limit
function
calculus
derivative
analytic geometry
absolute value
complex analysis

Hi people,
this is a problem from Spanish Community ....

Please put your solutions there throught this link:

 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53944

-----------------

Let a sequence $x_n$, such that:
$x_1 = 1$, $x_2=2$, $x_3=3$ and $x_4=4$.
Also, $4x_n = x_{n-1} + x_{n-2} + x_{n-3} + x_{n-4}$, for all $5 \leq n$.

Find $\lim x_n$.

-------------------

Carlos Bravo
Lima-Peru

This is a really interesting problem.

The answer is 3, and I think in general given a sequence with recurrence
\[ x_n = \frac{x_{n-1} + x_{n-2} + \cdots + x_{n-k}}{k} \]
the limit as $n\to \infty$ is
\[ \frac{x_1 + 2x_2 + \cdots + kx_k}{\frac{k(k+1)}{2}}. \]

To prove this first note that because sequences satisfying the (general) recurrence are additive, then if $\lim_{n\to \infty}x_n$ exists it is a linear function $w_1x_1+w_2x_2+\cdots+w_kx_k$ of the initial conditions.  Moreover, because we can officially "start" the sequence anywhere we like,
\begin{eqnarray*} w_1x_1+w_2x_2+\cdots+w_kx_k&=&\lim_{n\to \infty}x_n \\ &=&w_1x_2+w_2x_3+\cdots+w_kx_{k+1}\\ &=&w_1x_2+w_2x_3+\cdots+w_k\frac{x_1 + x_2 + \cdots + x_k}{k}. \end{eqnarray*}
So, comparing coefficients, $w_1=\frac{w_k}{k}$ and $w_i=\frac{w_k}{k}+w_{i-1}$ for $2\leq i\leq k$.  Solving this system, $w_i=\frac{i}{k}w_k$ for $1\leq i\leq k$.  It therefore suffices to find the limit $l$ (if it exists) of a recurrence with initial conditions \[ x_1=x_2=\cdots=x_{k-1}=0, x_k=1, \]
and the limit in the general case will then be
\[ l\frac{x_1 + 2x_2 + \cdots + kx_k}{k}. \]

The generating function $f(x)$ of the recurrence with these initial conditions satisfies
\[ \frac{f(t)-t^k}{t^k}=\frac{f(t)+\frac{f(t)}{t}+\cdots+\frac{f(t)}{t^{k-1}}}{k} \]
and so \[ (k-t-t^2-\cdots-t^k)f(t)=kt^{k}. \]
$(k-t-t^2-\cdots-t^k)$ factorises as $(1-t)[k+(k-1)t+\cdots+2t^{k-2}+t^{k-1}]$
so
\[ f(t)=\frac{kt^{k}}{(1-t)\(\displaystyle\sum_{i=0}^{k-1}(k-i)t^i\)} \]
\[ f(t)=\frac{2}{k+1}(\frac{t}{1-t}-\frac{t\displaystyle\sum_{i=0}^{k-2}(k-\frac{1}{2}i)(i+1)t^i}{\displaystyle\sum_{i=0}^{k-1}(k-i)t^i}). \]
This means that
\[ x_n=\frac{2}{k+1} + \displaystyle\sum_{i=1}^{k-1}c_ir_i^{n-1}, \]
where $r_1, r_2,\ldots r_{k-1}$ are the roots of
\[ kt^{k-1}+(k-1)t^{k-2}+\cdots+2t+1=0 \]
and $c_1, c_2,\ldots c_{k-1}$ are fixed coefficients.  If the absolute values of $r_1, r_2,\ldots r_{k-1}$ are all less than 1 then $x_n$ will tend to $\frac{2}{k+1}$ as $n\to\infty$, and in the general case $x_n$ tends to
\[ \frac{x_1 + 2x_2 + \cdots + kx_k}{\frac{k(k+1)}{2}} \]
as I claimed before.  For $k=1,2,3,4$ it is easy to check that the absolute values of the roots are less than 1, so we can certainly say that in the situation originally suggested ($k=4$) the answer is 3.  To complete the proof you'd need to prove that the roots have absolute value less than 1  for all $k$.  I'm not sure how to do this, but it might be helpful to note that $kt^{k-1}+(k-1)t^{k-2}+\cdots+2t+1$ is the derivative of $t^k+t^{k-1}+\cdots+t+1$ all of whose roots have absolute value exactly 1.

Anyone who knows some complex analysis able to complete the proof?

Actually, I can go a step further: we know that no roots of
\[ kt^{k-1}+(k-1)t^{k-2}+\cdots+2t+1=0 \]
have absolute value greater than 1, because if they did then some initial conditions would give sequences that grew without bound, whereas we know that the absolute value of each term $x_n$ is no more than the abolute value of the greatest term given in the initial conditions.  So it is only necessary to determine whether the equation ever has roots with absolute value exactly 1.

(If it does, then sometimes the sequence oscillates rather than tending to a limit.)

[quote="Agrippina"]in general given a sequence with recurrence
\[ x_n = \frac{x_{n-1} + x_{n-2} + \cdots + x_{n-k}}{k}  \]
the limit as $n\to \infty$ is
\[ \frac{x_1 + 2x_2 + \cdots + kx_k}{\frac{k(k+1)}{2}}.  \][/quote]

Indeed, this is just the particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=16861 post #8 for m = 1. Much said particular case, as the general case follows from this particular case by applying it to all coordinates of the points.

  darij

well ...!

Thank you Agrippina for your solution, ....

and thank you Darij for this information and the link...

Carlos Bravo
Lima-PERU


hi,
let $ S(i,j) \equal{} \sum_{k\equal{}1}^{n}{k^{i\plus{}j}}$, $ i,j \equal{} 1 ,...,n$. Calculate the determinant
\[ D \equal{} |S(i,j)|.\]

The same for question of me !
http://www.mathlinks.ro/viewtopic.php?t=262806
 [/hide]


Show that the polynomial $x^{n}+mx-b=0$ has a mutiple root provided $(\frac{m}{n})^{n}+(\frac{b}{n-1})^{n-1}=0$. Find this root.

Anyone? I think I have found the root.

[hide="Root"] $\frac{bn}{m(n-1)}$[/hide]

Is that right? Has anyone got any ideas on the first part of the question?

if $f(r)=f'(r)=0$, then $r$ is a double root

[quote="me@home"]So is this a conjecture of yours? It seems really strong and awkward also...[/quote]

This is not a conjecture of mine. It is an exercise from a book for the end of year exam. It is pretty strong.

This is actually a very simple problem now I come to think of it.

Let $p(x)=x^{n}+mx-b$ and let $\alpha$ be the multiple root.

$p(\alpha)=\alpha^{n}+m \alpha-b=0$

$p^{,}(\alpha)= n \alpha^{n-1}+m=0$

Thus $\alpha^{n-1}=\frac{-m}{n}$

Therefore we have $\alpha.\frac{-m}{n}+m \alpha-b=0$

Which gives us $\alpha=\frac{bn}{m(n-1)}$

Now we substitute this into $p^{,}(x)$ and obtain $b.(\frac{bn}{m(n-1)})^{n-1}+m=(\frac{b}{n-1})^{n-1}.(\frac{n^{n}}{m^{n-1}})+m=0$

Therefore $(\frac{b}{n-1})^{n-1}=-(\frac{m}{n})^{n}$

Is this the same as [b]felipesa's[/b] solution?


Let $ \alpha$, $ \beta \geq 1$ be real numbers. Prove that there are integers $ x, y \neq (0, 0)$ such that:

$ \beta \geq x \geq 0$
$ \alpha \geq y \geq 0$
$ \vert \alpha x \minus{} \beta y \vert \leq 1$.

Try $ \alpha\equal{}\beta\equal{}x\equal{}y\equal{}1$ So they exist :wink:
And this is an inappropriate forum :wink:

And also try $ \alpha\equal{}\beta\equal{}3,x\equal{}2,y\equal{}1$ so it doesn't hold for all numbers.

Who said that it holds for all $ \alpha, \beta, x, y$? The thesis says that for all $ \alpha, \beta$ we can find $ x, y$ such that it holds...

Bugi, the problem is to prove it for any given $ \alpha$ and $ \beta$, not just the ones you chose.
(Looks like TomciO beat me to it  :D )


Let ABC  be a triangle such that two cevians of equal length pass through the triangle´s fermat point. Prove that ABC  is isoceles.

WLOG, assume cevians BD and CE have equal lengths. Let F be the Fermat's point.
$\frac{AE}{\sin\angle ACF}=\frac{CE}{\sin A}=\frac{BD}{\sin A}=\frac{AD}{\sin\angle ABF}$.
$\frac{\sin\angle ABF}{\sin\angle ACF}=\frac{\frac{AF}{\sin\angle ACF}}{\frac{AF}{\sin\angle ABF}}=\frac{\frac{AC}{\sin\angle AFC}}{\frac{AB}{\sin\angle AFB}}=\frac{AC}{AB}$.
Thus, $\frac{AE}{AD}=\frac{AB}{AC}$, which means $DE//BC$.
Now the result should be trivial enough :)

[color=red][Moderator edit: A typo pointed out by Yptsoi was corrected.][/color]


factorial
number theory
prime factorization

What is the least positive integer $n$ such that 1560 divides $n!$?

o nvm i read the question wrong.....

 solution:

[hide] the prime factorization of 1560 is 2^3 * 3 * 5 *13.  The biggest factor here is 13.  Thus, [b] n= 13[/b] [/hide]

"nvm" =never mind.  the first time i answered incorrectly and edited my post.

yes it is 13

What about n! must be divisible by 798720?


I have a very nice problem

if   1+2+3+4+5+...+N+(N-1)+(N-2)+...+3+2+1 = 10 000

what is the value of N?

Good luck!

 :lol:

[hide]Hello,SuperSut.

$1+2+3+\cdots +(N-1)+N+(N-1)+(N-2)+\cdots +3+2+1=10000$

$\Longleftrightarrow 2(1+2+\cdots +N-1)+N=10000$

$\Longleftrightarrow 2\frac{(N-1)N}{2}+N=10000\Longleftrightarrow N^2=100^2\Longleftrightarrow N=100$[/hide]

[hide]factor things out first ;)

2(1+2+3+4+5+6...) + N = 10,000

2(1+2+3...+(N-2)+(N-1)) = $2 * (\frac{1}{2})(N-1) * N$ because we still have the 2... and we have the $\frac{1}{2} * (N-1)$ because we have to multiply the N (add the first and last terms, 2nd to last and 2nd terms, etc.) by 1/2 the number of terms.

This part of hte equation simplifies to N2 - N.

Now we have N2 - N + N = 10,000.

The N's cancel out so now we only have N2 = 10,000.

Now we have $ \pm N = 100$

We can't have a negative amount of numbers, so we have [b]N = 100. There's your answer.[/b][/hide]

In case i did something wrong, we can always check our work ;).

[hide]with the formula $\frac {n(n-1)}{2}$, we can have $\frac {(99)(100)}{2}$ which equals 4950. Multiply it by 2 (because the numbers go down, and we haev 9900. add 100 (becuz it's in the middle) and we have 10 000.[/hide]

kunny, nice use of LATEX

For this problem, the formula $\frac{n(n+1)}{2}$ is very important, unless there is a differrent way.
[/hide]

2*(N^2-N) /2 = N^2-N,   N^2-N + N= N^2=10000

n=100

i know i need to put aside some time to learn latex.....

just take the square root of "M" if it is

1+2+3+4+...(n)+1+2+3+4+...(n-1)=M


Find the functions wich satisfies $f(x+y)=[f(x)+f(y)](1+xy),\forall x,y\in \mathbb{R}$:)

Letting $y =-\frac{1}{x}$ we ascertain that $f(x-\frac{1}{x}) = 0$.

The range of that function is $\mathbb{R}$, so we conclude that $f(x) = 0$ is the only solution.

$\forall x\in R$, there exist $y\in R^{*}$,such that $y-\frac{1}{y}=x$
so $f(x)=f(y-\frac{1}{y})=[f(y)+f(-\frac{1}{y})](1-y.\frac{1}{y})=0$

edit: :blush: sorry [b]t0rajir0u[/b] I did not see your solution, because I have writing mine

sorry for may bad english


find all continuous functions f:R-->R such that g(g(x))=g(x)+2x for all real x.


superior algebra
superior algebra theorems

I will get directly to the point.   Is there any concept about splitting a long exact sequence ?

I'll appreciate your answers.

FS

a chain complex $ (C_n)$ is said to be split if there are maps $ s_n : C_n \to C_{n\plus{}1}$ satisfying $ d_n \equal{} d_n s_n d_n$ for all $ n$. if the complex is also acyclic, it's called split exact. this is the case if and only if it's contractible.


[quote="GameBot"]A fish tank weighs 80 pounds when $ 40\%$ full of water, and it weighs 140 pounds when completely full. How many pounds does the tank weigh when empty?[/quote]
60 percent of water is 60 pounds, so 40percent is 40 pounds.  80-40=40.


limit
trigonometry
calculus
calculus computations

Hey, this seems like it should be easy, but I don't see the trick or expansion. Any help would be greatly appreciated. Thanks!

\[\lim_{x \rightarrow 0}\frac{1}{(\tan x)^{2}}-\frac{1}{x^{2}}\]

Note that

$\frac{1}{\tan x}= \cot x = \frac{1}{x}-\frac{x}{3}+o(x)$

(Hint: consider the taylor series of $\frac{x}{\tan x}$.)

Then we have

$\frac{1}{\tan^{2}x}-\frac{1}{x^{2}}=-\frac{2}{3}+o(x)$

So the answer is $-\frac{2}{3}$.

Or applying L'hospital's rule to

$\lim_{x \rightarrow 0}\left( \frac{1}{\tan^{2}x}-\frac{1}{x^{2}}\right) = \lim_{x \rightarrow 0}\frac{x}{\tan x}\left( \frac{x}{\tan x}+1 \right) \cdot \frac{x-\tan x}{x^{3}}= 2\lim_{x \rightarrow 0}\frac{x-\tan x}{x^{3}}$,

we have

$2\lim_{x \rightarrow 0}\frac{x-\tan x}{x^{3}}= 2\lim_{x \rightarrow 0}\frac{1-\sec^{2}x}{3x^{2}}= 2\lim_{x \rightarrow 0}\frac{-\tan^{2}x}{3x^{2}}=-\frac{2}{3}$.


function
integration
calculus
derivative
trigonometry
calculus computations

These questions are relatively straightforward and not too challenging.

