articles
number theory
modular arithmetic

[quote="gauss1181"]What are the rules for numbers divisible by 11 and those divisible by 15?[/quote]
[hide="11"] Call the number $ {a_{1}a_{2}a_{3}a_{4}...}$ and say that it is divisible by 11.  Thus,
\[ a_{1}\minus{}a_{2}\plus{}a_{3}\minus{}a_{4}...\equiv\bmod 11\]
[/hide]

[hide="15"] The number must end with either 0 or 5 and it must have the sum of its digits divisible by 3.
[/hide]

as u can see quattomaster's one for 15 just used the fact that the number must be divisible by both 5 and 3, since $ 15|n\Longrightarrow (3\cdot 5)|p$ this can often be used for any number, especially those that do not have any particular rule. for example, if u have the a certain number n, and you want to check whether it is divisible by 18, just check if it is divisble by 2, 3, 6, 9.

Hope this helps!

That's not quite right. Take $ a \equal{} 209$, for instance. Rather than saying those two sums are equal, you must say that their difference is divisible by 11.

[quote="gauss1181"]What does mod11 mean, The Quatto Master 6000?[/quote]

$ \mod{11}$ means "modulo 11." The wiki has a nice article on [[modular arithmetic]] for more information.

Really basically, modulo returns the remainder when divided by the number specified.

[quote="gauss1181"]what are modular numbers?[/quote]
The clock (in military) time uses modular arithmetic. If it is 19:00 (7 p.m.) and 7 hours pass then it is 26:00, however there only 24 hours in a day so you must $ 26\mod24$ in order to obtain 02:00 (2 a.m.)

Hope that helps, it certainly helped me.

[quote="gauss1181"]I'm still confused-can anyone give a much simpler, concise way for divisibility by 11, just like you guys did for 15?[/quote]

The simplest way is to add alternating numbers and subtract:
[u][b]
Example:[/b][/u]
Check for divisibility for [color=red]1[/color][color=blue]9[/color][color=red]4[/color][color=blue]0[/color][color=red]2[/color][color=blue]4[/color][color=red]5[/color]

We simply find ([color=red]1+4+2+5[/color])-([color=blue]9+0+4[/color])=-1. Since this sum doesn't come out to be a multiple of 11, this number 1940245 is not a multiple of 11.

Don't forget, 0 is a multiple of 11...so...1111

$ (1\plus{}1)\minus{}(1\plus{}1)\equal{}0$, so 1111 is divisible by 11 :D

Take a number, say...8765...

It can be split into $ 8\cdot(10^{3})\plus{}7\cdot(10^{2})\plus{}6\cdot(10^{1})\plus{}5$

Since $ 10\equiv\minus{}1\mod{11}$, substituting in gives:

$ 8\cdot(\minus{}1^{3})\plus{}7\cdot(\minus{}1^{2})\plus{}6\cdot(\minus{}1)\plus{}5 (\mod{11})$...or
$ \minus{}8\plus{}7\minus{}6\plus{}5 (\mod 11)$. This means that for a number to be divisible by 11, the sum of its digits must be as well. In this case, 8765 is not divisible by 11.

For three, do the same thing with $ 10\equiv1\mod{3}$. Except in this case, you will see that the [i]sum[/i] of the digits must be divisible by 3. As for 5, just look at the last digit. If it's 0 or 5, the number is divisible by 5, so anything divisible by 5 and 3 is divisible by 15.


The circles  $ \Gamma_1$ and $ \Gamma_2$ intersect at the points $ A$ and $ B$. The tangent line to $ \Gamma_2$ through the point $ A$ meets $ \Gamma_1$ again at $ C$ and the tangent line to $ \Gamma_1$ through $ A$ meets $ \Gamma_2$ again at $ D$. A half-line through $ A$, interior to the angle $ \angle CAD$, meets $ \Gamma_1$ at $ M$, meets $ \Gamma_2$ at $ N$, and meets the circumcircle of $ \triangle ACD$ at $ P$. Prove that $ \vert AM\vert\equal{}\vert NP\vert$.

Inversion with center $ A$ and radius $ k$ takes the circles $ \Gamma_1$ and $ \Gamma_2$ into the lines $ \gamma_1$ and $\gamma_2,$ respectively parallel to the double lines $ AD$ and $ AC.$ Hence, $ A' \equiv \gamma_1 \cap \gamma_2,$ $ D' \equiv AD \cap \gamma_2,$ $C' \equiv AC \cap \gamma_1$ are the inverses of  $ B,D,C.$ Circle $ \odot (ACD)$ is taken into the line $C'D'$ $\Longrightarrow$ $ P' \equiv AP \cap C'D',$  $M' \equiv AM \cap \gamma_1,$ $N' \equiv AN \cap \gamma_2$ are the inverses of $ P,M,N.$ From [url=http://www.artofproblemsolving.com/viewtopic.php?t=301136]this topic[/url], we know that the circle centered at $ P'$ with radius $ P'A$ is orthogonal to the circle $ \odot(A'M'N')$

$\Longrightarrow  P'A^2 \equal{} P'M' \cdot P'N' \equal{} (AM' \minus{} AP')(AN' \minus{} AP')$

Using the power of inversion $ k^2,$ we obtain

$ \frac {k^4}{AP^2} \equal{} \left(\frac {k^2}{AM} \minus{} \frac {k^2}{AP} \right) \cdot \left( \frac {k^2}{AN} \minus{} \frac {k^2}{AP} \right)$

$ \Longrightarrow  \frac {1}{AP^2} \equal{} \left(\frac {1}{AM} \minus{} \frac {1}{AP} \right) \cdot \left( \frac {1}{AN} \minus{} \frac {1}{AP} \right) \Longrightarrow  AM \cdot AN \equal{} PM \cdot PN$

Which implies that $ P$ and $ A$ equidistant from the midpoint of $\overline{MN}$ $ \Longrightarrow$ $ AM \equal{} NP.$


[quote="silouan"]Let $ x,y,z > 0$ such that $ xyz \equal{} 1$ . Prove that

$ \frac {2 \minus{} x}{x^2 \plus{} 1} \plus{} \frac {2 \minus{} y}{y^2 \plus{} 1} \plus{} \frac {2 \minus{} z}{z^2 \plus{} 1}\geq\frac {3}{2}$[/quote]

It's true since $ \frac {2 \minus{} x}{x^2 \plus{} 1}\ge\frac1{2}\minus{}\frac3{4}lnx$  :)

[quote="dduclam"][quote="silouan"]Let $ x,y,z > 0$ such that $ xyz \equal{} 1$ . Prove that

$ \frac {2 \minus{} x}{x^2 \plus{} 1} \plus{} \frac {2 \minus{} y}{y^2 \plus{} 1} \plus{} \frac {2 \minus{} z}{z^2 \plus{} 1}\geq\frac {3}{2}$[/quote]

It's true since $ \frac {2 \minus{} x}{x^2 \plus{} 1}\ge\frac1{2} \minus{} \frac3{4}lnx$  :)[/quote]
Try $ x\equal{}0.1.$ :wink:

[quote="dduclam"]
It's true since $ \frac {2 \minus{} x}{x^2 \plus{} 1}\ge\frac1{2} \minus{} \frac3{4}lnx$  :)[/quote]

This is not true ... Take for example $ x\equal{}\frac{1}{10}$   :wink:

[quote="arqady"][quote="dduclam"][quote="silouan"]Let $ x,y,z > 0$ such that $ xyz \equal{} 1$ . Prove that

$ \frac {2 \minus{} x}{x^2 \plus{} 1} \plus{} \frac {2 \minus{} y}{y^2 \plus{} 1} \plus{} \frac {2 \minus{} z}{z^2 \plus{} 1}\geq\frac {3}{2}$[/quote]

It's true since $ \frac {2 \minus{} x}{x^2 \plus{} 1}\ge\frac1{2} \minus{} \frac3{4}lnx$  :)[/quote]
Try $ x \equal{} 0.1.$ :wink:[/quote]

My mistake  :( I'm sorry.

[quote="Vasc"]RCF-Theorem (corollary) works here.[/quote]

Vasc that was also my idea . But I am wondering if there is something elementary ... Note also that the inequality is equivalent to
$ \sum\frac{3a^2\minus{}2ab\minus{}b^2}{a^2\plus{}b^2}\geq 0$

[quote="silouan"]Let $ x,y,z > 0$ such that $ xyz \equal{} 1$ . Prove that

$ \frac {2 \minus{} x}{x^2 \plus{} 1} \plus{} \frac {2 \minus{} y}{y^2 \plus{} 1} \plus{} \frac {2 \minus{} z}{z^2 \plus{} 1}\geq\frac {3}{2}$[/quote]
We are not need RCF here. ;) Just AM-GM. :)
See my solution at: http://canhang2007.wordpress.com/2008/12/13/nice-inequality-of-silouan/ :)

Does anyone have another solution? I have
$ b^4c^4\minus{}2b^4c^3\minus{}2b^3c^4\minus{}2b^3c^3\plus{}5b^4c^2\plus{}5b^2c^4\minus{}2b^3c^2\minus{}2b^2c^3\plus{}6b^2c^2\minus{}2b^3c\minus{}2bc^3\minus{}2b^2c\minus{}2bc^2\plus{}c^2\plus{}b^2\minus{}2bc\minus{}2b\minus{}2c\plus{}5 \ge 0$. Can someone write this as a sum of squares?

Can-Hang, you have amazingly simple proof!
I have some ideas about this inequality too. However, i cannot finish  :noo:
Inequality is equivalent to proving that if $ xyz = 1$

$  2\sum_{cyc}{\frac {1}{x^2 + 1}} \geq \frac {3}{2} + \sum_{cyc}{\frac {x}{x^2 + 1}}$

After simple transformation and from Cauchy–Schwarz inequality , we have:

${ \frac {x}{x^2 + 1} = \frac {x}{x^2 + xyz} = \frac {1}{x + yz} \leq \frac {1}{4}(\frac {1}{x} + \frac {1}{yz}) = \frac {1}{4}(\frac {1}{x} + x)}$

Therefore, it is enough to prove stronger inequality:

$ 2\sum_{cyc}{\frac {1}{x^2 + 1}} \geq \frac {3}{2} + \frac {1}{4}\sum_{cyc}{(\frac {1}{x} + x)}$ or maybe we use the fact that for $ xyz = 1$ we have $ \sum_{cyc}{\frac {1}{x}} = \sum_{cyc}{xy}$ and maybe from there........
Anyone can finish it?
Thanks for attention. :)

Can_hang2007 , your solution is great and purely elementary  :lol:  Thanks for sharing

[quote="silouan"][quote="Vasc"]RCF-Theorem (corollary) works here.[/quote]

Vasc that was also my idea . But I am wondering if there is something elementary ... Note also that the inequality is equivalent to
$ \sum\frac {3a^2 \minus{} 2ab \minus{} b^2}{a^2 \plus{} b^2}\geq 0$[/quote]
$ \iff$
$ \sum_{cyc} (a\minus{}b)^2 \left ( \frac{9b\minus{}3a}{a^2\plus{}b^2} \plus{} \frac{\minus{}3b\minus{}c}{b^2\plus{}c^2} \plus{} \frac{a\plus{}3c}{c^2\plus{}a^2} \right ) \geqslant 0$

(If my calculations are right)

What about this? Is it nice enough for you, silouan?


number theory
greatest common divisor
modular arithmetic

Let $a,b,c$ and $n$ be positive integers such that $(a+bc)(b+ca)=5^{n}$. Prove that $n$ is even.

[hide] We know that the GCD of the two terms divide their difference, $(a+bc)-(b+ca) = (a-b)(1-c) \equiv 0 \pmod{5}$. Then $a \equiv b \pmod{5}$ or $c \equiv 1 \pmod{5}$. In each case, the expression $\pmod{5}$ is a square, which I believe is sufficient in determining that $n$ is even. [/hide]


How do I get Texnic Center out of overwrite mode?  i.e.  if I want to insert stuff back into a paragraph it replaces the other stuff.

[quote="joml88"]Ok, I have another unrelated question.  I don't think this affects my $ TeX $ document but how do you get rid of the bad boxes?  It says that I have "underfull hboxes".  this is just annoying me a little bit.[/quote]
That's a tricky question; many things can cause underfull (or overfull) hboxes. They don't matter unless your document doesn't look right.


calculus
integration
function
LaTeX
Putnam
real analysis
real analysis unsolved

For a continuous function f:[0,1]→R, let I = ⌠((x)^2)f(x) – x(f(x))^2 dx with limits 1 and zero respectively.

Find the maximum value of I and corresponding f(x).

PS: How do you use those math symbols in posts?[/hide]

The pretty math symbols are $ \text{\LaTeX}$. There's a subforum with various resources [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]here[/url].

This was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122456]Putnam 2006 B5[/url].

Beat me to the punch - I was about to link to Putnam 2006 B5.

