Brain Teaser No 1

Lots and lots
Countable ways.
I could count them, but it might take some time.
which I thought were not countable
I think you're right
that's what I wanted to write in the first place

So having danced a few times around the fire, the answer is.....

Cheers

Don

Uncountable.

Paul

I agree, there are an uncountably infinite number of ways.

Dan

quote:

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Originally posted by d marsh:
I agree, there are an uncountably infinite number of ways.

Dan

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Uncountably infinite or unaccountably infinte, it's all hyperbole to me.

An infinite number of ways.

cheers

Matthew

Matthew, it's all hyperbole to me.

Soory Matthew, no hyperbolas involved this time....!!

OK, you've all got the right idea. The essential elements are

The two straight lines must be at right angles
The two lines must be centred on the cross
Any angle will do provided the first two conditions are met
There are an infinite number of angles and hence an infinite number of solutions

Happy New Year and glad to see the old brains are getting back up to speed....

Cheers

Don

it's all hyperbole to me.

Sorry Matthew, no hyperbolas in this one.....!!

You've all got the right idea, Paul perhaps slightly more right. The essential elements are..

The two lines must cross at right angles
The two lines must be centred on the cross
Any angle will do
There are an infinite number of angles (even in 90 deg)
hence there are an infinite number of solutions

Well done guys, Happy New Year. Glad to see see the old brain boxes gradually comming back to life....

Cheers

Don

Sorry for the double post above. Thought I'd lost it so typed a second copy in, then found both versions appeared.

I didn't want to delete in case they both go. this new forum style is taking a bit of getting used to....old dogs.....new tricks....

Cheers

Don

A boxed set

I have three circular medals that I keep in a rectangular box. The smallest (radius 40mm) touches one side of the box. The mid-sized one (radius 90mm) touches two sides of the box. The largest touches three sides of the box Each medal touches both the others.

What is the radius of the largest medal ?

Cheers

Don

120mm

Matthew

Matthew,

The perfect answer.

Cheers

Don

Matthew,

I didn't want to spoil the above response with the usual....."but....."

Would you care to enlighten the other 2,000 members who might glance this way as to how they could deduce this perfect answer ?

I am not going to claim any "elegant" solution to this one, but I think I read this one in the Sunday Times some years back and I probably never saw their elegant answer (since I only ever buy the paper once every two years, (approximately)

Of course, others could publish their solutions if they wish....they will probably all be more elegant than mine!!

Cheers

Don

There is an elegant way of find the radius of the third coin in that the radius of all three and joined by the following equation.

radius of smaller two medals A and B
radius of larger medal C

(A x B) ^ 0.5 = 0.5 x C

Will let somebody else give else a chance to come up with a elegant solution before I post my solution which I would suggest is not that elegant (until you get to the equation above).

cheers

Matthew

OK,

Here is the method I used. Hopefully it is somewhat clear, every thing is based on triangles.

Matthew

Matthew

Hopefully it is somewhat clear,

Perfectly clear and very elegant.

I'm afraid on my first attempt, I introduced the known radii straight away. Gave a straightforward solution (120mm) but failed to yield the elegant C = 2(AB)^0.5 bit at the end. That bit only came some months later when I was daydreaming!

Neat

Cheers

Don

Matthew,
Nicely done! I had similar doodles, but didnt
follow it through as you have done,
cheers
Dan

The Law of Averages

X and Y are any two different positive numbers. Which is larger, (x^2 + y^2) or 2xy.

Cheers

Don

Don,
I'm not sure, what "y^2" means; but if it means "y*y" the correct answer should be:
(x*x+y*y)>2xy
Cheers

Christian

Hi Christian,

I'm not sure, what "y^2" means

Your assumption, y*y or (y squared) or (y to the power of 2) is absolutely correct.

And, of course, your answer is also absolutely correct.

Now the next part is the proof.....any takers?

Cheers

Don

Lynn,

factor the equation => x + y > 2

You've got me beat on this one........

As Cilla Black (one of our lovely singers from Liverpool) would say, "gissa clue" meaning 'would you mind awfully providing some small suggestion of a tiny hint'......

Perhaps, tonight, i'm being a bit thick....

Cheers

Don

Hi,

try this one:
Your wife sends you to the market with 100 pounds in your pocket. She wants you to pick exactly 100 animals spending exactly your 100 pounds. Rabbits, ducks and chicken are available. Prices as follows:
9 pounds per rabbit
4,50 pounds per duck
0,50 pounds per chicken.
How do you solve this "problem"? (No higher mathematics needed)
Regards

Christian

Lynn,

then by induction ........

Got it.......eventually!!!! Page 29 of my book by Robert Ainsley puts it in a nutshell :-

Proof by Induction
"A very important and powerful mathematical tool, because it works by assuming something is true and then goes on to proove that therefore it is true. Not suprisingly you can prove almost anything by induction so long as the proof includes three phrases like" :-

"Assume true for n; then also true for n+1 because......(write any semi-plausible but messy looking working out....)"

"But is true for n=0......(write a bit more semi-plausible but even more messy working out....)"

"So is true for all n. QED"

Well done Lynn.....BUT, since you invite alternative approaches, personnaly I prefer Proof by Assumption

Cheers

Dopn

Well done!
Lynn,
a messy kitchen indeed :-)

Don,
I have to agree with Lynn on the other problem.

Kind regards
Christian

Goats Anyone?

The head of the tribe is dying and he ask his wife to call together his three sons to tell them how he wants them to divide up his goat herd when he is gone. To his eldest son he leaves 1/2 the herd, to his middle son 1/3 of the goats and to the youngest 1/9. And with that he pops his clogs.

The sons go out into the field and could the herd and find 17 goats. Not wanting to kill any goats if they could help it, after much head scratching they go see to the village wise man. He lends them one of his goats to make 18. The eldest son then takes his 9 goats, the middle son his 6 and the youngest son his 2, leaving 1 goat which the wise man reclaims.

And everyone is happy ….. or are they ?

Clive

The head of the tribe is dying.......to his eldest son he leaves 1/2 the herd, to his middle son 1/3 of the goats and to the youngest 1/9. And with that he pops his clogs.

Now, as Lynn suggests above, he either wasn't very good at arithmetic and hence didn't realise he'd only accounted for 17/18ths of his herd, or he poped his clogs before he'd made his wishes known about the last 1/18th.

Either way, it seems his family was no better at arithmetic than he was and were quite happy with the wise-man's solution. The wise-man was probably a lot wiser than first impression suggests - he was probably something of a diplomat as well.

Let them continue in blisful ignorance like the rest of us....

Cheers

Don

The Law of Averages

Embarrasment and misunderstanding

Hi guys,

Could it be something to do with "two nations divided by a common language" or is it just me being completely thick. My money is on the latter.

I presume we are all agreed that (x*x + y*y) doesn't have any factors?

I presume that we are all agreed that we can make the proposition that (x*x + y*y) > 2xy and that this is what then has to be proven. Making the proposition is justified, because trial and error, with a few small numbers, suggests this looks promising.

If the proposition is true, then (x*x + y*y) - 2xy > 0

and consequently (x-y)(x-y) > 0 (this step might take a bit of thought)

or (x -y)^2 > 0

Whatever positive numbers are substituted (even if one of them happens to be zero) , this last expression, being a square, has got to be a positive number and hence > 0 QED

But I still can't see how x+y > 2 is a factor of (x*x + y*y) > 2xy !!!!!

or is it me just being completely thick ?? Polite replies only........please....

Cheers

Don