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Squares, Rectangles on a Chessboard


Date: 08/14/97 at 20:55:44
From: Henry Taur
Subject: Number problem

How many squares are there on a chessboard?


Date: 08/15/97 at 05:46:20
From: Doctor Anthony
Subject: Re: Number problem

Consider the lefthand vertical edge of a square of size 1 x 1.  
This edge can be in any one of 8 positions. Similarly, the top 
edge can occupy any one of 8 positions for a 1 x 1 square.  
So the total number of 1 x 1 squares = 8 x 8 = 64.

For a 2 x 2 square the lefthand edge can occupy 7 positions and 
the top edge 7 positions, 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.

We can summarize the results as follows:

   Size Of square      Number of squares
   ---------------     -----------------
      1 x 1               8^2 = 64
      2 x 2               7^2 = 49
      3 x 3               6^2 = 36
      4 x 4               5^2 = 25
      5 x 5               4^2 = 16
      6 x 6               3^2 =  9
      7 x 7               2^2 =  4
      8 x 8               1^2 =  1
                      ---------------
                      Total   = 204

There is a formula for the sum of squares of the integers 
1^2 + 2^2 + 3^2 + ...  + n^2

                n(n+1)(2n+1)
         Sum  = ------------
                     6

In our case, with n = 8, this formula gives 8 x 9 x 17/6 = 204.

As an extension to this problem, you might want to calculate the 
number of rectangles that can be drawn on a chessboard.

There are 9 vertical lines and 9 horizontal lines. To form a rectangle 
you must choose 2 of the 9 vertical lines, and 2 of the 9 horizontal 
lines. Each of these can be done in 9C2 ways = 36 ways. So the number 
of rectangles is given by 36^2 = 1296.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


From the discussion group geometry-puzzles:

Date: 10 Oct 98 12:15:05
From: Anonymous
Subject: chess puzzler

How many rectangles can be found on an 8x8 chess board (squares are 
rectangles)?


From: Anonymous
Subject: chessboard rectangles
Date: Sun, 11 Oct 1998 02:02:13

Someone asked the number of rectangles on a chessboard. The answer is
"Many, in fact 1296."

There are 64 one-by-one squares, 
49 two-by-two squares, ...
(8-n)^2 n-by-n squares, ...
1 eight-by-eight square; 

2 x (7x8)  one-by-two rectangles, 
2 x (6x8) one-by-three rectangles, ... 
2 x (1x8) one-by-eight rectangles; 
2 x (6x7) two-by-three rectangles, ... 
2 x (1x7) two-by-eight rectangles; ...; 
2 x (1x2) seven-by-eight rectangles.

This can all be simplified to find the sum total as the sum of the 
cubes of integers 1 to 8, which is 8^2 x 9^2 / 4 or 36^2 = 1296.

Have I missed any? Or added wrong? What if we reduce or augment the size
of the chessboard? Can we find a general formula? What is it? How do we
use induction to show that it is indeed general?

Mary Krimmel 


Date: Sun, 11 Oct 1998 14:13:23 -0400
From: Alan
Subject: Re: chessboard rectangles

I agree with your total (1296).   With a little rearranging your method
will produce the general formula for the number of rectangles on an n by n
chessboard.

Count the rectangles by rows.   

In row 1 there are n 1 by 1s , n-1 1 by 2s, ... two 1 by n-1s and 
one 1 by n. Row Total = 

   n + (n-1) + (n-2) + ... + 3 + 2 + 1 = n(n+1)/2 rectangles.

But of course there are n rows, giving n (row sum).

Now count all rectangles of height 2.  Start in the bottom row.  There are 
n 2 by 1s and n-1 2 by 2s and ...  and one 2 by n. Row Total =

   n  + (n-1) + (n-2) + ... + 3 +  2 + 1 = n (n+1)/2

The row total is the same, as it will be for all the rectangles of height
3, 4, ... n  because they all share the same bases at the bottom of the
board.  However, there are only n-1 rows of height 2 and n-2 rows of
height 3 etc.  Thus,

   number 1 by any totals:    n   [n(n+1)/2]
   number 2 by any totals:  (n-1) [n(n+1)/2]
   number 3 by any totals:  (n-2) [n(n+1)/2]   

                        ...

   number n by any totals:      1 [n(n+1)/2]

So the total number of rectangles is [n(n+1)/2](n + n-1) + ... + 3+ 2+1)

or   Total = [n(n+1)/2]^2  

which, as you mentioned, is the sum of the cubes from 1 to n.

Alan Lipp
    
Associated Topics:
High School Permutations and Combinations
High School Puzzles
Middle School Puzzles

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