Brain Teaser No 1

Don't feel too depressed about not solving Bam's ladder. The rest of us have been trying since before Xmas 01, ie best part of SIX MONTHS....who said "anoraks"

Cheers

Don

Matthew T,

I'm thinking, but getting all hot and bothered. Hope this is not adding to the greenhouse effect!

Cheers

Don

Looks elegant enough for me, although I haven't had time to check it out yet for accurcay !!

Have now checked and, of course, its perfect. Gives the value of d for a 'near horizontal' ladder. Is this what you had in mind when you wrote i should actually carry the plus or minus on the solution for d in term of x all the way -- but this doesnt make any difference to the solution


Cheers

Don

Don,

Answers...

Radius 4,3,2,1 gives 0.476220238080706

The fish bowl problem (I am very impressed with my spread sheet) gives 1.21543808788148

I will give you a little longer before I start giving some useful information to get a solution for the greenhouse gas question.

Are there any other brain teasers unsuccessfully answered?

Matthew

don, i had actually given up on the ladder problem. then, coincindentally, i was working on something involving hyperbolics -- and stumbled upon the the expression (x+sqrt(x^2-1)) a few times, then the penny dropped... the last clue by bam was quite useful as well.

i hope this finally lays this puzzle to rest.

enjoy

ken

Matthew, I can recall four.

Page 14 4th Jan Mattew T "Another question" about an airliner flying at altitude..........

Page 30 29th March Don A "What a mix-up" about people at the Bristol Hifi show.....

Page 31 18th April Don A "Beer circlers or larger circles" about the size of the shaded area......technically solved, but the ELEGANT solution not yet found. Matthew T offered to wait a while before publishing his.........

Page 35 23rd May Matthew T "Greenhouse gasses"

So, Matthew, a nice question...Are there any other brain teasers unsuccessfully answered?
. Looks like you hold the key to most of the answers.........fire away !!!!!

Cheers

Don

OK attached is the method I used to get the beer circles figured.

Matthew

Will have to further investigate the plane one.

Matthew

Will have to further investigate the plane one.


Matthew, you set the question ..........

Cheers

Don

Matthew, there is an elegant, non-mathematical (well, non-trig, non-Euclidian-geometric) solution, that demonstrates the shaded area is less than a quarter.

Your solution above is the equivalent of my solution to Bam's ladder, ie it works, but isn't elegant enough!. We are now looking for the elegant Ken C equivalent solution to Bam's ladder.

Remember, This is the NAIM brain teaser forum, NOT the Sony forum, where quite rightly your solution would be so elegant that the rest of their forum would resign in humble shame!!!!

Cheers

Don

Oooops! forgot to say "well done" on the (non-elegant)solution to the beer circle problem !

Apologies

Don

I am VERY impressed with Matthew T's co-ordinate geometry/spreadsheet solution to the five-kissing-balls problems.

HOWEVER, yes, you know what's comming, however, there is a really elegant solution available, really elegant

It looks elegant, and it ONLY requires a bit of VERY simple algebra (Let R=unknown radius!), fractions, and the famous quadratic solution "minus b, plus or minus the square root of b squared etc...."

Cheers

Don

My diagram became too complciated.......... I would need to write a table with all the info, which seems too laborious to bother with.

You could try excel !!

Not sure if I am really supposed to give a clue to solving someone else's puzzle.......

its so much more fun when different people give different clues............one man's clue could be another man's solution....

Cheers

Don

Using Matthew's diagram the assertion is,

3A + B < a/4, where 'a' is the area of a beer mat.

It's obvious that A + B = a/6, so the assertion reduces to,

2A < a/12, if 'r' is the radius of a beer mat then

2A < Pir^2/12

It's also clear that the area of the triangle in the centre of the intersection is r^2/2, so

6(A + r^2/2) = a = Pi r^2

So

2A = (Pi - 3)*r^2/3

Therefore the area of the intersection is less than 1/4 of the beermat if

(Pi - 3)/3 < Pi/12 or

Pi - 3 < Pi/4

Which is obviously true, so the area is obviously less than a quarter.... I'm sure this could be presented more elegantly, and I apologise for any errors.

(I note that I was supposed to use a quadratic equation, so I presume the above has an error. Your mission, should you choose to accept it, etc etc.)

Paul

Paul R,

(I note that I was supposed to use a quadratic equation, so I presume the above has an error. Your mission, should you choose to accept it, etc etc.)


The quadratic equation referred to the elegant solution to the 5 kissing balls teaser that Matthew T had been tackling.

The beer circles doesn't need quadratic equations for its elegant 'non-mathematical' solution.

