Why Stop at Three by One

'Why Stop at Three by One' printed from https://nrich.maths.org/

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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.

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 $\alpha_m$.