Age
16 to 18
| Article by
Alex Goodwin and Neil Donaldson
 | Published

Why stop at three by one

In this article Alex and Neil from Madras College give a generalisation of the Three By One problem. See also the article by the same authors, 8 Methods for 'Three by One' which, as the title suggests, brilliantly solves the same problem using 8 different topics in mathematics thus exemplifying the unity of the subject.



Why stop at Three by One
Given a rectangle of dimensions 1 by $n$, for each angle, $\alpha_m$ where $m$ is a positive integer, are there two other angles $\alpha_p$ and $\alpha_q$ whose sum is equal to $\alpha_m$?
Now αm=tan−1⁡(1/m)αp=tan−1⁡(1/p)αq=tan−1⁡(1/q) Using an exhaustive search for each $m$ from 1 to 12 the $(p,q)$ which satisfy this are:


1 2 3 4 5 6 7 8 9 10 11 12
( p , q ) pairs (2,3) (3,7) (4,13) (5,21) (6,31) (7,43) (8,57) (9,73) (10,91) (11,111) (12,133) (13,157)
(3,2) (5,8) (7,18) (9,32) (13,21) (11,50) (13,72)
(12,17)


Now, using the tan angle sum formula: αm=αp+αq⇒tan−1⁡(1/m)=tan−1⁡(1/p)+tan−1⁡(1/q)⇒tan⁡(tan−1⁡(1/m))=tan⁡(tan−1⁡(1/p)+tan−1⁡(1/q))⇒1m=(1/p)+(1/q)1−(1/pq)⇒1m=p+qpq−1⇒q=mp+1p−m If $p$ and $q$ satisfy this diophantine equation then these will be solutions to the above question for $\alpha_m$.


Conjecture 1:

For all $m$ there exists at least one solution where $p = m + 1$.


Proof:

To find a solution, use: q=mp+1p−m. When $p = m+1$ q=m2+m+1. Since $p$ and $q$ are integers there are two angles in the extended diagram which add up to $\alpha_m$.