salam
ye soale kheily jaleb
be adadi motanaveb migooyim ke ragham hahash yek dar miyan zoj va fard bashand
tamame adadi ra biyabid ke mazrabe motanaveb darand.
man ke khodam bahash kheily hal kardam.

salam
agha soale vaghean sadast ke shoma behesh javab nemidin?

fekr konam javabet ghalate
va javabesh tamame adad be joz 20 ast
(20k)
dar zemn tarahe in soalam aghaye jamali va yeki dige boodan

azize man age ye bar dige rahe halo bekhoni manam chizi gheir az nagoftam ke...
nazare asatatid chiye dar in mored?

yek ragham ghabl va yekl ragham bad az ashare adade zir ra biabid:
\[ (\sqrt{2}+\sqrt{3})^{1980}  \]

[quote="BaBaK Ghalebi"]sabet konid har adade tabii mazrabi darad ke tamame arghame $0$ ta $9$ dar an be kar rafteand[/quote]
bebin tamame adade $1$ ta $9$ ro har joor hal kardi benevis badesh be meghdare kafi sefr bezar va badesh ba hamneheshti be sadegi gavab mide choon az har n adade motavali yeki mazrabe n mibashad.(khaste nabasham  :rotfl: )

[quote="mohamadhosein"][quote="ehsan2004"]yaani to vaghean nemidooni ke in saale IMO 2004.  :P[/quote]
jedi migi :huh:
ama kheily asoontar az in harfha hal mishe pesar.
faghat adadi ra nemitavanim be in soorat benevisimke be soorate$2k*10^m$
bashand va bas .pas farz mikonim $n$ joze in daste nist.
zira agar donbaleye zir ra dar nazar begirim
$10,1010,101010,10101010,...$
va $n$ jomle az in donbale ra benevisim hatman ya  yeki az anha bar $n$ bakhshpazir ast ya tafazole tota az anha.yani jomleyi be soorate $1010...10100000...0$
ama chon $n$ joze an daste nist pas adadi mesle $101010101010$
bar n bakhshpazir ast va esbat tamam ast.
albate in rahe khodam bood :lol:[/quote]
shramande vali esbat roo havast
yeni joof zadi
chon hamoon joor ke gofti adadi be soorat $101010101000000$bar $n$ bakhsh pazire
vali fekr nakonam ke in dalili bar $n=2k*10^m$bashad
mesle adade $2$
sorry ;)

[quote="mohamadhosein"]khoob manzooreto nafahmidam vali tabloe ke adadi be sorate $2n*10^k,k>1$
nemitoonan mazrabe motanaveb dashte bashand bad man sabet kardam baghiye darand mesle adade 2  ;)  yani man farz kar damam n joze dasteye bala nist va bad
shon joze an daste nist pas mitame sefr haye samte rast be joz yeki ra hazf kard.[/quote]
bebin to gofti beine oon donbalat ya yeki az adad bar $n$bakhshpazire ya tafazole dotashoon
khob to gofti ke tafazole 2 tashoon be soorate$1010101000000$ast
badesh gofti ke chon farz kardim$n$joze oon daste nist pas halate avale
amma man migam ke tebghe gofteye to tafazole dotashoon bar $n$[b]bakhshpazir[/b]ast na barabar
vali bad to gofti chon $n$ozve oon daste nist pas $n\not=1010101000000$vali ma goftim ke adade$10101010000$
bar $n$bakhshpazir bashe na inke mosavie $n$ bashe
ya be ebarati momkene ye adad mazrabi be soorate $2k*10^m$dashte bashe vali khodesh be in shekl nabashe
doroste??

[quote="mohamadhosein"]be bin man goftam chon $n$ mazrabe $20k$ nist mishe hameye sefrhaye samte rastesho be gheir az yekishon hazf kard va baz ham mazrabe $n$ bemoone
[b]aya to in harf eshkali hast [/b]khob hala in mazrabe motanavabe dige azize man.[/quote]
bebin dari eshtebah mikoni yeni kam migi shayad man nemifahmam
vali age ye adad bar $n$bakhshpazir bashe dalili bar in nadare ke $n$mosavie oon adad bashe pas age $n$mazrabi be soorate $2k*10^m$dashte bashe dalili nadare ke $n$mosavie oon bashe
 :huh:  :!:

