Math Problems

[Fat middle]

Originally Posted by Mercutio

Now, theoretically I should be able to solve two equations simaltaneously, and then two others, and then those two answers…yadda yadda yadda… and get (x, y) of the centroid in terms of a, b, and c. But it ain’t happen’ baby.Did I miss something along the way?

As Janny says, the link doesn't work.

Anyway, if I've understand it correctly, as you say, you must solve two equations (RT and QS), then you solve two others (RT and PU) and then the other pair (QS and PU) to be sure that the three lines have the same common point.

All those three systems of two equations have the same solution: x = (a-c)/3; y = b/3 q.e.d.

Another, more correct IMHO, way for proving this would be to prove that the range of the matrix:

-b       (a + 2c)       bc
-2b      (c - a)      (c-a)b
-b       (-c - 2a)       0

is equal to 2. Have you studied matrixes?

[Mercutio]

I think I fixed the link

yes we've done matrices.

hmm... I think I'm a bit lost now-- How did you get the solution?

[Fat middle]

Originally Posted by Mercutio

I think I fixed the link

yes we've done matrices.

hmm... I think I'm a bit lost now-- How did you get the solution?

I just solved the equations. Since they're linear equations I've made it by the Gauss method, but you can solve it by other methods. What's exactly your problem solving them?

Anyway, since you have done matrices, I'll try to explain the way I'd prove the question.

I put above the matrix that represent the system of three equations. First column is for the coefficients of the "x", second column for the "y" coefficients, and third for the independant term (I'm sorry if my terminology sounds bizarre I'm "translating" from Spanish math terms). The theorem of Rouche-Frobenius says that the system will have an unique solution only if the number of ( I don't know how do you say in English "incógnitas" I mean, the X and the Y. enigma? ) incógnitas is the same as the range of the matrix of coefficients and the complete matrix of the system (what I put in the previous post). If we may prove that the determinant of the complete matrix is equal to zero, then the range of the matrix cannot be three. If then we find a submatrix 2*2 with range two (determinat non equal to zero) then we have proved that there's only one solution for the three equations, and so only one point in common for the three lines.

Sorry, just revised my calcs and seen I slipped an error. This is the matrix:

-b       (a + 2c)       bc
-2b      (c - a)      (c-a)b
-b       (-c - 2a)     -ab

The easier way to prove that the determinant is equal to zero is adding first line to third line, then you have:

-b       (a + 2c)       bc
-2b      (c - a)      (c-a)b
-2b      (c - a)      (c-a)b

Since this matrix have two identical lines, its determinat has to be equal to zero. The rest is easy.

I fear I'm not being too clear. And I've just reread the problem and have seen that no theorems were allowed, so perhaps we cannot use Rouche-Frobenius... Then the way you were solving it is the only one I can think about. Have you revised your calcs? What method are you using?

[Mercutio]

I tried a few different ways, including solving for one variable and plugging that expression into another equation, and putting two equal to each other and trying to simplify. It just got really complicated, and I could never get it very reduced. I think though, I just needed to solve two equations and then another two, and even if it wasn't reduced very far they should both have a new identical equation.

[Fat middle]

Originally Posted by Mercutio

I tried a few different ways, including solving for one variable and plugging that expression into another equation, and putting two equal to each other and trying to simplify. It just got really complicated, and I could never get it very reduced. I think though, I just needed to solve two equations and then another two, and even if it wasn't reduced very far they should both have a new identical equation.

Those methos are valid, but they're not the best in his case. As you say they got too complicated. Anyway, here's an example of how you can solve the variable x with the second method you've mentioned:

Solving RT&QS:

(bx+bc)/(a+2c) = (2bx)/(c-a) + b

(bx+bc)/(a+2c) = (2bx + b(c-a))/(c-a)

(c-a)(bx+bc) = (a+2c)(2bx + b(c-a))

bcx+bcc-abx-abc = (a+2c)(2bx+bc-ab)

bcx+bcc-abx-abc = 2abx+abc-aab+4bcx+2bcc-2abc

-3abx-3bcx = -aab+bcc

-3bx(a+c) = b(cc-aa)

-3x(a+c) = (c+a)(c-a)

-3x = (c-a)

x = -(c-a)/3

x = (a-c)/3

Now, to obtain the y variable I plug that expression into the equation of the RT line:

y = (b(a-c)/3 +bc)/(a+2c)

y = (ab-bc+3bc)/3a+6c

y = (ab+2bc)/3a+6c

y = b(a+2c)/3(a+2c)

y = b/3

hard, doesn't it?

So, IMHO, the best method is Gauss':

Solving RT&QS by Gauss' method:

RT: (a+2c)y = bx+bc     ->     -bx + (a+2c)y = bc

QS: (c-a)y = 2bx + b(c-a) ->  -2bx + (c-a)y = b(c-a)

RT can also be put as:        -2bx + 2(a+2c)y = 2bc

And now, doing QS-RT:            (c-a-2a-4c)y = bc-ab-2bc

(-3a-3c)y = b(-a-c)

y = b(-a-c)/3(-a-c)

y = b/3

and now you must plug that expression into RT or QS to obtain x. Much easier.

