# Fun little geometry problem (add yours if you want)
**URL:** https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312
**Category:** Miscellaneous and Personal Stuff I Must Share
**Created:** [December 8, 2023, 8:22pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312 "2023-12-08T20:22:40Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@Hampshire](https://boards.straightdope.com/u/Hampshire)
#### Post date: [December 8, 2023, 8:22pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/1 "2023-12-08T20:22:40Z")
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?
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### Author:  [@Kron](https://boards.straightdope.com/u/Kron)
#### Post date: [December 8, 2023, 9:35pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/2 "2023-12-08T21:35:18Z")
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…
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### Author:  [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)
#### Post date: [December 8, 2023, 9:45pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/3 "2023-12-08T21:45:28Z")
I got the same answer that @Kron did.
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### Author:  [@Dead\_Cat](https://boards.straightdope.com/u/Dead_Cat)
#### Post date: [December 8, 2023, 10:02pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/4 "2023-12-08T22:02:49Z")
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.
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### Author:  [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)
#### Post date: [December 8, 2023, 10:06pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/5 "2023-12-08T22:06:42Z")
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.
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### Author:  [@markn\_1](https://boards.straightdope.com/u/markn_1)
#### Post date: [December 8, 2023, 10:14pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/6 "2023-12-08T22:14:45Z")
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.
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### Author:  [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)
#### Post date: [December 8, 2023, 11:36pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/7 "2023-12-08T23:36:52Z")
> [@Dead\_Cat](#):
>
> calculate
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".
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### Author:  [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)
#### Post date: [December 8, 2023, 11:46pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/8 "2023-12-08T23:46:12Z")
> [@LSLGuy](#):
>
> The 3D Pythagorean is the answer for a rod of infinitesimal diameter.
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.
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### Author:  [@mixdenny](https://boards.straightdope.com/u/mixdenny)
#### Post date: [December 9, 2023, 5:46pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/9 "2023-12-09T17:46:00Z")
> [@markn\_1](#):
>
> 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’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.
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [December 10, 2023, 12:01am UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/10 "2023-12-10T00:01:13Z")
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.
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### Author:  [@beowulff](https://boards.straightdope.com/u/beowulff)
#### Post date: [December 10, 2023, 12:28am UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/11 "2023-12-10T00:28:32Z")
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).
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### Author:  [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)
#### Post date: [December 10, 2023, 4:09pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/12 "2023-12-10T16:09:19Z")
I’m with Amazon. Just tape the shipping label to the rod and send it off.
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### Author:  [@mixdenny](https://boards.straightdope.com/u/mixdenny)
#### Post date: [December 10, 2023, 5:45pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/13 "2023-12-10T17:45:43Z")
I’m a young guy. Just 3D print the damn thing to fit the box.
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### Author:  [@markn\_1](https://boards.straightdope.com/u/markn_1)
#### Post date: [December 11, 2023, 6:34pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/14 "2023-12-11T18:34:34Z")
I found [this 2006 paper](https://maa.org/sites/default/files/pdf/upload_library/22/Polya/jerrard93.pdf) 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.
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### Author:  [@John\_DiFool](https://boards.straightdope.com/u/John_DiFool)
#### Post date: [December 11, 2023, 8:31pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/15 "2023-12-11T20:31:47Z")
As a (former) SAT tutor, I found this one the other day:
> **[SAT-rotation-problem hosted at ImgBB](https://ibb.co/h7vCRcc)**
>
> Image SAT-rotation-problem hosted in ImgBB
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### Author:  [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)
#### Post date: [December 11, 2023, 8:34pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/16 "2023-12-11T20:34:18Z")
> [@John\_DiFool](#):
>
> 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
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### Author:  [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)
#### Post date: [December 11, 2023, 8:35pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/17 "2023-12-11T20:35:32Z")
> [@mixdenny](#):
>
> Just 3D print the damn thing to fit the box.
I’d 3D print the box to fit the thing.
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### Author:  [@Pleonast](https://boards.straightdope.com/u/Pleonast)
#### Post date: [December 11, 2023, 8:39pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/18 "2023-12-11T20:39:17Z")
> [@Hampshire](#):
>
> 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?
For extra credit:
Once the rod is in the box, what is the next longest rod that can be put into the box?
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [December 11, 2023, 8:43pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/19 "2023-12-11T20:43:01Z")
I’ll bet the problem writer thought it was C, though.
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### Author:  [@John\_DiFool](https://boards.straightdope.com/u/John_DiFool)
#### Post date: [December 11, 2023, 8:44pm UTC](https://boards.straightdope.com/t/fun-little-geometry-problem-add-yours-if-you-want/994312/20 "2023-12-11T20:44:17Z")
> [@Dr.Strangelove](#):
>
> > [@John\_DiFool](#):
> >
> > 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
Wayyy ahead of ya; I had to copy and save the image separate from all articles, which spoiled that little fact…
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