Find all real solutions (x,y,z) to the following system of equations.

x + y =  :sqrt: (4z - 1)
y + z = :sqrt:(4x - 1)
x + z = :sqrt:(4y - 1)

I know it's symmetric but I haven't gotten anywhere past squaring everything and adding to get 2(x :^2:  + y:^2: + z:^2:) + 2(xy + yz + xz) = 4(x + y + z) - 3.  From there though I'm stuck.  Any help'd be appreciated!  :-)

First thing I thought of was subtract the first two equations to get rid of y. However that didn't quite work so instead I [hide] let x+y = a and y+z = b. Then to get rid of square roots I took a^2 and b^2. So b^2 - a^2 = 4(x-z). But what does b-a equal?

[/hide]

WarpedKlown1335 wrote:Find all real solutions (x,y,z) to the following system of equations.

x + y = sqrt(4z - 1)
y + z = sqrt(4x - 1)
x + z = sqrt(4y - 1)

I know it's symmetric but I haven't gotten anywhere past squaring everything and adding to get 2(x^2 + y^2 + z^2) + 2(xy + yz + xz) = 4(x + y + z) - 3.  From there though I'm stuck.  Any help'd be appreciated!

[hide] square the first two => x:^2:+y:^2:+2xy=4z-1, y:^2:+z:^2:+2yx=4x-1, subtract => x:^2:-z:^2:=4(z-x)=> either x=z or x+z=4. We can do a similar analysis on other pairs of equations, so now we have either x=y=z OR x+y=4, x+z=4, y+z=4 => x=y=z=2 OR x=y, x+z=y+z=4 (w/ permutations among the variables in this case). The last case (the interesting one) gives us 16=4x-1, and 4x:^2:=4z-1 => x=y=17/4, z=293/16 (and permutations). The second case doesn't work. Period. The first case gives us 4x:^2:=4x-1=> x=y=z=1/2. So, I get all solutions as (17/4, 17/4, 293/16), (17/4, 293/16, 17/4), (293/16, 17/4, 17/4), (1/2, 1/2, 1/2). Correct?[/hide]

Ummm...(y+z) :^2:  isn't y:^2: + 2xy + z:^2:...sorry :( .  And the only solution I got when I solved it on my own time (didn't use your method) was (.5,.5,.5)  I didn't try out your answers yet though.

[color=cyan]Mystic, you should have checked the "interesting" solutions.[/color]

[quote="Arne"]What about this ?

If x < z then x + y < y + z so :sqrt:(4z - 1) < :sqrt:(4x - 1) so z < x.
Contradiction. So x, y, z must be equal.
Now we have 2x = :sqrt:(4x - 1) iff (2x - 1) = 0.
So the only solution is (x, y, z) = (1/2, 1/2, 1/2).[/quote]
That was ridiculously elegant. Wow.

[quote="Arne"]What about this ?

If x < z then x + y < y + z so :sqrt:(4z - 1) < :sqrt:(4x - 1) so z < x.
Contradiction. So x, y, z must be equal.
Now we have 2x = :sqrt:(4x - 1) iff (2x - 1) = 0.
So the only solution is (x, y, z) = (1/2, 1/2, 1/2).[/quote]

Hmm, is this a trick that is used a lot? That is, when given a bunch of equations that, when added, are symmetrical, trying to let one variable be < than another and then finding out that all of the variables are equal? Because I seem to recall a few other solutions that were previously posted around with forum using that strategy...I think...

Hey Arne, that is incredibly beautiful.  I got (1/2, 1/2, 1/2) also methodically, but it was a bit longer than what you did.  Eventually I figured out that x=y=z so then I solved the system by substitution to get (1/2, 1/2, 1/2).   :)

Well this trick can be very helpful when given a symmetric system.

Do you remember the system
x = 3y - 2
y = 3z - 2
z = 3x - 2 ?

This (selfmade) problem was solved using the same method.

Another idea when solving symmetric systems is using inequalities.
As an example, British Olympiad 1998 :

Determine all positive real solutions to
xy + yz + zx = 2
xyz = x + y + z + 2.


Here is the diagram again:

[asy]draw((-10,0)--(0,13)--(13,0)--cycle);
draw((0,0)--(0,13));
draw((7,0)--(0,13));
draw((7,0)--(11,2));[/asy]

It is easier to at first disregard the smallest line of the figure (appears to be cutting the rightmost triangle into two parts).  Once we can count the number of triangles without that little line, we can add back in the number of extra triangles resulting from the line.

The number of triangles in the diagram w/o the little line is actually C(4,2).  You may notice that the number of triangles will be the number of ways we can randomly choose 2 vertices of the total 4 on the bottom of the diagram since 1 vertex of the triangle is at the top.  (Triangles have 3 vertices).  Since the order in which we choose the vertices does not matter, we use Combination, not Permutations (P).

C(4,2) = [u]4![/u]
               2!2!

After calculating that out, you will eventually arrive at the answer 6.

But wait, 6 is not the answer!  Now we must count the number of extra triangles that little line creates!

The little line creates 2 more triangles by cutting the rightmost big triangle into 2 parts: a small part and a bigger part.

Thus, the answer is 6+2= [b][i][u]8[/u][/i][/b][/u][/i]

Oops.  C(4,2)=4!/(2!)(2!), just in case anyone gets confused.  I did it wrong.

4!=24, 2!=2, 24/(2)(2)=6.  There we go.

Answer is still 8.


[youtube]giAy8wEvoVsv[/youtube]
HAHA
This video is awesome

i saw this video so long ago
but is still funny
i love the asian mother part and the rapping part
*yo yo yo
my name's macbeth
my wife wants me to kill the king (etc.)*
 :rotfl:  :rotfl:  :rotfl:
is hilarious...


inequalities
function
inequalities proposed

Imagine a function f on the naturals.
Suppose n = akak-1...a2a1 in base 2.
Let f(n) be the sum of all indices i for which ai = 1. For example,
21 = (10101)2 => f(21) = 5 + 3 + 1 = 9.
29 = (11101)2 => f(29) = 5 + 4 + 3 + 1 = 13.
13 = (1101)2 => f(13) = 4 + 3 = 1 = 8.
Now prove that for any positive integer n:
f(3n) \leq 2f(n) + \sqrt(f(n))
and determine when equality occurs.


inequalities calculus derivative function AMC USA(J)MO USAMO

this has surely been posted on the forum before, because I have seen it :P it's trivial indeed using cauchy.

Lets denote with E that expression!

It's easy to see that we have: $ E \geq \frac{ (\sum a_{1})^{2}}{\sum a_{1}+ \sum b_{1}} =\frac{1}{2}$
So the minimum is $\frac{1}{2}$.

right?

We can easily generalize this:

Let n be a positive integer and $a_{1},a_{2}, \ldots,a_{n},b_{1},b_{2},\ldots, \b_{n}$ be 2n positive real numbers so that $ \sum a_{1}= \sum b_{1}=k$ with k>0 and q $\in N$ ($ q\geq2$).

Find the minimal value of:

$ \sum \frac{a_{1}^{q}}{(a_{1}+b_{1})^{q-1}}$.

In the same way we get that: $ \sum \frac{a_{1}^{q}}{(a_{1}+b_{1})^{q-1}} \geq \frac{( \sum a_{1})^{q}}{( \sum a_{1}+ \sum b_{1})^{q-1}}= \frac{k^{q}}{(2k)^{q-1}}$. for k=1 and q=2 we have this problem.

cheers! :D :D

This was problem 4...But a TST problem with an one-line solution, as noticed by Valentin, it is maybe really too easy... :(

Pierre.

[quote="pbornsztein"]Let $n$ be a positive integer, and $a_1,...,a_n, b_1,..., b_n$ be $2n$ positive real numbers such that
$a_1 + ... + a_n = b_1 + ... + b_n = 1$.

Find the minimal value of
$ \frac {a_1^2} {a_1 + b_1} + \frac {a_2^2} {a_2 + b_2} + ...+ \frac {a_n^2} {a_n + b_n}$.[/quote]

It's really easy. But let's try another solution.

Let $ \displaystyle X=\frac {a_1^2} {a_1 + b_1} + \frac {a_2^2} {a_2 + b_2} + \cdots + \frac {a_n^2} {a_n + b_n}$ and define it's dual $ \displaystyle Y=\frac {b_1^2} {a_1 + b_1} + \frac {b_2^2} {a_2 + b_2} +  \cdots + \frac {b_n^2} {a_n + b_n}$. Then

$\displaystyle X-Y = \sum_{i=1}^n \frac {a_i^2-b_i^2} {a_i + b_i} = \sum_{i=1}^n (a_i-b_i) = 0$. On the other hand

$\displaystyle X+Y = \sum_{i=1}^n \frac {a_i^2+b_i^2} {a_i + b_i} \geq \sum_{i=1}^n\frac{1}{2}(a_i+b_i) = 1$.

Therefore $X=Y \geq \frac{1}{2}$.

[quote="pbornsztein"]Yes, it has been solved like that by some of the contestants. :)
Most of the solvers did with Cauchy.
One uses partial derivatives for a 3 pages solution :D
Other uses convexity for a function with more than one variable to get a wrong solution. :maybe:
And many didn't solve it at all... :(

Pierre.[/quote]

LOL  :D  :D

By the way, how many are allowed to take TST? In US, only USAMO winners (max. 12) can take TST.

Look at problem 11 of http://www.bath.ac.uk/~masgcs/ukimo2004/team_sheet2.pdf

and at the top of the solutions  http://www.bath.ac.uk/~masgcs/ukimo2004/ts2.pdf

The same problem was being used for preparation last year by the top nation at IMO 2003

And it was being reused by UK and France for IMO 2004 preparation and TST!

Another solution:

It's easy to verify that $\displaystyle \frac{x^2}{x+y} \geq \frac{3}{4}x - \frac{1}{4}y$ for $x,y>0$.

So $\displaystyle \sum_{i=1}^n \frac {a_i^2} {a_i + b_i} \geq \frac{3}{4} \sum_{i=1}^n a_i - \frac{1}{4} \sum_{i=1}^n b_i  = \frac{1}{2}$.

Pierre, did anybody use this method? :-)

There were around 25 contestants (I do not remember exactly) :
The main part comes from our correspondance program. We add some from the 'Concours gnral' (which is now a big problem with a lot of questions for a lot of pages, and absolutely not an olympiad type competition), and some students from the lyce Louis-le-Grand, which is one of the best lycees in France (and because the test was organised there).

From what I've said, there is no surprise that each of the 6 members of the french Imo-team were coming from our correspondance program (the others had never seen an olympiad problem).

Pierre.

[quote="math_sipo"]a1               a1b1                 a1+b1
______=a1- ______>=a1- ________
a1+b1            a1+b1                 4

when u sum u obtain 1/2
so easy man
sorry guys[/quote] you can use latex to write up your stuff. Check out the latex tutorial in the latex help. :D

By Titu's lemma we have that:
$$\sum_{k=1}^{n} \dfrac{a_k^2}{a_k+b_k} \ge \dfrac{\left(\sum_{k=1}^{n} a_k \right)^2}{\sum_{k=1}^{n} a_k+\sum_{k=1}^{n} b_k}=\dfrac{1}{2}$$

This problem must be a joke. Using Cauchy-Schwarz we are done!
I wonder why this problem was problem 4.....


The last one for now  :D :

Suppose a, b, c > 0 such that ab + bc + ca  = 1. Prove that

(1 - a)/(1 + a) + (1 - b)/(1 + b) + (1 - c)/(1 + c) <= 3/2.

Have fun ! :D

A quick try, before grobber, Maverick or an other guy from the belgian-romanian very high-skilled gang comes back  :mrgreen:  :mrgreen:  :wink:

As said in a previous message, in a triangle
tgA+tgb+tgC = tgA tgb tgC.
if x = 1/tgA, y = 1/tgB, z=1/tgC,
xy + yz + zx = 1

After a little trigonometry what we have to prove is

3 - 2(cosA+cosB+cosC) <= 3/2

or (cosA+cosB+cosC) >= 3/4
or (sinA+sinB+sinC) <= 9/4

Alas, sin is [b]not[/b] convex   :evil:

C =  :pi: - (A+B)
sinA+sinB+sinC = sinA+sinB+sin(A+B)

Now I used sin(a+b), sin=1-cos, cos(a+b) to get

sinA+sinB+sinC = 2 + 2cosAcosBcosC

As shown before, cosA+cosB+cosC<=3/2
So by AM-GM, cosAcosBcosC <= (cosA+cosB+cosC)/3) :^3: <= 1/8

Phew !

A somewhat shorter way to prove that cos A + cos B + cos C >= 3/4 :

In every triangle we have
cos A + cos B + cos C
= 1 - 2 cos A cos B cos C
>= 1 - 2((cos A + cos B + cos C)/3)
>= 1 - 2((3/2)/3)
= 3/4

where cos A + cos B + cos C <= 3/2 follows from Jensen's inequality.
(The fact that
cos A + cos B + cos C + 2 cos A cos B cos C = 1
in every triangle is quite well - known)

Nice one huh ?  :D

Trigonometric substitution can be VERY useful !

[quote="Arne"]cos A + cos B + cos C + 2 cos A cos B cos C = 1 in every triangle is quite well - known)
[/quote]

Quite well known ?
I didn't know that just 4 hours ago  :!:

OK then ... You know, there are lots of useful identities like this.
It's no bad idea to study them for a while ...

That's also correct ... a non - trig solution ! Nice work !


Let $x,y,z,w >0$ positive numbers ;)

Such that:
$x^2 + y^2 + z^2 + w^2= 3(x + y + z + w).$

 Fin all solution:$(x,y,z,w)$.

-----------------
Fabiola Bravo    :rotfl:
Lima -PERU

Ani body ??