1. Calculate the Fourier transform of the following functions:
a. [size=200]$ f_{a}(x) = \{^{1 - \frac {|x|}{a}, |x| < a}_{0,\hspace{6 mm} |x|\geq{a}}$[/size]
b. $ f(x) = \frac {x}{(x^{2} + a^{2})(x^{2} + b^{2})}$

2. If $ f\in{G(R)}$, is $ \hat{f}\in{G(R)}$?

3. Solve the finite differential-difference equation $ y'(t) + y(t - 1) = t^{2}$, $ t > 0$, where $ y(t) = 0$ for every $ t\leq{0}$.

4. Apply the Laplace Transform to solve the initial value problem:
$ \Huge{\{^{y''(t) - y'(t) - 6y(t) = 0, t > 0}_{y(0) = 1, y'(0) = - 1}}$

Two notational questions:

1. Where is the $ 2\pi$ in your definition of Fourier transform? There are several widely used and viable conventions.

2. What is $ G(\Re)?$ I do not recognize that notation.

Sorry, about the $ G(\Re)$ notation. I didn't know how to get the blackboard bold symbol for real numbers. Also, I just realized I made a typo in that question so I'll edit. For the first question:
\[ \hat{f}(\omega) \equal{} \frac {1}{2\pi}\int^{\infty}_{ \minus{} \infty}{f(x)e^{ \minus{} i\omega{x}}dx}
\]

Kent asked about what $ G$ meant rather than about what $ \mathbb R$ meant...

Here's the first one. I do recognize this function, but starting from scratch anyway:

Note that $ \widehat{f'}(\omega) \equal{} \frac {1}{i\omega}\widehat{f}(\omega).$

Differentiate the function in #1 twice, and interpret the derivatives as weak (distributional) derivatives.

$ f''(x) \equal{} \frac1{a}\left(\delta(x \plus{} a) \minus{} 2\delta(x) \plus{} \delta(x \minus{} a)\right).$

Now take the transform of that:

$ \widehat{f''}(\omega) \equal{} \frac1{2\pi a}\left(e^{ia\omega} \minus{} 2 \plus{} e^{ \minus{} ia\omega}\right) \equal{} \frac1{2\pi a}(2\cos(a\omega) \minus{} 2)$

Then we divide by $ \minus{} \omega^2$ to get that (at least for $ \omega\ne 0$)

$ \widehat{f}(\omega) \equal{} \frac {2 \minus{} 2\cos(a\omega)}{2\pi a\omega^2} \equal{} \frac {4\sin^2\left(\frac {a\omega}{2}\right)}{2\pi a\omega^2}$

Of course, $ f,$ being compactly supported, has an analytic Fourier transform. We have that $ \widehat{f}(0) \equal{} \frac {a}{2\pi},$ which is just what it "should" be.

The original function was real, even, compactly supported, and continuous and with a bounded derivative, but the second derivative is not a function. We expect the Fourier transform to be be real, even, very smooth (in fact, entire), and decaying at infinity like a constant times $ \omega^{ \minus{} 2}.$ Our result meets those expectations.

---

And yes, fedja understood what I was asking. I still don't know what $ G$ is.


function
number theory unsolved
number theory

Find all possible positive integers $(x, y, z)$ to
$\frac{7}{x}+\frac{13}{y}+\frac{16}{z}=\frac{4}{5}$.

Clearly, $x=y=z=45$ is solution, does have other solutions?

[quote="Rust"]It had infinetely many solutions.
For example x=14,y=26,z=40.[/quote]

No, your example results in $\frac{7}{5}$

[quote="Rust"]I am sorry x=35,y=65,z=40. It is finit number of solutions.[/quote]
yes I agree,
we can find a constant M
such that $\min\{x,y,z\}\leq M$
for a fixed $x$(or$y,z$)the equality
$yz+ay+bz+c=0$have finit solution.


[quote="joao1"]For rationals a,b:
S(a,b)={a(x^2)+b(y^2): x,y are rationals}.
let a,b,c,d be rational numbers different form 0. Show that:

S(a,b)=S(c,d) iff ab/(cd) is the square of a rational number
and there is q (different form 0) in S(a,b) and in S(c,d).[/quote]
we have to prove that:
$ S(a,b) \equal{} S(c,d)\iff S(a,b)\cap S(c,d)\neq \{0\},\exists x\in\mathbb{Q}: \ ab \equal{} x^2cd$


A rectangular steel have sides of length $a,b,c$ and density $S$. Float it in water, with side of length $a$ vertical. Push it down a bit, making it move in simple harmonic. Find the frequency of the simple harmonic.

Well, it can't be solid steel because we want it to float. In other words, we need $S<1.$

Let's oversimplify the problem by neglecting all of the fluid dynamic forces needed to move the water that surrounds the floating box and concentrate only on the buoyancy force on the box itself.

The box has mass $abcS.$ When floating at a certain depth, it experiences no net force. If we lift it by a distance $y,$ its weight is the same as always, but the upward buoyancy force, which is equal to the mass of the displaced water, decreases by $bcgy$ If he box sinks below its equlibrium level, the buoyancy force increases by the same amount. From force equals mass times acceleration, we get

$abcSy''=-bcgy$ or $y''=-\frac{g}{aS}y.$

This gives simple harmonic motion with frequency $\frac1{2\pi}\sqrt{\frac{g}{aS}}.$

If it's a long skinny box, it bobs up and down slower if it's floating in an up-and-down configuration than if it's floating on its side.

In equilubrium,

$\rho_{0}y_{0}bcg=Sabcg$

If we move a small y distance,

$\rho_{0}bc(y_{0}+y)-Sabcg+Sabc\frac{d^{2}y}{dt^{2}}=0$

$\frac{d^{2}y}{dt^{2}}+\frac{\rho_{0}g}{Sa}y=0$

$\omega=\sqrt{\frac{\rho_{0}g}{Sa}}$

$f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{\rho_{0}g}{Sa}}$

$\rho_{0}$ would be the density of the fluid. In radio's original post, he called the fluid water, so I took $\rho_{0}=1.$ Hence my solution is the same as that of Namiqfizik.

What about this one:

A manometer, a U-shaped tube with uniform diameter throughout, contains mercury of density $\rho$ and total length $L$. What will its simple harmonic frequency be?

Make a new thread!! Do not continue old posts with new problems!  :mad:

I made one: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=854978#854978[/url]
 :P


algebra
polynomial
search
superior algebra
superior algebra solved

There is a well-known result in algebra that if $ f(x) \in \mathbb{Z}[x]$ and $ f(x \plus{} a)$ is irreducible in $ \mathbb{Z}[x]$, for some $ a \in \mathbb{Z}$, then $ f(x)$ is irreducible in $ \mathbb{Q}[x]$.  Is the converse true?, that is, if $ f(x) \in \mathbb{Z}[x]$, is irreducible over $ \mathbb{Q}[x]$, is there some $ a \in \mathbb{Z}$ such that $ f(x \plus{} a)$ is irreducible in $ \mathbb{Z}[x]$ ?

Thank you.

For every $ f(x) \in {\mathbb Z}[x]$, the following conditions are equivalent

a) $ f(x)$ is irreducible over $ {\mathbb Z}[x]$.
b) $ f(x)$ is irreducible over $ {\mathbb Q}[x]$.
c) $ f(x\plus{}a)$ is irreducible over $ {\mathbb Z}[x]$ for some $ a\in \mathbb Z$.
d) $ f(x\plus{}a)$ is irreducible over $ {\mathbb Z}[x]$ for all $ a\in \mathbb Z$.

The fact that (a) is equivalent to (b) is known as Gauss's Lemma ( http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial) ). Conditions (a), (c) and (d) are clearly equivalent.

So the answer to your question is "yes": (b) implies (c). In particular, if $ f(x)$ is irreducible over $ {\mathbb Q}[x]$ then $ f(x \plus{} 0)\equal{}f(x)$ is irreducible over $ {\mathbb Z}[x]$.

(Am I missing something?)

Probably the question you really wanted to ask is about the equivalence of a), c), and d).  This becomes clear when you look at contrapositives:  if $ f(x \plus{} a) \equal{} g(x) h(x)$ is a proper factorization then $ f(x) \equal{} g(x \minus{} a) h(x \minus{} a)$ is a proper factorization, and vice versa.

I totally messed it up.   My question should have been this:

Is there a $ c$ such that $ f(x\plus{}c)$ can be showed to be irreducible in $ \mathbb{Z}[x]$ by using Eisenstein's criterion?

Nope.  Showing that $ f$ is irreducible using Eisenstein's criterion relies on the fact that $ f(x) \equiv x^n \bmod p$ when such a proof is possible for some $ p$, but $ f$ ramifies at at most finitely many $ p$ (precisely the primes dividing its [url=http://en.wikipedia.org/wiki/Discriminant#Discriminant_of_a_polynomial]discriminant[/url]).

For example, $ x^3 \plus{} x \plus{} 1$ has discriminant $ \minus{} 31$, and $ \bmod 31$ we find that

$ \gcd(x^3 \plus{} x \plus{} 1, 3x^2 \plus{} 1) \equal{} \gcd(x^3 \plus{} x \plus{} 1, x^2 \minus{} 10) \equal{} \gcd(11x \plus{} 1, x^2 \minus{} 10) \equal{} \gcd(x \plus{} 17, x^2 \plus{} 21) \equal{} x \plus{} 17$

so indeed $ x^3 \plus{} x \plus{} 1$ has a repeated root of $ 14$, but of multiplicity $ 2$.  Hence Eisenstein's criterion cannot prove that $ x^3 \plus{} x \plus{} 1$ is irreducible.

[quote="Adriana"]I totally messed it up.   My question should have been this:

Is there a $ c$ such that $ f(x \plus{} c)$ can be showed to be irreducible in $ \mathbb{Z}[x]$ by using Eisenstein's criterion?[/quote]

The point is that Eisenstein polynomials correspond to totally ramified primes, but there are number fields with no such primes.
A web search should turn up expositions, e.g. http://google.com/search?q=eisenstein+totally+ramified

You may want to show that the only cyclotomic polynomials $ \Phi_n$ that can be proven to be irreducible using Eisenstein are those with $ n \equal{} p^s$ a prime power or $ n\equal{}2 p^s$.

[quote="Adriana"]I totally messed it up.   My question should have been this:

Is there a $ c$ such that $ f(x \plus{} c)$ can be showed to be irreducible in $ \mathbb{Z}[x]$ by using Eisenstein's criterion?[/quote]

Adriana your question is exactly [url=http://www.mathlinks.ro/viewtopic.php?t=307337]the same as mine[/url].
Basically I want to know how much to shift in order to apply Eisenstein.
I'm also interested if Eisenstein [b]can[/b] be applied to all polynomials(at least thosein $ Z[X]$), if not, I'd like to know specific rules for the polynomials to which it cannot be applied.

I think billwdg is right and also t0rajir0u is using the formula that [url=http://www.mathlinks.ro/viewtopic.php?t=307337]bill indicated to me here[/url].
I'd sure like to know where that formula comes from because I have no clue.

If you would have read my posts (or T0r's), you would have seen that it doesn't apply to all polynomials. Even some rules were given (the most general one being the degree of ramification).

If memory serves correct there's a very nice exposition of this and related topics in Filaseta's book on irreducibility testing - see my sci.math post [url]http://groups.google.com/group/sci.math/msg/749c4b448b406332[/url]


Let $a,b,c,d,e,g\in R$,a <> c <> 0.Find all $f:R\rightarrow R$ such that:
(i) f is continous
(ii)f(ax+b)+f(cx+d)=ex+g


calculus
integration
trigonometry
Euler
calculus computations

this looks like a good idea but i am not sure how to finish.
sub in $x=\tan(y)$ and we have $\int_{-\frac{\pi}{4}}^{0}\sec^{2}(y)\sqrt{\frac{\cos(y)+\sin(y)}{\cos(y)-\sin(y)}}dy=\int_{-\frac{\pi}{4}}^{0}\sec^{2}(y)\sqrt{\tan\left(y+\frac{\pi}{4}\right)}dy$

any ideas on how to finish?

nice i didnt even think about doing $y=\frac{1+x}{1-x}$

[quote="paladin8"]Multiply through by $\sqrt{\frac{1+x}{1+x}}$.

$\int_{-1}^{0}\frac{1+x}{\sqrt{1-x^{2}}}dx = \left[\arcsin{x}-\sqrt{1-x^{2}}\right]_{-1}^{0}= \frac{\pi}{2}-1$.[/quote]

paladin8, your answer is very natural. :)

[quote="didilica"]The integral equals: $\frac{\pi}{2}-1$. Let $I$ be the value of the integral.

Make the substitution $\frac{1+x}{1-x}=y$ and we get that

$I=2\int_{0}^{1}\frac{\sqrt{y}}{(1+y)^{2}}dy=-1+\int_{0}^{1}\frac{dy}{\sqrt{y}(1+y)}$, where the last equality holds in view of integration by parts formula.

Then make the substitution $y=t^{2}$ and we get that:

$I=-1+2\int_{0}^{1}\frac{dt}{1+t^{2}}=-1+\frac{\pi}{2}$.[/quote]

This is de second Euler substitution http://eom.springer.de/E/e036590.htm


I'm told that the answer to:

$ mgSin(z) \minus{} xy \equal{} ma$
$ y \minus{} mg Cos(z) \equal{} 0$

is:

$ a \equal{} g(Sin(z) \minus{} xCos(z))$

Can anyone tell me the intervening steps please?
(It's actually a physics prob. where x, y and z represent physics variables.)

$ y \minus{} mg \cos z \equal{} 0 \implies y \equal{} mg \cos z$

We substitute this value of $ y$ into $ mg \sin z \minus{} xy \equal{} ma$ to get $ ma \equal{} mg \sin z \minus{} mg x \cos z$

Factor out $ mg$ to get $ ma \equal{} mg ( \sin z \minus{} x \cos z)$

Finally, divide by $ m$ to get $ a \equal{} g (\sin z \minus{} x \cos z)$

What is the particular physics problem?

Oops, should have seen that; I was trying to solve it a different way
and got confused :)  Thanks very much anyway Zuton!

The prob is:
A pig slides down a 35 deg. slide in twice the time it would take to slide
down a frictionless 35 deg. slide.  What is the coefficient of kinetic friction
between the pig and the slide?           (Halliday Resnick)


Ahh, that's a good site. I use it mostly for AMC 10 and 12, because I usually get AIMEs from AoPS.

[quote="danb80"]Check out this site!

[url]http://www.math.ksu.edu/main/handbook/ProblemSets[/url]

(Yes, these may be done for bonus!)[/quote]

You. Are. Awesome.

[quote="danb80"]Check out this site!

[url]http://www.math.ksu.edu/main/handbook/ProblemSets[/url]

(Yes, these may be done for bonus!)

