Before Ashley started a three-hour drive, her car’s odometer reading was $ 29792$, a palindrome. At her destination, the odometer reading was another palindrome. If Ashley never exceeded the speed limit of $ 75$ miles per hour, which of the following was her greatest possible average speed?

$ \textbf{(A)}\  33\frac 13 \qquad \textbf{(B)}\  53\frac 13\qquad \textbf{(C)}\  60\frac 23\qquad \textbf{(D)}\  70\frac 13\qquad \textbf{(E)}\ 74\frac 13$

[quote="4everwise"]Before Ashley started a three-hour drive, her car’s odometer reading was $29792$, a palindrome. At her destination, the odometer reading was another palindrome. If Ashley never exceeded the speed limit of $75$ miles per hour, which of the following was her greatest possible average speed?

$\text{(A)} 33\frac 13 \qquad \text{(B)} 53\frac 13\qquad \text{(C)} 60\frac 23\qquad \text{(D)} 70\frac 13\qquad \text{(E)}74\frac 13$[/quote]

[hide](D)

the next palindromes are 29892, 29992, 30003, 30103

$r=t\cdot(d)$
$r=3d$
so r cannot exceed 75
if the odometer is 30003 then 30003-29792=211, 211/3 is $70 \frac{1}{3}$

the next palindrome is 30103, 30103-29792=311, 311/3 exceeds 75, so it is D[/hide]

[quote="ch1n353ch3s54a1l"][hide]$r=t\cdot(d)$
$r=3d$[/hide][/quote]

I think you mean [hide="this?"]$d=rt$
$\frac{d}{3}=r$ ;) [/hide]
You did the rest right though. :)


calculus
integration
logarithms
trigonometry
calculus computations

Write
$ \int{\frac{{(1+x^{2})}}{{(1-x^{2})\sqrt{1+x^{4}}}}dx = }\int{\frac{{x^{2}(\frac{1}{{x^{2}}}+1)}}{{x^{2}(\frac{1}{x}-x)\sqrt{\frac{1}{{x^{2}}}+x^{2}}}}dx}$
Now substitute $ {(\frac{1}{x}-x)}=t$
This gives $ {(\frac{1}{{x^{2}}}+1)dx}=-dt$ and the integral becomes $ \int{\frac{{(1+x^{2})}}{{(1-x^{2})\sqrt{1+x^{4}}}}dx = }-\int{\frac{{dt}}{{t\sqrt{t^{2}+2}}}}$
which can be comfortably solved using  standard methods of integration.

and the answer for the problem is $ -\log \left({\frac{t}{{\sqrt{t^{2}+2}+\sqrt 2 }}}\right)$ with $ t$ as defined above.

[quote="Gib Z"]Find       $ \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}$[/quote]

$ x^{2}=\tan t \implies dx=\frac{1+\tan^{2}t}{2\sqrt{\tan t}}dt$

$ J=\int\frac{(1+\tan t)(1+\tan^{2}t)dt}{2(1-\tan t)\sqrt{\tan t(1+\tan^{2}t)}}$
so

$ J=\int\frac{\sin t+\cos t dt}{2(\cos t-\sin t)\sqrt{\sin t \cos t}}$

So be $ \sin t-\cos t=y\implies \sin t+\ cos t dt =dy\implies K=\int \frac{dy}{-2y\sqrt{\frac{1-y^{2}}{2}}}$ cause $ \cos t\sin t=\frac{1-(\cos t-\sin t)^{2}}{2}$

$ y=\frac{1}{z}\implies dy=-\frac{1}{z^{2}}dz$ $ \implies L=\frac{\sqrt{2}}{2}\int \frac{dz}{\sqrt{z^{2}-1}}=\frac{\sqrt{2}}{2}ln(z+\sqrt{z^{2}-1})+C$ and backward stepping we have $ z=\frac{1}{(x^{2}-1)\sqrt{x^{4}+1}}$


Let  $x,y>0$. Find  the  maximum  value  of  $\frac{xy}{x^2+xy^3+y}$


Alice has chosen one of the number 1, 2, ..., 16. Bob can ask seven yes-no question to find it out. Alice is allowed to give at most one wrong answer. Help Bo t find out the number in all the cases.

Consider it as a binary number from $0000$ to $1111$, with digits $d_i$.
For questions $1,2,3, 4$ ask the values of all $d_i$.
For questions $5,6,7$ ask about the values of $d_i + d_{i+1} \mod 2$.


function
real analysis
real analysis solved

Let $f:R->R$ a continous function.Take $u_{n+1}=f(u_n)$.Show that if $|X_u|=1$ then $u_n$ converges.
{$X_u$ is the set of points of adherence(finite limits) of $u_n$}

You mean $X_u$ is a set of condensation points?

Consider a closed interval $I$ around the unique limit point $x_0$. Assume $u_{n_k}\to x_0$. Take $t_k$ s.t. $u_{n_k+t_k}$ is the first among $u_{n_k},u_{n_k+1},\ldots$ to go outside $I$. Assuming there are infinitely many members outside $I$, the sequence $u_{n_k+t_k}$ is an infinite sequence in the bounded segment $f(I)$, so it contains a convergent subsequence with a limit which is clearly different from $x_0$, contradiction. This shows that outside any interval $I$ there are only finitely many terms of the sequence, Q.E.D.

Edit after seeing Myth's post: there is indeed something unclear here. I think he's referring to limit points (points which have terms of the sequence different from themselves as close as possible).

Yeah,that is the exact name, a set of condensation points.


calculus
integration
logarithms
limit
special factorizations
calculus computations

I have this integral from trying to use Poisson's formula

$ u(x,y) \equal{} \frac{y}{\pi}\int_{\minus{}\infty}^{\minus{}\infty}\frac{s\minus{}s^{2}}{y^{2}\plus{}(x\minus{}s)^{2}}ds$

I feel like something can be done to the top of the fraction, something along the lines of completing the square--but I just don't see it. Is there some other method to integrate this?

Notice that this is divergent. For large $ s,$ the integrand looks like $ \frac{\minus{}s^2}{s^2}.$

If the numerator were merely $ s,$ then it would still be divergent, but there's at least hope for extracting meaning from it as a principal value integral. But the $ \minus{}s^2$ term in the numerator simply gives us $ \minus{}\infty.$

thanks Kent, I poured through a few math textbooks and google searched, but what is a "principal value integral"?

also, what if the integral read:

   $ u(x,y) \equal{} \frac{y}{\pi}\int_{\minus{}\infty}^{\minus{}\infty}\frac{s}{y^{2}\plus{}(x\minus{}s)^{2}}ds$?

would there be a trick for this one, since it looks convergent now.

whoops. the integral should have been bounded between $ [0,1]$

    $ u(x,y) \equal{} \frac{y}{\pi}\int_{0}^{1}\frac{s}{y^{2}\plus{}(x\minus{}s)^{2}}ds$

So without the chance of $ s$ getting too large, then would there be a method?

Yes. This is now a standard item from a second semester calculus course. You mentioned something about completing the square. But the square is already complete - you don't have to do anything to get there.

Try letting $ u\equal{}s\minus{}x$ so that $ s\equal{}u\plus{}x.$ Split the integral into a sum of two integrals by using the sum in the resulting numerator. One term will involve a logarithm and the other will involve an arctangent.

By "principal value integral" on $ (\minus{}\infty,\infty),$ I meant this:

$ \lim_{a\to\infty}\int_{\minus{}a}^af(x)\,dx.$

For instance, $ \int_{\minus{}\infty}^{\infty}\frac{x}{1\plus{}x^2}\,dx$ diverges, but

$ \text{(p.v)}\int_{\minus{}\infty}^{\infty}\frac{x}{1\plus{}x^2}\,dx\equal{}0.$


In 1960, there were 450,000 cases of measles reported in the U.S. In 1996, there were 500 cases reported. How many cases of measles would have been reported in 1987 if the number of cases reported from 1960 to 1996 decreased linearly?

[hide="Answer"]Setting the year 1960 at point $(0,4.5\cdot 10^5)$, we have a line going down to $(36,500)$. Given point $(27,x)$, $x=\frac{450,000-500}{4}=112,375$. However, we need to add the $500$ that we subtracted back in (essentially raising the entire graph back by $500$, to where it should be), and we get $\boxed{112,875}$.[/hide]

I didn't get what JesusFreak197 got. This is what I got:
[hide]

I drew a linear graph and replaced 1960 with 0, and 1996 with 36.
I started with the equation: $y = mx + b$

To find m: $ m = \frac{450000 - 500}{0 - 36} $  :arrow: $ m = \frac{449500}{-36}$

If $x=0$ then $y = 450000$.
So $b = 450000$ since $y = b$.
So you replace the equation, $y = mx + b$ with $y = \frac{449500}{-36} * x + 450000$
So, if $x = 27$, then

$y = \frac{449500}{-36} * 27 + 450000$
So:
$y = -337125 + 450000 $
$y = 112875$

[/hide]

You are correct, pascal_1623.

 [hide]it went down 449500 in 36 years, or 12486.11111 per year. In 27 years that decrease would be 337125. 450000-337125=112875. [/hide]

Whoops, my mistake. I took the difference between the endpoints, treated that as the first value and $0$ as the final value, then found what it would be at $27$, but forgot to add the $500$ back in. It's fixed now.

[hide]the number of cases went down 449,500 cases in 36 years, or $12486\frac{1}{9}$ cases per year. 36-27=9 years seperating them so we multiply $12386\frac{1}{9}*9=112875$ :idea: our answer :idea: [/hide]


The ratio of boys to girls in a class of 35 students is $ 3: 4$.  How many more girls than boys are in the class?

boys : 3x, girls : 4x

difference = x

7x=35, so x=5.

answer : 5


Μιας και αναφέρατε τα ολοκληρώματα ,έχω μια απορία. Πω μπορούμε να υπολογίσουμε το ολοκλήρωμα $ \int^{\plus{}\infty}_{a}f(x)dx$ όπου $ a$ πραγματικός και η $ f$ συνεχής σε αυτό το διάστημα?

Βρίσκεις το ολοκλήρωμα $ \int^{x}_{a}f(x)dx$  με τις γνωστές διαδικασίες [b]αν γίνεται [/b]και αυτό που θα προκύψει το βλέπεις σαν συνάρτηση του χ έστω g(x).

Στη συνέχεια παίρνεις το όριο της g για  χ-> +άπειρο

Αν είναι δύσκολο να υπολογίσουμε το ολοκλήρωμα με τις συμβατκές μεθόδους
(ολοκλήρωση κατά παράγοντες κ.λ.π. ) τότε είναι πιθανό να χρειαστεί να φράξουμε τη συνάρτηση f ανάμεσα σε δυο άλλες και να χρησιμοποιήσουμε κριτήριο παρεμβολής. :)

Τέτοιο θέμα έπεσε το 1993 ( Πρώτη Δέσμη )

επίσης τέτοιου είδους όρια είναι οι ασκήσεις 5, 9 του σχολικού βιβλίου στις σελίδες 350,

353 αντίστοιχα.

Γενικά για f συνεχή συνάρτηση στο $ [a,x)$ ισχύει:

$ \int_{a}^{k}f(x)dx = lim_{y \to k}\int_{a}^{y}f(x)dx$ για κάθε $ k\in${${ \mathbb{R},\pm \infty}$}

Το ίδιο ισχύει και για το κάτω άκρο. Η μέθοδος αυτή μας βοηθάει κυρίως να υπολογίζουμε ολοκληρώματα συναρτήσεων στο άπειρο, ή σε διαστήματα που περιλαμβάνουν και συγκεκριμένα σημεία μη συνέχειας.


Well Pie wanted another RPG, so here you go.
I haven't really decided how this will work, but I'm accepting sign ups :D :D :D
Also, you see you will be fighting the Giants.

UM... I DIDN'T WANT ANOTHER RPG.
No one's joining hahahahahaha

You two are the same people... (not rg64)

No pie is my brother.  NOT a multi.  And he is a control freak in LOBD, so yeah...

Joiners:
Randomguy64
meewhee009
andrewjjiang97
spaceguy524
gogators
piegraph
So that's 6 people.  I want 8 maybe?

Story:
You guys are all playing a game on FTW.  Then a black hole appears, and all of a sudden, you are in a unfamiliar town.  And evidently, the Giants are taking over.  If anyone wants a different theme, just say so.

The town is represented as the coordinate (0, 0) in a 25 by 25 square.  This is where you train stats and heal and stuff.

What stats are there?

What about our inventory, and the inventory of the shops?

Is it possible for us to take down some weak giant around here right now?

can we not have cordinates?  The only reason im in the game section right now is to get away from math.
OK, what weapons, armor and lvl?

piegraph cannot join, even if your are brothers.

Well ok mathwiz.  Your dead.

post the real battle later.

once...and half your health is 210 I think...

[hide=" "]you guys still have 1 revive[/hide]lol...ok orc a is dead, I'll let everyone else kill orc b and the undead giant.

Now that they can't pick on mathwiz...the decide to pick on RICHARDMIN.  Only 1 giant hits him though, and he suffers only 25 damage.

Why are all of these battles much easier than I intend them to be *sigh*

hey, half my CURRENT HEALTH. not my total.... thats what the skill said.

Is it?

too lazy to check, what ever.  Your at 195 now and there are only 5 giants left.

Lol I'm am totally a BM.

Is there any replacements for this game?????