The only thing I have to add is the FAQ link: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX.[/url]


trigonometry
real analysis
real analysis unsolved

The sequence $\{x_{n},\, n\ge1\}$ is defined as follows: $x_{1}=a$ and $x_{n+1}=3x_{n}-x_{n}^{3},\, n\ge1.$ Determine the set of real numbers $a$ for which the sequence is convergent.
(A. Kukush)

[hide="ignore"]we claim that $a\in [-2,2]$

Lemma 1:if $|x_{n}|\geq \sqrt{2}$ $\implies |x_{n}|\leq |x_{n+1}|$
Proof :
$|x_{n+1}|=|x_{n}\parallel 3-x_{n}^{2}|\geq |x_{n}\parallel 3-2|=|x_{n}|$

Lemma 2: if $|x_{n}|>2$ $\implies |x_{n+1}|\geq 2$
Proof : let $f(x)=3x-x^{3}$ then $f'(x)=3(1+x)(1-x)$ . So it is monotonic when $|x|>2$ . and $f(2)=-2$ , $f(-2)=2$ so it is obvious that $|f(x)|>2$ when $|x|>2$ and hence implies above lemma .

To prove the question , we observed that for $x=|a|\leq 2$ $\implies |f(x)|\leq 2$ where $f(x)=3x-x^{3}$ so the sequence $x_{n}$ converges in the interval .

when $x=|a|>2>\sqrt{2}$ , from lemma 1,2 we know that this interval is increasing , so $f(x)$ diverges.[/hide]

I can't understand your proof.

But If $a \in [-2,2]$,we let $a=2\sin x$
then $x_{n}=2\sin 3^{n}x$
will it always be convergent?

OK,I find the answer is:
$a=2\sin x$,For every $x=\frac{\pi}{4\cdot 3^{m}}+\frac{k\pi}{\cdot 3^{m}},k,m \in Z$ or $a=0$
And if Moubinool agree on my answer.
I will post my solution :P

i think that i only prove that $|a|>2$ then the series divergent . :blush:

if i followed your hint , i got

$a=2\sin x$ where $x=\frac{k\pi}{3^{m}}$ , $k,m\in\mathbb{Z}$

but what about $\sqrt 2$?
I think the answer may combine yours and mine :lol:


trigonometry
geometry
similar triangles
inequalities unsolved
inequalities

A, B, C are the angles of a triangle such that:

$ \sin A+\sin\frac{B}{2}+\sin\frac{C}{3}= \frac{3}{2}$.

Find value of A,B,C.

Sorry, my english is poor...

[quote="nkht-tk14"]A, B, C are the angles of a triangle such that:
$sinA+\sin\frac{B}{2}+\sin\frac{C}{3} = \frac{3}{2}$.
Find value of A,B,C.
[/quote]
For  example: $A=35.42842...^{\circ},$ $B=42^{\circ}$ and $C=102.57157...^{\circ}.$ ;)

he wants to prove that: A = pi/6, B = pi/3, and C = pi/2.

Who can prove that: $A = \frac{\pi}{6}, B = \frac{\pi}{3}$, and $C = \frac{\pi}{2}$ ???

Admins, this topic should be at Geometry. ;)

At the moment there is nothing to move. There are infinitely many non-similar triangles ABC satisfying $\sin A+\sin\frac{B}{2}+\sin\frac{C}{3}=\frac32$, so the problem makes no sense. (This is pretty much what Arqady wanted to say, too.)

  Darij


Euler
inequalities
logarithms
function
inequalities unsolved

Hi. This is interesting:

Let us denote
$\displaystyle \Phi=\{(x, y)\in\Re^+\times\Re^+\mid x^y\ge y^x\}$.
The program Derive6 generated the following graph:
http://www.petocerno.host.sk/FILES/flower.jpg
There we can see two main boundaries in this graph:
The first boundary is:  $y=x$ for $x>0$.
The second boundary I could not make out. It looks like hyperbola.
It is approximately:
$\displaystyle y=\frac{e}{x-(e-\sqrt e)}+(e-\sqrt e)$.
Where $e$ is an Euler number.

But the most interesting thing is that these
boundaries (in the graph of course) meets at point $(e,e)$.

And from this graph many interesting inequalities can be made out.

For instance prove that for every $x\in\Re^+$ the following
inequality holds:
$e^x\ge x^e$.

And for every $a\in\Re^+, a\neq e$ there are such numbers $b,c\in\Re^+$ that:
$a^b\ge b^a$,
$a^c < c^a$.

I would be grateful for any clear ideas or proofs you can post.  :)

Peter Cerno, SLOVAKIA

A proof for $e^x\geq x^e$ proceeds as follows:

$\frac {d}{dx}\left(\frac {x^e}{e^x}\right)=\frac {e^x(ex^{e-1})-e^x(x^e)}{(e^x)^2}$

So we need to consider $e^x(ex^{e-1})-e^x(x^e)$, which factors as $e^xx^{e-1}(e-x)$, which is strictly negative for $x>e$, hence we have the inequality for $x\geq e$.  So it remains to prove that $e^x\geq x^e$ for $x\geq e$.  Put $g(x)=x^e/e^x$.  Then $g(x)=0$, $g(e)=1$, and $g$ is differentiable on $(0,\infty)$.  If $g(x)\geq 1$, then we need to have $g(x)=1$ somewhere (by continuity).  If this is to happen, it must happen in $(0,e)$.  But then we can apply Rolle's theorem to $g$ over $[x,e]$ to conclude that $g'$ vanishes somewhere in $(x,e)$.  But $g'$ does not vanish at all, a contradiction.

The second part should be similar, so I'll leave it.

Hi, blahblahblah.

Thank you for your proof.
I am not sure, but I think that the simpler proof can be based on the
observation, that:
$e^x\ge x^e$
is equivalent with
$x\ge e\ln x$.
If we set $f(x) = x - e\ln x$, then
$f'(x) = 1 - \frac{e}{x}$ and
$f''(x) = \frac{e}{x^2} > 0$ for all $x\in \Re^+$
so for $x=e$ we have local (and also global) minimum.
Moreover $f(e) = 0$.

Note: This problem was inspired by another problem:
Decide which number is greater: $e^\pi$ or $\pi^e$.

But can you give me an exact formula for the second
boundary of the set $\Phi$ discussed above?
(with proof of course)  ;)

First consider the equation
$x^{x^{x^{x^{x^{x...\infty}}}}}=2$. This was published in VTM (that's a czech magazine, Peete, u probably know it). The solution is simple. With repacing into itself we get $x^2=2$, so $x=\sqrt 2$. In more general case
$x^{x^{x^{x^{x^{x...\infty}}}}}=n$ we get $x=\sqrt[n] n$. This is correct for n<e.
But if we take for instance n=4, we get  $x=\sqrt 2$, so the value of $x^{x^{x^{x^{x^{x...\infty}}}}}$ is equal to 2. In general case, it's equal to the lowest n for which $x=\sqrt[n] n$. For your boundaries holds $\sqrt[x]x=\sqrt[y]y$. If x>e the upper boundary is y=x and the lower  $z^{z^{z^{z^{z^{z...\infty}}}}}$, whwere
$z=\sqrt[x]x$. For 1<x<e you have to find x'>e, for which x' and x are boundaries. I think for x<1 there is only one boundary. I know this "proove" doesn't look very useful, but by evaluating the powers you will find it's true.

Thanks, Kondr.

So I suppose there is not an easy way how to
compute $z^{z^{z^{...}}}$ for $z=x^{1/x}$ ...  :)

For those who want to see a graph of
the function $f(x)=x^{1/x}$ just click
on the link below:
http://www.petocerno.host.sk/FILES/flower2.jpg

It is not so hard to show, that $f(x)$ reaches
it's global maximum for $x=e$ ...

Using Maple I found the second bound for x>e:
$-x \frac {LambertW(-\frac {ln(x)} x)} {ln(x)}$. I wanted to find more about Lambert function and found
http://mathworld.wolfram.com/LambertW-Function.html , where a lot of information about the aproximations of it can be found.

Another proof for $e^x\geq x^e$
Since $e^t=\sum_{k=0}^{\infty}\frac{t^k}{k!}$, we have $e^t\geq 1+t$. (Equality holds when t=0.)
Let $t=\frac{x-e}{e}$, then $e^\frac{x-e}{e}\geq 1+\frac{x-e}{e}$.
Then $e^\frac{x}{e}\cdot e^{-1}\geq \frac{x}{e}$, or $e^x\geq x^e$


Two astronauts A and B, each of mass M ,float freely in space at relative rest.

A throws a spanner of mass m to B ,who catches it and throws it back to A. Each throw results in the same relative velocity of spanner and thrower.

Show that spanner does not reach A unless 2m +M < M.$\sqrt5$

[quote]Show that spanner does not reach A unless 2m +M <Sqrt(5)[/quote]
 dimensions  do not match $\implies$  Does not make sense $\implies$  ... means there is a typo :)
 IOW the question should read: [quote] Show that spanner does not reach A unless  $2m + M < \sqrt 5 M$[/quote] )

I got something a bit different.
[hide]
When astronaut A throws the spanner, $M v_A = m v_s$, where $v_A$ is the velocity of astronaut A away from astronaut B, and $v_s$ is the velocity of the spanner towards astronaut B. Then when astronaut B catches the spanner, $m v_s = \left( M + m \right) v$. So then astronaut B throws the spanner at velocity $v_s$ away from himself, which is $v_s - v$ towards astronaut A in the original reference frame. This speed must exceed that of astronaut A.
$v_s - v = \frac{M}{M + m} v_s$

$v_A < \frac{M}{M + m} v_s$

$\frac{m}{M} < \frac{M}{M + m}$

$m^2 + m M < M^2$

$M^2 - m M - m^2 > 0$

$M > \frac{m \pm \sqrt{m^2 + 4 m^2}}{2}$

$M > \frac{m + \sqrt{m^2 + 4 m^2}}{2}$

$M > \frac{1 + \sqrt{5}}{2} m$
[/hide]

I’ll consider scalar variables, momentum and velocity, positive in the astronaut A’s direction of movement (that is also the  spanner’s final velocity direction, $\;\;v\;\;$) and negative in the opposite direction, except for the spanner's relative velocity,$\;\;v_{s}\;,\;$ that I considered  as absolute value.

 When astronaut A throws the spanner, momentum conservation of the two body system  tells us that
$\qquad Mv_{A}+m(v_{A}-v_{s})\,=\,0\;\;\;\Rightarrow\;\;v_{A}\,=\,{m \over{M+m}}\,v_{s}\;\;$, where$\;\;v_{A}\;\;$ is astronaut A’s velocity

The three body system , the two astronauts and the spanner, conserves its momentum, equal to zero all the time, so

$\qquad Mv_{B}+Mv_{A}+mv\,=\,0\,\,$ after the spanner was thrown by astronaut B,  i.e., $\;\;v_{B}\;\;$ and $\;v\;$ are the final velocities of astronaut B and of the spanner.

Putting $\;\;\;v_{B}\,=\,v+v_{s},\;\;\;$ together with $\;\;v_{A}\;\;$ from the beginig, in the expression above we get

$\qquad v\,=\,{\bigg{(}{M \over{M+m}}\bigg{)}}^{2}\cdot v_{s}\;\;\;$ and dividing it by $\;v_{A}\;,$  results

$\qquad{v \over v_{A}}\,=\,{M^{2}\over{(M+m)m}},\;\;$ that must be greater than 1. Then

$\qquad m^{2}+Mm-M^{2}\,<\,0\;\;$

Considering it as a quadratic trinomial in m, we find the positive root
as $\;\;m\,={M\over2}(\sqrt5\,-\,1)\;\;$ so, as the coefficient of the first degree of m is positive, $\;\;m\,<\,{M\over2}(\sqrt5\,-\,1)\;\;$ that leads to the desired answer.

For Whatever its worth,  Here is  a  "physicist way" :)
(The problem is not hard in any case, but still a physicist way is the lazy one)

Frame of reference  is  the space station (where everyone is at rest in the initial condition)  and consider the limiting case (Final velocity of spanner = Final velocity of A) (Direction of velocities will not be given as they would be obvious)

Let the spanner  be of mass $1$  and is thrown with a  velocity $1$  (We have freedom to choose our units :) ) and  man A  of mass $M$ will move with a velocity ($1/M$)
In the end, we have velocity of spanner=velocity of A= $1/M$
Relative velocity of A and (spanner initially thrown)  = $1+1/M$ =  relative velocity of B and final velocity of spanner($=1/M$)   this implies velocity of B=1.
Conservation of momentum gives:
 $M (1/M)+1(1/M) = M$ this gives the required result.
$M^{2}-M-1=0 \implies M = \frac{1+\sqrt 5}2$
(After changing M to (M/m)  and only taking the sign  which makes sense  to get the inequality etc...)


conics
ellipse
AMC
AIME
analytic geometry
special factorizations

I forgot how to graph them, but I just took a Mock AIME that had me draw an ellipse; I got it right. Is this the right method?

[hide="in here"]Let's take the ellipse defined by $ 16x^2\minus{}320x\plus{}25y^2\minus{}1500y\minus{}15900\equal{}0$. This is equivalent to $ (4x\minus{}40)^2\plus{}(5y\minus{}150)^2\equal{}200^2$. Now take the circle defined by $ (x\minus{}40)^2\plus{}(y\minus{}150)^2\equal{}200^2$ and "squash" the radii down; the vertical radii are now 1/4 of what they used to be and the horizontal radii are now 1/5 of what they used to be. This is the ellipse.[/hide]

If this is the right method, then do I have my vertical/horizontal radii backwards?