By the way, after you take the AMC 8, the best problems to prepare for MC are the AMC 10 and AMC 12.[/quote]
The last sentence is very true. If you notice this year's last sprint state question, it's the same as #20 on 10B.

So, if you can do great on 10/12, you will [u]most likely [/u]do well on State Mathcounts. Notice I underlined "most likely". THat's because the tests aren't the exact same type.

Anyway, most people recommend practicing AIMEs for National Mathcounts.

[quote="abacadaea"]isnt AMC 8 this tuesday?[/quote]

Yes, it is. At least for our school.

[quote="danb80"](Yes, these may be done for bonus!) [/quote]

eh.. I have a question. How would you give us an extra credit?

I am practicing on it,too.

It's a great web site.

Thanks. :lol:

DK:  Just turn them in as you would regular work.


If $S_{1}, S_{2}, S_{3}, ..., S_{n}$ are the sums of infinite geometric series whose first terms are $1, 2, 3, ... ,n$ and whose common ratios are $ \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ..., \frac{1}{n+1}$ respectively, then find the value of $\sum_{r=1}^{2n-1}S_{r}^{2}$.

Correct me if I'm wrong, but I think it's very easy and straightforward.
[hide]The first term of the series whose sum is $S_{r}$ is $r$ and the ratio is $\frac1{r+1}$, so $S_{r}=\frac{r}{1-\frac1{r+1}}=\frac{r(r+1)}{r+1-1}= r+1$. Thus $\sum_{1\leqslant r\leqslant 2n-1}S_{r}^{2}= \sum_{1\leqslant r\leqslant 2n-1}(r+1)^{2}= \sum_{2\leqslant r\leqslant 2n}r^{2}=$ $-1+\sum_{1\leqslant r\leqslant 2n}r^{2}= \frac{2n(2n+1)(4n+1)}{6}-1= \frac{8n^{3}+6n^{2}+n-3}{3}$. [/hide]


\[ \sum_{k\equal{}1}^{\infty}\frac{a_1c_1\plus{}a_2c_2\plus{}\cdots\plus{}a_kc_k}{k(c_1c_2\cdots
  c_k)^{1/k}} \equal{} \sum_{k\equal{}1}^{\infty} a_kc_k\sum_{j\equal{}k}^{\infty}
\frac{1}{j(c_1c_2\cdots c_j)^{1/j}}\]
$ a_k,c_k$ is positve real number.

This is a step in a proof .

 thanks

There is no r at the end of "formula".

This is just a rearrangement and a renaming of variables; write that sum in the numerator as $ \sum_{j\equal{}1}^k a_jc_j$. We have $ \sum_{k\equal{}1}^{\infty}\sum_{j\equal{}1}^k a_jc_j\cdot \frac1{k(c_1c_2\cdots c_k)^{1/k}}$
Switch the order, and it becomes $ \sum_{j\equal{}1}^{\infty}\sum_{k\equal{}j}^{\infty} a_jc_j\cdot \frac1{k(c_1c_2\cdots c_k)^{1/k}}$

The interchange is justified by Tonelli's theorem, since everything is nonnegative.


topology
Functional Analysis
real analysis
real analysis unsolved

Let $ E$ be a normed space and $ F$ closed subspace of $ E$. Define the norm on $ E/F$ by $ \parallel{}x\plus{}F\parallel{} \equal{} \inf_{y \in x \plus{} F} \parallel{}y\parallel{}$. It is well known that if $ E$ is Banach, then $ E/F$ is also Banach with the norm defined above.
Is the converse true, i.e. if $ E/F$ Banach, does this imply that $ E$ is Banach ?

What if we impose the condition that $ F$ is also Banach in in the norm, induced by the norm on $ E$.

[quote="eugene"]
Is the converse true, i.e. if $ E/F$ Banach, does this imply that $ E$ is Banach ?
[/quote]
Let $ E$ be a normed and non Banach space over $ \mathbb R$ or $ \mathbb C$ and $ F$ be any hyperplane, then $ dim E/F\equal{}1$ so complete.


algebra
polynomial
calculus
integration
function
real analysis
domain

[quote="computer_nuke"][quote="Inspired By Nature"]Find the integer m such that the polynomial $p^{3}+2p+m$ divides $p^{12}-p^{11}+3p^{10}-p^{2}+23p+30$.[/quote]

You can always use long division  :)[/quote]
I could, but won't that take a while? :D

[quote="lotrgreengrapes7926"][quote="t0rajir0u"][hide="Hint"] Plug in a few values of $p$ to find restrictions on $m$. [/hide][/quote]
[hide] Plugging in $p=0, 1$, we get $m|30$ and $2+m|55$, so $m=3$. [/hide][/quote]

What about -3?

[quote="Kurt Gödel"][quote="t0rajir0u"][hide="Hint"] Plug in a few values of $p$ to find restrictions on $m$. [/hide][/quote] Hmm, that's what I wanted to do in first place, but is it always true that if $P(x) = Q(x)R(x)$ and $P(x)$ and $Q(x)$ have integer coefficients, then $R(x)$ has integer coefficients as well? Could be...it looks logical, but is there a rigorous proof?[/quote]

I don't think that it is true that if $P(x) = Q(x)R(x)$ and $P(x)$ and $Q(x)$ have integer coefficients, then $R(x)$ has integer coefficients as well, but I think it is true that if $P(x) = Q(x)R(x)$ and $P(x)$ and $Q(x)$ have integer coefficients that are all relatively prim, then $R(x)$ has integer coefficients as well. I don't really know of a proof of it, but I'm sure there is one.

Also, can any tell me why -3 doesn't work? It divides 30 and when added 2, it divides 55, according to lotrgreengrapes7926

[quote="computer_nuke"]Also, can any tell me why -3 doesn't work? It divides 30 and when added 2, it divides 55, according to lotrgreengrapes7926[/quote]
It should be $3+m | 55$. So $m=-2$ or $m=2$...

Definition: Let $F$ be a field.  Then $F[x]=\{\text{polynomials in }x\text{ with coefficients in }F\}$.

Proof:  If $P(x)=Q(x)R(x)$, where $P,Q\in\mathbb{Z}[x]$, then clearly $R(x)\in\mathbb{Q}[x]$ (e.g, by the division algorithm).  By Gauss's Lemma, since $P$ is reducible in $\mathbb{Q}[x]$ (since $\mathbb{Z}[x]\subseteq\mathbb{Q}[x]$), $P$ is also reducible in $\mathbb{Z}[x]$.  Hence $R(x)\in\mathbb{Z}[x]$, which is what we want to prove.

[quote="computer_nuke"]I don't think that it is true that if $P(x) = Q(x)R(x)$ and $P(x)$ and $Q(x)$ have integer coefficients, then $R(x)$ has integer coefficients as well, [/quote]

This should be obvious from the process of either long or synthetic division.

blahblahblah:  $\mathbb{Z}$ isn't a field ;)

I'd prefer it if you included a proof of Gauss's lemma in your post, too  :P

It's not too hard, really.

[quote="blahblahblah"][quote="t0rajir0u"]blahblahblah:  $\mathbb{Z}$ isn't a field ;)[/quote]

Well, it's a ring, which is enough for our purposes here.[/quote]

What are fields and rings? :blush:

Fundamental structures of mathematics (after you get used to them, you will see them everwhere ;) ):

[b]Groups:[/b]

A set $G$ with an operation $*$ (meaning that for all $a,b \in G$ we have $a*b$ defined and in $G$) is called a "group" if the following properties hold:

- $*$ is "associative", meaning that $(a*b)*c=a*(b*c)$ (things in brackets computed first) for all $a,b,c \in G$; in other words: we can set brackets where and how we want.

- there is a neutral element $e \in G$ with $a*e=a=e*a$ for all $a \in G$.

- for all $a \in G$ there exists a so-called "inverse" $b$ of $a$, often denoted by $a^{-1}$, such that $a*b=e=b*a$.

The group is called "commutative" iff $*$ is commutative, so when $a*b=b*a$ for all $a,b \in G$ (this is not required to be so in the definition of a group).

Examples of groups:
- $\mathbb{Z}$ with $*$ being addition; the neutral element is what we denote by $0$ and the inverses are the negatives of the desired numbers.
- the nonzero rationals or reals with multiplication (note that $0$ would have no inverse).
- residues $\mod n$ also with addition.
- the coprime residues $\mod n$ with multiplication.
Till now all those are also commutative, so a noncommutative one now:
- the group $S_{n}$ of permutations of $n$ elements; the neutral element is the identity, the permutation that leaves all elements fixed; this group is in general not commutative, try $n=3$.

[b]Rings:[/b]

A set $R$ together with two operations $+$ (often called "addition" for the canonical reason) and $\cdot$ (often called multiplication) is called a "ring" if:

- $R$ together with $+$ is a commutative group; the neutral element is denoted by $0$ normally and the inverse of $a$ often by $-a$.

- the multiplication $\cdot$ is associative, so we have again $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ for all $a,b,c \in R$.

- the law of distributivity holds as we are used to it:
$a \cdot (b+c) = a \cdot b+a \cdot c$ and $(b+c) \cdot a = b \cdot a+c \cdot a$ for all $a,b,c \in R$.

Well, a lot of people add also two more conditions:
The ring is called commutative if the multiplication is so, thus when $a \cdot b = b \cdot a$ for all $a,b \in R$.
The ring is said to have unity if there is neutral element of multiplication, often denoted as $1$, so with $a \cdot 1 = a = 1 \cdot a$ for all $a \in R$.
In the previous posts and the rest of this one, ring always means commutative ring with unity till something different is said.

If $ab=0$ implies $a=0$ or $b=0$, then the ring is called an "integral domain".

Examples of rings:
- $R=\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$ with the standard operations we are used to.
- when $R$ is any ring, e.g. one of the above, then also $R[x]$, the set of polynomials with coefficients in $R$, is a ring.
- the rationals with denominator being a power of $2$.
- the rationals with odd denominator.
- the set $\mathbb{Z}^{2}$ consisting of the pairs $(a,b)$ of integers with addition by $(a,b)+(c,d): =(a+c,b+d)$ and multiplication $(a,b) \cdot (c,d): =(a \cdot c,b \cdot d)$; these definition is called "coordinatewise" addition/multiplication; another ring is archieved when defining $(a,b) \cdot (c,d): =(a \cdot c-b \cdot d , a \cdot d+b \cdot c)$.
- only for those who know them already: the arithmetic functions with pointwise addition and dirichlet convolution as multiplication build a ring also.

[b]Fields:[/b]

A commutative ring $F$ (operations $+$ and $\cdot$ again) with unity is called a field when:
All $r \in R \backslash \{0 \}$ have inverses $s$ (again often written as $r^{-1}$ then) in respect to multiplication, so with $r \cdot s = 1 = s \cdot r$.

Another definition:
A set $F$ with two operations as before is called a field if:
- $F$ builds a commutative group together with $+$.
- $F\backslash \{0 \}$ builds a commutative group together with $\cdot$.
- the law of distributivity given above holds.

Examples of fields:
- $\mathbb{Q}, \mathbb{R}, \mathbb{C}$.
-  residues $\mod p$ with $p$ prime.
- the numbers of form $a+b \sqrt{2}$ with $a,b$ rationals.
- the rational functions with coefficients in any field, e.g. the above ones; this is the corresponding to the polynomials for rings.

One reason to introduce them is the idea of generalization.
For example, try to prove the following statements just using the definitions:
If $G$ is a group with operation $*$ then:
- there is one and only one neutral element $e$.
- $(ab)^{-1}=b^{-1}a^{-1}$
- $(a^{-1})^{-1}= a$.
Let $R$ be a commutative ring with unity, operations as before:
- $(-a) \cdot b =-(a \cdot b)$.
- $0 \cdot a = 0$ for all $a \in R$.
- when $R$ is finite integral domain, then $R$ is already a field; use this to give an short proof that nonzero residues $\mod p$ prime are invertible.

Thanks, ZetaX! That's a nice summary  :)


Warm up 14

8. 3 digits are randomly selected w/o replacement from a set of two 2s, three 6s and one 9. What is the probability that the sum of the three digits is a multiple of 2? Common Fraction.

I got [hide]2/3[/hide] but answer is [hide]1/2[/hide]...

Warm up 16

10. In how many ways can two slashes be inserted between the letters of MATHCOUNTS to create three groups of letters? Each group of letters must contain each letter.

I've done this type of problem, but I can't remember the method. Hints please. :)

[hide="the primero uno"]
As long as the set contains no 9, the sum is even.
Probability of no 9 is
$\frac 56 \cdot \frac 45 \cdot \frac 34=\frac 36=\frac 12$
[/hide]
[hide="hint for the other one"]
Suppose that ^ denotes a possible space for a slash.
M^A^T^H^C^O^U^N^T^S[/hide]
[hide="solution for that one"]
Suppose that ^ denotes a possible space for a slash.
M^A^T^H^C^O^U^N^T^S
We have 9 spots for the slashes and two slashes to put
$\binom 92=36$[/hide]

So...

Was "Each group of letters must contain [i]each[/i] letter." supposed to be "Each group of letters must contain [i]a[/i] letter."?


What is the next number in the sequence $ 111, 21, 13, 12, 11, 10, 7, 7, ...$?

[hide=" Hint"]

number bases.

[/hide]

Each element is 7 in some base, starting with base 2 and increasing by 1 each time.  So every term to follow will be 7 since 7=7 in any base greater than 7.


Lauren solved the equation $ |x\minus{}5| \equal{} 2$. Meanwhile Jane solved an equation of the form $ x^2\plus{} bx \plus{} c \equal{} 0$ that had the same two solutions for $ x$ as Lauren's equation. What is the ordered pair $ (b, c)$?

First solve the absolute value equation.

$ |x \minus{} 5| \equal{} 2$

$ x \minus{} 5 \equal{} 2$

$ x \equal{} 7$

$ x \minus{} 5 \equal{} \minus{} 2$

$ x \equal{} 3$

$ x \equal{} 3, 7$

If these arre the roots of the quadratic, then the equation can be expressed as-

$ (x \minus{} 3) (x \minus{} 7) \equal{} 0$

$ x^2 \minus{} 10x \plus{} 21 \equal{} 0$

Ans: $ ( \minus{} 10,21)$


Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

\[ f(x + f(xy)) = f(x + f(x)f(y)) = f(x) + xf(y),\;\forall x,\;y \in \mathbb{R}.\]

I did find some properties (some of them are really important, like multiplicativity and injectivity), but I didn't manage to solve it.

It's clear that $f(x)=0\ \forall x\in \mathbb R$ works, so let's assume the function isn't this one. If there is $\alpha$ s.t. $f(\alpha)=0$ then replace $x$ with $\alpha$ and get $0=\alpha f(y)$. There is a $y$ s.t. $f(y)\ne 0$, so $\alpha=0$.

