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

Don

Its a great problem ! My first reaction was that you had forgotten to include the radius of the hole and/or the radius of the bead. But then I thought that was very unlikely !

The problem is a lovely extension of the annular ring problem that you set previously. The amazing (at least amazing to me) result is that the drilled bead has a volume equal to a solid sphere of the same height of the bead. So for this example:

Volume = 4/3 * Pi* R^3 which for a height of 6" is (36* Pi) cubic inches

The "proof" is a little tricky to write down - so I'll just outline the thought processes:

1) The drilled bead is essentially a stack of annular discs

2) The height of the stack in this example is 6". Let's call the height = 2 *H

3) The area of each annular ring can be represented by a solid disc of radius which corresponds to half the length of the  "tangent line" in your previous problem.

4) Let R = Bead radius

          r = Drilled hole radius in bead

5) Consider the annular ring which is at the equator of the bead (maximum radius). This slice has an equivalent radius R equiv given by:

R equiv ^2 = R ^ 2 - r ^ 2

6) From our previous annular disc problem we know R^2 - r^2 = H ^2. So this equatorial slice has an area which is equivalent to an equatorial slice within sphere of radius H.

7) The final part is to then show that the areas of the annular ring slices at any height (y) above or below the equator are equivalent to the same area of solid disc slices in the sphere of radius H.

8) The area of an annular ring slice in the bead at height y above the equator can be shown to be:

Area (y) = Pi (R^2 - r^2 - y^2)

(note: this can be proved by considering a point at on the circumference and recognising that the radius is R cos Theta and the height (y) = R sin Theta where Theta is angle subtended at the centre of the bead)

9) The corresponding area for the solid sphere disc at height y is given by:

Area (y) = H^2 - y^2

10) Now since in (6) we know that R^2 - r^2 = H^2, then the expressions in (8) and (9) are identical. So the annular ring can be mapped to an equivalent solid disc at the same height above the equator.

I hope folks can follow my logic. I thought about trying to post a diagram, but I failed miserably with my cut & paste. Seems that posting diagrams on the Forum is a little more complex than updating the firmware on my Uniti ?

Regards,

Peter

Peter,

That's a fair crack at this problem and a very original approach (btw, did you miss a "Pi" out in your equation No.9 ?)

Did you ever do Integration at school ? Most people who did "A" Level maths would have done so.

In this example, you would work out the area of a very thin (horizontal) slice of bead at a height "h" above your equator (all the radii cancel out).

The volume of this slice ("Delta"V,) is got by multiplying its area by "Delta"h (usually written as dV and dh respectively).

You then "Integrate" (ie add up the volumes of all the little slices) from -L to +L (where 2L is the length of the bead's hole)

and.....Voila ! you find that the volume of a "Bead" is the same as a sphere whose diameter is equal to the length of the hole though the bead

I'll post the method step by step tomorrow evening unless someone feels inclined to do so before then !

Well done Peter and sjbabbey

Yes I see where I went wrong. I've integrated the formula for the area of a circle (πrˆ2) instead of the formula for the surface area of a sphere i.e. 4πrˆ2. Schoolboy error.

So my answer should have been 36π cu ins or 113.143 cu ins.

I seem to recall that we used to use the rotation of the area under a curve to calculate the volume of a solid which I believe is the method you've described but it was a long time ago.

Don

Yes, there should be a "Pi" in point (9). So I lose a mark there !

I thought the problem could be done using calculus, so I look forward to seeing your solution (not that I will necessarily be able to follow it ?) I do still have a vague recollection of integration...but all in the dim & distant past.

Thanks again for setting a great problem with a lovely result.

Regards,

Peter

Guys remember, the title of the thread includes ....." ........or 50 years on !"

I guess most of us have this rather vague and somewhat dim recollection of things we did at school. In my case its all more than 50 years ago now.

I did, however, buy copies of Bostock & Chandler about 20 years ago so that I could help my three daughters with their A-Level maths and I have recently been browsing through Jean Holderness with her GCSE maths as my grandchildren seem to be growing up pretty fast now !!

Anyway, I do hope some of these problems get the old grey matter moving, without causing any angst or stress.................................

Cheers

Don

Picture first, text to follow............

FIG 1          FIG 2

X Section Thru' Bead          Plan on Slice

R   =  Radius of sphere

Ro =  (outer) Radius of thin slice

r_i    =  (inner) Radius of thin slice

l     =  ¹/₂ length of hole

A   =  area of thin slice

R_o^²  =  R^² - h^²

r_i^²    = R^² - l^²

R_o^² - r_i^²  =  l^² - h^²

A  =  π  (R_o² - r_i²)

Noting symmetry about x-x axis (ie thin slices above & below the equator hence 2xA)

δV =  2π ( R_o^² - r_i^² )  δh          (this is the volume of two thin slices, one above & one below the equator)

  V =  2π   Σ₀^l ( l^² - h^² )  ^dh/_dv          (accept Σ0l for the usual 'integrate from 0 to L' symbol)

      =  2π  [ hl^² - h^³/3 ]₀^{l          (that was the basic integration - remember ?)}

      =  2π  [ l^³ - (l^³⁾/3 ]          (I have substituted for the values h=L minus h=0)

      =  2π 2 (l^³⁾/3

      =  ⁴/₃π l^³          (the familiar volume of a sphere)

(ie Vol of bead is independent of sphere radius but equal to the volume of a sphere with Diameter equal to the length of the hole in the bead.

