Any maths teachers on this forum?

Looking forward to tomorrow.............

just remember George, sjbabbey is also on the case, as is "h" now that he's sober

57.735 mm

Space/beer is infinitely divisable but not infinitely addable.  Like the parable/riddle of going halfway to the finish line.... you never get there.

33.333

another good try and obviously in the right ball-park, but.........

.....it's not the answer I have,

and just to be on the safe side, I have done this calculation two different ways, one using George's idea of old-fashioned trigonometry - neat, but not a generically elegant technique, and the other using a rather elegant algebraic formula, specific to this type of problem (ie five touching spheres), that was known about a few hundred years ago, but only came to light again in the 1950's or 1960's (I think). - Both answers are the same.

Trig or "lost" formula ?

Trial and error.

I went down to Lidl first thing, armed with a tape., bought three 200mm diameter grapefruit's and a large bag of onions.

That Lidl bit of trial and terror certainly worked. 33.33' is spot-on. Well done that man !

but...............see my next post.........

Ok, a slight variation on the three oranges, especially for those who had a shot at the "flat table" version.

The "dead flat" table made it dead easy, so we'll substitute the table with a fourth orange, again with 100mm radius. So now we have four large, juicy (but firm) oranges. Each orange is a perfect sphere of radius 100mm. The four oranges just touch each other.

I have a bag of large cherries, not as big as the oranges of course, but firm, perfect spheres, none the less. I want to sit a single cherry within the group of four oranges, so that it just touches all four oranges.

What size (radius) cherry do I need? You can assume its the internal cherry that we are looking for, rather than the sphere that would just enclose all four oranges.

Hopefully this will flush out mista h's missing maths teacher

Very sneaky Don.

Don't try and pretend the timing of your post was an accident. You know very well Lidl close at 8pm. How am I going to work that out without a fourth grapefruit.

With no table the oranges will just fall on the floor and roll away.

Tesco and Asda stay open all night.

For the benefit of the guys who did have a go at the three oranges, I will do my best to outline the “trig” solution.

The centres of the 3 oranges sitting on the flat table form an equilateral triangle side 200mm. The centroid of this triangle is 115.4701mm (100/Cos30) from each apex.

The centroid lies directly above the centre of the cherry. The distance from the centroid to the centre of the cherry is 100-R (where R is the radius of the cherry).

The sloping distance from the centre of an orange to the centre of the cherry is 100+R.

The three sides described above form a right angled triangle in which old Pythag says

(100+R)^² = (100-R)^² + (100/Cos30)^²

substituting H=100/Cos30 gives (just to make this next bit easier to type !)

(100+R)^² = (100-R)^² + H^²

Expanding both sides and reducing gives 400R = H^²

Hence R = H^^2/400

Which can be written as R = (100 x 100 x 2 x 2)/ (400 x sqrt3 x sqrt3)      (Note H = Cos30 = [sqrt3]/2)

Hence R = 100/3 = 33.33’

19.4444.

Close, but not the same as mine. ( I will check mine again)

Trig ?, "lost" formula ? or Trial & Teror with Tesco (Aldi being closed)?

I used trig this time.

I've checked my calcs again, they seem OK. I'll get a definitive answer at the weekend after I've been to Lidl.

Just been making myself some supper, cheese triangles on crackers.

Made a three sided pyramid with the triangles and realised where I went wrong.

36.6mm (edit) but I looked at it again and I know it's not right...

Of course a line from the centre of the orange to the centroid would also have length 100cos30 and it is this line which would form the triangle.

Just so you don't get too far out, i've inserted a "division" line in your post that I think you omitted (in red)

Assuming that the spheres are packed in a pyramid and that their centres form an equilateral triangle in the vertical plane then the centroid of the pyramid would also be 100/cos30 from the centre of the top sphere which suggests that the radius would be (100/cos30)mm - 100mm i.e. 15.47mm.

However I'm sure the answer is more complicated than that.

Fatcat, did you manage to get that extra grapefruit?

apologies, I thought you were dealing with the three oranges on a flat table. The only way three oranges can sit on a flat table and just touch each other and the table, is in a flat triangular layout. The cherry then sits on the table, within the void between the three oranges and also just touches the three oranges.

In the four orange problem, the fourth orange can sit on top of the other three, thus forming a pyramid. In this case, the flat table plays no part of the maths.

CFMF, you are right, it's not right.