[quote="mohamadhosein"]salam
babak jan bebin man taze fahmidam eshtebahe shoma kojast man gofte boodam
tamame adadi ra peida konid ke [b]mazrabe[/b] motanaveb dashte bashand na khodeshoon motanaveb bashand.[/quote]
nanana eshtebah fahmidi man soalo dorost fahmidam
vali ye dor age rahe halleto vasse yeki tozih bedi eshkaleto mifahmi

[quote="mohamadhosein"]yani chi man in soalo baraye mr khezeli tozih dadam ishoon taiyid karde ke doroste.
bebin to ke ghabool dari ke $n$ mazrabi be sorate$1010...10$darad khoob in motanavebe dige ya to ghaboole inke $n$ mazrabe motanaveb dare shak dari ke oonam tabloe dige baba.[/quote]
midoonam vali agar $n$mazrabi be soorate$10101010100000$dashte bashe chi?? :?

azize man ,man ke nagoftam nadare akhe man kei hamchin harfi zadam.
manan migam hamchin mazrabi dare vali chon $n$ be sorate $20k$ nist mishe sefr hashe samte rastesho be gheir az yeki hazf kard .(khaheshan dige nagoo nemishe :D ).
agha aslan yeki dige biyad ye ghezavati beine man ba bak jan bokone.

adadi $1996$ raghami darim
dar har marhale adade yekan ra az baghie adad kam mikonim
sabet konid vaghti ba tekrare amale fogh be adadi $10$ raghami beresim adade hasel raghame tekrari darad

[quote="BaBaK Ghalebi"]farz konid donbale adade sahih va namanfi $a_1,a_2,...,a{}_1{}_9{}_9{}_7$ dar rabete:
$a_i+a_j\leq a_i_+_j\leq a_i+a_j+1$ baraye har $i,j\geq 1$ ba sharte $i+j\leq 1997$ sedgh mikonad
sabet konid adade haghighie $x$ vojood darad ke baraye har $i\leq n\leq 1997$ , $a_n=[nx]$ :?:
soalesh ghashange ;)[/quote]
baba yeki ye chizi bege :?  :huh:

[quote="mohamadhosein"]salam babak jan in hamoon soale [b]imo2004[/b]
$0$ khob bebin farz mikonim $n$ be sorate $20k$ nabashad
$1$ pas hade aksar 1 sefr dar samte rastash darad
$2$ donbaleye zir ra dar nazar bagir $10,1010,101010,. . .,1010...10$tedad 1 haye akhari $n$ ta ast
$3$ agar yeki az in adad mazrabe n bashad ke masale hal shde ast(chera?choon ke $n$
mazrabi motanaveb darad ke haman$1010...10$ ast.
$4$ agar hich kodam az $n$ adade donbaleye bala mazrabe $n$ nabashand pas tebghe asle lane kabootari do ta az anha be peimaneye $n$ hamneheshtand pas tafazoleshan
bar $n$ bakhshpazir ast.
$5$ ama tafazole anha be sorate $1010...100...0$mibashad
$6$ va chon $n$ hade aksar yek sefr dar samta rastash darad agar hameye sefr haye samte rastash be gheir az yeki az anha ra hazf konim adad baz ham mazrabe $n$
mimanad
$7$ dar in sorat niz adadi ke mazrabe n boode be shekle$1010...10$mishavd va in adad
motanaveb ast va masale sabet mishavad
babak jan har bandi ke fekr mikoni irad dare be man begoo. ;) [/b][/quote]
fekr konam doroste
vali kheili khoob neveshti
rahetam hatman doroste chon khodamam ino hamin tori halesh kardam
akhe to har dafe ke tozih midadi oonjai ke mitoonim 0 hasho hazf konimo dorost nemigofti
manam khodam hamino moshkel dashtam
vali khoob neveshti va in mozoo be marhale dovomet komak mikone ;)

[quote="mohamadhosein"]khoobe,be nazaret az$7$chand migire? :P[/quote]
fekr konam $7$
akhe man khodam ba in rahe dasto panje narm kardam vali age mosaheh hal nadashte bashe fekr nakonam az 6 paintar bede ;)  :D

[quote="amin_68"]so'ale sade'e nist.niyaz be baghimande chini dare va esteghra .kheyli ghashange,  sa'ay konid khoob roosh fekr konid :)[/quote]

