Hidden rectangles
Problem
Rectangles are considered different if they vary in size or have different locations.
How many different rectangles can be drawn along the lines on a chessboard?
How many of the rectangles are squares?
What happens on an n by n board?
Getting Started
What shape rectangles can you draw and in how many ways can these be fitted?
Student Solutions
A neat solution to the first part of the problem was received from Sana, Jenny, Chris and Rosion of Madras College, St. Andrews. I like the use of the lines rather than thinking about the rectangles in the same way as the squares (which is what I did). This idea can be generalised quite easily to an n x n chess board.
Other correct solutions were received from Andrei of School 205 Bucharest, Mary of Birchwood Community High School and Chen of The Chinese High School, Singapore.
Well done to all of you.
There are 1296 different rectangles on the chess board.
204 of these rectangles are squares.
First looking at the squares:
Consider placing a square of size 1 x 1 along the left hand edge of the chessboard. This square can be in any one of 8 positions (as there are 8 by 8 squares on a chessboard). Similarly, the square can be placed in any one of eight positions along the top edge. So the total number of
1 x 1 squares = 8 x 8 = 64.
A 2 x 2 square can occupy a 7 positions along the left hand edge and 7 positions along the top edge 7, giving 7 x 7 = 49 squares of size 2 x 2.
Continuing in this way we get squares of size 3 x 3, 4 x 4, and so on.
| Size of Square | Number of squares |
| 1 x 1 | 8 x 8 = 64 |
| 2 x 2 | 7 x 7 = 49 |
| 3 x 3 | 6 x 6 = 36 |
| 4 x 4 | 5 x 5 = 25 |
| 5 x5 | 4 x 4 = 16 |
| 6 x 6 | 3 x 3 = 9 |
| 7 x 7 | 2 x 2 = 4 |
| 8 x 8 | 1 x 1 = 1 |
So there are 204 squares.
Then looking at rectangles:
There are 9 vertical lines and 9 horizontal lines on the chess board.To form a rectangle you must choose 2 of the 9 vertical lines and 2 of the 9 horizontal lines.
For the two horizontal lines: the first line can be chosen in 9 ways the second in eight ways. This would imply that you could tell the difference between lines 1 and 3 say and 3 and 1, which is not the case so you need to divide 9 x8 8 by 2, making 36.
Similarly you can choose the two vertical lines in 36 ways.
So the number of rectangles is given by 362
362 = 1296
The following approach utilises several of the other solutions sent in.
A rectangle (or square) will have a height between 1 and 8 units and a width between 1 and 8 units. Tis can be represented by a table with each possible width represented by a column and each eight by a row.
The entries in the table below then indicate the number of each size rectangle on the chessboard (using similar arguments to first part of the problem above).
The final column of the table gives the total number of rectangles in each row.
| Height/Width | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Total |
| 1 | 8 x 8 | 7 x 8 | 6 x 8 | 5 x 8 | 4 x 8 | 3 x 8 | 2 x 8 | 1 x 8 | 8 x(1+2+3+4+5+6+7+8}= 8 x 36 |
| 2 | 8 x 7 | 7 x 7 | 6 x 7 | 5 x 7 | 4 x 7 | 5 x 7 | 2 x 7 | 1 x 7 | 7 x(1+2+3+4+5+6+7+8}= 7 x 36 |
| 3 | 8 x 6 | 7 x 6 | 6 x 6 | 5 x 6 | 4 x 6 | 3 x 6 | 2 x 6 | 1 x 6 | 6 x(1+2+3+4+5+6+7+8}= 6 x 36 |
| 4 | 8 x 5 | 7 x 5 | 6 x 5 | 5 x 5 | 4 x 5 | 3 x 5 | 2 x 5 | 1 x 5 | 5 x(1+2+3+4+5+6+7+8} |
| 5 | 8 x 4 | 7 x 4 | 6 x 4 | 4 x 4 | 5 x 4 | 3 x 4 | 2 x 4 | 1 x 4 | 4 x(1+2+3+4+5+6+7+8} |
| 6 | 8 x 3 | 7 x 3 | 6 x 3 | 4 x 3 | 5 x 3 | 3 x 3 | 2 x 3 | 1 x 3 | 3 x(1+2+3+4+5+6+7+8} |
| 7 | 8 x 2 | 7 x 2 | 6 x 2 | 4 x 2 | 5 x 2 | 3 x 2 | 2 x 2 | 1 x 2 | 2 x(1+2+3+4+5+6+7+8} |
| 8 | 8 x 1 | 7 x 1 | 6 x 1 | 4 x 1 | 5 x 1 | 3 x 1 | 2 x 1 | 1 x 1 | 1 x(1+2+3+4+5+6+7+8} |
| (1+2+3+4+5+6+7+8}x 36 = 362 |
Therefore the total number of rectangles in an 8 x 8 grid is (1 + 2 + 3 + ... + 8)2
The total number of rectangles in an n x n grid is (1 + 2 + 3 + ... + n)2 = (n2 (n+1)2 )/4. This uses the formula for the sum of the first n natural numbers, an arithmetic progression. For more details of this look at the proof sorter (link).
The number of squares is: (12 + 22 + 32 + ... n2 ) = (n(n+1)(2n+1))/6 ( for a proof see Telescoping series )