[quote="1=2"]I forgot how to graph them, but I just took a Mock AIME that had me draw an ellipse; I got it right. Is this the right method?

[hide="in here"]Let's take the ellipse defined by $ 16x^2 \minus{} 320x \plus{} 25y^2 \minus{} 1500y \minus{} 15900 \equal{} 0$. This is equivalent to $ (4x \minus{} 40)^2 \plus{} (5y \minus{} 150)^2 \equal{} 200^2$. Now take the circle defined by $ (x \minus{} 40)^2 \plus{} (y \minus{} 150)^2 \equal{} 200^2$ and "squash" the >> radii << down; the vertical radii are now 1/4 of what they used to be and the horizontal radii are now 1/5 of what they used to be. This is the ellipse.[/hide]

If this is the right method, then do I have my vertical/horizontal radii backwards?[/quote]

For an ellipse, they are not called "radii;" they have semiaxes instead.

I don't see your method working out.

Among different types of solutions:

Set 4x - 40 and 5y - 150, respectively, equal to 0 to find this ellipse's center:  (10,30).

Set (4x - 40)^2 = 200^2, with y = 30, and solve for the two values of x to get
the two coordinates of the ellipse at the extreme at the sides.

Set (5y - 150)^2 = 200^2, with x = 10, and solve for the two values of y to get
the two coordinates of the ellipse at the extreme at its top and bottom.

You can also graph this ellipse by rewriting it into standard form.
The standard equation of an ellipse is $ \frac{(x\minus{}h)^2}{a^2}\plus{}\frac{(y\minus{}k)^2}{b^2}\equal{}1$
[hide]
$ 16x^2\minus{}320x\plus{}25y^2\minus{}1500y\minus{}15900 \equal{}0 \\
16x^2\minus{}320x\plus{}25y^2\minus{}1500y \equal{}15900 $
Factoring out the 16 and 25 so the coefficients of  $ x^2$  and  $ y^2$  are 1:
$ 16(x^2\minus{}20x) \plus{} 25(y^2\minus{}60y) \equal{} 15900 $
Completing the square:
$ 16(x^2\minus{}20x\plus{}100)\plus{}25(y^2\minus{}60y\plus{}900) \equal{}15900 \plus{} (16)(100) \plus{} (25)(900) \\
16(x\minus{}10)^2 \plus{} 25(y\minus{}30)^2 \equal{}40,000 $
Dividing both sides by 40,000 so the equation is set equal to one:
$ \frac{16(x\minus{}10)^2}{40,000} \plus{} \frac{25(y\minus{}30)^2}{40,000} \equal{}\frac{40,000}{40,000} \\
\frac{(x\minus{}10)^2}{2500}\plus{}\frac{(y\minus{}30)^2}{1600} \equal{}1$
Now the equation is in standard form.
In standard form, the center is located at (h,k), so the center of this ellipse is (10,30).
$ a^2$ corresponds to the horizontal axis, and $ b^2$ corresponds to the vertical axis.
$ a\equal{}50$ and is the distance from the center to the left and right edges of the ellipse, and $ b\equal{}40$ and is the distance from the center to the top and bottom edges.
Using the values of $ a$ and $ b$, you can find four points to graph the ellipse. With $ a$, you can find one point directly to the left of the center and one point directly to the right. These points are $ (10\pm50,30)$, or (-40,30) and (60,30).
With $ b$, you can find one point directly above the center and one directly below.
These points are $ (10,30\pm40)$, or (10,70) and (10,-10).

Since $ a^2 > b^2$, the ellipse is stretched horizontally.

So the ellipse is centered at (10,30) and passes through the points (-40,30), (60,30), (10,70), and (10,-10).
[/hide]

[quote="maximos"]tornado was correct.  :lol:[/quote]

It's not no longer true.

Once tornado.adv4 got it correct, that user remained correct.

tornado.adv4 is correct.

yuh .. right! thanks for correcting my grammar. It [u]TOTALLY[/u] changed my life a lot . :w00tb:

Just kidding [b]Arrange your tan[/b].  :lol:

P.S.  [b]Arrange your tan[/b] I didn't mean you are wrong or anything on showing your solution. I'm so sorry if ever you got me wrong. Peace men. message me if you want, let's  both promote world peace. :omighty:

the way tornado did is the standard way. although the way you did it is pretty interesting.

Actually you can also think of this in terms of affine transformation... (say taking the circle $ (x \minus{} 40)^2 \plus{} (y \minus{} 150)^2 \equal{} 200^2$ and apply the transformation $ x\mapsto 4x',y\mapsto5y'$)  :)
(in fact this was what you basically did)


function
complex analysis
complex analysis unsolved

Prove that any entire function that tends to infinity at infinity is a polynomial. ($lim_{z \to  \infty} |f(z)|= + \infty$)

When you develop $g(z)=f\left(\frac 1z\right)$ as a Laurent series around $0$, you get something of the form $\frac{a_n}{z^n}+\ldots+\frac{a_1}z+a_0$ (and the result follows, of course).

This is because, on the one hand, the seris must be finite to the left, since $g$ has a pole at $0$, and, on the other hand, it cannot contain any positive values of $z$, because $g$ minus its principal part is a bounded entire function (so it must be constant).


How many lines in a three dimensional rectangular coordiante system pass through four distinct points of the form $(i,j,k)$ where $i,j,$ and $k$ are positive integers not exceeding four?

$\text{(A)} \ 60 \qquad \text{(B)} \ 64 \qquad \text{(C)} \ 72 \qquad \text{(D)} \ 76 \qquad \text{(E)} \ 100$

(D) 76
16 lines perpendicular to each plane parallel to one of the side of the cube for 3*16 plus the lines that pass through parallel edges separated by the cube's volume for 4*6 lines, plus the 4 lines passing through the cube's volume that connect opposite vertices for a total of 3*16 + 4*6 + 4 = 76.

We have four colinear points of the form $(i,j,k)$, where each of the three coordinates is between 1 and 4.

Case 1: Two of the three coordinates are the same for all four points. The third coordinate goes from 1 to 4. This can happen in 3*4*4 = 48 ways.

Case 2: One of the three coordinates is the same for all four points. The other two coordinates can either go in the same direction, or go in opposite directions. This can happen in 3*4*2 = 24 ways.

Case 3: No coordinate remains the same for all four points. Order the points in terms of the $i$ coordinate. Then the $j$ and $k$ coordinates will both be either increasing or decreasing. So this can happen in 2*2 = 4 ways.

So the total number of lines is 48 + 24 + 4 = 76.

TheAmazingOne's method can be extended to harder versions of this question.  For instance, what is the answer if we replace the second "four" in the question by "five?"  We could keep going like that, replacing it first with 5, then 6 and so on.  When we get to 8, however, it will get more complicated -- why?

When we get to 8 we'll have to start considering lines where the $i$ coordinate increases by two each time the $j$ coordinate increases by 1, etc, rather than only having to consider lines where coordinates increase at the same rate or remain constant. Actually, for 7 we would also have to consider some of those, such as the line through (1,1,1),(1,1,3),(1,1,5),(1,1,7).

Yes, you're right, I meant 7 instead of 8 -- $2n - 1$, in fact, where $n$ is the "first" 4 in the question.

While I'm here:  what is the answer to the original question if we replace [i]both[/i] instances of "four" in the question with $n$?  What if we replace the first with $n$ and the second with $n + 1$?

If we replace them both with $n$,

We have n colinear points of the form (i,j,k), where each of the three coordinates is between 1 and n.

Case 1: Two of the three coordinates are the same for all n points. The third coordinate goes from 1 to n. This can happen in 3*n*n = 3n² ways.

Case 2: One of the three coordinates is the same for all n points. The other two coordinates can either go in the same direction, or go in opposite directions. This can happen in 3*n*2 = 6n ways.

Case 3: No coordinate remains the same for all n points. Order the points in terms of the i coordinate. Then the j and k coordinates will both be either increasing or decreasing. So this can happen in 2*2 = 4 ways.

So the total number of lines is 3n² + 6n + 4.


Just wanted to know. (im on mathlinks.ro)

Im on Aops....but shouldnt this be on F&G or maybe The Round Table??

What were you searching for? "math" "message board" "art" "the area of a regular dodecagon = 3R 2 "? :lol:

[quote="Phelpedo"]good old google[/quote]

I didn't know you could see this message directly from the google site  :D  :D .

In my case, http://www.MathLinks.ro

I would say that this topic should be move to F&G.

If this post were in F&G or The Round Table, it would not measure how users of the MATHCOUNTS Forum accessed this site.  Therefore, I don't intend to move this poll.

At the time of this writing, the poll is 13 AoPS, 3 Mathlinks.  I think the poll confirms the common wisdom is that most US users and most pre-Olympiad users probably use the AoPS site.  (I am actually a little surprised there are so many mathlinks users lurking in this forum.)

[quote="Phelpedo"]i was searching for MC results and just found the site[/quote]
I did the exact same thing.

That's what I thought--just wanted to make sure.

[quote="biffanddoc"][quote="mathgeek2006"]I get on AoPS.  For the .ro at the end of mathlinks, what country is that?[/quote]Romania (where site creator was from)[/quote]*caugh* is from :P  :lol:

But as I hear from Mr. Rusczyk's blog, you are coming over here as soon as you can get a H1B visa? That way you will learn not to wander into the way of paintball-shooting thugs ;) !

Billy

[quote="solafidefarms"]But as I hear from Mr. Rusczyk's blog, you are coming over here as soon as you can get a H1B visa?[/quote]Probably ;) However that doesn't change the "is" into "was" :D

I use mathlinks because i think it looks way cooler.  :cool:

[quote="Valentin Vornicu"][quote="NoSoupForYou"]So the sites both use the same database?[/quote]Actually they are one and the same site (since last summer). Just that there are two ways to access it, and two skins which look differently but offer (almost) the same functionality.
The MathLinks skin is optimized for people interested in advanced maths and olympiads. The AoPS skin is optimized for the gifted math students or their parents/teachers searching for addition resources, not necessarily USAMO/MOSP-level.[/quote]

I suppose that MathLinks and Art of Problem Solving started as two separated sites (I suppose, I have been here just a few months and I haven't find out yet), nevertheless, the truth is that I have some idea of the amount of work that to build, update and administrate a site like this may be. I just wanted to make a stop in the way and say: [i]Guys, you really do a good job[/i]. I suppose the creators of the sites are the same current administrators, those are, Valentin, DPatrick, MCrawford, rrusczyk and vRusczyk (even when most of the people usually "blame" just Valentin for such a great job, actually I don't know exactly what part of the job is done by each of the Administrators) but anyway, I just wanted to say, speaking by myself, and I suppose that speaking by most of the users of the site: [b]We do appreciate all the work you do! Keep going[/b].

Best regards,

i totally agree with djimenez, thank you to the administrators  :D

same here...
and kingof21penguins, you said that there was nothing good about Aops forums but apparently you joined too! :)

magix for the last time i am not mark
who do u think i am

you would think that it would be obvious that i wouldnt mock my self and ill give u a nickel if ....


Here is a very challenging calc problem I was given... If anyone could help or post a solution to it that would be great!!

You have a metal cube that is 1 meter on each side and you drill a hole through
the center of the cube, starting at one corner and ending at the corner opposite. If the radius of
the drill is significantly less than 1 meter, what is the volume of the remaining material?
(“significantly less than” means “a very small fraction of.”)

Uh...1?  Depending on how big the drill is...Can mod move this to some other forum?


I proved the following ineq:(but I don't know the author?)
a)Let $a,b,c,d\in R$ such that :$\sum a^2=1$.Prove that
$\sum \frac{1}{1-ab}\le 8$
 And I have a generalization question:
b)Let $a_1,a_2,...,a_n\in R$ such that $\sum a_i^2=1$.Prove that :
$\sum \frac{1}{1-a_ia_j}\le \frac{n^2}{2}$
 ;)

Here it is

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=leq8&t=55410

Your link does not provide any solution pvthuan. The general solution is here:
[url]http://www.mathlinks.ro/viewtopic.php?t=299899&start=500[/url]
Post 509.


induction
number theory
greatest common divisor
Putnam
combinatorics proposed
combinatorics

Let $ S $ be a finite set of integers , each greater than 1 . Suppose that , for each integer n , there is some $ s \in S $ such that the greatest common divisor of $ s $ and $ n $ equals $ 1 $ or $ s $ . Show that there exist $ s,t \in S $ whose greatest common divisor is prime .

It's almost the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29019]this one[/url].

We can prove by induction on $n$ that given a set of $n$ elements, no two of which have a prime gcd, we can find a natural $N=N(n)$ which has gcd$>1$ with each one of the $n$ numbers, but, at the same time, is not divisible by any of them:

Suppose we've shown it for $n-1$ numbers $a_i,\ i\in\overline{1,n-1}$, and we add another number $a_n$ to the collection.

If $a_n$ has at least one prime factor which does not appear in the decomposition of the previous $n-1$ numbers, then we can choose either to multiply $N(n-1)$ with that prime factor to get $N(n)$ or to take $N(n)=N(n-1)$ unchanged, according to the situation. Either way we're done in this case. Assume, on the other hand, that all the prime factors of $a_n$ appear in the other $n-1$ numbers. If $N(n-1)$ is divisible by the number, then divide $N(n-1)$ by any prime factor of $a_n$ to get $N(n)$. If $(N(n-1),a_n)=1$, multiply $N(n-1)$ with any prime factor of $a_n$ to get $N(n)$. It's easy to see that these operations give the desired result due to our hypotheses.

