Brain Teaser No 1

Posted by: Don Atkinson on 16 November 2001

THE EXPLORER

An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey?

Cheers

Don

Posted on: 31 August 2004 by Nigel Cavendish
He's not flying upside down is he?

cheers

Nigel

Posted on: 31 August 2004 by Berlin Fritz
Can't take him anywhere, so we know who not to invite to the party next time, innit!


Cheers,
Fritz Von Knowanypunchlines Wink
Posted on: 31 August 2004 by Berlin Fritz
London Uniwersity's Queen Mary College for Good Eggs is presently deweloping a new application for a very ancient ratio with the value of (1.618 - 1). Any tokers for its name and original grounds for its creation ?


Fritz Von Notalotovpeopleknowthisorwanttoeitherforthatmatter Cool

We're Only Makin Plans For Nigel
Posted on: 31 August 2004 by Brian OReilly
Golden ratio ??

Building your very own pyramid ??
Posted on: 31 August 2004 by 7V
quote:
Originally posted by Berlin Fritz:
... rolling along at nearly Mach 2

A bit of a giveaway that, Fritz.

Steve Margolis
defy convention - make music
Posted on: 31 August 2004 by Berlin Fritz
quote:
Originally posted by Brian OReilly:
Golden ratio ??

Building your very own pyramid ??


Well done that man, though it relatesto human faces and art rather than early Swiss Toblerone developement, my mate's sisiter was a flake girl too, but she snuffed from pneumonia due to getting soaked all the time while paintin silly water colours when stoned ?

Fritz Von Rollinrollinrollin Big Grin
Posted on: 31 August 2004 by Don Atkinson
London Uniwersity's Queen Mary College for Good Eggs is presently deweloping a new application for a very ancient ratio

...any ideas about the new application....

Cheers

Don
Posted on: 07 September 2004 by Don Atkinson
Unhinged.....

Two equal rod, (AB, BC) are hinged at B and slide at A and C.

How should these rods be arranged to maximise the area ABCD ?

Cheers

Don
Posted on: 07 September 2004 by Don Atkinson
Un-hinged.........even bigger brain teaser

What happened to the "paper clip" for attaching pictures?

How do I load the image to illustrate these teasers?

Help.....

Cheers

Don
Posted on: 07 September 2004 by Paul Ranson
I believe Naim have disabled attachments to reduce storage requirements. I can store an image on my web space and link to it if that would help.
  B
 / \
A - C

or we could try 'ASCII' art.... The lengths AB and BC are equal, AC is variable. The solution is the angle ABC that maximises the area. I think the intuitively obvious solution is correct.

Paul
Posted on: 08 September 2004 by Don Atkinson
ASCII art....

D---C---X
|....|
|....|
A---B
|
|
|
Y

Hopefully, this will look like the picture I intended.

D is a right-angle
YAD is solid and straight
DCX is solid and straight
AB is solid and straight
BC is solid and straight
AB and BC are hinged at B
A can slide along DY
C can slide along DX
ABCD is drawn as a square simply because in ASCII that is the easiest shape to draw
Ignore the dots inside ABCD, they are only there to force the lines that form BC into the right place

The task is to define the arrangement of AB and BC that maximises the area ABCD
In other words, what is the shape of ABCD when Area ABCD is a maximum? (Bending of rods is not allowed).

I hope the question is clear

Cheers

Don
Posted on: 08 September 2004 by Don Atkinson
Remember....

AB and BC are equal (my picture wasn't as accurate as I had hoped....

Cheers

Don
Posted on: 08 September 2004 by Paul Ranson
That's a bit more interesting than my guess at what you meant...

Paul
Posted on: 08 September 2004 by Don Atkinson
That's a bit more interesting than my guess at what you meant...

...only "a bit" more interesting.....

OK you're right, it is only a bit more....

Cheers

Don
Posted on: 14 September 2004 by Don Atkinson
OK,

So the 'sliding rods' question was either too simple to be bothered with, or too difficult to start, or.....

I'll post a very simple one next, then we can come back to the rods

Cheers

Don
Posted on: 14 September 2004 by Don Atkinson
Commuter stares at neighbours....

