Haberdasher's Puzzle: Four Hinged Pieces, Triangle to Square

The puzzle

The haberdasher's puzzle is a dissection puzzle (one set of pieces, cut so it rebuilds more than one shape) that turns an equilateral triangle into a square. In Henry Dudeney's The Canterbury Puzzles (1919 edition), a haberdasher holds up a triangle of cloth and asks how "this piece of cloth may be cut into four several pieces that may be put together to make a perfect square."

Dudeney adds that it works "without turning over any piece." Compare the missing square puzzle, where two outlines built from the same pieces differ in area by one square (Missing Square Puzzle: Neither Shape Is Really a Triangle).

Show the solution

Where the cuts go

The 1919 edition's solution uses a ruler and compass. Put A at the top, B at the bottom left and C at the bottom right.

1. Mark D and E, the midpoints of AB and BC.
2. Extend AE past E to F, making EF equal to EB.
3. Draw a half circle on AF, and extend EB past B to meet it at H. EH is the square's side.
4. Swing an arc about E from H to AC, and call that point J. Mark K on JC with JK equal to EB.
5. Drop perpendiculars from D and K to EJ, meeting it at L and M.
6. Cut along EJ, DL and KM.

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Step 3 works because EH × EH equals AE × EF, the triangle's height times half its base, which is its area.

We reran the construction with a short program that keeps every square root exact. The four areas add up to exactly √3, the triangle's area for sides of 2. No pieces overlap in either shape. Every corner is in the CSV file: haberdashers-puzzle-pieces.csv (CSV; where this file comes from: source, version and license, JSON) Source data: Computed for this page from Dudeney's construction (The Canterbury Puzzles, 1919 edition, solution 26), with exact square roots

Swing the chain

Hold DBEL still. Give ADLJ a half turn about D, so A lands on B, and give ECKM a half turn about E, so C lands on B. KJM rides along with ECKM, then makes its own half turn about K. A half turn never turns a piece over.

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Dudeney describes a model "in polished mahogany with brass hinges" that closes one way into the triangle and the other way into the square. Our program turned the hinges at D, E and K together, checking every 1°: nothing overlaps if D turns counterclockwise while E and K turn clockwise. That is one path, checked at sampled angles, not a proof.

Repeat the triangle across the plane with half turns about D, E and K. Then do the same with the square. Each tiling repeats in two directions, like the patterns in Wallpaper Groups: Only 17 Symmetry Types for Two-Way Repeats. Laid together, the square copies cut every triangle along EJ, DL and KM.

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Who solved it first?

Dudeney posed the puzzle in his Weekly Dispatch column on 6 April 1902, according to Demaine, Kamata and Uehara. On 20 April he named C. W. McElroy of Manchester as a solver, and on 4 May he printed the four-piece answer. The three authors call it unclear whether Dudeney or McElroy found it first. Dudeney's 1919 text credits "Mr. C. W. M'Elroy" as the only reader who sent the correct solution.

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In his own account, Dudeney showed the problem to the Royal Society on 17 May 1905. The Canterbury Puzzles first appeared in 1907, and a second edition followed in 1919. The three-piece proof was first posted in December 2024 and revised in August 2025.

Why not three pieces?

Two pieces never work. Each piece must fit inside the square, whose widest span is its diagonal: about 1.861 when the triangle's sides are 2. The triangle's corners are 2 apart, so no piece can hold two of them, and two pieces cannot hold three corners. The 2024 proof's first step (Lemma 5.1) is the same argument for straight-edged pieces; it uses only distances, so this page extends it to flipped or curved pieces.

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Three pieces are the hard case. Theorem 1.1 of the paper reads: "There is no dissection with three or fewer polygonal pieces between a square and an equilateral triangle, when we forbid flipping pieces." Polygonal pieces have straight edges. The paper lists flipped and curved pieces as open problems, though the authors suspect the answer is still no.

Where people get stuck

How big should the square be?

Its side is about 0.658 times the triangle's side (the fourth root of 3, halved), so the areas match. A triangle with 12 cm sides needs a square with sides of about 7.9 cm.

Do any pieces get flipped over?

No. Dudeney's four pieces only slide and turn, and every move in this page's check is a half turn, which keeps a piece face up.

Can it be done in three pieces?

Not with straight cuts if no piece is flipped: Demaine, Kamata and Uehara proved that in a paper first posted in 2024. Flipped or curved pieces are an open question in that paper.