What is the value of $ a$ if $ 3^a2^a5^a \equal{} 0.0081$?

hello, we can write $ 30^a\equal{}\frac{81}{10000}\equal{}\frac{3^4}{10^4}$, taking the logarithm of both sides we have $ a\equal{}\frac{4\ln(3)\minus{}4\ln(10)}{\ln(30)}$.
Sonnhard.

This was a mathcounts question...logarithms are DEFINITELY not needed here.

hello, then solve this problem without logarithm, please. Are calculators allowed?
Sonnhard.

This was a chapter target.  Thus, you [b]were[/b] allowed to use a calculator.

There must be either:

1) Some intutively obvious way that we all missed.
2) A mistake in the problem.

hello, your topic (2) is the reason why i have solved this problem with logarithms. Topic (1) is meaningless.
Sonnhard.

I think it's 2, since the given number is 3^4*2^-4*5^-4.

Dr Sonnhard Graubner,

Did you get -4?
_________________
~pi man!! :rotfl:

According to ftw, the answer is -4.  How is that possible? 1/30^4 =1/8100!

hello, the solution of the equation
$30^{a}=\frac{81}{10000}$ is definitively
$a=\frac{\ln\left(\frac{81}{10000}\right)}{\ln(30)}$
Sonnhard.

Yeah I've seen this problem come up and I do believe the "FTW" answer is -4, which we know is grossly inaccurate.

[quote="ernie"]According to my calculator, it is $ \boxed{\approx \minus{}1.4159}$.[/quote]

According to my calculator and wolfram alpha, this is correct.


1) Prove that if $a|b$, then $2^{a}-1 | 2^{b}-1$. (What about the converse?)

2) Prove that if $2a|b$, then $2^{a}+1|2^{b}-1$. (What about the converse?)

[quote="Phelpedo"]1) Prove that if $a|b$, then $2^{a}-1 | 2^{b}-1$. (What about the converse?)

2) Prove that if $a|b$, then $2^{a}+1|2^{b}+1$. (What about the converse?)[/quote]
[hide="1)"]Notice that $2^{n}-1$ is a number built of $n$ ones in the binary system. Now we clearly see that 1) is true, for example $111_{(2)}\mid 111111111_{(2)}$ (the quotient is $1001001$). Also the converse is true, because $\underbrace{1\ldots 1}_{b}\equiv \underbrace{1\ldots 1}_{b\mod a}\pmod{\underbrace{1\ldots 1}_{a}}$ and $\underbrace{1\ldots 1}_{a}$ obviously does not divide $\underbrace{1\ldots 1}_{a'}$ for $1<a'<a$.
The solution of this problem is vey similar to the solution to problem 12 form 1st round of LVI Polish MO. The problem was
[color=blue][i]Given functions \[f(x)=2^{x}\text{ and }g(x)=f(f(f(f(f(f(f(x)))))))\] ($g$ is the seventh iteration of $f$). It it true that $g(2)-g(0)\mid g(3)-g(0)$?
[/i][/color][/hide]

Rzeszut, you proved 1) but kept referring to it as 2).

Sorry, it was a typo  :blush: . I will correct it.

Should be a - sign in the second half of 2. Fixed.

That's a cool solution, Rzescut. I can give another one for 1):
[hide]Let $b=ka$. Then $2^{b}-1 = 2{}^{k}{}^{a}a-1 = (2^{a})^{k}-1 = (2^{a}-1)((2^{a})^{k-1}+...+1)$

So now what about number 2 :lol: [/hide]

[quote="Phelpedo"]2) Prove that if $2a|b$, then $2^{a}+1|2^{b}-1$. (What about the converse?)[/quote]
[hide]
$\left(2^{a}+1\right)\left(2^{a}-1\right)=\left(2^{2a}-1\right)$
Set $k=2a$, then the problem becomes similar to 1).
[/hide]


What is the remainder when  $ 5^{81}$ is divided by 11.

(full solution plz)

[hide="A Possible Solution"]
We can see that
$ 5^{3}\equiv{4}\pmod{11}$
Now we can rewrite our expression as
$ (5^{3})^{27}\equiv{x}\pmod{11}$
Plug in the first congruency to get
$ 4^{27}\equiv{x}\pmod{11}$
Rewrite as
$ 2^{54}\equiv{x}\pmod{11}$
Observe that
$ 2^{5}\equiv{\minus{}1}\pmod{11}$
Rewrite our earlier congruency to get
$ (2^{5})^{10}*2^{4}\equiv{x}\pmod{11}$
Which finally simplifies to
$ 1*2^{4}\equiv{2^{4}}\equiv{16}\equiv{x}\pmod{11}$

Thus, the remainder is $ \boxed{5}$.[/hide]

There was probably an easier way, but don't ask.

mod is kinda like remainder (but not quite)
for the first line, it means 5^3 (125) and 4 have the same remainder when divided by 11

so in general, $ a \equiv b \pmod c$ for integers a, b, and c means a and b have the same remainder when divided by c or, equivalently, $ \frac{a\minus{}b}{c}$ is an integer

IS there an easier way to do this? btw this was on an algebra 1 test... (ciphering actually)

As a direct result of [b][url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's Little Theorem[/url][/b], $ a^{10} \equiv 1 \pmod{11}$.

$ \therefore \; 5^{81} \equiv (5^{8})^{10} \cdot 5 \equiv 5 \pmod{11}$

algebra 1 does NOT have mod arithmetic...

where did u get the 27, as a power earlier?

$ (a^b)^c \equal{} a^{bc}$
$ 5^{81} \equal{} 5^{3 * 27} \equal{} (5^3)^{27}$

You were referring to my solution, right?

[quote="DaReaper"]im in alg II/wtrig, and i have never heard of this, actually it was from alg II ciphering, my bad.

so does the exponent have to be prime? b/c 81 is not prime..[/quote]

Fermat's Little Theorem says that if $ p$ is prime and $ a$ is an integer, then the remainder when $ a^{p \minus{} 1}$ is divided by $ p$ is $ 1$. [b][color=red]Learn this for math competitions[/color][/b].

So let's consider $ a \equal{} 5^8$ and $ p \equal{} 11$. The remainder when $ a^{p \minus{} 1} \equal{} (5^8)^{11 \minus{} 1} \equal{} 5^{80}$ is divided by $ 11$ is $ 1$.

So, we can write $ 5^{80}$ as $ 11c \plus{} 1$ for some number $ c$.

Now, $ 5^{81} \equal{} 5 \cdot 5^{80} \equal{} 5 (11c \plus{} 1) \equal{} 55c \plus{} 5 \equal{} 11 (5c) \plus{} 5$

The remainder when the above is divided by $ 11$ is $ \boxed{5}$

[hide="Slightly Weirder Way, But It Does Work!"]Ok... so divide $ 81$ by $ 11$, to get $ 7 \text{R} 4$.

Then, add $ 7$ and $ 4$ for your new exponent.

Do this again.

$ 11 \div 11 \equal{} 1 \text{R} 0$

So $ 1$ becomes your new exponent.  Thus, the answer is clearly $ \boxed{5}$.[/hide]

I want to clarify what [b]ernie [/b]did up there...

Suppose you are working modulo some prime $ p$, and you have some number $ a^b$

We can write $ b$ in the form $ n \cdot p \plus{} m$, as a result of the division algorithm. So, $ a^b \equal{} a^{np \plus{} m}$

From Fermat's Little Theorem, we know that $ a^{p \minus{} 1} \equiv 1 \pmod{p}$, and multiplying by $ a$, we get $ a^{p} \equiv a \pmod{p}$

Therefore, $ a^{np \plus{} m} \equal{} (a^{p})^n \cdot a^m \equiv (a)^n \cdot a^m \equal{} a^{n \plus{} m} \pmod{p}$


LaTeX
integration
calculus
logarithms
trigonometry
derivative
calculus computations

can you solve this [t]\Bigint_{0}^{1}\frac{tan(x)dx}{x}[\t] i cant use latex but you can understand me ok good luck

Integration by parts yields:

$-\int_0^1 \ln x \sec^2 x \, dx$

OR

$-\ln \cos 1 + \int_0^1 \frac{\ln \sec x}{x^2} \, dx$

Substitution of $y = \tan x$ yields

$\int_0^1 \frac{y}{(\tan^{-1} y)(1+y^2)} \, dy$

Substitution of $y = \cos x$ yields

$\int_0^1 \frac{-1}{y \cos^{-1} y} \, dy$

Using a double integral that about the $\frac1{x}$ term

$\int_0^1 \int_{-\infty}^x \frac{-\tan x}{t^2} \, dt \, dx$

Maybe applying fubini will make that work.

I'm not sure how I would apply partial differentiation to this either.

Is this problem from a textbook or of your own creation?

I almost forgot taylor series representation of $\tan x$...nevermind it involves Bernoulli numbers.


MATHCOUNTS
geometry
perimeter
inequalities
arithmetic sequence
triangle inequality

I have a few problems that I need help with... some I do not get at all, others I always get a different answer from the correct one.

19. How many distinct, non-equaliteral triangles with a perimeter of 60 units have integer side lengths a,b, and c such that a,b,c is an arithmetic sequence?

14. The letters A,B,C and D represent four distinct integers from 0 through 9. When A is added to B the result is C. When B is subtracted from A the result is D. How many possible ways are there to assign values to A,B,C and D.

19. [hide]To be an arithmatic sequence and add up to 60, the middle number must be 20.
So we just find non-equilateral triangles like that. 19,20, 21 works, so does 18,20,22.... and so on all the way to 11, 20, 29. 10, 20, 30 does not work because it doesn't make a triangle. So it's one side from 11 to 19, and the number of triangles that is is 9.[/hide]

20. [hide]Just work out the number of ways you can get A and B.

3,1 works, 4,1 works, all the way to 8,1.   3,2 works, 4,2 works, all the way to 7,2.
4,3 5,3 6,3 work, and so does 5,4.

That totals 15.[/hide]

You're right, I counted 4 2 and 63 which repeat.

[hide="19"]Let a be the length of the smallest side and d be the difference in the aritmetic sequence.

We have $a+(a+d)+(a+2d)=60$
$\implies 3a+3r=60$
$\implies a+r=20$

We also know from Triangle Inequality that $a+(a+r) >a+2r$
$\implies 2a>a+r=20$

Thus, the smallest a can be is 11, the largest it can be is 19 (since the triangle is non-equilateral),
$19-11+1=\boxed{9}$ triangles[/hide]


function
probability
binomial coefficients
real analysis
real analysis unsolved

The generating function for the central binomial coefficients is

$ \sum_{n \ge 0} {2n \choose n} x^n \equal{} \frac {1}{\sqrt {1 \minus{} 4x} }$.

This function has the following product expansion:

$ \frac {1}{\sqrt {1 \minus{} 4x} } \equal{} (1 \plus{} 2x)(1 \plus{} 6x^2)(1 \plus{} 8x^3)(1 \plus{} 54x^4)(1 \plus{} 96x^5)(1 \plus{} 312x^6)(1 \plus{} 1152x^7)...$.

The sequence of coefficients in each factor is not in the OEIS.  What is it?  (It may or may not be more appropriate to place this question in Combinatorics.)

Really random question: where do products like that ever come up, except as generating functions for things like partitions?  I ask because once you've posed the problem, it sounds obvious, but it never occurred to me before :)  Also, we can obviously find the associated sequence for any given sequence -- and none of them seem to appear in the OEIS.  (In fact, looking at some obvious first choices -- Fibonacci, Catalan, Bell -- the only one that seems remotely easy to understand is the powers of 2.)

Edit: a related idea (inspired by "partitions with distinct parts" versus "partitions") is to look at products $ \prod_{i \equal{} 1}^\infty \frac {1}{1 \minus{} a_i x^i}$.  This is called the [i]Somos transform[/i], and so what you're looking at is something like a "inverse dual Somos transform."  Some inverse Somos transforms are in the OEIS, like [url=http://www.research.att.com/~njas/sequences/A006177]A006177[/url], the inverse Somos transform of the Catalan numbers.

Edit the second: It does seem like we should at least be able to write down some recursive summation formula for these, right?

A friend of mine asked me how I would go about writing a geometric distribution as a sum of Bernoulli distributions, and (unless I'm horribly wrong) this comes down to writing a generating function of the form $ \frac {\frac {x}{2} }{1 \minus{} \frac {x}{2} }$ as a product of terms of the form I gave above.  Here, the relevant product (again, unless I'm horribly wrong) is

$ \frac {\frac {x}{2} }{1 \minus{} \frac {x}{2} } \equal{} x \prod_{k \equal{} 0}^{\infty} \left( \frac {2^{2^k}}{2^{2^k} \plus{} 1} \plus{} \frac {x^{2^k}}{2^{2^k} \plus{} 1} \right)$,

that is, each Bernoulli distribution controls one bit of the output.  So that got me thinking about whether this idea generalizes, that is, whether there is a natural way to write distributions as a sum of Bernoulli distributions.  Unfortunately, I don't know enough probability to know whether this is a meaningful question to ask (or whether my interpretation of the original question is even correct), which is why I posed it here.

Belatedly: I don't have a clue :) (though I am still curious about this writing-of-generating-functions-as-products idea).  But this does sound like the sort of thing with the property that there are other people around here who [i]do[/i] have a clue about it.  (Apologies for the tortured sentence structure there.)  So perhaps one of them will take a look and see if there is any sense to be made out of it.

Edit: here's one sort of vague idea: let $ E(t) \equal{} \sum_{n \geq 0} e_n t^n$ where the $ e_i$ are the elementary symmetric functions.  Then $ E(t) \equal{} \prod_{n \geq 0}1 \plus{} x_i t$, so this is a natural-seeming relationship.  Since the $ e_i$ are algebraically independent, it's also well-suited to your situation.  But whether this is "useful" in any sense (e.g., for determining the coefficients of the sequence you mentioned), I have no idea.  This might be worth a trip to Professor Stanley's office ;)

A recursive relationship that can be written is:

$ \sum_{\frac{p}{i} \in \mathbb{N}}\frac{(\minus{}1)^{i\plus{}1}}{i}a^i_{\frac{p}{i}}\equal{}\frac{2^{2\,p\minus{}1}}{p}$

where

$ \prod_{k\equal{}1}^\infty(1\plus{}a_k\,x^k)\equal{}\frac{1}{\sqrt{1\minus{}4x}}$

(I hope I got it right..)