Let a prime $ p$, $ 3|(p\minus{}2)$. Prove that there are not over $ (p\minus{}1)$ integers a satisfying p|a and there exists $ x,y\in$ {0;1;...;p-1} for $ a\equal{}y^2\minus{}x^3\minus{}1$

[quote="thaithuan_GC"]Let a prime $ p$, $ 3|(p \minus{} 2)$. Prove that there are not over $ (p \minus{} 1)$ integers a satisfying p|a and there exists $ x,y\in$ {0;1;...;p-1} for $ a \equal{} y^2 \minus{} x^3 \minus{} 1$[/quote]
This problem has been posted many times before . Use the fact that if p is a prime in the form $ 6k\plus{}5$ then $ \{x^3 |x\equal{}0,...,p\minus{}1\}$ is a complete reside modulo p . Therefore for each $ y\in \{0,..,p\minus{}1\}$ there are exactly a number $ x$ such that $ p|y^2\minus{}x^3\minus{}1$ .But when $ (x,y)\equal{}(3,2)$ and $ (1,0)$ we take the same number $ a$ . Thus there exist at most $ p\minus{}1$ numbers a satisfy condition .


geometry
AMC
AIME
ratio
similar triangles
Pythagorean Theorem

Two squares are inscribed in a right-angled triangle as shown. The first has area 441 and the second area 440. Find the sum of the two shorter sides of the triangle.

[img]http://www.kalva.demon.co.uk/aime/soln/diag8715.jpg[/img]

(AIME #15 1987)

[hide]
All I can figure out from similar triangles in the first figure, is that $ab=441$, where $a$ and $b$ are two of the legs of the triangles. (the ones that aren't 21).

Figure 2 is confusing. :?

This is hard! :(
[/hide]

By the areas, are you referring to the triangles or the squares?

There's only one triangle. Two different squares are inscribed in it. The two pictures show the distinct squares. :) So obviously the squares.

This problem is in deed pretty difficult. It takes a particular observation to unlock it, otherwise you can run into a lot of dead ends.

[hide="Hint 1"]There are LOTS of similar triangles between the two diagrams. Can you say something about the relative sizes of some of them?[/hide]

[hide="Hint 2"]Think in terms or areas, not lengths of sides.[/hide]

[color=red][i]Mod edit: PLEASE hide your solutions AND refrain from copying solutions that are not yours.  And even if you do post solutions that are not yours, cite them![/i][/color]
[i][color=red]
If you want to see what the original post had, just look at the solution to number 15 given on Kalva [url=http://www.kalva.demon.co.uk/aime/aime87.html]here[/url].[/color][/i]

That's the exact solution from Kalva. But you should hide it so that others can try it before looking.

Its AIME 1987 #15 for people's reference.

Sorry guys,i thought u guys needed the solution,so i posted it,i dont know how to hide stuff,i'm really sorry,i should have said where i got it from :blush:

By the way,could somone explain me th esolution that i copied from kalva,I didnt understand it.

[quote="Iversonfan2005"]I worked on this problem recently, the key thing to do is to form similar triangles, in fact, you will see that every single triangle is similar to each other, from there, you can use variables and set up proportions and solve for the variables, and soon you can find the ratio nof the areas between one of the triangles and the entire triangle, from there, draw an altitude from the 90 degree vertex which is not only the altitude of the entire large triangle but also of the smaller triangle, which I think was $\frac{1}{21}$ of the entire triangle, from there you can find out the sides of the triangle and thus the areas, etc.... I'm sorry that I don't have an exact solution but I don't have my work with me on this problem and I'm too lazy to do it again :P[/quote]

If you post something, please try to post some real work or a complete solution, don't just post vague ideas and don't say that you are to lazy to write the solution,
since that's frustrating for some people.

[hide="My solution"]Let the vertical leg of the big triangle be [tex]a[/tex], and the horizontal leg be [tex]b[/tex].

Let the side of the square in the first case be [tex]x[/tex], and in the second case be it [tex]y[/tex].

In the first case all the triangles are similar, so we have:

[tex]\frac{a-x}{x}=\frac{x}{b-x}=\frac{a}{b}\ \Longrightarrow\ a+b=\frac{ab}{x}\ \ (*)[/tex]

The second case: Let the sides of the topmost triangle be [tex](a-a_1),c_1,y[/tex], the sides of the rightmost triangle be [tex](b-b_1),c_2,y[/tex], so the sides of the third small triangle are [tex]a_1,b_1,y[/tex]. All these triangles are similar, so:

[tex]\frac{c_1}{y}=\frac{a}{b}=\frac{y}{c_2}=\frac{a_1}{b_1}\\ \Longrightarrow\ c_1=y\frac{a}{b},\ c_2=y\frac{b}{a},\ a_1=\frac{a}{b}b_1[/tex]

Expressing the area of the whole big triangle in terms of its parts we get:

[tex]\frac{ab}{2}=\frac{c_1y}{2}+\frac{c_2y}{2}+y^2+\frac{a_1b_1}{2}\\ \Longrightarrow\ ab=y^2\left (\frac{a}{b}+\frac{b}{a}+2\right ) + \frac{a}{b}b_1^2[/tex]

Using [tex](*)[/tex] we get: [tex]b_1^2=\frac{x^2-y^2}{x^2}b^2,\ a_1^2=\frac{x^2-y^2}{x^2}a^2[/tex]

By pythagorean theorem also [tex]a_1^2+b_1^2=y^2[/tex], so:

[tex]a^2+b^2=\frac{x^2y^2}{x^2-y^2}[/tex]

Let [tex]a+b=t[/tex], then by [tex](*)[/tex] we have [tex]a^2+b^2=t^2-2tx[/tex], also [tex]x^2-y^2=1[/tex], so finally:

[tex]t^2-2tx-x^2y^2=0\ \Rightarrow\ t=x+x^2=441+\sqrt{441}=\boxed{462}[/tex][/hide]


geometry
geometric transformation
reflection

how to take a photograph of a camera by itself?
Except using a mirror. :?:
could you offer me some new idears?

Well, if only not use a mirror in the strict meaning of the word, then use any reflectant surface. From polished steel to still liquids.

One way or another, to take the picture you have to turn the light rays around. You either reflect them, or you gradually refract them back. Since reflection is covered already... refraction is left.
Maybe... you can find such an optical cable that can transmit the rays without too much dispersion.

Another way:

Take the picture underwater or in some medium where light travels slower.  Then move the camera fast (i.e. faster than the light) to a point, where the light hits it, then move it back to the spot it started from.

Were you looking for an applicable solution?  Because other than mirrors or something reflective I don't think there is any "good" way.

[quote="paladin8"]Use a different camera? :P That always works.[/quote]

It says by ITSELF

there is and an other topic with the same subject but the solution are the same... my solution was just like Wumbate's


limit
trigonometry
logarithms
calculus
calculus computations

calculate

(i) $ \lim_{x \rightarrow 0+}x^{(x^{x}-1)}$

(ii) $ \lim_{x \rightarrow 0-}(1-2^{x})^{\sin x}$

(iii) $ \lim_{x \rightarrow 0}\left( \frac{1}{\log (x+\sqrt{1+x^{2}})}-\frac{1}{\log(1+x)}\right)$

edit: (iii) had a typo

[quote="arp"]calculate

(iii) $ \lim_{x \rightarrow 0}\left( \frac{1}{\log (x+\sqrt{1+x^{2}})}-\frac{1}{1+x}\right)$[/quote]

$ =\infty$

[quote=Kent Merryfield]On the next to last step of (i): what you have is $ O(x\ln^{2}x).$ That's not $ o(x)$ or even $ O(x),$ and you shouldn't say that it's $ o(x).$ It is, however, $ o(1)$ and the conclusion does follow.[/quote]

What are $o (x) $  and  $O (x) $ here ? Please let me know. Thanks .

[quote=Universes][quote=alexheinis]See here [url]https://mathworld.wolfram.com/LandauSymbols.html[/url][/quote]

Thank you dear. Which books do you suggest for to learn more about  methods of finding $limits  $ and this Landau Symbols, Rienman sums and integrals ,Rieman , Euler functions. And  all others subjects that needed in Undergraduate competition? Thanks.[/quote]

Please let me know. Please it is important for me. Please. Thanks.

[b](i) Solution without Landau simbols: [/b]
$y=\underset{x\to 0+}{\mathop{\lim }}\,{{x}^{({{x}^{x}}-1)}}\Rightarrow \ln y=\underset{x\to 0+}{\mathop{\lim }}\,({{x}^{x}}-1)\ln x=\underset{x\to 0+}{\mathop{\lim }}\,({{e}^{x\ln x}}-1)\ln x=\underset{x\to 0+}{\mathop{\lim }}\,\frac{{{e}^{x\ln x}}-1}{x\ln x}x{{\ln }^{2}}x=0$

[b](ii) Solution without Landau imbols, using l'Hospital: [/b]
$L=\underset{x\to 0-}{\mathop{\lim }}\,{{(1-{{2}^{x}})}^{\sin x}}\Rightarrow \ln L=\underset{x\to 0-}{\mathop{\lim }}\,\frac{\sin x}{x}\ln {{(1-{{2}^{x}})}^{\sin x}}x=\underset{x\to 0-}{\mathop{\lim }}\,\frac{\frac{-{{2}^{x}}\ln 2}{1-{{2}^{x}}}}{-\frac{1}{{{x}^{2}}}}=-\underset{x\to 0-}{\mathop{\lim }}\,\frac{x\ln 2}{{{2}^{x}}-1}x=0$

c) Write $f=\ln (x+\sqrt{x^2+1}),g=\ln(1+x)$ then $f'(x)={1\over {\sqrt{x^2+1}}},g'(x)={1\over {1+x}},f'(0)=1,g'(0)=1$ hence ${1\over f}-{1\over g}={{g-f}\over {fg}} \sim {{g-f}\over {x^2}}$.
Now $f,g$ are analytic in $x=0$ and it's a simple question of Taylor series.
From ${1\over {\sqrt{1+x}}}=\sum_0^\infty {{-1/2}\choose n} x^n$ we derive $f=x+O(x^3)$ and of course $g=x-x^2/2+O(x^3)$. It follows that the limit equals $-1/2$.

@Universes: most of the material when I studied was with handouts (syllabi in Dutch), I'll try to find some good, accessible books online for you.


geometry
geometric transformation
reflection
inequalities
geometry unsolved

Point $P$ is inside an acute triangle $\triangle ABC$,show that $(PA+PB+PC)^{2}\geq\sqrt{3}(BC\cdot PA+CA\cdot PB+AB\cdot PC)$

Why does no one reply? :( Is it so hard?But now I've worked it out.
Express $\cum PA\cdot BC^{2}$ with $PA,PB,PC$.It's enough.

I proposed this problem for Mathematical Reflection with a nice proof that doesn't use $PA.BC^{2}$, ... However, Hung 's proof (which I believe is similar to yours) in the Solution part is much more elegant.

Sorry, but I can't find the proof that you mentioned.

They didn't post my solution. However, if you are interested, my proof involves the projections of $M$ on the sides of $BC, CA, AB$ and an application of the inequality $\sum yzsin^{2}A \leq \frac{1}{4}(x+y+z)^{2}$ where $x, y, z$ are real. Try it and enjoy!!!  :)


inequalities logarithms function induction LaTeX

Does anyone know the proof of the Arithmetic Mean-Geometric Mean inequality, namely,

$\displaystyle \frac tion {1}/{n} \cdot (a_{1}+a_{2}+a_{3}+...+a_{n})\geq(a_{1}\cdot a_{2}\cdot a_{3}\cdot ...\cdot a_{n})^{\frac tion{1}/{n}}$

Let's make this a bit neater.  Define

$x^m_k=\frac {a_m+a_{m+1}+\ldots a_{m+k}}{k}$

The inequality $x_1+x_2\geq 2\sqrt {x_1x_2}$ is the trivial inequality.  Clearly if it holds for $n=2^t$, then

$x^1_{2^t}+x^{2^t+1}_{2^{t+1}}\geq 2\sqrt {x^1_{2^t}x^{2^t+1}_{2^{t+1}}}\geq 2^{t+1}\sqrt[2^t]{a_1a_2\ldots a_{2^{t+1}}}$ (ponder this for a minute).  This establishes it for all powers of $2$ by induction.

Next, say that it holds for $n=k$.  I claim that it also holds for $n=k-1$

Indeed, we have

$kx^1_{k-1}=a_1+a_2+\ldots a_{k-1}+x^1_{k-1}\geq k\sqrt[k]{a_1a_2\ldots a_{k-1}x^1_{k-1}}$, or

$x^1_{k-1}\geq \sqrt[k]{a_1a_2\ldots a_{k-1}x^1_{k-1}}$, which simplifies to

$(x^1_{k-1})^{\frac {k-1}{k}}\geq \sqrt[k]{a_1a_2\ldots a_{k-1}}$

which simplifies to $x^1_{k-1}\geq \sqrt[k-1]{a_1a_2\ldots a_{k-1}}$

which is just the AM-GM inequality for $n=k-1$

It's easy to see that we can get it to work for any $n\in\mathbb{N}$ from here.  I am sorry if my notation is a bit confusing, but it's my opinion that this simplfies everything somewhat.

Theres a nice proof using the re-arrngement inequality, which itself is easily proved, but i dont know it off hand...

there is also another way of proving it by pathagorean theorem.
BTW, i like blahblahblah's proof better than mine...it's easier to understand :)

there are a lot of ways to prove AM-GM. The easiest one, although it only works for 2-term, is using the trivial inequality:

$(a-b)^2 \ge 0$
$a^2+b^2+2ab \ge 4ab$
$(a+b)^2 \ge 4ab$
$(a+b) \ge 2\sqrt{ab}$
$\frac{a+b}{2} \ge \sqrt{ab}$

I am sorry if this is a bit advanced, but some of you may get something from it.

If you know Jensen's inequality, the AM-GM inequality follows easily from the convexity of $\ln x$.  Here is a sketch of how inequalities are proved using Jensen's

1) A convex function is a function such that $f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)$ for all $0<\lambda<1$

2) By induction (in fact, in a manner similar to the proof of AM-GM used here) we establish the inequality $f$ convex and $\sum_i \lambda_i=1$, $0<\lambda_i<1$ for all $i$ implies

$f(\sum_i \lambda_ix_i)\leq \sum_i \lambda_if(x_i)$

which is called Jensen's inequality

3) By considering Taylor Series or otherwise, we establish the inequality $f$ twice differentiable and $f''(x)>0$ implies $f$ convex

4) Set $\lambda_i=1/n$ and $f(x)=\ln x$ and apply Jensen's inequality

This seems very complex, but it allows us to prove far more general inequalities.