This is some Putnam B6 question from a few years ago (1999?).

This is one the problem in last issue of the Iran's mathematic magazin  :lol:

[i]Solution:[/i]

Suppose that $n$ be the smallest number $\in \mathbb{Z}$ s.t. $gcd (n,s)>1$. It's obviously that for every $s\in S$ ,$n$ hasn't the repeated factor in prime divisor.It's obviously that there exist $s\in S$ s.t. $s\mid n$.In the other way if $p$ be the prime number s.t. $p\mid s$ and according to choosing $n$ , $\frac np$ is coprime some $t\in S$ and we know that $n$ is not coprime to $t$ ,so that $p\mid t$ but is not divisibe by the another prime divisor of $n$ then $gcd(s,t)=p$ and the proof is complete.  :D

what if S is a finite set of prime numbers?

I wanted to ask this as well, when I first saw the problem, but then I thought that maybe by "two elements whose gcd is prime" nttu didn't necessarily mean that the two elements must be distinct. So if the set has a prime member, then we're done, since it has a prime gcd with itself.


A company makes stadium seats. The company attaches the seats to a steel rod. The length of each rod needed depends on the # of seats attached to it. The distance from the center of 1 seat to the center of the next seat is 2.5 feet. The distance from the end of the steel rod to the center of the nearest seat is 2 feet.

A.	The company received an order to make 3 seats of stadium seats. The 1st set will contain 2 seats, 2nd set will contain 3 seats, and 3rd set will contain 4 seats. What is the length, in feet of each steel rod the company will use to make each set of stadium seats? Explain your steps.
B.	Write an expression that could be used to determine the length of a steel rod used to make a set of n stadium seats. Explain how you know your expression is right?

you can figure out that a seat is one foot long...

that gives you 2.5s-1.5, s is the number of seats...where did this question come from?

oops forgot the end of the rod, I thought it looked like this s-s-s-s-s

this doesn't look like mathcounts, what is it?

It doesn't matter how long the seat is
[hide]
6.5 feet for set 1
9 feet for set 2
11.5 feet for set 3
One seat must have two feet in each direction from the end of the steel rod making $4 feet$. Two must have another two and a half feet of spacing making $6.5 feet$ and so on.
Your expression is:
1.5 + 2.5s
[/hide]
Edit: This problem would be probably more suited for: http://www.artofproblemsolving.com/Forum/index.php?f=299


i need help proving that if quadrilateral ABCD is orthodiagonal and circumscribed around a circle, then (AB)(CD)=(BC)(AD).

[hide]
Prove

1. $AB+CD=BC+AD$

2. $(AB)^{2}+(CD)^{2}=(BC)^{2}+(AD)^{2}$

The result follows.
[/hide]


algebra
polynomial
number theory unsolved
number theory

If $ x$ is a real number such that $ x^2\minus{}x$ and $ x^n\minus{}x$ are integers for some $ n \ge 3$, prove that $ x$ is an integer.

If x is rational then it is trivially integer (just substitute x in lowest term in xⁿ-x). Now p(x)=k in Z, deg(p(x))=2 admit always solution of the form x=a+bt, with a,b in Z and t a real number with minimum polynomial quadratic(by definition).
If x is in R/Q then x in not in {0,1}, and since (xⁿ-x)/(x²-x) is in Q and is expressible by sum of power of x, we conclude that x is rational(t=0).


Hey guys,

I am in the process of making a mock AIME and I thought some of you guys who still have yet to take the AIME II or anyone might be interested. The test should be finished tonight or tomorrow so it should be posted sometime then.

Please post here if you are interested.

i may do it (depending on how much time i have).

[quote="Altheman"]I'll read over the problems for sure. I can check if you want (I'm not sure that I can find a three hour block but I still enjoy looking at these tests).[/quote]

That'll be great. I was hoping I could find someone to take a look at it before I give it out. I'll pm you about it when I get it finished.

I've already taken the AIME I, but I would love to take it.

I'll probably fail miserably, but I'll take it if I can find some time

I'll take it, especially because I am taking the AIME II.

[quote="mchoi815"][quote="davidyko"]I might take it just for the hell of it, even though I took the AIME I.[/quote]

same here

also, make it hard, really hard.  harder than aime i[/quote]

hehe i'll try  :P

I'll join the list of people who'll take it!

(Want to mix it up some, after reading 20 "I'll take it!" posts)

Okay everyone, the Mock AIME for 2008 is just undergoing some  edits by Altheman and not_trig so it should be out soon! (probably 1~2 hours)

Well, if it's not up yet, I guess I can still jump and an take it

It must be coming out right about now

Really good test.  Though I failed badly...
It seems a lot more difficult than the normal AIME, though...


get mining lvl up to 30 and mine coal and sell it for 100-200 gp each piece.

elvenchamp777

get mining and smithing up to high lvls, then mine for gold, mithril, coal etc, make stuff out of it, and sell it to other players or the merchants even to get some good cash

[quote="GoBraves"][quote="Jeffreywang"][quote="GoBraves"]I hate runescape, the tutorial takes too long( I never got past the tutorial becuse i died to a rat)[/quote]
hm...
i thought u culdn't die on tutorial island....
i tried to but once i got to 1 hp the rat kept doing 0's for a loooooong time till i gave up.
another way to make money is to sell nats (nature runes) they go to about 300gp each
or you can smith or craft and enchant ammy (amulets) and sell.
I'm a member and i sell ranaar weeds and unidentified herbs above that for lots of money  :D
grey wolf furs also help too, and pickpocketing gaurds is good too.
that's all i can think of...[/quote]

I left in on and left cuz i had to do something, and when i came back, well i was dead.[/quote]

Where did you appear at?

[quote="entropywins"]U sure you weren't just logged out? cuz u can't die on tutorial island[/quote]
that's wat i was thinking too
anyone here buying essense? (rune essense)

[quote="Jeffreywang"][quote="djimenez"][quote="Guard 125"]Do any of you have any advise for very good ways of making money (non-members) as i am new :([/quote]

Simple: find some market nest that is empty, and fill it. There are a bunch of them open, specially on the web, the hard part is to find them, because they are not that evident untill something start making money with them (Examples: Ebay, Google, E-Diet, E-Harmony, and other good bussiness in the web) unfortunatelly those examples are already filled, try to find one that is not.

Best,[/quote]
srry about the double post people...
but are you talking about runescape, cuz it doesn't sound like it....[/quote]

Sorry, I didn't know runscape is a game. Of course, I wasn't talking about it.

Best,

Well, I don't know the game, I have never played it, but even when I am not a fan of computer games, I prefer the old ones. I still have my 486 Laptop with Doom, Pacman, Prince of Persia, Digger and many more (all versions from 1994 or before :P  :D )

Best,

[quote="GoBraves"]Ok i dont know what really happened that day, it was like 2 years ago or something like that, but runescape is old, there's no point in the game, and has terrible graphics, so its not really my idea of fun. I find ragnarok, and gunbound, much better, but even those two dont compare to halo 2 or any nfl or nba games.[/quote]
Runescape is much better now, they made the graphics like 1000x better.  They added a ton of new stuff, and if you've only played it 2 years ago when they are always making it better you shouldn't diss it...you can diss the version from 2 years ago all you want, but never diss the current version.  Also, in response to post #1, i would suggest getting 40 crafting and good mining to go with it (at least 20, over 30 recommended) then mine silver in crafting guild-sell for 100-300gp each (the more at a time you have, the more per piece people sell.

Runescape is really repetative. The graphics still suck. There isn't really a point to the game, as gobraves said. Not only can it not even compare to games like Halo 2, Fable, and Morrowind, I don't even consider it that great of a free online game. I'd rather have a nice game of Texttwist any day.

I PLAYED TEXT TWIST IN 4th grade. Pretty entertaining until your eyes hurt. YEs RUNESCAPE is a piece of junk. Worse than all other XBOX games


<C=30 degrees.
From AC and BC , AD and BE are cut off respectively equal to AB ,
Prove that DE is perpendicular to IO as well as equal to IO, where I and O are respectively the incenter and circumcenter of triangle ABC .
[ i have a synthetic geometric solution, i am posting this as a challenge to solve the problem by complex-number or vector method or rigid transformations]

Can you post the synthetic geometry solution you got. :D

Contrary to the topic name, it holds for all triangles: http://www.mathlinks.ro/viewtopic.php?t=232923


Two little questions.

1) If $ F_m\equal{}2^{2^m}\plus{}1$ denotes the $ m$-th Fermat number. Prove that if $ F_m$ is not prime, then it's pseudo prime for the basis $ 2$ (i.e. $ 2^{F_m\minus{}1} \equiv 1 \mod F_m$).
2) If $ M_m\equal{}2^m \minus{}1$ with $ m$ not prime, is the $ m$-th Mersenne number. Prove that if $ d$ is a divisor of $ M_m$, dan is $ d$ pseudoprime for the basis $ 2$.

1)  It is easy to see that 2^(F(n)-1) = 1 (mod 2^(2^(2^n-k))+1) for all k. Let 2^n-k =n then 2^(2^(2^n-k))+1 = F(n) and 2^(F(n)-1) = 1 (mod F(n)) and if F(n) is composite then it is 2-pseudoprime.

2) I don't understand what u mean... is d prime or composite? and what we need to show to be 2-pseudoprime?

They don't mention what $ d$ should be, 'only a divisor'. I think they want us to show that $ 2^{d\minus{}1} \equiv 1 \mod d$ for $ d | M_m$

[quote="Halmos"]I think they want us to show that $ 2^{d \minus{} 1} \equiv 1 \mod d$ for $ d | M_m$[/quote]
I don't think so...For every integer k, M(n) = 0 (mod k) for some n. And we know that all composite numbers aren't 2-pseudoprimes...


inequalities
function
probability and stats

We can replace all of the inequalities with probabilities by [i]pointwise[/i] inequalities for the characteristic functions of the events.  I'll illustrate how it goes for a single case.  For example, say that we can show that
\[ f(\omega) : \equal{} 1_{\cup_i A} \leq \sum_i 1_{A_i} \minus{} \sum_{i < j} 1_{A_i\cap A_j} \plus{} \sum_{i < j < k} 1_{A_i\cap A_j\cap A_k} : \equal{} g(\omega)\]
If we can show that $ (f\minus{}g)(\omega)\leq 0$, then taking expectations on both sides gives you one case of Bonferroni's inequality.

So, say that $ 0\leq k\leq n$ of the $ A_i$'s satisfy $ 1_{A_i}(\omega) \equal{} 1$.  If $ 1_{A_i}(\omega) \equal{} 0$ for all $ i$, then both sides are $ 0$ when evaluated at $ \omega$, and there is nothing to prove.  If $ k\geq 1$, then
\[ f(\omega) \minus{} g(\omega) \equal{} \binom{k}{0} \minus{} \left(\binom{k}{1} \minus{} \binom{k}{2} \plus{} \binom{k}{3}\right) \equal{} \minus{} \binom{k \minus{} 1}{3} \leq 0\]
using a combinatorial identity and the convention that $ \binom{n}{k} \equal{} 0$ if $ 0\leq n < k$.

The general case follows in exactly the same way.


trigonometry
arithmetic sequence
geometric sequence

Given that $x,y,z$ forms an arithmetic sequence, $\sin{x}$, $\sin{y}$ and $\sin{z}$ forms a geometrical sequence, then find the common different $d$ of the arithmetic sequence.

[hide="Incomplete Solution"] We have the sequence $x,x+d,x+2d$. And the geometric sequence $\sin x, \sin (x+d), \sin (x+2d)=\sin x, \sin x\cos d+\cos x\sin d, \sin x\cos 2d+\cos x\sin 2d$.

If $\sin d=\sin 2d =0$ or $\cos x=0$, we have the sequence $\sin x, \sin x\cos d, \sin x\cos 2d$, thus, $\frac{\cos d}{1}=\frac{\cos 2d}{\cos d}\Longrightarrow \cos 2d=\cos^{2}d\Longrightarrow 2\cos^{2}d-1=\cos^{2}d\Longrightarrow \cos d=\pm1\Longrightarrow \boxed{d=k\pi}, k\in\mathbb{Z}$. This also satisfies $\sin d=\sin 2d =0$.

Also, if $\cos x=0\Longrightarrow x=k\pi+\frac{\pi}{2}\Longrightarrow\sin x=\pm1$, $d=k\pi$ still holds. [/hide]

General hint: When dealing with $3$-term arithmetic sequences, it's almost always the best to take $y-d,y,y+d$

[hide="Complete solution"]That gives

\begin{eqnarray*}\sin(y-d)\sin(y+d)=\sin^{2}y &\iff& \sin^{2}y\cos^{2}d-\cos^{2}y\sin^{2}d=\sin^{2}y\\ &\iff& \sin^{2}y(\cos^{2}d-1)-\cos^{2}y\sin^{2}d=0\\ &\iff&-\sin^{2}y\sin^{2}d-\cos^{2}y\sin^{2}d=0\\ &\iff& \sin^{2}d(\sin^{2}y+\cos^{2}y)=0\\ &\iff& \sin^{2}d=0\\ &\iff& \sin d=0\\ &\iff& d=n\pi,n\in\mathbb{Z}\end{eqnarray*}[/hide]


let $a,b,c,x,y,z > 0$ and $xyz=1$ [edit]( $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$) prove that is this inequlity true

$9(ax+by+cz) > = (a+b+c)(x+y+z)^2$
It might not be true but if it is that it is very helpfull.I'll now try to solve it :?

