The Missing Square, math that surprises you lesson

Four pieces drawn on a grid of unit squares, two right triangles and two stepped shapes, fit together into a figure that looks like a right triangle 13 squares wide and 5 squares tall. Slide the same pieces into new places and they fill what looks like the same figure, except that one grid square is now empty. No piece is cut, turned or stretched, so where does the extra square of space come from? This page compares the slopes of the two triangles, shows that the long edge of the figure is not straight, and accounts for the empty square exactly.

Key terms

Missing square puzzle

Definition The missing square puzzle is a dissection puzzle with four pieces on a grid of unit squares. Put the lower left corner of the grid at $(0, 0)$ , with $x$ measured to the right and $y$ upward. Two pieces are right triangles, a red one $8$ units wide and $3$ units tall and a blue one $5$ units wide and $2$ units tall. The other two are stepped pieces made of $7$ and $8$ unit squares. In the first arrangement the red triangle sits at the lower left, the blue triangle at the upper right, and the stepped pieces fill the $5$ by $3$ rectangle under the blue triangle. The pieces then fill the figure with corners $(0, 0)$ , $(13, 0)$ , $(13, 5)$ and $(8, 3)$ . In the second arrangement each piece is slid to a new place, without turning or flipping. The blue triangle sits at the lower left, the red triangle at the upper right, and the stepped pieces fill the $8$ by $2$ rectangle under the red triangle except for one unit square. The pieces then fill the figure with corners $(0, 0)$ , $(13, 0)$ , $(13, 5)$ and $(5, 2)$ , apart from that empty square. Both figures look like the same right triangle with legs $13$ and $5$ , but neither is a triangle. The long edge is made of the two hypotenuses, with slopes $3/8$ and $2/5$ , and since these slopes differ, the edge bends where the hypotenuses meet. In the first figure the edge bends inward and the area is $32$ , in the second it bends outward and the area is $33$ , and the pieces cover $32$ square units in both, which leaves exactly one square unit empty in the second.

In plain words The two hypotenuses almost line up, so the eye joins them into one straight edge. They do not line up, and swapping the triangles moves the bend from just below the straight line between the ends to just above it, which adds exactly one square unit of space.

Example The triangles have areas $\tfrac12 \cdot 8 \cdot 3 = 12$ and $\tfrac12 \cdot 5 \cdot 2 = 5$ , so all four pieces together cover

12 + 5 + 7 + 8 = 32

square units. Each figure splits into the two triangles and the rectangle under the upper triangle, so the areas of the first and second figures are

\begin{aligned} 12 + 5 + 15 &= 32, \\[0.4em] 5 + 12 + 16 &= 33. \end{aligned}

The stepped pieces contain $7 + 8 = 15$ unit squares. They fill the $5$ by $3$ rectangle of the first figure exactly, while the $8$ by $2$ rectangle of the second figure has $16$ squares, so one of its squares stays empty, the one with corners $(7, 1)$ and $(8, 2)$ . A true right triangle with legs $13$ and $5$ would have area $65/2 = 32.5$ , half a unit more than the first figure and half a unit less than the second.

Dissection puzzle

Definition A dissection puzzle asks for one figure to be cut into a finite number of pieces that can be fitted together into another figure. Each piece may be slid, turned or, if the puzzle allows it, flipped over, but it keeps its shape and size, and in each figure the pieces may not overlap or leave gaps. Moving a piece does not change its area, so two figures made from the same pieces in this way always have equal areas. When a rearrangement seems to gain or lose area, one of the figures is not what it appears to be, for example because pieces overlap, a gap is hidden, a piece is not the shape it seems to be, or an edge that looks straight is bent.

In plain words Cutting and rearranging can change a shape but never its total area. If the area seems to change, something in the picture is not what it looks like.

