OT - Maths question

London_Falcon

that solution is a thing of beauty, very nice

dhope

London_Falcon wrote:

that solution is a thing of beauty, very nice

Doesn't feel right though
One sec, I'll try and justify why

Edit: Apparently it is, fair play

SimonAH

Your answer is correct secret dancer - that would have been the don's equation. The beautiful thing here is that the carpetfitter, knowing the existence of the equation, does not need to know what the equation is to provide his answer to the lighthouse keeper.

kamiokande

It's a good maths puzzle, but, in reality, a carpet fitter would just tack on 50% to the chord length and get a square that big (15m x 15m) and then cut the ring out to fit it. If he follows that solution it works out at pi x 25 = ~ 80, so lets say he turns up with a piece of carpet 8m x 10m it's going to require an awful lot of cutting to get it to fit.

secretdancer

dhope wrote:

Doesn't feel right though
One sec, I'll try and justify why

But I showed my workings and everything

dhope

kamiokande wrote:

It's a good maths puzzle, but, in reality

:roll:

TheStone

secretdancer wrote:

A = π5^2

Almost makes me miss maths.

Aidy

Greg66 wrote:

London_Falcon wrote:

For you mathematicians out there:-

Given x = 1 and y = 1, then x = y
Multiplying each side by x gives x2 = xy
Subtracting y2 from each side gives x2 - y2 = xy - y2
Factoring each side gives (x + y)(x - y) = y(x - y)
Dividing out the common term, (x - y) results in x + y = y
Substituting the values of x and y, 1 + 1 = 1 or 2 = 1

Can anyone spot how this (doesn't) work??

Strictly, the problem arises in the emboldened line. At that point you introduce terms which mean that the equation becomes 0=0 (as y^2 = x^2). After that, you can multiply any number by zero to get zero (eg 54*0= 45*0) but you cannot divide both sides by zero (eg to get 54=45).

No, that's okay. 0 does equal 0.

The problem is trying to divide through by 0.
If you don't start from x = 1 and y = 1, then it's all okay until attempting to divide through by x-y.

In normal algebra, you'd put a disclaimer at this point of "for x != y".

SimonAH

OK, the carpetfitter's solution works by saying that if there is a formula it be a function of X and Y obviously.

So simply tend y to zero (ie have no central pillar). At this point your chord is your diameter.

Elegant aint it!

Wallace1492

dhope wrote:

walkingbootweather wrote:

I'd start by splitting 12 marbles into 3 groups of 4.
I'd weigh group 1 and group 2. If it balances then the irregular marble will be in group 3
... I'll let you do the rest

Wallace, Rick... WBW has the right idea.

Think about what you're not weighing too (with 8 marbles or 12)

Can get the solution for the irregular marble in 12 with 3 weighings, but cant quite get it for the irregular one in 8, unless you know if it is heavier or lighter.

dhope

Wallace1492 wrote:

dhope wrote:

walkingbootweather wrote:

I'd start by splitting 12 marbles into 3 groups of 4.
I'd weigh group 1 and group 2. If it balances then the irregular marble will be in group 3
... I'll let you do the rest

Wallace, Rick... WBW has the right idea.

Think about what you're not weighing too (with 8 marbles or 12)

Can get the solution for the irregular marble in 12 with 3 weighings, but cant quite get it for the irregular one in 8, unless you know if it is heavier or lighter.

With 8 then start by putting 3 on each side and consider the *3* possible outcomes

keyser__soze

Only works in max of two if you know for definite that the irregular marble is either heavier or lighter than the rest. If all you know is that it's different then you can't definitely solve in a max of two.

Wallace1492

dhope wrote:

Wallace1492 wrote:

dhope wrote:

walkingbootweather wrote:

I'd start by splitting 12 marbles into 3 groups of 4.
I'd weigh group 1 and group 2. If it balances then the irregular marble will be in group 3
... I'll let you do the rest

Wallace, Rick... WBW has the right idea.

Think about what you're not weighing too (with 8 marbles or 12)

Can get the solution for the irregular marble in 12 with 3 weighings, but cant quite get it for the irregular one in 8, unless you know if it is heavier or lighter.

With 8 then start by putting 3 on each side and consider the *3* possible outcomes

3v3 - they weigh the same, the odd one is in the other 2, weigh one of the 6 v one of the remaining 2, if they are the same odd one is the last unweighed one, but you dont know if it is heavier or lighter.