Any idea to solve this question??

Fabiola Bravo  :pilot:
LIma- PERU

Looks like number theory but indeed it's more algebra.

Wlog $x \leq y \leq z \leq w$
$3*x - x^2 \leq 2$ $\forall x \in 1,2,....$
So $w^2 \leq 3w+6$ or $w \leq 4$

I find 4 solutions with $x \leq y \leq z \leq w$


Diophantine equation
number theory proposed
number theory

Hey,no one is interested in my questions?It's a nice problem! :blush:

lemma 1: If $ v_2(3^n\minus{}1)$ is even then $ n\equal{}2^k$
Lemma 2: $ 3$ divides $ 2^{2^n}\minus{}1$ but $ 9$ does not.
Two lemmas above solve your problem!

[quote="Allnames"]lemma 1: If $ v_2(3^n \minus{} 1)$ is even then $ n \equal{} 2^k$
Lemma 2: $ 3$ divides $ 2^{2^n} \minus{} 1$ but $ 9$ does not.
Two lemmas above solve your problem![/quote]
No ,the first lemma is wrong . I have a solution for this problem but it 's quite  complex . It is very clear that $ n$ must be even ,if not then $ 2\parallel{}3^n \minus{} 1$ , so we can write in the form $ n \equal{} 2k$ . Suppose the contradiction this product is a perfect square ,then exist d,x,y such that :
\[ (3^k)^2 \minus{} 1 \equal{} dx^2 ; (2^k)^2 \minus{} 1 \equal{} dy^2\]
If d is a perfect square , it is not hard to show that $ k \equal{} 0$ ,which gives contradiction to the fact that $ n > 0$ . So d is not a perfect square and two equations show that $ 3^k$ and $ 2^k$ are solutions of Pell's equation  :
\[ X^2 \minus{} dY^2 \equal{} 1\]
Suppose the smallest solution of this Pell equation is $ (a,b)$ then all roots of this equation are :
\[ X_{0} \equal{} 1,X_1 \equal{} a ; X_{n \plus{} 2} \equal{} 2aX_{n \plus{} 1} \minus{} X_n (*)\]
Our purpose is showing that this sequence can't both contain $ 2^k$ and $ 3^k$ . Suppose exist $ i,j$ such that $ X_i \equal{} 2^k ,X_j \equal{} 3^k$ .
[u]Claim 1 [/u]: a must be an even number .
Proof : If not then by the relation (*) , all terms of this sequence are odd , which means that it doesn't contain a power of 2 .
[u]Claim 2 : [/u] $ X_{2k \plus{} 1}\equiv 1 (\mod 2 )$and $ X_{2k}\equiv 0 (\mod 2)$ for all positive integer $ k$ .
This fact is so obvious . Notice that we have $ X_i \equal{} 2^k,X_j \equal{} 3^k$ , by claim 2 we have $ i$ is even and $ j$ is odd .
[u]Claim 3 [/u]: $ X_{2n \plus{} 1}\equiv 0 (\mod a ),\forall n\geq 0$ and $ x_{2k}\equiv 1, \minus{} 1 (\mod a )$ .
Proof :We only need to notice that $ X_{n \plus{} 2}\equiv \minus{} X_n (\mod a)$ and $ X_1\equiv 0 (\mod a)$ .
By claim 2 and claim 3 we have $ x_i\equiv 0 (\mod a )$ or $ a|2^k$ ,which means that exist $ t\leq k$ such that $ a \equal{} 2^t$ .
[u]Claim 4 :[/u] 3 is a divisor of $ a$ . If not we consider two cases :
i) $ a\equiv 1 (\mod 2 )$ then the sequence $ x_n(\mod 3)$ is {1,1,1,..} ,and it doesn't contain any power of 3 .
ii) $ a\equiv 2 (\mod 3)$ then the sequence $ x_n(\mod 3)$ is {1,2,1,2...} , and it also doesn't contain any power of 3 .
So we  must have $ 3|a$ ,which gives contradiction to the fact that $ a \equal{} 2^k$ .

TTsphn has the right idea(very nice one, I must say) and almost the right proof (I think. Of course, the mistake is probably mine).

By claim 2 and claim 3 we have that $ a \equal{} 3^{t}$ and not $ a \equal{} 2^{t}$  (as $ a | X_{2n\plus{}1}$) which shortens the proof a little, as that immediately contradicts claim 1.

[hide="Edit, this doesnt work"] Nvm, I realized that my solution doesnt work.  :oops: [/hide]

[quote="shenyixin"]$ n$ is a positive integer.Prove that  $ (2^n \minus{} 1)(3^n \minus{} 1)$ isn't a  square.[/quote]
this problem is China TST 2002.
You can read offical solution(I have  :D ).

[quote="gb2124"]To math 10
Can you post your official soluton of it?I think it will be nice material.
Thanks.[/quote]
Yes,I will posst them in tomorrow.Becausse Now I verry busy :)
Please wait.

Ps.Offical solution be a nice proof :D

[quote="math10"]this problem is China TST 2002.
You can read offical solution(I have  :D ).[/quote]
I knew this problem from my teacher,with a solution different from TTsphn's.(TTsphn's is also a nice one :) )
Maybe it's just the official solution.

[quote="Allnames"]lemma 1: If $ v_2(3^n \minus{} 1)$ is even then $ n \equal{} 2^k$[/quote]
Actually，lemma 1 should be:$ v_2(3^n \minus{} 1)\equal{}v_2(n)\plus{}2$

While working on this problem I made a MAPLE-assisted observation: if p is a 'large' prime divisor of m^n - 1, it is VERY often the case that p - 1 is divisible by n.

Is this a well known theorem (with the addition of extra conditions -- like the obvious p > n -- as appropriate)?

[quote="shenyixin"][quote="Allnames"]lemma 1: If $ v_2(3^n \minus{} 1)$ is even then $ n \equal{} 2^k$[/quote]
Actually，lemma 1 should be:$ v_2(3^n \minus{} 1) \equal{} v_2(n) \plus{} 2$[/quote]

Hello shenyixin can you prove this lemma  :)

thank you !! :lol:

This is a direct consequence of the [url=http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf]lifting the exponent lemma[/url]. We have $ v_2(\frac{3^2 \minus{} 1}{2}) \equal{} 2$, so by the lemma $ v_2(3^n \minus{} 1) \equal{} 2 \plus{} v_2(n)$.

[quote="Zhero"]This is a direct consequence of the [url=http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf]lifting the exponent lemma[/url]. We have $ v_2(\frac {3^2 \minus{} 1}{2}) \equal{} 2$, so by the lemma $ v_2(3^n \minus{} 1) \equal{} 2 \plus{} v_2(n)$.[/quote]

Thank you "Zhero"  :lol:

Regarding #12 (not central to this discussion anyway) I got some [url=http://www.groupsrv.com/science/about443292.html]help[/url] at sci.math that might be of some interest.

[quote]The number $ (a^n \minus{} 1)(3^n \minus{} 1)$ is not a perfect square $ \forall a \geq 2,n\geq1$. [/quote]

Well, you also need $ a \neq 3$, for sure! In fact for your use of the Pythagorean triplets to work it is good to have $ a$ even ... but for sure it's a short and sweet proof for $ a \equal{} 2$ (the original problem). However, at second thought, I think that the co-primality of the elements of that Pythagorean triple also needs to be argued.


algebra
polynomial
quadratics
number theory
relatively prime

There is a polynomial $f(x)$ of degree $n$. The remainder is $1$ when $(x-1)$ divides it. When $(x-3)$ divides $f(x)$, the remainder is $3$. When $(x-5)$ divides $f(x)$, the remainder is $21$.

What would be the remainder when $f(x)$ is divided by $(x-1) \cdot (x-3) \cdot (x-5)$ ?

[quote="t0rajir0u"][hide="Method 1"] The remainder is some quadratic polynomial, and you can plug in $x = 1, 3, 5$ to solve for its coefficients. [/hide] [hide="Method 2"] The Chinese Remainder Theorem applies to polynomials. ;) [/hide][/quote]

[hide]Do you mean to say that the problem cannot be solved definitively? I tried with CRM, but could not complete it since $n$ is variable and unknown.  :( [/hide]

well...when you divide it by a first degree polynomial, the remainder is constant.  what does that say about division by a third degree polynomial?

edit:  tor already mentoined that its second degree

[quote="xxxyyyy"]I tried with CRM, but could not complete it since $n$ is variable and unknown.  :([/quote]

$n$ is irrelevant.  (In fact, you can assume WLOG that $n = 2$, which makes your life [b]very[/b] easy ;) )

The Chinese Remainder Theorem is a theorem about two things:

- A modulus (rather, a product of relatively prime moduli)
- A remainder (rather, a set of remainders)

Both are definite in this problem.

[hide]
Um, if I remember and if I have time, I'll type my solution in latex. Anyway, I just did what tor hinted at. I got $2x^{2}-7x+4$.[/hide]


If f(x)= m (subscript 1)x+b(subscript 1) and g(x)= m(subscript 2)x+b(subscript 2), prove (f(g(x)))'= f '(x) times g '(x).

I found out that (fog)'= f ' g(x) times g ' (x) and that f '(x)= m (subscript 1) and g '(x)= m (subscript 2) but not everything works out right.

Please help me. Thanks!

[quote]If $ f(x) \equal{} m_1 x \plus{} b_1$ and $ g(x) \equal{} m_2 x \plus{} b_2$, then prove that $ \left(f(g(x))\right)' \equal{} f'(x) \cdot g'(x)$.[/quote]
$ f'(x) \equal{} m_1$, $ g'(x) \equal{} m_2$, thus $ f'(x) \cdot g'(x) \equal{} m_1m_2$.

$ \frac{d}{dx}\left[f(g(x))\right] \equal{} f'(g(x))g'(x) \equal{} m_1 m_2$.

Note that $ f'(g(x)) \equal{} m_1$ as $ f'(x) \equal{} m_1$.

What's not working out?

wouldn't f'(g(x))= m(subscript 2)?

because it would be f'(m (subscrpt 2)x+b(subscript 2)) which then equals m (subscript 2)

If you don't want to use $ \text{\LaTeX}$, commonly accepted shorthand for "a subscript b" is a_b. Wrapping this in dollar signs gives you $ a_b$. Just for future reference.

Anyway, $ f(x)=m_1x+b_1$, so $ f'(x)=m_1$. This is a constant function. It is a horizontal line. Regardless of what value of $ x$ you put in there, you're going to get $ m_1$ as the output. Therefore, $ f'(g(x))$ is simply $ m_1$.

$ \frac{d}{dx}\left[f(g(x))\right] \neq f'(g(x))$


number theory
greatest common divisor
least common multiple
prime factorization

it can be proven that where (a,b) means the gcd of a and b, and [a,b] means the lcm of a and b

$(a,b)[a,b]=ab$

$(a,b,c)[ab,bc,ca]=abc$

$(ab,bc,ca)[a,b,c]=abc$

can you find more complex identites like these, or ones which generalize to arbitratily many variables?

[hide="Solution"] Let $e_{p}(n)$ denote the greatest $k$ such that $p^{k}| n$.

[b]Lemma:[/b]  $n = m \Leftrightarrow e_{p}(n) = e_{p}(m) \forall p \in \mathbb{P}$

[b]Lemma:[/b]  $e_{p}(nm) = e_{p}(n)+e_{p}(m)$

[b]Lemma:[/b]  $gcd(a, b, c, ...) = \prod_{p \in \mathbb{P}}p^{min(e_{p}(a), e_{p}(b), e_{p}(c), ...)}$

[b]Lemma:[/b]  $lcm(a, b, c, ...) = \prod_{p \in \mathbb{P}}p^{max(e_{p}(a), e_{p}(b), e_{p}(c), ...)}$

Hence the following gcd identities reduce to the simple identities

$min(x, y)+max(x, y) = x+y$
$min(x, y, z)+max(x+y, y+z, z+x) = x+y+z$
$max(x, y, z)+min(x+y, y+z, z+x) = x+y+z$

Where $x = e_{p}(a), y = e_{p}(b), z = e_{p}(c)$ for some unspecified prime $p$.  It is easy to see where these identities come from; simply assume WLOG $x \ge y \ge z$.

Similarly, arbitrary generalizations can be had in the following way:

Given $x_{1}\ge x_{2}\ge ... \ge x_{n}$ with sum $S$ we have the identities

$min(x_{i})+max(S-x_{i}) = S$
$min(x_{i}+x_{j})+max(S-x_{i}-x_{j}) = S$
...

This chain of identities is obvious because by our WLOG assumption we have

$min(x_{i}) = x_{1}, max(S-x_{i}) = S-x_{1}$
$min(x_{i}+x_{j}) = x_{1}+x_{2}, max(S-x_{i}-x_{j}) = S-x_{1}-x_{2}$
...

And so forth. [/hide]

i don't understand all that fancy notation  :blush:

can you write out the identities with the (a,b) and [a,b] notation i used?

They come out to what you expect.  For example, the four-variable identities are given by

$(a, b, c, d)[abc, bcd, cda, dab] = abcd$
$(ab, ac, ad, bc, bd, cd)[ab, ac, ad, bc, bd, cd] = abcd$
$(abc, bcd, cda, dab)[a, b, c, d] = abcd$

Edit:  Let me explain myself a little more.

[hide="Slower this time"] Let me explain the two-variable case using $e_{p}(n)$.