A question:

Suppose a function has infinitive number of zeros named $ z_k$ with multiplicity $ p_k$. Would it then be safe to say that this function can be written in a form similar to:

$ f(x) \sim \prod_{k} (x\minus{}z_k)^{p_k}$

That looks like the [url=http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem]Weierstrass factorization theorem[/url], although I don't think it applies directly here.

Your other recurrence looks almost like a Dirichlet convolution.  It might be feasible to extract a summation identity for $ a_n$ that way.

Yes, Weierstrass was my base idea.. But the idea it self should work for this..

It shouldn't work, because it's the wrong sort of convergence. The products t0rajir0u is looking at converge in the formal power series sense: the coefficient of $ x^n$ is eventually constant. A Weierstrass-type product doesn't do that; if things go right, it converges as an analytic function on some disk. Of course, in order to make things go right, you have to multiply by various nonvanishing entire functions in each term.


find x,y,z in positive integer such that z=4*q+3 and
        x^2+5=y^z. :D   ;)  :lol:   :lol:  :lol:  :lol:

no one answer my question  :(  :(  :(  :(  :(  :(

it is easy to prove that x is even and y is odd.
then x^2+4=y^z-1=(y-1)(y^(z-1)+..+1);
put A=y^(z-1)+...+1;it follow that A is odd such that y-1 divisible to 4
hence you have  the solution
good luck

it is easy to prove that x is even and y is odd.
then x^2+4=y^z-1=(y-1)(y^(z-1)+..+1);
put A=y^(z-1)+...+1;it follow that a is odd such that y-1 divisible to 4
hence you have  the solution
good luck


modular arithmetic
number theory proposed
number theory

Hope did not posted before :?
Show that if $a$ and $b$ are positive integers then:
$(a+\frac{1}{2})^n+(b+\frac{1}{2})^n$ is an integer for only finetely many positive integers $n$.

Take $n$ so large that $2^n$ does not divide $u-v$, where $u=(2a+1)^2,v=(2b+1)^2$. If we assume that $2^n|(2a+1)^n+(2b+1)^n$, then we have $2^n|(2a+1)^n+(2b+1)^n|u^n-v^n$.

Since the element $uv^{-1}$ is not the unit in the multiplicative group of invertible elements in $\mathbb Z_{2^n}$, it must have an order which is a power of $2$ greater than $1$ (because th group has $\varphi(2^n)=2^{n-1}$ elements), and this order must divide $n$ (because $(uv^{-1})^n=1$ in the mentioned group). However, it's easy to see that, in the given conditions, $n$ must be odd if it's larger than $2$ (because otherwise we would have $(2a+1)^n\equiv(2b+1)^n\equiv 1\pmod 4$, so $2^n\not |\ (2a+1)^n+(2b+1)^n$). We thus have a contradiction.


The word MATH is repeatedly written producing the pattern MATHMATHMATH… If the pattern is continued, what letter will occur in the 2009th position?

[hide] Since $ 2009\equiv1\pmod4$, the letter $ \text{M}$ will be in that position.[/hide]

[hide]dsince there are 4 letters we divide 2009/4 andget a re,inder of 1 so it is [b]M.[/b][/hide]


geometry
number theory proposed
number theory

Let R be a convex region symmetrical about the origin with area greater than 4.
Show that R must contain a lattice point different from the origin.

This is the 2-D case of Minkowski's theorem, right ?

How about the n-dimensional version ?

The n-dimensional version is : Given a convex region R symmetrical to the origin in the n-dimensional space.

Show that if R has volume greater than 2^n, then R contains a lattice point different from the origin.

The proof is the same. You reduce the whole thing to an area of 1 and then apply Blichfeldt's Lemma.


Let $ x,y\ge 0$. Prove that:

$ x^4y \plus{} x^2y^3 \minus{} 2x^3y^2 \plus{} x^4 \plus{} y^4 \plus{} xy^3 \minus{} 2x^2y^2 \minus{} x^3y\plus{}2x^3 \plus{} 2y^3 \minus{} 3x^2y\plus{}x^2 \minus{} xy \plus{} y^2\ge 0$

Fix $ x\plus{}y\equal{}2t$ and set $ x\equal{}t\plus{}\lambda$, $ y\equal{}t\minus{}\lambda$ ($ x\geq y$ is the worse case).

[quote="BG Yoda"]Fix $ x \plus{} y \equal{} 2t$ and set $ x \equal{} t \plus{} \lambda$, $ y \equal{} t \minus{} \lambda$ ($ x\geq y$ is the worse case).[/quote]

Can you complete your solution?


Find all $f: \mathbb Z \to \mathbb Z$ such that $f(x+y+f(y))=f(x)+2y$ for all $ x,y\in \mathbb Z$.

Hi :) !!!

$ f(x + y + f(y)) = f(x) + 2y$ $ (E)$

$ let$ $ f(0) = a \in \mathbb{Z}$

than $ let$ $ y = 0$:

$ f(x + a) = f(x)$

put $ x = - a$ we find $ f(0) = f( - a) = a$

in $ (E)$ put $ x = a$ and $ y = - a$ we find $ f(a - a + f( - a)) = f(a) - 2a %Error. "Rightlong" is a bad command.
f(a) = f(a) - 2a \Rightarrow a = 0$

than $ f(0) = 0$ we returd to $ (E)$:

$ let \ \ \ x = 0$:

$ (E)$ become's:

$ f(y + f(y)) = 2y$

$ f$ are bijective

$ let$ $ g(y) = f(y) + y$ than $ g(g(y)) = f(g(y)) + g(y) = 2y + g(y) = 3y + f(y) = 2y + g(y)$

than $ g^{[n]}(y) = \frac {(2)^n - ( - 1)^n}{3}f(y) + \frac {2^{n + 1} - ( - 1)^{n + 1}}{3}y$

than .....etc

in conclusion we can find:

$ f(x) = x$ and $ f(x) = - 2x$ $ \forall x \in \mathbb{Z}$

[quote="DCTPKTCSPHN"]Find all $ f: Z\rightarrow Z$ such that $ : f(x \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2y$  $ \forall x,y\in Z$[/quote]
Let $ P(x,y)$ be the equation $ f(x\plus{}y\plus{}f(y))\equal{}f(x)\plus{}2y$

Let $ a\in\mathbb{Z}$, let $ u\equal{}a\plus{}f(a)$ and $ v\equal{}2a$. We have $ f(x\plus{}u)\equal{}f(x)\plus{}v$ and so, with easy induction, $ f(x\plus{}nu)\equal{}f(x)\plus{}nv$ $ \forall n\in\mathbb{Z}$
For $ n\equal{}v$, we get $ f(x\plus{}uv)\equal{}f(x)\plus{}v^2$

Then, $ P(x\minus{}v,y\plus{}u)$ $ \implies$ $ f(x\minus{}v\plus{}y\plus{}u\plus{}f(y\plus{}u))\equal{}f(x\minus{}v)\plus{}2y\plus{}2u$, and so $ f(x\plus{}y\plus{}f(y))\plus{}v\equal{}f(x\minus{}v)\plus{}2y\plus{}2u$

So we have : $ f(x\plus{}y\plus{}f(y))\equal{}f(x\minus{}v)\plus{}2y\plus{}2u\minus{}v$
But : $ f(x\plus{}y\plus{}f(y))\equal{}f(x)\plus{}2y$
So : $ f(x)\plus{}2y \equal{} f(x\minus{}v)\plus{}2y\plus{}2u\minus{}v$
So : $ f(x)\equal{}f(x\minus{}v)\plus{}2u\minus{}v$ and so, with easy induction : $ f(x\plus{}nv)\equal{}f(x)\plus{}n(2u\minus{}v)$ $ \forall n\in\mathbb{Z}$
For $ n\equal{}u$, we get $ f(x\plus{}uv)\equal{}f(x)\plus{}u(2u\minus{}v)$

But : $ f(x\plus{}uv)\equal{}f(x)\plus{}v^2$

So $ v^2\equal{}u(2u\minus{}v)$

So $ 4a^2\equal{}2(a\plus{}f(a))f(a)$

So $ (f(a)\minus{}a)(f(a)\plus{}2a)\equal{}0$ $ \forall a\in\mathbb{Z}$ and so either $ f(a)\equal{}a$, either $ f(a)\equal{}\minus{}2a$

So, $ \forall x$, either $ f(x)\equal{}x$, either $ f(x)\equal{}\minus{}2x$

Assume now that there exist $ x\neq 0$ such that $ f(x)\equal{}x$ and $ y\neq 0$ such that $ f(y)\equal{}\minus{}2y$.

$ P(x,y)$ $ \implies$ $ f(x\minus{}y)\equal{}x\plus{}2y$
If $ f(x\minus{}y)\equal{}x\minus{}y$, we get $ x\minus{}y\equal{}x\plus{}2y$ and so $ y\equal{}0$
If $ f(x\minus{}y)\equal{}\minus{}2(x\minus{}y)$, we get $ \minus{}2x\plus{}2y\equal{}x\plus{}2y$ and so $ x\equal{}0$

So, either $ f(x)\equal{}x$ $ \forall x$, either $ f(x)\equal{}\minus{}2x$ $ \forall x$

And it's easy to verify that these two expressions both are solutions of the initial equation.

[quote="pco"][quote="DCTPKTCSPHN"]Find all $ f: Z\rightarrow Z$ such that $ : f(x \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2y$  $ \forall x,y\in Z$[/quote]
Let $ P(x,y)$ be the equation $ f(x \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2y$

Let $ a\in\mathbb{Z}$, let $ u \equal{} a \plus{} f(a)$ and $ v \equal{} 2a$. We have $ f(x \plus{} u) \equal{} f(x) \plus{} v$ and so, with easy induction, $ f(x \plus{} nu) \equal{} f(x) \plus{} nv$ $ \forall n\in\mathbb{Z}$
For $ n \equal{} v$, we get $ f(x \plus{} uv) \equal{} f(x) \plus{} v^2$

Then, $ P(x \minus{} v,y \plus{} u)$ $ \implies$ $ f(x \minus{} v \plus{} y \plus{} u \plus{} f(y \plus{} u)) \equal{} f(x \minus{} v) \plus{} 2y \plus{} 2u$, and so $ f(x \plus{} y \plus{} f(y)) \plus{} v \equal{} f(x \minus{} v) \plus{} 2y \plus{} 2u$

So we have : $ f(x \plus{} y \plus{} f(y)) \equal{} f(x \minus{} v) \plus{} 2y \plus{} 2u \minus{} v$
But : $ f(x \plus{} y \plus{} f(y)) \equal{} f(x) \plus{} 2y$
So : $ f(x) \plus{} 2y \equal{} f(x \minus{} v) \plus{} 2y \plus{} 2u \minus{} v$
So : $ f(x) \equal{} f(x \minus{} v) \plus{} 2u \minus{} v$ and so, with easy induction : $ f(x \plus{} nv) \equal{} f(x) \plus{} n(2u \minus{} v)$ $ \forall n\in\mathbb{Z}$
For $ n \equal{} u$, we get $ f(x \plus{} uv) \equal{} f(x) \plus{} u(2u \minus{} v)$

But : $ f(x \plus{} uv) \equal{} f(x) \plus{} v^2$

So $ v^2 \equal{} u(2u \minus{} v)$

So $ 4a^2 \equal{} 2(a \plus{} f(a))f(a)$

So $ (f(a) \minus{} a)(f(a) \plus{} 2a) \equal{} 0$ $ \forall a\in\mathbb{Z}$ and so either $ f(a) \equal{} a$, either $ f(a) \equal{} \minus{} 2a$

So, $ \forall x$, either $ f(x) \equal{} x$, either $ f(x) \equal{} \minus{} 2x$

Assume now that there exist $ x\neq 0$ such that $ f(x) \equal{} x$ and $ y\neq 0$ such that $ f(y) \equal{} \minus{} 2y$.

$ P(x,y)$ $ \implies$ $ f(x \minus{} y) \equal{} x \plus{} 2y$
If $ f(x \minus{} y) \equal{} x \minus{} y$, we get $ x \minus{} y \equal{} x \plus{} 2y$ and so $ y \equal{} 0$
If $ f(x \minus{} y) \equal{} \minus{} 2(x \minus{} y)$, we get $ \minus{} 2x \plus{} 2y \equal{} x \plus{} 2y$ and so $ x \equal{} 0$

So, either $ f(x) \equal{} x$ $ \forall x$, either $ f(x) \equal{} \minus{} 2x$ $ \forall x$

And it's easy to verify that these two expressions both are solutions of the initial equation.[/quote]
Pco: you have nice solution ,I agree with you :)


algorithm
function
combinatorics unsolved
combinatorics

On a blackboard there are $ n \geq 2, n \in \mathbb{Z}^{\plus{}}$ numbers. In each step we select two numbers from the blackboard and replace both of them by their sum. Determine all numbers $ n$ for which it is possible to yield $ n$ identical number after a finite number of steps.

I think the result holds for every n.
Lets go in reverse. Suppose we had n identical numbers on the board. Then , we could replace any two numbers such that the sum of the new numbers is equal to their initial occupants. Now here's a number set with the given property. This way, can't we say that every number n has the given property?
Have I misunderstood the problem? Or is the claim supposed to be proved for a given arbitrary set of numbers?

Hi, lajanugen, I think they require solution for arbitrary set of numbers.

My Solution:
[hide]
Claim:
It is always possible to equate all numbers [b]iff[/b] $ n$ [b]is even[/b].

First we prove that for all odd $ n$, there exists some $ n$-tuple such that it's impossible to equate all numbers.

