Tennis Ball Container

[Kyle]

Find the base circumference and height of the cone inside which fit four tennis balls (dia. 6.35cm) so that three balls form a triangular base and the fourth is nested between them.

[Kyle]

To edit the above, show general formulas for finding radius and height for four spheres of radius r.

[onta...@hotmail.com]

The centres of the tennis balls form the sides of a regular tetrahedron. If you go to
http://kjmaclean.com/Geometry/Tetrahedron.html you will find all the relevant calculations to easily solve this problem.

[Kyle]

Actually this was an extension of a classwork problem which my math department in an international school in Thailand gave our students. We had the students calculate the size of containers in the shapes of rectangular prisms, cylinders, and cones. The first two were obviously simple for them. For the cones, we gave them four tennis balls and recommended that they find the size and shape through experimentation and scale drawing. This project was for years 9 through 11.

I proposed the idea of having the year 12 and 13 students find the answer only through calculation. The other math teacher and I both worked on our plan for a few days, trying to find the simplest method we could. We both started with the idea of the tetrahedron. I stayed with the 3-d shape, using different triangles to find a constant of proportionality by which I could enlarge the tetrahedron to contain the balls and then spin it to form a cone. The other teacher used a 2-d cross section, and worked from congruent angles to enlarge the triangle and spin it.

We both got different answers. I haven't yet gone through the solutions above in depth, but so far it's looking like I'm gonna win.

[Philippe 92]

Kyle wrote :

> The other math teacher and I both worked on our plan for a few days,

> We both started with the idea of the tetrahedron.

Depends on which tetrahedron...
The one having as vertices the centers of the 4 balls (dense packing
of 4 balls in 3D space), then put the composite 4 balls solid inside
a cone. That's what is required.
Or a tetrahedron *containing* the balls ?

> I stayed with the 3-d shape, ... I could enlarge the tetrahedron
> to contain the balls

But packing *in* a tetrahedron is different of packing in a cone !

> The other teacher used a 2-d cross section,

That's seems the right method.

> We both got different answers.

Not surprising, if you started from a tetrahedron containing the
balls and he started with tetrahedron from the centers.

> ... but so far it's looking like I'm gonna win.

Who awards the prize ?

However, I agree with Avni's results, with a different expression as
sqrt( (sqrt(3)+1) / (sqrt(3)-1) ) (Avni's) can also be written as
(sqrt(3)+1) / sqrt(2) (mine)

Regards.

Note [OT] :
Also note that Avni's post and Ontadian's reply didn't propagate on
Usenet. Still stucked inside Drexel.
Just seen them because I went at Drexel site by chance, and for some
other reason.
I read/post to that forum through Usenet and my ISP's NNTP server.

--
Philippe Ch., mail : chephi...@free.fr
site : http://mathafou.free.fr/ (recreational mathematics)

[João Pedro Afonso]

hmmm... I confirm Avni and Philippe results (for the sake of discussion itself, I was hopping not :-).

Interesting problem, I used trigonometry to solve it, but we don't really need it, don't we? We only need to know some Tetrahedron properties like the height and the distance from the base center to one vertex, and from then on, it is only congruent triangles to play and achieve the solution.

I'm wondering, what is the maximum size a 5th ball can have, to be packed with the others in the same cone? How many can exist of the same size?

[onta...@hotmail.com]

An interesting variation of this puzzle is that of four solid spheres contained inside a hollow sphere.
If they are placed two over two at 90 degrees, will they fit exactly into the same diameter spherical cavilty as if they are one over three. If so, can it be proved that they can change configuration in situ.?