Brain Teaser No 1

Don,

you proved it very precisely:

1) x^2 + y^2 > 2xy
2) x^2 + y^2 - 2xy > 0
3) (x-y)^2 > 0
all squared numbers are positive => QED

Sorry, the term "factor" is a bit confusing for me. Our languages differ quite a bit indeed...

Cheers

Christian

Christian, Lynn, and any others out there !!!

Sorry, the term "factor" is a bit confusing for me. Our languages differ quite a bit indeed...

gulp ... and thanks for being easy-going.

Now for the Interesting bit !

Any ideas why I titled this teaser The Law of Averages ?

Nothing to do with misunderstanding, purely mathematical....

Cheers

Don

Don,

Your proof actually proves that all values of x and y (where they are real numbers) will give x^2 + y^2 > 2xy except for the case where x=y which gives x^2 + y^2 = 2xy.

cheers

Matthew

[This message was edited by Matthew T on MONDAY 20 January 2003 at 09:16.]

It appears that Don is.

As to the title:
Don, I´m afraid I don´t...

As to the language:
Lynn, would you please spell "DOH" and "OMG" for me?

Christian

NOW, can you figure-out whether the audio store owner to which I referred should be in jail?

Lynn, I have this feeling that you're going to tell us EACH tube was only worth $450 and the store owner claimed he had committed TWO offences.........neither of which carried the jail sentance...

As to whether I think he should be in jail....well that's another matter....

OTOH, Mick Parry could well be eyeing up his armory above the fireplace....

Cheers

Don

quote:

---

Originally posted by hlhopffgarten:

...what should happen to the owner of the store?

---

If a thief stole $900 cash but replaced it with, say, $500 cash it would be easy to say that s/he had actually stolen only $400. However, in this case if the store owner stole a $900 pair of tubes then the fact that s/he replaced them with a presumably inferior pair is irrelevant: it's a >$500 theft, so s/he should go to jail. But what if the original tubes failed, and the store owner simply replaced them? Clearly, that would not be a crime.

--J

Lynn, Don

Too sharp for the head of the tribe & his goats. Perhaps the surprise these days is, that someone didn't contest the will or sue the wiseman! Mind you, how far does 1/18th of a goat go?

In a game of poker, one of the hands of five cards had the following features: * There was no card above a 10.
~No two cards were of the same value.
~All four suits were represented.
~The total values of the odd and even cards were equal.
~No three cards were in sequence.
~The black cards totaled 10 in value.
~The hearts totaled 14.
~The lowest card was a spade. What was the hand?

D-4
H-9
H-5
C-8
s-2

?

I agree with Dobbin,

Nice puzzle!

Steve D

Lynn,

Going back to the 300Bs

I'm not sure that we really have enough facts to hang the store owner.....yet

We don't know who changed the tubes, only that they never left the store.

We don't know why they were loose.

We don't really know whether the replacement tubes were a better (lets say more expensive) set than those origionally supplied by the manufacturer.

We don't know whether the origional tubes failed during the 2 to 3 weeks of continuous use, ie there was a decent reason for the swap.

Too many unknowns at this stage for a safe conviction.

Any more relavent facts ?? we don't want too many.....that might cloud the issue or even lead to an aquittal......

Cheers

Don

The orange man is back!

This time the orange seller has two differnet size pyramids of oranges (regular tetrahedrons) and he claims that he made them from one pyramid with out any left over oranges.

How many oranges has he got?

Matthew

quote:

---

Originally posted by hlhopffgarten:
Matthew:
i think we have enough oranges for about 4 larges glasses of juice: 36.

a pyramid of 5 and 6 levels from his original 8 level regular tetrahedron. I can't "prove" there aren't other solutions ....

---

Now, a pyramid of 5 levels has 35 oranges...

Back to the drawing board.

Matthew

You grocer is telling the truth but I suspect he would be struggling to find enough oranges in his store to do it.

In fact it would be possible to take one layer off his pyramid and make a new one, he would need more oranges for this though!

Matthew

Oranges,

Been away for a few days, still busy but will start thinking about piles of oranges.

Initial thoughts are hovering around a (mis)conception that in a regular tetrahedron the number of oranges in the Nth layer is 0.5N(N+1)....