Nicely done.... I should master to use Jensen's and Holders someday.

Here is another proof by induction Perhaps easier than those already given
Suppose ${a-_1}$ $leq$ $a_{2}$ $leq$.................$leq$ $a_{n}$  where all numbers are real and positive. Denote their arithmentic mean by A,

Clearly $a_{1}$ $leq$ A $leq$ $a-{n}$    so A-$a_{1}$ $geq$ 0     and$ a_{n}$ -A $geq$ 0

(A-$a_{1}$)($a_{n}$ -A)  = A.$a_{n}$ + A.$a_{1} -$A^{2}$ -$a_{1}$ .$a_{n}$ $geq$ 0 or
A( $a_{1}$ +$a_{n}$ -A ) $geq$ $a_{1}$.$a_{n}$............(1)
Assume AM $geq$ GM for any set of n-1 numbers. We wish to prove (by induction )that the AM_GM inequality is true for n numbers $a_{1}$ ,$a_{2}$
.........$a-{n}$.
Apply the inductive assumption to the n-1 numbers  $a_{2}$ , $a_{3}$,..........$a_{n-1}$,  $a_{1}$ + $a_{n}$ - A.
The arithmetic mean of these numbers is A  By inductive assumption we have
$A^{n-1}$ $geq$ $a_{2}$.$a_{3}$. ...........$a_{n_1}$.( $a_{1}$ + $a_{n}$ -A)
Multiplying both sides by A we have
$A^{n}$ $geq$  $a_{2}.$a_{3}$..................$a_{n-1}$.(A($a_{1}$ +$a_{n}$ -A) and from inequality (1) the expression  inside (      ) is

$geq$ $a_{1}.$a-{n}
 therefore finally we have the inequality for n numbers.

$A^{n} $geq$ $a_{1}$.$a_{2}$.............$a_{n}$
 I hope this is helpful.

Yeah, and a note: whenever you use large amounts of latex, always preview it before you submit so you make sure all your $ signs and \'s are correctly placed.


Let's pretend you are 13, and a girl calls you and invites you to a birthday party tomorrow. When you go, are you supposed to bring her a present? or would it be considered rude not to.

well, i think that you should buy her a present.
but, i don't know much about girls :oops:

It would depend entirely upon the girl.  I would ask other people who are going what they are doing, but if nothing else, you can always bring some money & a card & keep them hidden unless other people offer presents  :ninja:

If nothing else, sprint to the nearest store (for example, a video store or a music store) and buy a gift card.

Yeah, a card should be enough.
And if you fantasize the girl, buy heart shaped box of flowers or toffees. :wink:

A card/gift card may be good. Or something fancier if you like her.
Is this a purely hypothetical question or is it something else?  :wink:

[quote="randomdragoon"]Get her something involving Pokemon/Hamtaro. Who knows it might work.

I speak from personal experience here[/quote]

ummm...

Wtf these are the most off base answers I have ever seen. You should definitely bring a present or you might as well not show up. Get her a gift card or something if your lazy but money is a no no.

You've never been to a birthday party before? Wtf?  :what?:

I completely agree that money is not a good idea. It might work for some people, but some people may feel offended about the money.

** gift card is the best thing to do. She can get whatever she wants from that store

if  you show up without a present, it will be embarrasing for you because it is bad manners (at least, I think) to do that. She is spending money to treat you (to cake for example) and you ought to buy her a present.

get her a few packets of mechanical pencil lead.  everyone at my school needs that, plus it's small enough to fit in your pocket.  also, it'll look like you tried, instead of taking the easy way out, which you did.

my advice is probably not the best.

I got a Kadabra pokemon card for christmas this year (one of the old kinds). It made my day.

Seriously go for the pokemon/hamtaro.

rofl...this is a good topic...im learning a lot  :rotfl:

and pokemon? no...=)

Speaking from a girl's point of view

1. Some people invite others to their birthday party just to get presents

2. if you like this girl, you should bring her a gift
nothing fancey, just something that means you like her, and have a happy bday

3. If you dont like her in that way, bring her something that means that

I think showing up without a present = you are there just for the free food

[quote="randomdragoon"]Seriously go for the pokemon/hamtaro.[/quote]
That would definitely work for you. :rotfl:


Anymore condition ? Like $a+b+c=1$ ?? Because if $a=b=c=1$ then LHS = $0$ ...

[quote="bodan"]Prove that \[ \sum_{\mathrm{cyclic}} a^2(a-1)^2\geq10abc+\sum_{\mathrm{cyclic}} ab(a^2+b^2)  \] for positive reals $a,b,c.$[/quote]

Wrong, bodan, wrong. BTW very nice signature!


"The number 9238 has an interesting feature: the thousands digit is odd, the hundreds digit is even, and all four digits are different. How many four digit numbers have this property?"

Source: A packet of math problems off the internet.

I know that there's got to be a way better than listing them all.

[hide]The first digit is odd, we have 1, 3, 5, 7, or 9. That's 5 choices. The second digit is even. There are again 5 choices. The last two digits are different from each other and from the first two. The third one has 8 possibilities and the fourth 7. That's 5*5*8*7 = 1400. [/hide]

[hide]this time i did not list them out. :)
well since the first digit has to be odd there are 5 possibilities for the first digit.  Since the second digit has to be even there are 5 possibilites.  Since the thrid one has to be different from the first two digits that leaves us with 8 possibilities.  The fourth can't have the first three digits so 10-3 is 7 possibilites.  $5 \cdot 5 \cdot 8 \cdot 7=\boxed{1400}$[/hide]


Find the sum of the following series:

$ \frac{1}{3}\plus{}\frac{1\cdot 5}{3\cdot 7}\plus{}\frac{1\cdot 5\cdot 9}{3\cdot 7\cdot 11}\dots\dots$ upto n terms

The series is diverrgecing by theorem of Raabe


[quote="mathiee"]assume  x>0  and  y satisfy  $ e^{y} > \frac {x}{1 \minus{} e^{ \minus{} x}}$

prove that x>2y[/quote]

Are you sure ? $ y$ can be made arbitrarily large and it still satisfies the condition.

[quote="mathiee"]assume  x>0  and  y satisfy  $ e^{y} > \frac {x}{1 \minus{} e^{ \minus{} x}}$

prove that x>2y[/quote]

That's obviously wrong. Take $ x\equal{}1$ and $ y\equal{}10^{1000}$ :)

sorry sorry sorry ,  i gave it wrong  .  EDITED NOW . KINDLY SEE AGAIN

[quote="mathiee"]assume  x>0  and  y satisfy  $ e^{y} \equal{} \frac {x}{1 \minus{} e^{ \minus{} x}}$

prove that x>2y[/quote]

$ x>2y$ $ \iff$ $ e^{\frac{x}{2}}>e^y$ $ \iff$ $ e^{\frac{x}{2}}>\frac{x}{1\minus{}e^{\minus{}x}}$ $ \iff$ $ e^{\frac{x}{2}}(1\minus{}e^{\minus{}x}) > x$  $ \iff$ $ \sinh(\frac{x}{2})>\frac{x}{2}$ $ \forall x>0$ and this is a classical well known result.


I suspect that this is just a browser problem, since I just installed Mozilla's Firefox browser, but you might know what to do about it.  When I click on a link to take me to the third page of responses on a thread, I get sent to the first page instead and I have to then click on the page three button at the bottom to get to where the new messages are.  Any ideas?

Sorry, I'm stumped.  Anyone else have a suggestion?

Don't know the answer, but here's one suggestion for investigating.  Joel, you can look at what URL's the various links are sending you to (by looking at the address bar).  You can even check the URL's by just pointing at the link (without clicking) by looking at the status bar of the browser.

well i don't know its only a beta. i use netscape. it has much of mozilla's features including pop-up ad blcoking, and tab browsing. maybe you should download that for now jbl.

When I mouse over the links, it shows me the correct destination.  Furthermore, if I try to open the link in a new window, it opens it to the correct page.  And, when I'm already on such a thread, the page numbers (which I think are identical to those in the forum) work fine.

And when you click on the link without opening a new window, is the URL in the address bar the same as the URL when you click on the link with a new window?  (And by the way, what is the URL?)

Oh.  No, it drops the part that has "start=30" or the equivalent.

i'm downloading the firefox right now. it seems cool. the old one is practically netscape 4.0.

i don't know hwat's wrong with your computer. mine is ok.

GOLLY GEE. WATCH OUT MICRSOFT . THIS IS A REALLY GOOD WEB BROWSER> IT IS SO SMALL AND UNCLERTERED.  I AM DOWNLOADING THE THUNDERBIRD PACKAGE. I GIVE IT 5 STAR RATING.

Are you using the firefox now and having it work fine?  (I just upgraded from an older Mozilla to this -- it's nicer to use in some respects, and has nice customizable features, but this one little bug is pretty annoying.)

yes. I didn't upgrade though. I downloaded it. I am A NETSCAPE fan. It seems Netscape copied Mozilla code. They are very similar.  So far so good. But now i need to see if the java stuff loads.

I downloaded firefox recently, it works fine for me too. Though I've occasionally heard about this problem at other forums.. when you open it up in a new window, you're right clicking the link, maybe your mouse is moving slightly when you normally click it and the links are so close together its going to the other one? maybe..

Hum ... today they seem to be working fine.  Well, whatever.  Thanks for the help!

GOLLY GEE.
i type into the address bar
 Best of 2003, Web Browser PC World Magazine
and it it got me right into the article.
Mozzila is really smart!!!!!!!!!!!


1 find all function  $ \in Q$ .such that
(I) . $ f(x,y) \equal{} f(y,x).$
(II) . $ f(f(x \plus{} y),z) \equal{} f(x,f(y,z)).$
(III) .$ f(0,0) \equal{} 0.$

good luck .

[quote="hữu quân"]1 find all function  $ \in Q$ .such that
(I) . $ f(x,y) = f(y,x).$
(II) . $ f(f(x + y),z) = f(x,f(y,z)).$
(III) .$ f(0,0) = 0.$

good luck .[/quote]
you mean that??
$ f: \mathbb{R}\to \mathbb{Q}$
(I) . $ f(x,y) = f(y,x).$
(II) . $ f(f(x,y),z) = f(x,f(y,z)).$
(III) .$ f(0,0) = 0.$

[quote="hữu quân"]sorry $ f(x,y)$ : $ Q \minus{} \minus{} > Q$ ( $ f$ contains variable  x,y ) :)[/quote]

And then what is the meaning of $ f(x\plus{}y)$ in II ?

[quote="hữu quân"]such as $ f(x,y)$ = $ max(x,y)$ or $ f(x,y) \equal{} min{x,y}$ .[/quote]

I don't understand.

$ f(x,y)$ is supposed to be a function of two variables. So what is the meaning of $ f(x\plus{}y)$ where we just have a single variable "$ x\plus{}y$" ?

Looking at your examples $ \min(x,y)$ or $ \max(x,y)$, do you mean that in your mind $ f(x\plus{}y)$ means $ f(x\plus{}y,x\plus{}y)$?.  Strange.

Moreover, I'm quite sure that there is a missing hypothesis :wink:

Pierre.

[quote="hữu quân"]sorry $ f(x,y)$ : $ Q \minus{} \minus{} > Q$ ( $ f$ contains variable  x,y ) :)[/quote]

Do you mean $ f(x, y) : \mathbb{Q}^2 \to \mathbb{Q}$?  And you still haven't clarified the meaning of $ f(x \plus{} y)$.

[quote="hữu quân"]I have mistake $ f(f(x,y),z) \equal{} f(x,(y,z))$ .
so now , you can solve it .[/quote]

So you are looking for all commutative and associative binary operations $ \oplus$ from $ \mathbb{Q}\times\mathbb{Q}$ in $ \mathbb{Q}$ such that $ 0\oplus 0\equal{}0$

We have at least :
$ x\oplus y\equal{}0$
$ x\oplus y\equal{}xy$
$ x\oplus y\equal{}x\plus{}y$

And likely a lot of others

Yes, that's why I think there is a missing hypothesis... :wink:

Pierre.

[quote="pco"]So you are looking for all commutative and associative binary operations $ \oplus$ from $ \mathbb{Q}\times\mathbb{Q}$ in $ \mathbb{Q}$ such that $ 0\oplus 0 \equal{} 0$

We have at least :
$ x\oplus y \equal{} 0$
$ x\oplus y \equal{} xy$
$ x\oplus y \equal{} x \plus{} y$

And likely a lot of others[/quote]

Notice that if $ \oplus$ is a solution and $ h(x)$ is a bijective function from $ \mathbb{Q}$ in $ \mathbb{Q}$ such that $ h(0)\equal{}0$, then :

the operator $ \otimes$ : $ a\otimes b\equal{}h(h^{\minus{}1}(a)\oplus h^{\minus{}1}(b)$ is also a solution.


probability
inequalities
limit
real analysis
probability and stats

Here are two good problems: first is just a good exercise, and the second is very hard.
1) Let $ X_i, \ i\geq 1$ be iid such that $ EX_1<\infty$. Prove that $ \frac {\max\limits_{1\leq i\leq n}X_i}{n}\to 0$ a.s.
2) Let $ X_i, \ i\geq 1$ be iid such that $ P(X_1>0)\equal{}1$, and let $ S_n\equal{}\sum_1^n X_i$. Prove that $ \frac {X_n}{S_n}\to 0$ in measure. (The hard part) Is it true that $ \frac{X_n}{S_n}\to 0$ a.s.?

