Brain Teaser No 1

I agree with Don.

However using bluetac I managed...

Matthew

BAM

I think we've been tortured enough.
Please publish elegant solution very soon!!

STEVE D

Fellow conjurers,
I love this thread. Don should receive high praise for creating it so long ago. I think Don was truely inspired when he innocently posted that first puzzle about the bloke walking for miles and ending up back where he started. Or perhaps this was a foretelling of how things would turn out? Isn't that first puzzle uncannily predictive of how events were to unfold in this thread - is Don now the victim of the monster he himself created? Mysterious forces may be at work.

I also really like the atmosphere of mutual discovery and respect that this thread embodies. It is an honour to be a participant and a joy to share the process of unravelling these puzzles. The sense of belonging and togetherness is deeply moving.

"oooops did I say last pint?"
"you should have a fairly soft landing. Any volunteers???(Bam?)"
"Wonder what sort of curve Bam would have followed from the 10th floor Presidential Suite of the Bristol Meridian........."
"I worked out an equation for the rope to hang BAM by using the sum of two sectors less the overlapping pair of Isoceles triangles."
"Lets hope we find that truly, stunningly, elegant formula for the ladder before you meet with any mishap!"
"I am visualising a further use for that rickety old ladder, the tea chest and that length of rope with a noose in it that is currently tied round that poor goat's neck......."
"As for BAM. I've got him on my list"
"You don't want to gain a reputation like Bam "
"so b$%^&#d bam can post it :-)"

I am really sorry if I am out of line here, but I do detect a certain trace of animosity directed at me among some of the postings with regard to the ladder problem. Now, I can be a little paranoid sometimes and I don't want to cause offense to anyone by suggesting that they may be expressing mild disharmony in any way within this esteemed thread. Oh, forget it - I'm probably just being silly.


It has been some four months now since the Sony engineers have been trying to work out the optimum position of that ladder. Let's review the history so far:

Page 13, Sat 21 Dec: Ladder problem posed.
page 16, Sun 13 Jan: Don posts a solution
page 16, Mon 14 Jan: bam prize announced for a more "elegant" solution.
page 16, Thu 17 Jan: A hint to Don "To win the champagne you need to use less conventional
maths."
page 17, Fri 18 Jan: Another hint to Don "I hope you are not angling for a hint, Don"
page 19, Tue 5 Feb: Paul Ranson and Ken C make a short-lived inroad in the right direction.
page 26, Wed 6 Mar: Another hint "And I should state for the record that although I went off on a bit of a tangent with my rant about TV documentaries I can assure you there are no hidden clues there."

A few false trails have arisen along the way. The solution is nothing to do with astrophysics. Nor is it to do with circles as such - fitting circles to the tragectories of parts of the ladder as bam falls of it, etc. "less conventional maths" means just that. It doesn't mean unconventional maths. Don's original answer has a basic algebraic form. Since the days of Pythagoras, mathematics has developed special functions for managing angular relationships. Incorporate some of these and the solution can be expressed more elegantly.
Enjoy!
BAM
(quickly ducks to avoid Don's right hook. Arrr, Jim lad.)

[This message was edited by bam on SATURDAY 04 May 2002 at 11:58.]

[This message was edited by bam on SATURDAY 04 May 2002 at 11:59.]

Now, I can be a little paranoid sometimes and I don't want to cause offense....(but)... I'm probably just being silly.

Just because you've got over the paranoia, doesn't mean they aren't still after you!!!

I only hope SteveD doesn't have a nervous breakdown.!!!!!

Cheers

Don

Don,
I forgot to say that the books must be one on top of the other, eg stacked.

And by the way, although I must admire the man for his sexual exploits, I feel I should point out that Mr Levinson hasn't had anything to do with Madrigal or "Mark Levinson" branded products for many, many years. He sold his name when he got out of the company.

Steve,
Deep breaths, deep breaths lad.

Mr Levinson hasn't had anything to do with Madrigal or "Mark Levinson" branded products for many, many years

Selling roses by all accounts at the moment. Bit of a downer from violins......or was it cellos?

Cheers

Don

BamI forgot to say that the ........

I am glad these words didn't appear in your last post about the ladder......otherwise it might well have been your last post.......and we would then have the grim satisfaction of hearing the last post played on a bugle.........


So going back to the books......Mathew T and I got the first version right? ......and now you would like us to try the second version? I seeeeee...........

