Table Mat Problem

[Guest]

The table mat shown is made of black and white circles joined so

that black circles are along the edge and white circles in the

interior. Twenty black and sixteen white circles are used.

There are two such table mats made with equal numbers of white and

black circles. What sizes are they?

see attachment for picture

[rookie1ja]

Could you upload the picture?

[Guest]

You can do an 8 x 6 grid. Makes the inside grid 6 x 4 = 24 whites, and the outside is 7 * 2 + 5 * 2 = 24 blacks.

[Guest]

8 X 6 is correct. The other answer is 12 X 5 = 30 Black & 30 White

[Guest]

So, I went back to the first, non-moved, post in this forum!

Lo, and behold, it was an interesting problem, but no one had posted a proof that there were only 2 unique solutions!

So, with time to spare and no one answering my three recent sequence posts, here we go!

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Let length = x and width = y

Inner area is (x - 2)*(y - 2). Using basic algebra, we get xy - 2x - 2y + 4.

Outer area can be represented by 2x + 2y - 4 (double counting the 4 corners).

Setting them equal and solving for y, we get:

xy - 2x - 2y + 4 = 2x + 2y - 4

xy - 4y = 4x - 8

y(x - 4) = 4x - 8

y = (4x - 8)/(x - 4)

We're looking for solutions for x and y that are positive integers. Quick substitution reveals that x = 1 gives a decimal answer for y, x = 2 and x = 3 yield non-positive answers for y, and x = 4 yields no solution. Basic graphical analysis would show us that there is a horizontal asymptote at y = 4, and therefore for x > 4, all answers are positive and greater than 4.

So, with more substitution:

x = 5 yields y = 12.

x = 6 yields y = 8

x = 7 yields y = 6.67 (no good, decimal answer).

x = 8 yields y = 6 (duh)

x = 12 yields y = 5 (duh)

x = 9, 10, 11 yields decimals between 5 and 6

x = 13 and higher yields decimals between 4 and 5.

[Guest]

Lo, and behold, it was an interesting problem, but no one had posted a proof that there were only 2 unique solutions!

excuse me I'll write later

Edited October 31, 2008 by nobody

[Guest]

Lo, and behold, it was an interesting problem, but no one had posted a proof that there were only 2 unique solutions!

There are 2 solutions:

Let a and b be the inner lengths.

The statement:

a*b = (a+2)*2 + (b+2)

a*b=2a + 2b + 4

ab - 2a - 2b = 4

a(b-2) -2b =4

a(b-2) - 2(b-2) -4 =4

(a-2)(b-2) = 8

8 has 2 set of multipliers: 8x1 and 4x2

The answers for the inners = (10x3) and (6x4)

I think this proves that there are 2 unique solutions!