Kodoom soal?? ;)  :)

so'ole 6 IMO 2004 :)
har adad mazrabi tanavobi darad be joz n=20k :)

:o  :!:
chi?baghimandeye  chini? manzooret cheye niazi be baghimandeye chini nadare.

chera dare ,alaki ke so'ale 6 IMO 2004 nashode .bayad khoob roosh fekr konid. :)


[b]i.)[/b] Calculate $x$ if \[ x = \frac{(11 + 6 \cdot \sqrt{2}) \cdot \sqrt{11 - 6 \cdot \sqrt{2}} - (11 - 6 \cdot \sqrt{2}) \cdot \sqrt{11 + 6 \cdot \sqrt{2}}}{(\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2}) - (\sqrt{\sqrt{5}+1})}  \]

[b]ii.)[/b] For each positive number $x,$ let \[ k = \frac{\left( x + \frac{1}{x} \right)^6 - \left( x^6 + \frac{1}{x^6} \right) - 2}{\left( x + \frac{1}{x} \right)^3 - \left( x^3 + \frac{1}{x^3} \right)}  \] Calculate the minimum value of $k.$


How od you evaluate sums? I know the basics, but how do you evaluate ones like $\displaystyle\sum_{i=0}^n i^5$. Sorry if I should have known this. :oops:

This shouldn't be in MC. The formula is $\frac{(n^2)(n+1)^2(2n^2+2n-1)}{12}$. There's a general formula to find $\sum_{i=1}^n{i^x}=n^{x+1}-\sum_{i=1}^n{(i^{x+1}-(i-1)^{x+1}-i^x)}$. Thats how you can derive the formulas for $\sum_{i=1}^n{i^x}$. Its a hassle when $x$ gets too big.

Ok....I don't get that at all and i don't think I'm goign to try to. I can do really simple sums but this one makes no sense whatsoever to me!

Well, to get the formula for your sum, you can make a proof by induction. :)

[quote="nat mc"]We can't use induction to derive the formula, only to prove it[/quote]

One practical way to derive the formula  is to use a general polynomial (one degree higher) and solve for the cofficients.
(This will also work, if you memory is not perfect and you want to check out a cofficent or so)

For  Example:  for sum of squares (You can use it for higher powers too)
Let   $\displaystyle\sum_{i=0}^n i^2 = An^3+Bn^2+Cn+D$
Now when n=0 gives D=0
n=1 gives (1^2 = 1 =A+B+C) => C =  1-A-B
n=2 gives (1^2+2^2 = 5 = 8A+4B+2C) => 6A +2B = 3
n=3 gives (14 = 27A+9B+3C) =>   etc  //
solving above gives A = 1/3; B=1/2; C= 1/6 so the answer is $ (1/6)(2n^3+3n^2+n) $

(BTW  for  $ \displaystyle\sum_{i=0}^n i^k $  will always start out with  $(1/(k+1)) n^{k+1} + (1/2)n^k  \dots $

Similary  another   way to  [b]prove [/b]    is just to show the formula is correct for n=0 to k+1

Eg  for   $\displaystyle\sum_{i=0}^n i^5 $ if the formula works for  n=0,1,2,...6$ it will work for all the integers.


Hope you like both two  :lol:
$1)$ Given $x,y,z>0,x+y+z=1,\alpha,\beta,\gamma>0$ prove that
$\alpha\sqrt{y^{2}+yz+z^{2}}+\beta\sqrt{z^{2}+zx+x^{2}}+\gamma\sqrt{x^{2}+xy+y^{2}}\ge \sqrt{3}(\sqrt{\beta\gamma}x+\sqrt{\gamma\alpha}y+\sqrt{\alpha\beta}z)$
when $\alpha=\beta=\gamma=1$ we get the well know
$\sqrt{y^{2}+yz+z^{2}}+\sqrt{z^{2}+zx+x^{2}}+\sqrt{x^{2}+xy+y^{2}}\ge\sqrt{3}$
$2)$ Given $x,y,z>0,x+y+z=1$ prove that
$\frac{1}{\sqrt[4]{y^{2}+yz+z^{2}}}+\frac{1}{\sqrt[4]{z^{2}+zx+x^{2}}}+\frac{1}{\sqrt[4]{x^{2}+xy+y^{2}}}\ge 2\sqrt{2}+\frac{2}{\sqrt[4]{3}}$
P/s I have a conjecture:
The express $(y^{2}+yz+z^{2})^{k}+(z^{2}+zx+x^{2})^{k}+(x^{2}+xy+y^{2})^{k},-1<k<0$ get min value when $(x,y,z)=(\frac{1}{2},\frac{1}{2},0)$