$e_{p}(n)$ is just a convenient notation for the exponents in the prime factorization of $n$.  That is, we have

$n = 2^{e_{2}(n)}3^{e_{3}(n)}5^{e_{5}(n)}7^{e_{7}(n)}...$

Given two numbers $a, b$ with some values of $e_{p}(a), e_{p}(b)$, $gcd(a, b)$ is the unique positive integer that satisfies

$e_{p}( gcd(a, b) ) = min(e_{p}(a), e_{p}(b))$

For each prime $p$ (because $x | y$ if and only if $e_{p}(x) \le e_{p}(y)$ for each prime; hence we require that $e_{p}( gcd(a, b)) \le e_{p}(a), e_{p}(b)$ for each prime).

On the other hand, and for the same reason, $lcm(a, b)$ is the unique positive integer that satisfies

$e_{p}( lcm(a, b) ) = max(e_{p}(a), e_{p}(b))$

When multiplying two numbers we simply add the exponents in their prime factorizations; in other words,

$e_{p}(xy) = e_{p}(x)+e_{p}(y)$

Hence, taking the identity

$(a, b) [a, b] = ab$

And comparing the exponents of the prime factorizations of both sides (which must be equal) we get the identity

$min(e_{p}(a), e_{p}(b))+max(e_{p}(a), e_{p}(b)) = e_{p}(a)+e_{p}(b)$

Which is easy to show for the reasons I gave in my first post, and easy to generalize.

The sequence of min-max identities corresponds to a sequence of $gcd, lcm$ identities. [/hide]


At least 5% of students participating in the 2007 AMC 12 A scored 97.5 or more, so the AIME qualifying score on the 2007 AMC 12 A is 97.5.

At least 1% of students participating in the 2007 AMC 10 A scored 117.0 or more, so the AIME qualifying score on the 2007 AMC 10 A is 117.0.

We still have much to do in the AMC office with the scoring and tabulating.  AMC 10 A and AMC 12 A school reports will go out to contest managers approximately Friday, Feb 23, 2007 assuming all continues to go well with no problems.

Please note:   This is an extremely busy time at the AMC office.  As a result, I will not answer any inquiries, specific or general at any time.  Any inquiry will be immediately deleted.  Your contest manager will receive your official score report when we  are able to send them.

Steve Dunbar
AMC Director

A couple more things regarding this:

- a thread for discussion of this announcement (though I'm not sure what there is to "discuss") is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133662]here[/url].

- When Mr. Dunbar says "Any inquiry will be immediately deleted", he means any inquiry that you send to the AMC offices.  You can still post questions here, and though I doubt Mr. Dunbar will have time to respond to them, an AoPS user or admin might know the answer.  But [b]please[/b], do at least a minimal amount of research before asking a question; for most questions that I see here, the answer is readily obtainable by visiting the [url=http://www.unl.edu/amc]AMC website[/url].

- If you doubt how busy the AMC is, check out [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-1-1012-dir,pic.shtml]these photos[/url].

By the way, thanks very much to the AMC Director for posting this information!


Find all $f(x)$ that map the reals to the reals such that $f(2x)=2f(x)-2x^{2}$.

[quote="Farenhajt"]Substitute $f(x)=g(x)-x^{2}$ to get $g(2x)=2g(x)$. Hence $g(x)$ is linear, giving $f(x)=ax-x^{2}, a\in\mathbb{R}$[/quote]

That is not necessarily true. Only if $g$ is continous at 0. Because the solution to the Cauchy's equation $f(x+y) = f(x)+f(y)$ has non-trivial (non-linear solutoions). If one accepts the axiom of choice. He can contruct something called the "Hamel Basis". It is the basis for the vector space of the reals over the rationals. So it does not need to be linear.

Let $h(x)$ and $j(x)$ be two functions that map $[0,1)$ to the reals.

Then, a generic solution is given by $g(x)=\left \{ \begin{array}{ll}0&{ \text{if}\ \ x = 0}\\ h( \{ log_{2}x \})x&{ \text{if}\ \ x>0}\\ j( \{ log_{2}(-x) \})x&{ \text{if}\ \ x < 0}\end{array}\right.$

$g(x)$ is equivalent to a linear function iff. $j(x)=h(x)=k$ for some real number $k$. However, the linear case is the only case in which $g$ (and therefore $f$) is continuous at $0$.


The incircle of an isosceles triangle $ABC$ with $AB = AC$ is centered at $O$ and touches $BC, CA, AB$ at $K, L, M$ , respectively. Lines $KM$ and $OL$ meet at $N$ , and line $BN$ meets $CA$ at $Q$. Let $P$ be the projection of $A$ onto $BQ$. Suppose that $BP = AP+2PQ$. Find all possible values of $\frac{AB}{BC}.$


We have recently made some changes on our servers, and while we tried to make sure we haven't missed anything, given the size of our site, we might have. So if you see any missing images (that were here before) or files, please let us know.

Also, the wiki is back online.

It looks like transcripts of recent Math Jams are missing.

(Separate point: On the transcripts page, it would be nice if the Jam Date listed the year in addition to the month and day.)


number theory
relatively prime
Diophantine equation
number theory proposed

For every positive integer $ n$, $ b(n)$ denote the number of positive divisors of $ n$ and $ p(n)$ denote the sum of all positive divisors of $ n$. For example, $ b(14)\equal{}4$ and $ p(14)\equal{}24$. Let $ k$ be a positive integer greater than $ 1$.

(a) Prove that there are infinitely many positive integers $ n$ which satisfy $ b(n)\equal{}k^2\minus{}k\plus{}1$.

(b) Prove that there are finitely many positive integers $ n$ which satisfy $ p(n)\equal{}k^2\minus{}k\plus{}1$.

(a) it is enough to take $ n\equal{}p^{k^2\minus{}k}$,where $ p$ is a prime number.
(b) Obviously if $ n>k^2\minus{}k\plus{}1$,then $ p(n)>n>k^2\minus{}k\plus{}1$,am i miss something?

[quote="Erken"](b) Obviously if $ n > k^2 \minus{} k \plus{} 1$,then $ p(n) > n > k^2 \minus{} k \plus{} 1$,am i miss something?[/quote]Why is $ n>k^2\minus{}k\plus{}1$? Did you notice that I wrote [b]finitely many[/b], not infinitely many? :wink:

[quote="Johan Gunardi"][quote="Erken"](b) Obviously if $ n > k^2 \minus{} k \plus{} 1$,then $ p(n) > n > k^2 \minus{} k \plus{} 1$,am i miss something?[/quote]Why is $ n > k^2 \minus{} k \plus{} 1$? Did you notice that I wrote [b]finitely many[/b], not infinitely many? :wink:[/quote]
I take enough large $ n$,in our case it is $ n>k^2\minus{}k\plus{}1$,so that our equality can not be true,so there is no $ n>k^2\minus{}k\plus{}1$,such that $ p(n)\equal{}k^2\minus{}k\plus{}1$,that's why there are finitely many $ n$.

[quote="Erken"][quote="Johan Gunardi"][quote="Erken"](b) Obviously if $ n > k^2 \minus{} k \plus{} 1$,then $ p(n) > n > k^2 \minus{} k \plus{} 1$,am i miss something?[/quote]Why is $ n > k^2 \minus{} k \plus{} 1$? Did you notice that I wrote [b]finitely many[/b], not infinitely many? :wink:[/quote]
I take enough large $ n$,in our case it is $ n > k^2 \minus{} k \plus{} 1$,so that our equality can not be true,so there is no $ n > k^2 \minus{} k \plus{} 1$,such that $ p(n) \equal{} k^2 \minus{} k \plus{} 1$,that's why there are finitely many $ n$.[/quote]Okay. Then why did you write 'am i miss something?'?

[quote="gagan_goku"]dont we have to show the existance of atleast one such n ?[/quote]Actually I think so. But the official solution does not mention it. Erken's solution is essentially the same as the official solution.

but if there are no such n , shouldnt we say there is no such n
dont know , you can also count 0 in finite numbers. anyways , someone please tell me how to increase my brain activity.
i feel really good when im able to solve some nice problem. but im too dumb  :(

$ 0$ is a finite number.

There are solutions for some $ k$:  $ p(p^2) \equal{} p^2 \plus{} p \plus{} 1 \equal{} (p \plus{} 1)^2 \minus{} (p \plus{} 1) \plus{} 1$ for $ p$ a prime.  There aren't solutions for every $ k$; I believe there are no solutions for $ k \equal{} 5$, for example.  (Additionally, $ p(k^2 \minus{} k \plus{} 1) \equal{} k^2 \minus{} k \plus{} 1$ whenever $ k^2 \minus{} k \plus{} 1$ is prime, but it's an open question whether infinitely many such primes exist.)

Edit:  Two other sporadic solutions.  $ p(36) \equal{} 91 \equal{} 10^2 \minus{} 10 \plus{} 1$ and $ p(16) \equal{} 31 \equal{} 6^2 \minus{} 6 \plus{} 1$.  If two consecutive values $ k, k \plus{} 1$ have (relatively prime) solutions, then so does $ k^2 \plus{} 1$.

Edit 2:  When $ n$ is a power of two, we are trying to solve the Diophantine equation $ 2^a \minus{} 1 \equal{} k^2 \minus{} k \plus{} 1 \implies 2^a \equal{} k^2 \minus{} k \plus{} 2 \implies 2^{a \plus{} 2} \equal{} 4k^2 \minus{} 4k \plus{} 8 \equal{} (2k \minus{} 1)^2 \plus{} 7$.  The solutions here are given by the solutions to [url=http://mathworld.wolfram.com/RamanujansSquareEquation.html]Ramanujan's Square Equation[/url], which here gives the rather large solution $ k \equal{} 91, n \equal{} 2^{a \minus{} 1} \equal{} 2^{13}$ (in addition to what we found above).  This is a difficult equation to solve, so it seems that even the case that $ n$ is a power of a prime is not easy.


Bump...

Although I'm not the original poster, I would very much like to know the solution. :)

The problem is equivalent to show $ (\frac {1}{2 \plus{} a})^3 \plus{} (\frac {1}{2 \plus{} b})^3 \plus{} (\frac {1}{2 \plus{} b})^3\ge 1/9$, where $ a,b,c\in R_ \plus{}$ and $ abc \equal{} 1$.
It remains to show $ 2(\frac {1}{2 \plus{} a})^3 \plus{} (\frac {1}{2 \plus{}1/a^2} )^3 \ge 1/9$  for $ a > 0$.

[quote="miwalin"]The problem is equivalent to show $ (\frac {1}{2 \plus{} a})^3 \plus{} (\frac {1}{2 \plus{} b})^3 \plus{} (\frac {1}{2 \plus{} b})^3\ge 1/9$, where $ a,b,c\in R_ \plus{}$ and $ abc \equal{} 1$.
It remains to show $ 2(\frac {1}{2 \plus{} a})^3 \plus{} (\frac {1}{2 \plus{} 1/a^2} )^3 \ge 1/9$  for $ a > 0$.[/quote]
It should be
$ (\frac {1}{2 \plus{} a})^3 \plus{} (\frac {1}{2 \plus{} b})^3 \plus{} (\frac {1}{2 \plus{} c})^3\ge 1/9$, where $ a,b,c\in R_ \plus{}$ and $ abc \equal{} 1$.
It remains to show $ 2(\frac {1}{2 \plus{} a})^3 \plus{} (\frac {1}{2 \plus{} 1/a^2} )^3 \ge 1/9$  for $ a > 0$.
Sorry for carelessness.

The inequality can be proved in the following way:
$ \large\begin{array}{rcl}
\frac{2}{(2\plus{}a)^3}\plus{}\frac{1}{\left (2\plus{}\frac{1}{a^2}\right )^3}&\ge &\frac{1}{9}\qquad\Leftrightarrow\\
& & \\
\frac{a^9\plus{}6a^8\plus{}12a^7\plus{}24a^6\plus{}24a^4\plus{}12a^2\plus{}2}{(2\plus{}a)^3(2a^2\plus{}1)^3}&\ge &\frac{1}{9}\qquad\Leftrightarrow\\
& & \\
9\left (a^9\plus{}6a^8\plus{}12a^7\plus{}24a^6\plus{}24a^4\plus{}12a^2\plus{}2\right )\minus{}(2\plus{}a)^3(2a^2\plus{}1)^3&\ge &0\qquad\Leftrightarrow\\
& & \\
a^9\plus{}6a^8\plus{}80a^6\minus{}150a^5\plus{}84a^4\minus{}73a^3\plus{}54a^2\minus{}12a\plus{}10&\ge &0\qquad\Leftrightarrow\\
& & \\
(a\minus{}1)^2\cdot (a^7\plus{}8a^6\plus{}15a^5\plus{}102a^4\plus{}39a^3\plus{}60a^2\plus{}8a\plus{}10)&\ge &0.
\end{array}$
But this is true for all $ a>0$.


Find all positive real solutions to x^3 + y^3 + z^3 = x + y + z, x^2 + y^2 + z^2 = xyz.

Such systems of equations having more variables than equations are usually solved using inequalities, especially when you are given the information that x, y, z must be positive.

In fact, from your two equations

$x^{3}+y^{3}+z^{3}=x+y+z$

and

$x^{2}+y^{2}+z^{2}=xyz$

it follows that

$\left( x^{3}+y^{3}+z^{3}\right) \left( x^{2}+y^{2}+z^{2}\right) ^{2}=\left( x+y+z\right) \left( xyz\right) ^{2}$.

But for any three positive numbers x, y, z we have

$\left( x^{3}+y^{3}+z^{3}\right) \left( x^{2}+y^{2}+z^{2}\right) ^{2}>\left( x+y+z\right) \left( xyz\right) ^{2}$,

since

$\left( x^{3}+y^{3}+z^{3}\right) \left( x^{2}+y^{2}+z^{2}\right) ^{2}$
$=\left( x^{3}+y^{3}+z^{3}\right) \left( x^{4}+y^{4}+z^{4}+2y^{2}z^{2}+2z^{2}x^{2}+2x^{2}y^{2}\right) $
$>\left( x^{3}+y^{3}+z^{3}\right) \left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\right) $
$=x^{3}\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\right) +y^{3}\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\right) +z^{3}\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\right) $
$>x^{3}y^{2}z^{2}+y^{3}z^{2}x^{2}+z^{3}x^{2}y^{2}=\left( x+y+z\right) \left( xyz\right) ^{2}$.