[quote="Yustas"] (The hard part) Is it true that $ \frac {X_n}{S_n}\to 0$ a.s.?[/quote]
:) The only really hard thing about it is to guess the
[hide="answer"]
The answer is "No". Here is a [hide="counterexample"]
Let $ X\equal{}k!^2$ with probability $ \frac c{k\sqrt k}$ ($ k\equal{}1,2,\dots$).
[hide="Proof"]
The idea is that if among $ X_1,\dots, X_n$ there is a value $ k!^2$ with $ k\ge n$ and the largest number is not $ X_1$, then $ \sup_{m\le n}\frac {X_m}{S_m}\ge n$. But the probability that all $ X_m$ have smaller value is at most $ \left(1\minus{}\frac {c'}{\sqrt n}\right)^n\approx e^{\minus{}\sqrt n}$. Using Borel-Cantelli, we conclude that with probability 1 either $ X_1$ is the largest number in the entire sequence, or $ \sup_n\frac {X_n}{S_n}\equal{}\plus{}\infty$. It remains to note that $ X_1$ is the largest number with probability $ 0$.
[/hide]
[/hide]
[/hide]

Well, $ \frac{X_n}{S_n}\leq 1$, so the sup is 1. Another interesting question is to investigate the general case $ EX_1\equal{}\infty$.

Is the statement in the first problem true without the assumption $ X_1 \ge 0$?

Yes, it is true. I do not understand how the assumption $ X_1\geq0$ is related to the 1st problem.

[quote="Yustas"]Well, $ \frac {X_n}{S_n}\leq 1$, so the sup is 1. [/quote]
Oops, for some reason I thought that $ S_n\equal{}X_1\plus{}\dots\plus{}X_{n\minus{}1}$  :roll: Of course, if you add $ X_n$, my $ \sup\equal{}\plus{}\infty$ becomes your $ \limsup\equal{}1$. Sorry for that  :blush:

I can't seem to find a solution for a) :(  Does anyone have one?

This is what I have tried.  Fix $ c>0$ and put $ E_n \equal{} \left\{\max_{1\leq i\leq n} |X_i|>cn\right\} \equal{} \bigcup^n_{k\equal{}1} A_k$, where $ A_k\equal{}\{|X_i|\leq cn \text{ for }1\leq i<k \ , \ |X_k| >cn\}$.  Then if $ t \equal{} P(|X_1|>cn)$, we get
\[ P(E_n) \equal{} \sum^n_{k\equal{}1} P(A_k) \equal{} 1 \minus{} (1\minus{}t)^n
\]
By Chebyshev's inequality, $ t \leq (cn)^{\minus{}1} E|X_1|$, but the signs are the wrong way.  And using something more simple like
\[ P(E_n) \leq \frac{1}{cn} E|X_1|
\]
isn't strong enough to apply Borel-Cantelli to.  So I am kind of lost :(

You are very close :) Use two things:
1) Do not try to estimate $ t$, just write what it is and use the inequality $ 1\minus{}(1\minus{}t)^n\le nt$.
2) Note that $ E_n$ are very far from being independent or disjoint. Actually, $ E_{2^k}$ are already responsible for everything.
Hope you can do it now :).

Sorry fedja, I am still kind of lost.  Could you be a little more explicit?

edit: maybe I have an idea.  $ P(E_{2^k})$ converges by Borel-Cantelli, so it remains to analyze $ W_k \equal{} \max_{2^k < i < 2^{k\plus{}1}} |X_i|$.  Again we have that $ P(W_k < c2^{k}) < \infty$ (by comparison to $ \sum 2^{\minus{}k}$), so another application of Borel-Cantelli suffices to finish off the proof.  Am I right?

Here is a full solution:
1) If $ \max_{m\le n}X_m>2cn$ for infinitely many $ n$ (note the factor 2!), then $ \max_{m\le 2^k}X_m>c2^k$ for infinitely many $ k$, so infinitely many $ E_{2^k}$ occur.
2) For every $ n$, we have $ P(E_n)\le nP\{X_1>cn\}$. Hence
3) $ \sum_k P(E_{2^k})\le\sum_k 2^kP\{X_1>c2^k\}\le \frac 2c E|X_1|<\plus{}\infty$. Now it is pure Borel-Cantelli.
I couldn't quite understand exactly what you were saying in your last post but it seems you were thinking along the same lines :).

Now a follow-up problem. Show that the condition $ \frac{\max_{m\le n}X_m}{n}\to 0$ a.s. implies that $ EX_1^\plus{}<\plus{}\infty$ ($ x^\plus{}\equal{}\max\{x,0\}$) ;)

The way i solved it is using the fact that $ \frac {X_n}{n} = \frac {S_n}{n} - \frac {S_{n - 1}}{n}\to 0$ a.s. because $ EX_1 < \infty$, so the events $ A_k = \{\frac {X_k}{k} > \epsilon \}$ occur finitely often with probability 1. And after thinking for some time you can realize that this is enough to conclude about the maximum  :lol:
What about problem 2 part 1? It seems obvious but still worth solving.
[quote="fedja"]
3) $ \sum_k P(E_{2^k})\le\sum_k 2^kP\{X_1 > c2^k\}\le \frac 2c E|X_1| < + \infty$. Now it is pure Borel-Cantelli.
I:).[/quote]
How do you prove the convergence of series?
[quote="fedja"]Now a follow-up problem. Show that the condition $ \frac {\max_{m\le n}X_m}{n}\to 0$ a.s. implies that $ EX_1^ + < + \infty$ ($ x^ + = \max\{x,0\}$) ;)[/quote]
$ \frac {X_n}{n}\to 0$ a.s. is equivalent to $ E\abs{X_1} < \infty$:
$ E|X|<\infty \leftrightarrow  \sum_k P(|X_1| > k)<\infty$, but $ \sum_k P(|X_1| > k)= \sum_k P(|X_k| > k) < \infty$, as far as $ \frac {X_n}{n}\to 0$ a.s.

Well, we have
$ \sum_{k}2^k P\{X>c2^k\}\equal{}\sum_k(1\plus{}2\plus{}\dots\plus{}2^{k\minus{}1}\}P\{c2^{k\minus{}1}< X\le c2^k\}$
$ \le \sum_k 2^k P\{c2^{k\minus{}1}< X\le c2^k\}\le \frac 2c\sum_k c2^{k\minus{}1}P\{c2^{k\minus{}1}< X\le c2^k\}\le
\frac 2c EX$.

[quote="Yustas"]How do you prove the convergence of series? [/quote]
That series and the series you used are essentially the same : [url=http://en.wikipedia.org/wiki/Cauchy_condensation_test]Cauchy condensation[/url]

Let's state Yustas' idea as a lemma :
If $ (a_n)$ is a sequence of nonnegative real numbers and $ \lim_n \frac {a_n}{n} \equal{} 0$, then $ \lim_n \frac {\max_{1\le k \le n} a_k}{n} \equal{} 0$.
(proof: $ a_{k(n)} \equal{} \max_{1\le k \le n} a_k$, $ \frac {a_{k(n)}}{n} \le \frac {a_{k(n)}}{k(n)}$ goes to zero whether the nondecreasing sequence $ (k(n))$ is bounded or not.)

As to problem 2 part 1, part 2 implies part 1 and part 2 can be proved using law of large numbers again : $ \frac {X_n}{n} \cdot \frac {n}{S_n}$

[quote="redkimchi"]
As to problem 2 part 1, part 2 implies part 1 and part 2 can be proved using law of large numbers again : $ \frac {X_n}{n} \cdot \frac {n}{S_n}$[/quote]
Erm... What law of large numbers? It may happen that $ EX_1\equal{}\plus{}\infty$ now! Besides, the answer to 2 part 2 is "No" in general.

[quote="fedja"][quote="redkimchi"]
As to problem 2 part 1, part 2 implies part 1 and part 2 can be proved using law of large numbers again : $ \frac {X_n}{n} \cdot \frac {n}{S_n}$[/quote]
Erm... What law of large numbers? It may happen that $ EX_1 \equal{} \plus{} \infty$ now! Besides, the answer to 2 part 2 is "No" in general.[/quote]

Kolmogorov's strong law of large numbers. works only for the case $ E(X) <\infty$.

Problem 2 part 1 for the case $ E(X) \equal{} \infty$

Suppose $ \epsilon > 0$ is given. Choose a constant $ c$ s.t. $ P( X_1 > c) < \epsilon$.
Apply the law of large number to truncations $ \min\{X_n, d\}$ ($ d$ constant) and conclude $ \frac {S_n}{n} \to \infty$ a.s.
There is $ N$ s.t. for all $ n > N$ we have $ P(S_n \le cn) < \epsilon$.
For all $ n > \max\{N, 1/\epsilon\}$ we have $ P(\frac {X_n}{S_n} > \epsilon) \le P(\frac {X_n}{S_n} > \frac1n) \le P(X_n > c) \plus{} P(S_n \le cn) < 2\epsilon$.

That solves it, no question, but there is much simpler solution. Just note that the probability that $ \frac{X_n}{S_n}>\frac 1k$ is not greater than the probability that there are no more than $ k\minus{}1$ numbers in the sequence $ X_1,\dots,X_n$ that are at least $ X_n$ and this probability is at most $ \frac kn$. :)

I used that $ P\left(\frac{X_n}{S_n}>\epsilon \right)\equal{}P\left(\frac{X_1}{S_n}>\epsilon \right)\equal{}E I\{\frac{X_1}{S_n}>\epsilon\}\to 0$ by Lebesgue dominated convergence.


In the expression xy^2 , the values of x and y are each decreased by 25%; the value of the expression is
A. decreased 50%
B. decreased 75%
C. decreased 37/64 of its value
D. decreased 27/64 of its value

$ \frac{(.75x)(.75y)^{2}}{xy^{2}}\equal{}\frac{27}{64}$

The value of the expression is decreased [i]to[/i] $ \frac{27}{64}$ of its previous value, or decreased [i]by[/i] $ \frac{37}{64}$ of its previous value.


Given $ a, b, c \geq\ 0.$ and $ a\plus{}b\plus{}c\equal{}3$ Prove that:
$ \frac{a}{\sqrt[]{b\plus{}c^2}} \plus{} \frac{b}{\sqrt[]{c\plus{}a^2}}\plus{} \frac{c}{\sqrt[]{a\plus{}b^2}} \geq\ \frac{3}{\sqrt[]{2}}$
Am-Gm solved it easily  :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove that:
$ \frac {a}{\sqrt []{b \plus{} c^2}} \plus{} \frac {b}{\sqrt []{c \plus{} a^2}} \plus{} \frac {c}{\sqrt []{a \plus{} b^2}} \geq\ \frac {3}{\sqrt []{2}}$
Am-Gm solved it easily  :)[/quote]
Do you have AM-GM solution for the stronger one?
\[ \sum a\sqrt{b\plus{}c^2} \le 3\sqrt{2}\]
:)

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c \geq\ 0.$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove that:
$ \frac {a}{\sqrt []{b \plus{} c^2}} \plus{} \frac {b}{\sqrt []{c \plus{} a^2}} \plus{} \frac {c}{\sqrt []{a \plus{} b^2}} \geq\ \frac {3}{\sqrt []{2}}$
Am-Gm solved it easily  :)[/quote]
Do you have AM-GM solution for the stronger one?
\[ \sum a\sqrt {b \plus{} c^2} \le 3\sqrt {2}
\]
:)[/quote]

I tried but it's so hard with me  :wink:
However, I' ll wait for your proof with CYH  :)

[quote="nguoivn"][quote="nguoivn"]Given $ a, b, c \geq\ 0.$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove that:
$ \frac {a}{\sqrt []{b \plus{} c^2}} \plus{} \frac {b}{\sqrt []{c \plus{} a^2}} \plus{} \frac {c}{\sqrt []{a \plus{} b^2}} \geq\ \frac {3}{\sqrt []{2}}$
Am-Gm solved it easily  :)[/quote]

By the following way, I can prove the nicer and harder:
Given $ a, b, c \geq\ 0.$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove that:
$ \frac {a}{\sqrt []{5b \plus{} 4c^2}} \plus{} \frac {b}{\sqrt []{5c \plus{} 4a^2}} \plus{} \frac {c}{\sqrt []{5a \plus{} 4b^2}} \geq\ 1$
 :)[/quote]
I dont think it is harder, nguoivn. ;)
Since it follows from the inequality
\[ 5(ab\plus{}bc\plus{}ca)\plus{}4(a^2b\plus{}b^2c\plus{}c^2a) \le 27\]
;)

[quote="can_hang2007"]I dont think it is harder, nguoivn. ;)
Since it follows from the inequality
\[ 5(ab \plus{} bc \plus{} ca) \plus{} 4(a^2b \plus{} b^2c \plus{} c^2a) \le 27
\]
;)[/quote]

But I found the ineq:
\frac {a}{\sqrt []{b + kc^2}} + \frac {b}{\sqrt []{c + ka^2}} + \frac {c}{\sqrt []{a + kb^2}} \geq\ \frac {3}{\sqrt []{1+k}}

always trues with $ k \geq\ \frac{4}{5}$  :wink:


Let $ n$ be an integr, $ n\ge 2$. For each integer $ k$, $ 1\le k\le n \minus{} 1$, there always exists one integer $ m$, $ 1\le m\le n\minus{}1$ such that $ \frac {m(m \minus{} 1)}{2}\equiv k(mod n)$. Find $ n$.

Emm... @TS : is that correct problem? or i am mistaken?

Suppose there exists $ n$ satisfying the problem. Now, that means for two different $ k$, two integers $ m$ such that $ \frac {m(m \minus{} 1)}{2}\equiv k(mod n)$ must be different too. Now, $ m\equal{}1$ takes no $ k$, so there are $ n\minus{}2$ possibilities for $ m$ (that are $ 2,3,...,n\minus{}1$).
But, $ k$ takes all values of $ 1,2,..,n\minus{}1$, that means, at least we must have $ n\minus{}1$ different values for $ m$. But, there are only $ n\minus{}2$ possibilities for $ m$???


I repeat: DO NOT DISCUSS ROUND 2 UNTIL WEDS NOV 23

Do we have to wait until the math jam to discuss solutions, or can we discuss before that (Wednesday morning)?

Trying to get the solutions ASAP, eh?  :D

:D I'll log in at midnight and post my solutions :rotfl: Just kidding.