Consider the $ n$-tuple $ (2,2,\cdots ,2,2,1)$ (total odd numbers)

Suppose to the contrary that it is possible to yield equal numbers.
i.e. $ m,m,\cdots ,m,m,m$

Clearly $ m>2$

Note that if the operation $ (a,b)\to (a\plus{}b,a\plus{}b)$ can get $ (2,2,\cdots ,2,2,1)\to(m,m,\cdots ,m,m,m)$
Then the reverse operation $ (p,p)\to (q,p\minus{}q)$ can get $ (m,m,\cdots ,m,m,m)\to(2,2,\cdots ,2,2,1)$

Note that the operation $ (p,p)\to (q,p\minus{}q)$ decreases both numbers.
The operation must act on each $ m$ in the set $ (m,m,\cdots ,m,m,m)$

Also this reverse operation requires both numbers to be equal.
Thus it will always act on the $ m$s in pairs.
But there are odd number of $ m$s.
Hence there will remain one $ m$ on which this reverse operation never acts.

Contradiction.

Now it remains to prove that for all even $ n$ every $ n$-tuple can be transformed into equal numbers.

Lets treat the $ n$-tuple as a $ m$-tuple of pair of elements of the $ n$-tuple. ($ n\equal{}2m$)
That is $ (a_1,a_2,\cdots ,a_n)\equal{}((a_1,a_2),(a_3,a_4),\cdots ,(a_{2m\minus{}1},a_{2m}))$

Lets do the operation on the 2 elements of each pair.
Let $ b_i\equal{}(a_{2i\minus{}1}\plus{}a_{2i},a_{2i\minus{}1}\plus{}a_{2i})$
$ ((a_1,a_2),(a_3,a_4),\cdots ,(a_{2m\minus{}1},a_{2m}))\to (b_1,b_2,\cdots ,b_m)$
(From now the two elements of each pair will never be different)
Suppose $ b_i\equal{}(a,a),b_j\equal{}(b,b)$
$ (b_i,b_j)\to (b_i\plus{}b_j,b_i\plus{}b_j)$ is equivalent to $ ((a,a),(b,b))\to ((a\plus{}b,a\plus{}b),(a\plus{}b,a\plus{}b))$.
which is the composition of 2 original operations.

Actually now we treat the $ n$-tuple as an $ m$-tuple.
also note that we now have a extra operation: we can multiply some $ b_i$ by $ 2$ whenever we like, because $ (a,a)\to (2a,2a)$

Thus at some point we can multiply powers of $ 2$ to some $ b_i$s so that the largest power of 2 dividing $ b_i$ are same for all $ b_i$

Let $ o(x)$ denote the largest odd divisor of $ x$.
If $ b_i\not \equal{}b_j$ have the same largest power of $ 2$ then note that
$ o(b_i\plus{}b_j)<\max(o(b_i),o(b_j))$
(It's not hard to see, because the sum of odd numbers is even)

At some particular state of the blackboard define $ l,s$ such that for all $ i$, $ o(b_l)\ge o(b_i)\ge o(b_s)$

Now the strategy of obtaining all equal numbers is:

$ \boxed{}$ operate pairs of the $ n$-tuple and make equal to get it into $ (b_1,b_2,\cdots b_m)$ form.
$ \boxed{}$ get all numbers into equal largest powers of $ 2$
$ \boxed{}$ operate on $ (b_l,b_s)$
$ \boxed{}$ get all numbers into equal largest powers of $ 2$
$ \boxed{}$ operate on $ (b_l,b_s)$
$ \boxed{}$ get all numbers into equal largest powers of $ 2$
$ \boxed{}$ operate on $ (b_l,b_s)$

$ \cdots$

Now it remains to prove that this strategy will yield equal numbers within finite number of steps.

After all "operate on $ (b_l,b_s)$" steps, the-largest-odd-number-dividing-at-least-one-$ b_i$ decrease unless $ o(b_l)\equal{}o(b_s)$

But the-largest-odd-number-dividing-at-least-one-$ b_i$ can't decrease indefinitely.
Hence, after finite number of steps, $ o(b_l)\equal{}o(b_s)$
But $ o(b_l)\equal{}o(b_s)$ implies $ o(b_i)$ is same for all $ i$

And then we can equate the largest power of $ 2$ of all $ b_i$
Now we have all equal numbers.

Example:

$ 3,10,7,9,4,6$
$ 13,13,16,16,10,10$
$ 26,26,16,16,20,20$

$ \cdots$

$ 2^4\cdot 13,2^4\cdot 13,2^4\cdot 1,2^4\cdot 1,2^4\cdot 5,2^4\cdot 5$
$ 2^4\cdot 14,2^4\cdot 14,2^4\cdot 14,2^4\cdot 14,2^4\cdot 5,2^4\cdot 5$
(from now there won't be any unnecessary repeatations.)
$ 2^5\cdot 7,2^5\cdot 7,2^5\cdot 5$
$ 2^5\cdot 7,2^5\cdot 12,2^5\cdot 12$
$ 2^7\cdot 7,2^7\cdot 3,2^7\cdot 3$
$ 2^7\cdot 10,2^7\cdot 10,2^7\cdot 3$
$ 2^8\cdot 5,2^8\cdot 5,2^8\cdot 3$
$ 2^8\cdot 5,2^8\cdot 8,2^8\cdot 8$
$ 2^{11}\cdot 5,2^{11}\cdot 1,2^{11}\cdot 1$
$ 2^{11}\cdot 6,2^{11}\cdot 6,2^{11}\cdot 1$
$ 2^{12}\cdot 3,2^{12}\cdot 3,2^{12}\cdot 1$
$ 2^{12}\cdot 3,2^{12}\cdot 4,2^{12}\cdot 4$
$ 2^{14}\cdot 3,2^{14}\cdot 1,2^{14}\cdot 1$
$ 2^{14}\cdot 4,2^{14}\cdot 4,2^{14}\cdot 1$
$ 2^{16}\cdot 1,2^{16}\cdot 1,2^{16}\cdot 1$
(Note that it might have terminated before reaching all powers of $ 2$, but that doesn't matter)[/hide]

[b]The solution might look big. It looks big because I have tried to explain things in detail.
Please someone check if my solution is understandable/true.
And reply.[/b]

[quote="Akashnil"]Please someone check if my solution is understandable/true.[/quote]  It looks good to me.

Thanks for checking. It took me a lot of time to type it. :roll:

P.S. What's "MEMO, Team"  :?:

MEMO = Middle European Math Olympiad

and Team means its from the team competition.

[hide="Solution"]If $ n$ is odd, then start with the numbers $ (2,2,1,\dots,1)$. Because we always have either a new maximal value which occurs twice or the total number of maximal values increases by two, there is always an even amount of maximal values, so there can't be $ n$ equal numbers ($ n$ is odd).

If $ n$ is even, group the numbers in pairs like Akashnil did. Now we define the permutation $ \sigma$ as follows: If $ a_i$ is even, then $ \sigma_i \equal{} i$, and if $ a_i$ and $ a_j$ are both odd, then $ a_i > a_j \iff a_{\sigma_i} < a_{\sigma_j}$ (that is, the odd numbers are sorted in the opposite way). Note that $ \sigma\circ\sigma$ is the identity, so all cycles have length of no more than 2. Now, if $ a_i$ is odd, then replace it and $ a_{\sigma_i}$ by $ a_i\plus{}a_{\sigma_i}$ (obviously, this is done only once for each pair). Then all numbers are even, and dividing by 2 won't change the result in any way. Now set $ f(a) \equal{} \sum_{i\equal{}1}^m a_i^2$ and identify the algorithm with a function $ \phi$. Then

$ f(a)\minus{}f(\phi(a)) \equal{} \sum_{i\equal{}1}^m a_i^2 \minus{} \left(\sum_{a_i\text{ even}} \frac{a_i^2}{4}\plus{}\sum_{a_i\text{ odd}} \left(\frac{a_i\minus{}a_{\sigma_i}}{2}\right)^2\right) \equal{} \frac{3}{4}\sum_{a_i\text{ even}} a_i^2 \plus{} \sum_{a_i\text{ odd}} \left(\frac{a_i\minus{}a_{\sigma_i}}{2}\right)^2 \ge 0$

shows that iteration of $ \phi$ will make $ f$ decrease. However, $ f$ must be nonnegative; so $ f$ must be constant from some point on, and this is only the case if there are no even numbers and $ a_{\sigma_i} \equal{} a_i$, but by the definition of $ \sigma$, this means that minimum and maximum must be equal, and thus that all numbers must be equal.[/hide]

This problem, where the numbers are real numbers instead of positive integers, and where the operation is addition instead of multiplication is also true - came out in a national TST.


MATHCOUNTS geometry AMC USA(J)MO USAMO FTW

Answer the question.

I slept at around 2:30.

I didn't sleep. Instead I played fruits and vegetables (which failed), swam in the pool for a while, etc. Then I completely crashed when I came back home.  :D

Why are you not sleeping? It's not good for studying!

I didn't need sleep...I drank this like powerful energy drink, did random stuff like swam in the pool and threw water balloons at people...

I did crash at like 2 AM, but I somehow was able to stay awake..

I stayed up the whole night with sjcamath!!!  :D

Although I did start dozing off in DBR's dorm at around 8 AM...  :sleep2:

[quote="alexhhmun"]I decided to go to sleep at around 1:00 and got a whole 4 hours of sleep.  I don't know anyone who slept more than I did.  HA!   :P[/quote]

I went to sleep at 2:30 and got up at 7:30... that's one hour more than you!!!!! HA!  :P  :P  :P

oh yeah all of you guys have been owned i slept from 1 to 7 !!! and i woulda slept more if i hadnt been woken up by like 5 people to catch my shuttle bus which was scheduled at 6:50.

I slept like 4 hours because I was cleaning up after people when they were leaving in the middle of the night...

Well... it was easier for all of you guys to get more sleep since your shuttle left at 8:30 or something. Since my shuttle left around 5:30 it was a bit more difficult... I think not sleeping for a whole night had lasting effects. I think I'm still more tired than usual. :sleeping:

I got 6 hours, which seems to be the current record.

I tied you zephyredx (almost)
I slept from 3:30 to 9:00
My bus was supposed to come at 9:00, so that's when I got up.
Except the bus was an hour late.
So I could have slept longer instead of having to sleep on the plane.  :mad:

haha people actually slept on the last day?!?
WHATT?!??
youre supposed to hang out&make the final memories with one anotherrrr!!
:O

i was in someone else's dorm the entiiireee night lol
watching movies online :P

[quote="sjcamath"]Instead I played fruits and vegetables (which failed)[/quote]

My impression of Ice actually beat meenamathgirl! :D

I tried to but I finally gave up and had some half-hour snoozes at the pool. I was packed and ready to go before midnight (  :o  )

No sleep woot. but I did randomly venture into the pool area... and coincidentally met jeff and COUGH i wonder there... lol. i think they left their room after i randomly walked in and started watching that korean soap opera with them

Overachievers? You haven't seen Gunn...teachers here are on crack, yet there are people who have half A+s half As, and taking all APs. I'm probably the biggest slacker, my grade report generally looks like: A- A- A- A- A- A- B+ (when an 90.1 accidentally slips :()

That is a lot of caffeine. When I was at Mathcounts State a long time ago, I sucked really bad at math, but I somehow got much better than I ever could have without caffeine. In the morning, as soon as I got off the car I drank one, to cure my carsickness+sleepyness. Then before the contest, I got really thirsty and had no water, so I drank another. Then during the contest, I finished that one and got thirsty again, and drank a third. Halfway through the third I wasn't really thirsty anymore, but it felt cool to be sipping that kind of junk.

The crash happened at 4am in the morning, 20 hours later :O

I'm pretty sure my caffeine saved me yesterday, since I had 3 finals and I slept 2 hours...

*take out gun* jkjkjk. lolll.
just try and cut the caffiene?
lol serialk11r, is there a diff. between A-, A, A+ ??
how bout the entire state of cali consists of hecka crazyyy asian overacheivers. and to relieve stress. move out to WI. XP
wellll thxfully my 'rents ban caffeine/i cant drive/its too cold to walk so... no starbucks/energy drinks for me lol XP ive learned to live w/o it -.-

They misspelled my AoPS username in the yearbook.  It's bpgbcg, not bpglrg.  (My handwriting is terrible, which is probably what caused the mistake.)

i tried to live without it but my parents drink coffee quite frequently too, so there's infinite supply of coffee stuff at my house! :D
yayyy ahahahahhahaa xP
so bad.
but yeah.
dude..90 degrees todayy x.X when i got in my car after my prep courses..it was BLAZING HOT in there. yikkkes. im so glad i dont have black interior [first of all it would look FUGLY with baby blue exterior ew ew ew]

bpgbcg: i found u in the yrbk!! lol. i just had to know -.-

lol. it got warmer today!!! :D

Oh dang 48% is pretty crazy...Gunn is only around 35% I think. We also have 10% Jews, and a few racist Jewish teachers to go with them. However down in Cupertino there's Monta Vista which is 75% asian :)
At Gunn, there is no difference between A-, A, A+ GPA wise, but I know hella people who get disappointed when they get an A-. On person who lives on my street (Jeff also lives on my street actually...my street has hella Gunn kids) has parents who didn't let her go to homecoming dance because she got an A- in bio...

=_=''
then thats just stupiddddd. asian parents -.- srsly.
an A's an A. good god. -_-;; lol i have an increased appreciation for my 'rents now lol XP
...
crazy. amt. of. asians. xO

[quote="sooozyy"]hahahahahaa dude i agree that Gunn's pretty insane too..
but try surviving at a school with asian population of 48%. :O  at least Gunn isn't infected with asian to this extent. [but i gotta admit gunn has some hella smart matheletes. my school doesnt since they all graduated last year]
we're all loaded with APs...who isnt? [me! but thats only cuz i take the courses outside of school at a community college to save some of my sanity...]

haha i dont crash on daily basis but i crash like twice a week. my stress will just stack up to the point where i cant take it anymore and BOOM.
like last night when i almost fell asleep on jeff while talking on the phone xP

HAHAHA I LOVE CAFFEINE. [shelly32494, dont shoot me! T_T roflmao xD][/quote]

ONLY 48%!!!!!!!!!!!!

