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 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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Create a free website or blog at

%d bloggers like this: