Is there an easier way of solving this problem?

Question

Take a look at this figure:

What's the length of the longest 1-dimensional rod which can pass through this corridor horizontally? I can solve this problem mathematically using first and second derivative tests, but is there an easier way, similar to what feynman does in many of his lectures? One approach might be to take advantage of the symmetry of the situation.

The longest such rod will make an angle of pi/4 with the original direction when it (almost) touches the inner corner. I need a valid logical explaination for that.

I've edited the question accordingly. I'm voting to move this to math SE — , Aug 20 '18 at 13:07
Strangely, what you want is not the longest pipe that makes it around the corner, but the shortest pipe that can't! — DJohnM, Aug 20 '18 at 22:06

score 1 · Answer 1 · answered Aug 20 '18 at 13:10

You can get an intuitive explanation by considering the extreme cases. Bring the rod into the corner parallel to the first corridor, and it can be infinitely long. Start to turn it around the corner, and it must be shorter and shorter to fit between the two outer walls (the lower one and the right-most one). When you have passed the 45 degree point, symmetry shows that this same logic repeats: the rod can now be longer and longer until it can be infinitely long when parallel with the second corridor. The shortest moment was obviously at the 45 degree point.

If the corridors weren't equal in width, it'd quickly get too complicated to intuitively figure it out this way.

Mathematically, I think I would parameterize a straight line (something like $s=\left(a\cos(v),a\sin(v)\right)$, with some proper x- and y-interval restrictions) and equate it with the parameterized corridor area and with the corner point. Solving those two equations simultaneously should give you the length $a$ which you can then minimize. But I would advice you to ask this on Mathematics SE if you seek more in-depth mathematical answers.

score 0 · Answer 2 · answered Aug 20 '18 at 13:44

I’ve seen a similar problem like this before, many years ago. The horizontal and vertical outside walls of the corridor where the ends of the rod touch them will form a right triangle, with the rod as the hypotenuse when it is up against the inside corner. Assume the rod is moving to the right. In order for the rod to clear the corner without jamming as it pivots about the inside corner, the gain in the length of the vertical side and the loss in the length of the horizontal side of the triangle with time must be equal. This is only possible with an angle of $\frac {\pi}{4}$ for maximum length of rod.

score -1 · Answer 3 · answered Aug 20 '18 at 17:49

This is a "common sense" argument, which could be turned into rigorous math using calculus if you want to do that:

The angles $\dfrac \pi 4 - \alpha$ and $\dfrac \pi 4 + \alpha$ give the same length, for any angle $\alpha$. This is obvious from the symmetry of the situation - just draw a picture.

It follows that angle $\dfrac \pi 4$ corresponds to either a minimum or a maximum value for the length of the rod, and common sense says it can't be a maximum - therefore it is the minimum length.

Is there an easier way of solving this problem?

3 Answers3