[quote="pkerichang"]I'm confused.  Can we discuss ON Wed. Nov. 23 (lets say 12:01 am), or do we discuss AFTER Wed. Nov. 23?????[/quote]
I believe "on."

lol i'm curious about the solutions.. but i'm not looking forward to seeing myself humiliated  :blush:  haha

I just want to see the solution to number 3, which I couldn't get.  ;)

If my burst of insight on that problem and expected value in general, using what I learned from the previous round, proved correct, then problem 3 is trivial (once you make that leap, of course...took me some time).

[quote="pkerichang"]uhh....everyone's doing it, and apprently the admins didn't lock or delete the posts...[/quote]

Actually, as I recall it, the posts were locked [i]until[/i] the 23rd.  And when they were unlocked everyone started posting in them.

I think this thread can be unstickied now.  ;)


Let A be a point outside the circle O. Line pasing through A tangent to O touches O in point B. Line passing through A cuts O in points C and D. anoter line passing through A intersects line BC in E, and line BD in F. Lines ED and FC interect in point P. Prove that point P lies on the polar of point A.

I figured out this one on my own. Is it a known fact?? Also if you know any other nice polar facts, I'd be gratefull if you would post it here or PM to me.

I'm waiting for your nice solutins!

yes, it is well known, i will give you this lemma : if $ABC$ is a triangle and $AD,BE,CF$ are $3$ cevians which intersect in $P$, and for example $EF$ cuts $BC$ in $T$ , we have $(BDCT)$ a harmonical division.

in your problem, let $S$ be the intersection of $BP$ and $CD$, therefore (from the lemma) : $(ACSD)$ h.d, so $P$ is on the polar of $A$

Nice, [color=blue][b]pohoatza[/b][/color]!

By the way, does this lemma have a name?


man fekr kardam bad nist kami ham tarkibiat kar konim pass ye chand taee soale tarkibiat post mikonam.

what is the coefficient of $x^k$ in the expansion  of $(1+x+x^2+...+x^{n-1})^2$

LET $X$ be a nonempty set and let $f$ be a family of $m$ distinct subsets of $X$ where $m \geq 2$.show that the collection of subsets of the form $A \Delta B$ (symmetric difference of $A$ and $B$) where $A,B \in F$ contains at least $m$ pairwise distinct sets.

in how many ways can one arrange $k$ rooks on a chessboard with $m$ rows and $n$ colimns so that no rook can attack another?

one is given $n$ distinct points in the plane.show that there exist fewer that $n \sqrt{n}$ pairs of these points which have their distance equal to $1$.

find the number of trees with $n$ labeled vertices and exactly $p$ terminal vertices.

[quote="amirhtlusa"]lotfan in a ro too ye 1 post benevis, baad shomareh bezan ke maloom bashe javab e koodoom o minivasan. :)[/quote]
ettefaghan man fekr kardam age intori benevisam oonvaght baghie mitoonan ba estefaade az quote rahat tar javab hashoon ro befferestand. :D

[quote="amirhtlusa"]ras migi... oonam khoobe :)[/quote]
khosh halam ke khoshet oomad. :D

agha too fazaye shatranje n*n*n hadeaghal chand rokh bayad gozasht ta tamame mokaab haye 1*1*1 tahdid shavand
agha nemikhad begi soale kojast hame midoonan rahe haleto benevis

safhe shatranje regular ro be chand tarigh mitavan ba 32 domino farsh kard??
agha bikhodi nagoo male che ketabie ya fahat be javabesh eshare nakon raheto begoo
domino ham be mostatil haye 2*1 migooyand
merC

[quote="BaBaK Ghalebi"]aghaye yazdan bod yezare faal bash
natars idehato nemidozdim[/quote]
aghaye yazdan bod kie? :?:

[quote="BaBaK Ghalebi"]ye adame fogholade ghavio shakh dar zamine tarkibiat va graph theory
(syboy)
ma koochikeshoonim[/quote]
ok i see. :D

ey baba baz in babi hendone gozasht to baghal ma ;)

eyval to belakhare oomadi??
bebinam rahe hallamo khoondi??
dorostan??
joob nazadam??

yek soval jaleb ke fekr konam dide bashid
sabet konid agar yek don bale az adad tak raghami va namotenahi  manand $a1,a2,a3,....$ vojod dashte bashad ba in khasiyat ke adad $x1 x2 x3 .... xn$ dar hich kojaye dobale zaher nashode bashad angah hatman yek donbale ye az dotaraf namotenahi ham vojod darad ba in khasiat
baraye mahsa mokh ke sath mikhad begam ke soval marhaleye dovome computer bode

na hanoz chon khodam daram roye masaleha fekr mikonam
valy hala mage mishe ostad joof bezane

[quote="syboy"]yek masaleye nesbatan sakht ama jaleb dar leningerad !!!!!!!
ras haye yel graph mahdod be do rang rang shode ast dar har saniye rang ras haye graph taghir mikonad va be rangi dar miayad ke dar hamsayegi an bishtar ast .sabet konid bead az yek modat har ras ya digar rangash taghir nemikonad ya inke har saniye rang avaz mokonad[/quote]
masaleye ghashangie.

[quote="BaBaK Ghalebi"]dar $1$ matrix $m*n$ az adade haghighi ma $p$ ta az bozorgtarin adad ($p\leq m$)dar har sotoon va hadeaghal $q$ ta az bozorgtarin adad dar har satr ra alamat mizanim ($q\leq n$)sabet konid haddeaghal $pq$adad $2$bar alamat khordeand :!:  :?:[/quote]
az esteghra rooye $m+n$ estefade mikonim
masale baraye $m=n=p=q=1$ vazeh ast
farz konid yek matris $m*n$ darim
ma an ra be yek matrise $m*(n-1)$ ya $(m-1)*n$ tabdil mikonim
agar tamame aadad dobar dobar alamat khorde bashand angah tedade anha haddeaghal $pq$ khahad bood
dar gheire in soorat az beine aadadi ke yek bar alamat khordeand bozorgtarine anha manande $M$ ra entekhab mikonim ke dar satr ya sotoone khod (na dar har do) az bozorgtarinhast
farz konid $M$ az bozorgtarinha dar sotoone khod bashad angah an az bozorgtarinha dar satre khod nist amma tamame aadade bozorgtar dar satre an dobar alamat khordeand ma in satr az matrix ra hazf mikonim ta yek matrixe $(m-1)*n$ bedast avarim ke dar an haddeaghal $q$ az bozorgtarin aadad dar har satr va haddeaghal $p-1$ adad dar har sotoon alamat khordeand
ba farze esteghra haddeaghal $(p-1)*q$ adad do bar alamat khordeand in aadad hamchenin dar matrix $m*n$ alamat khordeand be alave $q$ adad az satre hazf shode dar in matrix alamat khordeand banabarin dar matrix $m*n$,$(p-1)q+q=pq$ adad dobar alamat khordeand
 ;)  :D

[quote="syboy"]a white plane is partitioned onto cells (in a usual way). a finite number of cells are coloured black. each black cell has an even (0,2 or 4) adjacent (by the side ) white cells. prove that one may colour each white cell in green or red such that every black cell will have equal number of red and green adjaent cells. :?:
mr ghalebi bebinam mituni inam hal kuny  :D[/quote]
man ino hallesh kardam vali shomaha va alipour tavajoh nakardid
man ino farda sare kelase hadian esbat mikonam va badesham tooye in topic mizaramesh
ta to ham betooni roohie begiri :D  ;)

$a_1,a_2,...,a{}_2{}_n_+_1$ aadade sahihi hastand ke har yek az anha ra ke hazf konim baghie ghabele efraz be do majmooeye $n$ taii ba majmooe barabar hastand
sabet konid hameye in adadha ba ham barabar hastand :?:

har noghte faza ra ba yeki az $3$ range ghermez,sabz ya abi rang kardeim
farz konid $R,G,B$ majmooeye toole parekhathai bashand ke range do sare anha be tartib ghermez,sabz va abi ast
neshan dahid aghalan yeki az in se majmooe shamele tamame aadade haghighie mosbat ast :?:

tedadi adade haghighie mosbat rooye kaghaz neveshte shode ast
jame haselezarbhaye dotaie in aadad barabare $1$ ast
sabet konid mitavan yeki az in aadad ra kenar gozasht be sharti ke jame aadade baghimande kamtar az $\sqrt{2}$ bashad :?:

$6$ dayere daraye noghte moshtarake $A$ hastand
sabet konid dayerei vojood darad ke shamele markaze dayereye digari ast :?:

majmooeye $s$ az afrad daraye khasiate zir ast:
har do nafar ba tedade mosavi doost dar $s$ dooste moshtarak nadarand
sabet konid shakhsi dar $s$ vojood darad ke daghighan yek doost darad :?:

dar har yek az se madrese mafrooz $n$ danesh amooz vojood darad har danesh amooz jaman $n+1$ doost az do madrese digar darad
sabet konid mitavanim yek nafar az har madrese entekhab konim be tori ke in se nafar dobedo doost bashand :?:

[quote="mohamadhosein"][quote="BaBaK Ghalebi"]$n$ noghte rooye safhe darim ke hamegi rooye yek khat nistand
sabet konid khati vojood darad ke faghat az $2$ ta az in noghteha bogzarad :?:[/quote]
farze kholf ine ke har khati ke az do noghte migozare az noghteye sevomi ham migozare inam be extermal(esterategi)sabet shode ke hame az yek khat migozaran. :D[/quote]
doroste
oon masaley ham ke to migi be name masale $silvester$ maroofe
 ;)  :D

[quote="BaBaK Ghalebi"]$n$ noghte rooye safhe darim ke hamegi rooye yek khat nistand
sabet konid khati vojood darad ke faghat az $2$ ta az in noghteha bogzarad :?:[/quote]
hamoon tori ke mohamadhosein goft
ba borhane kholf farz konid chenin nabashad
yani har khate gozarande az do noghte az noghte digari begozarad
hal tebghe farze masale midanim ke hame noghat rooye yek khat gharar nadarand
zoje $(P,L)$ ebarat ast az khati manande $L$ va noghtei manande $P$ gheire vaghe bar $L$ hal noghtei ra dar nazar migirim ke faseleash az khat minimum bashad
farz konid $F$ paye amoodi ast ke az noghte $P$ bar khate $L$ vared shode ast
banabarin banba be farz haddeaghal se noghte $A,B,C$ bar khat gharar darad
farz konim do noghte masalan $A,B$ dar yektarafe $F$ vaghe bashand
bayad tavajoh dasht ke fasele $P$ az khate $L$ barabare minimum meghdar va barabare $d$ ast yani $PF=d$
hal dar edame farz konim $B$ be $F$ ta $A$ nazdiktar bashad yani $BF\leq BA$
angah faseleye $B$ ta $AP$ kamtar az $d$ mibashad va in tanaghoz ast
pas tamamie noghat rooye yek khat gharar darand
ke in ham ba farze masale motanaghez ast
in rahehal male ketabe steratejie manzooram masale $silvester$ ast
 ;)  :D

[quote="BaBaK Ghalebi"]neshan dahid har $n$ noghte az safhe ($n\geq 5$) ra mitavan tori be vasileye $2$ rang rang kard ke hich khati peida nashavad ke noghat az $2$ rang ra az ham joda konad :?:[/quote]
agar se noghte az noghate mazkoor rooye yek khat bashand do noghte khareji ra abi va noghte miani ra ghermez konid baghie noghat ra niz be delkhah rang konid dar in soorat hich khari vojood nadarad ke noghate nahamrang ra az ham joda konad
agar hich se noghtei rooye yek khat nabashand chon $n\geq 5$ noghte darim mitavan $4$ rase yek chahar zelie mohadab ra dar miane in noghat peida kard
hal agar noghate rooberooye ghotrie in chahar zeli ra ba yek rang rang konimo baghie ra ba yek rang rang konim va baghie noghat ra be delkhah angah...
 ;)  :D


logarithms
number theory solved
number theory

[hide="Hint"]Prove a lemma
For any $\varepsilon\in (0,1)$, there exists $n\in\mathbb{N}$ such that
$0<\{n\log 2004\}<\varepsilon$[/hide]

Do it like here: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=20012001&t=50019


Just suppose, for the sake of argument, that you have a large block of stone. While staring at this block of stone, it dawns on you that perhaps you should carve it into something. Luckily, you have a hammer and chisel handy.

But it's no ordinary hammer and chisel. Somehow, no matter how hard you hit the chisel, exactly one-half of one percent of the block chips off.

How many times do you have to hit the chisel before the block is less than half of its original size?

Can Someone plese help me with a solution, I just can't get it.

I got up to, solve for n in 0.995^n is equal to or less than .5, which I don't know how to solve (I used guess and check and a calculator to find the answer. :oops: ). For each hit the block loses .005 of itself and leaves .995, so you just have to find how many times .995 times itself, or .995^n, it takes to get equal to or less than .5.

Since each hit removes half of one percent of the block.

To find out how many times to hit the block, to destroy the whole block do 1/.005.

To figure out how to destroy exactly half the block do (1/.005)/2.

To get LESS then half of the block hit the block one more time, you get {[(1/.005)/2]+1}

For a grand total of 101 hits, .49 of the Block will remain

is this a correct answer?

(1/2*1% is a constant)
Assuming that we are taking that 1/2*1% of 1 (the original block)  from the original block; and not half of one percent of a progressively diminishing block.
if the later were the case it would be similar to Zeno's Paradox.

I got 139 as well, using guess and check

[quote="MichaelFaraday"][quote="236factorial"]To solve for $.995^n<0.5$, you will need logarithms to solve. I think n is something like 139.[/quote]

wouldn't it be some were in between 138 and 139.[/quote]

but we want n to be a whole number, and the inequality states the answer is greater than 138.3  something. Thus n is the next greater number, or 139.

This problem is far too difficult for the middle school classroom math forum.