If $x=y$ then $f(x+f(x^2))=f(x+f^2(x))=f(x)(1+x)$. If $x=-1$ then we find that $f(f(1)-1)=0$, and from the result in the previous paragraph we find $f(1)=1,f(0)=0$. By making $x=1, y=1$ (one at a time) we find two more results which might be useful: $f(x+f(x))=x+f(x),f(1+f(x))=1+f(x)\ (*)$.

Now rewrite $f(x+f(xy))=f(x)+xf(y)$ as $f(x+f(a))=f(x)+xf(\frac ax),\ x\ne 0$. If there are $a\ne b$ s.t. $f(a)=f(b)$, then we plug $b$ instead of $a$ in the previous equality and we find $f(\frac ax)=f(\frac bx)$. This means that if $f(x)=f(rx)$ for some $x\ne 0$ then it's true for all $x\in \mathbb R$, and we thus find that there is $\alpha\ne 1$ s.t. $f(\alpha)=1$. We now make $x=\alpha$ in the equality $f(x+f(xy))=f(x)+xf(y)$. We now know that $f(\alpha y)=f(y)$, so $f(\alpha +f(y))=1+\alpha f(y)\Rightarrow f(\alpha +2)=1+2\alpha$ because from $(*)$ we find $f(2)=2$. On the other hand, we also have $f(\alpha +1)=\alpha +1$, and by applying $(*)$ again we find $f(\alpha+2)=\alpha+2$. This means that $\alpha+2=1+2\alpha\Rightarrow (\alpha-1)(2-1)=0\Rightarrow \alpha=1$, which is a contradiction, so $f$ is injective, so $f(x+f(xy))=f(x+f(x)f(y))\Rightarrow f(xy)=f(x)f(y)$.

It really isn't hard to prove another property, namely that $f(r)=r,\ \forall r\in \mathbb Q$.

I don't know how to continue for now, although I did prove a few other things as well, like $\forall x\in \mathbb R,\ f(f(x))\in \{\pm f(x)\}$.

i remember having solved that one in german imo-training this year, using only the first equation($f(x+f(xy)=f(x)+xf(y) (*)$).
grobber already showed $f(0)=0, f(1)=1$ the same way we did. it is now easy to show by $f(1+f(x))=1+f(x) (**)$ that $f(n)=n$ for all natural n. putting $x=y=-2$ in (*) we see that $f(-2)=-2$ so $f(-1)=-1$. we get $f(-1+f(y))=-1-f(-y) (***)$. from (**) it follows that if x is in the image of f, then x+1 is, and from (***) it follows that if x is in the image, then -x-1 is. therefore it follows that if x is in the image, then -x-1 is, so -x is, so -x+1 is, so x-1 is. therefore (**) now reads $f(f(x))=f(x)$. i remember that it wasn't hard from here anymore, but i don't remember the argument right now, let me think of it for a minute offline.

Peter

I'm not sure right now, but I don't think that my proof of the injectivity of $f$ used the second equality (only the proof of the multiplicativity did), so $f(f(x))=f(x)\Rightarrow f(x)=x$.

now here it is(of course i could use grobber's injectivity, but that's too easy :D ).
put y=-1 in (*) to get $f(blabla)=f(x)-x$, so $f(x)-x$ is in the image of f, so $f(f(x)-x)=f(x)-x$. thus,
$f(x)=f(f(x))=f(x+f(x)-x)=f(x+f(f(x)-x))=f(x+f(x(\frac{f(x)}{x}-1)))=f(x)+xf(\frac{f(x)}{x}-1)$,
so if $x\neq 0$, $f(\frac{f(x)}{x}-1)=0$, thus $\frac{f(x)}{x}-1=0$, thus $f(x)=x$.
this holds for $x=0$ too, as shown above.

Peter

i had the argument after 1 or 2 minutes but my mother phoned that long... :?

execuses, execuses... :)
Nice work though. :)


geometry
rectangle
calculus
integration
IMO Shortlist
number theory

Let $ R$ be a rectangle that is the union of a finite number of rectangles $ R_i,$ $ 1 \leq i \leq n,$ satisfying the following conditions:

[b](i)[/b] The sides of every rectangle $ R_i$ are parallel to the sides of $ R.$
[b](ii)[/b] The interiors of any two different rectangles $ R_i$ are disjoint.
[b](iii)[/b] Each rectangle $ R_i$ has at least one side of integral length.

Prove that $ R$ has at least one side of integral length.

[i]Variant:[/i] Same problem but with rectangular parallelepipeds having at least one integral side.

One way is to color the big rectangle in a chess-board like pattern with distance $ \frac 12$, starting at the lower left corner. Then show that every subrectangle $ R_i$ having an integer side contains the same amount of each color, so in total the big rectangle $ R$ has the same amount of each color. This contradicts if $ R$ doesn't have a side of integer length.

PS: moved to combinatorics (wouldn't call this number theory ;) ).

Stan Wagon compiled a series of 14 proofs of this result in his article "14 proofs of a result about tiling a triangle" (American Mathematical Monthly, Aug-Sep 1987)

i think the easiest solution is to consider this $ ABCD$ rectangle as a $ XOY $ coordinate system and the bottom  $ A $ vertex was $ (0 ; 0) $ point($AB$ and $AD$ are on the $ OY $ and $ OY$,respectively.).Now let us consider two sets,$1st$ set contains the vertices which have at least one integral coordinate,and $2nd$ set contains the rectangles.Now let us connect each point from $1st$ set to the rectangles in $2nd$ set,if this rectangle consists of this point.Which means the amount of lines going out from $1st$ set to $2nd$ set is even,if we won't consider lines with $A,B,C,D$ points.This is true,since 1 point is mutual for $0$,$2$ or $4$ rectangle,if this point isn't $A,B,C,D$ (then it is mutual for 1).And the lines,coming out from $2nd$ set must be also even,but since (0 ; 0) is in $1st$ set,it means the lines coming out from $1st$ set is odd,if this set doesn't contain at least one more vertex from $ABCD$ rectangle.However,this set contains $B$ or $C$ point.Hence $ABCD$ has at least one integer side.$\blacksquare$

[quote="enescu"][quote="Blitzkrieg97"]i think the easiest solution is ...[/quote]

So, finally, you succeeded in reposting the problem ( http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=610014, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=609286, and maybe more :) ).

Here is one link to an article quoted in this thread: [url]http://qcpages.qc.cuny.edu/~zakeri/mat208/wagon.pdf[/url][/quote]
thanks for reminding


It can be rewritten $(a+b)^2 + c^2 \geq 2(a+b)c$ which is just AM >= GM with equality iff $a+b = c$

umm... yes. Some people call it $(a+b-c)^2$ as well. :P

Since $a,b,c >0$, multiplying throughout by $abc$ we have $a^2+b^2+c^2 \geq 2bc+2ac-2ab \iff [(a+b)-c]^2 \geq 0$, trivially true. Equality occurs iff $(a+b)-c=0$, i.e. $a+b=c$.

[quote=Peter]Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\] and determine when equality occurs.[/quote]
Proof of Zhangyunhua:By AM-GM,$$\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} =\frac{(a+b)^2}{abc}+\frac{c}{ab}-\frac{2}{c}\ge\frac{2(a+b)}{ab}-\frac{2}{c}= \frac2a+\frac2b-\frac2c.$$

can we use cauchy schwarz inequality here?

By AM-GM, $(a+b)^2+c^2 \geq 2ac+2bc$. Simplifying, we get $a^2+b^2+c^2 \geq 2bc+2ac-2ab$, and dividing by $abc$ yields the desired inequality.

$(a+b-c)^2\Rightarrow a^2+b^2+c^2\ge2bc+2ac-2ab\Rightarrow\frac a{bc}+\frac b{ac}+\frac c{ab}\ge\frac2a+\frac2b-\frac2c$

[hide=Solution]Rewrite as $a^2+b^2+c^2 \geq 2bc+2ac-2ab$, or $(a+b-c)^2 \geq 0$, which is true from the Trivial Inequality. All steps are reversible, so we're done.[/hide]

[quote=e61442289][quote=Peter]Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\] and determine when equality occurs.[/quote]

$\frac {a} {bc} + \frac {b} {ac} + \frac {c} {ba} = \frac {a ^ 2 + b ^ 2 + c ^ 2} {abc}$ and  $\frac {2} {a} + \frac {2} {b} + \frac {2} {c} = \frac {2 (bc + ac-ab)} {abc}$
then:
$\frac {a ^ 2 + b ^ 2 + c ^ 2} {abc}$  is greater than  $\frac {2 (bc + ac-ab)} {abc}$  and this is true if and only if  $a ^ 2 + b ^ 2 + c ^ 2$  is greater than or equal to  $2 (bc + ac-ab)$  which is true if and only if  $(a + b - c) ^ 2$ is greater than or equal to $0$ which is true.[/quote]

FTFY. For the fraction $\LaTeX$, it starts with "\" not "/".
Edit: What does $\mathbb{R}^+_0$ mean in the question? Sorry I am not familiar with the notation.

We have $\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} = \frac{a^2+b^2+c^2}{abc}$ and $\frac2a+\frac2b-\frac2c = \frac{2bc + 2ac - 2ab}{abc}$. They both have the same denominators so we need just prove that $a^2+b^2+c^2\geq2bc+2ac-2ab$.

Note that $a^2+b^2+c^2 - 2bc - 2ac + 2ab = (a + b - c)^2\geq0$, showing the given inequality. Equality occurs when $a + b = c$.

Since $a,b,c>0$ we multiply all by abc
So we are left with:
$a^2+b^2+c^2\ge 2bc+2ac-2ab$
$\rightarrow (a+b)^2+c^2\ge 2c(a+b)$
which is true for $AM-GM$

[quote=Peter]Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\] and determine when equality occurs.[/quote]

[hide=Solution]
Since $a, b, c > 0,$ we multiply by $abc.$ The expression translates to $a^2 + b^2 + c^2 \geq 2bc + 2ac -2ab = a^2+b^2+c^2+2(ab-bc-ac).$ This simplifies to
\[(a+b-c)^2 \geq 0,\]
trivial from there.
Equality:
IFF $(a+b-c)^2 = 0,$ or $a+b=c.$

[quote=e61442289][quote=Peter]Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\] and determine when equality occurs.[/quote]

$/ frac {a} {bc} + / frac {b} {ac} + / frac {c} {ba} = / frac {a ^ 2 + b ^ 2 + c ^ 2} {abc} and  $/ frac {2} {a} + / frac {2} {b} + / frac {2} {c} = / frac {2 (bc + ac-ab)} {abc}$
then:
$/ frac {a ^ 2 + b ^ 2 + c ^ 2} {abc}$  is greater than  $/ frac {2 (bc + ac-ab)} {abc}$  and this is true if and only if  $a ^ 2 + b ^ 2 + c ^ 2$  is greater than or equal to  $2 (bc + ac-ab)$  which is true if and only if  $(a + b - c) ^ 2$ is greater than or equal to 0 which is true.[/quote]

LATEX

[quote=e61442289][quote=Peter]Prove that for all $a,b,c \in \mathbb{R}^+_0$ we have \[\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac2a+\frac2b-\frac2c\] and determine when equality occurs.[/quote]

$\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ba} = \frac{a^2 + b^2 + c^2}{abc}$ and  $\frac{2}{a}+ \frac{2}{b} + \frac{2}{c} = \frac{2 (bc + ac-ab)}{abc}$
then:
$\frac {a ^ 2 + b ^ 2 + c ^ 2}{abc}$  is greater than  $\frac{2 (bc + ac-ab)}{abc}$  and this is true if and only if  $a^2 + b^2 + c^2$  is greater than or equal to  $2 (bc + ac-ab)$  which is true if and only if  $(a + b - c) ^ 2$ is greater than or equal to 0 which is true.[/quote]

I fixed the LaTeX but idk about this proof

I will present mathmax12's solution (which he shared with me and gave me permission to show.):

 Multiplying by $abc$ gives you $a^2+b^2+c^2 \ge 2bc+2ac-2ab$, so $((a+b)-c)^2 \ge 0$, which is always true by the trivial inequality, equality holds when $\boxed{a+b=c}.$


Bicycle license plates in Flatville each contain three letters. The
first is chosen from the set $ \{C,H,L,P,R\}$, the second from
$ \{A,I,O\}$, and the third from $ \{D,M,N,T\}$.
\\ \\
When Flatville needed more license plates, they added two
new letters. The new letters may both be added to one set or one
letter may be added to one set and one to another set. What is the
largest possible number of ADDITIONAL license plates than can be
made by adding two letters?

At first, there are 12 letters distributed into 3 sets, and then after adding the 2 new letters, we need to add them to sets to maximize the product, since this product is the number of license plates that can be formed.  Since the sum of the letters in the set remains constant (14), to maximize the product, we make the numbers as close to each other as possible.  This occurs when 2 sets have 5 letters and the other has 4, so we add the two letters to the set $ \{ A,I,O \}$
Using the new sets, there are $ 5\cdot 5\cdot 4\equal{}100$ license plates that could be formed.
Before, there were $ 5\cdot 4\cdot 3\equal{}60$ license plates, so there are $ 100\minus{}60\equal{}\boxed{40}$ more license plates

unless am-gm fails me again...

That's correct.

And AM-GM usually isn't required in the AMC 8 :wink:

Not required, but very helpful anywhere :wink:

The AM-GM is [hide=pizza and coke]very rare ok rare[/hide] in the [hide=patio]Amc 8[/hide] but great explanation!


Two bags contain some numbers, and the total number of numbers is prime.

When we tranfer the number 170 from 1 bag to bag 2, the average in both bags increases by one.

If the total sum of all numbers is 2004, find the number of numbers.

Please, what the meaning of "amount" ?!
 :(

Hey Guest, have a look (because of "amount")  :P ..

Let S1,S2 be the sum of numbers for each bug, and let a1,a2 be the amount of numbers  for each bag. We can write :
(S1 - 170)/(a1 - 1) = S1/a1 + 1   ==> S1 = a1*(169 + a1)
(S2 + 170)/(a2 +1) = S2/a2 + 1  ==> S2 =  a2*(169 - a2)
So S = S1+S2 = (169 + a1 - a2)*(a1+a2) = 2004 = 167*3*4
a = a1+a2 must be prime so a =  2,3 or 167 .
If a= 2 or 3 then  169 + a1-a2  < 167*2 , so S < 2004 (contradiction)
So we have a1 + a2  = 167 and  169 + a1 - a2 = 3*4
Finally we obtaint a1 = 5 , a2 = 162 , S1 = 5*174 , S2 = 162*7
 :lol:

nice short&straight solution, Diogene :)


Calculate the sum $ S=\sum^n_{i=0}\frac{4^i}{1+cos(2^i)}$

Does anyone spend any minutes for this sum?

