On squared paper lies a 13 × 5 triangle made of four pieces: a red and a blue triangle and two shapes made of squares. The pieces swap places and form the same 13 × 5 triangle. Only now one square inside is empty. The same pieces, the same outline. Where does the hole come from? Lock in your answer, then check it.
Neither figure is a triangle. The red piece rises 3 squares over 8, the blue one 2 over 5. The slopes, 0.375 and 0.4, differ, so the long side of each figure is bent, and the bend is so slight that the eye sees a straight line. In the first figure it sags inwards and the area is 32 squares. In the second it bulges outwards and the area is 33. The pieces fill 32, so one square stays empty.
This puzzle is credited to Paul Curry, an amateur magician from New York, and is also called Curry's paradox. Everything is in plain sight: the grid, the pieces, squares you can count.
The corners of the pieces sit on grid points.
The red triangle on the left, the blue one at the top right, the two square shapes under the blue one. Together they make a right triangle with legs of 13 and 5 squares.
The pieces are only slid, not turned or flipped: the blue one is now on the left, the red one at the top right. The outline is again 13 × 5, but one square inside is empty.
The same pieces, the same outline. Where did the empty square come from?
Figure 1 at the top, figure 2 below. The empty square has a dashed outline.
"Same pieces, so the same area. There can't be a hole"
Yet there it is, plain to see. The picture is cheating somewhere. Where exactly?
Pick the explanation you think is right. Your answer counts as soon as you click.
You can change your mind, but the first choice goes into the stats.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Count the squares in each piece and add them up. And how many squares does a right triangle with legs 13 and 5 hold?
Compare the red and blue triangles: how much does each rise for one square across?
Hold a ruler from the bottom-left corner to the top-right one. Does it pass exactly through the corners (8, 3) and (5, 2)?
The next rounds show where the square went: pieces you can move yourself, a ruler and a magnifier, a million random points and the explanation. Answer round 1 first, or go on?
You chose to always show all rounds.
Drag the pieces with a finger or the mouse; they snap to the grid. Build both figures yourself or with the buttons, then switch on the corner-to-corner ruler and the magnifier at the bend.
The ruler joins the corners (0, 0) and (13, 5) of a true triangle. The magnifier shows the spot where the red and blue pieces meet, six times larger.
The eye is easy to fool; counting is harder. The computer throws random points into the 13 × 5 rectangle and counts what share of them land under the long side of each figure. That share times the rectangle's 65 squares is the figure's area.
The sizes 2, 3, 5, 8, 13 are neighbouring Fibonacci numbers, where each is the sum of the two before. Take the next ones and the trick repeats: the sliver between the two bent lines gets ever thinner, yet its area is exactly one square every time.
Once you answer, the simulation code appears here so you can run it yourself.
The same simulation, short and with no libraries. Run the JavaScript in your browser console (F12) or in Node; the Python needs only the standard library.
// Curry's triangle: throw random points into a 13 × 5 rectangle and measure the areas under three lines
const N = 1e6;
const fig1 = x => x <= 8 ? 3 * x / 8 : 3 + 2 * (x - 8) / 5; // red 8 × 3, then blue 5 × 2
const fig2 = x => x <= 5 ? 2 * x / 5 : 2 + 3 * (x - 5) / 8; // blue, then red
const line = x => 5 * x / 13; // the true diagonal
let a1 = 0, a0 = 0, a2 = 0;
for (let i = 0; i < N; i++) {
const x = Math.random() * 13, y = Math.random() * 5;
if (y < fig1(x)) a1++;
if (y < line(x)) a0++;
if (y < fig2(x)) a2++;
}
const area = k => (65 * k / N).toFixed(3);
console.log('figure 1 ', area(a1), 'exact 32');
console.log('triangle ', area(a0), 'exact 32.5');
console.log('figure 2 ', area(a2), 'exact 33');
console.log('sliver between them', area(a2 - a1), 'exact', 8 * 2 - 3 * 5);
# Curry's triangle: throw random points into a 13 × 5 rectangle and measure the areas under three lines
import random
N = 1_000_000
fig1 = lambda x: 3 * x / 8 if x <= 8 else 3 + 2 * (x - 8) / 5 # red 8 × 3, then blue 5 × 2
fig2 = lambda x: 2 * x / 5 if x <= 5 else 2 + 3 * (x - 5) / 8 # blue, then red
line = lambda x: 5 * x / 13 # the true diagonal
a1 = a0 = a2 = 0
for _ in range(N):
x, y = random.random() * 13, random.random() * 5
a1 += y < fig1(x)
a0 += y < line(x)
a2 += y < fig2(x)
area = lambda k: f"{65 * k / N:.3f}"
print("figure 1 ", area(a1), "exact 32")
print("triangle ", area(a0), "exact 32.5")
print("figure 2 ", area(a2), "exact 33")
print("sliver between them", area(a2 - a1), "exact", 8 * 2 - 3 * 5)
The area doesn't change, and the pieces are cut exactly. What changes is the outline: the "hypotenuse" of each figure is made of two segments with different slopes. The difference is so small that the brain fills in a straight line.