Example The parallelogram with corners $(0, 0)$ , $(5, 0)$ , $(7, 3)$ and $(2, 3)$ has base $5$ and height $3$ . Cut it along the vertical segment from $(2, 0)$ to $(2, 3)$ and slide the triangle with corners $(0, 0)$ , $(2, 0)$ and $(2, 3)$ five units to the right. It lands on $(5, 0)$ , $(7, 0)$ and $(7, 3)$ and completes the rectangle with corners $(2, 0)$ , $(7, 0)$ , $(7, 3)$ and $(2, 3)$ , so the parallelogram has the rectangle’s area $5 \cdot 3 = 15$ . In the missing square puzzle the pieces fit without overlaps in both figures and leave no gaps apart from the empty square, so the empty square has to be explained by a difference between the two outlines.

Slope

Definition The slope of the line through two points $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 \ne x_2$ is

m = \frac{y_2 - y_1}{x_2 - x_1},

the rise divided by the run. A straight line has the same slope between any two of its points, and lines with different slopes have different directions. Two fractions $a/b$ and $c/d$ with positive denominators can be compared through products, because $a/b < c/d$ exactly when $ad < bc$ .

In plain words Slope measures steepness as the change in height for each unit moved to the right.

Example The red hypotenuse rises $3$ over a run of $8$ and the blue one rises $2$ over a run of $5$ . Over a common run of $40$ these slopes are $15/40$ and $16/40$ . The straight segment from $(0, 0)$ to $(13, 5)$ has slope $5/13$ , which lies between them, since $3 \cdot 13 = 39$ is less than $8 \cdot 5 = 40$ and $5 \cdot 5 = 25$ is less than $13 \cdot 2 = 26$ . So

\frac{3}{8} < \frac{5}{13} < \frac{2}{5}.

In decimals the three slopes are $0.375$ , about $0.385$ and $0.4$ . The blue edge is steeper by only $1/40$ , a difference that is hard to see at the size of the drawing.

Collinear points

Definition Points are collinear if one straight line passes through all of them. Three points $(x_1, y_1)$ , $(x_2, y_2)$ and $(x_3, y_3)$ are collinear exactly when

\begin{aligned} &(x_2 - x_1)(y_3 - y_1) \\[0.4em] &\quad = (x_3 - x_1)(y_2 - y_1). \end{aligned}

When $x_1 < x_2 < x_3$ , this says that the segment from the first point to the second and the segment from the second point to the third have the same slope. When the two sides differ, half of their difference, taken as a positive number, is the area of the triangle with the three points as corners.

In plain words A path made of two straight pieces is one straight segment only if the second piece continues in exactly the direction of the first. Otherwise the path has a corner, and between the path and the straight segment joining its ends there is a triangle, which can be very thin.

Example For $(0, 0)$ , $(8, 3)$ and $(13, 5)$ the two sides are $8 \cdot 5 = 40$ and $13 \cdot 3 = 39$ . They differ, so the corner $(8, 3)$ of the first figure is not on the segment from $(0, 0)$ to $(13, 5)$ . At $x = 8$ that segment has height $40/13$ , so the corner lies $1/13$ of a unit below it, and the thin triangle between them has area $\tfrac12 (40 - 39) = \tfrac12$ . That is the half unit the first figure lacks. In the second figure the corner $(5, 2)$ lies $1/13$ of a unit above the segment and adds another thin triangle of area $\tfrac12$ . Together the two thin triangles form the long strip with corners $(0, 0)$ , $(8, 3)$ , $(13, 5)$ and $(5, 2)$ between the two bent edges. Its area is $1$ , the same as the empty square, but spread thinly along the whole edge. By contrast the points $(0, 0)$ , $(8, 3)$ and $(16, 6)$ are collinear, since $8 \cdot 6 = 48 = 16 \cdot 3$ .

Check yourself

In the second arrangement the corner of the long edge is at $(5, 2)$ . Use the collinearity test to show that $(0, 0)$ , $(5, 2)$ and $(13, 5)$ are not collinear, and find how far the corner lies above the straight segment from $(0, 0)$ to $(13, 5)$ .