If anyone can find a closed form value for this $S$, I'll give you a "Manly" kiss as a reward  :D


I can't seem to remember how to factor this;

36x(squared)-25x(squared)

MOD: Moved to High School Basics.

Problems like this probably belong [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]here[/url].

You probably meant $25y^{2}$, since as written your question is just $11x^{2}$.

Rewrite it as $(6x)^{2}-(5y)^{2}$.


$\lim_{x \rightarrow 0^{-}}{\frac{\sin{(2x)}}{\sqrt{2-2\cos{(x)}}}}$=?

[color=red]Latex edited by mod.[/color]

(click the smile)
[hide=" ;) "]
$-2$

[/hide]


function
real analysis
real analysis unsolved

Show that the following statements are equivalent :

(a) $ f$ is continuous on $ \mathbb{\mathbb{R}}$

(b)$ \forall\lambda,$ $ A_{\lambda}$ $ \equiv$ $ \{x\mid f(x)>\lambda\}$,$ B_{\lambda}$ $ \equiv$ $ \{x\mid f(x)<\lambda\}$ are both open set.

(c)$ \forall\lambda,$ $ C_{\lambda}$ $ \equiv$ $ \{x\mid f(x)\geq\lambda\}$,  $ D_{\lambda}$ $ \equiv$ $ \{x\mid f(x)\leq\lambda\}$ are both closed set.

thanks~

Here's a more long-winded proof, but it's a bit more general.

[b]Theorem[/b] [i]A function $ f$ is upper semicontinuous if and only if $ \{f(x)\geq a\}$ is closed for all finite $ a$.  A function $ f$ is lower semicontinuous if and only if $ \{f(x)\leq a\}$ is closed for all finite $ a$.[/i]

[i]Proof[/i] It suffices to prove the first part since $ f$ is usc if and only if $ \minus{}f$ is lsc.  Suppose that $ f$ is usc.  Given $ a$, let $ x_{0}$ be a limit point of $ \{f(x)\geq a\}$.  Then there exist $ x_{k}$ such that $ x_{k}\to x_{0}$ and $ f(x_{k})\geq a$.  Since $ f$ is usc at $ x_{0}$, we have $ f(x_{0})\geq\limsup_{k\to\infty}f(x_{k})$.  Therefore, $ f(x_{0})\geq a$, so that $ x_{0}\in\{f(x)\geq a\}$.  Hence $ \{f(x)\geq a\}$ is closed.

Conversely, let $ x_{0}$ be a limit point.  It $ f$ is not usc at $ x_{0}$ then $ f(x_{0})<\infty$ and there exists $ M$ and $ (x_{k})$ such that $ f(x_{0})<M$, $ x_{k}\in\mathbb{R}$, $ x_{k}\to x_{0}$, and $ f(x_{k})\geq M$.  Hence $ \{f(x)\geq M\}$ is not closed since it does not contain all its limit points.

A function is continuous if and only if it is both usc and lsc.


Let $x_1$ be a real number such that $0<x_1<1$.

Define $x_{n+1}= \frac{1}{x_n}-\lfloor \frac{1}{x_n} \rfloor $ for $x_n$ non-zero and $0$ for $x_n$ zero.

Show that $x_1 + x_2 + \cdots + x_n < \frac{F_1}{F_2}+\frac{F_2}{F_3}+\cdots +\frac{F_n}{F_{n+1}}$,

where $F_n$ is the Fibonacci sequence: $F_1=F_2=1$, $F_{n+2}=F_{n+1}+F_n$.

What shortlist does this problem come from, if I may ask?


How to find the number of n-element permutations with exactly k inversions?? (n,k are given)

A pair of indices (i,j), 1 <= i<= j <= n, is an inversion of the permutation A if ai>aj.

It doesn't appear to be a particularly simple problem, see:

http://www.lacim.uqam.ca/~plouffe/JIS/inversions.pdf

OK - I see... But what if we take 1<=n<=12 and 0<=k<=98 .... Is there any formula for getting the number of n-element permutations with exactly k inversions??

If $[n] := 1 + q + \dots + q^{n-1} $  then
$[1][2]....[n] $   is a generating function for what you want.
(Note that for $q=1$ we recover $n!$ as the number of permutations,
so this is a q-analogue of the factorial.)

Good luck if you want to go beyond that...


college
geometry
calculus
real analysis
complex analysis

[img]http://images.amazon.com/images/P/0387949348.01._SCLZZZZZZZ_.jpg[/img]

This book is a compilation of approximately nine hundred problems, which have appeared on the preliminary exams in Berkeley over the last twenty years. It is an invaluable source of problems and solutions for every mathematics student who plans to enter a Ph.D. program. Students who work through this book will develop problem solving skills in areas such as real analysis, multivariable calculus, differential equations, metric spaces, complex analysis, algebra, and linear algebra. The problems are organized by subject and ordered in an increasing level of difficulty. This new edition contains approximately 120 new problems and 200 new solutions. It is an ideal means for students to strengthen their foundation in basic mathematics and to prepare for graduate studies.

[url=http://muebook.blogspot.com/2008/02/berkeley-problems-in-mathematics.html]Download here[/url]

Definitely not High-School Contest content (PhD!!!) but it is a worthy book to have in your collection.


Would you need to add quotation marks around thoughts and what you said when writing in a journal?  :maybe:  (this is for school)

For ex:  "Good Luck," I whispered.    or should it be Good Luck, I whispered.

and thoughts:  "How great that would be," I thought.  or How great that would be, I thought.

Thanks!!

[quote="lotrgreengrapes7926"]People here are good at math, not English. :roll:
I believe that the whisper would have quotation marks, since it was said aloud. But the thought does not have quotation marks.[/quote]

He said journal, do the same priciples apply for books and journals?

Do whatever works best for you.  It will probably be clearer if you use quotation marks around both thoughts and spoken phrases.  One thing you can do that avoids this question is the following (although I wouldn't recommend it in general):

"You're just a dumb mathematician, so bug off," he said.

becomes

He called me a dumb mathematician and told me to bug off.

For me, (I don't know if it's correct) I italicize thoughts.

I agree with SplashD. I ususally see italicized thoughts

From my experince:

I thought, "This is amazing!"

I thought, [i]This is amazing![/i]

I personally italicize usually, but that is kind of hard if you're handwriting in a journal.  But anything spoken does get quotation marks.  I'm sure of that.


Consider a group of $2n-1$ people (where $n \geq 2$) such that each of these persons belongs to a group of $n$ people who are pairwise friends.

For which $n$ does it necessary imply that there is someone in the whole group who is friend with everyone?

Pierre.

Sorry, but what does pairwise mean ? I've looked for it in dictionary but I haven't found anything  :blush:

Pierre means that  in each group of those n persones, everyone is friends with everyone else :) Pairwise would most probably be described as "each pair" ...

the problem can be reformulated as the following:(i am using a few terms without writing their definitions but the context should make the definition clear)
consider the set S={1,2,...,2n-1}.is it possible to find a subset A of n-sets of S that covers all the singletons both not all the pairs incident with any i?(to see the equivalence,the n-sets represent the cliques of mutual friends.then since A covers all the singletons,every i is in some n-set,i.e.in a group of n people who are mutually friends.however if the collection does not cover all pairs incident with any i then it means that no person is everyone's friend).
  if n>4,then such a collection can be found as follows:
   take the sets {1,2,....n},{n-1,n,...2n-2},
{1,2,...[(n-1)/2],n+1,n+2,...,n+(n-1-[(n-1)/2]),2n-1}.(i.e.take 'half' as many from
{1,2...n-2} and roughly 'half' as many from {n+1,n+2,...n+(n-2)},along with the missing 2n-1). clearly each set has n members and every i in S belongs to some set in this collection.note that when n>4,the choices as above will still leave behind some elements in both of {1,2...n-2} and {n+1,n+2,...n+(n-2)}.and this ensures that for no i can all pairs (i,j), i,j distinct ,be covered by the 3 sets above.
   the case 2 will necessarily imply that such a collection of 2 sets is not possible in a set of 3 elements(trivial).the cases n=3 and n=4 should also necessarily force the existence of a 'common friend',i think...(i guess it is easy to argue casewise if necessary;i haven't actually tried it out).


$ a,b,c > 0$.Prove that:
$ \frac {a^3}{2a^2 \plus{} b^2} \plus{} \frac {b^3}{2b^2 \plus{} c^2} \plus{} \frac {c^3}{2c^2 \plus{} a^2}\geqslant \frac {a \plus{} b \plus{} c}{3}$

 $ a,b,c > 0$.Prove that:
$ \frac {a^2 \plus{} 2bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2ab}{a \plus{} b}\geqslant \frac {3(a \plus{} b \plus{} c)}{2}$

[quote="fesha3t"]

 $ a,b,c > 0$.Prove that:
$ \frac {a^2 \plus{} 2bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2ab}{a \plus{} b}\geqslant \frac {3(a \plus{} b \plus{} c)}{2}$[/quote]
$ LHS\minus{}RHS\equal{}\sum \frac{1}{2}(a\minus{}b)^2*\frac{a^2\plus{}b^2\plus{}ab\minus{}c^2}{\prod (a\plus{}b)}$
assume that:$ a\geq b\geq c$
we have $ S_{b},S_{c}\geq 0$
and $ S_{b}a^2\plus{}S_{a}b^2 >(a^2\minus{}b^2)^2)\geq 0$
Q.E.D

[quote="aauqen"]$ 1) \sum \frac {a^3}{2a^2 \plus{} b^2} \geq \frac {(\sum a^\frac {3}{2})^2}{3\sum a^2} \geq \frac {\sqrt {3} \cdot (\sum a^2)^\frac {3}{2}}{3\sum a^2} \equal{} \sqrt {\frac {\sum a^2}{3}} \geq \frac {a \plus{} b \plus{} c}{3}$

by Cauchy and inequalities between power means.[/quote]

Correct me if I'm wrong but I think $ \frac{1}{2} < \frac{2}{3}$ therefore

\[ \sqrt[\frac{1}{2}]{\frac{\sum (a^3)^\frac{1}{2}}{3}} \le \sqrt[\frac{2}{3}]{\frac{\sum (a^3)^\frac{2}{3}}{3}}\]

[quote="fesha3t"]
 $ a,b,c > 0$.Prove that:
$ \frac {a^2 \plus{} 2bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2ab}{a \plus{} b}\geqslant \frac {3(a \plus{} b \plus{} c)}{2}$[/quote]

This one can be done by brute force too:
First note that $ (a^2\plus{}2bc)(a\plus{}b)(a\plus{}c) \equal{} a^4 \plus{} (b\plus{}c)a^3 \plus{} 3bc a^2 \plus{} 2bc(b\plus{}c)a \plus{} 2b^2c^2$
And that $ (a\plus{}b)(a\plus{}c)(b\plus{}c) \equal{} 2abc \plus{} \sum a^2b$
(All summation signs are symmetric)

So after clearing the denominators, we have
$ \frac{1}{2} \sum a^4 \plus{} \sum a^3b \plus{} \frac{7}{2} \sum a^2bc \plus{} \sum a^2b^2 \geq (2abc \plus{} \sum a^2b) \frac{3(a\plus{}b\plus{}c)}{2} \equal{} \frac{3}{2} \sum a^3b \plus{} 3 \sum a^2bc \plus{} \frac{3}{2}\sum a^2b^2$

So finally our inequality becomes:
$ \sum a^4 \plus{} \sum a^2bc \geq \sum a^3b \plus{} \sum a^2b^2$
But using Schur:
$ \sum a^2(a\minus{}b)(a\minus{}c) \geq 0 \iff \sum a^4 \plus{} \sum a^2bc \geq 2 \sum a^3b \geq \sum a^3b \plus{} \sum a^2b^2$ (the second inequality is from AM-GM)

[quote="hendrata01"]
So finally our inequality becomes:
$ \sum a^4 \plus{} \sum a^2bc \geq \sum a^3b \plus{} \sum a^2b^2$
[/quote]
You have a mistake in calculus. Notice that this ineq is symetric, not cyclic  :wink:
[quote="fesha3t"] $ a,b,c > 0$.Prove that:
$ \frac {a^2 \plus{} 2bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2ab}{a \plus{} b}\geqslant \frac {3(a \plus{} b \plus{} c)}{2}$[/quote]
We can prove it easily by CS and Voni Schur  :)
Besides, the stronger of Nguyen Van Thach is very hard:
$ \sum\ \sqrt {\frac {a^2 \plus{} 2bc}{b \plus{} c}} \geq\ \frac {3}{\sqrt {2}}\sqrt {a \plus{} b \plus{} c}$

[quote="nguoivn"][quote="hendrata01"]
So finally our inequality becomes:
$ \sum a^4 \plus{} \sum a^2bc \geq \sum a^3b \plus{} \sum a^2b^2$
[/quote]
You have a mistake in calculus. Notice that this ineq is symetric, not cyclic  :wink:
[/quote]

Nope, I think I'm right. The summation signs are all symmetric (as stated in my post above). That means $ \sum a^4$ = $ 2(a^4\plus{}b^4\plus{}c^4)$ (that's how it's usually understood).


geometry
power of a point
radical axis
geometry unsolved

let E be a point and (C) be a circle . and EC and ED are tangent to (C). let F and G be the midpoints of EC and ED.

let Q be a variable point on line FG. QA and QB are tangents to (C). and AB intersects line GF in point P.

prove that :

                angle PEQ=90º

Ashegh ,you should post this in the Proposed problems ,cause our teacher has solved it for us like this:
[hide="Solution"]

Line $PQ$ is the Radical axis of $E$ and the circle $C$ .

So $QA=QB=QE$.

We will show that $PE$ is tangent to circle $D=(Q,QE)$. which is equal to $PEQ=90$.

$P$ lies on the radical axis of $D$ and $C$ and it lies on the radical axis of $E$ and $C$ too . Which means

$PE^2=\text{Power of Q wrt D}$ which means $PE$ is tangent to $D$,hence result follows.
[/hide]

that  was the most beautifull problem that i have ever seen .
and i'm going to solve it as soon as possible...
 :D  :)

ee... bacheha che bahal to in safhe hame az iran post kardan. :roll:  :P

yes i am agree with pooyan in everything and i will think about this problem either and i'll post my solution as soon as i can(my powerful solution)
and another more thing,Ashegh wher do you get these beautiful problems? :D

you know it better jensen.from u teacher Mr.vakily ;)

[quote="lomos_lupin"]Ashegh ,you should post this in the Proposed problems ,cause our teacher has solved it for us like this:
[hide="Solution"]

Line $ PQ$ is the Radical axis of $ E$ and the circle $ C$ .