Cheers

Don

Back on page 14 Mattew T asked
Another Question

When an airliner is flying at an altitude of 10 km, the temperature of the air outside may be as low as -50ºC. One might think that this would require the use of heaters inside the cabin, but in fact an aircraft flying this high must use air-conditioners. Why?

Even at 10km altitude, the atmosphere is dense enough to heat an airliner through skin friction to quite high temperatures. Concorde for example at 1500mph heats up to approx 100 deg C and expands in length by about 1 ft. This in turn heats the interier cabin atmosphere.

Engine power is then partially used to cool, or air-condition, the airliner's cabin atmoshere.

Perhaps.

OTOH, Bam might have been right about sub-sonic friction not being too great and it all being down to gossip by passengers. Or Kieth M's sunlight might have had a ray of truth in it.....

Perhaps

Don

quote:

---

The beer circles doesn't need quadratic equations for its elegant 'non-mathematical' solution.

---

Doh.

I'll get back in my post-Euclidean box now.

Paul

Logical beer circles

In the above the red areas might be 'R' and the blue 'B'. The area of a beer mat is 6(R + B), the area of the overlap of three mats is 2R + B. The assertion is that that overlap is less than a quarter of the area of a single mat.

So,

4(2R + B) < 6(R + B)

8R + 4B < 6R + 6B

2R < 2B

R < B

As can be seen from the additional cyan outline, the green area is the same as the red. It also fits within the blue. Therefore R < B and QED.

Paul

Airplane

This isn't the definitive answer but the major factos keeping the plane warm more to do with have a sealed insulated tube filled with people then anything else.

Some useful info for the Greenhouse gas problem...
Atmospheric Preesue = 100kPA
Avragardos number = 6.0221367x10^23
Assume atmosphere = 80% N2 (Mass 14 g/mole), 20% O2 (Mass 16g/mole)

Matthew

Paul,

your red/blue/green/cyan drawing is a real nice try...........but.......you can develop it a bit more and then make the explanation more simple......but if I tell you too much, it will take away that real sense of satisfaction when you 'see' the elegance of it, for yourself.

Cheers

Don

Matthew T,

Insulated tubes with hot sweaty bodies = aeroplanes that need to be airconditioned........I can just about understand

Avogadro's Number...I can vaguely recall from school physical chemistry....and just about remember how to spell, I think ?

But linking Avogadro, nitrogen 14, oxygen 16, carbon dioxide and its inter-molecular (or inter-atomic) absorption properties of a spectrum of electro-magnetic radiation to the greenhouse effect leaves me feeling about as optomistic of finding a solution as recovering from a hole in my head...........ah-ha !!!.....or a hole in the ozone layer caused by insulated flying tubes......

Let me describe the look of amused disbelief on the faces of the regulars down at 'The Wheatsheef' when I casually sought a bit of light hearted discussion on the matter......or perhaps not.....

More ice, please ! More ice !

Cheers

Don

[This message was edited by Don Atkinson on FRIDAY 31 May 2002 at 23:55.]

I've just got back from listening to a load of Mana, so now it is obvious.

'B' becomes the area of the blue bit in my diagram bounded by two reds and a green. In that case the area of a beer mat is

6(2R + B )

and the area of the intersection is

3R + B

The assertion is then that

4(3R + B ) < 6(2R + B)

which I think I can safely say is obviously true.

Off tomorrow to do some motorsport, I expect an exuberance of elegance to await me on my return.

Paul

I expect an exuberance of elegance to await me on my return.


There, I KNEW you would feel better !!

Cheers

Don

Here's my solution FWIW

A circle contains two basic shapes, bananas (B) and deltas (D. A full circle comprises 12 x (B) plus 6 X (D).
Hence a quarter circle would comprise 3 x (B) plus 1.5 x (D)

The overlapping portion from the original question, shown shaded again, comprises 3 x (B) plus 1 x (D) which is smaller than a quarter circle by 0.5 x (D)

We must have all drawn six-petalled roses inside circles when we were chirdren ??

Cheers

Don

OK, so the World Cup is dominating the newspapers and the Forum.

We already know that the maximum number of footballs, cricket balls, golf balls or marbles that we can stack so that each one just touches each of the others, is five, and that applies to any combination of footballs, cricket balls, marbles etc as well. And Matthew T has clearly got a spread sheet for working out the size of the fifth one, given the size of the other four.

One more chance for anybody to publish a simple and elegant formula, before I publish my version in the next day or so.

Cheers

Don

[This message was edited by Don Atkinson on SUNDAY 09 June 2002 at 17:43.]