Show the answer

The two sides of the test are $5 \cdot 5 = 25$ and $13 \cdot 2 = 26$ . They differ, so the three points are not collinear. The straight segment has slope $5/13$ , so at $x = 5$ its height is

\frac{5}{13} \cdot 5 = \frac{25}{13},

slightly less than $2$ . The corner is higher by $2 - 25/13 = 1/13$ of a unit. This is the outward bend that gives the second figure half a square unit more than a true triangle with legs $13$ and $5$ .

Suppose the red triangle were $10$ units wide and $4$ units tall, while the blue triangle stays $5$ by $2$ . Build both arrangements as before, with one triangle at the lower left and the other at the upper right on top of a rectangle filled by stepped pieces. Could a square go missing now?

Show the answer

No. The slopes are now $4/10 = 2/5$ and $2/5$ , which are equal. With the red triangle first, the test for $(0, 0)$ , $(10, 4)$ and $(15, 6)$ gives $10 \cdot 6 = 60$ and $15 \cdot 4 = 60$ . With the blue triangle first, the test for $(0, 0)$ , $(5, 2)$ and $(15, 6)$ gives $5 \cdot 6 = 30$ and $15 \cdot 2 = 30$ . In both arrangements the corner lies on the straight segment, so both figures are the same true triangle with legs $15$ and $6$ and area $45$ .

The rectangle under the upper triangle is $5$ by $4$ in one arrangement and $10$ by $2$ in the other, and both have area $20$ , so in each case the area is $20 + 5 + 20 = 45$ . Stepped pieces that fill one rectangle have total area $20$ , so if they fit into the other rectangle without overlapping, they fill it completely and no square goes missing. For example, four strips of $5$ unit squares fill both rectangles. The surprise depends on the two slopes being different.

The side lengths $2$ , $3$ , $5$ , $8$ and $13$ are consecutive Fibonacci numbers, in which each number is the sum of the two before it, and the next one is $21$ . Make a larger figure of the same kind from a right triangle $8$ wide and $3$ tall and a right triangle $13$ wide and $5$ tall, so that the long edge runs from $(0, 0)$ to $(21, 8)$ , with either triangle at the lower left and a rectangle under the upper triangle. Find the areas of the two figures and how far each corner lies from the straight segment.

Show the answer

The triangles have areas $\tfrac12 \cdot 8 \cdot 3 = 12$ and $\tfrac12 \cdot 13 \cdot 5 = 32.5$ . With the smaller triangle at the lower left, the corner is at $(8, 3)$ and the upper triangle stands on a $13$ by $3$ rectangle, so the area is $12 + 32.5 + 39 = 83.5$ . With the larger triangle at the lower left, the corner is at $(13, 5)$ and the upper triangle stands on an $8$ by $5$ rectangle, so the area is $32.5 + 12 + 40 = 84.5$ . A true triangle with legs $21$ and $8$ has area

\frac{21 \cdot 8}{2} = 84,

so the two figures again differ by exactly one square unit, half a unit on each side of the true triangle.

The straight segment has slope $8/21$ . At $x = 8$ its height is $64/21$ , so the corner $(8, 3)$ lies $1/21$ of a unit below it, and at $x = 13$ its height is $104/21$ , so the corner $(13, 5)$ lies $1/21$ of a unit above it. The collinearity test gives $8 \cdot 8 = 64$ against $21 \cdot 3 = 63$ , and $13 \cdot 8 = 104$ against $21 \cdot 5 = 105$ , so each thin triangle has area $\tfrac12$ . The bend is now even smaller than the $1/13$ of the original puzzle, while the difference in area stays one whole square.

The Missing Square

Video 1, chapter 1

Key terms

Missing square puzzle

Dissection puzzle

Slope

Collinear points

Check yourself