So $ QA \equal{} QB \equal{} QE$.

We will show that $ PE$ is tangent to circle $ D \equal{} (Q,QE)$. which is equal to $ PEQ \equal{} 90$.

$ P$ lies on the radical axis of $ D$ and $ C$ and it lies on the radical axis of $ E$ and $ C$ too . Which means

$ PE^2 \equal{} \text{Power of Q wrt D}$ which means $ PE$ is tangent to $ D$,hence result follows.
[/hide][/quote]
I don't know what is radical axis? Can you explain for me?
Here is my solution:
Let H be the intersection of OE and CD (O is the center of (C)), (Q,QE) intersects (O) at A' and B'
We have $ OA'^2\equal{}OB'^2\equal{}OD^2\equal{}OH.OE$ and H lies on (Q) so OB and OA are two tangents of (Q), it's equivalent to $ \angle QB'O\equal{}\angle QA'O\equal{}90^o$ or $ A\equiv A', B\equiv B'$
  Let J,K be the intersection of AB and CD, QO and AB
We wish to have $ \angle QEP\equal{}\angle QHP\equal{}90^o$
Because $ PF//CD$ then $ \angle QPK\equal{}\angle KJH\equal{}\angle KOH (1)$
On the other side, $ QK.QO\equal{}QB^2\equal{}QH^2$ or $ \frac{QK}{QH}\equal{}\frac{QH}{QO}$
Hence $ \Delta OQH$~$ \Delta HQK$ then $ \angle QOH\equal{}\angle QHK (2)$
From (1) and (2) we get $ \angle QHK\equal{}\angle QPK$ therefore QKHP is cyclic
So $ \angle QHP\equal{}\angle QKP\equal{}90^o$

$ FG$ is radical axis of the said circle $(O)$ and degenerate circle $ E$ $ \Longrightarrow$ $ PE^2 \equal{} PB \cdot PA$ and $ QE \equal{} QB \equal{} QA.$ But power of $ P$ to the circumference with center $ Q$ passing through $ A,B$ is $ PB \cdot PA \equal{} PQ^2 \minus{} QA^2 \equal{} PQ^2 \minus{} QE^2$ $ \Longrightarrow$ $ EQ^2 \equal{} PQ^2 \minus{} PE^2$ and the conclusion follows.

[b]Remark:[/b] If $ M \equiv{} AB \cap CD,$ then we have $ OM \parallel PE.$


Alex can bike one lap of a racecourse in 30 minutes at 25 miles per hour   How many minutes does it take Lance to bike the same lap at 30 miles per hour?

If it takes Alex 30 minutes at 25 miles an hour, that means the racecourse is $ \frac{25}{2}$ or $ 12.5$ miles long. So, at 30 miles an hour:

$ \frac{12.5}{x} \equal{} \frac{30}{60}$

$ 750 \equal{} 30x$

$ 25 \equal{} x$

Thus, $ \boxed{25}$ minutes.

If we are trying to do this quickly (like in a fast FTW game) we could take a shortcut and notice,

$ r\equal{}\frac{d}{t}$ where $ r$ is rate $ d$ is distance, and $ t$ is time

We can multiply by $ t$ to get $ d\equal{}rt$

Since $ d$ is a constant (the length of the racetrack), the product $ rt$ will always be the same.

Plugging in the numbers:

$ r_1t_1\equal{}r_2t_2$
$ 25\cdot{30}\equal{}30\cdot{t_2}$
$ 25\equal{}t_2$

This may same also very long, but once you recognize that with a constant distance, rate and time vary inversely (or whatever its called), you can do this problem and problems like it in sub-5 seconds.


http://www.zaba.ru/cgi-bin/tasks.cgi?tour=russia.festival.X.final

It`s from Russian math festival Adler but I have seen this problem in many small math olympiads in belarus and bulgaria.

[quote="Beat"]Prove that for any natural number n we have $\sqrt [n]{(n+1)!} - \sqrt [n]{n}\geq 1$

Excuse my terrible mistake (Kalimdor) and the others who tryed this problem.

By the way $\sqrt [n]{(n+1)!} - \sqrt [n]{n!}\geq 1$ is that true, too?[/quote]
FOR the first one.I think it is easy to prove.
just prove $(n+1)! \geq (\sqrt [n]{n}+1)^n$

$n\geq2$ is natural number. Then $(\sqrt[n]{n!})^2\geq\sqrt[n+1]{n+1!}\sqrt[n-1]{n-1!}$

Can somebody prove this?
And zhaobin please use more details I didn`t understand what you mean.

I mean $\sqrt [n]{(n+1)!} - \sqrt [n]{n}\geq 1$
is equal to $(n+1)! \ge (\sqrt [n]{n}+ 1)^n$
and $(n+1)!=n!+\cdots+n!$ there have n+1 $n!$
and expand $(\sqrt [n]{n}+ 1)^n$
then you are easily see the result.
sorry I can't explain it more clear :(

now i understand, thanks. Can you proove the second one?  :)


#5

The perimeter of a sector of a circle is the sum of the two sides formed by the radii and the length of the included arc. A sector of a particular circle has a perimeter of 28 cm and an area of 29 sq. c.m. What is the length of the arc of this sector?

There's a simple formula for the area of a sector: $ A\equal{}\frac{rl}{2}$, where r is the radius and l is the length of the arc. So if perimeter is 2r+l=28 and area is rl/2=29, plug in l=-2r+28 into the second equation.
It becomes -r^2+14r-29=0, and when you solve for r, and finally l, you get $ l\approx 5.06.$

I think you typoed the problem. It should be that the area of the sector is 49 sq. cm if I remember correctly. In that case, you get an integer answer as the problem implies. :wink:

[hide="solution"]Let the fraction of whole circle that the sector is be $ x$. We have $ 2{\pi}rx\plus{}2r\equal{}28$ and $ {\pi}r^2x\equal{}49$. We wish to find $ 2{\pi}rx$.

Rearranging the first eqaution, $ 2{\pi}rx\equal{}28\minus{}2r\implies{2{\pi}r^2x\equal{}28r\minus{}2r^2}\implies{49\cdot{2}}\equal{}28r\minus{}2r^2\implies{r^2\minus{}14r\plus{}49\equal{}0}\implies{r\equal{}7}\implies{2{\pi}rx\equal{}28\minus{}2r}\implies{2{\pi}rx\equal{}14}$[/hide]

I remember this problem, at the competition, I saw 2 multiples of 7, so I just guessed 14 and got it right :D


geometry
angle bisector
geometry proposed

Let $ P$ be a point inside $ \triangle ABC$ such that $ \angle BPC \equal{} 60^\circ \plus{} \angle A$, $ \angle BPA \equal{} 60^\circ \plus{} \angle C$, and $ \angle APC \equal{} 60^\circ \plus{} \angle B$ (For simplicity, each of the angles of $ \triangle ABC$ is less than $ 120^\circ$). Prove that angle bisector of $ \angle BAC$ and $ \angle BPC$ meet on $ BC$.

it's an old problem and u just need:$ \angle BPA \minus{} \angle ACB\equal{}\angle CPA \minus{} \angle CBA$
it implies that the bisector angle of $ ACP$ and $ ABP$ intersect on $ AP$.
 it means that $ \frac{AB}{BP}\equal{}\frac{AC}{CP}$
which implies $ \frac{BA}{AC}\equal{}\frac{BP}{PC}$.DONE!

it's an adaptation of my previous post. http://www.mathlinks.ro/viewtopic.php?t=296623


$ x,y,z$ are positive reals such that $ xyz\equal{}1$ and $ x,y,z \not\equal{} 1$. Prove that
$ \frac{z^2}{(x\minus{}1)^2}\plus{}\frac{x^2}{(y\minus{}1)^2}\plus{}\frac{y^2}{(z\minus{}1)^2} \ge \frac{(1\minus{}x)^2 (1\minus{}y)^2 (1\minus{}z)^2}{(1\minus{}x)^4\plus{} (1\minus{}y)^4 \plus{}(1\minus{}z) ^4}$


[quote="beta"][quote="tongos"]the question was this:
The function f satisfies f(x) + f(x-1) = x2 for all x. If f(19) = 94, find the remainder when f(94) is divided by 1000.

he said this:
 f(94)=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)= ... = (94^2-93^2) + (92^2-91^2) +...+ (22^2-21^2)+ 20^2-f(19) = 94+93+...+21+400-94 = 115·37+306 = 4561.

how is 94^2-93^2 equal to 94 ????

i get 701 as my answer.
perhaps im wrong[/quote]

94^2-93^2 = (94+93)(94-93)=94+93, although this wasn't clear in his solution.

Although sometimes kalva solutions are not elegant and hard to understand, but I don't think kalva had ever had a wrong answer.[/quote]

sorry to sound nitpicky, but for the 2004 AIME1, he rounded $100(4-\pi)=858$ when it is supposed to be $86$

also, since kalva tried to avoid copyright problem, he changed the wording, changing the answers... but some of them he left the original aanswer....

also, his answers are 50% times not integers between 0-999


What is the remainder if x^81 + x^27 + x^9 + x^3 + x is divisible by x-1?
[size=59](Sorry if I didn't use LaTeX, I dunno how to use it and can't be bothered to learn at the moment.)[/size]

I don't quite get the question... can someone enlighten me please? Thank you!

[quote="isabella2296"]Do you mean:

What is the remainder [b]when[/b] $ x^{81} \plus{} x^{27} \plus{} x^9 \plus{} x^3 \plus{} x$ is [b]divided[/b] by $ (x \minus{} 1)$?[/quote]

I was thinking that, but the question was really "What is the remainder IF ... IS DIVISIBLE ..."  :?:

[b]Problem Restatement:[/b] Let $ P(x) \equal{} x^{81} \plus{} x^{27} \plus{} x^9 \plus{} x^3 \plus{} x$. Then, we can write $ P(x)\equal{}Q(x) \cdot (x\minus{}1) \plus{} c$ for some polynomial $ Q(x)$ and some constant $ c$. What is this number $ c$?

[b]Answer:[/b] Plug in $ x\equal{}1$. You find that $ P(1) \equal{} Q(1) \cdot (1\minus{}1) \plus{} c \equal{} Q(1) \cdot 0 \plus{} c \equal{} c$.

Therefore, $ c \equal{} P(1) \equal{} 1^{81} \plus{} 1^{27} \plus{} 1^9 \plus{} 1^3 \plus{} 1 \equal{} 5$

[quote="The Zuton Force"][b]Problem Restatement:[/b] Let $ P(x) \equal{} x^{81} \plus{} x^{27} \plus{} x^9 \plus{} x^3 \plus{} x$. Then, we can write $ P(x) \equal{} Q(x) \cdot (x \minus{} 1) \plus{} c$ for some polynomial $ Q(x)$ and some constant $ c$. What is this number $ c$?

[b]Answer:[/b] Plug in $ x \equal{} 1$. You find that $ P(1) \equal{} Q(1) \cdot (1 \minus{} 1) \plus{} c \equal{} Q(1) \cdot 0 \plus{} c \equal{} c$.

Therefore, $ c \equal{} P(1) \equal{} 1^{81} \plus{} 1^{27} \plus{} 1^9 \plus{} 1^3 \plus{} 1 \equal{} 5$[/quote]

Thank you! :)

This is a very common thing to find/do when dealing with polynomials. It's called the Remainder Theorem.

Note that it can only be done when dividing by a linear.

When dividing a polynomial by (x-a), just plug in a for all x's and simplify. (A more succinct version of what The Zuton Force said.  :wink: )

[quote="The Zuton Force"][b]Problem Restatement:[/b] Let $ P(x) \equal{} x^{81} \plus{} x^{27} \plus{} x^9 \plus{} x^3 \plus{} x$. Then, we can write $ P(x) \equal{} Q(x) \cdot (x \minus{} 1) \plus{} c$ for some polynomial $ Q(x)$ and some constant $ c$. What is this number $ c$?

[b]Answer:[/b] Plug in $ x \equal{} 1$. You find that $ P(1) \equal{} Q(1) \cdot (1 \minus{} 1) \plus{} c \equal{} Q(1) \cdot 0 \plus{} c \equal{} c$.

Therefore, $ c \equal{} P(1) \equal{} 1^{81} \plus{} 1^{27} \plus{} 1^9 \plus{} 1^3 \plus{} 1 \equal{} 5$[/quote]
This doesn't work because you're dividing by (1-1)=0

The division algorithm on polynomials is best phrased, not in terms of the expression $ \frac{P(x)}{x \minus{} 1} \equal{} Q(x) \plus{} \frac{c}{x \minus{} 1}$ (which is probably what you're thinking of, and is how we express division by integers), but exactly as it is written - there exists a constant $ c$ and a polynomial $ Q(x)$ such that $ P(x) \equal{} (x \minus{} 1) Q(x) \plus{} c$ for [b]every[/b] real $ x$, and that includes $ x \equal{} 1$.  There is no division by zero because nowhere are we actually dividing by $ x \minus{} 1$ when $ x$ has a value.

[quote="isabella2296"]What do you mean? We're not dividing by $ (1 \minus{} 1)$...[/quote]

 :blush: ooops, i thought...
[hide]
nvm
[/hide]

So zuton force is right! :blush:


calculus
derivative
vector
analytic geometry

Find equations for the planes tangent to $ z\equal{}x^{2}\minus{}6x\plus{}y^{3}$ that are parallel to the plane $ 4x\minus{}12y\plus{}z\equal{}7$.  I need a solution that makes use of partial derivatives.

There is a calculus forum- you should have posted this there, and you should expect a moderator to move this there.

How do you find a tangent plane?
How do you find out whether two planes are parallel?

I hope I did this correctly

[hide="Solution"]Just like in 2-D we had the equation $ \Delta y \equal{}\frac{dy}{dx}\Delta x$ for a tangent line, in 3-D we have:
$ \Delta z \equal{}\frac{\partial z}{\partial x}\Delta x\plus{}\frac{\partial z}{\partial y}\Delta y$

Alternatively, we can write $ \vec{n}: \equal{}\left(\minus{}\frac{\partial z}{\partial x},\minus{}\frac{\partial z}{\partial y}, 1\right)\implies\vec{n}\cdot (x,y,z) \equal{}\vec{n}\cdot (x_{0},y_{0},z_{0})$.

Then, $ \vec{n}$ is a normal vector to the plane.

Two planes are parallel if a normal vector from one is a scale multiple of the normal vector for another.