Thank you, Gemath. I've just waken up and here is my solution,
For [1] first. It is pretty easy, Gemath.
\[\sum x\sqrt{\beta\gamma}\le \frac{1}{2}\sum x(\beta+\gamma)=\frac{1}{2}\sum \alpha(y+z).\]
Notice that
\[\frac{y+z}{2}\le \frac{y^{2}+yz+z^{2}}{\sqrt{3}}.\]
So we have done.

For [2], I think there is something wrong. The inequality should be

\[\sum\frac{1}{\sqrt[4]{y^{2}+yz+z^{2}}}\ge 3\sqrt[4]{3}.\]

For your conjecture, it is false for $k=-1$ so it can not be true.

ok that's right I only note that $1)$ is Erdos-Mordell inequality for equilateral triangle with weight!
Sorry the second it must be
$2)$ Given $x,y,z>0,x+y+z=1$ prove that
$\frac{1}{\sqrt[4]{y^{2}+yz+z^{2}}}+\frac{1}{\sqrt[4]{z^{2}+zx+x^{2}}}+\frac{1}{\sqrt[4]{x^{2}+xy+y^{2}}}\ge 2\sqrt{2}+\sqrt[4]{\frac{4}{3}}$ or it has minimum at $(\frac{1}{2},\frac{1}{2},0)$
For my conjecture with $k=1$ I think it is still correct, or can you show me conuter example
note that $f(\frac{1}{3},\frac{1}{3},\frac{1}{3})\ge f(\frac{1}{2},\frac{1}{2},0)$

Ok I see it is false for $k=-1$ sorry I made a mistake but now I can assert the express $f(x,y,z)=(y^{2}+yz+z^{2})^{k}+(z^{2}+zx+x^{2})^{k}+(x^{2}+xy+y^{2})^{k},-\frac{1}{2}\le k\le 0$ get min value when $(x,y,z)=(\frac{1}{2},\frac{1}{2},0)$ and orther permution!


algebra
polynomial
induction
real analysis
real analysis unsolved

I suppose this is going to be rather easy, but I thought it was nice:

Given a polynomial $f\in\mathbb C[X_1,\ldots,X_n]$, prove that the subset of $\mathbb C^n$ on which $f$ is non-zero is connected.

[quote="julien_santini"]It suffices to show this set is pathwise-connected. Taking a = (a_1, ... , a_n) and b = (b_1, ... , b_n) such that f(a) <> 0 and f(b) <> 0, and fixing the (n-1) first components of a and b, it is clear (a_1,...,a_n) can be continuously linked to (a_1,...,b_n) (because we work in C, and the polynomial P(a_1,...,a_(n-1),X) has at n complex roots (which is finite)). By induction it is then easy to show a can be continuously linked to b, whence the result.[/quote]

When i say continuously linked, I mean of course through a path with values which are not roots of P (that's all the interest of the problem).


let $ a,b,c\ge 0$ and  $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ prove that  $ \frac {c}{c^3 \plus{} ab} \plus{} \frac {b}{b^3 \plus{} ac} \plus{} \frac {a}{a^3 \plus{} bc}$>3

[quote="dgreenb801"]The given is equivalent to $ \sum \frac {z(x \plus{} y \minus{} 1)}{xy \plus{} z} > 0$, where $ xy \plus{} yz \plus{} xz \equal{} 1$. Can anyone take it from there?[/quote]
I don't know
x,y,z  :| why?


A rectangular sheet of paper has dimensions 11 inches by 12 inches. A triangular piece is cut from the corner to leave a pentagon with side lengths 11, 12, 7, 5, and 9 inches. How many square inches are in the area of the pentagon?

I KNOW ALL THE ANSWERS TO THE CHALLENGING QUESTIONS, SO DON'T THINK I'M TRYING TO MAKE YOU ANSWER AN ASSIGNED PROBLEM BY MY MATH THEACHER!

