Chaitanya's Random Pages

September 17, 2011

The shortest broken line of fixed angle joining three parallel lines

Filed under: mathematics — ckrao @ 8:10 am

My previous post was inspired by the following problem I found at the 8foxes.com geometry problems website.

Given three parallel lines separated by distances a, b as shown, what is the minimum length of x+y as a function of $\theta$?

The quantity to minimise is $\displaystyle x+y = \frac{a}{\sin \alpha} + \frac{b}{\sin \beta}$, which is the same expression we needed to minimise in the previous post, except that this time we have $\alpha + \beta = \theta$ instead of $\alpha + \beta + \theta = \pi$. We simply replace $\theta$ with $\pi -\theta$ in the answer given there, and find that the minimum length is

$\displaystyle d = \frac{1}{\sin \theta}\left[3x^2 - \frac{a^2b^2}{x^2} + 4(a + b \cos \theta )x + a^2 + b^2 + 4ab \cos \theta\right]^{1/2},$

where $x$ is the positive root satisfying the cubic equation

$\displaystyle x^3 + (b \cos \theta)x^2 - (ab \cos \theta)x - ab^2 = 0.$

Indeed, if we rotate the top line by $\pi - \theta$ about the “hinge” we obtain the same picture as last time and see that a and b represent the same quantity in both cases:

The only difference is that the first problem allows for the case of $\theta = \pi$, in which case the minimum distance is clearly a+b.

It is interesting how these seemingly different problems are identical, with a simple rotation connecting them.