I can't understand what you mean here.
But I have substituted $a=b=1$, $x=1$, $y=1/c$, $z=c$. And the inequlity becomes wrong for sufficiently large $c$.


geometry
rectangle
combinatorics unsolved
combinatorics

Prove that for $ 5\times16$ table and we colored all of its cells by 3 colors : Blue Green Red.
Prove that there are rectangles with vertices of the same colors and with side parallel to the sides of the rectangle.

Is it too hard? Anyone can solve it? Help me! Thanks

It is an immediate result of Pigeonhole.

Count the number of pairs of grids with same color in each of the 16 columns. For example, bbrrg gives 2 such pairs and gggbr gives 3 such pairs. It is easy to see that in each of the 16 columns there are at least 2 such pairs, give a total of at least 32 pairs.

According to the row number (1 to 5) of grids in each pair we can assign each pair into one of the 10 sets. (row 12, row 13,..., row 45) So there will be a set contains at least 4 such pairs. Two of the 4 pairs must have same color, thus gives a desired rectangle.

How do we colour each cell in 5x15 table by R,G and B such that there are no rectangle with vertices of the same colors and with side parallel to the sides of the rectangle?  :)  Thanks.


calculus
derivative
algebra unsolved
algebra

Let $ w$ be a primitive $ 2007$th root of unity.

Calculate $ \sum_{k \equal{} 1}^{2006}\frac {1}{2 \minus{} w^k}$.

let $ f(x) \equal{} \prod_{i \equal{} 1}^{2006}(x \minus{} \omega ^i) \equal{} \frac {x^{2007} \minus{} 1}{x \minus{} 1}$
then $ g(x) \equal{} \frac {f'(x)}{f(x)} \equal{} \sum_{i \equal{} 1}^{2006}\frac {1}{x \minus{} \omega ^i}$
So you are looking for $ g(2)$ which is a straightforward derivative computation.

The derivative is even simpler if we use the definition
\[ f(x) \equal{} \prod_{k\equal{}0}^{2006} (x \minus{} w^k) \equal{} x^{2007} \minus{} 1.
\]
We can always subtract the $ k \equal{} 0$ term from the sum at the end.


function
limit
irrational number
real analysis
real analysis unsolved

A cute problem :):

Prove or disprove the existence of a differentiable function $f:\mathbb R\to\mathbb R$ which takes rationals to rationals and irrationals to irrationals, but which is not linear on any interval.

Since there is a rational number between any two irrational numbers and there is an irrational number between any two rational numbers, I think we can just let $f(x) = x$ if $x$ is rational and $f(x) = -x$ if $x$ is irrational.

Does this work?

[quote="blahblahblah"]I would like to see you differentiate that.  :D[/quote]

Whoops...I missed that part...  :oops:

[hide="Nevermind"]If $f$ is differentiable the it is continous. So if we have a sequence of converging rationals to a irrational number $\lim f(r_{n})=f(\lim r_{n})$. Does this mean that a rational equals an irational?[/hide]

Any solutions?

@spix: *I thinhk that is not true...It only means that u have a sequence of rational numbers converging to an irrational one :wink:
@grobber: That site requires special access? I can not open it, but i think the link is broken or something :(

[quote="grobber"]A cute problem :):

Prove or disprove the existence of a differentiable function $f: \mathbb R\to\mathbb R$ which takes rationals to rationals and irrationals to irrationals, but which is not linear on any interval.[/quote]

f(x) = x/(1+|x|)

[quote="trichoplax"][quote="grobber"]A cute problem :):

Prove or disprove the existence of a differentiable function $f: \mathbb R\to\mathbb R$ which takes rationals to rationals and irrationals to irrationals, but which is not linear on any interval.[/quote]

$f(x) = x/(1+|x|)$[/quote]How about a $C^\infty$-function?


Hi everyone,

Could someone please tell me if this is correct?

The number x is prime only if x is odd.

P: X is odd
Q: The number x is prime

P arrow Q
The negation form of this is
 :huh:

Therefore, it would be read as "The number x is prime and x is not odd"

Thank you very much

Yes I think it is correct.

The negation for $ p\Rightarrow q$ is $ p\wedge\sim q$.

You can verify it by constructing the tables for both.

 :)


An arch is in the shape of a parabola with a vertical axis.  The arch is 15 ft high at the center and the 40 ft wide at the base.  At which height above the base is the width 20 ft?

$ V(0,15)$ so my equation becomes:  $ (x\minus{}0)^2\equal{}\minus{}4p(y\minus{}15)\rightarrow x^2\equal{}\minus{}4p(y\minus{}15)$

$ F(0,15\minus{}p)$ & $ D: y\equal{}15\plus{}p$

Base = 40 ft.  So at the axis of symmetry, it's 20 ft to the left & right.  So that means the x value we're at for a total of 20 ft is actually x = 10.

I can substitute 10 into my equation, but I still can't grasp how to find p.

Just a hint please :)

[quote="Carbon57"]You can plot the point (20,0) to get p.[/quote]Keep going :p

[quote="JBL"]First off, your 10 is wrong: you've split 40 into two halves, and then you inexplicably split it again.  You've correctly figured out what value of $ x$ you want to be plugging in -- what you haven't figured out is what the associated $ y$ value is.  Once you figure that out, you'll be able to plug both in and solve for $ p$.

A similar topic was http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=544884669&t=146085[/quote]Hmm, I don't see why my x value is wrong.

The x-intercepts are $ I_x(\pm20,0)$, which gives my Arch a base of 40 ft.  Wouldn't I have to split it again?  I need to have a base of 20 ft, so my points for a 20 ft base are $ (\pm10,y)$.

Now let me go look at that thread you linked, thanks.

The equation for a parabola is:

$ y\minus{}k\equal{}\frac{1}{4p}(x\minus{}h)^2$

Right?

lol! I GOT IT :) Thanks! I have my x-intercepts but yet I wasn't using them :p

ok i found it, isnt the equation of a parabola:
$ y \equal{} a(x\minus{}p)^2\plus{}q$ ?

[quote="neurosurgeon1118"]sorry, but what is the equation of a parabola when its easy to find the vertex and the axis of symmetry?
isnt it something like $ y \equal{} a(q \minus{} p)^2$ ?
also is the answer?
[hide]
15
[/hide][/quote]No.  I'm using the equation of the Parabola: Foci version, lol ...

$ x^2 \equal{} 4py$ etc

Answer is 11.25 ft


Let
\[ w \equal{} \frac { \minus{} 1 \plus{} i\sqrt {3}}{2}
\]
where $ i \equal{} \sqrt { \minus{} 1}$.

Compute the product $ (1 \minus{} w \plus{} w^2)(1 \plus{} w \plus{} w^2)$

is there an easier way to do this than just pluggin' and chuggin'?

i got 0 by just working it out, but i was wondering if there is something that i should have noticed and just circled 0. :ninja: .

[quote="DaReaper"]Let
\[ w = \frac { - 1 + i\sqrt {3}}{2}
\]
where $ i = \sqrt { - 1}$.

Compute the product $ (1 - w + w^2)(1 + w + w^2)$

is there an easier way to do this than just pluggin' and chuggin'?[/quote]
[hide="Cis!"]Simplifying that out we now have $ w^{4} + w^{2} + 1$. And notice that $ \frac { - 1 + i\sqrt {3}}{2}$ is the same as ${ - \frac {1}{2} + \frac {\sqrt {3}}{2}}i$, which is the same as $ cis\frac {2\pi}{3}$. Then $ (cis\frac {2\pi}{3})^{4} + (cis\frac {2\pi}{3})^{2} + 1 = cis\frac {8\pi}{3} + cis\frac {4\pi}{3} + 1 = - \frac {1}{2} + \frac {\sqrt{3}}{2}i - \frac {1}{2} - \frac {\sqrt{3}}{2}i + 1 = -1 + 1=0$[/hide]

[hide="Hint"]
Recognize that $ w$ is a third root of unity, so $ w^3\equal{}1$. Try some manipulation with this equation.
[hide="Giveaway"]
$ (w\minus{}1)(w^2\plus{}w\plus{}1)\equal{}0$, and $ w\ne1$, so...
[/hide]
[/hide]

You should immediately recognize $ \omega\equal{}\frac{\minus{}1\plus{}i\sqrt{3}}{2}$ as a third root of unity (at least I do after having seen it so many times).

In general, [hide]
if $ \omega$ is an $ n^{\text{th}}$ root of unity, $ \sum_{k\equal{}0}^{n\minus{}1}\omega^{k}\equal{}0$. Why? Use the geometric series closed form to simplify that and it should be obvious. Or look at the post above mine ;)
[/hide]


Seriously, what were the results? Also who got what role?

It was a game in a *Missouri room*...you expect it to be legit?

*cough* I'd still like results, you know.

MASS ROLECLAIM EVERYBODY!

I was townie game 1 and doctor game 2 and would-be mafia game e (which never happened?)

Yeah, I remember that one. I only played that one game, in which I was mafia. I recall Calvin kept kicking me in the back, so I moved from the foot of his bed to the couch.

Around five minutes into the game.
Me: *taps Max on the shoulder for water bottle*
Max: Huh? Oh. (hands me water bottle)
Me: Max is jester. (tosses water bottle. calvin catches.)
Max: (WITHOUT the water bottle, I must note) No, we, uh, decided Bob was jester, remember?
Everyone else: Yeah.
(five seconds later)
Everyone else: We vote to lynch max.
Max: Haha, losers, I was jester. I win.

Bleh? Max was being pretty obvious about it - he claimed doctor (no one counterclaimed, surprisingly. I think calvin was doctor?) and was in general talked more than anyone else about who was who - a "clear" sign of mafia guilt. I knew he wasn't mafia, but also knew he wasn't doctor (I heard John come over to my side of the room to pick doctor, and Max was on the other side), so he had to be jester.

That was a fail game.

Well that was cuz I thought I was jester since [i]someone[/i] tapped me when he was like "I am tapping the jester" (ahem John...)

aaw now i'm too old for this sort of thing *cries*


One way is using the residue theorem from complex analysis...
$\sum_{k=1}^{+\infty}\frac{1}{k^{4}+k^{2}+1}=\frac{\pi\sqrt{3}}{6}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)-\frac{1}{2}$

EDIT: I see now that a finite sum was intended, but in view of the above result, I suspect no nice closed form for the finite form...


how many people do they take to nationals?

there is also 7 other territories/d.c. and department of defense. so 228 people + coaches go to nationals.


limit
integration
inequalities
function
real analysis
real analysis unsolved

Compute $ \underset{n\to \infty }{\mathop{\lim }}\,\int_{0}^{1}{\frac {dx}{\left( 1 \plus{} x^{n} \right)\sqrt [n]{1 \plus{} x^{n}}}}.$

I think we can apply the DCT here but don't know how to find a bound for the integrand.

$ g(x)\equal{}1$ on $ [0,1]$ makes a perfectly good dominating function.

Okay, then do you imply that is $ f\le g,$ since the integrand is positive.

But my huge question is how do you find that bound!! :(

Bernoulli's inequality implies:
$ \left(1 \plus{} x^n\right)^{1 \plus{} \frac{1}{n}} > 1 \plus{} x^n \left(1 \plus{} \frac{1}{n} \right) > 1$
Now take the reciprocals and the inequalities reverse, giving you the dominant function you were after.

You don't have to be precise about that sort of thing- just that 1+something is bigger than 1, an nth root of something bigger than 1 is bigger than 1, multiplying keeps us there, and then dividing by the whole thing flips it to being less than 1.
It should be obvious; it took me about a second to see the bound. Be bold and make crude estimates; you can always refine later if they aren't good enough.

[Edit- Bernoulli's inequality? Serious overkill.]

Ahh well, reading it as you phrased it, then I have $ 1\plus{}x^n\ge1$ and $ \sqrt[n]{1\plus{}x^n}\ge1$ and then that produces faster our result.

My problem was that we have $ 0\le x\le1$ and then I was complicating myself on finding a bound with that requirement, but now it's clear.


calculus
derivative
quadratics
algebra
polynomial
geometry
perpendicular bisector

You are walking north on a straight path. Ahead of you and on the east side of the path is
a beautiful billboard 54 feet across. The billboard is aligned perpendicularly to the path,
and the west edge of the billboard is 18 feet from the path. At what point on the path
would you see the billboard from the widest possible angle?

This is for computations and tutorials.

Let $x$ be the distance between you and the point on the path that is closest to the billboard. The angle in question is $\mathrm{arccot}(x/72)-\mathrm{arccot}(x/18)$ (why?) -- take derivative and equate it to zero.

I am still stuck on isolating x once I set it equal to 0.

Don't you get a quadratic equation for $x$? And of the simplest kind, without linear term.