"I KNOW ALL THE ANSWERS TO THE CHALLENGING QUESTIONS, SO DON'T THINK I'M TRYING TO MAKE YOU ANSWER AN ASSIGNED PROBLEM BY MY MATH THEACHER!"

...
that makes me think you're trying to make us answer assigned problems by your math teacher.

[quote="mathkid"]that makes me think you're trying to make us answer assigned problems by your math teacher.[/quote]

Probably not, considering the simplicity of the question.

I don't know. Maybe this is some sort of extra credit assignment given by a math teacher. As he said, he is in Algebra I, and my friends in Algebra I can do this problem with a little time. Also, It's pretty unusual for somebody to just come onto the forums and immediately posts 3 problems. The sig fig one is especially suspicious. The conic section one is equally suspicious, seeing as it could easily be figured out with a little knowledge of conics.

I, for one, definitely think these are homework problems and should be deleted.

draw a rectangle and cut out a triangle

the right triangle should be a 3-4-5 rt triangle.

the side lengths of the pentagon are 11,12,9,7,5

the 11 and 12 are untouched by the cutting, the 3 is chopped off from the 12-side, leaving us with a 9 for a length of the pentagon. the 4 is chopped from the 11-side, leaving us with a 7,

the 5 is derived from pythagoreon theorem. 3 2 +4 2 = 5 2

so the answer is 11*12 - 1/2*3*4 = 126

you could also find the area of the triangle and subtract from the total area of the rectangle, which gives you 126

[quote="yif man12"][quote="marucho8d7"][quote="soulzmischief"]LOL yo are you from hopkins? if you are, say hello to ur tutor :P[/quote]

if you know i'm from hopkins, tell me who you are, and tell them that THESE ARE NOT HOMEWORK PROBLEMS[/quote]

Do you know Mr. MacArthur?[/quote]

no, soryy


function
integration
vector
limit
calculus
probability and stats

I actually need something slightly different but let me ask this first:

Assume that $ f(x)$ ($ x\in X$) is a real Gaussian random process with mean $ 0$ and uniformly bounded variances, $ \mu$ is some finite measure on $ X$, and $ \psi$ is a reasonable strictly convex function on $ \mathbb R$. Show that the random variable $ \int_X \psi(f(x))\,d\mu(x)$ is not normal.

Any ideas are welcome :).

The only idea which comes into my mind is to consider the excess (kurtosis) - fourth central moment divided by squared variance minus three - and try to prove it's positive.

I have no idea if this will be useful, but I have just tried to see what was happening in the simpler case where $ X$ is finite. This may be completely rubbish, but I find the problem very interesting, so here it goes:

let $ G: \Omega \to \mathbb{R}^n$ be a Gaussian random vector, $ \mu_1, \mu_2, \ldots, \mu_n >0$ some positive numbers, and $ \phi$ a convex function. I will assume too that $ 0<d^\minus{} \equal{} \lim_{\minus{}\infty} \phi'(x)<d^\plus{}\equal{}\lim_{\plus{}\infty} \phi'(x)<\plus{}\infty$ (so that $ \phi(x) \approx d^{\pm}x$ near $ \pm \infty$) and define $ \Phi(x)\equal{}\sum_i \mu_i \phi(x_i)$ where $ x\equal{}(x_1, x_2, \ldots, x_n) \in \mathbb{R}^n$. Hence, one would like to prove that $ \Phi \circ G: \Omega \to \mathbb{R}$ is not a random variable. In order to do that, let's compute an approximation of the density of $ \Phi \circ G$:
\[ P[ \Phi \circ G \in (a;a\plus{}da)] \equal{} K \int_{\mathbb{R}^n} 1_{ \Phi(x) \in (a;a\plus{}da)} \exp\{ \minus{}\frac{1}{2}(x, Lx) \} dx\]
where $ K  \exp\{ \minus{}\frac{1}{2}(x, Lx) \} dx$ is the density of the random vector $ G$. One thus assumed that $ L$ is invertible, hence $ L \in S^{\plus{}\plus{}}(\mathbb{R}^d)$.