Therefore, we get contradiction, and your system does not have any positive solutions.

  Darij


Stanford
college
function
geometry
algebra
floor function

yea i did, took the general, didn't stick around for the awards and solutions though, so i don't know what i got

yeah, i put n^2 instead of m^2 too...grrrr... and on the floor function one... i did the exact opposite function and got x to equal 3 instead of 2. how uncool. i was surprised i actually got 10th though....

i bombed geometry too... haha.. but we won't go into that...


probability
calculus
calculus computations

Q1 Find  probability of getting 3 consecutive sixes in 10 throws of a die.

Q2 A die is thrown again and again until we obtain 2 consecutive sixes are  obtained .find the probabiltiy of getting 2 consecutive sixes

Neither of these are calculus problems, and the second is nonsensical.  For the first, a standard attack with recursions or equivalently state diagrams works nicely to compute the probability of the complement. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190619]here[/url] for one example of the method; searching for the words "state diagram*" (e.g., in the High School Forums) should find more.


algebra
polynomial
superior algebra
superior algebra open

This is a shameless reposting from MathOverflow but the question is extremely beautiful and I really want to know what mathlinkers can make of it. I have already spent 3 days thinking of it myself but with very little progress :wallbash:.

Do there exist an infinite field $ k$ and a polynomial $ p\in k[x]$ such that the image of $ k$ under the mapping $ x\mapsto p(x)$ is not all $ k$ but its complement in $ k$ is finite?

I can only show that if $ \text{char\,} k\ne 2,3$, then we must have $ \text{deg\,}p\ge 4$ and handle a few very special polynomials of degree $ 4$ and $ 5$. :(

I agree it's a great question!  Just for the record the MO question is [url=http://mathoverflow.net/questions/6820/can-a-non-surjective-polynomial-map-from-an-infinite-field-to-itself-miss-only-fi]here[/url] and the comments there are certainly helpful, even if they don't give a complete answer.


counting
derangement
complementary counting

hi again,

Last time i posted a qstn on a derangement problem. BUt how about a variation like this?. I have two vatying answers from two different methods. Can anyone confirm either of them?

Letters A,B, C D E F are arranged in order such taht A is not in the 1st positon, B is not in the 2nd position, and F is not in the 6th position. How many possible ways are there of arranging the letters?

[hide="Method of attack"]For each of C, D and E, consider whether it is in its "natural place."  This gives you eight (or four with an obvious simplification) cases (representative example: C and E are in their natural places, but D is not), and each is an instance of the derangement problem for some number of letters.[/hide]
A more general version of this problem is the following:

Suppose we have $ n$ letters with $ k$ special letters.  How many permutations are there of these $ n$ letters such that none of the $ k$ special letters occurs in its natural place?
[hide="Answer to the general version of this problem"]$ \sum_{j \equal{} 0}^{n \minus{} k} \binom{n \minus{} k}{j} D_{k \plus{} j}$.  Here $ j$ counts the number of non-special letters that don't go to their natural place.  $ \binom{n \minus{} k}{j}$ chooses every possible subset of them, while $ D_{k \plus{} j}$ arranges them and the $ k$ special letters so that none goes to the natural place.  (The other $ n \minus{} k \minus{} j$ letters have only one choice each, their natural places.)[/hide]
Does this answer simplify?

The way I'm reading the problem (which is most likely wrong, considering this is in the pre-olympiad forum  :P) is that we could fix A,B,F and check the number of permutations that way ($ 3!$) and subtract that from the total number of permutations. So our answer would be $ 6!\minus{}3!$. What is it that I'm missing that leads JBL to his solution?

[quote="max_tm"]we could fix A,B,F[/quote]

Nope.  The complement of the case that none of $ A, B, F$ are in their natural position is that [b]at least one of[/b] $ A, B, F$ is in its natural position, not all of them.

Sorry torajirou but could you elaborate on that a bit? It's not coming as immediately intuitive to me.

With no conditions, there are $ 6!$ permutations.  We are considering the subset of these permutations that satisfies [b]all[/b] of the following three conditions:

- $ A$ is not in the first position.
- $ B$ is not in the second position.
- $ F$ is not in the sixth position.

The number of permutations fixing $ A, B, F$ in the first, second, and sixth positions is the number of permutations that satisfies [b]none[/b] of the above three conditions.  However, your logic completely ignores the permutations that satisfy [b]some, but not all[/b] of the above three conditions.

Ah, I see. Could we patch up my logic? As in counting:

- When A is fixed
- When B is fixed
- When C is fixed
- When A,B is fixed
- When B,C is fixed
- When A,C is fixed
- When all three of A,B,C are fixed

.... Although I would never be proud of a solution like that  :P.

A slight modification of that idea (to avoid overcounting) works, but each of those individual cases is just as hard as the original problem (I think, except for the last).  Look at JBL's idea.

max_tm's idea should work for a perfectly respectable solution by inclusion-exclusion.  If we interpret "A is fixed" as "A is fixed and we permute the others however we like," then each individual case is simple and inclusion-exclusion takes care of the overcounting.  It might even result in a simpler expression for the final sum -- give it a try!

yehh guys. i think i have it. I seem to be getting 426 with 3 different approaches.

As you guys mentioned, one method is complementary counting. It is not as complicated as it seems. Working through each of the cases, we want to subtract all subsets with at least one of A B or F in their correct position (ie, only A, B, F is in their correct position, only, AB, AF, and BF are in their correct positions, and when all A E F are in their correct positions).

1.  So now, for the case where all ABF are in correct position is simple; if we treat positions 1,2 and 6 as one group and then the rest as another, we get that there is only one way to permute  ABF in positions 1,2 ,6 such that they are all in the correct position, and then there are 3! ways of permuting the others in the other 3 positions).

2.   Now, when only one of A E or F are in their correct positions, there are 3 ways of choosing which to be in their correct positions only. Now for the other two who are not to be in their fixed position (ie say i chose A to be in its fixed position, so B and F  can either be in the group positon 1,2 6 and not in their position in various permutations)
i) when all 2 others that are not to be in their fixerd positions are not to be in the 1,2 6  position group, the number of ways is 3*2*3!. (since two of the other 3 letters C D or E must be placed in the remaining 2 unoccupied positions in 1,2 6)
ii) when only 1 other that are not to be in their fixed positions are not to be in 1,2 6 position group, the  number of ways is 2*3*3! (this time, the 2 meaning there are two choices of which of the 2 unfixed numbers are placed in the 126 positonal group)
ii) when all the other 2 unfixed letters are to be placed in the 1,2,6 group --> 3! (since there is only one way of derranging the 2 unfixed letters such that they dont appear in their forbidden positions)

Now for this case, (from i), ii), iii)), you get 3(2*3*3! + 2*3*3! +3!)

3.  Now, when all 2 other unfixed letters are to be in their correct positions, then there would 3(3*3!) ways

Hence the total number of arrangements are  6! - 3! - 3(3*3!) - 3(2*3*3! + 2*3*3! +3!) = 426.

I understand this is a rather longwinded and elaborate method, which is not very desirable. But a better method is applying the Principle of Inclusion and Exclusion. and u simply get:  6! - (3C1)(5!) + (3C2)(4!) - (3C3) (3!) = 426.

A THIRD METHOD, can follow along the lines of the first method, but not do complementary counting, but rather look at the 4 cases, again treating positions 1,2 and 6 as one group and the other positions as another (ie. when all A B F are in the 126 group, when only 2 of them are in the 126 group, when only 1 of them are in the 126 group, and when all of them are out of the 126 group). THis method givse:

2*3! + (3!)(3!) + (3C1)(2)(3*2)(3!) + (3C2)(3)(3)(3!) = 426

I would assume from 3 different methods, the answer should be 426. Funnily enough, from my first attempt, i achieved answers of 372 from my third method, and 480 for my 1st method. And then i did my second, and got 426, and realised how my previous 2 methods, the first was down by 54 and the other was up by 54. I thought this could not be a coincidence, and so i thought since one method (the first) was subtracting cases, and the last was adding cases (as methods 1 and 3 are mirrors of each other), I realised i must have made the same mistake for both! ....and eventually i found it LOL!).

Just for your satisfaction (if you need it), your answer is definitely right :)

[hide][code]p = Permutations[{1, 2, 3, 4, 5, 6}];
Total[Table[Which[(p[[n]][[1]] - 1)*(p[[n]][[2]] - 2)*(p[[n]][[6]] - 6) == 0, 0,True, 1], {n, 1, Length[p]}]][/code][/hide]


function
integration
calculus
calculus computations

[color=darkred]$ 1\blacktriangleright a\in\mathcal R$ and a continue monoton function $ f: [0,1]\rightarrow\mathcal R$ $ \implies$ $ \int_{0}^{1}\left|f(x)\minus{}a\right|\ dx\ge\int_{0}^{1}\left|f(x)\minus{}f\left(\frac{1}{2}\right)\right|\ dx$ ([b]level ****, i.e. difficult ![/b]).
$ 2\blacktriangleright$ A continue function $ f: [a,b]\rightarrow\mathcal R$ $ \implies$ exists $ c\in (a,b)$ so that $ \int_{a}^{c}f(x)\ dx\plus{}\int_{b}^{c}f(x)\ dx\equal{}(a\plus{}b\minus{}2c)\cdot f(c)$ ([b]level **, i.e. easy ![/b]).[/color]

$ 1.$ First we assume $ f$ is an increasing function (if $ f$ satisfies $ \rightarrow\minus{}f$ satisfies the conditions with $ \minus{}a$).  Then it's easy to see that it's enough to prove the statement for $ a\in[f(0),f(1)]$. Then let $ f(b) \equal{} a$, and suppose $ b\ge\frac{1}{2}$ (for $ b <\frac{1}{2}$ it's the same). Then we have:
$ \int_{0}^{\frac{1}{2}}f(b)\minus{}f(x)dx\plus{}\int_{\frac{1}{2}}^{b}f(b)\minus{}f(x)dx\plus{}\int_{b}^{1}f(x)\minus{}f(b)dx\ge$
$ \int_{0}^{\frac{1}{2}}f(\frac{1}{2})\minus{}f(x)dx\plus{}\int_{\frac{1}{2}}^{b}f(x)\minus{}f(\frac{1}{2})dx\plus{}\int_{b}^{1}f(x)\minus{}f(\frac{1}{2})dx$
This is equivalent with $ (b\minus{}\frac{1}{2})f(b)\ge\int_{\frac{1}{2}}^{b}f(x)dx$, but $ \int_{\frac{1}{2}}^{b}f(x)dx \equal{} (b\minus{}\frac{1}{2})f(c), c\in(\frac{1}{2},b)$, so we get
$ f(b)\ge f(c)$ which is obviously true($ b\ge c$).

$ 2.$ We apply Rolle to the following function:
$ G(t) \equal{} (a\plus{}b\minus{}2t)(\int_{a}^{t}f(x)dx\plus{}\int_{b}^{t}f(x)dx)$

The first problem can easily be generalized:
$ f: [a,b]\rightarrow\mathbb{R}$,$ a < b$ continuous and monotonic function,$ c\in\mathbb{R}$ then:
$ \int_{a}^{b}\left|f(x)\minus{}c\right|\ dx\ge\int_{a}^{b}\left|f(x)\minus{}f\left(\frac{a\plus{}b}{2}\right)\right|\ dx$

This can easily be reduced to the original problem with the substitution $ t \equal{}\frac{x\minus{}a}{b\minus{}a}$.
It's a well-known problem, but thank you for reminding me of it. Va multumesc! :)


Show in the neatest possible way that \[\frac{c(a - b)^3 + a(b - c)^3 + b(c - a)^3}{c^2(b - a) + a^2(c - b) + b^2(a - c)} = a + b + c.\]

Another way to factor the numerator: consider the polynomial $P$ whose roots are $a-b,b-c,c-a$. Since they add up to zero, $P$ has the form
\[P(x)=x^3+mx+n,\]
where $n=-(a-b)(b-a)(c-a)$.
Because $cP(a-b)+aP(b-c)+bP(c-a)=0$, we obtain
\[\sum {c(a-b)^3}+m\sum {c(a-b)}-(a+b+c)(a-b)(b-c)(c-a)=0.\]
Obvously, $\sum {c(a-b)}=0$, hence the result.


geometry
circumcircle
geometric transformation
homothety
search

Prove that $ E$ is midpoint of major arc $ \stackrel{\frown}{AB}$.

What is $ E$?

Ah, that $ E$ looked like an $ F$ to me

[quote="Ubemaya"][hide="Hint"]
Draw perpendicular bisector of AB, let it hit the circle at F, and prove that E and F coincide.[/hide][/quote]

Somebody can show how to do that? No other approach?

Maybe the figure helps.

If $ l$ is the line that is tangent to the circumcircle of triangle $ ABE$ at $ E$, then $ l//AC$ since a homothety center $ D$ maps one circle (and tangent line) to another. Then let $ A'$, $ B'$ be on $ l$ so $ A'XB'$ is a line, and $ AB$, $ A'B'$ are similarly oriented.

Then $ \text{arc}\, AE\equal{}\angle ABE\equal{}\angle BEB'\equal{}\text{arc}\, BE$.

Thank you for your answer.
I would like to know if there are another solution. The book where i got this problem don't say anything about homothetic center.