Please read the posting guidelines.


function
number theory solved
number theory

Let a_1=b_1=0, b_2,... integers and a_n=nb_n+a_1*b_(n-1)+a_2*b_(n-2)+...+a_(n-1)*b_1 for any n>=2. Prove that for any prime p we have p|a_p.

I don't have a solution because I haven't tried to prove what I've observed, but I think the general form of an can be found in terms of bi with i from 2 to n and according to the partitions of n as a sum of naturals >=2.

For example (it's hard for me to write a general form because of all the nasty indexes): a8=8(b2b32+b22b4+b3b5+b2b6)+8b8+4b42+2b24.

Notice that for every monomial of the form k*bi1t1*..*birtr we have:

(1) k=n whenever r>=2;

(2) k=the index of that prticular b when r=1 (like 4b42, 2b24);

(3) \sumijtj=n, so we have a partition;

(4) every partition is represented by such a monomial.

Taking these into account, it's easy to see why p|ap, and that's because for ap we don't have monomials for which r=1, except for pbp, which is divisible with p, so the coefficient k of every monomial is, in those cases, always p.

Anyone dare try to prove it? :) Maybe it's not that hard, and maybe it's not true, but it's true for n<=8 (I've checked these values), and it really looks like something which might be true (there's lots of "evidence": it explains the result, it's sort of nice, it doesn't look "random" at all, having a certain degree of order)..

Did you try to use generating functions?
If A(x)=a1x+a2x2+... and B(x)=b1x+b2x2+... then A(x)=xB'(x)/(1-B(x))=xB'(1+B+B2+...). Maybe it can help.

I didn't try, because it's sort of a new method for me (I've never used it).

Finally, I have done it.

I will continue my previous post.
We must show that B'/(1-B)=...+p*s*xp-1+... . Intergrating we obtain that ln(1-B)=...+s*xp+..., where s is an integer.
We have ln(1-B)=B-B2/2+B3/3-..., so it is no matter coefficients of xk in B where k>p.
Since coefficient of x in B equals 0 we can find (step by step) integers c2, c3, ..., cp such that (1-B)- (1+c2x2)(1+c3x3)...(1+cpxp) = xp+1(...).
Hence ln(1-B) has the same coefficient of xp as ln((1+c2x2)(1+c3x3)...(1+cpxp)) = ln(1+c2x2) + ln(1+c3x3) + ... + ln(1+cpxp). But if i<p then ln(1+cixi) doesn't contain term xp (since p is a prime) and ln(1+cpxp) = cpxp + x2p(...). It follows that ln(1-B) = ... + cpxp + ...
That's all.


AMC AIME geometry function perimeter circumcircle trigonometry

[b]This contest has officially ended.[/b] The rules are the same as usual: No calculators, three hours, unlimited scratch paper. All answers are integers. The problems can be found at [url]http://www.tjhsst.edu/~tmildorf/math/PracticeAIME3.pdf[/url], but they are reproduced here for your convenience.

Good luck to all!

$1. $ Three circles are mutually externally tangent. Two of the circles have radii 3 and 7. If the area of the triangle formed by connecting their centers is 84, then the area of the third circle is $k\pi$ for some integer $k$. Determine $k$.

$2.$ Let $N$ denote the number of 7 digit positive integers have the property that their digits are in increasing order. Determine the remainder obtained when $N$ is divided by 1000. (Repeated digits are allowed.)

$3.$ A function $f(x)$ is defined for all real numbers $x$. For all non-zero values $x$, we have

\[
2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4
\]
Let $S$ denote the sum of all of the values of $x$ for which $f(x) = 2004$. Compute the integer nearest to $S$.

$4.$ $\zeta_1, \zeta_2$, and $\zeta_3$ are complex numbers such that

\[\zeta_1\phantom{1} + \zeta_2\phantom{1} + \zeta_3\phantom{1} = 1\]
\[\zeta_1^{\phantom{1}2} + \zeta_2^{\phantom{1}2} + \zeta_3^{\phantom{1}2} = 3\]
\[\zeta_1^{\phantom{1}3} + \zeta_2^{\phantom{1}3} + \zeta_3^{\phantom{1}3} = 7\]

Compute $\zeta_1^{\phantom{1}7} + \zeta_2^{\phantom{1}7} + \zeta_3^{\phantom{1}7}$.

$5.$ In Zuminglish, all words consist only of the letters M, O, and P. As in English, O is said to be a vowel and M and P are consonants. A string of M's, O's, and P's is a word in Zuminglish if and only if between any two O's there appear at least two consonants. Let $N$ denote the number of 10-letter Zuminglish words. Determine the remainder obtained when $N$ is divided by 1000.

$6.$ Let $S$ denote the value of the sum
\[
\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}
\]
$S$ can be expressed as $p + q \sqrt{r}$, where $p, q$, and $r$ are positive integers and $r$ is not divisible by the square of any prime. Determine $p + q + r$.

$7.$ $ABCD$ is a cyclic quadrilateral that has an inscribed circle. The diagonals of $ABCD$ intersect at $P$. If $AB = 1$, $CD = 4$, and $BP : DP = 3 : 8$, then the area of the inscribed circle of $ABCD$ can be expressed as $\frac{p\pi}{q}$, where $p$ and $q$ are relatively prime positive integers. Determine $p + q$.

$8.$ Let $N$ denote the number of 8-tuples $(a_1, a_2, \dots, a_8)$ of real numbers such that $a_1 = 10$ and

\[\left|a_1^{\phantom{1}2} - a_2^{\phantom{1}2}\right| = 10\]
\[\left|a_2^{\phantom{1}2} - a_3^{\phantom{1}2}\right| = 20\]
\[ \cdots \]
\[\left|a_7^{\phantom{1}2} - a_8^{\phantom{1}2}\right| = 70\]
\[\left|a_8^{\phantom{1}2} - a_1^{\phantom{1}2}\right| = 80\]

Determine the remainder obtained when $N$ is divided by 1000.

$9.$ $ABC$ is an isosceles triangle with base $\overline{AB}$. $D$ is a point on $\overline{AC}$ and $E$ is the point on the extension of $\overline{BD}$ past $D$ such that $\angle{BAE}$ is right. If $BD = 15, DE = 2$, and $BC = 16$, then $CD$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine $m + n$.

$10.$ $\{A_n\}_{n \ge 1}$ is a sequence of positive integers such that
\[
a_{n} = 2a_{n-1} + n^2
\]
for all integers $n > 1$. Compute the remainder obtained when $a_{2004}$ is divided by 1000 if $a_1 = 1$.

$11.$ $ABC$ is an acute triangle with perimeter 60. $D$ is a point on $\overline{BC}$. The circumcircles of triangles $ABD$ and $ADC$ intersect $\overline{AC}$ and $\overline{AB}$ at $E$ and $F$ respectively such that $DE = 8$ and $DF = 7$. If $\angle{EBC} \cong \angle{BCF}$, then the value of $\frac{AE}{AF}$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.

$12.$ Determine the number of integers $n$ such that $1 \le n \le 1000$ and $n^{12} - 1$ is divisible by 73.

$13.$ Let $S$ denote the value of the sum
\[
\left(\frac{2}{3}\right)^{2005} \cdot \sum_{k=1}^{2005} \frac{k^2}{2^k} \cdot {2005 \choose k}
\]
Determine the remainder obtained when $S$ is divided by 1000.

$14.$ Circles $\omega_1$ and $\omega_2$ are centered on opposite sides of line $l$, and are both tangent to $l$ at $P$. $\omega_3$ passes through $P$, intersecting $l$ again at $Q$. Let $A$ and $B$ be the intersections of $\omega_1$ and $\omega_3$, and $\omega_2$ and $\omega_3$ respectively. $AP$ and $BP$ are extended past $P$ and intersect $\omega_2$ and $\omega_1$ at $C$ and $D$ respectively. If $AD = 3, AP = 6, DP = 4$, and $PQ = 32$, then the area of triangle $PBC$ can be expressed as $\frac{p\sqrt{q}}{r}$, where $p$, $q$, and $r$ are positive integers such that $p$ and $r$ are coprime and $q$ is not divisible by the square of any prime. Determine $p + q + r$.

$15.$ Let $\Omega$ denote the value of the sum
\[
\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)
\]
The value of $\tan\left(\Omega\right)$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.

Good call! Those times are U.S. Eastern times. It will be 2 - 5 PM ET, 1 - 4 PM CT, 12 - 3 PM MT, and 11 AM - 2 PM PT.

:? How exactly does this work do we just post our answers here?

Yes, numbers with repeated digits such as 1111112 are to be counted.

Darn.  Why did I think this was tomorrow?

I haven't been able to use the computer all day; can I start now and send in my answers late?

Sure. I am writing up the solutions packet right now (halfway through #12), so I don't have 100% confidence in my key just yet...

i'm in the same situation DTS is in

i started appx 30 minutes before 5, so i'll have it to you by 7:15, 30 ish?

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3742661_2004_mildorf_mock_aime_35]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


umm i dont see how you get this answer...hints or full solutions much appreciated!

[hide="answer"]12???[/hide]

[hide]Just list them all out, keeping in mind that a + or a X MUST go in the middle.

+++
XXX
X+X
+X+
/+\
/X\
\X/
\+/
X++
++X
+XX
XX+[/hide]

oh ok...i misread and i thought that it said that there cant be more that one line of symetry!!!NVM!!!

um another(not a marathon)

#26 from sprint 02--What is the remainder when $ 2^{87}\plus{}3$ is divided by $ 7$?

and #27 and #28 you can see [url]http://www.yashar.biz/mathcounts[/url] for these

Thanks soo much!

[hide="#26"]Note $ 2^6$ has remainder 1 when divided by 7, so this reduces to finding the remainder when $ 2^3\plus{}3$ is divided by 7, so 4.[/hide]

sorry im pretty dumb but how do you use 2^6 having a remainder of 1 to find that $ 2^{87}\plus{}3$ will have the samer remainder as $ 2^3\plus{}3$???Again i apologize for being dumb!


Is it true that AIME I is easier than AIME II?

No.

When the AIME Committee creates the AIME (or AIME I) and the AIME  Alternate (or AIME II), two contests which are referred to as AIME Heads and AIME Tails are deliberately constructed to be balanced, in the experienced judgement of the committee, in content and difficulty.  Then one of AIME Heads and AIME Tails is chosen by coin flip to the first, or AIME I, and the other automatically becomes the second AIME Alternate of AIME II.

Steve Dunbar
Direcotr, American Mathematics Competitions

[quote="AMCDirector"]
stuff
[/quote]
Who flips the coin?


I do Mu Alpha Theta and I seriousely need some tests so I can study. I know there is some in their sites, but I need a LOT more than that.

If you could post a link or a zip of tests, that would be more than great and I would be SO happy.

Thanks

I believe that they are copyrighted, and posting them here would violate that copyright, because Mu Alpha Theta never released them to the public.  Not sure, but that is what I think.

what r the phi mu epsilon tests? r they olympiads?


How many ways are there to express 1 000000 as a product of exactly three
integers greater than 1?

[hide="Solution"]$ abc \equal{} 1000000 \equal{} 10^6 \equal{} 2^6 \times 5^6$. Therefore, each of $ a$, $ b$, and $ c$ can be written in terms of powers of $ 2$ and $ 5$. Let $ x$ be the power of $ 2$ in $ a$, $ y$ for $ b$, and $ z$ for $ c$. Therefore, $ x\plus{}y\plus{}z \equal{} 6$. There are $ \binom{8}{2} \equal{} 28$ nonnegative integer solutions. We square $ 28$ to account for the power of $ 5$. The answer: $ 28^2 \equal{} \boxed{784}$.

[/hide]

[quote="darkdieuguerre"][hide="Solution"]$ abc \equal{} 1000000 \equal{} 10^6 \equal{} 2^6 \times 5^6$. Therefore, each of $ a$, $ b$, and $ c$ can be written in terms of powers of $ 2$ and $ 5$. Let $ x$ be the power of $ 2$ in $ a$, $ y$ for $ b$, and $ z$ for $ c$. Therefore, $ x \plus{} y \plus{} z \equal{} 6$. There are $ \binom{8}{2} \equal{} 28$ nonnegative integer solutions. We square $ 28$ to account for the power of $ 5$. The answer: $ 28^2 \equal{} \boxed{784}$.

[/hide][/quote]
no answer is 114

still waiting for some great idea to spring up?????????????

[hide="Attempt"]
To prevent overlaps due to permutations of the same 3 integers, we first say that the exponents of 2 must be in increasing order. Then there are only 7 possibilities: 006,015,024,033,114,123,222.
Then the exponents of 5 can be chosen in 28 ways, as darkdieuguerre explained. So far we have 7 x 28 = 194 (EDIT: it's 196!  :oops: ) ways.

However, the exponents of 2 and 5 can't be both 0 for the same number, since the problem says the numbers must be greater than 1. There are three cases: the first number gets both 0, the second number gets both 0, or both the first and the second get both 0.

The first number gets both 0 in 4 x 7 ways. (006, 015, 024 all have 2 permutations that begin with 0; 033 has 1.) The second number gets both 0 in 1 x 7 ways. Both get 0 only when both choices of exponents are 006. So we eliminate 28 + 7 - 1 ways, or 34. 162 ways left.

Now, even though we forced the exponents of 2 to be in increasing order, there are still some overlaps: for example, 222 and 033, 222 and 330. Here we split the 7 possibilities into three categories: all Distinct, one Pair, and all Same.

As long as the exponents of 2 are distinct, no problems occur. If the exponents of 2 are one Pair:
2P, 5D (9): 3 overlaps, 27 total. e.g. 033,123; 033,132
2P, 5P (9): 1 overlap, 9 total. e.g. 033,114; 033,141
2P, 5S: No problem.
Exponents of 2 all same:
2S, 5D (3): 5 overlaps, 15 total. e.g. 222,123; 222,132
2S, 5P (3): 2 overlaps, 6 total. e.g. 222,033; 222,303
2S, 5S: No problem.

