## Dissection of a Rectangle into Two Chessboards

Can you cut a 10×13 grid rectangle into pieces from which you'll be able to assemble two 8×8 chessboards? Andy Nisbet can and he shows how:

An explanation by Vladimir Zajic.

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny10·13 = 130 = 64 + 64 = 128 ?

Very good. However, the exact rearrangement would be possible only if the rectangle lattice points E14, G9, H6, and J1 were collinear (i.e., on the same line or on the same cut) and only if the square lattice points E9, G4, and H1 in both squares were also collinear. When 3 points A, B, and C are collinear (point B being in between), the distance AC = AB + BC. If this is not so, the points A, B, and C make a triangle and AC < AB + BC. Let's see what is true for the rectangle lattice points, for example E14, G9, and J1. By Pythagorean theorem, we have:

d(E14, J1) = Ö(5^{2} + 13^{2})
= Ö(25 + 169) =
Ö194

d(E14, G9) = Ö(2^{2} + 5^{2}) =
Ö(4 + 25) =
Ö29

d(G9, J1) = Ö(3^{2} + 8^{2}) =
Ö(9 + 64) =
Ö73

Since Ö29 + Ö73 ¹ Ö194, it must be Ö29 + Ö73 > Ö194. The points E14, G9, and J1 make a triangle with an area greater than zero. Similarly, we can show that the rectangle lattice points E14, H6, and J1 are not collinear. For the square lattice points E9, G4, and H1 we have

d(E9, H1) = Ö(3^{2} + 8^{2}) =
Ö(9 + 64) =
Ö73

d(E9, G4) = Ö(2^{2} + 5^{2}) =
Ö(4 + 25) =
Ö29

d(E4, H1) = Ö(1^{2} + 3^{2}) =
Ö(1 + 9) = Ö10

Again, since Ö29 + Ö10 ¹ Ö73, it must be Ö29 + Ö10 > Ö73. The points E9, G4, and H1 make a triangle with an area greater than zero.

Now, observe that the rectangle lattice
parallelogram E14-G9-H6-J1 with area
A_{P} > 0 is left over after cutting
the rectangle. On the other hand, triangles
E9-G4-H1 with areas A_{T} > 0 in both
squares are covered twice. So the above
graphical equation really is

130 - A_{P} = (64 + A_{T}) +
(64 + A_{T})

But what are the parallelogram and triangle
areas A_{P} and A_{T}? For the
simplest answer, use Pick's theorem at https://www.cut-the-knot.org/ctk/Pick.shtml:
Let P be a lattice polygon. Assume there are I
lattice points in the interior of P, and B
lattice points on its boundary. Let A be the
area of the polygon P. Then

A = I + B/2 - 1

Since the rectangle lattice parallelogram E14-G9-H6-J1 has 4 border points and no interior points, its area is

A_{P} = 0 + 4/2 - 1 = 1

Since
the square lattice triangles E9-G4-H1 have 3
border points each and no interior points, their
areas are

A_{T} = 0 + 3/2 - 1 = 1/2

and the modified graphical equation is

130 - 1 = (64 + 1/2) + (64 + 1/2)

129 =
129

Arithmetics has been saved, by George!

Regards, Vladimir

### Dissection Paradoxes

- Curry's Paradox
- Dissection of a 10×13 Rectangle into Two Chessboards
- A Faulty Dissection
- John Sharp's Paradox
- Langman's Paradox
- Popping A Square
- Sam Loyd's Son's Dissection

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny71745217