[hide="Alternate Solution"]
Call the center of the circle containing chord $ AB$ to be $ O$ and that containing point $ C$ to be $ Q$.  So, we call $ \angle AOB \minus{} 2\theta$.  Thus, we have that $ \angle OAB \equal{} \angle OBA \equal{} 90 \minus{} \theta\rightarrow \angle OBC \equal{} 90 \plus{} \theta$.  Now, we call $ \angle QCD \equal{} \phi \equal{} \angle QDC$, so $ \angle BCD \equal{} 90 \minus{} \phi$ and $ \angle CDO \equal{} 180 \minus{} \phi$.  Thus, $ \angle BOD \equal{} 360 \minus{} (\angle OBC \plus{} \angle BCD \plus{} \angle CDO) \equal{} 2\phi \minus{} \theta$.  Also, $ \angle CDQ \equal{} \angle ODE \equal{} \phi \equal{} \angle OED$, so $ \angle DOE \equal{} 180 \minus{} 2\phi$.  Thus, $ \angle AOE \equal{} 360 \minus{} (\angle AOB \plus{} \angle BOD \plus{} \angle OED) \equal{} 180 \minus{} \theta$.  Also, $ \angle BOE \equal{} \angle BOD \plus{} \angle DOE \equal{} 180 \minus{} \theta \equal{} \angle AOE$, so arcs $ AE$ and $ BE$ have the same angle, and thus $ E$ is the midpoint of major arc $ AB$.   [/hide]
EDIT: Oh, sorry, that should be $ \angle QCD \equal{} \phi \equal{} \angle QDC$, edited.

[quote="The QuattoMaster 6000"]Now, we call $ \angle QCD \equal{} \phi \equal{} \angle ODC$  [/quote]

Maybe $ \angle QCD \equal{} \phi \Longrightarrow \angle ODC\equal{}90^{\circ}\plus{}\phi$?


are the phase curves of the following differential equation closed?

$ x''\plus{}100x\equal{}0$

Yes. The general solution is $ x(t) \equal{} \dots$, and the parametric equation $ t\mapsto (x(t),x'(t))$ describes an ellipse.

There are more challenging questions of whether phase curves are closed. For instance, this well-known classic:

$ x'\equal{}ax\minus{}bxy$
$ y'\equal{}\minus{}cy\plus{}dxy$

with $ a,b,c,d>0.$ And the only phase curves we care about are in the first quadrant, $ x>0,y>0.$

Topic-changing replies to this have been moved into the Round Table forum.


1)Let T:V-->V be any linear map and V* be its dual.Let W be a k-dimensional subspace of V.(V is finite dimensional).Then prove that there exits a Linearly independent subset {f1,f2,...f(n-k)} subset of V* such that W=inetsection of Kernel(fi), where i=1,2,...,n-k.

2)Let T* denote the adjoint of T. If T is surjective then show that T* is injective

1) $ dim(V)\equal{}n$. Let $ \{w_1,\cdots,w_k,v_1,\cdots,v_{n\minus{}k}\}$ be a basis of $ V$ s.t. $ \{w_1,\cdots,w_k\}$ is a basis of $ W$. We define $ f_i$ as follows: if $ x\equal{}\sum\lambda_rw_r\plus{}\sum\mu_jv_j$ then $ f_i(x)\equal{}\mu_iv_i$.
2) "Let $ V^*$ be its dual" is a nonsense. You must define a basis of $ V$: $ \{e_1,\cdots,e_n\}$.
If $ T$ is onto then $ T$ is one to one. Then the correct question is: if $ T$ is invertible then $ T^*$ is invertible.
Assume $ T^*(f)\equal{}0$. Then for all $ i$, $ T^*(f)e_i\equal{}0\equal{}f(T(e_i))$. Thus $ f(Im(T))\equal{}0$ or $ Im(T)\subset{k}er(f)$. Conclusion: $ f\equal{}0$ and $ T^*$ is invertible.

I wrote to quickly. There are 2 corrections:
1) Replace "let $ V^*$ be its dual " is a nonsense with "Let T* denote the adjoint of T" is a nonsense. Indeed there do not exist any canonical isomorphisms between $ V$  and $ V^*$.
2) Replace $ f_i(x)\equal{}\mu_iv_i$ with $ f_i(x)\equal{}\mu_i$.

Third correction !
The adjoint, or the transpose, of $ T$ is well-defined. -I thought about the transpose of a matrix (Too much champagne during the festivals !)-
If $ T: V\rightarrow{W}$ then $ T^*: W^*\rightarrow{V^*}$ is defined by $ T^*(f)\equal{}foT$.
Assume that $ T$ is onto and $ T^*(f)\equal{}0$ that is $ foT\equal{}0$.
Then $ f(W)\equal{}0$ and $ f\equal{}0$ and $ T^*$ is one to one.


logarithms
pigeonhole principle
continued fraction
absolute value

Prove or disprove that any positive real number $ n$ may be represented to an arbitrary degree of precision (so, any degree of precision as high as you desire - pi could be represented to 10 digits, 100, 1000...) as $ 2^j*3^k$, where j and k are integers.

Maybe I'm understanding this wrong, but $ 5\neq2^k3^j$ because the only prime factor is 5.

$ k,j$ are not necessarily positive integers, and by "degree" sponge means "degree of precision", i.e. that for any $ \epsilon>0$, you can choose $ j,k$ so that $ |2^j3^k\minus{}n|<\epsilon$.

Thanks for the clarification, Temperal; I seem to be remarkably bad at writing down mathematical thoughts translated to English. (fixed the first post)

You can rewrite the relation as

$ a\log_2 2\plus{}b\log_2 3\equal{}\log\plus{}2 n$ where $ a,b\in\mathbb{Z}$.

Consider all $ m^2\minus{}1$ pairs $ (a,b)$ with $ a,b\in \{1,2,...,m\}$

Now $ 1<a\plus{}b\log_2 3\le m\log_2 6$.

Now break the interval $ (1,m\log 6]$ into $ m^2\minus{}2$ intervals. By the pidgeonhole principle, there must be two pairs $ (a_1,b_1)$ and $ (a_2,b_2)$ that lie in the same interval of width $ \frac{m\log_2 6}{m^2\minus{}2}<\frac{3m}{m(m\minus{}1)}\equal{}\frac{3}{m\minus{}1}$ for sufficiently large $ m$.

Then it follows that $ (A_m,B_m)\equal{}(a_1\minus{}a_2,b_1\minus{}b_2)$ is in the interval $ (0,\frac{3}{m\minus{}1}]$. (The difference cannot be zero $ a\plus{}b\log_2 3\ne 0$ for any $ a,b$ as $ \log_2 3$ is irrational. ).

Now if we consider $ (l A_m,l B_m)$ it is easy to show that we get to $ n$ with an arbitrary degree of accuracy where $ l$ is an integer (I'm being lazy...)

[hide]Lemma: $ \log 2$ and $ \log 3$ are linearly independent over the rationals.
Proof: If this were not true, then $ a\log 2 \plus{} b\log 3 \equal{} 0$ or, by the properties of logs, $ \log 2^a3^b \equal{} 0$.  We take the antilog of both sides to get $ 2^a3^b \equal{} 1$ which is absurd (unless $ a \equal{} b \equal{} 0$ which is the trivial solution).  So the lemma is proven.

We take the log of $ n\approx 2^k3^j$ and get $ log n\approx k\log2 \plus{} j\log3$ where $ \approx$ means to whatever degree is needed.  Now we need to express $ \log n$ as a linear combination of $ \log 2$ and $ \log 3$ to arbitrary degree.  We can do this by taking the continued fraction expansion of $ \frac {\log 3}{\log 2}$ to arbitrary degree: let $ \frac {\log 3}{\log 2}\approx \frac {p}{q}$.  Then $ |q\log 3 \minus{} p\log 2| \equal{} \epsilon$ where $ \epsilon$ is very small.  Now let $ \frac {\log n}{\epsilon} \equal{} r$.  ($ \epsilon$ is never 0, because the definition of $ \epsilon$ and the fact that $ \log 2$ and $ \log 3$ are linearly independent over the rationals combine to show this.)  Then $ \log n \equal{} r\epsilon \equal{} r|q\log 3 \minus{} p\log 2|$.  WLOG we can drop the absolute value signs: if $ p\log 2\geq q\log 3$, then switch the two in the following.

We have $ \log n \equal{} (rq)\log 3 \plus{} ( \minus{} rp)\log 2$.  We round $ rq$ and $ \minus{} rp$ to the nearest integers $ s$ and $ t$.  Take the antilog of both sides and get $ n\approx 3^s2^t$.  Note that the higher the degree of $ p$ and $ q$, the higher the degree of accuracy of $ s$ and $ t$.  Thus $ s, t$ are the integers we are looking for.[/hide]
Edit: Ninja'd again.


quadratics
number theory unsolved
number theory

Let $p = a^2 + 4b^2$ be a prime. Show that $(\frac ap) = +1$

Look at the equation $\mod a$:
We get that $\left( \frac{p}{a} \right) = 1$, and since $p \equiv 1 \mod 4$ it follows by quadratic reciprocity (for Jacoby symbols here) that $\left( \frac{a}{p} \right) = 1$.


algorithm
number theory proposed
number theory

Show that there exists infinite triples $(x,y,z) \in N^3$ such that $x^2+y^2+z^2=3xyz$.

Well, I think it is an algorithm which generates ALL solutions. In fact it resembles the more general problem to find all numbers such that $\frac{x^2+y^2+z^2}{xyz}=S$ is integer. And $S=3$!

[quote="Ramanujan"]First we see that $(x,y,z)=(1,1,1)$ is a solution.

Now let $(a,b,c)$ is a solution of the equation with $1 \leq a\leq b \leq c$.

Then $a$ is a root of the equation $x^2-3bcx+b^2+c^2=0$ so
$x^2-3bcx+b^2+c^2=(x-a)(x-(3bc-a))=0 \Longrightarrow (3bc-a,b,c)$
 is also a solution.

Since every permutation of a solution is also a solution we take that
permutation of $(3bc-a,b,c), \, (a_{1},b_{1},c_{1})$ with $a_{1} \leq b_{1} \leq c_{1}$
and we do the same with this solution.

If we begin with the solution $(1,1,1)$ and apply the above algorithm
we take infinite distinct solutions .[/quote]
Did you apply the algorithm?
First we have $(1,1,1)$.Then we get $(2,1,1)$ and after that we'll have $(1,1,1)$ again!

No, [color=#0000FF]Atlas[/color]. After you get $(2,1,1)$, you permute to $(1,2,1)$ before getting the next, which will be $(5,2,1)$, then permute again, etc. Funny (and pity) the famous [b]Markoff equation [/b]found its way in such a recent Brazilian test.

using quadratic equation formula in solving x this is solved when you will find (x,y,z) = (1,1,1) is a solution


function
real analysis
real analysis unsolved

Let $ f: [a,b] \to \mathbb{R}$ a function continuous on $ [a,b]$ and diferentiable on $ (a,b)$ with $ f(a)\equal{}f(b)$.Prove that $ \exists c \in (a,b)$ such that \[ f'(c)\cdot (c\minus{}a)\equal{}f(a)\minus{}f(c)\]

Let $ g: [a,b] \to \mathbb{R}: \forall x\in[a,b]: \ h(x)\equal{}(f(x)\minus{}f(a))(x\minus{}a)$
$ g$ is a  continuous function on $ [a,b]$ and diferentiable on $ (a,b)$ with $ h(a) \equal{} h(b)\equal{}0$
then $ \exists c\in(a,b): \ h'(c)\equal{}0$ and $ h'(c)\equal{}f'(c)(c\minus{}a)\plus{}f(c)\minus{}f(a)$  :wink:


inequalities
vector
analytic geometry
induction
geometry
Putnam
algebra

Hi!  I recently derived a relatively simple proof of the Cauchy Inequality, but it relies on the fact that the triangle inequality is true for any pair of vectors in n-space.  I know that it works in the coordinate plane, and it seems logical that it works in all dimensions.  Is this true?

So you want

$\|v\|+\|w\| \geq \|v+w\|$?

If you saw my previous two attempts at a post, they were wrong lol.

isn't that obvious from vector addition (i think)?  im used to calling that the triangle inequality for complex numbers.  i used that to get the form $(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+...+y_{n}^{2})\ge(x_{1}y_{1}+x_{2}y_{2}+...+x_{n}y_{n})^{2}$
isn't that what people are referring to when they say "cauchy?"

Well, the way I've seen it written is $(u \cdot v)^{2}\leq{\|u\|}^{2}{\|v\|}^{2}$.

The statement "the product of the sum of the squares is greater than or equal to the square of the sum of the products" follows directly from this, by looking at this inequality in n-space and working out the dot product/length.

The triangle inequality does hold in n-space, and this can easily be shown with Cauchy-Schwarz as expressed in vector form (although this proof is useless to you, as you can't use Cauchy-Schwarz to prove the Triangle Inequality after using the Triangle Inequality to prove Cauchy-Schwarz!).

Look here:
[url]http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality[/url] :roll:

oh, so basically what I didn't know was that the Cauchy inequality is essentially the same as the triangle inequality.

[quote="daermon"]wait a minute... isn't the fact $|v|+|w|\ge|v+w|$ obvious from vector addition and the triangle inequality?
so then my proof [i]would[/i] work  :maybe:
am i missing something?[/quote]

That is the Triangle Inequality for vectors.

[quote="rjt"]you can't use Cauchy-Schwarz to prove the Triangle Inequality after using the Triangle Inequality to prove Cauchy-Schwarz!.[/quote]

A subtle point that is not always easy to grasp.  daermon:  this implies that we need to show Cauchy-Schwarz by other means to even define the concept of angles (through the dot product) in n-space.