“In plain words” is hidden on all puzzles.
The red triangle is half of an 8 × 3 rectangle, so 12 squares. The blue one is half of 5 × 2, so 5. With the 7 and 8 square shapes that makes 32. But a true 13 × 5 triangle is half of a 13 × 5 rectangle: 32.5 squares.
So the pieces can't make an exact 13 × 5 triangle: they are half a square short. This count shows there is a trick, but not yet where it is. The next card does that.
The red hypotenuse rises 3 squares over 8 squares across, 0.375 per square. The blue one rises 2 over 5, 0.4 per square. The blue one is slightly steeper, so placed end to end they make a bent line, not a straight one.
In figure 1 the gentler red segment comes first, then the steeper blue one: the line sags inwards. In figure 2 it is the other way round: the line bulges outwards. The slopes differ by only 0.025, an angle of about 1.25° between the segments.
Lay the outlines of the two figures on top of each other. Between the sagging and the bulging line there is a thin sliver shaped like a very long parallelogram. Its area is exactly 1 square, but it is stretched along the whole diagonal, almost 14 squares long and at most 1/8 of a square thick. On paper with 1 cm squares that is 1.25 mm at its widest, and towards the ends of the diagonal the sliver tapers to nothing. In the picture above the sliver's height is magnified tenfold.
When the pieces go from figure 1 to figure 2, the outline "inflates" by this sliver. The outline's area grows from 32 to 33, while the pieces still cover 32. The difference gathers into one visible empty square inside.
“Rigorous” is hidden on all puzzles.
| Figure 1 | Triangle | Figure 2 | |
|---|---|---|---|
| bend at | (8, 3) | none | (5, 2) |
| height at x = 8 | 3 | 40/13 ≈ 3.077 | 3.125 |
| height at x = 5 | 1.875 | 25/13 ≈ 1.923 | 2 |
| area under the edge | 32 | 32.5 | 33 |
| covered by pieces | 32 | – | 32 + 1 |
Between x = 5 and x = 8 both bent lines run parallel, and the vertical distance between them is 1/8 of a square throughout. The true diagonal runs between them: 1/13 of a square above the corner (8, 3) and 1/13 below the corner (5, 2).
Figure 1: there is nothing under the red triangle, and under the blue one stands a 5 × 3 rectangle made of the two square shapes. Area 12 + 5 + 15 = 32.
Figure 2: under the red triangle stands an 8 × 2 rectangle. Area 5 + 12 + 16 = 33. But the two square shapes have only 7 + 8 = 15 squares, so one square of the 8 × 2 rectangle stays empty.
The whole trick is that 5 × 3 = 15 while 8 × 2 = 16. The rectangles under the triangles differ by one square, while the triangles merely swap places.
Three points lie on one line if and only if the slopes are equal. Here they differ, so the "hypotenuse" is bent.
The sliver is the parallelogram spanned by the side vectors (8, 3) and (5, 2). A parallelogram's area is the absolute value of the determinant of their coordinates.
So each figure's area differs from the true triangle by half the sliver.
“Deeper” is hidden on all puzzles.
All the sizes come from the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21…: red 8 × 3, blue 5 × 2, outline 13 × 5. The ratios 2/5 = 0.4, 3/8 = 0.375 and 5/13 ≈ 0.385 close in from both sides, in turn, on one number, 1/φ² ≈ 0.382, where φ ≈ 1.618 is the golden ratio. So all three slopes are very close, and none equals another.
And the sliver's area is always exactly 1. For neighbouring Fibonacci numbers Fn+1·Fn−2 − Fn·Fn−1 = ±1, for example 8 · 2 − 5 · 3 = 1 and 13 · 3 − 8 · 5 = −1. It is a close relative of Cassini's identity Fn−1·Fn+1 − Fn² = ±1. The bigger the numbers, the thinner the sliver, but its area stays one square.