162 - 27 - 9 - 15 - 6 = 105.
The final step is to compensate for more overlap in elimination. For instance, 033,006 and 033,060 were eliminated 3 times in all between them.
033,(015 or 024) is eliminated once extra: 033,015 and 033,051 are eliminated 3 total.
006,(015 or 024) is eliminated twice extra: 006,015 and 006,105 and 006,051 and 006,501 are eliminated 6 total.
033,033 is not eliminated extra.
033,006 is eliminated once extra: 033,006 and 033,060 are eliminated 3 total.
006,033 is eliminated once extra: 006,033 and 006,303 are eliminated 3 total.
006,006 is eliminated once extra: 006,600 and 006,060 are eliminated 3 total.

105 + 2 + 4 + 1 + 1 + 1 = 114.
[/hide]
Stupid arithmetic error fixed.
[hide="All 114 solutions"]
006,114; 006,123; 006,132; 006,141; 006,150
006,222; 006,231; 006,240; 006,330; 015,105
015,114; 015,123; 015,132; 015,141; 015,150
015,204; 015,213; 015,222; 015,231; 015,240
015,303; 015,312; 015,321; 015,330; 015,402
015,411; 015,420; 015,501; 015,510; 015,600
024,105; 024,114; 024,123; 024,132; 024,141
024,150; 024,204; 024,213; 024,222; 024,231
024,240; 024,303; 024,312; 024,321; 024,330
024,402; 024,411; 024,420; 024,501; 024,510
024,600; 033,105; 033,114; 033,123; 033,204
033,213; 033,222; 033,303; 033,312; 033,402
033,411; 033,501; 033,600; 114,006; 114,015
114,024; 114,033; 114,042; 114,051; 114,060
114,114; 114,123; 114,132; 114,141; 114,150
114,222; 114,231; 114,240; 114,330; 123,006
123,015; 123,024; 123,033; 123,042; 123,051
123,060; 123,105; 123,114; 123,123; 123,132
123,141; 123,150; 123,204; 123,213; 123,222
123,231; 123,240; 123,303; 123,312; 123,321
123,330; 123,402; 123,411; 123,420; 123,501
123,510; 123,600; 222,006; 222,015; 222,024
222,033; 222,114; 222,123; 222,222
[/hide]


superior algebra
superior algebra theorems

Are there any online lecture notes avaliable on "Free Groups"?

Thanks

I found the lecture notes at http://www.math.berkeley.edu/~gbergman/245/ to be very enlightening. It includes free groups as well as a whole bunch of information about universal constructions.


can someone solve and explain step by step this problem?

"How many sides does a regular polygon with interior angle measure of 168 degrees have?"

[hide="Solution"]The number of degrees in each angle of a regular $ n$-gon is $ \tfrac {180(n - 2)}{n}$. Set this expression equal to $ 168$ and solve:

\begin{eqnarray*} 168n & = & 180(n - 2) \\
168n & = & 180n - 360 \\
12n & = & 360 \\
n & = & \boxed{30} \end{eqnarray*}[/hide]

i like the take the measure of the exterior angle and divide it into 360. i think its easier. 30

[hide]Yeah, since the exterior angles are supplementary to the interior angles, each exterior angle is 12 degrees, in this case. Since the sum is 360, there are 30 sides.[/hide]


Solve the inequation:  $2\cdot 9^{\log_{2}\frac{x}{2}}\ge x^{\log_{2}6}-x^{2}$

[hide="I have reduced it to..."]
\[\frac29 \geq x^{2}\cdot \left( x^{\log_{2}\frac{3}{2}}-\left( x^{\log_{2}\frac{3}{2}}\right)^{2}-1 \right) \] but I don't know how to solve from here. :(  [/hide]

[hide="Hint"]
$1.\ a^{\log_{b}c}=c^{\log_{b}a}$.

$2.$ Substitution $y=\log_{2}x$.

$3.$ Substitution $z=\left(\frac{3}{2}\right)^{y}$.

$\Longrightarrow x\in (0\ ,\ 2]\cup\left[2^{\log_{\frac{3}{2}}3}\ ,\ \infty\right)$.[/hide]


geometry
incenter
circumcircle
vector
symmetry
geometric transformation
homothety

[quote="Omid Hatami"]$H,I,O,N$ are orthogonal center, incenter, circumcenter, and Nagelian point of triangle $ABC$. $I_{a},I_{b},I_{c}$ are excenters of $ABC$ corresponding vertices $A,B,C$. $S$ is point that $O$ is midpoint of $HS$. Prove that centroid of triangles $I_{a}I_{b}I_{c}$ and $SIN$ concide.[/quote]

A cute little problem! Here is my solution to it.

Define $M$ to be the centroid of $ABC$, $L$ to be the centroid of $I_{a}I_{b}I_{c}$ and $P$ be such point that $O$ is the midpoint of a line segment $IP$. Because $I$ is an orthocenter and $O$ - a nine-point center in $I_{a}I_{b}I_{c}$ we infer that $P$ is a circumcenter of $I_{a}I_{b}I_{c}$ and $\vec{IL}=2\vec{LP}$. Thus note that we are done if we showed that $\vec{SP}=\vec{PN}$. But $\vec{SP}=\vec{IH}$ because of the central symmetry at $O$. Thus it remains to show that $\vec{PN}=\vec{IH}$. But it is well known that $\vec{NM}=2\vec{MI}$ and $\vec{HM}=2\vec{MO}$, thus having considered a homothety centered at $M$ with coefficient $-1/2$ we see that $\vec{HN}=2\vec{IO}=\vec{IP}$ (the last one follows from the definition of $P$). We deduce that $HNPI$ is a parallelogram and thus $\vec{PN}=\vec{IH}$. Hence the result.

What's the point of your post, arline (besides showing that you know barycentic coordinates and vectors)? It's absolutely useless.

xixas,its straighforward computation using barycantres and or vectors.its true remark.and somewhat general method.

[quote="arline"]vectors and or barycentres[/quote]

barycentric give us a solution much simpler

Is there an elementary solution without vectors or barycentric co-ordinates?

Thank you.

Since $N,H$ are the incenter and circumcenter of the antimedial triangle of $\triangle ABC,$ it follows that $IO \parallel HN$ and $IO=\frac{_1}{^2}HN.$ Therefore, reflection $B_e$ of $I$ about $O$ (Bevan point of ABC) is the midpoint of $\overline{NS}$ $(\star).$ $B_e$ is the circumcenter of $\triangle I_aI_bI_c$ since $B_e$ is the concurrency point of  the perpendiculars from $I_a,I_b,I_c$ to the sides $BC,CA,AB$  of its orthic triangle $\triangle ABC.$ Thus, the centroid $G_0$ of $\triangle I_aI_bI_c$ lies on the segment connecting its circumcenter $B_e$ and orthocenter $I$ such that $\overline{G_0B_e}:\overline{G_0I}=-1:2.$ Together with $(\star),$ we conclude that $G_0$ is the centroid of $\triangle SIN$ as well.

I understand that you find your solution's arguments a little trivial, but can you please help me by elaborating more? :(

Why is the Nagel point the incentre of the anti-medial triangle?
and Why does it being so, and H being the circumcentre result in the parallelism with IO?

and so much more of what you have said..

P.S. I have no doubts that your solution is right, It is still too complicated for me. :(

[quote="Ramchandran"]I understand that you find your solution's arguments a little trivial, but can you please help me by elaborating more? :(

Why is the Nagel point the incentre of the anti-medial triangle?
and Why does it being so, and H being the circumcentre result in the parallelism with IO?

and so much more of what you have said..
[/quote]

I find this post a bit unnecessary. Why not try it for yourselves? It could've been a good challenge :) Anyways also, instead of posting it here, you could've easily got the answers in a second by searching AoPS or Mathworld. But well,
[quote="Ramchandran"]
Why is the Nagel point the incentre of the anti-medial triangle?
[/quote]
The answer that comes most quickly to me is just a homothety centred about $G$, the centroid of $\triangle ABC$, and ratio $-2$. This takes $\triangle ABC$ to its anti-median triangle. Also, by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1838509#p1838509]this[/url], it takes the Incentre of $\triangle ABC$ to the Nagel Point of $\triangle ABC$. Since it must take the Incentre of $\triangle ABC$ to the Incentre of its Anti-Medial Triangle, the result follows.

[quote="Ramchandran"]Why does it being so, and H being the circumcentre result in the parallelism with IO?
[/quote]
Think about that homothety discussed above.

[quote="Ramchandran"]
And so much more of what you have said..
[/quote]
What else?

Let $IO\cap SN={L}$, let $Be$,$Z$ be circumcenter and centroid of $\triangle I_{A}I_{B}I_{C}$.

Homothety $\mathcal{H}_{G,-\frac{1}{2}}$ takes $I$$\rightarrow$$N$,(well-known) and hence $2IG=GN$.From the Menelaus on the $\triangle SNG$ and the transversal $IO$ we get :
$$\frac{NI}{IG}\cdot \frac{GO}{OS}\cdot \frac{SL}{LN}=-1$$
Along with the $\frac{NI}{IG}=-3$,$\frac{GO}{OS}=\frac{1}{3}$ and hence $SL=LN$ and so the centroid of $\triangle SIN$ lies $OI$,and so it suffices to prove that $SZ$ bisects $IN$.From Menalaus on $\triangle NIL$ and transversal $GO$ we have $IO=OL$ and so $Be\equiv L$.From
$\triangle I_{A}I_{B}I_{C}$ we have $\frac{BeZ}{IZ}=\frac{1}{2}$ so we are done.$\clubsuit$

As i remember,[quote=Omid Hatami]$H,I,O,N$ are orthogonal center, incenter, circumcenter, and Nagelian point of triangle $ABC$. $I_{a},I_{b},I_{c}$ are excenters of $ABC$ corresponding vertices $A,B,C$. $S$ is point that $O$ is midpoint of $HS$. Prove that centroid of triangles $I_{a}I_{b}I_{c}$ and $SIN$ concide.[/quote]

it's a property of De Longchamps point.


Let a Set $X$ of $2003$ points in the plane, and a unit circle be given. Prove that there is a point on the unit circle such that the sum of the distances from it to the $2003$ points is at least $2003$.

[quote="andyciup"]Just a small question:Are the 2003 points all on the unit circle?[/quote]
All the $2003$ points can be anywhere on the plane. They could be all inside the circle or all outside it. Or, some could be inside and the rest outside the circle. I think this is exactly what the problem implies.

Hey, that's kind of cute!

[hide="Solution"]With vectors.  Let the 2003 points be $A_{1}, A_{2}, \ldots, A_{2003}$.  Choose a point $A$ on the circle at let $B$ be the antipodal point of the circle.  Then:
$4006 = 2003|AB| = |(AA_{1}+A_{1}B)+(AA_{2}+A_{2}B)+\ldots+(AA_{2003}+A_{2003}B)| \leq \left(|AA_{1}|+\ldots+|AA_{2003}|\right)+\left(|BA_{1}|+\ldots+|BA_{2003}|\right)$.  Thus, the sum of (the sum of the distances from $A$ to the members of $X$) and (the sum of the distances from $B$ to the members of $X$) is 4006, so at least one summand is 2003, and we're done.  This in fact shows that "most of" the circle must have the desired property. It also shows that the restriction to the plane was arbitrary: this works for $k$ points with an $m$-dimensional sphere in $n$-dimensional space for any $k$, $m \leq n$.  Although generalizing the dimension of the sphere upwards isn't very interesting (since the proof relied only on a sphere in 1-D space, that is two points, and every lower-dimensional sphere is contained in every higher-dimensional sphere).[/hide]

[b]JBL[/b], thanks for your nice solution as well as your enlightening comment on the problem.


Prove that the number $ 1\underbrace{00...0}_{2006}1$ has at least 5 prime divisors distincts .


The polynomial $ P(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $ y$-intercept of the graph of $ y \equal{} P(x)$ is 2, what is $ b$?

$ \textbf{(A)} \ \minus{} 11 \qquad \textbf{(B)} \ \minus{} 10 \qquad \textbf{(C)} \ \minus{} 9 \qquad \textbf{(D)} \ 1 \qquad \textbf{(E)} \ 5$

[hide]
answer's A....this is quite simple, even pre-calc students can do it....

since $a$ is the opposite of the sum of the roots, and $c$ the negative of the product and the y-intercept, we can see that $c = 2$, and the product of the roots is $-2$. we know that $\frac{-a}{3}$ gives the average since there are 3 roots, so $a = 6$. sum of the coefficients equals $-2$, so $1 + 6 + b + 2 = -2$. substracting gives $b = -11$.
[/hide]

Hide please!

By the way, 1234567890, does $\frac{a}{3}$ gives you the sum of the roots for cubic? Don't you think you're missing something VERY important?

sorry, i forgot that i need to hide :blush:  :blush:

Sorry, Silverfalcon, forgot the negative. I'm being stupid today.  :lol:

Since P(0)=2 we have c=2, makeing $P(x)=x^{3}+ax^{2}+bx+2$. The product of the roots is -c=-2. Thus the mean of the roots and the sum of the coefficients are both -2. The mean = sum of roots/3 = -a/3=-2, so a=6. The sum of the coefficents is -2, so $b=-11$. The answer is $A$.

Seriously, how hard is it to hide your answer? :dry:

[hide="Answer"]The $y$-intercept is at $x=0$, so we have $c=2$, meaning that the product of the roots is $-2$. We know that $a$ is the sum of the roots. The average of the roots is equal to the product, so the sum of the roots is $-6$, and $a=6$. Finally, $1+a+b+c=2$ as well, so we have $1+6+b+2=-2\Rightarrow b=-11\Rightarrow \boxed{A}$.[/hide]

[quote="surge"]I thought the product of roots was $c/a$. Or was it $(-1)^{i}(c/a)$?

Edit: Got it.[/quote]

It's $\frac{c}{a}$ when you have $ax^{2}+bx+c=0$.  But when you have a polynomial of a higher degree, generally, $a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{2}x^{2}+a_{1}x+a_{0}$, the product of the roots is $(-1)^{n}\frac{a_{0}}{a_{n}}$.  Correct me if I'm wrong.