In this case, the near end of the billboard is 18 feet from the path and the far end is 72 feet from the path. The correct answer should be $x=36$ feet.

For the generalized version, in which the near end is distance $a$ from the path and the far end distance $b,$ the $x$ that maximizes the viewing angle works out to be $\sqrt{ab},$ the geometric mean of $a$ and $b.$ (Among other things, that implies that from that point, if you look at a 45 degree angle, you'll always be looking into the interior of the billboard.)

It's a fairly staightforward calculus problem, assuming you know how to differentiate arccot (or arctan, if you do the labeling a little differently) and can handle the resulting polynomial algebra. But I'm interested in the non-calculus solution. Can you arrive at this conclusion using only concepts in Euclidean geometry?

[quote="Kent Merryfield"]But I'm interested in the non-calculus solution. Can you arrive at this conclusion using only concepts in Euclidean geometry?[/quote]
[hide]
Let $A,B$ the endpoints of the billboard.
Take $P$ be a point on the line $d$ (the road that is). Consider the circle of center $O_{P}$ which passes through $A,B,P.$
Since $\measuredangle A O_{P}B = 2 \measuredangle A P B$, it suffices to maximize $\measuredangle A O_{P}B.$
We know that $O_{P}$ moves on the perpendicular bisector of $AB$ as $P$ varies. The closer $O_{P}$ is to $AB$, the bigger the angle $\measuredangle A O_{P}B$ is. But keep in mind the fact that the circle $\left( O_{P}\right)$ must cut $d.$
A bit of intuition will help you see that the minimum corresponds exactly to the case when the circle is tangent to $d$, in a point which we'll call $X$. Now let $AB \cap D = \left\{ W \right\}.$ Then, by the power of a point theorem, $WA \cdot WB = WX^{2}$, so that
\[WX = \sqrt{WA \cdot WB}= \sqrt{ab}. \]
[/hide]

perfect_radio's post is exactly the argument I was looking for.


[quote="makar"]If $ a,b,c$ are three positive real numbers, prove that   $ \frac {a^{2} \plus{} 1}{b \plus{} c} \plus{} \frac {b^{2} \plus{} 1}{c \plus{} a} \plus{} \frac {c^{2} \plus{} 1}{a \plus{} b}\ge 3$[/quote]
$ \sum \frac{a^2\plus{}1}{b\plus{}c} \ge \frac{2a}{b\plus{}c} \ge 3$ (nebsit inequality)

[quote="makar"]If $ a,b,c$ are three positive real numbers, prove that   $ \frac {a^{2} \plus{} 1}{b \plus{} c} \plus{} \frac {b^{2} \plus{} 1}{c \plus{} a} \plus{} \frac {c^{2} \plus{} 1}{a \plus{} b}\ge 3$[/quote]
The inequality is obviously true because $ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \geq (a \plus{} b)(b \plus{} c)(a \plus{} c)$ holds for all positive real numbers $ a,b$ and $ c$.

That is easily solved by AM-GM and cs
 ∑(a^2+1)/(b+c)≥∑2a/(b+c)≥∑2a^2/(ab+ac)≥2(∑a)^2/2∑bc≥3

can also be done very easily using titu's lemma

[quote=Zarif]This is very easy problem.[/quote]
Applying AM-GM and Nesbitt`s inequality we have

Take $x = b + c, y = c + a, z = a + b$.\\
Thus, $a = \dfrac{y+z-x}{2}, b = \dfrac{z+x-y}{2}, z = \dfrac{x+y-z}{2}$.\\

Therefore, $\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b} = \dfrac{y+z-x}{2x}+ \dfrac{z+x-y}{2y}+ \dfrac{x+y-z}{2z}$\\
$\hspace{5.5cm} = \dfrac{1}{2}\left[ \left(\dfrac{x}{y} + \dfrac{y}{x}\right) + \left(\dfrac{y}{z} + \dfrac{z}{y}\right) +  \left(\dfrac{z}{x} + \dfrac{x}{z}\right)\right]   $\\
$\hspace{5.5cm}\geq \dfrac{1}{2}[2 + 2 +2] = 3.$

[quote=makar]If $ a,b,c$ are three positive real numbers, prove that   $ \frac {a^{2}\plus{}1}{b\plus{}c}\plus{}\frac {b^{2}\plus{}1}{c\plus{}a}\plus{}\frac {c^{2}\plus{}1}{a\plus{}b}\ge 3$[/quote]

Sorry for reviving this thread for such a mere solution of mine, but its a little different.

By Titu's Lemma,
$ \frac {a^{2}\plus{}1}{b\plus{}c}\plus{}\frac {b^{2}\plus{}1}{c\plus{}a}\plus{}\frac {c^{2}\plus{}1}{a\plus{}b}$
$\geq \frac{(a+b+c)^2}{2(a+b+c)} + \frac{3^2}{2(a+b+c)}$
Write $x=\sum_{a,b,c} a$
Therefore, we have to show that $\frac{x^2+9}{2x} \geq 3$
$\implies x^2+9 \geq 6x$
$\implies (x-3)^2+6x \geq 6x$
$\implies (x-3)^2 \geq 0$ and that's trivial as any square is $\geq 0$.

By Titu inequality and AM-GM inequality,

$\sum_{cyc}\frac{a^2+1}{b+c}\geq \frac{(a+b+c)^2}{2(a+b+c)}+\frac{3^2}{2(a+b+c)}\geq 2\sqrt{\frac{a+b+c}{2}\cdot \frac{9}{2(a+b+c)}}\geq 3$

:D

[quote=WolfusA]$ \frac {a^{2}+1}{b+c}+\frac {b^{2}+1}{c+a}+\frac {c^{2}+1}{a+b}\ge 2\left(\frac {a}{b+c}+\frac {b}{c+a}+\frac {c}{a+b}\right)\ge 3$
First by AM-GM, second by Nesbitt's
Equality iff $a=b=c=1$[/quote]

Brilliant, oh no, there was analogous proof before.

[quote=makar]If $ a,b,c$ are three positive real numbers, prove that   $ \frac {a^{2}\plus{}1}{b\plus{}c}\plus{}\frac {b^{2}\plus{}1}{c\plus{}a}\plus{}\frac {c^{2}\plus{}1}{a\plus{}b}\ge 3$[/quote]

A different solution:
\[\sum \frac{a^2+1}{b+c} \geq \sum \frac{a^2+1}{\sqrt{(b^2+1)(c^2+1)}} \geq 3 \sqrt[3]{\prod \frac{a^2+1}{\sqrt{(b^2+1)(c^2+1)}}}=3\]

[quote=janitor]

Brilliant, oh no, there was analogous proof before.[/quote]

I didn't read it though.

This can be solved by Jensen's inequality taking the function as f(x)=(x²+1)/((a+b+c)-x)

$LHS=\frac{a^2}{b+c}+\frac{1}{b+c}+\frac{b^2}{a+c}+\frac{1}{a+c}+\frac{c^2}{a+b}+\frac{1}{a+b}\geq\frac{(a+b+c+3)^2}{4(a+b+c)}$ by Titu lemma
Now, let $a+b+c=k$
We need to prove $\frac{(k+3)^2}{4k}\geq 3$
Or, $k^2+6k+9\geq 12k$
Or, $(k-3)^2\geq 0$ which is trivial [s]by trivial inequality[/s]

[quote=Leooooo][quote=xzlbq][quote=makar]If $ a,b,c$ are three positive real numbers, prove that   $ \frac {a^{2}\plus{}1}{b\plus{}c}\plus{}\frac {b^{2}\plus{}1}{c\plus{}a}\plus{}\frac {c^{2}\plus{}1}{a\plus{}b}\ge 3$[/quote]
$x,y,z>0$,prove that:

\[{\frac {1+x}{\sqrt {y+z}}}+{\frac {1+y}{\sqrt {z+x}}}+{\frac {z+1}{
\sqrt {x+y}}}\geq 3\,\sqrt {2}\][/quote]

[b]Proof[/b] [b]:[/b] Since $$\sum \frac{2x}{\sqrt{y+z}}\ge \sum \sqrt{y+z}$$ $$\Leftrightarrow \sum \frac{(x-y)^2}{\sqrt{(x+z)(y+z)}(\sqrt{x+z}+\sqrt{y+z})}\ge 0$$ thus $$\sum \frac{x+1}{\sqrt{y+z}}\ge $$ $$\ge \frac{1}{2}\sum \sqrt{y+z}+\sum \frac{1}{\sqrt{y+z}}\ge 3\sqrt{2}$$[/quote]

Nice!

2-var hold.

$x,y>0$,prove that

\[{\frac {1+x}{\sqrt {y}}}+{\frac {1+y}{\sqrt {x}}}\geq 4\]

[quote=xzlbq]$a,b,c,d>0$,prove

\[{\frac {1+a}{\sqrt {b+c+d}}}+{\frac {1+b}{\sqrt {a+c+d}}}+{\frac {1+c}
{\sqrt {d+a+b}}}+{\frac {1+d}{\sqrt {a+b+c}}}\geq \frac{8}{3}\,\sqrt {3}\][/quote]

By Holder,need to prove

${\frac { \left(  \left( 1+a \right) ^{2} \left( b+c+d+1 \right) +
 \left( 1+b \right) ^{2} \left( c+d+a+1 \right) + \left( 1+c \right) ^
{2} \left( d+a+b+1 \right) + \left( 1+d \right) ^{2} \left( a+b+c+1
 \right)  \right) ^{3}}{ \left( b+c+d \right)  \left( 1+a \right) ^{4}
 \left( b+c+d+1 \right) ^{3}+ \left( a+c+d \right)  \left( 1+b
 \right) ^{4} \left( c+d+a+1 \right) ^{3}+ \left( d+a+b \right)
 \left( 1+c \right) ^{4} \left( d+a+b+1 \right) ^{3}+ \left( a+b+c
 \right)  \left( 1+d \right) ^{4} \left( a+b+c+1 \right) ^{3}}}$
$\geq {
\frac {64}{3}}$

[quote=xzlbq]$x,y,z>0$,prove that

\[{\frac {yz+1}{\sqrt { \left( z+x \right)  \left( x+y \right) }}}+{
\frac {xz+1}{\sqrt { \left( x+y \right)  \left( y+z \right) }}}+{
\frac {xy+1}{\sqrt { \left( y+z \right)  \left( z+x \right) }}}\geq 2\,
\sqrt {2+2\,{\frac {xyz}{ \left( y+z \right)  \left( z+x \right)
 \left( x+y \right) }}}\][/quote]

By Holder,need to prove

\[{\frac { \left( {\it \sum} \left(  \left( yz+1 \right) ^{2} \left( 3\,
x+2\,y+2\,z \right)  \right)  \right) ^{3}}{{\it \sum} \left(  \left(
z+x \right)  \left( x+y \right)  \left( yz+1 \right) ^{4} \left( 3\,x+
2\,y+2\,z \right) ^{3} \right) }}\geq 8+8\,{\frac {xyz}{ \left( y+z
 \right)  \left( z+x \right)  \left( x+y \right) }}\]

Use Tittu Lemma, after separating 1 from each numerator and we can apply it on 3 terms for 2 time and at last apply AM-GM...

Let $ a,b,c$ are three positive real numbers. Prove that
$$ \frac {a^{2}+1}{a+b}+\frac {b^{2}+1}{b+c}+\frac {c^{2}+1}{c+a}\ge 3$$
[quote=makar]If $ a,b,c$ are three positive real numbers, prove that   $$ \frac {a^{2}\plus{}1}{b\plus{}c}\plus{}\frac {b^{2}\plus{}1}{c\plus{}a}\plus{}\frac {c^{2}\plus{}1}{a\plus{}b}\ge 3$$[/quote]
$$\sum \frac {a^2+1}{a+b}\ge  \frac {1}{2}\sum \frac {(a+1)^2}{a+b}\ge  \frac {1}{2} \frac {(\sum a+3)^2}{2 \sum a}\ge 3$$
$$\sum \frac {a^2+1}{b+c}\ge  \frac {1}{2}\sum \frac {(a+1)^2}{b+c}\ge  \frac {1}{2} \frac {(\sum a+3)^2}{2 \sum a}\ge 3$$
[url]https://math.stackexchange.com/questions/2452783/prove-that-a2-1b2-1c2-1-ge-a-bb-cc-a-for-a-b-c?noredirect=1[/url]
$$(a^2 + 1)(1+b^2) \ge (a + b)^2$$

Let $ a,b,c$ are  real numbers. [url=https://math.stackexchange.com/questions/2452783/prove-that-a2-1b2-1c2-1-ge-a-bb-cc-a-for-a-b-c?noredirect=]Prove that[/url]
$$(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$$
[url=https://math.stackexchange.com/questions/1955141/if-a-b-and-c-are-three-positive-real-numbers-prove-that-fraca21b?rq=1]h[/url] [url=https://artofproblemsolving.com/community/c6h383100p2124305]h[/url]
$$\implies$$
Let $a,b,c>0 $. Prove that$$\frac{a^2+1}{b+c} +\frac{(b^2+1)(c^2+1)}{(c+a)(a+b)}\geq 2$$
$$\frac{a^2+1}{a+b} +\frac{(b^2+1)(c^2+1)}{(b+c)(c+a)}\geq 2$$


function
geometry
geometric transformation
algebra
domain
complex numbers
AMC

A function f(z)  is defined on the complex numbers by f(z)=(a+bi)z where a  and b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin.