$ \vec{v}: \equal{} (4,\minus{}12,1)$ is a normal vector to the given plane. We need it to be a scale multiple of $ \vec{n}\equal{} (\minus{}2x\minus{}6,\minus{}3y^{2}, 1)$

Comparing third coordinates, clearly the scale factor is $ 1$. Then, $ 4 \equal{}\minus{}2x\minus{}6\implies x \equal{}\minus{}5$ and $ \minus{}12 \equal{}\minus{}3 y^{2}\implies y \equal{}\pm 2$. This yields two points: $ (\minus{}5,\minus{}2, 47)$ and $ (\minus{}5, 2, 63)$.

Plugging back into our equation gives $ \boxed{\vec{v}\cdot (x,y,z) \equal{} 51}$ as one plane and $ \boxed{\vec{v}\cdot (x,y,z) \equal{} 19}$[/hide]


This person needs to be banned immediately: [url]http://www.artofproblemsolving.com/Forum/search.php?search_author=sherry023[/url]

[quote="Temperal"]A link to his profile would be more useful... or is he already banned?[/quote]

He's been banned. He was the guy posting the ads in the USAMTS forum (and others), if you missed it.


A right triangle is drawn with legs of lengths 3 on the x-axis and 4 on the y-axis lying along the positive coordinates. Find the coordinates of the midpoint of the hypotenuse.

[hide]The two endpoints of the hypotenuse are (0,4)and (3,0). The midpoint formula tells us that the hypotenuse is (1.5, 2)[/hide]

[hide="Solution"]The coordinates are $(3,0)$ and $(0,4)$

Use midpoint formula-

$\frac{3+0}{2}$

and

$\frac{4+0}{2}$

so the coordinates are $(1.5,2)$[/hide]

If you didn't know, the midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

In other words, it's the average of the x and y in the coordinate pair.

[quote="4everwise"]That is correct. I don't think i shouid have put this in the mathcounts forum, oops.[/quote]

We do all make this kind of mistakes. It's sort of hard to estimate the levels of all the problems correctly at first.


Suppose that $a$ and $b$ are natural numbers such that \[p=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}\] is a prime number. What is the maximum possible value of $p$?

If $ b$ is odd, then it is easy to check that $ p$ is not an integer. So, $ b$ is even.
Let $ b \equal{} 2c$ for some positive integer $ c$. Then,
          $ \left(\frac{2p}{c}\right)^{2}\equal{}\frac{a\minus{}c}{a\plus{}c}$
Let $ \frac{n}{m}\equal{}\frac{2p}{c}$ with $ gcd(m,n) \equal{} 1$ and let $ g \equal{} gcd(a\plus{}c,a\minus{}c)$ Then,
          $ a\plus{}c \equal{} gm^{2}, a\minus{}c \equal{} gn^{2}$
Or,
          $ 2a \equal{} g(m^{2}\plus{}n^{2}), 2c \equal{} g(m^{2}\minus{}n^{2})$
So,
          $ 4p \equal{}\frac{2cn}{m}\equal{}\frac{ng(m^{2}\minus{}n^{2})}{m}$
Since $ gcd(m,n) \equal{} 1 , gcd(m,m^{2}\minus{}n^{2}) \equal{} 1$ , we have $ m | g$ , say $ g \equal{} mk$ for some positive integer $ k$.
Thus, $ 4p \equal{} nk(m^{2}\minus{}n^{2})$
Note that $ p$ is a prime and so we can cosider the following two cases.

Case 1. $ m , n$ are both odd.

Since $ 4p \equal{} nk(m^{2}\minus{}n^{2})\equiv 0 (\mod 8)$, we obtain
$ p \equal{} 2$ and hence $ nk\times\frac{m^{2}\minus{}n^{2}}{8}\equal{} 1$ i.e., $ nk \equal{}\frac{m^{2}\minus{}n^{2}}{8}\equal{} 1$
Thus, $ n \equal{} k \equal{} 1, m \equal{} g \equal{} 3, a \equal{} 15,c \equal{} 12, b \equal{} 24$ i.e., $ (p,a,b) \equal{} (2,15,24)$

Case 2. $ m , n$ have different parity. i.e., $ (m,n) \equal{} (odd,even)$ or $ (even, odd)$

Since $ m^{2}\minus{}n^{2}$ is odd and $ 2c \equal{} g(m^{2}\minus{}n^{2})$ , we have $ 2 | g$ , say $ g \equal{} 2s$ for some positive integer $ s$.
So, $ 2p \equal{}\frac{cn}{m}\equal{}\frac{ns(m^{2}\minus{}n^{2})}{m}$
Since $ gcd(m,n) \equal{} 1 , gcd(m,m^{2}\minus{}n^{2}) \equal{} 1$ , we have $ m | s$ , say $ s \equal{} mt$ for some positive integer $ t$.
Thus, $ 2p \equal{} nt(m\plus{}n)(m\minus{}n)$ $ \cdots$ (*)
Note that the lefthand side of (*) has at most 2 nontrivial factors $ 2, p$.

① $ n \equal{} 1$
          $ 2p \equal{} t(m\plus{}1)(m\minus{}1)$
since $ m\minus{}1 < m\plus{}1$ , $ m\minus{}1 \equal{} 1$ or $ 2$.
If $ m\minus{}1 \equal{} 2$ , then $ p \equal{} 4t$ , a contradiction.
So, $ m\minus{}1 \equal{} 1$ and hence $ 2p \equal{} 3t$ , $ p \equal{} 3, t \equal{} 2$
Thus, $ g \equal{} 8, a \equal{} 20, c \equal{} 12, b \equal{} 24$. i.e., $ (p,a,b) \equal{} (3,20,24)$

② $ n > 1$
Since $ m\plus{}n > n > 1$, we have $ m\plus{}n \equal{} p$ .
So, $ n \equal{} 2 , t \equal{} m\minus{}n \equal{} 1$
Thus, $ m \equal{} 3, p \equal{} 5 , g \equal{} 6 , a \equal{} 39, c \equal{} 15, b \equal{} 30$. i.e., $ (p,a,b) \equal{} (5,39,30)$

Therefore, the maximum possible value of $ p$ is $ 5$.

I approached this in a slightly different manner.

If $b$ is odd then $2a-b$ is odd so henceforth $\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$ cannot be an integer.  We consider two cases:   $b=4k$, or $b=2m$ where $m$ is odd.

[b] Case 1:  [/b]  $b=4k$

 \begin{eqnarray*} p=k\sqrt{\frac{2a-4k}{2a+4k}}=k\sqrt{\frac{a-2k}{a+2k}}\\ \left(\frac{p}{k}\right)^2=\frac{a-2k}{a+2k} \\ p^2a+2p^2k=k^2a-2k^3 \\ a(k^2-p^2)=2k(k^2+p^2) \\ a=\frac{2k(k^2+p^2)}{k^2-p^2} \end{eqnarray*}

We know that $a$ must be an integer.  Consider the cases $p|k$ and $p\nmid k$.  The first case gives us $k=mp$.  Therefore \[a=\frac{2mp(m^2p^2+p^2)}{m^2p^2-p^2}=\frac{2mp(m^2+1)}{m^2-1}\]  We have $\gcd(m, m^2-1)=1$ therefore we must have
 \begin{align} \frac{2p(m^2+1)}{m^2-1}=2p\left(1+\frac{2}{m^2-1}\right)\in \mathbb{Z}\\ \implies \frac{4p}{m^2-1}\in \mathbb{Z} \end{align}

When $m$ is even we must have $(m^2-1)\mid p\implies p=m^2-1=(m-1)(m+1)$ since $p$ is a prime.  Therefore $m-1=1$ which gives us $(p,m)=(3,2)$.  When $m$ is odd we must have $\frac{m^2-1}{4}\mid p$ therefore $p=\frac{m^2-1}{2}$ or $p=\frac{m^2-1}{4}$.  The first case renders no solutions while the second renders the solution $p=2, m=3$.

$p=2$ gives the solution $(a,b,p)=(15, 24, 2)$ and $p=3$ gives us $(a,b,p)=(20,24,3)$.

Now when $p\nmid k$ we must have $a=\frac{2k(k^2+p^2)}{k^2-p^2}$ be an integer.  However $\gcd(k, k^2-p^2)=1$ therefore we must have $\frac{2(k^2+p^2)}{k^2-p^2}=2+\frac{4p^2}{k^2-p^2}$ be an integer.  However $\gcd(p^2, k^2-p^2)=1$ therefore we must have $\frac{4}{k^2-p^2}$ be an integer.  However, this gives $\begin{cases} k^2-p^2=1 \\ k^2-p^2=2 \\ k^2-p^2=4 \end{cases}$ of which gives no positive solution for $p$.

[b] Case 2:  [/b] $b=2m$ where $m$ is odd

\begin{eqnarray*}  p=\frac{m}{2}\sqrt{\frac{2a-2m}{2a+2m}} \\ p=\frac{m}{2}\sqrt{\frac{a-m}{a+m}} \\  \left(\frac{2p}{m}\right)^2=\frac{a-m}{a+m} \\ 4p^2a+4p^2m=am^2-m^3 \\ a(m^2-4p^2)=m(4p^2+m^2) \\ a=\frac{m(4p^2+m^2)}{m^2-4p^2}=\frac{m(m^2+n^2)}{m^2-n^2} \end{eqnarray*}

Where $n=2p$.  Since $\gcd(m,2)=1$ we have two cases to consider:  $p\mid m$ and $\gcd(m,n)=1$.

When $\gcd(m,n)=1$ however we have $\gcd(m, m^2-n^2)=1$ therefore we must have $\frac{m^2+n^2}{m^2-n^2}=1+\frac{2n^2}{m^2-n^2}$ be an integer.  $\gcd(n^2, m^2-n^2)=1$ therefore we must have $\begin{cases} m^2-n^2=1 \\ m^2-n^2=2 \end{cases}$ which gives no valid solutions.

Therefore $p\mid m$.  Let $m=pk$.  We arrive at \[a=\frac{pk(4p^2+p^2k^2)}{p^2(k^2-4)}=\frac{pk(4+k^2)}{k^2-4}\]  Keeping in mind the fact that $k$ must be odd we arrive at $\gcd(k(k^2+4), k^2-4)=1$.  Therefore we must have $\frac{p}{k^2-4}$ be an integer or henceforth $k^2-4\mid p$.  Since $p$ is prime we arrive at $p=k^2-4$.  This is only solved when $k=3$ which gives $p=5$.  This gives the solution pair $(a,b,p)=(39,30,5)$.

In conclusion the answer is $p=5$.


factorial
number theory solved
number theory

Let (x,y,z) \in Z^3

(1) Prove that there is only one solution to x!+y!=z! .

(2) Prove that the equation x!y!=z! has infinitely many solutions with x,y,z > 1.  (x,y,z) = (1,y,y) is considered to be a trivial solution for all y.

1) Clearly, we must have z > max{x,y}. Thus x,y  \leq z-1. It follows that x!+y!  \leq 2(z-1)! < z! for z > 2.
For z  \leq 2, it is easy to check that (1,12) is the only solution.

2) An infinite family of non-trivial solutions is (n, n! - 1, n!) where n  \geq 1.

Pierre.


Hi!

I don't know how to use math symbols in this bulletin.

Isn't there any one help me?

I want to present 1/x+y+3 like a image like any others

[url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]LaTeX[/url] is the way to go  :wink: . Simply enclose dollar signs around your math code and you produce the product.

[code]$\frac{1}{x}$[/code]

is

$ \frac{1}{x}$.

JBL, can you please move this to LaTeX and Asymptote? Thanks.


Consider a circle $C$ with diameter $AB=1$. A point $P_0$ is chosen on $C$, $P_0 \ne A$, and starting in $P_0$ a sequence of points $P_1, P_2, \dots, P_n, \dots$ is constructed on $C$, in the following way: $Q_n$ is the symmetrical point of $A$ with respect of $P_n$ and the straight line that joins $B$ and $Q_n$ cuts $C$ at $B$ and $P_{n+1}$ (not necessary different). Prove that it is possible to choose $P_0$ such that:

[b]i[/b]  $\angle {P_0AB} < 1$.

[b]ii[/b] In the sequence that starts with $P_0$ there are $2$ points, $P_k$ and $P_j$, such that $\triangle {AP_kP_j}$ is equilateral.

How does an angle less than 1? You meant 1 degree? Impossible. 1 in radian?


I am thinking about taking the Aime problem series. About how much did it help you? And about how much did it raise your score?

I'm planning to take it in this fall along with AMC-12 Problem Series.

As far as raising your score, I'm pretty sure it will because AIME Problem Series will probably cover the AIME-level problems topic by topic so you get use to the concept.

But remember that the main thing is how hard you work toward it. Getting in AIME Problem Series doesn't mean it will raise your score. You have to work on it.

Hope this helps!

Since students have not yet commented about how the class helped their scores, you might use the search button at the top of the forums to look for discussion about the class.  I suspect you'll find some discussion.

In general, the class surveys have been very positive in regards to all our problem series classes, though obviously you'd like to hear it for yourself from some students.

To your first question, I would have to agree with Silverfalcon. I was one of those students who took that class two years ago, that didnt do much. I thought taht being in class would help me. It didnt raise my score much.

The fact comes down to this: It is in your effort. I self-studied last yr, harder than when i took the class, and easily raised my score. Sure the class is great (and I dont mean to downplay all the things this forum has done to me) but if you dont do much, it will be a waste of 100 some dollars. So TRY YOUR BEST, whether you decide to take the class or not.

[quote="Calculus"]About how many problems both message board and homework problems will there be altogether? And also about what % of these are from past AIME tests?[/quote]

There are around 200 problems.  There are no exams in our problem series.  It all class time and message board -- and you get access to the Intermediate chat room.


calculus
integration
function
real analysis
real analysis theorems

Hi
I'm a beginner in integral calculus...i'm not able to integrate [{log(1+x)}/x]dx where log is taken to the base e...cud someone help me please??
Shyam[/img]

actually there is no cute formula for this

see http://en.wikipedia.org/wiki/Polylogarithm

using the fact $Li_{1}(z)=-ln(1-z)$ you can calculate your integral

for $s=1$ and $x =-z$

$Li_{s+1}(z) = \int_{0}^{z}\frac{Li_{s}(t)}{t}dt$

In other words, this antiderivative is not an elementary function. It's well studied, but can only be expressed as an integral or series in terms of functions you know. One nice form:
for $|x|\le 1,$ $\int_{0}^{x}\frac{\ln(t+1)}{t}\,dt=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n}}{n^{2}}$


calculus
integration
logarithms
derivative
limit
real analysis
real analysis theorems

I'm currently in Calc BC, and I know the basic derrivatives and integrals and such, but I would like to know how exactly they got all of these (such as e^x, lnx, cosx, sinx, etc)...
Does anyone have any links where I could find this? I've looked on MathWorld, but I can't find where to get the proofs...