Hope this is clear ?

Cheers

Don

Nice work. It is quite counter-intuitive as to why the bead should be exactly equal in volume to a sphere whose diameter is the same as the length of the bead.

Thank you winky. as you say, its quite counter-intuitive and I have always felt this, and a few other problems make you stop and think with a more open mind, rather than simply applying a suitable formula and turning the "sausage-machine" handle.

Cheers

Don

"A" Levels still seem to include a few Probability questions, and so far these haven't caused much of a problem here. We also covered Combinations and Permutations in my "A" Level courses which can sometimes be useful when evaluating probability..

A nice easy one to start..........

One card is drawn at random from a pack of 52 playing cards. What is the probability that it is an Ace ?

Now a slightly harder one..........

Four cards are drawn at random from a pack of 52 playing cards. What is the probability that all four cards are Clubs ?

and a third one for good luck............................

Four cards are drawn at random from a pack of 52 playing cards. What is the probability that at least one of them is an Ace ?

11/4165 for 4 Clubs.

Is each card placed back into the deck after drawing, or kept aside?

If they're kept aside, I agree with sjabbey:

1/4 x 12/51 x 11/50 x 10/49 = 1320/499,800 = 33/12,495 = 11/4165 = 0.2641056%

Again, not placing the cards back in the deck between draws:

1/13 + 4/51 + 4/50 + 4/49 = 31.69871%

(I'm using Excel and can't bothered doing the addition of fractions properly). 

A harder question is "What is the probability that exactly one of them is an ace?"

I used the same method as Winky and got odds of 3.1547:1 or 31.6987%

I' m guessing that it would be 1/13 x 48/51 x 47/50 x 46/49 i.e. 15.6525 : 1 or 6.3888%

Apologies. Good point.

None of the cards are replaced in the pack.

PS. I now notice that both you and sjbabbey have gone this route and get the correct answer. Great.

again apologies. None of the cards selected at random are returned to the pack, but...............

................I don't get 0.317. I get a slightly lower probability..............0.28 to 2dp

hmmmmm both you and winky agree

I went down the route of determining the probability of four cards containingno aces.....and got 0.72

Then deducting that from 1.0 gave me 0.28 to 2dp

But I could be wrong, so I will check more carefully tomorrow.

Cheers

Don

I agree with winky and sjbabbey for the first two questions. For number three (at least one ace) my thoughts align more with Don.

The odds for only drawing only other cards are 48/52 * 47/51 * 46/50 * 45/49 = 0,7187. All other scenarios contain at least one ace. So 1 -  0,7187 = 0,2812

I checked my own working and am convinced I have the correct answer. Its also good to have an endorsement from Mulbery. But who knows, we might both be wrong !

My thinking was as follows -

• The number of Combinations of drawing four cards (without replacement) that do not contain an Ace is 48C4 ie (48x47x46x45)/(1x2x3x4)
• The total number of Combinations of drawing four cards (without replacement) from a pack of 52 is 52C4 ie (52x51x50x49)/(1x2x3x4)
• The probability of drawing four cards at random that are not Aces is therefore (48C4)/(52C4) ie (48x47x46x45)/(52x51x50x49) = 0.72 to 2 dp
• Hence the probability of at least one of the four cards being an Ace is 1-0.72 = 0.28 to 2 dp

we divide the pack into 2 sets, 4 aces and therefore 48 non-aces.

we can select 1 card from the 1st pack in 1!4 (or equivalently 4C1 using combinatoric notation) ways

we then select the remaining 3 from the 48 non-aces in 3!48 ways

hence the probability is:  ((1!4)*(3!48) ) / 4!52  = 0.255551

In fact this result can be used to check that the "at least 1 ace" probability i indeed 0.28 because the "at least 1 ace" case is exactly the sum of probably of the following 4 events:

- exactly 1 (as above)

- exactly 2

- exactly 3

- all 4 are aces

when i worked this out, the answer as indeed 0.28 so we can be confident that this is indeed correct.

enjoy

ken

I have drawn a Probability Tree for the last puzzle.

Abstracted from the Tree I have the following _

Probability of zero aces  = 0.718737

Exactly one ace          = 0.255551

Exactly two aces          = 0.025

Exactly three aces          = 0.000709

Exactly 4 aces          = 0.000004 (0.0000038 something)

The sum of these is 1 (as expected...........subject to rounding errors)

I will try to photograph my scruffy Tree and the associated Excel spreadsheet that I used to calculate the various probabilities.

I think the Probability Tree helps to make the various options incredibly clear. But i'm somewhat busy at work until Tuesday and i'm sure Mrs D has a few jobs lined up for Tuesday as well, so it might take a day or so to get around to this task.