3v3 - unbalanced. You do not know if odd is on lighter or heavier side, so you cannot determine in 1 more weighing.

dhope

Keyser__Soze wrote:

Only works in max of two if you know for definite that the irregular marble is either heavier or lighter than the rest. If all you know is that it's different then you can't definitely solve in a max of two.

Fair point. Fractionally heavier then

Sorry Wallace, misread your post at first.

London_Falcon

Wallace1492 wrote:

dhope wrote:

Wallace1492 wrote:

dhope wrote:

walkingbootweather wrote:

I'd start by splitting 12 marbles into 3 groups of 4.
I'd weigh group 1 and group 2. If it balances then the irregular marble will be in group 3
... I'll let you do the rest

Wallace, Rick... WBW has the right idea.

Think about what you're not weighing too (with 8 marbles or 12)

Can get the solution for the irregular marble in 12 with 3 weighings, but cant quite get it for the irregular one in 8, unless you know if it is heavier or lighter.

With 8 then start by putting 3 on each side and consider the *3* possible outcomes

3v3 - they weigh the same, the odd one is in the other 2, weigh one of the 6 v one of the remaining 2, if they are the same odd one is the last unweighed one, but you dont know if it is heavier or lighter.

3v3 - unbalanced. You do not know if odd is on lighter or heavier side, so you cannot determine in 1 more weighing.

8 marble puzzle in 2 we have to know whether is is heavier or lighter to begin with, it cannot be either

Wallace1492

London_Falcon wrote:

Wallace1492 wrote:

dhope wrote:

Wallace1492 wrote:

dhope wrote:

walkingbootweather wrote:

I'd start by splitting 12 marbles into 3 groups of 4.
I'd weigh group 1 and group 2. If it balances then the irregular marble will be in group 3
... I'll let you do the rest

Wallace, Rick... WBW has the right idea.

Think about what you're not weighing too (with 8 marbles or 12)

Can get the solution for the irregular marble in 12 with 3 weighings, but cant quite get it for the irregular one in 8, unless you know if it is heavier or lighter.

With 8 then start by putting 3 on each side and consider the *3* possible outcomes

3v3 - they weigh the same, the odd one is in the other 2, weigh one of the 6 v one of the remaining 2, if they are the same odd one is the last unweighed one, but you dont know if it is heavier or lighter.

3v3 - unbalanced. You do not know if odd is on lighter or heavier side, so you cannot determine in 1 more weighing.

8 marble puzzle in 2 we have to know whether is is heavier or lighter to begin with, it cannot be either

Quite. Think I may have advised this in the post following the question!

Wallace1492

dhope wrote:

Keyser__Soze wrote:

Only works in max of two if you know for definite that the irregular marble is either heavier or lighter than the rest. If all you know is that it's different then you can't definitely solve in a max of two.

Fair point. Fractionally heavier then

Sorry Wallace, misread your post at first.

And at second and third...... :P

However, it is a good puzzle. People tend to do 4v4, then 2v2 and cant contenplate stepping outside the box.

rolf_f

dhope wrote:

kamiokande wrote:

It's a good maths puzzle, but, in reality

:roll:

That's actually not fair - kamiokande is sort of right though not elegantly done.

In reality, you use Secretdancers method to solve for X and the answer is that you order squares of carpet X metres by X metres multiplied by the number of 360 degree rotations of the staircase. If the staircase a partial rotation then a smaller section might be needed (or an offcut from the centre).

Otherwise you'll just get a load of ugly joins everywhere!

walkingbootweather

Learning that my six year old had been doing some maths at school I asked him "what is 5 oranges plus 3 oranges?"

"Dunno", says he. "The teacher only taught us how to do sums with apples".....

SimonAH

If.you have six.apples in one hand and eight apples in the other hand what have you got?

"very big hands Sir"

rjsterry

Rolf F wrote:

dhope wrote:

kamiokande wrote:

It's a good maths puzzle, but, in reality

:roll:

That's actually not fair - kamiokande is sort of right though not elegantly done.

In reality, you use Secretdancers method to solve for X and the answer is that you order squares of carpet X metres by X metres multiplied by the number of 360 degree rotations of the staircase. If the staircase a partial rotation then a smaller section might be needed (or an offcut from the centre).

Otherwise you'll just get a load of ugly joins everywhere!

Well actually, carpet comes on rolls of a standard width - 12', 13'6", 15' - and you order by length, rather than area, so you'd need to work out the most efficient way of fitting the squares onto a standard roll as well.