[Janny]

Hey Mercutio! You're in a maths team: What's 45 squared? Chop chop!

Or better, do you know the trick which lets you solve multiplication like 45 x 45 or 33 x 37 or 61 x 69 really quickly?

[Mercutio]

Originally Posted by Janny

Hey Mercutio! You're in a maths team: What's 45 squared? Chop chop!

Or better, do you know the trick which lets you solve multiplication like 45 x 45 or 33 x 37 or 61 x 69 really quickly?

hmm?

[Janny]

I'm such a clear writer...

There's a neat trick to do some multiplications in your head - provided that the 'tens' are the same, and the units add up to ten.

[inked]

And the trick would be? ... use a calculator of the electronic variety?

[Janny]

Very wise master Inked.

Ok... the two numbers have the same tens ('t') and their units ('u') add up to 10, ie. u and 10 - u.

Multiply these two numbers t + u and t + 10 - u

.............t................u
.............t................10 - u
---------------------------------------------
.............10t - tu........u(10-u)
t (sq)......tu...............0
---------------------------------------------
t (sq).....10t............u(10-u)

So the answer is 100 x (t squared) + 10 x 10t + u(10-u)
Or: 100 t(t+1) + u(10-u)

That is, you multiply the two unit numbers and stick the answer in the 'tens' and 'units' column. Then multiply the 'tens' of your original numbers by (itself +1) and stick the answer in the 'thousands' and 'hundreds'.

It works because the 'tu's cancel and the 'tens' column will always not have a value from the '10t'.

Clear as mud.

[sun-star]

Is Janny showing off because he went to a lecture?

[Janny]

Nope, the lectures at the Royal Institute (where they hold the televised Christmas Lectures ) were about science. This came up in my maths class.

[Elanor's Angel]

Ok, for all of thoes who are 'Less than brilliant' (like I am on some days), can someone give us a problem we can follow?

[Mercutio]

Originally Posted by Janny

Very wise master Inked.

Ok... the two numbers have the same tens ('t') and their units ('u') add up to 10, ie. u and 10 - u.

Multiply these two numbers t + u and t + 10 - u

.............t................u
.............t................10 - u
---------------------------------------------
.............10t - tu........u(10-u)
t (sq)......tu...............0
---------------------------------------------
t (sq).....10t............u(10-u)

So the answer is 100 x (t squared) + 10 x 10t + u(10-u)
Or: 100 t(t+1) + u(10-u)

That is, you multiply the two unit numbers and stick the answer in the 'tens' and 'units' column. Then multiply the 'tens' of your original numbers by (itself +1) and stick the answer in the 'thousands' and 'hundreds'.

It works because the 'tu's cancel and the 'tens' column will always not have a value from the '10t'.

Clear as mud.

Hmm...I think I prefer to write it out the elementary way! (or cheat my brain and use a handy dandy calculator )

[Beren3000]

This has probably been done before, but I'll post it anyway:

Three friends went to a diner, and decided to split the bill between them. The bill was for $30 so they each paid $10. However, the diner's owner decided to charge them for $25 only since one of them was his friend. So he sent one of his waiters back to the table with these $5. However, the waiter wasn't as honest as he had hoped and took $2 for himself and gave them back the remaining $3, a dollar each. So each had paid $10 in the first place, each got a dollar back so each has now paid $9. 3x9 = 27 add the two dollars the waiter took and that's $29, where did the missing dollar go?

[Count Comfect]

There's a problem with logic. You don't add the two dollars the waiter took, you subtract them to make the $25 that they were charged. There is no reason to add positive money the waiter has to negative money the diners paid. Instead, you add the 27 dollars they paid (25+2 the waiter took) and 3 that they have back to make the 30. That is, 25 went to the owner, 3 to the diners, and 2 to the waiter.

[Beren3000]

Yes!

[Janny]

I have recently purchased a charming little book "A Notebook in Pure Mathematics". In the logs exercises there is a question I can't get right. Since I'm on summer holidays, I'm asking for a little help.

Expand: [ln (1+x)] / (1-x) to the term x^3.

Expanding the log:
[x + (1/2)x^2 + (1/3)x^3 + (1/4)x^4 +...] / (1-x)

Yes, little help. The answer in the book is

x + (1/2)x^2 + (5/6)x^3 {No term independent of x?!}

Thanks

[Beren3000]

It's because your expansion of the log is wrong. It should be expanded as:
x - 1/2x^2 + 1/3x^3 - 1/4x^4 +....

[Janny]

Yes, that is true. But I've done it with the correct expansion and not got it.
Since it's over (1-x) you should expand up to the fourth term?

So:
ln (1+x) = x - 1/2(x^2^) + 1/3(x^3^) - 1/4(x^4^) +...

=> {-x/12 . [3x^3^ - 4x^2^ + 6x - 12] } / (1 - x)
=> {x/12 . [3x^3^ - 4x^2^ + 6x - 12]} / (x - 1)
=> x/12 . [ (3x^2^ - x + 5) - 7 / (x - 1) ]

Which doesn't yield the answer. Is there some playing with logs to be done beforehand? If I were ln (1+x)^(1/(1-x)) I would rearrange it to how the question is phrased though...