In an exam room, triangles are better than squares
The Sunday Nation
Nairobi,
15 November 2015
Isaac Langat is a school teacher and he is wondering how to arrange desks in an examination room in order to fit the largest number of pupils. He writes: “The Kenya National Examinations Council specifies the minimum distance between candidates on all sides as 1.22m (or about 6 feet). I was wondering how to achieve the maximum number of candidates in a typical classroom measuring 6m by 8m.”
First of all, Isaac, 1.22m is not equivalent to 6 feet; it is 4 feet. Secondly, it is clearly specified that this is the minimum distance between candidates; not between the desks. That simplifies the problem significantly.
It is better to work this problem in feet because 1.22m is a cumbersome quantity while 4ft is a nice round figure. So, we convert the classroom dimensions from 6m by 8m to 20 feet by 26.25 feet. We must leave about one foot from the walls so we are left with a room measuring 18ft by 24.25ft.
Now the science of crystallography reveals that a square or pattern is NOT the most efficient way of fitting objects on a flat surface. A triangular pattern is better. Let’s see how they compare.
We start by placing candidates along the 18ft width ensuring that they are at least 4ft apart. One row will accommodate 5 pupils. Along the 24.25ft length, we can fit 7 students. So we have 7 rows with 5 pupils each; that is, 35 in all. This is the “normal” square grid arrangement.
To make a triangular pattern, we start with the 5 pupils in the first row along the 18ft width. Let’s label these with letters A, B, C, D, and E. Candidate F has to go to the second row, but we do not place her directly in front of A. Instead, we position her in the middle of A and B, but 4ft from each of them.
This forms an equilateral triangle with candidates A, B, and F at the corners. In the same way, we place number G between B and C; H between C and D; and I between D and E.
We notice that the second row has four pupils while the first one has five. However, the two rows are closer to each other than they were in the rectangular grid. We apply Pythagoras’s theorem to work out the distance between these two rows. The answer is 3.464ft (or 3ft and 5.5inches).
The third row will also be 3.464ft from the second one, but, its arrangement will be the same as that of the first. If we repeat this pattern to fill up the room we will end up with eight rows.
The 1st, 3rd, 5th, and 7th rows will have five pupils each making a total of 20. The other four (2nd, 4th, 6th, and 8th) will have four to make 16 candidates. So in total we shall have 36 students in the hall.
This triangular grid accommodates one more student than the square pattern. Is it worth the trouble? Crystallography shows that, if you are using a large hall, you can increase the capacity by about 14 per cent.
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