My school's about 80% asian...hecka awesomeness

lol 80% is insane!

which school?

i know that uni high in irvine is close to 60%?   [Dai, correct me if im wrong?]

anyways irvine is asian...WE HAVE A KOREAN MAYOR.
that just blows it allllllll.

is dai that guy who sang,'everybody likes problem-solving'??becuz if he is,im still mad @ him for making briam do that stupid sing-a-long for 'everytime we touch'.i mean,EVERYTIME WE TOUCH?!?!?!?!WTF!!!!!!!!!o,and WAYYYYYYYYYYYYY bak on the 'did u sleep' i stayed up with katherine and sumun and carl and um............o yea sasha playing volleyball..........erik and meena were somewhere else...........o, and this is steve

omg. steveeee!!!!!
yes that is that dai.
nd omg. hey. i liked everytime we touch it was a good song. and basham was amazinggg at it :)
lol. wow we got rly off topic.

hey . well. my school's like. 80% white. ahaha.

lol typical us.
digressing :P

OMG HI STEVE! :)))))))

 agreed with shelly32494, i liked that song! =]

Carbon57 goes to Lynbrook.
At Gunn there's a senior who's the nephew of the new Secretary of Energy :D
Bay Area is definitely like, the most asian place in the world outside of asia.

lol...Yeah...Lynbrook have A- or A+, just As, and nobody cares about it, since it doesn't go on our semester grades, but our school's still insanely competitive though...too many asians trying to get 4.0 and shiz

Yup...bay area is asian since it's the closest place (except Hawaii and Alaska) to Asia in America...and a lot of tech jobs


superior algebra
superior algebra unsolved

Let A be a commutative ring with 1.  We want to show that the Jacobson radical is an ideal in A.  Also how would I show A/R (where R=nilradical of A) has no nonzero nilpotent elements.  Furthermore, how can I show that an element a can be in the Jacobson radical iff 1-ab is invertible for all b in A.  Also I want to show that the nilradical is contained the Jacobian.  Any help would be great :)

What's your definition of the Jacobson radical? By the usual definition as intersection of all maximal left/right ideals, it's without question an ideal, therefore I ask. But anyway, you can find the proofs here :-)

http://www.mathreference.com/ring-jr,intro.html

Besides, what's the Jacobian radical? I have not found any definition by google  :?

A Jacobson radical is the intersection of all maximal ideals of a commutative ring.  I am still not clear on how to show an element a in the Jacobson(A) iff 1-ab is invertible for all b in A.

Just open the link. There's the proof.

Do you mean with Jacobian the same as Jacobson? :-) Then the nilradical is of course in it because the nilradical is the intersection of all prime ideals.


How could 'change' the way we think of 'cardinals' but even our 'universe' (sets) if we defined alternatively when 2 sets have are equinumerous? Do such definitions exist? I think that such a definition could 'work' for finite sets and the 'new' results could appear only when talking abut infinite sets. Any ideas?

In the future, please don't post the same topic in multiple forums, and see my comment in your other thread.  Another comment:

I think one of the problems with finding an alternate definition of cardinality is that, from a purely set-theoretic point of view, and ignoring any other properties of a set, cardinality (as it is usually defined) is the [i]only[/i] way we have to distinguish two abstract sets.  So any alternate (in particular, any stricter) definition of cardinality would need to be specific to certain types of sets.


algebra
polynomial
abstract algebra
real analysis
real analysis solved

Prove that the polynomial Xn + aXn-1 + ... + aX - 1, with n  \geq 2 and a \in Z\{0} is irreducible over Z[X].

well I've used this problem in MathLinks contest, 1st Edition, Round 3, problem 2 (the well-known Round 3 :D) ... :P

If I'm not wrong, a particular case of this problem (a=-1) appeared in AMM! So, imagine its difficulty!

Suppose WLOG a<0, b=|a|.
Let p(x)=xn-bxn-1-bxn-2-...-bx-1.
Consider f(x)=(x-1)p(x)=xn+1-(b+1)xn+(b-1)x+1, g(x)=xn+1-(b+1)xn.
Let t>0 be a small number, z be complex number and |z|=1+t. We have |f(z)-g(z)|=|(b-1)z+1| \leq (b-1)(1+t)+1. In the other hand |f(z)|+|g(z)| \geq |g(z)|=(1+t)n(a-t). So |f(z)-g(z)|<|f(z)|+|g(z)| for all sufficiently small t.
Due to Rushe's theorem f(z) and g(z) have equal number of roots inside |z|<1+t. It follows that f(z) has exactly n roots in |z|\leq 1.
It is easy to show that z=1 is only root of f(z) on circle |z|=1, so p(z) has exactly (n-1) roots inside |z|<1 and single root outside of |z|<1.

Suppose p(x)=u(x)v(x), where u,v \in Z[x]. Then -1=u(0)v(0) ==> u(0),v(0)=\pm 1 ==> both u(z) and v(z) have root in |z| \geq 1==> p(z) has at least two roots outside of |z|<1 - contradiction.

There are common theorem:
Let a1 \geq a2 \geq ... \geq an \geq 1, n \geq 2. Then f(x)=xn-a1xn-1-...-an-1x-an is irreducibale over Z[x].

For a polinomial $p=\sum\limits_0^na_iX^{n-i}\in Z[X],$ we define $q=\sum\limits_0^na_iX^i\in Z[X],$ and $h=\frac{p+q}{2}, k=\frac{p-q}{2}.$
I will use this facts:
1. Every zero of p which lies on the unit circle is a common zero for h and k.(easy to prove)
2.If $p_{\lambda}=X^n+a_1(\lambda)X^{n-1}+...+a_n(\lambda)$ is a family of polynomials with the property that $p_{\lambda}(z)\neq 0,$
for z in the unit circle, then $p_{\lambda}$ has the same number of zeros in the inetrior of the unit circle.(a little harder.)
Back to problem. We consider a>0.
$p=x^n+aX^{n-1}+...+aX-1.$ We consider the family of polynomials:
$p_t=x^n+(at+1-t)X^{n-1}+...+(at+1-t)X-1, p_1=p,$ and
$p_0=x^n+X^{n-1}+...+X-1.$ We show that if |z|=1 $p_t\neq 0.$ Indeed, if exists z such that |z|=1 and $p_t(z)=1,$ with 1., we obtain thar z is a common zero for $h_t, k_t$ But
$h_t=(at+1-t)(X^{n-1}+...+x=(at+1-t)X(X^{n-2}+...+1), and$
$k_t=x^n-1.$ So z is a root of order n-1 fespectiv n of the unity. Because n and n-1 are relative prime we obtain that z=1, which is not a solution for $p_t.$ We can apply 2. to conclude that $p=p_1$ have the same number of roots in the interior of the unit circle as $p_1.$ I'll prove that $p_1$ have one zero in the interior of the unit circle.
Consider the family $q_t=tx^n+X^{n-1}+...+x-1, q_1=p_1, q_0=X^{n-1}+...+x-1.$ In the same way we can prove that on yhe unit circle $q_t$ have no zeros, so the number of zeros in the unit circle for $p_1=q_1$ is the same with the number of zeros in the unit circle for $q_0.$ Apply the same reasoning n_2 times we conclude that the number of zeros in the unit circle for $p_1$ is the same with the number of zeros in the unit circle for $x^2+x-1$ which is 1.
So qur polynomial have n-1 roots with supraunitary module. If p is reducibile, one of the factors will have all zeros supraunitary in module, which is imposibile because the free term is one. :lol:


Euler
geometry
angle bisector
power of a point
radical axis
geometry proposed

Mmm ! A really nice problem ! Please draw a gifure.

 Let $QH$  meets $CD$ at $M$  and $OQ$ meets $CD$ at $N$ , where $O$ is the center of the great circle. Let line $QO$ ,  produced , meets the circle $O$ at $E$. Then $OHMN$ is inscribed and :

      $QM \cdot QH = QN\cdot QO = \frac{1}{2} QN \cdot QE = \frac{1}{2} QC^2 = \frac{1}{2} QH^2$

 Hence

    $QM \cdot QH = \frac{1}{2} QH^2$ or $QM = MH$

[u]Babis[/u]

[quote="Valentin Vornicu"]From a point $Q$ on a circle with diameter $AB$ different from $A$ and $B$, we draw a perpendicular to $AB$, $QH$, where $H$ lies on $AB$. The intersection points of the circle of diameter $AB$ and the circle of center $Q$ and radius $QH$ are $C$ and $D$. Prove that $CD$ bisects $QH$.[/quote]

Actually is rather simple provided inversion is applied... Center $Q$ and radius $QH$. The big circle turns into line $CD$ and viceversa. If $X$ is the other point where $QH$ meets big circle and $Y$ is the point of intersection of $CD$ and $QH$ then it must be
$QY \cdot QX = QH^2 \Rightarrow \frac{QY}{QH} = \frac{QH}{QX} = \frac{1}{2}$

Daniel

[quote="M4RI0"][quote="stergiu"]
$\frac{1}{2} QN \cdot QE = \frac{1}{2} QC^2$ [/quote]

Don't understand this part, could someone explain me? :blush:[/quote]

The triangle $ECQ$  is right and  $CN$ is its height . In Greece we called it as <<theorem of  projections>>  and you can prove it by similar triangles.

[quote="M4RI0"][quote="stergiu"]
$\frac{1}{2} QN \cdot QE = \frac{1}{2} QC^2$ [/quote]

Don't understand this part, could someone explain me? :blush:[/quote]

The similarity of the triangles gives us this relation. Since $QN$ is the diameter of the first cirlce we have $CN$ perpendicular to $CQ$. Aslo we know that the diagonals of kites are perpendicular to each other which is : the diagonals of a kite $QCOD$ are $QO$ and $CD$ are perpendicular to each other. Therefore we have a similarity of the triangles $QCP$ and $QNC$. In other words, $\frac{QC}{QN}= \frac{QP}{QC}$, quivalent to $QC^2=QN \times QP$

Note that the typos are according to my own figure
- $O$ is the centre of the first circle
- $Q$ is the centre of the second circle
- $N$ is the intersection of point of $QO$ with the first circle
- $P$ is the intersection point of $CD$ with $QO$

[color=red][Maybe they are the same but i have no time check, sorry][/color]

Davron

Let V be the intersection point of OH with circle O and let N be the intersection point of DC and QH.
Denote angle QDN with x and angle HDC with y.
Since QH=QD we have angleDQH=180-2(x+y) so angleDCH=1/2angleDQH=90-(x+y).
 ALso we have angleDCV=angleDCH+angleHCV=90-(x+y) + angleHCV so angleHCV=90-(x+y) so this means that HC is angle bisector of angleNCV in triangle NCV and from this we conclude that CN/CV=NH/HV or NH=CN/CV*HV (1)
Also we have angleQVC=angleQDC=angleQCD=x and since angleNQC=angleVQC it follows that triangles QNC and QVC are similar so QN/NC=QC/VC or QN=NC/VC*QC (2)
Point H is a feet of perpendicular from O to VQ in an  triangle VOQ (VO=OQ) so QH=HV.
Also QH=QC (radius of circle with radius QH) so we conclude that HV=QC (*)
Now, from (1), (2) and (*) we conslude that NH=NQ so it means that CD bisects QH. q.e.d. :lol: [/i]

[quote="delegat"]Let $V$ be the intersection point of [b][color=red]QH[/color][/b]   with circle $O$  - the center of the first circle - and let $N$  be the intersection point of  $DC$ and $QH$.
Denote  $\angle QDN = x$ and $\angle HDC = y$.
Since $QH=QD$  we have
 $\angle DQH=180-2(x+y)$ so $\angle DCH= \frac{1}{2} \angle DQH=90-(x+y)$.
 Also we have $\angle DCV=angleDCH+ \angle HCV=90-(x+y) + \angle HCV$  , so
  $\angle HCV=90-(x+y)$ .This means that $HC$ is angle bisector of  $\angle NCV$ in triangle $NCV$  and from this we conclude that
  $\frac{CN}{CV} = \frac{NH}{HV}$  or  $NH= \frac{CN}{CV} \cdot HV$  [b](1)[/b]
Also we have
  $\angle QVC= \angle QDC= \angle QCD=x$
  and since  $\angle NQC= \angle VQC$ ,
    it follows that triangles $QNC$  and $QVC$  are similar so
  $\frac{QN}{NC}= \frac{QC}{VC}$  or $QN= \frac{NC}{VC} \cdot QC$  [b](2)[/b]

Point $H$ is a feet of perpendicular from $O$  to $VQ$ in  $\triangle VOQ , (VO=OQ)$ so $QH=HV$ .

Also $QH=QC$  (radius of circle with radius $QH$) so we conclude that $HV=QC$ [b](*)[/b]
Now, from  $(1), (2)$  and [b](*)[/b] we conslude that
  $NH=NQ$ , which  means that $CD$ bisects  $QH$. q.e.d. :lol: [/i][/quote]

[b] I simply made a little correction(...typo)  to this elementary and nice solution![/b]
[u]Babis[/u]

[quote="stergiu"]
$QM \cdot QH = \frac{1}{2} QH^2$ or $QM = MH$
[u]Babis[/u][/quote]

It follows from Euler theorem in right triangles
Since $QE$ is diameter  so triangle QCE is right at $\angle C$ and foot of altitude from $\angle C$ is $N$
 So $CQ^2= QN\times QE$

Let us denote the the circle with centre Q and radius QH by W. Let CD intersect QH at M. Extend QH to meet the circle with diameter AB at M'. (Note that since QH is perpendicular to the diameter AB, H is the midpoint of QM'.)

Now, consider inversion with respect to the circle W. Circle with diameter AB passes through Q, the centre of W. Therefore that circle inverts into a line. Further, the points C and D are fixed under inversion because they lie on W. This implies that the circle with diameter AB inverts into line CD.

Thus the inverse of point M' is M. Now ,
QM' = 2QH.
QM * QM' = (QH) ^ 2   (M is inverse of M')
So, QM = 1/2 QH
Thus M is midpoint of QH.