Edit:  The algebraic expression of the Triangle Inequality in n-space is known as Minkowski's Inequality, and alternately you can prove that inequality algebraically (through induction?).

There are several things to realize:

most (meaning all I can think of, ie no exception in mind) inequalities used in algebra (not number theoretic, etc.) are a consequence of sums of squares. That is the essence of Cauchy; it is merely a heuristic tool which tells us when a nice sum of squares exists without having to do all that work.

Also, any proof of Cauchy utilizes this in some way (except possibly the interesting geometric area proof I once saw). An induction proof (used to prove many famous inequalities) are another way to hide messiness in sum of squares but that is all that is being done...

sorry if this doesn't make sense or I've said something blatantly incorrect (btw a putnam problem once asked to prove that for any polynomial always greater than zero, a sum of squares exists to prove this... basically, it is well known that all classical inequalities have a sum of squares solution)


Place 2n points arbitrarily in the plane such that no three are collinear. Color each point with one of k colors. For each color, there must be at least one point which is painted in that color. A "good line" is a line that goes through two points of the same color such that there are equal number of points in either one of the half-planes, created by the line. Show that for every positive integer 1<=k<=2n, there exists a coloring such that the number of good lines is even.  :)

no one? guess everyone is caught up with the IMO...understandable. :)

I think I have an idea though it might be a little hard to explain.

Firstly take some random good line (it is obvious that some must exist) - we will construct our set inductively. Call the endpoints of the chosen line $a_1$, $b_1$ - we have two half-planes. Take one point from the one half-plane and one from the other half-plane such that they have the property that line through them creates the smallest angle between $a_1b_1$. As no 3 points are collinear it is clearly possible - call this two ponts $a_2$ and $b_2$. Analogically name all the points as to $a_n$, $b_n$ so that there are two halfplanes - $a_i$'s belong to one while $b_i$'s to other. Easy observation leads to the fact that only $a_ib_i$ can be a good line (if colored properly) - lines $a_pb_q$ where $p \not = q$ divide set into two unequal subsets. So we have (almost  :P ) full control of how many good lines we want to have  - anyway obtaining even number of good lines is clearly possible.

[quote="Megus"]I think I have an idea though it might be a little hard to explain.

Firstly take some random good line (it is obvious that some must exist) - we will construct our set inductively. [/quote]

hmm, my confidence about your proof already dropped to 50% at the first sentence, when I made up the problem, I purposely considered a lot of cases where nothing else is possible EXCEPT to have 0 good lines, which is even. :D For a trivial example, take k = 2n, then clearly you have 0 good lines. For sufficiently large k, it is only possible to prove that we can have 0 good lines. So it's not at all obvious to me why good lines must exist.

Think about construction above in other way - you pick up not exactly good lines but "could-be" good lines - at first you don't color them. After you have all such "could be" good lines you begin coloring of endpoints. If $k$ is in range $n \leq k \leq 2n$ we can easily create no good lines - just color (using different color for each endpoint) randomily all $a_i$'s and few (if $n < k$) $b_i$ and then color remaining (if there are any) $b_i$'s such that each of $b_i$ has different color than corresponding $a_i$. For $k<n$, create $k$ good lines, if $k$ is odd, add one more (it is possible since $2k \leq 2n-2$) Remaining endpoints color in such way that corresponding $a_i$ and $b_i$ have different colors.

Does it make you more confident about my proof?   ;)


what is cyclic notation ?
i read about it in Count Down  when they were explaining Ian Le's solution when he used Jensen's  Inequality.

Cyclic notation is a shorthand for writing down the sum of a bunch of terms that "cycle" through the variables. For example, $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n$ can be written as $ \sum_{cyc} x_1$.

A more rigorous explanation is at [[Cyclic sum]].


function
probability
geometry
geometric transformation
calculus
derivative
algebra

I would like to check that I am understanding this correctly.

The Laplace Transform for the function $ g(t) \equal{} H(t\minus{}\frac{\pi}{6}) \ \sin(t\minus{}\frac{\pi}{6})$ is simply $ \frac{e^{\frac{\pi}{6}s}}{s^2\plus{}1}$, where $ H(t)$ is the Heaviside function.

Is that the correct Laplace Transform?

Thanks.

That should be $ \frac{e^{\minus{}\frac{\pi}6}s}{s^2\plus{}1}.$

Other than that, you're on the right track.

Thanks for catching that mistake!  I misread my notes.

By the way, are Laplace Transforms used for methods other than solving differential equations?  That's the only use I've found so far.

Thanks again!

There are a whole family of transforms that are closely related: Fourier transform (including the mapping from periodic functions to Fourier series coefficients and the discrete Fourier transform), the Laplace transform, the Z-transform, the moment generating function and characteristic function of probability theory, and many of the ideas surrounding generating functions.

The Laplace transform is one-sided (that is, you usually consider the interval $ [0,\infty)$) and it has provisions for incorporating initial conditions. That makes it the member of this family directed most towards solving initial value problems for ODE's.

There are a number of identities or principles that hold broadly across this family, wherever they make sense:

* Translation on one side gives multiplication by an exponential (a "character") on the other side. This is the principle you're using in this particular problem.

* Differentiation on one side gives multiplication by a variable on the other; more complicated constant-coefficient differential operators on one side give multiplication by polynomials on the other side. (In the usual way Laplace transforms are used, there's an initial condition adjustment to this.)

* Pointwise multiplication on one side gives convolution on the other side.

The whole family of transforms has a vast array of uses. Laplace transforms are the member of the family optimized for ODE uses.

My favorite Laplace transform identity: Suppose $ A$ is a square matrix. Then the Laplace transform of $ e^{tA}$ is $ (sI\minus{}A)^{\minus{}1}.$


$ x^3\plus{}px^\plus{}qx\plus{}72 polynom divided by x^2\plus{}ax\plus{}b and x^2\plus{}bx\plus{}a so find x^3\plus{}px^\plus{}qx\plus{}72 polynom root$

Restated
$ x^{3} \plus{} px^{2}\plus{}qx \plus{} 72$ polynomial is divided by $ x^{2} \plus{} ax \plus{} b$ and $ x^{2} \plus{} bx \plus{} a$
 so find$ x^{3} \plus{} px^{2} \plus{} qx \plus{} 72$ polynomial roots
Frankly, I still don't understand.
Is it
$ x^{3} \plus{} px^{2}\plus{}qx \plus{} 72$ polynomial [b]could be[/b] divided by $ x^{2} \plus{} ax \plus{} b$ and $ x^{2} \plus{} bx \plus{} a$?


function
set theory
real analysis
real analysis solved

Since nobody has answered for a month and since it is a good problem, I'll post a
[hide="hint"]
The answer is "Yes".
[/hide]

It seems to me that I found a solution, if I understand the problem correctly.

First of all, denote by $ \Gamma $ the set of dyadic rationals, and by $ \mathbb{Q} $
the set of all rationals. Then from a theorem in set theory, there exists a bijection
$ \mathbb{Q} \rightarrow \Gamma $, which is monotonic increasing. Then, it is easy to
show, that this function extends to a continuous function $ \mathbb{R} \rightarrow
\mathbb{R} $. Therefore it is enough to solve the problem, taking instead set
of rationals $ \mathbb{Q} $, the set of dyadic rationals $ \Gamma $ ( just after finding
the desired sequence of continuous functions for $ \Gamma $, compose them with the function
we have found).

Now, denote by $ <x> := dist(x, \mathbb{Z}) $ for $ x \in \mathbb{R} $. Then define
$ f(x):= max(1 - 4<x>, 0) $. Then $ f $ is of course a continuous function
$ \mathbb{R} \rightarrow \mathbb{R} $, $ f(x) = 1 $ for $ x \in \mathbb{Z} $, and
$ f(x) > 0 \Leftrightarrow dist(x, \mathbb{Z}) < \frac{1}{4} $. Now define
$ f_{n}(x) := nf(2^{n}x) $ for $ n \in \mathbb{N} $.
Then of course for every $ x \in \Gamma $, we have for $ n \in \mathbb{N} $, that
$ 2^{n}x \in \mathbb{Z} $, so for $ n $ large enough, we have $ f_{n}(x) = n $, so
$ lim_{n\rightarrow \infty} f_{n}(x) = \infty $. Now suppose that we have a number
$ x \in \mathbb{R} $, and $ lim_{n\rightarrow \infty} f_{n}(x) = \infty $.
Then in particular there exists $ n_{0} $, such that for every $ n \geqslant n_{0} $,
we have $ f_{n}(x) > 0 $, so $ f(2^{n}x) > 0 $, so for $ n \geqslant n_{0} $ we have that
$ dist(2^{n}x, \mathbb{Z}) < \frac{1}{4} $. Now it is easy to see that if we have
a real number $ y \in \mathbb{R} $, such that $ dist(y, \mathbb{Z}) < \frac{1}{4} $,
then $ dist(2y, \mathbb{Z}) = 2dist(y, \mathbb{Z}) $. Therefore for $ n \geqslant n_{0} $
we have that $ dist(2^{n+1}x, \mathbb{Z}) = 2 dist(2^{n}x, \mathbb{Z}) $, but on the
other hand, all $ dist(2^{n}x, \mathbb{Z}) $ are bounded by $ \frac{1}{2} $, so the only
conclusion is that $ dist(2^{n_{0}}x, \mathbb{Z})  = 0 $, so $ 2^{n_{0}}x \in \mathbb{Z} $,
so $ x \in \Gamma $. Therefore we see that $ lim_{n\rightarrow \infty} f_{n}(x) = \infty $
if and only if $ x \in \Gamma $.

:(  :(  :(  :(
All my work is vain.

The complete answer for this question is in http://www.mathlinks.ro/Forum/viewtopic.php?t=22177

Ops... I solved [i]slighly[/i] another problem in that topic.

I feel Leva's proof is very illuminating to say the least....

I have erased everything I had written, because it was stupid :(.

[quote="Leva1980"]Myth, what another problem did you solved?[/quote]
Follow the link I wrote above.
I proved that if $f_n(x)$ is unbounded for all rationals, then there is irrational $s$ s.t. $f_n(s)$ is unbounded too. :?

[quote="fleeting_guest"][quote="eugene"]Does there exist the sequence of continious real functions $f_{n}(x)$ s.t. $f_{n}(x)\to+\infty$ ($n\to\infty$) at rational points $x$ only.[/quote]

For an enumeration of rationals $q_i$, let $f_n(x)$ be the n'th partial sum of $a_1|x - q_1| + a_2|(x-q_1)(x-q_2)| + a_3|(x-q_1)(x-q_2)(x-q_3)| + ...$.
It is possible to choose $a_i$ and the enumeration to guarantee that the sum is infinite at irrational points.[/quote]
Why did you said that?

[quote="eugene"]Does there exist the sequence of continious real functions $f_{n}(x)$ s.t. $f_{n}(x)\to+\infty$ ($n\to\infty$) at rational points $x$ only.[/quote]

I think the following works as well:

If $(q_n)_{n\ge 1}$ is an enumeration of the rationals, we take $f_n$ to be $n$ in $q_1,q_2,\ldots,q_n$, and $0$ outside $A_n=\bigcup_{k=1}^n(q_k-\varepsilon_n,q_k+\varepsilon_n)\ (*)$. If $\varepsilon_n>0$ is small enough s.t. the intervals in $(*)$ are disjoint for every $n$, and $(\varepsilon_n)_n$ decreases (to $0$) fast enough s.t. $A_n\cap A_{n+1}=A_{n+1}\setminus(q_{n+1}-\varepsilon_{n+1},q_{n+1}+\varepsilon_{n+1}),\ \forall n$, then every irrational will lie outside infinitely many $A_n$'s:

Let $t\in\mathbb R$ and $n_0$ is large enough s.t. $t\in A_n,\ \forall n\ge n_0$. Since $t\in A_n$, it lies in some $(q_k-\varepsilon_n,q_k+\varepsilon_n)$. $t\in A_{n+1}$ as well, so, from the conditions imposed on the $A_n$'s, we must have $t\in(q_k-\varepsilon_{n+1},q_k+\varepsilon_{n+1})$ (the same $k$ as before). We repeat this for $n\to\infty$, and we get $t=q_k$ because $\varepsilon_n\to 0$, so $t$ is rational.
Is there anything wrong with this? :?


logarithms
number theory unsolved
number theory


Mods, can this be put to Intermediate or High School Basics?


For those of you who don't know, Bezout's Lemma states that for any two integers $ a$ and $ b$, there exist integers $ m$ and $ n$ such that $ am \plus{} bn \equal{} (a, b)$, where $ (a, b)$ denotes the greatest common factor of $ a$ and $ b$.

How can we show that there exist infinitely many ordered pairs $ (m, n)$ satisfying $ am \plus{} bn \equal{} (a, b)$?

$ am \plus{} bn \equal{} a(m \plus{} bk) \plus{} b(n \minus{} ak)$.

Thanks. Part two:

Let's say $ (m, n)$ is one solution to the equation. Then do all other solutions $ m'$ and $ n'$ necessarily take the form $ m \plus{} bk$ and $ n \minus{} ak$ for some integers $ k$? If so, prove it. If not, give a counterxample.

Thats incorrect, but it is true if you write $ m\plus{}\frac{b}{(a,b)}\cdot k$, etc

We prove this by saying that $ (m,n)$ is a solution so $ am\plus{}bn\equal{}(a,b)$.

Then suppose that $ (x\plus{}m,y\plus{}n)$ is another solution...

$ a(x\plus{}m)\plus{}b(y\plus{}n)\equal{}(a,b)$
$ ax\plus{}by\plus{}(am\plus{}bn\minus{}(a,b))\equal{}0$
$ ax\equal{}\minus{}by$.