Plus, I was trying to find how they found $\frac{d}{dx}\ln x$ using the basic derrivative rule $\lim_{h\to\infty}{\frac{f(x+h)-f(x)}{h}}$
I came to this:
$\lim_{h\to\infty}ln \left(1+\frac{h}{x}\right)^{\frac{1}{h}}$

NOTE: Are there too many equations?

\[ \frac{d}{dx}(\ln x)=\lim_{h\rightarrow0}\frac{\ln(x+h)-\ln(x)}{h}=\lim_{h\rightarrow0}\ln\left(\frac{x+h}{x}\right)^{1/h}. \] Set $n=\frac{1}{h}$, so that \[ \frac{d}{dx}(\ln x)=\lim_{n\rightarrow\infty}\ln\left(1+\frac{1}{nx}\right)^{n}. \] Hence, \[ \frac{d}{dx}(\ln x)=\ln\left(\lim_{n\rightarrow\infty}\left(1+\frac{1}{nx}\right)^{n}\right)=\ln\left(e^{1/x}\right)=\frac{1}{x}. \]

Alternatively, you could say $y=\ln x\implies x=e^{y}$.  Take the derivative of both sides with respect to $x$ and use the chain rule to get

\[ \frac{d}{dx}x=\frac{d}{dx}e^{y}\implies 1=e^{y}\frac{dy}{dx}\implies \frac{dy}{dx}=\frac{1}{e^{y}}=\frac{1}{x} \]


Under plan A, Corine can be paid a monthly salary of $ \$600$ plus a commission of $ 5\%$ on total sales. Under plan B she will receive $ \$800$ plus a commission of $ 6\%$ on total sales in excess of $ \$10,000$. What is the least amount of sales for which plan B would be equal to or better than plan A if sales are always more than $ \$10,000$?

Once you get 10,000 sales, commission rate under A is 1100 vs 800 in B.

This is a lag of 300 dollars.
But B catches up to A at the rate of 1 dollar per hundred she sells, so she needs 300*100=30,000 more sales to get more money under B.

But we had 10,000 earlier, so answer is 40,000.

Suppose the total sales are $ x$.  Then, we must have \[ 600\plus{}.05x\equal{}800\plus{}.06(x\minus{}10000) \Rightarrow x\equal{}\boxed{40000}\]


geometry
circumcircle
ratio
geometric transformation
rotation
geometry solved

Let $ABC$ be a triangle . Consider triangles $A_1B_1C_1$ satisfying : $\triangle A_1B_1C_1$ is same-direction - similar to $\triangle ABC$ . Put $\frac{AB}{A_1B_1}=k >0$ , $k$ does not equal 1 . Suppose $AA_1.BC =BB_1.AC=CC_1.AB$ . Find the locus of the circumcenter of the circumcircle $\triangle A_1B_1C_1$

This is standing around here for too long. Let me solve it in a slightly modified writeup:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle, and k a positive real. Let $A_{1}B_{1}C_{1}$ be a triangle directly similar to triangle ABC with ratio $B_{1}C_{1}: BC=C_{1}A_{1}: CA=A_{1}B_{1}: AB=k$ and satisfying $AA_{1}\cdot BC=BB_{1}\cdot CA=CC_{1}\cdot AB$. Let O be the circumcenter of triangle ABC, and let $I_{1}$ and $I_{2}$ be the two isodynamic points of triangle ABC (that is, the two points in the plane satisfying $AI_{1}: BI_{1}: CI_{1}=\frac{1}{BC}: \frac{1}{CA}: \frac{1}{AB}$ and $AI_{2}: BI_{2}: CI_{2}=\frac{1}{BC}: \frac{1}{CA}: \frac{1}{AB}$). [Note that if triangle ABC is equilateral, only one of these two isodynamic points $I_{1}$ and $I_{2}$ exists; in this case, ignore all assertions about the non-existing one.]
Assume that triangle ABC is non-equilateral or $k\neq 1$ (the "or" is a mathematical one, i. e. non-exclusive).
Show that the circumcenter of triangle $A_{1}B_{1}C_{1}$ lies on the circle with center $I_{1}$ and radius $k\cdot I_{1}O$ or on the circle with center $I_{2}$ and radius $k\cdot I_{2}O$.[/color]

[i]Proof.[/i] Since the triangles ABC and $A_{1}B_{1}C_{1}$ are directly similar, there exists either a translation or a spiral similitude that maps the points A, B, C to the points $A_{1}$, $B_{1}$, $C_{1}$. If there would exist a translation mapping the points A, B, C to the points $A_{1}$, $B_{1}$, $C_{1}$, then we would have $AA_{1}=BB_{1}=CC_{1}$, what, together with $AA_{1}\cdot BC=BB_{1}\cdot CA=CC_{1}\cdot AB$, would yield BC = CA = AB, so that triangle ABC would be equilateral, and also it would yield $B_{1}C_{1}=BC$ (since translations preserve distances), so that $k=B_{1}C_{1}: BC=1$; but this would contradict with the condition of Theorem 1 that triangle ABC is non-equilateral or $k\neq 1$. Thus, there must exist a spiral similitude that maps the points A, B, C to the points $A_{1}$, $B_{1}$, $C_{1}$. Denote by P the center of this spiral similitude; the factor of this similitude must be k, since $B_{1}C_{1}: BC=k$, and the rotation angle of this spiral similitude is $\measuredangle APA_{1}=\measuredangle BPB_{1}=\measuredangle CPC_{1}$. Now, since k is the factor of this similitude, we have $PB_{1}: PB=k$ and $PC_{1}: PC=k$, so that $PB_{1}: PB=PC_{1}: PC$. Together with $\measuredangle BPB_{1}=\measuredangle CPC_{1}$, this yields that triangles $BPB_{1}$ and $CPC_{1}$ are similar. Thus, $BB_{1}: CC_{1}=BP: CP$. But $BB_{1}\cdot CA=CC_{1}\cdot AB$ yields $BB_{1}: CC_{1}=\frac{1}{CA}: \frac{1}{AB}$. Hence, $BP: CP=\frac{1}{CA}: \frac{1}{AB}$. Similarly, $CP: AP=\frac{1}{AB}: \frac{1}{BC}$. Thus, $AP: BP: CP=\frac{1}{BC}: \frac{1}{CA}: \frac{1}{AB}$. Consequently, the point P is one of the two isodynamic points $I_{1}$ and $I_{2}$ of triangle ABC.

Therefore, if we succeed to show that the circumcenter of triangle $A_{1}B_{1}C_{1}$ lies on the circle with center P with radius $k\cdot PO$, then Theorem 1 will follow. Now, since a spiral similitude with center P and ratio k maps the points A, B, C to the points $A_{1}$, $B_{1}$, $C_{1}$, it must map the circumcenter O of triangle ABC to the circumcenter $O_{1}$ of triangle $A_{1}B_{1}C_{1}$; thus, $PO_{1}=k\cdot PO$. Thus, the point $O_{1}$, that is, the circumcenter of triangle $A_{1}B_{1}C_{1}$, lies on the circle with center P and radius $k\cdot PO$. This proves Theorem 1.

  Darij


Find the remainder when 7^2008 is divided by 19.

I have seen a lot of problems with divisibility by 19. Is there any trick or rule to work with when doing these questions?

hello, we have $ 7^3\equiv1 \mod 19$ and $ (7^3)^{669}\equal{}7^{2007}\equiv1\ mod 19$, from here we get $ 7^{2008}\equal{}7^{2007}\cdot7\equiv7 \mod 19$.
Sonnhard.

The "rule" you want to apply to these questions applies to every prime, not just $ 19$; it's known as [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url].  Also of interest when the modulus is not prime is [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's totient theorem[/url].


When I try to graph on my calculator...I get a ERROR 13 DIMENSION
can someone tell me what can be done in order for my calculator to graph?!? thank you =]

I don't have an 86, but I might have an idea.  Have you checked your window settings?

My window is what its supposed to be
-10
10
1
-10
10
1
So i dont think its the window
but thank you anyways

Well, according to the manual, you get an Error 13 Dimension when:

¨  You entered an argument with an inappropriate dimension.
¨  You entered a matrix or vector dimension < 1 or > 255 or a noninteger.
¨  You attempted to invert a matrix that is not a square matrix.

So it doesn't look like to me it has anything to do with graphing   :?

Try with a simple graph. Then you can see, if it is the window settings. Do a simple graph work?

You have picked plot1 without information for the plot. Up arrow to the plot1 and press F5 to unselect it.  Then it will graph for you.


Hi

I don't know if vectors belong in geometry, but here it is.

For vectors in 3-D real space, establish the identity

$[a \times (b-c)]^{2}+[b \times (c-a)]^{2}+[c \times (a-b)]^{2}=(b \times c)^{2}+(c \times a)^{2}+(a \times b)^{2}+(b \times c+c \times a+a \times b)^{2}$

I understood the first step into the problem.

Let $a \times b=d$, $a \times c=e$, $b \times c=f$.

So, $a \times (b-c)=a \times b-a \times c=a \times b+c \times a=d+e$, is the next step that he stated.

What I don't get now is why does that become true?  Could someone explain?

Is it just known that in 3-D space that for vectors a and c, that $a \times c=-c \times a$?

Thanks

[quote="mathgeniuse^ln(x)"]Is it just known that in 3-D space that for vectors a and c, that $a \times c=-c \times a$?[/quote]

This is one of the [i]very[/i] basic properties of the cross-product.

  Darij

Oh, I was confused with dot product, not the cross product.
Never mind

Thanks


[quote="236factorial"]A soccer ball is constructed of 32 polygons, 12 of them regular pentagons, and the rest of the regular hexagons. A seam is sewn whenever edges meet. How many seams are there?

No idea on how to solve this.[/quote]

[hide]I think I saw this on another thread, but here goes anyway.

There are 12 Pentagons and 20 hexagons.

The 12 Pentagons have a total of 12x5=60 edges, while the 20 hexagons have a total of 20x6=120 edges.

So there are 60+120 = 180 edges

Since 2 edges are always sewn together to make a seam, there are 180/2 = 90 seams.

Hope that helps.[/hide]

Ahh, this was the last state sprint I took before my real state. I didn't get it then, but here's how to do it now:

[hide]  12 pentagons means 20 hexas. Pentas have 60 edges, hexas have 120. 180 edges overall. Divide 180 edges by 2 to get the answer (two edges make a seam). Thus, there are 90 seams :) [/hide]

[hide="I think this is how you do it..."]
So...
12 pentagons=60 edges
20 hexagons=120 edges
120+60=180
Each seam will have [b]two[/b] edges so 180/2=90[/hide]

[quote="hockeyadi23"]Different approach here:
Euler's Theorem states that F+V-2=E. You know that there are 32 faces, and there is a vertex at every corner of the 12 pentagons (12*5=60). 32+60-2=90. This is a quick way to solve this problem.[/quote]
im pretty certain that that is Euclid's theorem

[quote="hockeyadi23"]Different approach here:
Euler's Theorem states that F+V-2=E. You know that there are 32 faces, and there is a vertex at every corner of the 12 pentagons (12*5=60). 32+60-2=90. This is a quick way to solve this problem.[/quote]

What happens if you use the hexagons instead of the pentagons?

[quote="cobra"][quote="hockeyadi23"]Different approach here:
Euler's Theorem states that F+V-2=E. You know that there are 32 faces, and there is a vertex at every corner of the 12 [b]pentagons[/b] (12*5=60). 32+60-2=90. This is a quick way to solve this problem.[/quote]

What happens if you use the hexagons instead of the pentagons?[/quote]

Yes, One has to be a little careful with this. One way to think is to count all the corners :
 $12 \times 5  + 20 \times 6 = 180$ and since every corner is counted three times the answer would be $60$

(Again one must be more careful because[b] if we did not have pentagons or hexagons, a corner may be 'counted' by more than 3 times! )[/b]

Counting edges ( divide by 2 as each is counted twice) may be easier.

BTW , If the original question was;

[quote] soccer ball is constructed of 32 polygons, $10$   of them regular pentagons, and the rest of the regular hexagons[/quote]

The answer would be ..[b] NO Answer[/b] !!! as you can't make a soccer ball , if  fact if you are making a soccer ball with pentagons and hexagons you have to have exactly 12 pentagons... Can you, now,  prove why?

[hide="another appraoch"]every pentagon in a soccer ball is NEVER next to another pentagon, but instead is surrounded by hexagons, so its just the total perimemter of the pentagons.
12*5=60
>>60<<[/hide]
[hide="something like proof"]if you do hexagons on a flat plane, you can use ONLY hexagons, and no pentagons, but this is a 3d surface, not 2d. If you take you honeycomb and try to dome it, the hexagons overlap each other. Instead of a hexagon in the middle, you need a pentagon to account for the extra degrees, as each vertice has more degree, hence room for the inward folding. and you also need a ratio of 3/8, otherwise it just wont work, I think. (this is where I need the help on)[/hide]

[quote] every pentagon in a[b] soccer ball is NEVER next to another pentagon,[/b] but instead is surrounded by hexagons, so its just the total perimemter of the pentagons.
12*5=60
>>60<<[/quote]

[b]Why is that? [/b] (No where in the problem it tells you that! " that pentagons are surrounde by hexagons" ..)

You can have 12 pentagons, and the number of corners can be  just  $20$  (when there are $0$ heaxagons)[img]http://www.merrimack.edu/~thull/combgeom/bucky/soccer.gif[/img]

 or   a higher number...if you use more hexagons.

Its [i]just so happens [/i]that the  number  of $60$ occurs ONLY when there are  exactly 20 hexagons...It is true that the normal  soccer ball is like that  but you have to be carful when you make these assumtions...

Hope this helps.[img]http://gwydir.demon.co.uk/jo/solid/football.jpg[/img]

Isn't it possible, if it's a truncated icosahedron?

[quote]if it's a truncated icosahedron?[/quote]
Yep!  [img]http://gwydir.demon.co.uk/jo/solid/buckyball.gif[/img]
Most likely shape.
(The point in previos post(s)  was, that one can not assume that that's the only shape possible.. specially when there is an easy alternative - like count the edges directly - available)

Also [quote]m pretty certain that that is Euclid's theorem[/quote]
E+2 =F+V  is known as Euler's Theorm .

Edited later:
BTW  there is a story   about   how a scientist was trying to explain  a new form of carbon  molecule   ("  ..a bit like a ball, but made of hexagons and pentagons etc" was told by his mathematician friend .." you mean like a  a soccer ball." ..:)   This shape is called a buckyball

I see that the topic has grown quite complicated, but the answer is 90, and I did think of what was discussed about 6 or so posts before this one.

I lost my soccer ball somewhere, so I couldn't just count edges  :lol: