Brain Teasers ? or 50 Years On........... ?

It depends upon the relative size of d & L.

If d=L/2 (ie the hoop diameter is half the square length) then probability = 0.75. If the hoop diameter is 0.9L, then the probability goes up to 0.99. If the diameter of the loop is equal to the square length, then the probability is 1.

Note that the equation only holds for d less than or equal to L. Obviously if d>L then the probability has to be 1.

Lionel, Adam, good starts.................

"d" and "L" are labeled but not defined.

If "d" is > than "L", then the probability that the hoop will lay across more than one square is 1 ie a dead cert.

Put aside the special case where d = L

If "d" is < than "L" then the probability depends on the ratio of "d" to "L", It's that relationship that we're looking for. (And the picture was drawn with this in mind.)

Hope this helps ?

Ah !

I started my reply above to Lionel and Adam, Mrs D called me out for diner.

I came back, completed my half-finished post and pressed the "Send" button, only to find that Sophiebear has been very active !!

Nice formula Sophiebear.

I left mine in the 1 minus state ie Prob = 1 - [(L - D)^2 / L^2]

I prefer yours.

by way of explanation :...........

when d < L draw  all the hoops that can be drawn entirely within a single square AND which touch the boundary of that square either at one edge (ie at one point) or in a corner (ie at two points) . You will see that these hoops have centres that lie on a SMALLER square exactly in the middle of the larger square.

For a hoop of diameter "d" the side of this smaller square is (L - d)

You will then see that any hoop whose centre is inside the smaller square does NOT touch the edge of any large square. The probability of such a hoop having its centre within the smaller square is the ratio of the small to large squares ie (L - d)^2 / L^2

Subtract this from1 (unity) and you have the formula above

Is that the answer? If so can your translate into numbers?

Edit: no you cannot because it depends on the size of the hoop?

That's right, As Sophiebear mentioned, you simply substitute the actual size of the hoop and the actual size of the squares, into the formula.

So if d = 1 unit and L = 2 units (could be inches, meters or miles etc) you wouls get a probability of 0.75 (75%)

if d = 1cm and L = 4cm the probability drops to 7/16

The beauty of the formula is it means you can work out the probability for any combination of d and L.

Try 12.5 mm and 50 mm, for example

If anybody is interested but "stuck", the following diagram and concepts might help you get started.........

The diagram is not drawn to scale, but with a sharp pencil, ruler and pair of dividers or a circle-drawing compass, you could do a scale drawing - it would work !

Don

I'll have a stab at 3 hours...but I don't really trust my working out !!

Regards,

Peter

Peter, You would be justified in having more confidence in your working out 

Well done that man !

Having got this far, the required heading shouldn't be too much of a problem - really !

Thanks Don

I think the sine of the angle to the horizontal works out to be SQRT (3/7) ? Sadly I don't have my school books any longer to look that up as an angle in degrees !! However it must be less than 45 degrees because the tangent of the angle = (SQRT 3)/2.....and I still remember TAN 45 = 1. Well at least it was when I was at school (which is many years ago !!)

Best regards,

Peter

Its about 41 degrees making the heading about 270 - 41 = about  229 degrees.

Well done again.

Cube within a sphere

Ok, back to more basic pure maths - probably at "O" Level this time.........

What's the volume of the largest cube that fits entirely within a sphere of unit diameter ?

Don, withdrawn my answer and I'll have a rethink later 

Simon, You were getting close, don't be embarrassed, this is a friendly thread ! It takes time to put the clock back to our school days !!

Your answer actually made me go back and re-check my calcs last night, and I gave up and had to restart again this morning. Your answer might have been incorrect but it made me think very hard

Don - yep it takes you back - so I forgot to think accurately in three dimensions - so here goes

Sphere Diameter = D =1

The side of the cube = L

Now the diameter D is a three dimensional hypotenuse of the cube of side L

Therefore

D = √(L² + L² + L²  )= L√3

As D = 1 (unit sphere) then L =1/√3

The volume = L³ = (1/√3)³ = 1/(3√3) units cubed

Its been bugging me all day - and I might have it still wrong...

Simon

Well done Simon,

We're either both right, or both wrong. And i'm sure you're right !

Hope you sleep well tonight, with a clear mind ! aka do not read the next geometry question...................

Cube within a sphere II

Ok, back to more basic pure maths - probably at "As" Level this time.........

What's the volume of the largest cube that fits entirely within a sphere of unit volume ?

Don

I think it works out as 1 / (Pi * sin 60) which is around 0.3676

Regards,

Peter

Spot-on Peter,

I wrote the result as 2/(Π√3)    (two, divided by Pi root 3)

Hope this helps Simon to sleep tonight Simon ?

I find that trying to do these in my head either has me off to sleep in less than  60 seconds, or awake until the early hours !

I'd agree with this result. A more interesting question is perhaps "What is the probability that the disc overlaps MORE THAN 2 squares?".

Don, building on the earlier equations, assuming they were correct and knowing the volume of a sphere is 4/3 ∏(D/2)³

then I think the volume of the largest cube is 2/(√3∏ )

That is L = 2/√3 . ³√(3/(4∏ ))

ie L³ = 8/(3√3) . 3/(4∏ )

Winky - that's a nice extension to the original problem.

I think you can still apply the same logic as Don outlined in his original solution. Here however the hoop will be bounded in a rectangle (dimensions L*2L) rather than a square.

Working through using the same method, I come up with the probability that the hoop crosses more than 2 squares is given by: d/2L * (3 - d/L)

This will result in a lower probability than was calculated for crossing any number of boundaries. So for example in the initial instance when the hoop diameter was half the square length (d = L/2) the probability = 0.75.  With the new criterion of needing to cross at least 2 squares, then the probability reduces to 0.625.

I hope this makes sense...and I hope that its correct ?

Regards,

Peter

Hmmmm,

My initial thought was simply d^2 / L^2.

But now i'm not sure and will have to come back later when I have more time and draw things out, or...........

get my Chess Board out, together with some brightly coloured discs that can be "flipped" at random called .......wait for it !.................Tiddly Wink(y)s.........

I'll get me coat......................!

Don was I right? Have you a puzzler for today    ?

Apologies Simon,

I totally overlooked the fact that I hadn't responded. Yes, you are spot-on

I actually posted my answer at the bottom of the previous page - in response to  Sophiebear who also gave the correct answer but in a different format. I think I must have assumed I had posted that response for you as well. Apologies.

I have been rather busy today and i'm back to work tomorrow (work a funny week !) so I haven't had time to type up or produce any diagrams for my next one quite yet.

However, winky set us all a neat puzzle at the top of this page and both Sophiebear and myself have had a shot at it - with completely different results so far. I haven't found time to get my tiddly wink(y)s set out yet to re-evaluate my initial thoughts.