Then it is easy to see that $ \frac{b}{(a,b)}$ must divide $ x$, and such...


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14780[/img]

 :(

Let's solved with your figure.
Let M,N be on AC such that AM=MN=NC.
S(AHDB)=S(ABM)=S(ABC)/3, so S(BDH)=S(BMH), thus BH is parallel to DM. H have just been constructed.
The line through N and parallel to BH meets BC at I. S(IHN)=S(IBN), so S(IHC)=S(BNC)=S(ABC)/3=S(CEKH), which follows that S(IHE)=S(KHE), hence IK is parallel to EH.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14791[/img]

[b]"The line through N and parallel to BH meets BC at I" not correct

"The line through N and parallel to AB meets BC at I"   correct[/b]

 :thumbup:


I have trouble understanding this... (excerpted from The Art and Craft of Problem Solving by Paul Zeitz, pg. 224 - 225)

[quote](d)If $ g|a$ and $ g|b$, then $ g|ax \plus{} by$, where x and y can be any integers. WE call a quantity like ax + by a linear combination of a and b. [...] 7.1.5 An important consequence of the division algorithm, plus 7.1.3 (d), is that [i]The greatest common divisor of a and b is the smallest positive linear combination of a and b.[/i] [...] Let's prove 7.1.5; it is rather dry but a great showcase for the use of the extreme principle pls arguement by contradiction. Define u to be the smallest positive linear combination of a and b and let g := (a,b). We know from 7.1.3 (d) that g divides any linear combination of a and b, and so certainly g divides u. This means that $ g \le u$. We would like to show that in fact, $ g\equal{}u$. We will do this by showing that /u is a common divisor of a and b. Since u is greater than or equal to g, and g is the greatest common divisor of a and b, this would force g and u to be equal.
Suppose, on the contrary, that u is not a common divisor of a and b. Without loss of generality, suppose that u does not divide a. Then yb the division algorithm, there exists a quotient $ k /ge 1$ and the positive remainder $ r < u$ such that \[ a \equal{} ku \plus{} r\]
But then $ r \equal{} a \minus{} ku$ is positive and less than u. [b]This is a contradiction, because r is also a linear combination of a and b, yet u was defined to be the smallest positive linear combination![/b] Consequently, u divides a, and likewise u divides b. So u is a common divisor; thus $ u \equal{} g$. [/quote]

Can anyone please explain the bolded part - namely, "This is a contradiction, because r is also a linear combination of a and b, yet u was defined to be the smallest positive linear combination!"

I mean, can anyone explain in detail how r is a linear combination of a and b???????

I thank you for your time and patience.

just for purposes of clarification: I understand everything up to the bolded point. Thanks.

$ u$ is a linear combination of $ a,b$; hence so is $ a \minus{} ku$.

sorry, but apparently i can't think striaght today for some reason  :(

Can you explain in detail? (some algebra would be nice)

$ u$ is a linear combination of $ a,b$; hence we may write $ u \equal{} ax \plus{} by$, where $ x,y$ are integers. Then $ r \equal{} a \minus{} ku \equal{} a \minus{} k(ax \plus{} by) \equal{} (1 \minus{} kx)a \plus{} (\minus{}ky)b$, where $ 1\minus{}kx, \minus{}ky$ are obviously integers.


geometry
3D geometry
LaTeX
Olimpiada de matematicas

[color=green]disculpen la molestia, si alguien me pudiera proporcionar los enunciados de la olimpiada de mayo 2005 y 2006 se los agradeceria mucho, he pasado miles de horas buscando en internet y he encontrado hasta la 2004, y los estoy necesitando ahorita. gracias de antemano x su ayuda![/color] :)

[color=blue]Ya me pidieron también algunos chicos, y en ningún lado se consiguen, así como los enunciados de la XVI Olimpiada Matemática del Cono Sur.
¿Quién podría explicar el motivo?[/color]

aqui tem tanto da cone sul, como da olimpiada de maio, mas está em português!  :blush:

http://www.obm.org.br/frameset-provas.htm

muchas gracias a todos x su ayuda :lol: se los agradezco mucho, la verdad q estos examenes no son muy faciles de encontrar :)

Buenas tardes,

aki tengo las fuentes de latex q generaron este archivo ..

claro si alguien las desea para editar otros exams de este formato ..

saludos,

[color=blue]Aviso que mis "discípulos" realizarán el examen de la mencionada olimpiada este sábado 11 de mayo.
Es bien sabido que está prohibido divulgar hasta el mes siguiente.[/color]

Buenas noches,

eso es cierto!!

No publicar los exams de Mayo no al menos dentro de 1 mes

Gracias!!

[quote="tipe"]Para ser más exactos no se puede divulgar las pruebas hasta el 25 de Mayo.

$Tipe$[/quote]

Perfecto Jorge,

Gracias por la aclaracion...

Saludos,

[quote="tipe"]Para ser más exactos no se puede divulgar las pruebas hasta el 25 de Mayo.

$Tipe$[/quote]

Ya es 26 de mayo :P , ¿alguien tiene los enunciados? Gracias y saludos!

yo los tengo  :)  solo q para tener un poco mas de orden mejor los voy a poner en topicos separados! en un momento los subo


I was doing team round problems, I could do all but 7 and 8, could some 1 help?

http://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_Team.pdf

Edit: I found out 7

Next time, please post the problems

[hide="hint for number 8"]
Use Pythagorean Theorem (or memorize some triples) to find AG, and then AG=AC because they are both radii and then figure it out from there
[/hide]

Ok so um heres some hints.
For number 7, use the fact that the radius of the circle is the same as a line from the center of the hexagon to one of it's vertexes (Plural form.... :| ), which means that it is also the length of one of the sides of the hexagon.
For number eight, just use the Pythagorean Theorem to find out the radius of the circle, and subtract 7 from it.


college
geometry
Stanford
Berkeley Math Circle

Hi,
I'm at UC Berkeley right now for the summer (until august 14) for the summer college program.
I'm a high school student, and I'm hoping that I can get involved in some math clubs or activities nearby.
If there's any program or meeting held during the summer near the UC berkeley campus, and if it's not too late, can anyone provide me with information regarding nearby math circles etc.?
(My goal for this is to prepare for upcoming olympiads next year.)

Thanks

David Lee

I'm also going to be around UC Berkeley for the summer. There's Berkeley Math Circle on campus, but that won't start until September.

The only math circle in the area that has summer sessions is the Stanford Math Circle http://www.stanfordmathcircle.org/Main_Schedule.html It's from 7-9 on Tuesday nights - It's fairly far from Berkeley - if you want to get there by public transportation you'll have to take BART then transfer to Caltrain. Then its a walk to campus.

Sounds fun, Looks fun, Seems fun!
But, yea.. I don't think I'll be able to make it..
Still, I'll try my best to get there if I can, if I have time and no writing homework :D


I want some easy inversion problem,if anybody have plz send. :D

[quote="hucht"]Have you proved that The inverse of a line that doesn't pass by the inversion center is a circunferencia thought the center?[/quote]
of course. :D

A problem of mine is a example of using inversion to solve problems easily.

See

http://www.mathlinks.ro/Forum/viewtopic.php?t=55206

http://www.mathlinks.ro/Forum/viewtopic.php?t=14958
http://www.mathlinks.ro/Forum/viewtopic.php?t=56658

  Darij


vector
function
geometry
parallelogram
search
linear algebra

A norm on a vector space U is a function from U to nonnegative real such that ||u||>=0, ||au|| = |a|||u|| for all a in F and all u in U, and ||u+v|| is less than or equals to ||u||+||v|| for all u,v in U. Prove that a norm satisfying the parallelogram equality comes from an inner product(that there exists an inner product < , > that ||u||=$ < u,u > ^\frac {1}{2}$ for all u in U)

[quote="t0rajir0u"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1961927883&t=224293]here[/url].[/quote]

I got confused, the problem want to show that the function is the norm of an inner product space, but the first step of the proof just say $ {\parallel{}x\plus{}y\parallel{}}^2\equal{}<x\plus{}y,x\plus{}y>$, which I think only holds if the function is  the norm of an inner product space. Is this a logical error?

You're reading the .pdf file downloaded from that, right? What you're quoting is from a "discovery" phase of the project: suppose there were a valid inner product. How would it be related to the norms?

The proof then starts about halfway down the page, with the statement (at least for the real case) that "Our goal is to show that $ \langle x,y\rangle\equal{}\frac{\|x\plus{}y\|^2\minus{}\|x\minus{}y\|^2}{4}$ does indeed define an inner product."

I'm sorry, but I'm not able to prove the continuity of <ax,z> for a.(last step for proving real case) Can anyone help?

At this point we've already proved the linearity for all real and homogeneity for all rational. So we can use these results.

The last step is Use Parallelogram identity to prove that , for any x,z, the function a - <ax,z> is a continuous function of a. (a is real)

Help me please :maybe:


Let n be a natural number grater or equal to 2. Determine the numbers a_k naturals and different from 0, k={1,2,...,n}, distinct two by two, knowing that
$ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}\dfrac{n^2\plus{}n\plus{}2}{2}$

[quote="St. Elena"]Let n be a natural number grater or equal to 2. Determine the numbers a_k naturals and different from 0, k={1,2,...,n}, distinct two by two, knowing that
$ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} \dfrac{n^2 \plus{} n \plus{} 2}{2}$[/quote]

$ a_i\equal{}i$ $ \forall i\in[1,n\minus{}1]$
$ a_n\equal{}n\plus{}1$


I bought Barron's SAT II Math IIC book and like people said, it was much harder than some of the practice tests I took. But how much harder is it? Will questions like: what is the 3rd term of (8x-y)^(1/3) be in the test?

Thanks for any answer.

Ok let's put it this way. Under practice-test (meaning extremely distracted) conditions, not bothering to double-check, etc, I got a best score of 45 out of 50 on the Barron's practice tests (I took all 6).

I went into the test thinking I would get a 750 at best.

I got an 800 and thought it was extremely easy.

Barron's is notorious for making their prep books harder than the real test to "over-prepare" you because they assume that there's no way that you can know all the material in a review book anyways.

I find the practice tests in the SparkNotes Math 1C and 2C books to be pretty close in level to the real thing.

Out of curiosity, how do you have a third term for $\sqrt[3]{8x-y}$?

haha, i think he probably meant to the third power, not third root.

It has some errors, but they aren't too much of a burden. However, it doesn't parallel the actual test at all - nowhere on the test you'll have to provide multiple answers for a trigonometric equation, for example. It's actually nice for improving one's knowledge of Pre-Calc concepts.

Hmmm I always thought those books were routine, but I guess I did learn Algebra from Barron's "Algebra the Easy Way."  I have to say from all that I've seen/heard, Barron has some good books.

[quote="eryaman"][quote="Ravi B"]Or it could have been referring to the infinite [url=http://mathworld.wolfram.com/BinomialSeries.html]binomial series[/url], which is indeed too advanced for SAT II math.[/quote]Wow I never knew that.[/quote]

Discovered by Newton. :lol:

Don't stress too much over the IIC.  It's not that hard.

best of luck to everybody including my brother (he's taking bio) :)

[quote="bubala"]If you don't make careless errors--read the question--and go at the right pace, then you will get an 800 easily.[/quote]

I'm sure that there're a lot of people who do those things, but don't get an 800. What's easy to you may not be easy to everyone else. You have to respect people's individuality.

well, I think that most, if not all people, on THIS FORUM in particular, are capable of getting an 800 or close to that.

I wouldn't say something like that to all my friends in school, for example.

It depends on the environment.

However, I apologize for such comments because the above statement may not indeed be true.

I took the IIC on May 7th and scored an 800....it was extremely easy, and I'm not surprised that 800 is only 90th percentile.

If you are in even the second highest level of 11th grade math offered at your school, this test should be a breeze, no book required

[quote="ffdbzathf"]im taking it this saturday, june 4th.  i've been using the princeton review though.[/quote]

i used the princeton review book too.  i just toke it 1hr and 15mins ago :D

[quote="anhnguyen"]What is the lowest raw score that guarantees a 800?[/quote]

you can miss like 5 or 6 and still get a 800 on IIC

Each administration of the test is scaled differently, so the number of questions that equals an 800 will vary from administration to administration.

So it's been concluded that a question like the oneproposed in the topic will not be present in the SAT II Math IIC?

If not, how would you work it out? I saw the link, but something doesn't sink in unless I see it being done.

No it will not be in the SAT II. I asked the question and I just took the test this morning. It was much easier than Barron's. Don't worry about those types of questions. They are too advance to be in the test.


AMC 1976 #30.

How many distinct ordered triples (x,y,z) satisfy the equations
x+2y+4z=12
xy + 4yz + 2xz = 22
xyz = 6

Show your work.

[hide] x+2y+4z=12, 2xy+8yz+4xz=44, and x(2y)(4z)=48, so polynomial with roots x, 2y, and 4z is x^3 -12x^2 +44x +48= 0[/hide]

more later gotta run.

Since x+2y+4z=12, xy+4yz=12y-2y∧2
Since xy+4yz+2xz=22, xy+4yz=22-2xz
So 12y-2y∧2=22-2xz
	6y-y∧2=11-xz
	y∧2-6y=9=xz-2
	(y-3) ∧2=xz-2	since xyz=6
so	(y-3) ∧2=6/y-2
	y∧3-6y∧2+9y=6-2y
	y∧3-6y∧2+11y-6=0
	y∧3-y-(6y∧2-12y+6)=0
	(y-1)(y-2)(y-3)=0
so y can be 1, 2, or 3
substitute the values of y into the original equation and get the answer


[color=blue][b] Problem.[/b]

  Let $a, b, c\in (0,1)$. Prove that the following inequality holds true:

                        \[ \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq\frac{3\sqrt [3] {abc}}{1-\sqrt [3] {abc}}. \]

  [/color]

$\\\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\\ =\sum_{k=1}^{\infty}(a^k+b^k+c^k)\geq 3\sum_{k=1}^{\infty}\sqrt[3]{abc}^k$
It is suffiient enough, due to symmetry, to prove that $a^k+b^k+c^k\geq 3\sqrt[3]{abc}^k$, which is true by AM-GM.
Q.E.D.

