Thought this one up to ask my teenager to see how he might go about solving it:
You want to buy a 1/2" diameter iron rod in the longest length you can fit in a crate you have.
The inner dimensions of the crate are 12" x 16" x 15". The iron rod is sold in even in lengths only (2",4",6"…) What is the longest rod you can fit in the box? How did you go about solving this?
24".
Pythagoras on 15 x 16 to get longest two-dimensional length; the Pythagoras using the first result and 12 for the sides of the three-dimensional triangle. Comes out at 25, so for the longest even-number you can fit, it is 24.
I didn’t realize it at first, but I made it more difficult than needed, because it’s really just SQRT(a^2+b^2+c^2).
It would have been more difficult if the answer came out to an even length, because I would have had to factor in the diameter of the rod after rotation…
I got the same answer that @Kron did.
Without reading the spoiler, this was my way - it may not be the most efficient, is there a shortcut I’m missing?
First we need to calculate the length of the diagonal of one side of the box. Conveniently, 12^2 + 16^2 = 20^2, so that diagonal of that side of the box is 20". Then we can use that to calculate the biggest internal diagonal of the box, which will be the square root of 20^2 + 15^2. Which is 25. So the answer to the puzzle is the rod of length 24".
ETA: looks like Kron initially did a slightly harder calculation, but then found a useful shortcut.
That problem would get massively uglier if you actually wanted to compute the absolute max rod length given its diameter. The two rod ends fit sorta kinda into the two 3D corners at awkward angles and once the rod has non-zero diameter the centerline doesn’t intersect either corner and …
Whole lotta trig or calculus needed to make that work. The 3D Pythagorean is the answer for a rod of infinitesimal diameter. I can sorta vaguely see the rough shape of the necessary parametric equations to compute the length as a function of rod diameter. But they’re nasty IMO. There’s probably a formula in the CRC book, but I haven’t owned one of those in a long time … a long time.
Hope one of the real engineers or math folks will stop by.
I’m an engineer. My solution is, calculate the length by the Pythagorean theorem assuming the rod has zero diameter, and then bend the rod to make it fit.
I know what you’re saying, but I am too lazy to work it out. But I did for the 2-dimensional version. The longest 1/2" wide stick you can fit into a 12 x 16 rectangle is about 19.52". The full diagonal is 20".
Yeah, but without any complicated math, it’s easy to see there can’t be more than ~0.433" of “slop” on each end. And the problem is carefully designed to give 1" of margin total, so the rod easily fits.
I’m a carpenter. My solution is, calculate the length by the Pythagorean theorem assuming the rod has zero diameter and then carve out the corner of the box to make it fit.
I would have sharpened the ends of the rod.
And like @Dead_Cat , I used the two 3-4-5 triangles to shortcut the computation. I’ve been telling my students for weeks now to keep an eye out for Pythagorean triples.
I’m in the shipping department.
Get a 28” rod, jam it into the box, and then put a lot of tape over the end poking out of the box.
(At least, that’s the way we’ve always done it).
I’m with Amazon. Just tape the shipping label to the rod and send it off.
I’m a young guy. Just 3D print the damn thing to fit the box.
I found this 2006 paper which discusses the general problem of fitting a cylinder diagonally into a box. It derives an equation relating the cylinder length to the input conditions (box dimensions m_1,m_2,m_3 and cylinder radius r), but it turns out to be quite complex, and the paper concludes
Unfortunately, equation (7) is too complicated to solve explicitly for t in terms of r and (m_1,m_2,m_3). This makes determining the maximum length l as an explicit function of the given geometric data out of the question.
Equation 7 is this fourth degree equation
2rt \sum_{i=1}^3{\epsilon_i m_i \sqrt{r^2+t^2-m_i^2} - (2r^4-t^4+r^2t^2 + \frac{1}{4}(t^2-r^2)d^2) = 0}
t is half the cylinder length, d is twice the space diagonal and each \epsilon_i is +1 or -1.
Despite not coming up with an algebraic solution, the analysis in the paper is interesting.
As a (former) SAT tutor, I found this one the other day:
Not a direct answer, but important:
None of the given answers (A-E) are correct
I’d 3D print the box to fit the thing.
For extra credit:
Once the rod is in the box, what is the next longest rod that can be put into the box?
I’ll bet the problem writer thought it was C, though.
Wayyy ahead of ya; I had to copy and save the image separate from all articles, which spoiled that little fact…