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.
