Squares, Rectangles on a ChessboardDate: 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 |
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