"Vanishing area" tricks on squared paper are older than this triangle. In 1774 William Hooper, in his book "Rational Recreations", turned a shape of area 32 into a rectangle of area 30. In 1868 Oskar Schlömilch published the cutting of an 8 × 8 chessboard into a 5 × 13 rectangle, "64 = 65", and in 1879 Victor Schlegel linked such tricks to the Fibonacci numbers.
Martin Gardner credited the triangle with a hole to Paul Curry, an amateur magician from New York, and dated it 1953. Curry is also known for the card trick "Out of This World". What was new in his version is that the extra square doesn't vanish into sliver-thin gaps but appears as a visible hole inside a figure with the same outline.
The same one-square sliver turns a chessboard into a rectangle of 65 squares. And "proofs" that look flawless often break at a single step, just like our hypotenuse.
An 8 × 8 chessboard is cut into two 3 × 8 triangles and two trapezoids with sides 3 and 5. Rearranged, they make a 5 × 13 rectangle, and 64 squares become 65. The rectangle's diagonal is bent again: it is made of the same vectors (8, 3) and (5, 2) as our "hypotenuse", and a sliver of area 1 is left between the pieces.
It works for any square whose side is a Fibonacci number Fn, because Fn−1·Fn+1 − Fn² = (−1)n. A square of 8 × 8 = 64 gives 5 × 13 = 65, a square of 13 × 13 = 169 gives 8 × 21 = 168: here the pieces overlap slightly and a square "vanishes". A 21 × 21 = 441 square gives 13 × 34 = 442.
A sophism is an argument that looks right but reaches a false conclusion. It is built like the triangle with a hole: almost every step is flawless, and the error hides in one unremarkable place. In the triangle it is a 1.25° bend; in "proofs" it is a division by zero, a square root that forgets the sign, or a drawing where a point ended up on the wrong side.
Hence the method. If the conclusion is false and the steps seem right, don't look for the error "in general". Check each step on its own, and check the drawing with numbers, the way we checked the slopes 3/8 and 2/5.
Two segments follow each other: the first goes a squares right and b up, the second c right and d up.
The cheapest "is it straight?" check is to compare the slopes as fractions or work out a·d − b·c. Zero means the three points are on one line. Any other whole number means a bend, and that number is exactly the area of the sliver that rearranging the pieces can "hide". For Curry's triangle it is 8 · 2 − 3 · 5 = 1.
Computational geometry, which drawing software is built on, checks whether three points lie on one line the same way: it computes the same expression a·d − b·c (called the orientation test) rather than looking at the picture. Try it by hand: from (0, 0) to (8, 3) and on to (13, 5): 8 · 2 − 3 · 5 = 1, so it is bent.
All these tricks rest on one thing: two segments with very close slopes that the eye reads as one straight line. Check each row: multiply the sides.
| Trick | Before → after | Where the difference went |
|---|---|---|
| Hooper, "Rational Recreations", 1774 | 32 → 30 | The pieces in the rectangle actually overlap slightly. The drawing in the first edition was inaccurate and was corrected in the 1782 edition. |
| Chessboard, Schlömilch, 1868 | 8 × 8 = 64 → 5 × 13 = 65 | A sliver of area 1 along the rectangle's diagonal. |
| A 13 × 13 square | 169 → 8 × 21 = 168 | The pieces overlap along a strip of area 1. |
| Curry's triangle, 1953 | 32 → 32 + 1 | The outline inflates by 1 square, which becomes a hole inside. |
William Hooper described a trick called "the geometric money", in which an area of 32 becomes 30. The drawing in the first edition of "Rational Recreations" was itself inaccurate and was corrected only in the 1782 edition.
Oskar Schlömilch printed the cutting of a chessboard into a 5 × 13 rectangle in a paper called "Ein geometrisches Paradoxon" in the Zeitschrift für Mathematik und Physik.
Victor Schlegel generalised the chessboard cutting in the same journal and showed that the pieces' sizes should be neighbouring Fibonacci numbers.
Charles Dodgson, the author of "Alice", looked for all the squares that can be recut into a rectangle with a "gain" of one square. He was apparently unaware that Schlegel had already done it. Warren Weaver described his notes in 1938.
Paul Curry (1917–1986) was an amateur magician from New York. Martin Gardner dated his triangle to 1953 and wrote about it in his 1956 book "Mathematics, Magic and Mystery".
The red and blue hypotenuses are tilted at ≈ 20.56° and ≈ 21.80°, a difference of ≈ 1.25°. On a drawing with 1 cm squares, the corner (8, 3) is off the true diagonal by only 1/13 cm ≈ 0.77 mm.
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