Any maths teachers on this forum?

We are obviously talking at cross purposes. You will see that my post acknowledges that the oranges are arranged in a pyramid and I have not mentioned the table at all. With or without a table surely the centres of the 3 lower oranges must still be in the same (horizontal) plane in order to form the base of a pyramid.

Yes, that's right. In the 4xOrange problem, the oranges are arranged in a pyramid with the cherry in the centre of the pyramid. The lower three oranges can be considered as forming a horizontal triangle, but that is purely optional, the pyramid could have any orientation (but it does help describe the problem and the solution if you define the orientation)

22.644

Very, very close to my figure

My answer is 100/(2+sqrt6) which I evaluated as 22.47449mm.

I presume you started with a triangular pyramid, side 200mm and distance to the centroid 100+R

I didn't use trigonometry.

Checked using trig and still get 22.4745mm

Don

Rechecked figures again. I now also get 22.47455 and found my mistake.

I can't find my scientific calculator and obtained values for trig functions from the internet. I copied tan 30 as 0.557735027 instead of 0.57735027.

Don't you just hate it when that happens?   

You mean you don't have Excel ? =Tan(30*Pi()/180)

Anybody sorting the four oranges and a cherry without access to trig tables or excel deserves a round of applause,

For “h”, George, (and anybody else who might still be following this thread and be even remotely interested in arithmetic/algebra/trig/geometry) I outline below my trig solution to the four oranges and cherry problem. I imagine this is how Fatcat solved the problem, but only he knows how he did it ! ( but an incredible effort when having to use the internet for trig functions !!)

• 4 oranges must be arranged in a triangular pyramid if they are all to touch each other.
• The centres of the oranges therefore form a tetrahedron in which all sides are 200mm long. (For simplicity this can be considered 2 decimetres per side).
• The 4 corners of the tetrahedron can be labelled A, B. C, D.
• Let X be the centroid of the tetrahedron ie the centre of the cherry
• AX = BX = CX = DX = 1+R where R = radius of the cherry )
• For convenience, assume A is the top of the Pyramid and B, C, D form the base.
• Let L be the mid-point of BC and AL be the sloping “height” of triangle ABC
• From Pythag, AL = sqrt3 (decimetres)
• Let M and N be the mid-points of BD and CD respectively.
• DL, CM and BN meet at “O”, the centroid of the base.
• OL = (sqrt3)/3 – (there are several ways of calculating this, but it is basically 1/3 of DL which in turn is the same as AL)
• The vertical height of the tetrahedron AO can be deduced by pythag
• AO^2 = AL^2 – OL^2 = 3 – 1/3
• AO = 2*sqrt(2/3)
• Remember X is the centroid of the tetrahedron, which is ¼ of the vertical height (you can look this up in the internet, or use calculus to prove it !)
• AO = AX + XO and, more importantly, AX = three quarters of AO
• AX = AO*3/4 = sqrt(3/2) = 1.224745 decimetres = 122.4745mm
• Deduct 100m radius of orange, leaves 22.4745mm = radius of cherry.

BTW, this is NOT the way I did it initially. There is a more simple method.

Dear Don,

I have been following this. but have not had the energy to actually work through the problems you set, because of other things, like a visit to the consultant this coming week with eye problems.

Sorry to leave it alone, but such little challenges are lovely IMO!

ATB from George

No worries George, hope you keep following the thread, you've only seen the easy ones so far..................

I thought i'd save the next one for "h" since I just "know" he is sharpenning up his pencil, and didn't want to waste his efforts of the above "starters for 10" 

I hope the eye problem is soon put right. take care.

Don

Don

As you described, I calculated using trig the height of the tetrahedron to be 163.299455mm.

The fourth orange is 100mm radius, so the distance from the top of the cherry to the base of the tetrahedron is 63.299455mm.

I then used the same method as I used in the 3 oranges on the table puzzle. From that puzzle I know the distance from an orange centre to a vertical line through the centre of the cherry is 115.4701mm.

So.

(100+r) squared = 115.4701 squared + (63.299455 - r) squared.

Neat, I like it ! 

PS, I calculated the height of the thetrahedron as sqrt(8/3) decimetres which seems to evaluate as 163.299316mm

But I kept to the vulgar fractions (I don't like these new-fangled decimals ) and eventually reduced the Radius of the Cherry to sqrt(3/2) decimetres or 22.47448714mm

By now it should be obvious to the average "O-Level" trainee, that in the case of the four oranges, not only could a small sphere (cherry) be placed within the spaced formed by the oranges, but that a sphere (bubble) could be placed arround the oranges and it would have a radius of 122.47449mm.

In general, its possible to find a fifth sphere to just fit within four touching spheres - which can each be a different size, and get a sixth sphere to enclose the four spheres.

Try it with spheres of 200mm, 300mm and 400mm radius sitting on the outside of a globe radius 1,200mm all just touching each other.

Perhaps Richard could persaude Hoop.la to give an extra Community Point to the provider of the first correct answer

OK, nobody is paying attention ! 3 hours is long enough !

There is a deliberate mistake in the above post. More specififically in the first paragraph starting "By now it should be obvious....."

I bet Fatcat is the first to spot it

The deliberate mistake was 122.47449 mm should have been 222.47449 mm

Try it with spheres of 200mm, 300mm and 400mm radius sitting on the outside of a globe radius 1,200mm all just touching each other.

For the benefit of anybody trying this one my answers are 82.63mm and 3443.63mm

Answers on a postcard pls.

Are we assuming that all the drive belts remain in place and that all the cogs are actually well lubricated and free to turn and that the lid on the box is not rusted in place ?

In that case, I'll just refill my glass and get back to this problem in a 'mo.

Struth, kudos to you sir!   '60 years on and BODMAS still springs to mind'.....

I'm just about to complete a maths course and sit the exams 25 years after leaving school.  I wish I could have remembered that just 25 years on and I probably wouldn't be doing this all again now!

Oh, and I reckon wind the handle as directed and it opens the box.......?

Meshed gears           = Reversed direction

Direct drive pulley      = Same direction

Figure 8 pulley           = Reversed direction

Input gear = No1 = Forward Rotation F (Clockwise)

Anti-clockwise = Backward Rotation B

Sequence

1  F          6  F          11  B

2  B          7  B         12  F

3  B         8  F          13  B

4  F          9  B          14  F

5  B         10 F          15 B

Output gear = No15 = Backward (Anti-clockwise) rotation = Opens Box

I reckon Jason was right