Why Stop at Three by One

Age 16 to 18
Article by Alex Goodwin and Neil Donaldson

Published June 1998,February 2011.

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.




Diagram of rectangle
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 \begin{eqnarray} \alpha_m & = & \tan^{-1}(1/m)\\ \alpha_p & = & \tan^{-1}(1/p)\\ \alpha_q & = & \tan^{-1}(1/q) \end{eqnarray} 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: \begin{eqnarray} \alpha_m = \alpha_p + \alpha_q & \Rightarrow &\tan^{-1}(1/m)= \tan^{-1}(1/p)+ \tan^{-1}(1/q)\\ & \Rightarrow &\tan(\tan^{-1}(1/m))= \tan(\tan^{-1}(1/p)+ \tan^{-1}(1/q))\\ & \Rightarrow & {1\over m} = {(1/p)+ (1/q)\over 1 - (1/pq)}\\ & \Rightarrow & {1\over m} = {p + q \over pq - 1}\\ & \Rightarrow & q = {mp + 1 \over p - m} \end{eqnarray} 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: $$q = {mp+1\over p - m}.$$ When $p = m+1$ $$q = m^2+ m +1.$$ Since $p$ and $q$ are integers there are two angles in the extended diagram which add up to $\alpha_m$.