Cantr II Discussion Forum

Here's one I was given recently. It wasn't too easy to solve.

You have 4 spherical jugling balls of radius 1 inch and you want to find the volume of the smallest possible regular tetrehedral box that will fit the jugling balls inside. (For the non-Americans it can be centimeters... but then they would have to be marbles or something.)

You get as many tries as you want. If you get the right answer, I'll... um... I can't think of a good prize that I would actually be able to carry out, but you'll have my respect.

Ummm... tetrehedral???

Not that I would solve it, I've always been terrible at math.
luckily we have GPS's and computers nowadays.

Yeah....what is tetrehedral?? LOL

tetrehedral = in a tetrehedron shape. the tetrehedron is the simplest euclidean solid. 4 triangular faces, 6 edges, 4 vertices. you know, a trianglar pyramid.

and in this case it's regular so those triangles are equilateral.

Why did't you say that right away?

And it seems more like a logical than a math. question. The real question is: How will you stuff the balls in the pyramid. The rest is basic maths i think. ummmm... that's what I guess, i don't have a clue how to calculate the volume of a pryamyd. And in geometry i had the only good ratings in maths in all my life.
Yep, that's how lousy I was.

I could totally do this problem, if only I didn't hate math...than I might sit down and try to work it out...

I got 10.3816, which is clearly wrong because it's less that the total volume of the spheres. . .

oh come on, you gotta give me an exact answer!

it's around 40, so yeah, that's a bit off.

Maybe they're those soft juggling balls and you can squeeze them into a really small volumed pyramid.

Hmmm, Squeezability, 's that maths too?

I got 2.59. Which is really small. Let me try again.

Well there was a few problems with my first calculations. 1 was that I just noticed it says radius is 1 inch not diameter and 2 I just noticed the ball wouldn't fit in the area I wanted. I just had geometry last year so this should be fresh in my head and I know I wll be abe to figure it out.

Yes! Yes! I think I got it!
Exact answer (for those people who, like me, are sticklers for accuracy and know Scheme): (/ (+ (* 12 (sqrt 3)) (* 38 (sqrt 2))) 3)
Approximate answer (for everybody else): 38.697981480888. . .
Am I deluding myself again, or am I right?

As far as I can tell, the easy part is deciding how to put them in, and the hard part is calculating it. Phew!

yay! sho is right! now you have to put it up on the board!

And you have to give me a prize! No excuses!

How I got the answer (with some skipped reasoning that kroner will provide):

The optimal configuration of the spheres is stacked so that the centers form a regular tetrahedron and each sphere is tangent to all of its neighbors and three of the faces of the tetrahedron.
The proof:
Hey look! A distraction! *points the other way*
Q.E.D. See?

First, you look at the tetrahedron in such a way that two of the spheres overlap each other. The whole thing should now look like an isoceles triangle with three circles in it. Label the vertex that the line of symmetry passes through A, and the other two B and C. Draw the perpendicular from B to AC. Label the pedal point D.
Triangle BAD is a right triangle. AB is the same length as an altitude of one of the equilateral triangles making up the tetrahedron. AC is the same, and AD is one third of that, because D is the center of a face of the tetrahedron, and the center of an equilateral triangle trisects the altitudes. Therefore sin ABD is (/ 1 3), and the angle ABD is arcsin (/ 1 3).
Now, take the circle closest to B. Call it and its center E. It is tangent to AB but not to BC. Call its point of tangency to AB F. BEF is another right triangle. EF is 1. BF is (/ 1 (tan (arcsin (/ 1 3)))). This is equal to (* 2 (sqrt 2)).
Now, we view the tetrahedron along one of its altitudes. It now looks like an equilateral triangle with four circles and some lines in it. Call the vertices G, H and I. Take the centers of the circles near the vertices and call them J, K and L, with J being near G, K near H and L near I. JKL is another equilateral triangle. Call the center M.
We know that GJ is equal to (* 2 (sqrt 2)). We know that a side of JKL is equal to 2. JM is equal to (/ (* 2 (sqrt 3)) 3). GM is equal to GJ+JM. The height of GHI is (/ (* 3 GM) 2), and a side is that times (/ 2 (sqrt 3)). The volume of a tetrahedron with side n is (/ (exp x 3) (* 6 (sqrt 2))), so that's it.

Maybe I'll make some diagrams someday. Until then, think of this as a drafting project or an exercise in imagination.

Yay! *grabs kroner's lunchbox and runs off* Meeg!