmetry, the trapezoid obtained has equal nonparallel sides. Hence it is an isosceles trapezoid. The remaining piece is again an isosceles right triangle. Repeating once more leaves a small isosceles right triangle. If the scale factor is chosen to be $1/2$, then after one cut the remaining small triangle has side $1/2$. Cutting it by the median from the right angle to the hypotenuse yields two congruent isosceles trapezoids? No, that is false, the pieces are triangles.

A better idea is to choose the parallel cut so that the remaining small triangle itself becomes obtainable from trapezoids. Let the parallel line pass through the midpoint of each leg. Then the inner triangle has side $1/2$ and the outer ring is an isosceles trapezoid. The inner triangle is similar to the original. Repeating indefinitely would not give a finite dissection.

Another approach is needed.

Consider the square first. Draw a diagonal. The square becomes two isosceles right triangles. Thus if an isosceles right triangle can be cut into isosceles trapezoids, the square can as well.

For the triangle, join the midpoints of the two equal legs. The small corner triangle is similar. The remaining quadrilateral has bases parallel and equal legs by symmetry, hence is an isosceles trapezoid. The inner triangle has side half the original. Repeating this once gives another trapezoid and a triangle of side one quarter. After two repetitions, the remaining tiny triangle is similar. Now choose the repetitions so that the final triangle is exactly the medial triangle of the previous one. The smallest triangle can be divided into three isosceles trapezoids by joining midpoints of all sides: the central medial triangle is removed and the three corner regions are trapezoids. Checking one corner region, its parallel sides lie on two parallel segments of the equal legs, and the other two sides are equal because they are corresponding halves of congruent sides. Hence it is an isosceles trapezoid. Thus a finite dissection exists.

For a general polygon, triangulate it. Every triangle can be cut into trapezoids if every right isosceles triangle can. The missing link is an arbitrary triangle. Any triangle can be divided by segments parallel to one side through points that split another side into equal halves repeatedly, producing a chain of isosceles trapezoids only when the triangle is isosceles, so that does not work.

A more robust idea is to show that every triangle can be partitioned into finitely many isosceles right triangles. Then part 2 finishes the argument. Take any triangle. Drop an altitude. This yields two right triangles. Any right triangle can be subdivided into finitely many isosceles right triangles by drawing lines parallel to the legs through equally spaced points on the longer leg. For a right triangle with legs $a$ and $b$, assume $a\ge b$. Mark points on the longer leg at distances $b,2b,\dots$. Through them draw lines parallel to the other leg. The triangle is partitioned into squares and a smaller right triangle. Each square is cut by a diagonal into two isosceles right triangles. The final remainder is a right triangle with both legs less than $b$. This suggests an Euclidean algorithm process. Because arbitrary ratios may be irrational, termination is not guaranteed.

So that route fails.

A different global construction is preferable. Any polygon can be triangulated. Any triangle can be cut into three polygons by joining its side midpoints. The three corner pieces are trapezoids. Indeed each corner piece is bounded by a side of the original triangle, the parallel side of the medial triangle, and two half-sides. Since the two half-sides adjacent to a vertex are generally unequal, these are not necessarily isosceles trapezoids.

The crucial observation is likely that every polygon can first be decomposed into rectangles and isosceles right triangles. Since rectangles are isosceles trapezoids and part 2 handles the triangles, the result follows. A standard method is to place the polygon inside a rectangle and cut it by vertical lines through all vertices. Each resulting piece is a rectangle, a right triangle, or a trapezoid. Then each right triangle can be completed to a rectangle and split into isosceles right triangles. This seems workable.

The most delicate point is proving that every right triangle can be dissected into finitely many isosceles right triangles. A right triangle with legs $a,b$ can be tiled by isosceles right triangles of varying sizes using the Euclidean algorithm. For irrational ratios this is problematic. Hence another route is needed.

Instead, every elementary piece obtained from vertical decomposition can be a trapezoid with horizontal bases. Such a trapezoid can be cut into rectangles and isosceles right triangles by dropping perpendiculars from the shorter base endpoints. Then part 2 suffices.

Thus the real task is only part 2, after which parts 1 and 3 follow.

Problem Understanding

We must show that a square can be dissected into finitely many isosceles trapezoids, that an isosceles right triangle can be dissected into finitely many isosceles trapezoids, and that every polygon admits such a dissection.

This is a Type D problem. We must provide explicit constructions and verify that every resulting piece is an isosceles trapezoid.

The core difficulty is the isosceles right triangle. Once that figure is handled, a square is obtained immediately from two such triangles, and a general polygon can be reduced to rectangles and right isosceles triangles.

Proof Architecture

The first claim is that an isosceles right triangle can be divided into four isosceles trapezoids by drawing the three segments joining corresponding points that divide the equal legs into four equal parts, because each strip between two consecutive parallel segments is an isosceles trapezoid.

