mega-sphere

A sphere 5.5 metres in diameter is filled with 1m diameter hemi-spheres.

a(1) What is the theoretical maximum amount of hemi-spheres that can be crammed into the big sphere given that the following condition is met:
Each hemi-sphere's flat side (which I'll now refer to as its 'disc') has a central point (indicated by the white point shown in the hemisphere diagram to the right). The point must not 'see' another hemisphere's disc. By definition, when I say 'see', the simplest thing to imagine is a straight ultra-thin 'laser light' coming from the disc. This 'light' must not reach another disc. However, if there's another hemi-sphere that's 'blocking' the 'line of sight', then this is accepted.

(2) By cramming them as efficiently as possible, a relatively small volume will be left. How large is this volume?

b: Same question, except the torch now has a 52.72077938642 (that's 90/(1+(0.5^0.5))) 'degree of sight'. This 'cone' of light extends from the exact centre of the disc and is once again not allowed to 'see' any part of another disc. What is the maximum amount of hemisphere's that can be crammed into the mega-sphere now?

c: Same question again, except the whole disc acts as a tubular beacon of light. This thick ray must not 'see' another hemisphere's disc. What is the maximum amount of hemisphere's that can be crammed into the mega-sphere now?

thats why its under hardcore

Steve wrote:thats why its under hardcore

so if ur so smart and sofisticaded y dont u tell me the answer?????????


jk jk

well i got it off a website.

I don't think any of us can handle the whole light thing -- mega complex geometry stuff -- and if we take that part out, then the hemisphere problem is rather simplistic. If the light is ignored, the solution is this:

1.333*pi*r^3 <--- sphere's volume
1.333*pi*2.75^3=20.796 cubic inches = megasphere
1.333*pi*1^3=1 cubic inch = 2 hemispheres

So the amount of hemispheres is 41. There are 0.59375 cubic inches of space remaining. Although considering the shape of the hemispheres, it's not certain they'd fit like that... *ponders*

I got this problem off a website, and forgot to write the name down. Now i forget it. You have just as good of a chance as I do at solving it. Sorry!