QED

Please see post 3 , by Daniel  ;)

   [u]Babis[/u]

[quote="xeroxia"]Is it so simple or am I doing something wrong?

Let $M$ be the intersection of $CD$ and $QH$.
$QD = QC \Rightarrow \angle QDC = \angle QCD = \angle QRC \Rightarrow QH^2 = QC^2 = QM\cdot QR = QM\cdot 2QH \Rightarrow QH = 2QM  $[/quote]

This solution is correct,  very nice and simple too, but probably not easier than the solution with inversion !

 This solution has two important points :

- To take the whole circle and produce $AH$

- To prove that $QC$ is tangent with the circle $(R,M,C) $

Of course it is not an national or IMO level  olympiad problem, it is easy for senior students but is still an ellegant and intreresting problems for juniors.

 Thank you for sending your solution !

Babis

[hide="Picture"][img]http://i.snag.gy/Mcw4T.jpg[/img][/hide]

Let $M$ be the midpoint of $QH$, and let $QH$ cut $(AQB), (CHD)$ for a second time at $P,N$ respectively. Now $MQ\cdot MP=(\frac{1}{3} MN)(3 MH)=MN\cdot MH$, so $M$ has equal powers with respect to both circles. Thus $M$ lies on the radical axis $CD$.

[quote="xeroxia"]It is a team selection test problem. I think the proposers missed the easy solution.[/quote]

Yes !

In my notes I see that it is a problem in some Bulgarian olympiad, probably tst 2004.

But it is olympiad problem for some other country too !

Your proof is original and I have not seen it before.

Babis


function
calculus
algebra
polynomial
functional equation
algebra unsolved

Find all functions $f : N \Rightarrow N$ such that
$f(f(f(n))) + 6f(n) = 3f(f(n)) + 4n + 2001$; $n \in N$

[quote="Kondr"]Well I don't have much time so only the idea....
define sequence a_i as a_1=n, a_2=f(n), a_3=f(f(n)) etc.
From the equation we'l get a recursive formula with characteistic polynom which has roots
$1-\sqrt 3i$,$1+\sqrt 3i$  and twice 1, so $a_i=k.2^i.sin(i.\pi/3+\phi)+li+m$. For k not null it diverges very fast so k=0. Pluging $a_i=li+m$ into the equation
4l+m + 6(2l+m) = 3(3l+m) + 4l+4m + 2001,
3l  =  2001, so l=667.
Because of that
f(n)-n=a_2-a_1=667,
f(n)=n+667.[/quote]

I really don't understand you .

Please post it clearier.

Do you use calculus to do it?

Please , Post you solution without calculas!

Where did you stop?
I mean, what can't you understand? The whole thing, or just from a certain point?

Ok, then, let's follow and detail [b]Kondr[/b]'s reasoning :
We have : $f(f(f(n))) + 6f(n) = 3f(f(n)) + 4n + 2001$.
Now, like he says :
[quote="Kondr"]
define sequence a_i as a_1=n, a_2=f(n), a_3=f(f(n)) etc.[/quote]
So, our functional equation becomes :
$a_4 - 3a_3 + 6a_2 - 4a_1 - 2001 = 0$
So, our recursive formula is :
$a_{n+3} - 3a_{n+2} + 6a_{n+1} - 4a_n - 2001 = 0$.
The characteristic polynomial is thus :
$r^3 - 3r^2 + 6r - 4 = 0$
And, from this, if we solve in $\mathbb{C}$, we get :
$r = 1 - \sqrt{3}i$ or $r = 1 + \sqrt{3}i$ or $x=1$(1 being a double root).

So far, do you understand?
If yes, then I'll go on. ;)

Thank you very very very much ,mathmanman. You are very kinds of me!

But Ok!I don't anything know about characteristic polynomial.

Is it hard or long ? Give me some definitions about that please.

Ps.Thank you very much

Ok! :)

Then, I suggest you this :
http://www.mathpages.com/home/kmath003/part3/sympart3.htm

And, even better :
http://mathcircle.berkeley.edu/BMC3/Bjorn1/Bjorn1.html

You're welcome. ;)


It was a problem from TTT 2 in Vietnam :D


inequalities
function
inequalities unsolved

Let $a,b,c$ be the sides of a triangle. Prove that:
$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$

I made a part of the solving but cant get to its end :(. Here it is:
$a=y^2+z^2;b=x^2+z^2;c=x^2+y^2$
Now
$x\sqrt{2}+y\sqrt{2}+z\sqrt{2}\leq\sqrt{x^2+y^2}+\sqrt{x^2+z^2}+\sqrt{y^2+z^2}$
$2xy+2yz+2xz\leq\sqrt{(x^2+y^2)(x^2+z^2)}+\sqrt{(x^2+z^2)(y^2+z^2)}+\sqrt{(x^2+y^2)(y^2+z^2)}$
Since
$\sqrt{(x^2+y^2)(x^2+z^2)}+\sqrt{(x^2+z^2)(y^2+z^2)}+\sqrt{(x^2+y^2)(y^2+z^2)}\geq\sqrt{2xy(x^2+z^2)}+\sqrt{2xz(y^2+z^2)}+\sqrt{2yz(x^2+y^2)}$
Somebody :roll: just has to prove that:
$\sqrt{2xy}(\sqrt{2xy}-\sqrt{x^2+z^2})+\sqrt{2xz}(\sqrt{2xz}-\sqrt{y^2+z^2})+\sqrt{2yz}(\sqrt{2yz}-\sqrt{x^2+y^2})$

$xy+yz+zx\leq x^2+y^2+z^2$ but I think this isnt enough to solve that problem  :huh:

Here is a nicer solution for this one: We only use Cauchy-Schwarz inequalilty in the form $\sqrt{x}+\sqrt{y}\leq\sqrt{2(x+y)}$.Applying this inequality we get $\sqrt{b+c-a}+\sqrt{c+a-b}\leq\sqrt{2\cdot 2c}$ and we write the similar ones and by adding them we get the desired result. ;)

[quote="Gohan"]Let $x,y,z$ be the sides of a triangle. Prove that:
$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$

I made a part of the solving but cant get to its end :(. Here it is:
$a=y^2+z^2;b=x^2+z^2;c=x^2+y^2$
  [/quote]

Let's go from the beginning, nice and easy.... what of these are sides of a triangle $a,bc$ or $x,y,z$

 :)  :)  :)  :)

bye......

[quote="cezar lupu"]We only use Cauchy-Schwarz inequalilty in the form $\sqrt{x}+\sqrt{y}\leq\sqrt{2(x+y)}$[/quote]

Huh .. :huh: I wish I knew all tha formulas :P.

Its  APMO  1996 . An  easy solution is to consider  [tex] a\geq b\geq c[/tex] . But  then  [tex](a+b-c,c+a-b,b+c-a)>(a,b,c).[/tex] Aply Karamata  for  function  [tex] f(x)=\sqrt x [/tex]

Maybe,you shouldn't call it Karamata's inequality because it isn't his. Actually this is Hardy-Littlewood-Polya inequality. ;)


Cho m,n,x,y,z là các số thực dương.Chứng minh rằng:
             (mx+ny)/(mz+nz)+(my+nz)/(mx+nx)+(mz+nx)/(my+ny)>=3
 Xin lỗi các bác em khôgn biết đánh kí hiệu nên nó hơi khó đọc. Bác nào dạy em cái!!

Bất đẳng thức đã cho tương đương với:
m( x/z + z/y + y/x) + n( y/z + z/x + x/y) >= 3(m + n)
Áp dụng BĐT Côsi (vì x, y, z, là các số thực dương) ta được :
(x/z + z/y + y/x) >= 3 và (y/z + z/x + x/y) >= 3
Vậy:
m( x/z + z/y + y/x) + n( y/z + z/x + x/y) >= 3m + 3n = 3(m + n)
Đẳng thức xảy ra khi và chỉ khi x = y = z

[quote="tmathst"]
 Xin lỗi các bác em khôgn biết đánh kí hiệu nên nó hơi khó đọc. Bác nào dạy em cái!![/quote]
Cố thử dịch cái này nhé bạn http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php


What is the positive difference between the two solutions of $ |x \plus{} 5| \equal{} 20$?

one solution is if x+5=20, so x=15.  The other solution is if x+5= -20, so x=-25.  The positive difference ie absolute value equals 15--25=40.

You can generalize and say that for any equation of this type $ |{mx \plus{} a}| \equal{} n$, the postive difference between the solutions is $ |\frac {2n}m|$. This is because:

$ |{mx \plus{} a}| \equal{} n$
so
$ mx \plus{} a \equal{} n$
$ x \equal{} \frac {n \minus{} a}m$
or
$ mx \plus{} a \equal{} \minus{} n$
$ x \equal{} \frac { \minus{} n \minus{} a}m$
Finding the difference gives
$ \frac {n \minus{} a}m \minus{} \frac { \minus{} n \minus{} a}m \equal{} \frac {2n}m$
So positive difference is $ |\frac {2n}m|$

So the answer here is $ |\frac {2*20}1| \equal{} 40$

$ \left|\frac{2n}{m}\right|$ looks nicer


A farmer plants seeds for a $ 75$-acre field of yellow sweet clover. A $ 25$-pound bag of seed costs $ \$24$. How much would it cost, in dollars, to seed the field if $ 12$ pounds of seed were used per acre?

[hide]
Multiply $ 75$ by $ 12$ to find how many pounds were used then divide by 25 to find the amount of bags used and then multiply by 24.
\[ \left(\frac{75\cdot12}{25}\right)24\equal{}36\cdot24\equal{}\boxed{864}
\]
[/hide]


Prove that there is NOT exist function at $R$ such that   $f(f(x))=-x^3$ for each $x\in R$

Are you sure that you do not miss the continuity?

Pierre.

Ok, you are right. I thought that any orbit was infinite but this is not true...
Assume that such a function does exist. Note that clearly $f$ is bijective from $\mathbb {R}$ onto itself.
Let $a=f(1).$
Then $f(a)=-1,f(-1)=-a^3,f(-a^3)=1,f(1)=a^9.$
Therefore $a=a^9$ which forces $a=0,-1$ or $1$.

But, if $a=1$ then $f(1)=1$ so that $1=f(f(1))=-1$, a contradiction.
If $a=-1$ then $f(-1)=f(f(1))=-1.$ This leads to $f(1)=f(-1)$, a contradiction.
If $a=0$ then $f(0)=f(f(1))=-1$ and $f(-1) = f(f(0))=0$ so that $f(1)=f(-1)$, a contradiction. And we are done.

Pierre.


function
algebra
polynomial
induction
number theory proposed
number theory

Let $f(n)=\sum_{k=0}^{n-1}x^ky^{n-1-k}$ with, $x$, $y$ real numbers. If $f(n)$, $f(n+1)$, $f(n+2)$, $f(n+3)$, are integers for some $n$, prove $f(n)$ is integer for all $n$.

Nice!! Observe that $ f(n+1)f(n-1)-f^2(n)=-(xy)^{n-1}$ and so it follows that $ (xy)^n$ and $ (xy)^{n+1}$ are integers. Thus xy is rational and algebraic integer and so it is integer. Next, observe that you can write $f(n)$ as a polynomial with integer coefficients in xy and x+y (or, at least, it seems it is true by cheking small cases, so it must be true  :D ) such that $ (x+y)$ appears with maximal power alone (not multiplied with xy) and with coefficient 1. It thus follows that x+y is algebraic integer and rational (since  f(n), f(n+1), f(n+2) and xy are integers) and thus it is integer. The conclusion follows. This, if the claim with the polynomial is true.  ;)

Sorry, I was idiot, the assertion with the polynomial comes by induction from the fact that $ f(n+1)=(x+y)f(n)-xyf(n-1)$ and xy integer. The assertion is: for all n, f(n) has the form $ f(n)=(x+y)^{n-1}+P(x+y,xy)$ where P is a polynomial with integer coefficients and such that all powers of x+y has exponent at most n-2.

This was also used in the 1995 Bulgarian olympiad.


What is the greatest positive integer $n$ such that $n^2$ is less than 95?

i remember answering a question like this...
[hide]so 8^2=64
9^2=81
10^2=100
so 9 is the answr.[/hide]

[quote="venkat3848"]its 9. :D[/quote]
I know this is pretty obvious but please say WHY it's [hide="~"]9[/hide]!


If $x,y,z>0$  and $x+y+z=1$  prove that

$xy+yz+zx\geq 4(x^2y^2+y^2z^2+z^2x^2)+5xyz$

[quote="sqing"]Serbia MO 2008:
If $x,y,z>0$  and $x+y+z=1$  prove that\[x^2+y^2+z^2+3xyz\ge\frac{4}{9}\].[/quote]

The problem is equivalent to:
$  x,y,z>0,x+y+z=3\Rightarrow x^2+y^2+z^2+xyz\ge 4 $
Following result is known:
(see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=369554)
$ x,y,z>0\Rightarrow x^2+y^2+z^2+xyz+5\ge 3(x+y+z) $
Using this result we obtain QED.
___________
Sandu Marin

Inequality is equivalent to:
1) $ x,y,z>0,x+y+z=3\Rightarrow 9(xy+yz+zx)\ge4(x^2y^2+y^2z^2+z^2x^2)+15xyz $
$ xy+yz+zx=s\&(x,y,z>0,(x+y+z)^2\ge 3(xy+yz+zx))\Rightarrow 0<s\le 3 $
After calculations and simplifications (1) be rewritten as:
2)  $ 9xyz\ge 4s^2-9s $
and must be true for $ x,y,z>0,0<s\le 3 $ ;obviously (2) is true for $ 0<s\le\frac{9}{4} $,
there remains a likely for $ \frac{9}{4}\le s\le 3 $
Applying Schur's inequality, we obtain:
$ (x+y+z)^3+9xyz\ge 4(x+y+z)(xy+yz+zx)\Leftrightarrow 9xyz\ge 12s-27 $
We show that:
$ \frac{9}{4}\le s\le 3\Rightarrow 12s-27\ge 4s^2-9s\Leftrightarrow 4s^2-21s+27\le 0\Leftrightarrow (s-3)(4s-9)\le 0 $
Since $ \frac{9}{4}\le s\le 3\Rightarrow (s-3)(4s-9)\le 0 $
proof is complete!!
_____________
Sandu Marin


I recently read about proofs involving Ramsey numbers, and I understand a basic definition of Ramsey's Theorem, but I am not sure how to apply it. Could any of you please enlighten me as to where Ramsey Numbers become useful?