Let's take $ a$ very large, positive:
Since the Gaussian density is decaying quite fast, the main contribution of the integral is in the neighbourhood of $ x^{*}$, solution of:
\[ \min (x,Lx) \qquad \text{under the constraint} \qquad \Phi(x) \equal{} a\]
Since for large $ x$ (here I mean that every component is a large number): $ \nabla \Phi(x) \approx (\mu_1 \cdot d^{\pm}, \mu_2 \cdot d^{\pm}, \ldots, \mu_n \cdot d^{\pm}) \equal{} \mu$ (where $ \mu$ does not depend on $ a$) and $ \nabla (x \mapsto (x,Lx)) \equal{} 2Lx$, one can see that $ x^* \approx \lambda v$ (Lagrange Multipliers condition), where $ v \equal{} \pm L^{\minus{}1} \mu$ and $ \lambda > 0$.
Moreover $ a \equal{} \Phi(x^*) \equal{} J(\lambda) \approx J'(\infty) \cdot \lambda$ where $ J(\lambda) \equal{} \sum \mu_i \phi(\lambda v_i)$. Notice that $ J$ is a convex function and $ J'(\minus{}\infty) \neq J'(\plus{}\infty)$. This gives the approximation $ \lambda \approx \frac{a}{J'(\plus{}\infty)}$.

Finally one can approximate $ P[ \Phi \circ G \in (a;a\plus{}da)]$ by $ \frac{da}{\|\mu\| \approx \| \nabla \Phi(x^*)\|}$ times the integral of $ K  \exp\{ \minus{}\frac{1}{2}(x, Lx) \} dx$ on the hyperplane $ \mathcal H_a$ going through $ x^*(a)$ and perpendicular to $ x^*(a)$.That would mean that $ P[ \Phi \circ G \in (a;a\plus{}da)]$ is approximatively equal to:
\[ \frac{K}{\| \mu \|} \int_{y \in \mathcal H(a)}   \exp\{ \minus{}\frac{1}{2}(y, Ly) \,dy\]
but it is not hard to see that this integral is equal to $ C_1\exp(\minus{}C_2 \lambda^2(a))$, where $ C_1,C_2$ are constants that only depend on $ L$ and $ \mu$.

But one knows that $ \lambda(a) \approx \frac{a}{J'(\plus{}\infty)}$ for very large positive $ a$, and similarly $ \lambda(a) \approx \frac{a}{J'(\minus{}\infty)}$ for very large negative $ a$. Because $ J'(\minus{}\infty) \neq J'(\plus{}\infty)$, that would show that the density of $ \Phi \circ G$ does not decrease at the same rate near $ \plus{}\infty$ and $ \minus{}\infty$.

What do you think about such an approach ?

ps: reading it again, I have just seen a flaw: I said that $ \nabla \Phi(x^*(a))$ is almost constant for large $ a$. I think that this is true for all large positive $ a$, and for all large negative $ a$. But it may not be the same value for positive/negative $ a$  :maybe:

Another idea:
$ \minus{} f$ has the same distribution as $ f$. Then if $ \int\psi(f)d\mu$ is Gaussian, then so is $ \int\psi( \minus{} f)d\mu$.

Now let $ \xi$, $ \eta$ be two Gaussian variables (not necessarily jointly Gaussian). Does $ \xi \plus{} \eta \ge 0$ imply that $ \xi \plus{} \eta$ is constant? I feel that yes, it does. You can assume that the distributions of $ \xi$ and $ \eta$ are identical if you need.

UPD I don't fell that "yes, it does" anymore :)


If a round 8-inch diameter pizza serves two students, how many students should two 12-inch diameter pizzas serve?

a. 9 b. 8 c. 6 d. 4 e. NG

[hide] If we assume that area is the reason for increase(as opposed to circumference), then we have $ 2\times{2}\times{(\frac32)}^2 \equal{} \boxed{9}$. Hence, the answer choice is A. [/hide]

[hide]So you can cross multiply.
8=2
12=?
and then you get 3...Then you have to multiply that by 2, because there are 2 of those pizzas, and I got 6, which is C.[/hide]

thesd0, I believe your solution is incorrect.
[hide="solution"]You're assuming that the number of people served is directly proportional to the radius of the pizza. However, the number of people is proportional to the AREA of the pizza, which is proportional to the SQUARE of the radius.
Therefore, like gaussintraining said, since $ 8$ inch serves 2, then $ 12$ inch, which is $ \frac{3}{2}$ times as much will serve $ 2\cdot\left(\frac{3}{2}\right)^2\equal{}\frac{9}{2}$ and there are $ 2$ pizzas making it $ \boxed{9}$ people.

