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: 24 March 2002 by bam
Does the radius of the larger 4th circle = 6?
Posted on: 26 March 2002 by Don Atkinson
Ladders, goats, damsels and kissing circles. Its getting more like exam time every day!!!!

There are three couples having dinner together before the concert. They have been discussing whether the concert will sound as good as their Naim hi-fi systems.

Mick is older than Michelle's husband. Vic's wife is older than Nadia and Lynette's husband is older than Mick.

Michelle is not Zoran's wife. The oldest man is married to the youngest woman.The oldest woman paid the bill. Who was this?

All characters are entirely ficticious and any similarity with persons on this forum either living or dead is improbably intentional....

Cheers

Don

Amended to remove the contadiction noted by Paul R. Ooops!

[This message was edited by Don Atkinson on TUESDAY 26 March 2002 at 23:19.]

Posted on: 26 March 2002 by Paul Ranson
quote:
Mick is older than Michelle's husband. Vuk's girlfriend is older than Nadia and Lynette's husband is older than Mick.

etc

Minor contradiction here? Implication of both marriage and girlfriend. A very open relationship...

Paul

Posted on: 27 March 2002 by bam
Well, what a tangled love triangle this may well be. Now let's see if logic can be applied to male-female relationships.

MICK--->older than---> Michelle's husband
Vic's wife ---> older than ---> NADIA
Lynette's husband -->older than-->MICK

This leaves ZORAN, MICHELLE, VIC and LYNETTE
Michelle's husband is not Zoran (the problem says so) so VIC must be Michelle's husband. Since Lynette's husband is older than Mick and Mick is older than Vic, Zoran must be Lynette's husband. And since the youngest woman is married to the oldest man, Lynette must be the youngest.

ZORAN -->older than--> MICK -->older than--> VIC
MICHELLE -->older than--> NADIA -->older than--> Lynette

Michelle, being the oldest woman, paid the bill.

BAM

Posted on: 27 March 2002 by Paul Ranson
That's what I thought.

Who's 'Vic'?

Paul

Posted on: 27 March 2002 by bam
Who's Vuk? wink
Posted on: 27 March 2002 by Don Atkinson
Oh! BTW, there is a standard way and an elegant way to solve this little problem. The normal way will do for most, but in the case of Bam………..

Now I have a hunch that Bam has remembered the famous formula (if not the rhyming ditty that goes with it)for solving four circles. (We shall come on to five balls later! and 'n' dimensional problems after that)

Does the radius of the smaller 4th circle = 6/23? - Yes

Does the radius of the larger 4th circle = 6? - Yes

Now don't go despairing just 'cos Bam can turn out the answers. The 1,2,3 circles are not that difficult to sort out. cool

I'll post the answers after Easter, just to give all those on holiday a chance (my name is not Zoran, so we won't be having a poll etc etc)

Cheers

Don

Posted on: 27 March 2002 by Don Atkinson
Who's 'Vic'? Who's Vuk?

This is the REAL Who's Who. Not that pretentious little booklet that sits on library shelves around the UK and which is clearly very parochial.

Vic is Vuk's friend, who's sister is Michelle. Vuk's girfriend is also called Michelle and Michelle is also a friend of Vic. Vic is married to Vuk's sister, Michelle.

Perhaps you can now sympathise with my little mistake. Or perhaps you can see straight through the unholy blunder and carelessness and this pathetic attempt at a cover up?

Mustn't confuse people on this forum with characters in the puzzles........

Cheers

Don

Posted on: 28 March 2002 by Matthew T
Don,

Using my probablility method these are the results that I generated
1 0.282896825396825
2 0.365793650793651
3 0.398690476190476
4 0.398253968253968
5 0.372817460317460
6 0.327380952380952
7 0.265277777777778
8 0.188888888888889
9 0.100000000000000

For further consideration. The daughters are not all so beautiful, in fact you know that 3 are not at all pretty and you wouldn't marry at any cost whereas one of the daughters is absolutely beautiful and the other seven, though not as beautiful, certainly inspire devotion. What is your best course of action (any of the daughters could have the highest dowry)!

Matthew

Posted on: 29 March 2002 by Don Atkinson
The Story
During the recent Bristol Hifi Show, five forum members were sat next to each other in the front row. They were Zoro, Steve, Vic, Joe, and Mick. Their jobs, in no particular order, are safety advisor, pilot, comic, construction worker and artist. They had travelled to the show from homes in Wrexham, Reading, Yorkshire, Scotland and Canada. In the general chin-wag after the 15.30 forum special demo, holidays were mentioned with Blackpool, the Canaries, Poland, Italy and Hawaii featuring this year. After the show, one was planning to listen to some new vinyl on his LP12, one was going to the dogs, one to listen to some new CDs on his CDSii, one to listen to his new NAP500 and the other was going to spend a quiet night in doing forum brainteasers.

The Clues
Mick was sat on the extreme left
The guy who enjoyed dog racing was sat in the middle
Zoro is a safety adviser
Steve is from Wrexham
Vic enjoys his cdsii
Joe is planning a holiday in Hawaii
The pilot loves vinyl especially on his LP12
The artist is off to the Canaries for inspiration
The Yorkshire man is off to Blackpool for his hols
After listening to his NAP500 our happy member if off to Italy
The pilot sat next to the comic
The guy off to Poland sat next to the guy from Canada
The guy from Reading sat next to the guy off to the Canaries
Mick was sat next to the construction worker

The Questions
Where does Joe live?
What does the brainteaser do for a living?

Cheers

Don

Posted on: 29 March 2002 by Don Atkinson
Nice work, I haven't checked but I presume the probs are the same as Paul R's who has explained why his computer figures are precise!