This isnt the whole entire question, but is the image of each point f(z) and when it says its equidistant, is it equidistant to z and (0+0i)?

If it is, why is f(z) an image?

That's what it means. Or another way,

$ |f(z) \minus{} z| \equal{} |f(z) \minus{} 0|$

Yes, the number 0 is viewed as the origin in the complex plane.  In a general setting, "zero" and "the origin" are usually interchangeable terms.  Sometimes, we denote the complex number $ a\plus{}bi$ as $ (a,b)$, emphasizing that the complex plane really is the Euclidean plane.

It's a standard mathematical convention to speak of functions as "mappings" which take objects to their "images" under the mapping.  You may be more familiar with this language in a geometric setting---if you apply a translation to a circle, then your original circle's [i]image[/i] under this translation (a certain type of function) is a congruent circle in a different location.  So in general, if $ f$ is a function, and $ z$ is an object in the domain of $ f$, we speak of $ f(z)$ as the image of $ z$ under the mapping $ f$.


prove that

$ (\forall a,b \in \mathbb{R}^{*\plus{}});(1\plus{}a^2)(1\plus{}b^2) \ge a(1\plus{}b^2)\plus{}b(1\plus{}a^2)$

See here : http://www.mathlinks.ro/Forum/viewtopic.php?t=318272   :wink:


For a permutation $ \sigma\in S_n$ with $ (1,2,\dots,n)\mapsto(i_1,i_2,\dots,i_n)$, define
\[ D(\sigma) \equal{} \sum_{k \equal{} 1}^n |i_k \minus{} k|
\]
Let
\[ Q(n,d) \equal{} \left|\left\{\sigma\in S_n : D(\sigma) \equal{} d\right\}\right|
\]
Show that when $ d \geq 2n$, $ Q(n,d)$ is an even number.

[hide="Hint"]A very interesting idea is to consider the determinant
$ \Delta(x)\equal{}|a_{ij}|$ where $ a_{ij}\equal{}x^{|i\minus{}j|}$.
Therefore $ \Delta(x)\equal{}\sum_{\sigma \in S_{n}}(\minus{}1)^{Inv(\sigma)}x^{D(\sigma)}$. So $ Q(n,d)$ has the same parity as the coefficient of $ x^d$ in $ \Delta(x)$.
Computing $ \Delta(x)$ we get $ (1\minus{}x^2)^{n\minus{}1}$ so for $ d\geq 2n$ the coefficient of $ x^d$ is 0 in $ \Delta(x)$ so $ Q(n,d)$ is even.[/hide]

An alternative method:

The involution $ \sigma \leftrightarrow \sigma^{\minus{}1}$ preserves $ D(\sigma)$, so the parity of $ Q(n,d)$ is the same as that of $ R(n,d)$, where $ R(n,d)$ only counts permutations which are their own inverse (the remaining permutations all pair off).  Every permutation counted in $ R(n,d)$ is the product of disjoint 2-cycles.  We next perform the following involution:

Given a permutation $ \sigma$, let $ (i,j)$ be the smallest pair lexicographically (assuming one exists) satisfying $ \max\{i,j\}<\min\{\sigma(i), \sigma(j)\}$.  We replace the cycles $ (i,\sigma(i))$ and $ (j,\sigma(j))$ by $ (i,\sigma(j))$ and $ (j, \sigma(i))$.  Again this preserves $ D(\sigma)$, so again we can pair off and reduce to computing the parity of $ S(n,d),$ where $ S$ is the collection of permutations that are their own inverses such that no such $ (i,j)$ exists.

Consider such a permutation, and let $ i<\sigma(i)$ be any point which is not fixed by it.  If $ i<j<\sigma(i)$, then $ \sigma(j)$ must equal $ j$ for the involution above not to work.  In other words, any permutation counted towards $ S$ can be written as $ (i_1, \sigma(i_1))(i_2,\sigma(i_2)),\dots(i_k,\sigma(i_k))$, where for each $ m$ we have $ \sigma(i_m)<i_{m\plus{}1}$.  But this immediately implies $ D(\sigma) \leq 2n\minus{}2$.

It follows that for $ d \geq 2n$, we have $ S(n,d)\equal{}0$, so $ R(n,d)$ and $ Q(n,d)$ must be even.


Has anybody here been in the high school division of the San Diego Math Circle while in 9th grade?

If so, please message me about how difficult the material is. :?

You should contact the user mcrawford for information about the San Diego Math Circle.


If $ a, b, c$ are numbers positive and distinte, prove or disprove that

$ \sum[\frac {a^3b \plus{} 2(b^2c^2 \plus{} c^2a^2 \minus{} a^2b^2) \plus{} b^3a}{(a^2 \minus{} b^2)^2}]\ge \frac {5}{2}$ .

 :?:  :?:

It's obivously trues because:
$ \sum\ \frac{a(b\plus{}c)}{b^2\plus{}c^2} \geq\ 2 > \frac{5}{4}$
Equality doesn't occur. Indeed, I don't like it  :wink:


invariant
Functional Analysis
combinatorial geometry
advanced fields
advanced fields unsolved

$C$ a convex, closed, bounded subset of a Hilbert space (not compact ..). Suppose that $f: C\to C$
satisfies $|f(x)-f(y)| \leq |x-y|$ for any $x,y \in C$. Prove that $f$ has at least
a fixed point.

I've recently learned about a generalization. First some terminology:

We say that a point $x$ of a bounded subset $K$ of a normed space is [i]diametral[/i] if $\sup_{y\in K}\|x-y\|=\rho(K)$, the diameter of $K$. Naturally, a non-diametral point will simply be a point which is not diametral.

Let $K$ be a closed, convex, bounded subset of a normed space. We say that $K$ has [i]normal structure[/i] if every closed, convex subset of $K$ which has at least two points has a non-diametral point.

Notice that every closed, convex, bounded subset $K$ containing at least two points of a uniformly convex Banach space $B$ has a non-diametral point, and hence every closed, convex, bounded subset of such a Banach space has normal structure. Indeed, let $x,y$ be two distinct points of $K$. The definition of uniform convexity easily shows that $\frac{x+y}2$ cannot be diametral.

The generalization I was talking about is this:

$(*)$ Let $K$ be a weakly compact convex subset of a Banach space $B$, having normal structure. Then every map $f: K\to K$ such that $\|f(x)-f(y)\|\le\|x-y\|,\ \forall x,y\in K$ (i.e. a so-called non-expansive map) has a fixed point.

Because a Hilbert space is reflexive, a closed, convex, bounded subset is weakly compact. On the other hand, because a Hilbert space is uniformly convex, by the observation above, a closed, convex, bounded subset has normal structure. Hence the hypotheses of $(*)$ are satisfied, so the problem posted by alekk is indeed a consequence of $(*)$.

Now for a proof of $(*)$:

Consider the collection $\mathcal C$ of all $f$-invariant closed, convex subsets of $K$. All such sets are weakly compact, so a descending chain has non-empty inetrsection. This means that $\mathcal C$, partially ordered by inclusion, satisfies the hypotheses of Zorn's Lemma, and must thus have minimal elements. Let $\widetilde K$ be such a minimal element. $\widetilde K$ either consists of one point only, in which case we are done, or has at least two points. The aim will now be to show that the latter cannot occur.

Since $K$ has normal structure, $\widetilde K$ has a non-diametral point $x_{0}$. From now on we forget about $K$ and only work inside $\widetilde K$.

There is a $t>0$ such that $\sup_{y\in\widetilde K}\|x_{0}-y\|\le\rho\left(\widetilde K\right)-t$. Now consider the (non-empty, because it contains $x_{0}$) set $L=\left\{x\in\widetilde K\ |\ sup_{y\in\widetilde K}\|x-y\|\le\rho\left(\widetilde K\right)-t\right\}$. $L$ is closed, because the map $x\mapsto\sup_{y\in\widetilde K}\|x-y\|$ is continuous. Also, notice that $L$ is convex: if $x_{i}\in L,\ i=1,2$ and $y\in\widetilde K$, and $\lambda\in[0,1]$, then $\|\lambda x_{1}+(1-\lambda)x_{2}-y\|\le\lambda\|x_{1}-y\|+(1-\lambda)\|x_{2}-y\|\le\rho\left(\widetilde K\right)-t$. Finally, $L$ is $f$-invariant: $\widetilde K$ is minimal $f$-invariant, so an arbitrary $y\in\widetilde K$ is in the closed convex hull of $f\left(\widetilde K\right)$. This means that for an arbitrary $x\in L$ we have $\|f(x)-y\|\le\sup_{z\in\widetilde K}\|f(x)-f(z)\|$, which, in turn, by the non-expansivity of $f$, is at most $\sup_{z\in\widetilde K}\|x-z\|\le\rho\left(\widetilde K\right)-t$. This proves that $f(x)\in L$, as desired. The definition of $L$ implies that it cannot be equal to $\widetilde K$ so we have found a closed, convex $f$-invariant set strictly smaller than $\widetilde K$. This contradicts the minimality of $\widetilde K$, and finishes the proof.


What are all ordered triples of real numbers (x,y,z) which satisfy

(x+y)(x+y+z) = 120
(y+z)(x+y+z) = 96
(x+z)(x+y+z) = 72?

I'll also post this problem in other forums.

Good thinking.  You're on the right track so far.  That's what I did, but I want to see if anybody else here can get this correct.  It's not that hard anyways.

Arne wrote:Summing these equations :

[(x + y) + (y + z) + (z + x)](x + y + z) = 288

=> x + y + z = 144 or x + y + z = -144

It is trivial from here on.

[hide]Don't we have

(2x+2y+2z)(x+y+z) = 288

so

(x+y+z)(x+y+z) = 144

(x+y+z) = 12 or -12[/hide] (as opposed to 144 or -144)?

[hide]If x+y+z = 12, then

x+y=10

y+z=8

x+z=6

So z=2, x=4, y=6.

If x+y+z = -12, then

x+y=-10

y+z=-8

x+z=-6

So z=-2, x=-4, y=-6

Hence the ordered triples are (4,6,2) or (-4, -6, -2).[/hide]


calculus
integration
trigonometry
limit
calculus computations

does
$ \int_0^\infty sin x dx$ exist?
if so, what is it's value and the method to solve it?

i think that it will become a limit of n tending to infinity of the integral $ \int_0^n sinx dx$and then we evaluate the integral and take $ \cos\infty$ = 0 as its a random number  :maybe:

No, the it cannot exist, as the integral doesn't converge.

The integral doesn't converge in the classical sense. On the other hand, we can try some summation method. The easiest one to use will be Abel-Poisson, which amounts  to finding the limit $ \lim_{r\to 0\plus{}}\int_0^\infty e^{\minus{}rx}\sin x\,dx$. Writing $ \sin$ as a difference of exponents, integrating and taking the limit of the resulting expression, you get $ 1$.

[quote="fedja"]The integral doesn't converge in the classical sense. On the other hand, we can try some summation method. The easiest one to use will be Abel-Poisson, which amounts  to finding the limit $ \lim_{r\to 0 \plus{} }\int_0^\infty e^{ \minus{} rx}\sin x\,dx$. Writing $ \sin$ as a difference of exponents, integrating and taking the limit of the resulting expression, you get $ 1$.[/quote]
yes that is what i did. and integral of cosx =0.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi

Suppose p>2 is a prime and $x^{2}+1$ is reducible in $\mathbb{F}_{p}$[x]. Prove that the multiplicative group of $\mathbb{F}_{p}$ contains an element of order 4. Deduce that p is congruent to 1 modulo 4.

What exactly do we have to do with this question?

Thnx!

Show that any zero of $x^{2}+1$ has order 4.

Now, what's the order of the multiplicative group $F_{p}^{*}$? What must be true if there is an element of order 4?

The converse is also true; whenever $p\equiv 1\mod 4$, there is an element of order 4 and $x^{2}+1$ factors. This follows from the fact that any finite multiplicative subgroup of a field is cyclic.

we can use Legendre's symbol, and euler's criterion (sorry i'm not that specific, i'm in a hurry this morning :oops: )


[quote="langtupanda"]Co bac nao biet cho nao co tai lieu toan tren dai hoc khong cho minh xin voi, Cam on nhieu!!!!![/quote]
 hinh` nhu ban o nuoc' ngoai`
  ban co' the vao` day
      http://vietnamnet.vn/giaoduc/tuyensinh/


How many positive integers less than 1000 have a 1 in their base 16 expansion?
Please include everything you used to sole this problem.

Thanks

[hide]
For integers between 1 and 15 (1 "digit" in base 16), there is only one: 1.

For integers between 16 and 255 (2 "digits" in base 16): there are
16 that are 1x and 15 that are y1, but 11 is counted twice, so there are only 30.

For integers between 256 and 1000 (3 "digits" in base 16): there are 256 that are 1xy.  For numbers in the 200-2FF range, there are 31 that contain a 1.  (It is the same as the combined number for 1 and 2-digit base 16 numbers)  For numbers in the 300-3FF range, there are also 31 numbers that contain a 1, but 1000 in base 10 is less than 3FF.  Since 400 (base 16) is 1024 (base 10), we counted one extra number in base 16: 3F1.  It is pretty evident that 3E1 is the last number containing a 1 that is less than 1000.