[/hide]


advanced fields
advanced fields unsolved

Prove that in any vectorial space with norm, any sequence  absolutely convergent is convergent if and only if that space is Banach. ;)

uhm... sorry, what do you mean by "absolutely convergent sequence"?

Are you looking for a proof or posing it as a question for others?  It's a rather well-known result, so I'll only post the solution if you need an answer.

that's absolutely convergent _series_, not sequence...
cezar, could you please tell us? :)
(merry christmas everybody, by the way)

take $S_n = \sum_{k=1}^{n} a_k$, $T_n = \sum_{k=1}^{n} |a_k|$.
since $T_n$ is convergent, it's cauchy sequence.
but $|S_m - S_n| = \left|\sum_{k=m+1}^n a_k\right| \le \sum_{k=m+1}^n |a_k| = |T_m - T_n|$, hence $S_n$ is convergent.
same as real case  ;)

[quote="cezar lupu"]Blahblahblah, do yopu a have a proof for this one? :)[/quote]

Yup.  If $X$ is complete and $\sum_i \|a_i\|<\infty$, put $S_n=\sum^n__{i=1} a_i$.  Then
\[ \|S_n-S_m\|\leq \sum^{n}_{i=m+1}\|a_i\|  \]
and the latter sum goes to zero as $m,n\to\infty$ (since every absolutely convergent series in $X$ converges)

For the converse, let $(a_i)_i$ be a Cauchy sequence.  Take a subsequence $(a_{i_j})_j$ such that $\|a_m-a_n\|<2^{-k}$ for $m,n\geq i_k$

Put $b_1=a_{i_1}$ and $b_j=x_{i_j}-x_{i_{j-1}}$ for $j>1$.  Then
\[ \sum^k_{j=1} y_j=a_{i_k}  \]
so that
\[ \sum^{\infty}_{i=1} \|y_i\|\leq \|y_1\|+1<\infty  \]
so the subsequence $(a_{i_j})_j$ converges to a limit $L$.   Because $(a_i)$ is Cauchy, it also converges to $L$; hence $X$ is complete.


The International Forums can be viewed by everyone whereas the American ones can only be viewed by people in that Usergroup.  You can still see the American ones but you cannot see how many posts or the content of the posts.

[quote="JBL"]When I am not logged in, all of the American communities purport to have no posts.  This is not true of the international forums.  Any particular reason for the difference?[/quote]Most American local communities wanted to have a certain degree of privacy for certain reasons, so accesing the topics in those forums can only be done if you are the member of the respective usergroup.

The other national forums at this point are designed to a "friendly welcome" place for those math lovers that can't speak english. We already have an automatic language mechanism which detects your language and displays the controls of the forum in your language (for now just working for Romanians and Germans, but will be expanded to all the other languages presented there). So new international users are encouraged to join the community, even if, in the beginning they will spend more time in their own national forums, and as soon as they can master sufficient english, they move on to the main forums.

I have two questions:
1) How do I join a community?
2) Can I join multiple communities?

Sorry I'm new to this new feature :blush:

[quote="Valentin Vornicu"]There is a global announcement in this section which explains what the national and local comm. are all about http://www.mathlinks.ro/Forum/viewtopic.php?t=22123

For the second question, I'm not sure if the moderator of a state community will accept your submission if you are already a member of another state community (you have to talk with him on private message).[/quote]
Thanks, I'll contact with the moderators and see if I'm allowed to join multiple communities.


probability
calculus
geometry
rectangle
induction
search
real analysis

Randomly pick three points on a plane. What is the probability that the triangle with these points as the vertices is an acute triangle?

I don't think this is quite in the spirit of the Calculus Computations and Tutorials section, so I'm moving it "up" to Calculus Unsolved Problems.

This problem isn't exactly well-defined, becuase there is no natural sense in which the plane $\mathbb{R}^2$ is a probability space.

As an interpretation of what the meaning most likely is, let's say we let choose the three points (independently) according to a uniform distribution on the square $[-N,N]\times[-N,N],$ find the probability of the acute triangle, then take the limit as $N\to\infty.$

Or we could as the same question about picking the three points uniformly on the circle centered at the origin of radius $R$, then taking the limit as $R\to\infty.$

I'm afraid that the answer may depend on the shape we adopt.

