What would be the vollume be ?

[onta...@hotmail.com]

Suppose - (instad of a square with 10 units side) we have a cube with a 10 units side, and each of its 8 corners is the centre of a 10 unit radius sphere, can we find the volume that is common to all of the spheres ?

[João Pedro Afonso]

Outgh... this is too tough to start a morning! Lets start with something much more easier, shall we? Instead of a cube, let's try a 10 unit side hypercube in 4 dimensions. What is the volume of the intersection of the hyper-spheres of radius 10 centered on the 8 hypercube vertices's?

:-)

I'll think on the cube problem but I think my solutions are going to be very messy.

Cheers,
Joao Pedro Afonso

[João Pedro Afonso]

Hmmmm... no one answered my little teasing with hypercubes... but it should have be more much more easy than this problem.

Anyway, my candidate solution to the problem above is 15.2055, if I did things right.

Cheers,
Joao Pedro Afonso

[onta...@hotmail.com]

Joao,
I believe that you have missed a simple fact that makes the answer => 105.2486562.......

[onta...@hotmail.com]

Whatever simple method I had, in retrospect, was wrong, so I leave this puzzle in other capable hands.

[João Pedro Afonso]

> > Suppose - (instad of a square with 10 units side)
> > we have a cube with a 10 units side, and each of
> > its 8 corners is the centre of a 10 unit radius
> > sphere, can we find the volume that is common to
> > all of the spheres ?
>
> Outgh... this is too tough to start a morning! Lets
> start with something much more easier, shall we?
> Instead of a cube, let's try a 10 unit side hypercube
> in 4 dimensions. What is the volume of the
> intersection of the hyper-spheres of radius 10
> centered on the 8 hypercube vertices's?

Oops!..... Sorry, Sorry, Sorry, Sorry, Sorry!... It's not 8 hypercubes vertices but 16. I hope no one was fooled about and thought in the corrected picture (I could have asked 8 vertices, but then I had to say which).

Cheers,
João Pedro Afonso

PS.: I really think this problem is much easier than the others, you should give it a try. I'm serious. :-)

[João Pedro Afonso]

> Joao,
> I believe that you have missed a simple fact that
> makes the answer => 105.2486562.......

Bill,

I believe that you have missed a simple fact that

makes the answer < 37.211.......

(Sorry, I couldn't resist :-) Anyway, I repeated the Volume Integral calculation and got the same as before. I'll stand for my result, for now. To explain the above limit, think on the sphere covering the pretended region)

João Pedro Afonso

[João Pedro Afonso]

> Whatever simple method I had, in retrospect, was
> wrong, so I leave this puzzle in other capable hands.

Care to share what was your idea? Perhaps it doesn't work here but it could work on another place (Or perhaps I'm completely wrong, and the method is right after all).

Bill, redeem yourself and answer my hypercube variation of the problem, before I start to have serious doubts over my earlier appreciation of it (say the one who didn't published yet the solution to the pendulum problem he promised :-)

Cheers,
João Pedro Afonso

[onta...@hotmail.com]

Enough of frivolity
Sober second thoughts.
Let us consider a sphere of sqrt(10^2+10^2+10^2)/2 radius and from an infinite number of points (with its surface as a centre) we draw spheres of 10 radius.We will have drawn a central overlapped sphere of volume 10.07293380. If we gradually remove the these satellite spheres can the "common" volume increase ? (It can't be less).

[onta...@hotmail.com]

As Shakespeare might have said, "Toutest thy successes, but be sure that thy failures are buried very deepeth; goeth not to the confessional with them ".
As for hypercubes,I don't know what Shakespesre would have advised, so I would have to consult H.G. Wells about 4D and "time"-travel.

[João Pedro Afonso]

Good reasoning. That number marks indeed a lower limit to the common volume. If you start to remove spheres, the common volume shared by all the others but not by the ones removed, turns eligible then, and adds to the number you got. So it is a lower limit.

About my reasoning for a upper limit, the common volume from the 8 spheres is defined by sphere parts in each octant, sphere those centered on the opposite vertex to the octant. ... they can be surely covered by a sphere of radius sqrt(10^2-2*5^2)-5=2.071 and volume 37.211. So this a upper limit.

My result (15.2055) was between those two, so, it is not disproved by them.

Cheers,
Joao Pedro Afonso

[Robert May]

The hypervolume would be 0. The radius of a hypercube is equal to the length of its edge, so the 16 hyperspehres of radius 10 centered at the vertices would all meet in a point.

[João Pedro Afonso]

Yep! It was easy as I said, wasn't? :-)

João Pedro Afonso