You are in a long line of commuters, randomly spaced out along a platform awaiting a train. If each commuter is asked to look at the person nearest them, what is the probability that the commuter to your left AND the commuter to your right will BOTH be looking at you.

Cheers

Don
Posted on: 15 September 2004 by Don Atkinson
Naim Double Paul Darwin's Salary.....

Paul Darwin received his salary for last year in a single lump, last Friday. It was delivered to his desk by Paul S as £ coins in a wheel barrow. (yes, it was an exact number of pounds)

Now Paul D set to work immediately arranging his salary into nine unequal piles of coins, but triumphantly realised he had created a 3x3 'magic square'.

Paul S was impressed, but pointed out that there wasn't a single prime number in any of the piles.

But Paul D wasn't finished. "Ahah!" he shouted, "give me another £9" (which Paul S did) "if I add one coin to each pile, I create another 3x3 magic square, but this time ALL the piles are prime numbers!"

Paul S was even more impressed and offered to double Paul D's salary if he created another 3x3 magic square of primes.

9 seconds later, Paul D had removed £2 from each pile to create another 3x3 magic square of prime numbers.

Paul S was really impressed and, sticking to his word, immediately doubled Paul D's salary.

What was Paul D's initial salary (which has now doubled).

Cheers

Don
Posted on: 16 September 2004 by Lo Fi Si
unhinged

I don't think I can ASCII the answer so

The angle DCB (call it c) is 67.5° for maximum area. I assumed that the solution would be symmetrical about DB so BDC is 45 °. The area is then proportional to sin(c)*sin(135-c) which has a maximum at c=135/2.

Simon
Posted on: 16 September 2004 by Don Atkinson
Hi Simon,

Well spotted. You certainly have the right answer.

My description would be that internal angle ABC = 135 deg with AD = AC

ABCD is then one quarter of an octagon.

Cheers

Don
Posted on: 01 October 2004 by Don Atkinson
A couple of easy ones for the weekend....

Cheers

Don

But don't forget the Paul Darwin salary one from the previous page....
Posted on: 01 October 2004 by Don Atkinson
Matthew & Mick go cycling together....

Young Matthew has taken his mountain bike (he lives in London!) down to Swindon for the weekend. Mick is riding around Swindon in gentleman fashion on his ex post-office shopper.

Matthew stopped at a set of traffic lights which have just turned green and he is ooooffff....., passing the green light at 1m per sec and accelerating at 2m per sec per sec.

Mick has been approaching the same lights at a steady speed, but on recognising Matthew has put on a bit of a dash, and passed the green light 4 seconds after Matthew at 16m per sec, accelerating at 1m per sec per sec.

Haw far past the green light will Mick overtake Matthew?

Cheers

Don
Posted on: 01 October 2004 by Don Atkinson
Number flushes

Can you write the numbers 1 to 9 in such a sequence such that it isn't possible to extract from the sequence ANY four-digit rising or falling subsequence?

eg 1 2 3 5 4 6 9 7 8 ISN'T acceptable because 1 2 3 5 is a rising subsequence as is 1 2 3 4 and 1 2 3 9 etc etc. But there are no four-digit falling sub-sequences in this particular sequence.

Cheers

Don
Posted on: 01 October 2004 by rodwsmith
Surely just alternating does the trick?

9, 1, 8, 2, 7, 3, 6, 4, 5

Perhaps I am missing something?

Rod
Posted on: 01 October 2004 by Don Atkinson
Rod

Perhaps I am missing something?

unfortunately you are missing somthing, but perhaps I need to clarify the question.

Your sequence contains (amongst others) 1 2 3 4 as a sub-sequence. (The numbers don't have to be adjacent to form the sub-sequence.)

You also have 8 7 6 5 as a descending sequence, (again with non-adjacent numbers).

So, the sequence I am looking for, mustn't contain ANY four-digit, rising or falling, subsequences, (using adjacent or non-adjacent digits).

Hopefully this helps clarify?

Cheers

Don
Posted on: 01 October 2004 by Don Atkinson
3 8 7 2 4 9 1 6 5

this doesn't qualify either, because there is a descending run 8 7 6 5

Again, hope this helps

Cheers

Don