Cheers

Don

I have a dead flat table. (probably the same make as Bam's with his three books). Sitting on this table I have three, large, juicy (but firm) Jaffa oranges. Each orange is a perfect sphere of radius 100mm (well I did say large!). The three 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 on the table, within the group of three oranges, so that it just touches the table and the three oranges.

What size (radius) cherry do I need?

Bam will hold his guns for the next version, where the three oranges are replaced by one orange, an apple and a lemon and they all sit on a globe!

Cheers

Don

At last, I've cracked it, THE solution to Bam's ladder

So Bam's main criteria include

Less maths
Tangents
Angles

Let A=Angle between ladder and wall
Then Tan A=d (where d=distance of ladder toe from crate)

Hence THE EQUATION is *** d=tan A ***

Now this is VERY elegant etc etc

Oh! excuse me ! did you say something? What do you mean......how do you find tan A??????

Tan A = 0.5{[L^2 + 1]^0.5 - 1 - (([L^2 + 1]^0.5 - 1)^2 - 4)^0.5}

L of course = length of ladder.

Now where have I seen that before!!!!

Cheers

Don

Bam asked how far three identical books could be got to cantilever beyond the edge of a flat table. The books to be sandwiched, and no glue allowed.

Well, I tried experimenting with three table mats which, appart from the pictures, are identical. Each one is 210mm long.

Obviously, simply stacked one on top of each other, they can be cantilevered 105mm out from the table.

However, pulling the top pair back a little, makes the whole thing more stable. Consequently you can push the bottom one out a bit further than 105mm. In fact I managed to push the bottom one out to 175mm with the top pair just balanced centrally on the the back edge of the bottom one, thus still formming a sandwich.

This means the bottom mat is projecting 0.8333L from the edge of the table, where L= length of mat (or book).

I wonder if I can improve on this?

Cheers

Don

I wonder if I can improve on this?

Tried again with the 210mm long table mats. This time I got 193mm or thereabouts.

Will explain later

cheers

Don

As I said above, I managed to get 3 table mats to project 193 mm when each mat was 210 mm long.

This was done by cantilevering out the first (bottom) mat by 35 mm, the second (middle) mat by a further 53mm beyond the bottom mat, and the third (top) mat a further 105 mm beyond the middle mat.

The total projection was 35 + 53 + 105 = 193 mm.

Now this was an experimental result and subject to small errors. First I couldn't always get the mats dead square to the table or each other and second, I couldn't measure the projections better than +/- 1mm. Also, the mats were on the point of balance and swaying gently. I therefore reduced the projections slightly, to keep the mats stable and compensated by rounding the measurements up, to the nearest mm.

Of course, although I have described the end result as "…. cantilevering out the first (bottom) mat by 35 mm…." this isn't how I did it.

I actually started with the top mat and balanced it (105 mm) on the middle mat, then slid the middle and top mats out until they were just on the point of toppling off the bottom mat. I then slid all three mats out until they were just on the point of falling of the table.

Well, I couldn't stop at three mats……..but that’s another story.

Now I wonder if I can improve on this?

Cheers

Don

Where have all the mathematicians gone?

It should be possible to do some maths with the sum of the centres of gravity of the mats/books, assuming they're about to topple, and therefore have to be in balance about a pivot, which is equivalent to the edge of the table.

If you have one mat there is only one balance, if you have two there are an infinity of arrangements, but the maximum overhang of the upper mat from the pivot is 3/4 of a mat length.

Three isn't immediately obvious.

I have a strong recollection that a complete overhang is possible with sufficient mats.

Paul

Well, I couldn't stop at 3 mats could I? We have, after all, got the usual 6 mats in the set. So I kept on going; 105mm; 53mm; 35mm; 26mm; 21mm; and 18mm. Then I plotted them on a graph.....and got a nice smooth curve.

Then the brain-wave struck! Ok, Ok, I know it takes a long time at my age for the penny to drop! Smooth curve usually = simple formulae.

Then I read Paul Ranson's note about the mathematicians. Well I'm no mathematician so I can't prove the next bit.....its more like a physics experiment.

L = length of book (table-mat) = 210mm
y = L/2 = 105mm
n = number of the mat (1 = top mat)

By inspection of the smooth curve, the cantilevered lengths (beyond the mat below) in terms of y are (pretty dam close!)

y/1; y/2; y/3; y/4; y/5; y/6......y/n

When y = 105mm, the calculated projections, using the above conjecture are:-

105mm; 52.5mm; 35mm; 26.25mm; 21mm; 17.5mm…..

The first three terms add up to 192.5mm, or 11/12*L, which is pretty close to my first 193mm experimental result!

