## Shading Fraction of a Rectangular and Koeller Ratio – Flavour of Galois 2017

Another interesting problem in Galois 2017 is the following one, which considers the Koeller Shading Ratio. The problem is as below.

__ Problem.__ A Koeller rectangle:

- is an m by n rectangle where m,n are integers with m>=3 and n>=3,
- has lines drawn parallel to its sides to divide it into 1 by 1 squares, and
- has the 1 by 1 squares along its sides unshaded and the 1 by 1 squares that do not touch its sides shaded (as in figure).

An example of a Koeller rectangle with m = 6 and n = 4 is shown. (The cell marked S is shaded, the blank cell is unshaded.)

S | S | S | S | ||

S | S | S | S | ||

For a given Koeller rectangle, let r be the ratio of the shaded area to the unshaded area.

Now are the questions:

(a) [Answer only] Determine the value of r for a Koeller-rectangle with m = 14 and n = 10.

(b) [Full solution required] Determine all possible positive integer values of u for which there exists a Koeller-rectangle with n = 4 and r = (u/77).

Well, readers are advised to try first before reading the solution below, followed by some remarks.

**Solution: **

First, develop some formulae.

Unshaded region = m + m + n + n – 4 = 2 (n+m-2)

Shaded region = (m-2) (n-2) = mn – 2m – 2n +4

Ratio r (of shaded to unshaded)

r = (m-2) (n-2) ⁄

_{ (2 (n+m-2)) }

For (a), m = 14, n = 10, therefore

r = (12) (8) ⁄ _{ (2 (22)) }= (24) ⁄ _{ (11) }.

For (b), since n = 4, r = (m-2) ⁄ _{ (m+2)}

As its said that r = (u / 77), so:

u = 77 r = (77) (m-2) ⁄ _{ (m+2) }(*)

As (m-2) ⁄ _{ (m+2)} = 1 – [4 ⁄ _{ (m+2)}], only 2 or 4 divides both m+2 and 4 (besides 1 as trivial factor).

In order that u be an integer, m+2 must contains a factor of 77, as well a factor 2 or 4.

As 77 = 7 x 11, so we will consider the cases

(a) m+2 = 7, 14, 28 –> m = 5, u = 33; m = 12, u = 55; m = 26, u = 66;

(b) m+2 = 11, 22, 44 –> m = 9, u = 49; m = 20, u = 63; m = 42, u = 70;

(c) m+2 = 77, 154, 308 –> m = 75, u = 73; m = 152, u = 75; m = 306, u = 76.

Therefore, possible values of u are __33, 55, 66, 49, 63, 70, 73, 75, 76__ (total nine solutions).

*Remarks: *

(1) For Koeller shading ratio *r*, we recommend to work with formula from the very beginning; it is clearer and reduce repetitious work. The formula links r with n, m.

(2) In question (b), given 77 as the common denominator of a series of ratios, it asks to find all possible integers u to appear in the numerator – this requires an analysis of factors – as in reduction of fractions.

(3) As u is an integer, an explicit expression of u – as in (*) — is convenient to work with.

Finally, note that has 3 x 2 x 2 = 12 factors. But m+2 cannot be 1, 2 or 4 (as m>=3). —

Therefore, the other 12 – 3 = 9 factors correspond to the nine solutions.

This is just a step-by-step work with algebra, plus some analysis of factors. A neat and clear presentation is what is asked by the problem.