So:
1 + 30 + 256 + 31 + 31 - 1 = 348.

[/hide]

There is probably an inclusion/exclusion method for this problem as well going something like this:
total = {# that have 1 in the one's digit} + {# that have 1 in the 16's digit} + {# that have 1 in the 256's digit} - {# that have 1 in the one's digit and 16's digit} - {# that have 1 in the one's digit and 256's digit} - {# that have 1 in the 16's digit and 256's digit} + {# that have 1 in all three digits}[/hide]

Do you have access to the SDMC sub-forum?  We could just make a new topic to discuss solutions.


ΝΔΟ
$ a^2 b \plus{} b^2 c \plus{} c^2 a \geq ab \plus{} bc \plus{} ca$
οταν $ abc\equal{}1$ και $ a,b,c >0$

Παραλίγο γιαυτήν να πάρουμε με το Βαφείδη 10 ευρώ στις Βρυξέλλες  :P  Από ΑΜ-ΓΜ έχουμε
$ 2a^2b\plus{}b^2c\geq 3\sqrt[3]{a^4b^4c}\equal{}3\sqrt[3]{a^3b^3}\equal{}3ab$ Προσθέτοντας κυκλικά παίρνουμε η ζητούμενη  :)

Στα υπ' όψιν .Δουλεύεται εύκολα και με Lagrange όπου καταλήγεις να δείξεις ότι a=b=c=1 για να έχεις min

Λύνεται και με Muirhead αμέσως, πολλαπλασιάζοντας το δεξί μέλος με $ \sqrt[3]{abc}$


let $ a_1,a_2,a_3,...    and   b_1,b_2,b_3,...$ are 2 sequnce such that :
 \[ \forall n \geq 2 :  (a_n \minus{} a_(n\minus{}1))(a_(n\minus{}1)  \minus{}  a_(n\minus{}2))\plus{} (b_n \minus{} b_(n\minus{}1))(b_(n\minus{}1)  \minus{}  b_(n\minus{}2))\equal{}0\]
prove that there is a natural number  like k such that : \[ a_k \equal{} a_(k\plus{}2008)\]

stvs_f , You might have made some typing mistakes.
Firstly, does a(k+2008) on the last line mean a_(k+2008)
   If it does , then your statement is wrong because a_k can be made any number as b_k varies .

Also, do the two sequences contain natural numbers only?

the question is for USATST 2008.
all of themare index
and $ a_i and b_i \in Z$

You reposted it even knowing the source?

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1247503#p1247503

[i]Please stop[/i] reposting problems that can already be found on this forum; all the discussion for one problem should stay collected in one topic. If you want someone to take another look at it or have a question, revive the topic yourself.


People take pride in different accomplishments. Some prefer having the ability of Countdowning well, possibly because they like that they can show their talents off. Others are proud of being able to score well on written tests. Which holds more value with you?

Personally, I valued countdown a lot more last year. I looooved FTW (I still do) but my skills didn't show in written. This year, my last year, I feel like solid written scores is a better attribute, I think because it's my last shot at this and I want to make it as far as possible. And, It seems whichever I enjoy more I do better at (Chapter last year: 5th written, 1st countdown. This year: 1st written, 2nd countdown.)

I used to be a more countdown oriented person as well, but now I've lost most interest for speed (which I should probably change :maybe: ).

Hmm. Are you referring to importance to somebody or the importance according to the rules? Either way, for me, it's written.

I'm wayy better at countdown, but written definitely is more important.  Countdown trophies are bigger, though :)

Written for sure. Probably because I stink at countdown... I focus a lot more of my time to written as well.

I think written is more important... besides cd is like 2 or 3 questions.
And our written trophies are bigger than cd ones :P

Written is definitely more important, but countdown is definitely more fun  :P

Point is, you first have to do well to even MAKE it to countdown, but from then on it's just a popularity show.

And countdown trophies aren't much bigger than regular trophies in my chapter.  At states, we didn't even get countdown trophies -- we got these cheap copper medals with stickers on them that didn't even say "MathCounts"  :( ...

For me, definitely written. I used that to gauge how well I was doing in the math problems. The countdown tested how awake and lucky I was.

Although my chapter/state's countdown is unofficial, I somehow got 2nd at countdown chapter.
So I kinda value both of them the same even though I suck at CD...but fortunately I'm realizing that I'm actually getting faster at solving problems mentally.

I generally value written more since it gives you a score to aim for. But still, I get pretty upset when I do badly at CD.

I haven't done a  single MC test ALL YEAR!!! woohooo! anyway, both for me.

Actually, in all perspective, i think they are equally valued.. I'm kind of an average guy, like i get 3rd place in chapter and then i move up to countdown and get 1st.. This is a concept one has too learn!!! The only way to get into countdown is with written!!! I know this sixth grader that is totally beastly at countdown and possibly may be better than people at nationals, but his written test is not that good... that's why he doesn't get better. My mom is always telling me, "SHOW ME YOUR BALANCE!!!" and I know she's right.. So I improved. And now I'm pretty leveled out. It's really helpful to be balanced!  :D

Countdown, much more intense. Also, at state, if you pwn, everyones like OMG :O


Find all integers $x,y,z,t$ such that $x^y+y^z+z^t=x^{2005}$.
Can we replase 2005 by n?

[quote="Beat"]Find all integers $x,y,z,t$ such that $x^{y}+y^{z}+z^{t}=x^{2005}$.
Can we replase 2005 by n?[/quote]
One solution $y=1, t=1, z=x^{n}-x-1$, were $x$ is any integer.


How many different integers can be expressed as the sum of
two different numbers in the set $ \{1,2,3,4,5,6\}$?

The minimum is $ 3$, and the maximum is $ 11$, so our answer is $ 11 \minus{} 3 \plus{} 1 \equal{} \boxed9$.


Let f,g be two functions from the positive integers to the positive integers.
Prove there exist distinct p,q naturals such that

f(p)=<f(q)   and   g(p)=<g(q).

Bomb

If $f(p)=f(q)$ for $p \neq q$ then we are done. Hence we can assume $f$ is injective. But then $\exists n, \forall k \geq n, f(k) \geq f(1)$. Similarly, $\exists m, \forall k \geq m, g(k) \geq g(1)$. Now consider $max(n,m)$.

Just do case-bash....injective or not...

Bomb


I'm having issues compiling with \documentclass{letter}. I can't seem to use any \usepackage{} commands.

Any help?

Here's my log
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File: latex209.def 1998/05/13 v0.52 Standard LaTeX file

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("C:\Program Files\MiKTeX 2.7\tex\latex\base\letter.cls"
Document Class: letter 1999/04/29 v1.2z Standard LaTeX document class
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File: size10.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
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LaTeX Info: Redefining \cal on input line 391.
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)

! LaTeX Error: LaTeX2e command \usepackage in LaTeX 2.09 document.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
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l.3 \usepackage
               {amsfonts}
? x

Here is how much of TeX's memory you used:
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No pages of output.
[\code][/code]

The problem is caused by using an older version of LaTeX which doesn't support the packages:
[quote]          Entering LaTeX 2.09 COMPATIBILITY MODE
 *************************************************************
    !!WARNING!!    !!WARNING!!    !!WARNING!!    !!WARNING!!  [/quote]
Are you using \documentstyle{letter} which causes this instead of \documentclass{letter}?
If it's not that please post a minimal example of your tex document.

Hi,

thelog file contains the answers:

[quote="1/(ln x)"]
 New documents should use Standard LaTeX conventions and start
 with the \documentclass command.
...

! LaTeX Error: LaTeX2e command \usepackage in LaTeX 2.09 document.
[/quote]

Perhaps have a look also at the modern alternative letter class scrlttr2, I'm using it and I'm very satisfied with it.

Stefan


We have just started to encourage students at the Metroplex Math Circle to look into the Linguistics Olympiads.

http://metroplexmathcircle.wordpress.com/2008/11/28/naclo-2009/

Are these contests very popular outside the US, and do the problem solving skills of math competitions lend themselves to success in these contests?

Hi!  I'm a linguistics graduate student, and even though I've never participated in any linguistics competitions, I can certainly address your questions about linguistics and mathematical problem solving.

So, I looked over some problems from past linguistics competitions (NAMCLO 2007 and IOL6) and the questions seem to range from logic problems having to do with words to problems that are along the lines of "real linguistics" (in fact, one of my professors gave us a problem he'd written for the ILO as an exercise the first day of class).  I imagine that strong mathematical problem solving skills are as good a preparation as any for the "logic problems having to do with words" kinds of problems.  For the "real linguistics" types of problems, again, I think that mathematical problem solving would be as good a preparation as any for the problem-solving skills, but it would also be important to have a good familiarity with linguistics to know the framework within which linguistics problems work (eg, so you'll know how phonology problems in general work).  It's not completely false that the best linguists are the ones that have looked at the most languages.

Anyway, I'm not too much help on how linguistics competitions themselves work, but if you'd like a reading list for real linguistics, or have any questions about linguistics in general, feel free to PM me.

Thank you, Osud!  I will PM you but I'll post here as well in case anyone else has a similar interest.

I am very much a lay person but I have always been fascinated by linguists and by the fact that so many linguists make great contributions outside of their field (like Pinkner, Chomsky, Tolkien, Nietzche, etc.)  I'm sure it was very basic, but I also enjoyed the Teaching Company series by McWhorter (http://www.teach12.com/ttcx/CourseDescLong2.aspx?cid=2270)

While I'm too old to be anything but a linguistics fan, computational linguistics might be something that would combine my son's interests.  He certainly enjoys all of his AOPS classes and Latin, but the one thing he would do all day long is his Logic class (http://www.eimacs.com).

I guess that by the fact you are on these forums you have more than a passing interest in mathematics  :) and any resources or advice you could give would be greatly appreciated.

I would definitely agree with Osud, though I would emphasize even more the connection between linguistics and mathematics. Many of the "real linguistics" problems, as far as I've found, really only require mathematical reasoning; the "good familiarity with linguistics" is mainly what one gets after doing many of these problems, not by studying linguistics from outside sources. Just to note, I'm a high school student who participated in the NACLO. The website http://webscript.princeton.edu/~ahesterb/index.php is extremely good for linguistics problems.


invariant
combinatorics
IMO Shortlist
game
termination

$ 1994$ girls are seated at a round table. Initially one girl holds $ n$ tokens. Each turn a girl who is holding more than one token passes one token to each of her neighbours.

a.) Show that if $ n < 1994$, the game must terminate.
b.) Show that if $ n \equal{} 1994$ it cannot terminate.

A harder variant of this problem can be discussed [url=http://www.mathlinks.ro/viewtopic.php?p=7623#7623]here.[/url]

Denote the girl originally holding all coins $A$. If $n<1994$, we just remove the girl opposite $A$ and consider the girls sitting in a row with A in the center. We use a string to represent the coins in each girl's hand. I'll just write out the numbers to the right of $A$.

0 ...
20...
02...
101...
201...
011...
111...
211...
021...
202...

Clearly, the string cycles between 20202020.... and 1111111... (in each cycle, another 20 and 11 appear)

Thus, if $n<1994$, we get 111..... ($A$ is left with no coins if $n$ is even, and one coin if $n$ is odd), so the girl opposite $A$ never gets a coin.

If n=1994, the 0's and 2's switch back and forth, i.e.

20202020.....
02020202.....

so the game does not end.

@Above Sorry, your proof is probably very wrong - a person can have way more than $2$ cards.

[hide=A "symmetric" invariant for (a) ] Start from any girl, and number them $1, \cdots, 1994$, and denote $C[j]$ by the number of tokens of $j^{th}$ girl. The quantity $$ \sum_{j \; odd} C[j]  \mod 2$$ is always $0$ [/hide]

[hide= IMO Compendium "assymetric" invariant for (a) ] The quantity $$\sum_{0<j<1995} j*C[j] \mod 1994 $$ is always $0$, following the notation of above. [/hide]

[hide=Solution to part b] We use invariants here. Let the 1994 girls have labels $G_1, G_2, ...G_1994$. Assign a weight of $I$ to each coin $G_i$ holds. Let $G_1$ start with $1994$ tokens originally. Notice that in each turn, the sum of the weights is invariant mod $1994$. However, in order for the game to terminate, each girl must have 1 token. The sum of the weights in this case is $1+2+..+1994$, which is NOT congruent to 0 mod 1994. Hence, the game can never terminate. [/hide]


Who is more beast?

Greasers or Socs?

Ponyboy or Bob?

Johnny or Randy?

Who is more Missourian????

RIP Johnny  :(

P.S.  I first though Socs was pronounced "Socks".  HAVE OODLES OF FUN!!!

Greasers are better but being a Soc helps with MathCounts  :D

Ponyboy everytime

Johnny, he's a hero  :D

uhhh I dont know the last part...I'm not from Missouri  :P

Outsiders? Bleh. I had to read it for English. But, I'd say Greasers, dunno, Johnny.

omg!
i'm reading this in language arts right now

That book sucked. The movied sucked more.

[quote="abacadaea"][quote="Mewto55555"]It all sucked. That's 8th grade English![/quote]

7th grade for me[/quote]

Dude it is all the same.