Please post some challenging problems, and tell me (as well as the entire AoPS community) what Ramsey Numbers exactly are. Post some interesting facts if you'd like :D

Suppose you have a regular n-gon with all the diagonals drawn in. You want to color each diagonal with one of k colors so that you don't have an a1-gon with all diagonals in color 1, a regular a2-gon with all diagonals in color 2, etc. What is the largest possible value of n?

That is the standard Ramsey Theory problem. You can make variations on it (for example, what if you only want the sides and not the diagonals to be not monochromatic).

Ramsey Theory problems tend to be incredibly hard. Here's a seemingly very simple problem that is actually unsolved (at least, as of a few years ago, and I doubt anyone has solved it since):

What is the smallest value of n so that if you color all the sides and diagonals of a regular n-gon red or blue, you are guaranteed to have a regular pentagon with all diagonals in either red or blue?

In space, there are given p_n=[en!]+1 points, where [x] is the least integer function
Each pair of points is connected by a line, and each line is colored with one of n colors.  Prove that there is at least one triangle with sides of the same color

Lol. Ramsey numbers...brings back some pretty weird memories.

I remember at MathPath 2003 we were introduced to Ramsey numbers. One of my friends and I spent about a week trying to come up with some formula to express them (the dichromatic Ramsey numbers, anyway.) I remember we came up with some pretty fair approximations.


So I skimmed through Mathematical Olympiad Treasures on Amazon and my jaw dropped.  Could I get a recommendation on a textbook for harder Algebra problems such as the ones I saw in MOT book.

[quote="t0rajir0u"]Mathematical Olympiad Treasures is a problem-solving book.  The subjects it handles are not meant to be approached in a "textbook" manner: you won't find any large theorems from which calculations can be done mechanically, such as in, say, differential calculus.

That having been said, [u]Problem-Solving Strategies[/u] by Arthur Engel is a good book that approaches Olympiad subjects in a somewhat more cohesive manner.[/quote]Aw, that's sad to know.  I am currently in Calculus 2 but I find myself encountering Algebra problems (student tutor) that I've never seen before and usually I can solve them but some definitely give me a hard time and some I can't even solve.  I just love problem-solving, thanks!

Apparently, you already know some problem-solving, so I would recommend you Problem-Solving Strategies too... after all the book is written by a German, so it has to be great :P

[/snobism]


Prove that if a sequence $ \{G(n) \}_{n\equal{}0}^{\infty}$ of integers satisfies: $ G(0)\equal{}0, G(n)\equal{}n\minus{}G(G(n))\quad (n\equal{}1,2,3,...)$ then:
$ (a)$ $ G(k) \ge G(k\minus{}1)$ for any positive integer $ k$
$ (b)$ no integer $ k$ exists such that $ G(k\minus{}1)\equal{}G(k)\equal{}G(k\plus{}1)$.

is easy to pove for all $ n > 0 G(n) < > 0$ because if not $ G(n) \equal{} 0 \equal{} n \minus{} G(G(n)) \equal{} n \minus{} G(0) \equal{} n$ so $ n \equal{} 0$ absurde

and for all $ n > 0 n G(n) \equal{} n \minus{} G(G(n)) \le n \minus{} 1$

1) by recurence (fort) ( tha for all $ n$ $ G(n)\geq G(n \minus{} 1) \geq ....\geq G(0)$

for $ n \equal{} 1$  $ G(1)\in N$ so $ G(1) \geq 0 \equal{} G(0)$

suppose that is true for  $ n$ and prove that is true for $ n \plus{} 1$

$ G(n \plus{} 1) \equal{} n \plus{} 1 \minus{} GG(n \plus{} 1)$  you have $ G(n \plus{} 1) \le n$ so  $ G(n) \geq G(G(n \plus{} 1))$ so

$ G(n \plus{} 1) \equal{} n \plus{} 1 \minus{} GG(n \plus{} 1) \geq n \plus{} 1 \minus{} G(n) \equal{} G(n) \plus{} 1 > G(n)$

2) if $ G(n) \equal{} G(n \minus{} 1)$ ===> $ n \minus{} G(G(n)) \equal{} n \minus{} 1 \minus{} G(G(n \minus{} 1)) \equal{} n \minus{} 1 \minus{} G(G(n))$ ===>$ 1 \equal{} 0$ absurde so

$ G(n)>G(n\minus{}1)$

[quote="pco"][quote="mohamed-01-01"] ... $ G(n \plus{} 1) \equal{} n \plus{} 1 \minus{} GG(n \plus{} 1) \geq n \plus{} 1 \minus{} G(n) \equal{} G(n) \plus{} 1 > G(n)$ ...[/quote]
Hello mohamed-01-01!

Why is "$ n \plus{} 1 \minus{} G(n) \equal{} G(n) \plus{} 1$" true ?[/quote]

oh sorry my solution is wrong but I think there is a mistake in this probleme

because if $ G(n) \geq G(n \minus{} 1)$

and you have for all $ n$ $ G(n) < > G(n \minus{} 1)$ because if $ G(n) \equal{} G(n \plus{} 1)$ ==>

$ n \plus{} 1 \minus{} G(G(n \plus{} 1)) \equal{} n \minus{} G(G(n)) \equal{} n \minus{} G(G(n \plus{} 1))$ so $ 1 \equal{} 0$

so $ G(n \plus{} 1) > G(n)$ ===>$ G(n) \geq n$

so $ G(n) \equal{} n \minus{} G(G(n)) \le n \minus{} G(n) \le n \minus{} n \equal{} 0$ ???

[quote="mohamed-01-01"] oh sorry my solution is wrong but I think there is a mistake in this probleme

because if $ G(n) \geq G(n \minus{} 1)$

and you have for all $ n$ $ G(n) < > G(n \minus{} 1)$ because if $ G(n) \equal{} G(n \plus{} 1)$ ==>

$ n \plus{} 1 \minus{} G(G(n \plus{} 1)) \equal{} n \minus{} G(G(n)) \equal{} n \minus{} G(G(n \plus{} 1))$ so $ 1 \equal{} 0$

so $ G(n \plus{} 1) > G(n)$ ===>$ G(n) \geq n$

so $ G(n) \equal{} n \minus{} G(G(n)) \le n \minus{} G(n) \le n \minus{} n \equal{} 0$ ???[/quote]

There is a little mistake in your demo but your conclusion seems correct to me : $ G(n)\ne G(n\plus{}1)$ only for $ n>0$, so maybe $ G(1)\equal{}G(0)$ So, if the required non decreasing assertion is true, we have $ G(n)\geq n\minus{}1$ (and not $ G(n)\geq n$). So $ G(G(n))\geq n\minus{}2$ and $ n\equal{}n\minus{}G(G(n))\leq 2$ !
And so it seems indeed that there is a mistake in this problem ...  :huuh:


geometry
incenter
circumcircle
conics
ellipse
perpendicular bisector
geometry unsolved

Let AD  be the altitude of a triangle  ABC  and  E , F  be the incenters of the triangle  ABD  and  ACD , respectively. line  EF  meets  AB  and AC at  K  and L. prove tht AK=AL if and only if AB=AC or A=90

Let I be the incenter of the triangle $\triangle ABC$. We have to show that the bisector AI of the angle $\angle A$ is perpendicular to EF iff either AB = AC or the angle $\angle CAB = 90^\circ$ is right. The bisectors $AI, BI \equiv EI, CI \equiv FI$ are concurent cevians of the triangle $\triangle AEF$. Let $X \in AF, Y \in AE$ be the feet of EI, FI in this triangle.

$\angle EXA \equiv \angle BXA = 180^\circ - \frac{\angle B}{2} - (90^\circ - \angle B) - \frac{90^\circ - \angle C}{2} =$

$= \frac{90^\circ + \angle B + \angle C}{2} = 90^\circ + \frac{90^\circ - \angle A}{2}$

$\angle FYA \equiv \angle CYA = 180^\circ - \frac{\angle C}{2} - (90^\circ - \angle C) - \frac{90^\circ - \angle B}{2} =$

$= \frac{90^\circ + \angle C + \angle B}{2} = 90^\circ + \frac{90^\circ - \angle A}{2}$

Hence, $\angle EXA = \angle FYA$. If the angle $\angle A = 90^\circ$ is right, EX, FY are 2 altitudes of the triangle $\triangle AEF$, I its orthocenter and $AI \perp EF$ its remaining altitude. Assume now that the angle $\angle A \neq 90^\circ$ is not right and that we still have $AI \perp EF$, i.e., AI is the A-altitude of the triangle $\triangle AEF$, but I is no longer its orthocenter. Since $\angle EXA = \angle FYA \neq 90^\circ$, then also $\angle EXF = \angle FYE \neq 90^\circ$, i.e., the quadrilateral EFXY is cyclic. Let (P) be its circumcircle, which is centered on the perpendicular bisector of the segment EF. Using parallel projection, we can project the triangle $\triangle AEF$ into a triangle $\triangle A'EF$, so that the point I is projected into its orthocenter I'. Then the cevians EX, FY are projected into its altitudes EX', FY', i.e., the quadrilateral EFX'Y' is also cyclic, with the circumcircle (O') centered at the midpoint O' of the segment EF. Thus the points X, Y lie both on the circle (P) and on an ellipse $o$ with the main axis EF (which is projected into the circle (O') in our parallel projection), not identical with the circle (P). Since both the ellipse $o$ and the circle (P) are symmetrical with respect to the perpendicular bisector of the segment EF, so are their intersections X, Y, i.e. EX = FY and EY = FX. Hence, the triangles $\triangle EFX \cong \triangle FEY$ are congruent and it immediately follows that the triangle $\triangle AEF$ is isosceles with AE = AF. This and $AD \perp BC$ implies that the incircles $(E) \cong (F)$ are congruent, which means that the triangle $\triangle ABC$ itself is isosceles with AB = AC.


Out of curiousity, do you all take Senior A?

Any idea why they're written at so many times of the year?  I mean, Calgary doesn't write them for another month or so.

What is this?  Have a link?  Thanks.

-The Living in Closet all life person

[quote="billzhao"]Just a reminder that the problems should be not posted or discussed before they are officially posted on the website, since other cities may write the exam at a later date;[/quote]

I have no idea when Edmonton is writing this. :? Does anyone know? :D

Oh, William, you've never heard of Tournament of Town? I guess they don't hold it at Waterloo. Here is link for T of T in Toronto, [url]http://www.math.toronto.edu/oz/turgor/news.php[/url]. It's basically a 4/5 hour contest primarily organized by some Russian math commitee and hold in a few cities in Europe and North America. I think the contest is rather an informal one, but in a sense, make Olypiad level problems more approachable to beginner like me. The contest is fairly nice. O-Level is not that challenging, A-Level is a bit hard for me...

BTW, Alex, I did horrible today. I solved only two problems and BSed for couple others.  I wrote 3 pages of trigs for #5, and couldn't prove it in the end.... Was you there today? I didn't even see you.

yes i was there; an hour into the end i left because i knew i couldnt break a max-O-level score which I am sure that i could get later anyway.

btw, solving 2 problems and bsing #5 is very good.  Isn't this close to a perfect score? (assuming "solving 2 problems" doesn't include #1)

I had very strong ideas on how to solve all, but couldn't bring them out.

[quote="TheDreamer"]Is this the same thing as mathbattle?

Either way, I'm looking for activities for our math club to join as a team.  Maybe this, mathbattle, and MCTIC (CARML) would be good candidates?

BTW Alan good to hear from you again!  Haven't seen you in so long  :)[/quote]

I'm not sure what exactly mathbattle is, but I'm quite sure that it's different.
   This competition gives you 5~7 problems(I don't remember the exact number) and 4~5 hours(I don't remember this either) to solve them.  Each problem is weighted differently and they only take the best three(right?) problems that you solved. There are two rounds in a year, fall and spring round. There are also two levels, O level for grade 10 and below and A level for grade 11 and above(but they have some overlapping problems). This is an individual competition. There are no prizes(except for a certificate for doing ok), but the problems are [i][b]very[/b][/i] beautiful.

[quote="lightrhee"]There are also two levels, O level for grade 10 and below and A level for grade 11 and above(but they have some overlapping problems).[/quote]

You got it confused there.  There are 2 levels, Junior for grade 10 and under, and Senior for grade 11 and over.  There are two rounds, O-Level written first, the easier of the two, and A-Level which is harder.

Yea, this is different from mathbattle. This is more or less in a format similar to
CMO(although much longer). T of T is individual contest, whereas mathbattle is a team competition. MathBattle is a lot more fun though. It's like you have a set of problems and your team will work on them for like 2.5 hrs. And you and your opposing team take turn in showing solution to problems by the other team. You will have rebuttals, judges too. It's strategic sometimes too, not always depend on how many prolbems you solved.  There are more info. on the website I provided in the above post. Oh, we get free pizza and pop too. So free lunch, if that is an incentive for anyone to come.

BTW, William you are going to have a team from your school come to Toronto to join the MathBattle? Well, Waterloo is not that far from Toronto after all. I would be glad to see you there. I knew couple others people from your school also.

Hmm that would be really cool.  I'll see how much interest there is in our math club.  We wouldn't really stand a chance though :s, so it may not be such a good idea, making ppl lose confidence lol

How large is your Math Club? We only have 10 members at most.

We had 25 the first time, with another 25 who said they'll come next time.  Although prolly some of them will just keep procrastinating and won't come at all, so i'm guessing in the end we'll have around 30, which is about right i think


geometry
perimeter
trigonometry
function
complex analysis
parameterization
parametric equation

A mark is placed on the perimeter of a wheel where it touches the road. The wheel is rolled along the road without skidding though less than half a rotation. The mark has advanced horizontally from its starting position by x. Given x and the radius of the wheel (r) what is the height (y) of the mark above the road?