Does this qualify in the eyes of Mr Ransom as an acceptable solution?

Now I wonder if there's a better way than this? Bam?

Cheers

Don

[This message was edited by Don Atkinson on FRIDAY 17 May 2002 at 19:01.]

Now I wonder if there's a better way than this?

Use beer mats???

Cheers

Don

I don't understand how the maximum overhangs of the first three mats will support further overhanging mats when you extend to 6

I've obviously misunderstood something?

It should be possible to write down an equation stating something like 'for each mat, the cumulative CoG of the mats stacked above must be within its width' and then maximise for the overhang of the top mat.

For two mats, 0 and 1, stacked 0 then 1, of width 2L, where 'xn' is the overhang of mat n, then

(x0 - L ) + (x1 - L) = 0

and

x1 - x0 < L

Values of x0=0.5 and x1=1.5 produce the maximum overhang (or value of x1) boundary condition.

Presumably for 3,

(x0 - L) + (x1 - L) + (x2 - L) = 0

x1 - x0 < L
x2 - x1 < L

TBC?

Paul

There's nothing like saying you want to do one thing, and then doing quite another. The above contains some complete bollocks. Please treat with the contempt it deserves.

Paul

You suggested It should be possible to do some maths with the sum of the centres of gravity of the mats/books, assuming they're about to topple, and therefore have to be in balance about a pivot, which is equivalent to the edge of the table.

A few posts above I tried the following experiment

.....However, pulling the top pair back a little, makes the whole thing more stable. Consequently you can push the bottom one out a bit further than 105mm. In fact I managed to push the bottom one out to 175mm with the top pair just balanced centrally on the the back edge of the bottom one, thus still formming a sandwich.

This means the bottom mat is projecting 0.8333L from the edge of the table, where L= length of mat (or book).


Now this can be mathematically demonstrated to be stable by taking moments about a convenient point, such as the edge of the table.

I think it illustrates the maximum projection possible based on a sort of 'conventional' balance/counterbalance theory.

But bam is far too subtle to post a 'straightforward' teaser! So i started looking at other possibilities by experimenting.

It should be possible to 'prove' my y/1; y/2; y/3....conjecture is 'just' stable by taking moments about the edge of the table, or some other convenient point. What it won't do, is prove whether this solution gives the maximum possible projection, which for three books, is probably L.

Confused? I am

Cheers

Don

A couple of weeks ago I posted the following teaser:-

I have a dead flat table. (probably the same make as Bam's with his three books). Sitting on this table I have three, large, juicy (but firm) Jaffa oranges. Each orange is a perfect sphere of radius 100mm (well I did say large!). The three 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 on the table, within the group of three oranges, so that it just touches the table and the three oranges.

What size (radius) cherry do I need?


Not a single reply. Not even a hint of a reply.
Is it too boringly simple, too difficult or is my paranoid feeling of being entirely unloved and rejected, comming true???????

Careful with the answers, please

Cheers

Don

33 and a third mm seems likely.

Paul

Paul,

Many thanks for the sanity restoration therapy.

Dead right....I thought 33 and a third had a nostalgic, vinyl ring to it?

Now for the fruit cocktail that I indicated earlier :-

What size (radius) cherry do I need?

Bam will hold his guns for the next version, where the three oranges are replaced by one orange, an apple and a lemon and they all sit on a globe!


So the orange is 4" radius, the apple 3", the lemon 2", the globe 12"

And the cherry is ?

Cheers

Don

That cherry is going to require some though. BAM should find it trivial though.

Anyway popping more than one cherry a night seems somewhat extravagant for a man of my age.

Paul

one.....a night seems somewhat extravagant for a man of my age.

For a man of your age, not yet 18 months, they obviously have a good maths teacher at that kindergarden/nursery school that your mum takes you to!

Cheers

Don

A few post back you said

It should be possible to do some maths with the sum of the centres of gravity of the mats/books, assuming they're about to topple, and therefore have to be in balance about a pivot, which is equivalent to the edge of the table.

If you have one mat there is only one balance, if you have two there are an infinity of arrangements, but the maximum overhang of the upper mat from the pivot is 3/4 of a mat length.

Three isn't immediately obvious.

Have a look at the attachment and let me know what you think.

Cheers

Don

I have a strong recollection that a complete overhang is possible with sufficient mats.


Using my conjecture that
the total projection is y/1 + y/2 + y/3...+ y/n
where y = L/2

then 4 books should give an overhang > L

In fact it should be 25L/24

I think!

Cheers

Don