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This question was inspired by the following question in quora:

The lines $y = 1/3 x$ and $y = 13/9 x$ are drawn in the coordinate plane. What is the slope of the line that bisects the angle these lines make?

It turns out that the slope of the bisector is 3/4 (and, of course, -4/3).

This suggested the question:

Which rational slopes have angle bisectors with rational slopes?

There are a number of related questions, but I have not seen this particular variation.

Here's what I have so far.

If the slopes are $a$ and $b$, by using $\tan(u-v) = (\tan(u)-\tan(v))/(1+\tan(u)\tan(v)) $, we can show that the slope $x$ of the bisector satisfies $x^2+2rx-1 = 0 $ where $r = (1-ab)/(a+b) $.

From this, $x =-r \pm \sqrt{1+r^2} $.

For this to be rational, $\sqrt{1+r^2} $ must be rational.

It turns out that $1+r^2 =\dfrac{(1+a^2)(1+b^2)}{(a+b)^2} $ so a necessary and sufficient condition is that $(1+a^2)(1+b^2)$ is a rational square.

If $a = \dfrac{s}{t}, b=\dfrac{u}{v}$, this is equivalent to $(s^2+t^2)(u^2+v^2) $ beinga square, or, using the standard identity, $(su\pm tv)^2+(sv\mp tu)^2 $ is a square.

What I do not know is necessary and sufficiaent conditions on $s, t, u, v$ which make $(s^2+t^2)(u^2+v^2) =(su\pm tv)^2+(sv\mp tu)^2 $ a square.

In the original problen, $a = 1/3, b = 13/9$, so $s^2+t^2 = 10, u^2+v^2 = 250 $ and $(su\pm tv)^2+(sv\mp tu)^2 =(13\pm27)^2+(9\mp 39)^2 =40^2+30^2 =50^2$ or $14^2+48^2 =4(7^2+24^2) =4(25^2) =(50)^2 $.

I haven't been able to take this any further.

The tags could use improving, too.

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  • $\begingroup$ Reduce to the case $a=0$, so $u^2+v^2$ is a square, which is of course the classification of primitive pythagorean triples $u=A^2-B^2, v=2AB$ or the other way depending on which one is even. $\endgroup$ – user10354138 Aug 5 at 5:27

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