It seems a simple problem but as far as I can gather there is no solution to this question that does not involve numerical aproximation methods.

[quote="mlok"]Imagine that the wheel was squeaky clean at the beginning, and now a part of it (i.e. a circular arc) is covered with road dirt. The length of this arc is $ x$, therefore ...

Also, this is not complex analysis, and not even calculus.[/quote]

You have mis-interpreted the question.

The dusty arc represents the distance the wheel the has rolled along the road and this distance (call it z) is given by $ R\theta$ wher $ \theta$ is the angle the wheel has turned through. The x in the question is the distance the mark has traveled along the x axis and this is given by $ x\equal{}r(\theta\minus{}\sin(\theta))$ or $ x\equal{}z\minus{}\sin(\theta)$ and is not the same as the distance the wheel has rolled. (The mark traces out a cycloid). The parametric equation for a cycloid defines y as $ R(1\minus{}\cos(\theta))$ but I am looking for y in terms of x when $ \theta$ is not given.

If you think that is an easy question I'll give $ \$$50 to a nominated charity or to whoever answers the question (if that is allowed) if you provide an equation for y in terms of x where the angle the the wheel has turned through $ (\theta)$ is not given.

Rules:

The answer must be given in the form  of a cartesian equation with y  and only y on the left hand side and where $ \theta$ or y do not appear on the right hand side of cartesian equation.

By the way, if you know y then there is a cartesian equation for x that does not include $ \theta$:

$ x\equal{}\pm\arccos\left({R\minus{}y \over R}\right)R\pm\sqrt{2Ry\minus{}y^2}$

All you have to do is solve that equation for y ;)

Would have been nice if you had actually read the question properly before moving it.

Don't waste your time looking. The inverse function $ y(x)$ is not an elementary function. There is no reason to expect transcendental equations to be solvable in elementary terms.

In practice, you can do anything you want with the parametric form. Numerical methods also work well here.


real analysis
modular arithmetic
function
real analysis unsolved

A cute result :):

Let $T$ be a measure preserving invertible transformation on the unit interval $I$ (with the usual Lebesgue measure), such that for almost all $x\in I,\ T^{n}x=x$, and $x,\ Tx,\ \ldots,T^{n-1}x$ are distinct.

Prove that there is a measurable subset $E$ of $I$ with measure $\frac{1}{n}$ such that $E,\ T(E),\ \ldots,T^{n-1}(E)$ are pairwise disjoint.

While we're all trying to prove this, let me state, trivially, that the topic is not vacuous. It is possible to construct such a $T.$

The simplest such example I can think of is to let $T(x)\equiv x+\frac1n\pmod{1}.$

(And in that case, $E=\left[0,\frac1n\right)$ can be the set we want.)

Here's an attempt.  Let $J$ be the set of points $x$ that have period $n$.  By the hypothesis, $J$ has measure 1.

Say that a point $x$ is [i]early[/i] if $x \le T^{k}x$ for all integers $k$.  Let $E$ be the set of early points in $J$.  Then $E$ is measurable because, um, well, I don't really know why.  (How do I even show that the set of $x$ such that $x \le Tx$ is a measurable set?)  But I will assume that $E$ is measurable.

By the given property, it is easy to see that $E$, $TE$, $\ldots$, $T^{n-1}E$ form a partition of $J$.  All these $n$ sets have the same measure because $T$ is measure-preserving.  Thus $E$ has measure $\frac{1}{n}$.

Neat :). My proof was a bit more involved. Anyway, regarding your question:

One possible definition for the product $\sigma$-algebra of Lebesgue measurable subsets of $I\times I$ is: the smallest $\sigma$-algebra that will make both projections $\pi_{i},\ i=1,2$ from $I\times I$ onto $I$ (i.e. the projections on the two factors) measurable. This implies that a map $S: I\to I\times I$ is measurable iff $\pi_{i}\circ S$ are both measurable self-maps of $I$.

Now let $Sx=(x,Tx)$. By the above characterization of measurable functions from $I$ to $I\times I,\ S$ is measurable. Your set $\{x\ |\ x\le Tx\}$ is the preimage of the (measurable, of course) triangle $\{(x,y)\ |\ x\le y\}$ through $S$, and is thus measurable. And when you do this for every $T^{k}$ and take intersections as $k$ goes from $1$ to $n-1$, you find the set of early points to be measurable, as desired.


modular arithmetic
polynomial
number theory unsolved
number theory

If $ a^2\plus{}ab\plus{}b^2$ is a prime$ (a,b \in Z)$,prove that $ a^2\plus{}ab\plus{}b^2\equiv 1 \pmod 3$

Try $ p \equal{} 3$.

$ p \equal{} a^2 \plus{} ab \plus{} b^2 \Leftrightarrow$
$ 4p \equal{} (2a \plus{} b)^2 \plus{} 3b^2 \implies$
$ (2a \plus{} b)^2 \equiv \minus{} 3b^2 \bmod p \implies$
$ \left( \frac {2a \plus{} b}{b} \right)^2 \equiv \minus{} 3 \bmod p$

which is possible if and only if $ p \equal{} 3$ or $ p \equiv 1 \bmod 3$.  (In the other direction:  $ p \equiv 1 \bmod 3 \implies \exists x : p | x^2 \plus{} 3 \equal{} (x \plus{} \sqrt { \minus{} 3})(x \minus{} \sqrt { \minus{} 3})$, which implies that $ p$ is not prime in the [url=http://mathworld.wolfram.com/EisensteinInteger.html]Eisenstein integers[/url].)

:arrow: If $ 3|a \equal{}> a\equal{}3$.
We have $ A \equal{} b^2\plus{}3b\plus{}9$
If $ b \vdots 3 (b\equal{}3)$, $ A\equal{}27$, which is not a prime.
So b is not divisible by 3, then $ b^2 \equiv 1 (mod 3) \equal{}> A \equiv 1 (mod 3).$

 :arrow: If a and b are not divisible by 3, $ a^2\plus{}b^2 \equiv 1\plus{}1\equal{}2 (mod 3).$
$ ab \equiv 1 or 2 (mod 3)$
Then, $ A \equiv 0 or 1 (mod 3)$.
Because A > 3, $ A \equiv 1 (mod 3)$.

Hence, $ A \equiv 1 (mod 3)$ if A is a prime.

Another problem

If $ 3k\plus{}1$ is a prime , prove that it can be a form like "$ a^2\plus{}ab\plus{}b^2(a,b\in Z)$"

l'm sorry.My English is poor,so it's hard for me to translate the problems from Chinese into English. :fool:

Already solved that problem in the last line.  See also my comments on the corresponding theorem for $ a^2 \plus{} b^2$ [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].

[quote="skywalkerJ.L."]If $ a^2 \plus{} ab \plus{} b^2$ is a prime$ (a,b \in Z)$,prove that $ a^2 \plus{} ab \plus{} b^2\equiv 1 \pmod 3$[/quote]

if $ a\equal{}b\equal{}1$ ? then $ a^2\plus{}b^2\plus{}ab\equal{}3$ !!!!!!!!


So, what are some formulas that are usefull?

The reason nobody's responding to your post is that there aren't just a few formulas that are useful. Anyway, formulas are not the important thing. They really don't help much when solving problems, except for the formulas that we all know and use commonly.

Search Mathworld if you want to know something.  It's much easier than memorizing obscure formulae.


There is something about matrices that I have found out... and I attempted to see if my idea has ever been thought of before, and I found nothing.
I need someone to help me find out if this idea was pre-existant and I just came up with something that already existed.

What its about: Matrices

Uses principles presented in: Guassian Elimination, Gauss-Jordan Elimnation

Here it is... Im posting it up

Basic idea:

Given any sets of three matrices where AX=B, one may perform any identical Gaussian elimination operations to both A and B and the equation will remain equal.

What this can be used for:

Instead of needing to find the inverse of Matrix A, and multiply both sides by A, one can simply manipulate both sides until A is the identity.  :P

Yes, this is well known. If you look at it closely you will see that it's just a slight extension of the Gauss-Jordan method for solving systems of equations.

The good thing about it is that it doesn't just work for equations of 2x2 or nxn size, but also for matrix equations of any size (something that Cramer's Rule doesn't do). It also ultimately turns out to be the fastest and most efficient way of solving matrix equations for large matrices.

You've re-discovered a really great thing here! See if you can use it to try and solve equations of non-square matrices.


inequalities
induction
rearrangement inequality
inequalities proposed

The inequality to prove rewrites as $ A = (\sum x_k^2y_k)(\sum y_k) - (\sum x_ky_k)^2\ge B = \frac14(\sum y_k)(\sum y_kx_kx_{k - 1})$. From the identity $ (\sum a_i^2)(\sum b_i^2) - (\sum a_ib_i)^2 = \displaystyle\sum_{i < j}(a_ib_j - a_jb_i)^2$, we have $ A = \sum_{i < j}y_iy_j(x_i - x_j)^2$. We are going to resort now to induction.

For $ n = 1,2$ the inequality can be verified easily. For the step of the induction, it is sufficient to prove that
$ y_n\displaystyle\sum_{k = 1}^{n - 1}y_k(x_n - x_k)^2\ge \frac {y_nx_nx_{n - 1}(y_1 + \ldots + y_{n - 1} + y_n)}4 +$ $ y_n\cdot\frac {y_2x_2x_1 + \ldots + y_{n - 1}x_{n - 1}x_{n - 2}}4$.

Since $ x_k\le x_n$, it is enough to prove the last inequality for $ x_1 = \frac {x_2}2 = \ldots = \frac {x_n}n$. Actually we can assume $ x_i = i$, $ i = 1,2,\ldots,n$. Hence we are to prove
$ \displaystyle\boxed{\sum_{k = 1}^{n - 1}y_k(n - k)^2\ge\frac {n(n - 1)(y_1 + \ldots + y_n)}4 + \frac {y_2(2^2 - 2) + \ldots + y_{n - 1}[(n - 1)^2 - (n - 1)]}4}$.
Let ${ Y = y_1 + \ldots + y_{n - 1}}$.
From $ y_n\le y_{n - 1}\le\ldots\le y_1$ and rearrangement inequality we deduce the following:

[b]A.[/b] $ y_1 + \ldots + y_n\le \frac n{n - 1}Y$.

[b]B.[/b] $ y_2 + \ldots + y_{n - 1}\le \frac {n - 2}{n - 1}Y$.

[b]C.[/b] $ \displaystyle\sum_{k = 1}^{n - 1}y_k(n - k)^2\ge \frac {Y[1^2 + 2^2 + \ldots + (n - 1)^2]}{n - 1} = \frac {n(2n - 1)}6Y$.

[b]D.[/b] $ E = y_2(2^2 - 2) + \ldots + y_{n - 1}[(n - 1)^2 - (n - 1)]\le$ $ \frac {(y_2 + \ldots + y_{n - 1})[1^2 - 1 + 2^2 - 2 + \ldots + (n - 1)^2 - (n - 1)]}{n - 2}$.

[b]E.[/b] $ E\le\frac Y{n - 1}[\frac {(n - 1)n(2n - 1) - 3n(n - 1)}6]$.

Hence, it would be enough to have that $ \frac {n(2n - 1)}6Y = \frac {n^2}4Y + Y[\frac {n(2n - 1) - 3n}{24}]$, which is an identity.


logarithms
induction
algebra unsolved
algebra

given the array $ x_{n}$: that:
$ x_{1} \equal{} c$ ,c is an integer
$ x_{n \plus{} 1} \equal{} cx_{n} \plus{} \sqrt {(c^2 \minus{} 1)(x_{n}^2 \minus{} 1)}$
prove that $ x_{n}$ is the array of integers

Let $ c\equal{}ch(a)$, were $ a\equal{}\ln {(c\plus{}\sqrt{c^2\minus{}1})}$. Then by induction $ x_n\equal{}ch(na)\equal{}P_n(ch(a)$, were
\[ P_n(x)\equal{}\sum_{k\equal{}0}^{[n/2]}C_n^{2k}x^{n\minus{}2k}(x^2\minus{}1)^k\]- polinom with integer coefficients.
Therefore \[ x_n\equal{}\sum_{k\equal{}0}^{[n/2]}C_n^{2k}c^{n\minus{}2k}(c^2\minus{}1)^k\] - is integer.

thanks Rust
but ,I need a nice solution
 :) any idea???

[quote="xxp2000"]If $ c\le \minus{} 1$, the sequence is periodic $ c,2c^2 \minus{} 1,c,2c^2 \minus{} 1,\cdots$.