For further consideration. The daughters are not all so beautiful, in fact you know that 3 are not at all pretty and you wouldn't marry at any cost whereas one of the daughters is absolutely beautiful and the other seven, though not as beautiful, certainly inspire devotion. What is your best course of action (any of the daughters could have the highest dowry)!

Stick with the original plan, if you get the dowry but she isn't the most beautiful go for the divorce (taking half the dowry) then go for the good looking one. OTOH, if Anna is reading this then you can only go where the heart leads......

Cheers

Don

Posted on: 07 April 2002 by Paul Ranson
with the lead piping in the library?

Paul

Posted on: 08 April 2002 by bam
Here's the official solution (who's source I've disguised) to the dowry puzzle.
BAM
Posted on: 08 April 2002 by bam
Surely someone has a single, simple PC application that I can download from somewhere that will enable me to take a bitmap image, change its resolution and then save it as a jpeg. Is this too much to ask?

I have an official solution that is short and on one page of an MS Word document. The .doc file is 26kB. I cannot upload it directly to the forum. So I copy it to MS Paint which makes it into a 1.7MB bitmap. I then save this and upload it into Adobe PhotoDeluxe (which came free with my PC) and then export it as a jpeg. The jpeg is 136kB.

Surely this is a puzzle worthy of this thead: how can I convert a 26kB .doc file into a .jpeg of less than 30kB?

I realize I may well be stretching the bounds of technology beyond what is reasonably achievable in the 21st century. mad mad mad

The brief answer is that the number of daughters to be skipped is 0,1,1,2,2,2,3,3,3,4,4,5,5,5... for n=1,2,3... daughters. For n=10 you should skip the first 4 daughters and then choose the first with a dowry higher than the first 4.

Posted on: 08 April 2002 by Andrew L. Weekes
quote:
Surely someone...

My first opportunity to contribute something to this thread that's within my spare mental capacity wink

I have what I believe you need - take a peek here: -

http://www.serif.com/photoplus5/index.asp

and download the fully featured, no nags software that has a very easy to use image exporter, that will hopefully do just what you want.

I'm currently using v7, but 5 has the same feature AFAIK, as an image editing program I prefer Photoshop / Photopaint, but as an exporter of images it's a good freebie.

Andy.

Posted on: 08 April 2002 by bam
Andy,
You probably deserve a bottle of Asti Spumanti for that. wink
BAM
Posted on: 08 April 2002 by bam
Omer,
I'm still single so I'm hardly the right person to ask about successful marriage strategies.

The solution comes from a reputable source of well-known puzzles in mathematical circles so I assume there is a good reason why this particular strategy is considered best. But I cannot prove it.

BAM

Posted on: 09 April 2002 by Don Atkinson
Just before Easter, I set a puzzle involving four circles and said I would post the answer after Easter. Here it is.

A maximum of four circles can be made to 'just' touch one another

Curvature (c) is the reciprocal of radius (r), hence c=1/r
Curvature is positive when circles touch externally
Curvature is negative when circles touch internally
Four circles A,B,C and D have curvatures a,b,c and d respectively

2(a^2 + b^2 + c^2 + d^2) = (a + b + c + d)^2

If the radii corresponding to a,b,c are 1,2,3 then the radius corresponding to d will be either 6/23 or 6. (as Bam correctly answered!)

Take care with the arithmetic......eg b^2 = 1/4.....it does work. Let me know if you have difficulty!

Cheers

Don

Posted on: 09 April 2002 by Don Atkinson
Three circles radii 2,3,10 just touch. What are the sizes of the fourth circles that just touch the first three?

Trivial, I know, (well it wasn't trivial last week, but it is now!!) but at least it will help check out that new calculator!!

Cheers

Don

[This message was edited by Don Atkinson on TUESDAY 09 April 2002 at 23:20.]

Posted on: 10 April 2002 by steved
LADDER PUZZLE

I've been on holiday for the past couple of weeks.

What is the situation regarding the ladder puzzle?

I've admitted defeat, but would like to see the solution.

Have I missed it? Could I have some idea when all will be revealed?

STEVE D

Posted on: 11 April 2002 by Don Atkinson
Steved says

I've admitted defeat, but would like to see the solution.

Have I missed it?

Steved, if you missed it, then I guess I did too. Oh dear!!!!!!!!

Cheers

Don

Posted on: 11 April 2002 by Paul Ranson
BAM now has the technology!

Can't wait.

Paul

Posted on: 11 April 2002 by Don Atkinson
BAM now has the technology!

I think you'll find Bam's technology is limited to squeezing a mere 256kb of Word data into a miserable 30kb of Naim forum GIF data.

Hardly adequate to do justice to this enormous ladder problem we have!!

Cheers

Don

Posted on: 12 April 2002 by Matthew T
No elegant and exact solution to the ladder problem yet (he says with hope!).

Matthew

Posted on: 13 April 2002 by Don Atkinson
Young Paul Darwin has been driving the Naim Delivery Vehicle since 7am.

Naim have taken my advice. They have a new warehouse stacked with Supercaps, XPSs, Hicaps, NAP500s etc etc all powered up, run-in and ready for delivery. Your order (via your local demonstration consultant.......hifi shop to you and me!) is personally delivered and set up by a Naim Installation Team direct from Salisbury. They use fully equipped crew vans (or aeroplanes if you live across the pond!)and your equipment is kept fully powered up in transit.

Paul and the Naim Installation Team (no acronyms allowed!) will have their morning snack in twenty minutes time when Paul's driving time will be three times the time from now until 10am.

What time is it now?

Cheers

Don