Masoud Zargar

[quote]$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c} =\sum_{k=1}^{\infty}(a^k+b^k+c^k)$[/quote]
Can you please tell me why this is true?

I think I also have a proof for the inequality:

By Chebyshev inequality, we obtain: \[ \displaystyle\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c} \geq \frac{a+b+c}{3}.\left(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\right) \]

By AM-HM and AM-GM, we know that \[ \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \ge \frac{9}{3-(a+b+c)} \ge \frac{3}{1-\sqrt[3]{abc}} \]
Using this, we get \[ \displaystyle\frac{a+b+c}{3}.\left(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\right) \ge \frac{a+b+c}{1-\sqrt[3]{abc}} \ge \frac{3\sqrt[3]{abc}}{1-\sqrt[3]{abc}} \]

$\mathbb{QED}$

[quote="Jan"][quote]$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c} =\sum_{k=1}^{\infty}(a^k+b^k+c^k)$[/quote]
Can you please tell me why this is true?
[/quote]
He used $\frac{1}{1-x}=1+x+x^2+x^3+\cdots =\sum_{k=0}^\infty x^k$ for any real number $\in (0,1)$. Hmm... strange solution. :)

Well, since $a,b,c\in (0,1)$ and we have $\frac{a}{1-a}$ and such terms, I thought of using geometric series; $\sum_{k=1}^{\infty} x^k=\frac{x}{1-x}$ :)

[hide]Divide out the numerators: $\frac{a}{1-a}=\frac{1}{1-a}-1$ (similar for the others) and the right side is $\frac{3}{1-\sqrt[3]{abc}}-3$.

So this inequality is equivalent to $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\geq \frac{3}{1-\sqrt[3]{abc}}$.

By AM-HM, $\frac{\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}}{3}\geq \frac{3}{(1-a)+(1-b)+(1-c)}$.

This becomes $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\geq \frac{9}{3-(a+b+c)}=\frac{3}{1-\frac{a+b+c}{3}}$.

Now, from AM-GM, $\frac{a+b+c}{3}\geq \sqrt[3]{abc}\Rightarrow 1-\frac{a+b+c}{3}\leq 1-\sqrt[3]{abc}\Rightarrow \frac{3}{1-\frac{a+b+c}{3}}\geq \frac{3}{1-\sqrt[3]{abc}}$

(where the last inequality is true because $a,\ b,\ c\in (0,1)\Rightarrow \frac{a+b+c}{3},\ \sqrt[3]{abc}<1$.

Since $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\geq \frac{3}{1-\frac{a+b+c}{3}}\geq \frac{3}{1-\sqrt[3]{abc}}$, the inequality is true.[/hide]


Esmeralda has a bottle with 9 liters of a mixture that has 50% alcohol and 50% water. She wants to put water in the bottle so that only 30% of the mixture is alcohol. How many gallons of water will put it?

Portuguese question (if necessary):
Esmeralda tem uma garrafa com 9 litros de uma mistura que tem 50% de álcool e 50% de água. Ela quer colocar água na garrafa de tal forma que apenas 30% da mistura seja de álcool. Quantos litros de água ela irá colocar?

How to do ?  :|

We see there are 9 liters of the mixture and half of it is water, so there's $ 4.5$ liters of water.

Let $ x$ be the amount of water in liters to produce a 30% alcohol (70% water) solution.

$ \frac{4.5\plus{}x}{9\plus{}x}\equal{}\frac{7}{10}$

Cross multiplying givexs

$ 45\plus{}10x\equal{}63\plus{}7x$
$ 3x\equal{}18$
$ x\equal{}6$

So $ 6$ liters, or $ \frac{6}{3.78541178} \approx \boxed{1.585}$ gallons need to be added.

Bom dia luiseduardo.

Suppose she puts in $ x$ litres of water. Then she has $ 4.5$ litres of alcohol and $ 4.5 \plus{} x$ litres of water.

$ \frac{4.5}{9\plus{}x} \equal{} \frac{3}{10}$

$ 27 \plus{} 3x \equal{} 45$

$ x \equal{} 6$

Hello Andrew and Gauss, Thank you very much ;)


this is similar to an USAMO problem i did

Prove that:

$a^ab^bc^c \ge a^bb^cc^a$

for $a,b,c >0$

it seems kinda like rearrangement.
if u know the prove of rearrangment, here is the idea:

$a\ge b$
$a^{a-b}\ge b^{a-b}$
$a^a\cdot b^b\ge a^b\cdot b^a$

;) im 2 lazy to write up the whole prove but u only have to look at 2 cases because WLOG $a\ge b\ge c$ or $a\le b\le c$ so u can do the thing above first with $a$ and $b$ and then with $b$ and $c$

If you take the log of both sides, it is direct rearrangement.  Actually if you apply your inequality cyclically you get the USAMO inequality.

Here is a nice proof of the USAMO inequality
It comes from John Scholes -his solution to Question 2 Canadian MO 1995.

Assume  a>= b >= c.

Then a/c, a/b  and b/c  are all at least 1 and

(a-c)/3 , (a-b)/3  and (b-c)/3  are all positive.

Therefore  (a/c) ^([a-c]/3) ,  (a/b)^([a-b]/3)  and ( b/c)^([b-c]/3) are all at least 1.
Hence their product is at least 1, which leads to the required inequality.

[quote="gotztahbeazn"]heh nice job

this is pretty similar to the USAMO problem:

$a^ab^bc^c \ge (abc)^{(a+b+c)/3}$

for $a,b,c>0$

i think thats the problem at least[/quote]
I got the generialization from this USAMO inequality.  :P
http://www.mathlinks.ro/Forum/viewtopic.php?t=46247


superior algebra
superior algebra unsolved

Prove that a left annihilitor of a right ideal of a ring $ R$ with unity is generated by an idempotent

I have no idea what you could mean,
but take $ R\equal{}\mathbb{Z}/4\mathbb{Z}$ and ideal $ I\equal{}2\mathbb{Z}/4\mathbb{Z}$.


1. A man had a book that was worth $\$40,000$. He threw it in the furnace, reducing it to a pile of soot. Why did he do this?
[hide="hints"]
A) He is not mentally disordered or retarded. He did this for a very good reason.
B) He does care about money.
[/hide]
2) The police arive at a house. There is a fresh corpse on the floor. Other than that, there is only a tape recorder in the room. When the police played it, they heard this: "I can't go on. I have nothing to live for" After this, they hear a gunshot. How did they know the person had been [i]murdered?[/i]

My educated guess for the 2nd situation:
[hide]Those quoted words are the words of the killer.  He recorded them (his own words) and the gunshot's sound on purpose.  He turned off the tape recorder right after that so that sounds following that would not be heard on the playing of the tape.[/hide]

not quite. Youre on the right track though.

[hide="My Thoughts on the situation"][hide="1)"]He did not care about money, so he just used the book to burn and keep himself warm.[/hide][hide="2)"]The question reads [quote="SoxFan34"]How did they know the person had been killed?[/quote]Uhhh...maybe because he's lying there dead? :rotfl: [/hide][/hide]

[quote="kstan013"][hide="My Thoughts on the situation"][hide="2)"]The question reads [quote="SoxFan34"]How did they know the person had been killed?[/quote]Uhhh...maybe because he's lying there dead? :rotfl: [/hide][/hide][/quote]

nice thought on #2 :rotfl: ...but the person can be dead even if it was suicide. :ninja:

[hide="2"]
If it had been a suicide, the tape would never have stopped. The fact that it was not only stopped but also [b]rewound[/b] indicates that someone else was involved.
[/hide]

Contributing to the first one:

[hide] Perhaps there were only two copies of the book in the entire world, and he had both...so by destroying one, he made the other significantly more valuable?[/hide]
Or
[hide]In the form of soot, it turns out to be more valuable than in its original form[/hide]

[hide="1"]the book was worth and insured for 40,000 dollars but he payed a lot less for it[/hide]

I made a topic exactly like number 2 with the tape recorder[/hide]

[hide]

K81o7's first solution is right; the man destroys it because he has two, and by only having one, the value goes way up.
Kstan, mathmeister22, youre both wrong, but that's partly my fault... :rotfl:   Mathnerd is right: If he'd commited suicide, then the tape recorder wouldve kep going. It was Rewound. [/hide]


I took the AMC 8 on tues. and I was wondering when we get our "Offical" scores. Does anyone know?

My teacher says that our official scores don't come out until like Late January/Early February.

Oh they come out normally around Dec. 10th.

The AMC office will begin mailing official scores and reports in early to mid-December, precise start of mailing depending on the scoring and processing.  Most schools in the US should have results before the end-of-year holidays.

Please also read Section VI, Report of Results, page 5 in the 2009 AMC 8 Teachers' Manual,
http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC8/2009-amc8TM.pdf

Steve Dunbar
Director, MAA American Mathematics Competitions

Since the teachers are allowed to open the solutions on November 25, and they can keep the actual test booklet, aren't we able to know unofficial results as early as next week? I remember circling every answer on the test booklet, so that should be enough, right?

yes
when I took the AMC8, my math coordinator checked the answers on our test booklets with the official answers, so we had the unofficial results within a week

[quote="AlphaBetaTheta"]is that even allowed?[/quote]

Yes, if you only discuss within the school. My teacher did that too.

[quote]is that even allowed?[/quote]

Teachers' Manual, Section VI, Item 4. page 5.  Pay attention to items 4.a. and 4.b. too.

Steve Dunbar
Director, MAA American Mathematics Competitions

aren't the teachers unable to discuss the answers until next week.
i mean she said that she couldn't tell us the answers yet because other people take it at different times.

i'm not certain, but don't all competitors take the test at the same time?
that's how we did it at my school, anyway

there is an AMC8B that some people may take, but it should be different enough from AMC8A that discussion of the answers should not affect the score of those taking AMC8B that much

I'm pretty sure they are the same test, but there is an alternate date for taking it or something like that, in case of conflicts.

really?
well, idk
my coordinator never ordered the second one =)

i would think there would be some difference tho
otherwise, couldn't they just give one test with a range of dates to administer them?

Go to the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC8/2009-10-AMC8-bro.pdf]AMC8 brochure[/url], the second page, at the top it says "Administration". That explains it a little.

How it works is there is a single test that is mailed to all the proctors. They are supposed to open them for the first time and give the tests to the students on Tuesday, November 17. If a school has a conflict or problem with that date, they can request special permission to administer the test on another date. Solutions (which are mailed to proctors with the tests) should not be opened until November 25, when everybody should have taken the test. Also, students may not discuss the test with anyone but the people who took the test at the same place and time (you signed something at the bottom of the answer sheet pledging that). On November 24th, the proctors mail the answer sheets back to the AMC headquarters, where the official grading is done. Since the actual question pamphlets don't have to be mailed, if the person who took the test circled all the answers he/she had in the test pamphlet, the proctor can compare the solutions with the circled answers and give an unofficial grade. The only time when this is different than the official grade is if you bubbled incorrectly or your proctor graded incorrectly.


quadratics
algebra
quadratic formula
special factorizations

I need help with the steps turning the standard form into the quadratic formula can someone please help me with the proof and the steps involved. Thanks

One way to prove the quadratic formula by completing the square - since the general composition of a quadratic equation is of the form $ ax^2 + bx + c$, we can divide both steps of the equation by $ a.$

So far we have:

$ ax^2 + bx + c = 0$
(Note that $ a\neq0.$)
$ x^2 + \dfrac{bx}{a} + \dfrac{c}{a} = 0$
$ x^2 + \dfrac{bx}{a} = - \dfrac{c}{a}$
Completing the square, we add $ \dfrac{b^2}{4a^2}$ to both sides of the equation.
We now have the following:
$ x^2 + \dfrac{bx}{a} + \dfrac{b^2}{4a^2} = - \dfrac{c}{a} + \dfrac{b^2}{4a^2}$.
$ (x + \dfrac{b}{2a})^2 = \dfrac{ - 4ac}{4a^2} + \dfrac{b^2}{4a^2}$.
$ (x + \dfrac{b}{2a})^2 = \dfrac{b^2 - 4ac}{4a^2}$
Therefore,
$ x + \dfrac{b}{2a} = \pm \sqrt {\dfrac{b^2 - 4ac}{4a^2}}$
$ x + \dfrac{b}{2a} = \pm \dfrac{\sqrt {b^2 - 4ac}}{2a}$
$ x = - \dfrac{b}{2a}\pm \dfrac{\sqrt {b^2 - 4ac}}{2a}$
Adding the two like terms we get $ \dfrac{ - b\pm \sqrt {b^2 - 4ac}}{2a}$.

Did you understand my steps?  If you need any help, just ask.

Of course though, $ a\neq0.$ Otherwise the expression wouldn't be a quadratic and you'd be dividing by 0. The rest is okay.

You should have also explained why the step $ \sqrt {4a^2} \equal{} 4a$ is valid. Remember that in general $ \sqrt {x^2} \equal{} |x|$. Using a bit of casework ($ a < 0$ and $ a > 0$),you'll notice that regardless of the sign of $ a$, the final expression will ultimately remain the same.

Which step are you talking about?  $ \sqrt{4a^2} \neq 4a$.

He meant $ \sqrt{4a^2}\equal{}2a$. Also, remember that since were working with variables, $ a$ could be negative.