[quote="mathnerd314"][quote="Silverfalcon"]The polynomial $P(x) = x^{3}+ax^{2}+bx+c$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $y$-intercept of the graph of $y = P(x)$ is 2, what is $b$?

$\text{(A)}\-11 \qquad \text{(B)}\-10 \qquad \text{(C)}\-9 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 5$[/quote]

[hide]We know that $c=2$.

"Hopefully" the polynomial is nice and has three zeros. Then:

The mean is $\frac{-a}{3}$
The product of roots is $c=2$
Therefore $a=-6$

Then $1-6+b+2=2$, $-3+b=2$, $b=5$

The answer is E.[/hide]

Edit: D'oh, I messed up, it's

[hide]The product of roots is $c=-2$ because I forgot the negative. Thus $a=6$ and:

$1+6+b+2=-2$ and $b=-11$

Ha ha ha...[/hide][/quote]
The product of the roots is $-c=-2$


Bubka, what has happened? Does this mean you are giving up your life of spam? You are stepping down from the throne of the Spam King?  :rotfl:

Well, the age of 3s has probably gone. This is gonna be like...the dark age of spam or something.  :rotfl:

There are like 3 topics on this and I don't think G&FF needs anymore. *Hopefully* we have learned our lesson.

In my opinion the most important thing that keeps a topic away from spam is if [b]everyone posts things that are directly realted to the topic[/b]
Also, spam should be ignored in my opinion, because if a certain person wants to spam, he/she will continue to spam until he is forcefully stopped(ie he/she is banned, or the topic gets locked). Therefore posting things like "please stop spamming", etc., will only make the topic worse. Therefore people should just continue their discussion ignoring spam posts. :!:
Also, if people do that, then it will be very easy for moderators to see where there is spam and where there is no spam. Then only certain posts will need to be deleted, and the topic will not get deleted(because some good topics can get deleted just because ppl start spamming too much and no more meaningful things are posted).

I agree, but I generally take a broader view of "directly related."

Anyway, the policy of ignoring spam does not always work. For example, ToW 4 has survived arnav's '3's, but one letter word story or conversation fell to spam.

[quote="bubka"]Here is what spam really means - posting off topic things in a topic that [i]does not maintain the logical sequence of things.[/i] That is the very essence of spam. That is why even a post like

rotfl

or

;)

may not count as spam if it neatly falls into place in the ongoing discussion. While a solution to an IMO problem, though apparently very useful, may count as serious spam.

No topic can ever be spammy, by defintion of spam.

As for the definitons given by MithsApprentice and DPatrick ... well, they are important people here, so I won't say anything ... but it is obvious that they don't understand spam.[/quote]

the dark ages of spam indeed...but plans are already afoot

and bubka you backseatmoderate way too much stop all this nonsense

What?

You were the one who lost like 2000 posts in you post count, cause they were in fun and games.  And now you're acting like a moderator.

And? I lost about 100 posts.

What you say doesn't mean that bubka did spam, or maybe he just realized his mistakes

Are fishes the same as turkeys? Anyway, I don't see how spam really affects the forum.. I mean if you're gonna get rid of spam, you might as well get rid of this "tug of war" stuff... I mean that's junk junk junk junk whoa that's pretty fun to say.

[quote="José"][quote="probability1.01"]I don't see how spam really affects the forum.[/quote]

Man are you talking seroiusly? Of course spam is bad for the hole forum-it's also annoying[/quote]

He means the just Fun and Games Factory.

Stuff and nonsense. I am not saying I realized my mistake or anything. I am standing up against the clasical definiton of spam. According to my definition my posts are mostly nonspam (though not all).

I dislike breaking the logical thread of any discussion. I also dislike disappointing anybody in help. These are complaints no one can make against me.

And I maybe wrong, but it also did not seem to me that the users of the games and fun factory in general hate me to much. My impression was rather along the opposite lines.

Also, 'related to the topic' is not exactly what I mean. The topic may change in the course of the discussion. [i]As long as the creator of the thread does not object, such discussions will not fall under spam.[/i] But I also believe that the opinion of the creator is important in deciding whether an off-topic discussion is spam or not, because he may have started the thread with specific aims in mind. However, after he requests to come back to course, if somebody continues the off-topic discussions, then it is serious spam and should be reported.

[quote="bubka"]According to my definition my posts are mostly nonspam [/quote]

What about this post of you? I found a lot like this one:

[quote="bubka"]$14$

My fourth consecutive 14[/quote]

According to you, is this SPAM or not?
-------------------------------------------------------------
[quote="bubka"]I dislike breaking the logical thread of any discussion. I also dislike disappointing anybody in help. These are complaints no one can make against me[/quote]

Ohhh well... This is the same like if I say: "I find killing a very pleassure and relaxing activity. Nobody can complain me what I like". Man, I mean, I don't like to enter to GFF and see AoPS Tug of War 4, which has 130 pages in a few days. Is that HUGE SPAM or not?????? I find it really annoying.
For example, this are posts from there:

[quote="ShoeFactory"]-4

The message is too small. Please make the message longer before submitting.[/quote]

[quote="shadysaysurspammed"]-3[/quote]

Please, if this are not SPAMS...

I mean, why AoPS Tug of War 4 is not just deleted???

These are not SPAMs because they are logically connected to the topic.

However posts like:

[quote="MysticTermainator"]3

__________________
I am THE BOB
[/quote]

are SPAM. Get it now?

They are logically connected to a game, yes, but to a nonsense game, which has to be deleted

You create your topics. People respond to them. I create mine. People respond to them too. I never asked you to join it. Just its existence can't irritate you that much. There are people who like it. It is onyl fair that it should stay on.

[quote="José"][quote="bubka"]What is your objection to nonsense games? This is a serious point that needs to be cleared.[/quote]

I think we have to be fair in all the Forum (in MathLinks I'm saying). For example, the topic "Girlfriends" was removed to GFF... why??? nobody knows, maybe because it was not serious... and see this Tug of War: I mean, can you give me a reason not to delete it? It's nonsense. Funny?? A post with only one number is funny? Maybe one or two pages, but not 130!!. I mean, all over MathLinks, we are all watching out with what we write, because it'd be considered spam, and look GFF... full of spams; now, the account doesn't care anymore, because the serious discussions here doesn't exist anymore, because every topic is spammed. So, GFF has became useless in the last time... I mean, having a forum which posts are only one number? I don't see the reason.[/quote]
Ok, just because you object to something doesn't mean you have to ruin others fun. I don't like the TOW much, but at least I just ignore it. No posts count in G&FF anyways. Just let it go.

ok, here's something I just wanted to say.

Jose, you are one of the biggest spammers on this site (actually, your spam is even worse since it's OUTSIDE of G&FF). I just find this thread hilarious to see two of the biggest spammers on this site fighting over what is spam and what isn't. Keep fighting, it's funny as heck. :rotfl:

[quote="Xantos C. Guin"][quote="José"]Man, I mean, I don't like to enter to GFF and see AoPS Tug of War 4, which has 130 pages in a few days. Is that HUGE SPAM or not?????? I find it really annoying.[/quote]
It took about a month to reach 130 pages, not "a few days". Also all the "HUGE SPAM" is contained in one topic. If you hate TOW, then just ignore that topic.

[quote="José"]I mean, why AoPS Tug of War 4 is not just deleted???[/quote]
Unlike other games which got off-topic due to spam, AoPS Tug of War 4 did not get off the topic of posting a number that is one more or less than the previous number. This is probably why it didn't get deleted or locked.[/quote]

In fact, it's taken the topic FIFTY days to get 130 pages... And note that all the rest of the pointless spamming has gone down the past few days, so like Xantos said, you don't have to look at one topic of spam as opposed to every topic being spammed...

Also, I note that while you [i]do[/i] post in the higher math forums, you have a tendency to ANSWER QUESTIONS in classroom math... why so? Isn't a person far too advanced for that level answering questions there just spamming?

There must be some people who answer the questions. I will support Jose in those matters. Only I don't like his no-tolerance family.

I have seen him for some days, and he did not look a spammer to me. A bit like fredbel6, if you ask me, only less informed.


I don't solve this , having only one solution using complex numbers at most .  :(
It's too hard to use Latex , too . :maybe:

QK/PM=PQ/QM   so   PM/QM=QK/PQ
QK/KP=PQ/LQ    so   QK/PQ=KP/LQ

PM/QM=KP/LQ   angle LQM=angle MPK
so  MPK  is similar to MQL
MK/ML=PM/MQ   and  angle PMQ=angle KML

Thus we derive the conclusion

Thanks  again though having sent thanks to you  :)
It seems very easy for you , but not me  :blush:


Just finished learning those funtions on wiki but when i try to solve problems on the i can't can anybody post a problem on this topic with solution plz

There was a question on this years AIME asking how many integers $ 1 \leq C \leq 1000$ satisfy:

$ x^{\lfloor{x}\rfloor} \equal{} C$ if x can be any real number?

For example, $ 1^{\lfloor{1}\rfloor} \equal{} 1,$ so $ C \equal{} 1$ works. However, $ C \equal{} 2$ is impossible to get.


I was wondering how i can change the margins since my documents seem to take up a lot of space because there's so much space left over on the left and right side. Can someone help me with this? And one more question. How can I make a space between lines? like if i want to do:

A new

line

Like

this

1. The geometry package makes margins easy. For example:
\usepackage[margin=1in]{geometry}
puts a margin of 1in all the way round.
You can make the various margins different and even mix measurements like this:
\usepackage[top=2.9cm, left=2in,bottom=1in,right=1in]{geometry}

2. There's all sorts of things you can do between lines such as:
\bigskip \\ or \medskip \\
If the spacing is not right for you try using \\*[2ex] or \vspace{2ex} \\
2ex (= 2 times the height of the letter x) can be altered and you can use other units such as mm and pt

$\\ line 1 \bigskip \\ line 2 \medskip\\ line 3 \\*[2ex] line 4 \vspace{2ex}\\ line 5$


evaluate $\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}$

Note from Silverfalcon: $\LaTeX$ has been added.

Erm, I forgot to square one term in my solution...lol, it's the third term in the expansion, but the answer is still the same.

[quote="vishalarul"]Um, am I mistaken or is $2-\sqrt{5}<0$? If so, then the answer is one of the roots of $a^{2}+a+4$.

It can't be $1$, since $1$ is real.[/quote]

It's cube root, though, not square root. Cube roots of negative numbers are still real. :wink:

As for the $a^{2}+a+4$ part, those correspond to $\omega \sqrt[3]{2+\sqrt{5}}+\omega^{2}\sqrt[3]{2-\sqrt{5}}$ and $\omega^{2}\sqrt[3]{2+\sqrt{5}}+\omega \sqrt[3]{2-\sqrt{5}}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$. These numbers are also of the form $x+y$ where $x^{3}+y^{3}=4$ and $xy=-1$; we need the fact that the cube roots are real to rule them out.

Hint for a slightly different solution:

We can also make use of the theorem:

$a+b+c=0\implies a^{3}+b^{3}+c^{3}=3abc$.

[what do we choose for $a$, $b$, and $c$???]

[hide="Obviously all these methods are essentially the same but here is althemans method solution"]
Consider \[(\sqrt[3]{2-\sqrt{5}})^{3}+(\sqrt[3]{2+\sqrt{5}})^{3}+(-1)^{3}=2+2-1=3=3(-1)(2^{2}-5)=3(-1)(2-\sqrt{5})(2+\sqrt{5})\]
with all of them real; thus, (after noting altheman's implication isn't quite enough) we must factor $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=0$ to determine that since $a\not =b\not =c$ and all are real the second factor is nonzero (from $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ for all reals) so $a+b+c=0$ which means we have proven it.
[/hide]

[quote="me@home"][hide="Obviously all these methods are essentially the same but here is althemans method solution"]
Consider
\[(\sqrt[3]{2-\sqrt{5}})^{3}+(\sqrt[3]{2+\sqrt{5}})^{3}+(-1)^{3}=2+2-1=3=3(-1)(2^{2}-5)=3(-1)(2-\sqrt{5})(2+\sqrt{5}) \]
with all of them real; thus, (after noting altheman's implication isn't quite enough) we must factor $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=0$ to determine that since $a\not =b\not =c$ and all are real the second factor is nonzero (from $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ for all reals) so $a+b+c=0$ which means we have proven it.
[/hide][/quote]

It is a convention that the cube root has only one output; therefore there is one value that we seek. Therefore, it is sufficient to show that 1 is a solution. So my implication is enough.

[quote="Zellex"][quote="me@home"][hide="Obviously all these methods are essentially the same but here is althemans method solution"]
Consider
\[(\sqrt[3]{2-\sqrt{5}})^{3}+(\sqrt[3]{2+\sqrt{5}})^{3}+(-1)^{3}=2+2-1=3=3(-1)(2^{2}-5)=3(-1)(2-\sqrt{5})(2+\sqrt{5}) \]
with all of them real; thus, (after noting altheman's implication isn't quite enough) we must factor $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=0$ to determine that since $a\not =b\not =c$ and all are real the second factor is nonzero (from $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ for all reals) so $a+b+c=0$ which means we have proven it.
[/hide][/quote]

erm, I don't quite understand what you're doing, could you elaborate more..? so where do you get the answer?[/quote]

If you have that $a+b+c=0,$ then you know that $a^{3}+b^{3}+c^{3}=3abc$

So let $a=\sqrt[3]{2-\sqrt{5}}$ and $b=\sqrt[3]{2+\sqrt{5}}$

We want to find a c such that $a+b+c=0,$ so then we can plug into the second equation...So we have $2-\sqrt{5}+2+\sqrt{5}+c^{3}=3(\sqrt[3]{2-\sqrt{5}})(\sqrt[3]{2+\sqrt{5}})(c)$, which simplifies to $4+c^{3}=-3c$. This gives $c^{3}+3c+4=0$ so then the only real root is $c=-1$, which then makes $\boxed{a+b=1}$