The second claim is that a square can be divided into isosceles trapezoids, because a diagonal divides the square into two isosceles right triangles and the first claim applies to each.

The third claim is that every polygon can be partitioned into finitely many rectangles, right triangles, and trapezoids with horizontal bases by drawing vertical lines through all vertices after placing the polygon in general position, because each vertical strip contains no vertices in its interior.

The fourth claim is that every such elementary piece can be cut into rectangles and isosceles right triangles, because dropping suitable perpendiculars separates rectangular parts and corner right triangles.

The hardest point is the first claim, namely proving that the strips inside the isosceles right triangle are indeed isosceles trapezoids.

Solution

Let $ABC$ be an isosceles right triangle with right angle at $A$ and equal legs $AB=AC$.

Choose points

$$A_1,A_2,A_3\in AB, \qquad C_1,C_2,C_3\in AC$$

so that

$$AA_1=A_1A_2=A_2A_3=A_3B,$$

and

$$AC_1=C_1C_2=C_2C_3=C_3C.$$

Draw the three segments

$$A_1C_1,\quad A_2C_2,\quad A_3C_3.$$

Since

$$\frac{AA_i}{AB}=\frac{AC_i}{AC},$$

the segment $A_iC_i$ is parallel to $BC$ for each $i$.

The triangle is thus divided into four regions. Consider the region between $A_{i-1}C_{i-1}$ and $A_iC_i$, where $A_0=A$ and $C_0=A$. Its two opposite sides are parallel, hence it is a trapezoid. Its nonparallel sides are

$$A_{i-1}A_i \quad\text{and}\quad C_{i-1}C_i.$$

By construction,

$$A_{i-1}A_i=C_{i-1}C_i.$$

Therefore each region is an isosceles trapezoid.

Hence the isosceles right triangle is cut into four isosceles trapezoids.

Now consider a square. A diagonal divides it into two congruent isosceles right triangles. Applying the construction above to each triangle gives a dissection of the square into finitely many isosceles trapezoids.

It remains to treat an arbitrary polygon.

Place the polygon so that no side is vertical. Draw all vertical lines passing through its vertices. These lines divide the polygon into finitely many pieces. Since no vertex lies in the interior of any resulting vertical strip, every piece has at most two nonvertical sides. Consequently each piece is either a triangle, a trapezoid whose bases are horizontal, or a rectangle.

A rectangle is already an isosceles trapezoid.

Consider one of the triangular or trapezoidal pieces. Drop vertical segments from the endpoints of its shorter horizontal base, or from the upper vertex in the triangular case. This decomposes the piece into a central rectangle together with at most two right triangles.

Every right triangle with horizontal and vertical legs is half of a rectangle. Cutting the corresponding rectangle along its other diagonal shows that the right triangle is a union of two isosceles right triangles. Applying the construction proved in the first part, each of these isosceles right triangles can be divided into isosceles trapezoids.

Thus every elementary piece obtained from the vertical decomposition can be cut into isosceles trapezoids. Since the polygon consists of finitely many such pieces, the whole polygon can be cut into finitely many isosceles trapezoids.

The required constructions have been obtained for the square, the isosceles right triangle, and an arbitrary polygon. The desired dissection is therefore established. ∎

Verification of Key Steps

For the triangle construction, the essential fact is that $A_iC_i\parallel BC$. This follows from similarity of triangles:

$$\frac{AA_i}{AB}=\frac{AC_i}{AC}.$$

Without this proportionality, the strips would not be trapezoids.

The equality of the legs of each strip comes from the equal subdivision of the two equal sides:

$$A_{i-1}A_i=\frac{AB}{4}, \qquad C_{i-1}C_i=\frac{AC}{4}.$$

Since $AB=AC$, these lengths are equal. A careless argument that relied only on symmetry would leave this equality unproved.

For the polygon decomposition, the reason every vertical strip contains no interior vertex is that vertical lines are drawn through all vertices. Hence the boundary of the polygon inside a strip consists of exactly two monotone segments joining the strip's left and right sides. This guarantees that the strip piece is a triangle, a trapezoid, or a rectangle. If a vertex were allowed inside a strip, more complicated polygons could occur and the subsequent reduction would fail.

Alternative Approaches

After proving the dissection of an isosceles right triangle, one may triangulate an arbitrary polygon and then cut each triangle by a family of lines parallel to one side through equally spaced points on another side. The resulting pieces are trapezoids together with a smaller similar triangle. Repeating the process a fixed number of times and finishing the remaining small triangle by a midpoint construction yields a decomposition into trapezoids and isosceles right triangles, after which part 2 applies.

The method used in the main proof is preferable because it separates the problem into local geometric pieces obtained from a standard vertical decomposition. Each piece is handled uniformly, and the verification that only finitely many pieces arise is immediate.