Don't expect too much effort, too soon.....I haven't listened to any music for a few days either.

Cheers

Don

Matthew, Lynn, Oranges

Just seeking clarification for the moment.

Are we talking, for example about a pile of oranges, say, 15 layers high to start with, containing about 680 oranges. Then dividing this heap into two heaps, one comprising say, 14 layers with about 560 oranges in it and the other heap having say, 8 layers with about 120 oranges in it ?

Would this sort of idea satisfy the tetrahedron problem ?

I haven't looked at the pyramid one yet.

I'm away for the rest of this weekend, back Sunday night.

Hope i'm working along the right lines......I fancy a few grapes before bed..

Cheers

Don

Well Don's numbers work, but are there any
other solutions?

If the total number of oranges in a tetrahedon of base n is f(n) = n(n+1)(n+2)/6 then we're looking for n,o,p such that
f(n) = f(o) + f(p), n>o>=p>0.

Is n=15,o=14 and p=8 the only solution?

-Dan

Oranges,

Is n=15,o=14 and p=8 the only solution?

Not according to my preliminary calculations.

I tried looking for 'whole number solutions' based on the number of oranges in a tetrahedron being n(n+1)(n+2)/6 along the lines suggested by Dan. But I decided that in the short term a simple spreadsheet would generate a few answers with n<1,000. I developed the spreadsheet with just n<100 and up popped a few whole number solutions, with 8;14;15 being the lowest. Of course, 3;3;4 also works, but Matthew wanted DIFFERENT sized piles.

Cheers

Don

Soldiers, and sailors and ORANGES

What, might you ask, do these have in common, Well, in doing some reading this evening, I found the following:-

"Such problems had more than a mere curiosity value to soldiers. As late as the eighteenth century in Europe, manuals of mathematics showed how the number of cannon balls in a pile could be calculated from the length of one side, depending on whether they were piled in a triangular or a square pyramid. Armed with these formulae, the spy with a telescope could estimate the amount of ammunition the enemy had available"

Unfortunately, it doesn't go on to say how many little piles could be made out of one big pile.....

Cheers

Don

Back to oranges

For initial triangular base piles, up to 100 layers high, I have only found the following combinations that work :-

15 = 14/8
55 = 54/20
58 = 55/30
74 = 70/39

And 4 = 3/3 seems to be the only one that divides into two equal piles!!.

Unless somebody knows better!!.?

Cheers

Don

BTW it should be easy to check the validity of answers, now that that the formulae are available ie n(n+1)/2 and n(n+1)(n+2)/6.....PROVIDING you believe the formulae.

BTW (again) I didn't use either of these formulae in my spreadsheet

You are right on the tetrahedron pyramid with 15 levels is the lowest possible solution, there are others with more oranges.

I have a solution for the square based pyramid but I will allow you a little longer to find an answer.

cheers

Matthew

Oranges and pyramids

Hi guys, especially Matthew who was giving us all achance to catch up !!!!

So, a quick flick of the wrist on the old Dell spreadsheet and the tetrahedron was transformed into a pyramid.

A quick run through the first stack of oranges, up to 100 layers high reveals two results.

Now my recollection is that we were looking for the case where only the bottom layer is removed to form the third pyramid.

There is only one solution in the range 0<n<100 that satisfies this criteria. If n= the height of the initial pyramid, the two resulting pyramids are n-1 and n-46

The other result forms two pyramids with heights n-10 and n-13. Of course this doesn't satisfy the set criteria.

BTW "n" is different for the two results.

Now I might be wrong, I hope not, but I'm sure you'll let me know if I am !!!!

And in the "one layer removed" solution he needs 116,795 oranges to perform the trick, in the other solution he only need 56,980 oranges

Cheers

Don

Lynn

Why, therefore, do I doubt his word?>>

So you would like to know how high a 70 layer stack of spheres each 3" diameter is ?

Its past my bedtime tonight ! Tomorrow.

Unless someone gets there before me.

Of course, its somewhat less than 70/4=17.5 feet

Cheers

Don

I think your grocer would find getting 69 layers of oranges onto his table a little difficult, now 68 could be managed and he would about half an inch to spare whilst rolling the top orange on.

Matthew