Yehao Wang

MT191 Fundamental Mathematics I

Professor Chi-Keung Cheung

May 5, 2014

Art of cutting and assembling graph

When we are given two fixed polygons, can we cut one

polygon into several pieces and assemble them into

another polygon with the same area?

When I first see the problem, I have no idea, but I can

use several steps to get a conclusion.

Most of us may have seen the problem below:

A graph is formed by 5 small squares with same area,

please cut it to three part and assemble into a big

square.

Suppose that the area of each small square is 1. We may

notice that when we assemble it into a new square, the

area is still 5, so the length of a side is √5. As a

result, we will try to get some line segments with length

√5, and then the answer is easy to come.

Inspired by this, we can cut any two squares into 5

pieces and assemble into a big square. (shown below)

So we get a common conclusion in graph cut and assemble

problems.

Conclusion 1: Any two squares can be but and assembled

into a big square.

An interesting follows, if there is three squares, can

you cut them and assemble into a big square? The answer

is yes. With conclusion 1, we can get a new square by either

two squares and then use conclusion 1 again to from a big

square. Obviously, no matter how many squares are there

at the beginning, the method is suitable: continue to cut

and assemble two squares into a new square until there is

only one big square left. So we can draw a stronger

conclusion.

Conclusion 2: Any number of squares can be cut and

assembled into a big square.

In fact, to prove conclusion 2, we unconsciously use some

basic qualities of graph cutting and assembling. Now, I

want to provide two obvious conclusions without proves

because it’s not difficult to understand.

Conclusion 3: If graph A can be cut and assembled into

graph B, and graph B can be cut and assembled in to graph

C, so graph A can be cut and assembled into graph C.

Conclusion 4: If graph A can be cut and assembled into

graph B, then graph B can be cut and assembled into graph

A.

Then we think further, can we cut and assemble any

rectangle to a square?

Shown as picture below, in rectangle ABCD, length AB=a,

width BC= b, so the length of side in the square which

has the same area is √ab. We cut out a segment DG=√ab on

DC, and a segment BH=√ab on BA. Then connect and extend

CH, intersect with the extended line of DA on E. After

that, through G, we makes an vertical line of DC and

through E, we makes an vertical line of DE. The two

vertical lines intersect on F. Obviously DEFG is a

rectangle.

We can easily find that ΔAHE and ΔGCM are congruent, and

ΔEFM and ΔHBC are congruent. So when we cut ABCD into

three pieces with CH and GM, we can assemble them into

rectangle DEFG. Because one side of DEFG is √ab and the

area is ab, the length of another side is also √ab. So

DEFG is a square.

However, there is a restricted condition: the length

cannot be 4 times longer than the width, or some part of

ΔGCM would be out of the rectangle like picture below

shows.

So we get an imperfect conclusion.

Conclusion 5: If the scale of length and width is smaller

than 4:1 in a rectangle, the rectangle can be cut and

assembled into a square.

Can we cut and assemble a rectangle into a square when

the scale of length and width is larger than 4:1? The

answer is yes, too. As picture below shows, we cut the

rectangle into two same parts with a line that is

perpendicular to the long line. Then we can get a new

rectangle if we put one piece on another. We can repeat

the operation until we get a rectangle in which the scale

of length and width is smaller than 4:1.

Conclusion 6: Any rectangle with a scale of length and

width which is larger than 4:1 can be cut and assembled

into a rectangle with a scale of length and width which

is smaller than 4:1 finally.

We can combine the two above conclusions, and also with

conclusion 3, we get a perfect conclusion.

Conclusion 7: Any rectangle can be cut and assembled into

a square.

Then we may think that for any parallelogram, we can

change it into a rectangle as in the picture below.

So we have:

Conclusion 8: Any parallelogram can be cut and assembled

into a rectangle.

Also, we have got that any rectangle can get a square. So

any parallelogram can be cut and assembled into a square.

When we cut a triangle with its median line, we can also

find that any triangle can become a parallelogram (like

picture below).

So we have a new conclusion.

Conclusion 9: Any triangle can be cut and assembled into

a parallelogram.

From the conclusion, we can learn that a triangle can

also be cut and assembled into a square. At the same

time, we notice that a quadrilateral can always be

divided into two triangles, each triangle can be cut and

assembled into a square and the two squares can become a

big square, so we can get conclusion 10.

Conclusion 10: Any quadrilateral can be cut and assembled

into a square.

In fact, not only quadrilateral, any polygon can be

divided into several triangles and each triangle can be

cut and divided into a square. Any number of squares can

be cut and assembled into a big square like picture below

presents.

As a result, we get a crazy conclusion.

Conclusion11: Any polygon can be cut and assembled into a

square.

Any two given polygons A and B can be cut and assembled

into squares from conclusion 11. If the two polygons have

the same area, they can get a same square. From conclusion

4, we know that the square can also be cut and assembled

into polygon A or polygon B. Then combine conclusion 3, we

can get a method that the two polygons can transform

mutually: Polygon A ↔ Square↔ Polygon B. So it seems that

we get another amazing conclusion.

Conclusion 12: Any given two polygons with the same area,

they can be cut and assembled into each other!

The theorem is called Bolyai–Gerwien Theorem, which is

proven by mathematicians Farkas Bolyai and Paul Gerwien

in 1833 and 1835.

Reference

Sen Gu, “Pleasure of thought, matrix 67’s math notes”.