If $ c \equal{} 0$, the sequence is periodic $ 0,1,0,1,\cdots$

If $ c\ge 1$, the sequence is $ x_{n \plus{} 1} \equal{} 2cx_n \minus{} x_{n \minus{} 1}$. It can be shown by induction.[/quote]
thanks
 :)


let $ a,d\geq 0$,$ b,c>0$, such that $ a\plus{}d\leq b\plus{}c$

find the minimum of

$ \frac{b}{c\plus{}d}\plus{}\frac{c}{a\plus{}b}$

hello, after my conjecture is ${ \frac{b}{c+d}+\frac{c}{a+b}\geq\frac{1}{2}(2\sqrt{2}-1)}$. The equal sign will obtain for $ a=\frac{1}{2}+\frac{\sqrt{2}}{2},b=\frac{1}{2}(\sqrt{2}-1),c=1,d=0$.
Sonnhard.


prove that doesnot exist the function f: R -> R satisfy   f(f(x)) = x^2 - 2008

[quote="caothujjj"]prove that doesnot exist the function f: N -> N satisfy   f(f(x)) = x^2 - 2008[/quote]

$ x\equal{}1$ implies $ f(f(1))\equal{}\minus{}2007<0$, which is impossible, since $ f(n)$ is from $ \mathbb{N}$ in $ \mathbb{N}$

[quote="pco"][quote="caothujjj"]prove that doesnot exist the function f: N -> N satisfy   f(f(x)) = x^2 - 2008[/quote]

$ x \equal{} 1$ implies $ f(f(1)) \equal{} \minus{} 2007 < 0$, which is impossible, since $ f(n)$ is from $ \mathbb{N}$ in $ \mathbb{N}$[/quote]

  sorry, i have repaired it


Let $ a,b,c$ be positive number. Prove:
$ \frac{(a\plus{}b)^2}{a\plus{}b\plus{}2c} \plus{} \frac{(b\plus{}c)^2}{b\plus{}c\plus{}2a} \plus{} \frac{(c\plus{}a)^2}{c\plus{}a\plus{}2b} \ge \sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}$

[quote="......"]Let $ a,b,c$ be positive number. Prove:
$ \frac {(a \plus{} b)^2}{a \plus{} b \plus{} 2c} \plus{} \frac {(b \plus{} c)^2}{b \plus{} c \plus{} 2a} \plus{} \frac {(c \plus{} a)^2}{c \plus{} a \plus{} 2b} \ge \sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})}$[/quote]
Easy SOS:
$ \sum_{cyc}\frac {(a \plus{} b)^2}{a \plus{} b \plus{} 2c}\geq\sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {(a \plus{} b)^2}{a \plus{} b \plus{} 2c} \minus{} c\right)\geq\sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})} \minus{} a \minus{} b \minus{} c\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a \plus{} b)^2 \minus{} c^2 \minus{} c(a \plus{} b \plus{} c)}{a \plus{} b \plus{} 2c}\geq\sum_{cyc}\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c \plus{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a \plus{} b \plus{} c)(a \plus{} b \minus{} 2c)}{a \plus{} b \plus{} 2c}\geq\sum_{cyc}\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c \plus{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \plus{} b \plus{} c)\left(\frac {b \minus{} c}{a \plus{} b \plus{} 2c} \minus{} \frac {c \minus{} a}{a \plus{} b \plus{} 2c}\right)\geq$
$ \geq\sum_{cyc}\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c \plus{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \plus{} b \plus{} c)\left(\frac {a \minus{} b}{a \plus{} c \plus{} 2b} \minus{} \frac {a \minus{} b}{b \plus{} c \plus{} 2a}\right)\geq$
$ \geq\sum_{cyc}\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c \plus{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2\left(\frac {a \plus{} b \plus{} c}{(a \plus{} c \plus{} 2b)(b \plus{} c \plus{} 2a)} \minus{} \frac {1}{a \plus{} b \plus{} c \plus{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\right)\geq0.$
Let $ a\geq b\geq c.$ Hence,
$ \sum_{cyc}(a \minus{} b)^2\left(\frac {a \plus{} b \plus{} c}{(a \plus{} c \plus{} 2b)(b \plus{} c \plus{} 2a)} \minus{} \frac {1}{a \plus{} b \plus{} c \plus{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\right)\geq$
$ \geq\sum_{cyc}(a \minus{} b)^2\left(\frac {a \plus{} b \plus{} c}{(a \plus{} c \plus{} 2b)(b \plus{} c \plus{} 2a)} \minus{} \frac {1}{2(a \plus{} b \plus{} c)}\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {(a \minus{} b)^2(c^2 \plus{} ac \plus{} bc \minus{} ab)(a \plus{} b \plus{} 2c)}{2(a \plus{} b \plus{} c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)(a \plus{} b \plus{} 2c)}\geq$
$ \geq\frac {(a \minus{} b)^2(c^2 \plus{} ac \plus{} bc \minus{} ab)(a \plus{} b \plus{} 2c)}{2(a \plus{} b \plus{} c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)(a \plus{} b \plus{} 2c)} \plus{}$
$ \plus{} \frac {(a \minus{} c)^2(b^2 \plus{} ab \plus{} bc \minus{} ac)(a \plus{} c \plus{} 2b)}{2(a \plus{} b \plus{} c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)(a \plus{} b \plus{} 2c)}\geq$
$ \geq\frac {(a \minus{} b)^2((c^2 \plus{} ac \plus{} bc \minus{} ab)(a \plus{} b \plus{} 2c) \plus{} (b^2 \plus{} ab \plus{} bc \minus{} ac)(a \plus{} c \plus{} 2b))}{2(a \plus{} b \plus{} c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)(a \plus{} b \plus{} 2c)} \equal{}$
$ \equal{} \frac {(a \minus{} b)^2(b^3 \plus{} c^3 \plus{} b^2a \plus{} c^2a \plus{} 2b^2c \plus{} 2c^2b)}{(a \plus{} b \plus{} c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)(a \plus{} b \plus{} 2c)}\geq0.$

:( you said it easy...For such terrible caculation...
Any nice method to this problem?

We can solve it easily by Am-Gm, my friend  :)

Can you post your solution for this problem, nguoivn ?

Using the well-known result: $ \sum\ \frac {x^2}{y \plus{} z} \geq\ \frac {5\sum\ x^2 \minus{} 2\sum\ xy}{2(x\plus{}y\plus{}z)}$
(we can prove easily this result by SOS).
Let $ x \equal{} a \plus{} b$;... we obtain:
$ \sum\ \frac {(a \plus{} b)^2}{a \plus{} b \plus{} 2c} \geq\ \frac {2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c} \equal{}$
$ \equal{} \frac {1}{2}[a \plus{} b \plus{} c \plus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}] \geq\ \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}$  (by Am-Gm)
We have done  :)


floor function
irrational number
algebra proposed
algebra

Here is something for everyone: part (i) --- for beginners; part (ii) --- for intermediate; part (iii) --- for advanced problem-solvers (I haven't yet found a solution for (iii); may be very difficult, but may as well be trivial, in case I missed something obvious).

[b](i)[/b] (easy; many would perceive it as a joke  :) ). Let $\alpha > 1$ be a real number such that $n \mid \lfloor n\alpha^n \rfloor$ for all $n \in \mathbb{N}$. Must $\alpha$ be an integer?
[b](ii)[/b] (intermediate) Does there exist an irrational $\alpha > 1$ such that $n \mid \lfloor \alpha^n \rfloor$ for infinitely many $n$?
[b](iii)*[/b] (very difficult, I think; may be an open problem) Does there exist an $\alpha > 1$ such that $n \mid \lfloor \alpha^n \rfloor$ for [b]all[/b] $n \in \mathbb{N}$?

Perhaps a little surprisingly, I think that the answer to (iii) should be 'no.' I'd find quite unnormal the existence of an integer sequence $\{a_n\}$ such that $\big( (a_{n+1}-1)^{\frac{1}{n+1}}, (a_{n+1} + 1)^{\frac{1}{n+1}} \big) \subset \big( (a_n-1)^{\frac{1}{n}}, (a_n + 1)^{\frac{1}{n}} \big)$ AND $n \mid a_n$, because the length of the interval $\big( (a_n-1)^{\frac{n+1}{n}} + 1, (a_n + 1)^{\frac{n+1}{n}} - 1 \big)$ (from which we have to choose $a_{n+1}$) is $o(n)$ (if I did the estimates correctly). Any ideas are welcome!  :lol:

Besides, enjoy  :D !

--Vesselin

(i) Hmm... $\alpha = \frac 1 2 $  so that $0 \leq n \alpha^n < 1$ for all $n$.
I like problems I can solve in 10 seconds  :P

Pierre.

What exactly does that mean? :? He mentioned $\alpha>1$ in all three problems.

Aaargh...10 seconds is not enough for me to understand the problem :blush:

Pierre.

(i) Hint: Construct [b]explicitly[/b] an irrational number $\alpha > 1$ such that $\lfloor n\alpha^n \rfloor = n \lfloor \alpha^n \rfloor$ for all $n$  :lol: . This is why I said the problem is a 'joke,' but it's not [i]absolutely[/i] trivial --- though this hint itself is possibly the most non-trivial part  :) .


Suppose that $ a$ and $ b$ are positive real numbers such that:

$ a^{3}\equal{} a\plus{}1$ and $ b^{6}\equal{} b\plus{}3a$

Show that $ a > b$.


algorithm
function
algebra
polynomial
rational function
complex analysis

Hello everyone! I am beginner, I have a questions(may be easy). Help me please.

Let $ P(z)\equal{}\displaystyle\sum_{k\equal{}1}^{n}a_kz^k$ be a polynomial.
i)Express $ P(z)$ as a power series around some point $ u\neq 0$.
ii)Express the following rational map explicitly as a power series around $ z\equal{}2$: $ f(z)\equal{}\frac{3z^4\plus{}z^3\plus{}2z^2\plus{}7}{z^2\minus{}z}$.
ii)Give an algorithm of expressing as a power series an arbitrary rational function $ f\equal{}\frac{P(z)}{Q(z)}$.

Step 1: Substitute $ w\equal{}z\minus{}u$ everywhere and expand. It's just easier to see things that way.
In the rational function cases, you should definitely do the long division to separate out the polynomial part. For the proper rational part, there are two basic approaches:
- Build the coefficients recursively, getting the equations by clearing the denominator and matching coefficients. You don't usually get an explicit closed form this way, but it's pretty easy to calculate as far as you want.
- Partial fractions. $ \frac1{(1\minus{}az)^n}\equal{}\sum_{k\equal{}0}^{\infty}\binom{n\plus{}k\minus{}1}{k}a^k z^k$; use these for an explicit closed form for the coefficients.


linear algebra
matrix
function
Euler
algebra
polynomial
linear algebra unsolved

Let $ \mathbf{P}$ denote the "infinite matrix"

$ \left[ \begin{array}{ccccc} 1 & 0 & 0 & 0 & \hdots \\
1 & 1 & 0 & 0 & \hdots \\
1 & 2 & 1 & 0 & \hdots \\
1 & 3 & 3 & 1 & \hdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{array} \right]$

with entries $ \mathbf{P}_{ij} \equal{} {i \minus{} 1 \choose j \minus{} 1}$ and let $ \mathbf{I}$ denote the "infinite identity matrix."  Compute the inverse of $ \mathbf{P} \plus{} \mathbf{I}$.

[hide="Remark"] The computation of this inverse has apparently been the subject of several published papers, one of which has the thread title, even though it is relatively straightforward if looked at from the proper perspective. [/hide]

[hide="Hint 1"] $ \mathbf{P}$ acts on $ \mathbb{C}[[x]]$ (regarded as the ring of formal exponential generating functions over $ \mathbb{C}$) by multiplication by $ e^x$. [/hide]

[hide="Hint 2"] Generalize to multiplication by $ e^{tx}$. [/hide]

[hide="Hint 3"] It suffices to compute the power series of $ \frac{1}{1 \plus{} e^x}$.  This computation can be done using [url=http://mathworld.wolfram.com/EulerPolynomial.html]Euler polynomials[/url] or you can write it as $ \frac{1}{2} \cdot \frac{1}{1 \plus{} \frac{e^x \minus{} 1}{2}}$ and expand directly. [/hide]


Find a recursion for $ a_{n}$, where $ a_{n}$ is the number of ways you can tile a $ 2\times n$ rectangle with $ 1\times 1$ squares and L-trominoes.

I know problems like this have been posted, but for some reason my answer doesn't agree with JBL's from a post of his in 2004.

[hide="My solution"]
The standard procedure is to consider tiling the leftmost side. Using only $ 1\times 1$ squares, there is 1 way. Using only L-trominoes, there are 2 ways.

$ a_{n}=a_{n-1}+2b_{n-1}$,

where $ b_{n-1}$ is the number of ways to tile a $ 2\times (n-1)$ rectangle with a corner missing.

$ b_{n-1}=a_{n-2}+a_{n-3}$.

So $ a_{n}=a_{n-1}+2a_{n-2}+2a_{n-3}$.

JBL's answer was $ a_{n}= a_{n-1}+4a_{n-2}+2a_{n-3}$.[/hide]

I calculated the first four cases and got $ a_{1}= 1, a_{2}= 5, a_{3}= 11, a_{4}= 58$, which fits JBL's recursion.

But your solution makes sense too so I don't know what happened.

Oh I just realized I might have forgotten the case where you use 1 $ 1\times 1$ square and 1 L-trominoe.

So $ a_{n}=a_{n-1}+2b_{n-1}+2a_{n-2}$.

Then $ b_{n-1}=a_{n-2}+a_{n-3}$.

$ a_{n}=a_{n-1}+4a_{n-2}+2a_{n-3}$, like JBL's answer.


algebra polynomial function LaTeX calculus

Hey, this is a total shot in the dark, but... I just joined because I need help with a math problem [i]now[/i].  I have been working on it all night long and my brain is finally shot.  I can't seem to do anything else with it.  If you could show me how to solve for x I'd be forever indebted to you, but if you can't, I'll live and still thank you for reading this.

[921,600x^10+1,105,920rt(3)x^8-2,525,184x^6-2,112,306rt(3)x^4+4,357,416.96x^2-1,901,076x^(-2)+268,738.56x^(-6)+597,196.8rt(3)]/[146,966,400x^4+62,985,600rt(5)x^4]=-(2/3)x^2+(36/25)x^(-2)-(6rt(3)/15)

rt=square root, and I know it's long and messy, but I hope you could follow it...

How can I solve for x?

:o

Where on Earth is this problem from?

Well, just for screams and giggles, I'll translate this into latex and you can tell me if I got it right. Ok?
$\frac{[921600x^{10}+1105920\sqrt{3}x^{8}-2525184x^{6}-2112306\sqrt{3}x^{4}+4357416.96x^{2}-1901076x^{-2}+268738.56x^{-6}+597196.8\sqrt{3}]}{146966400x^{4}+62985600\sqrt{5}x^{4}}=$
$=-\frac{2}{3}\cdot x^{2}+\frac{36}{25}\cdot x^{-2}-(\frac{6\sqrt{3}}{15})$

Is that right?

Well, unless you happen to have mathematica, I think you're hopelessly screwed. However, it's late at night, and this might be one of those trick questions that work out nicely. However, like Torajirou said... Where did this come from???

Yep, that's it.  The problem came from my own awful mind.  I had to create a story problem for AP Calculus, and I had about six pages of work to get to that point where I just totally blanked and couldn't solve anymore.  I guess I'll just have to make an easier problem before third period today.