[quote="Kent Merryfield"]
I'm afraid that the answer may depend on the shape we adopt.[/quote] It surely does. If we choose a long and narrow rectangle, the probability gets close to $0$. So the question should be posed somewhat differently. I can offer the following version: find the supremum of the probability of an acute triangle over all continuous probability distributions on the plane. This is a well-defined problem and I suspect that the answer is not $1$. I'm afraid only that, with this formulation, it will be eventually moved one more level up to Calculus Open Questions... :)

I'm not so sure the $sup$ is less than 1.  The idea would to use a Cantor-like construction to divide up the square into smaller squares as follows:

Let $k$ be a fixed odd number

$S_0$ the unit square

$S_{n+1}$ is formed from $S_n$ by taking each square of $S_n$ and dividing it into $k^2$ equal squares, then only taking the bottom left, upper middle, and right middle squares (below shows $S_1$ for $k=2$)

[code]
OOXOO
OOOOO
OOOOX
OOOOO
XOOOO
[/code]

Since the triangle formed by $(0,0), (0.5, 1),$ and $(1,0.5)$ is acute, by taking $k$ to be sufficiently large we can make it so the only way three points in $S_1$ can form an non-acute triangle is if they are located in the same smaller square.  Induction now gives that the same is true for $S_n$.

Therefore if we take $P_n$ to be the uniform distribution on $S_n$, the probability 3 points taken from $P_n$ form a non-acute triangle is at most ${1 \over 9^n}$.  $P_n$ isn't continuous, but we can approximate it by continuous distributions so that the probabilities don't change much.

[quote="kevinatcausa"]so the only way three points in $S_1$ can form an non-acute triangle is if they are located in the same smaller square.  [/quote]
Not quite: you will have plenty of non-acute triangles whose 2 vertices are located in one square and the third vertex in another square. My suspicion that the answer is less than 1 is based on a simple observation that for every 4 points on the plane some 3 of them form a non-acute triangle.

A Google search gave me this: http://mathworld.wolfram.com/ObtuseTriangle.html

I had only thought about that last bit (which was why I originally placed this post under "calculus computations and tutorials"), but I guess it's much more complex. You really learn something new every day on AoPS.  Thank you for the replies. :)

[quote="litoxe2718281828"]Randomly pick three points on a plane.[/quote]Just as a side note, how do you define "random point" on a plane? random x and y? random r and theta? random ...? (often the probability is different)


I heard that each blank would count for only 1.5 points this year.  Is the AMC commitee going to make the problems slightly easier to make up for this? or is it going to be the same difficulty? oh and do you think getting an 800 on the sat math is equivalent to getting like a 100 on the amc 12? thanks.

There's no way to know what the difficulty will be until the test happens- the AMC can't test their exams' difficulty against large numbers of students. If they get the difficulty wrong, there's always the possibility of changing the qualifying standards.

As for the SAT question- SAT math scores and AMC scores are on different scales. An 800 SAT doesn't mean much of anything for an AMC score- you don't have to know problem-solving tricks to get there.

Hm, I had an 800 SAT last year, and I wouldn't have gotten a 100 on the AMC 12 (if the new scoring method applied.)  An 800 is a good start, but it will take more practice.

Why don't you try an old AMC 12 to see where you stand?

[quote="1337h4x"]Are you talking about SAT I or SAT II? Either way, I think an 800 is more difficult. Last year my school had around 10 AIME qualifiers and maybe half of them had 800's. Bear in mind that you're really comparing apples to oranges.[/quote]

Come on, getting an 800 on the SAT is a lot easier than getting a 100. You have to be more creative on the AMC 12. The last questions on the SAT I are comparable to a #8 on the AMC 12.
All the SAT I really looks for, is if you know basic geometry, arithmetic and algebra 1.

yea ...each blank is worth 1.5points ... :( they changed it

I think a 100 this year is probably more difficult, but in years past, you only needed to get 11 questions right. I think that scoring an 800 is slightly more difficult than that (one careless error screws you over). Maybe I'm just pissed off that I didn't get an 800 but did get >100. Whatever...

It's probaby as easy or as hard to get 11 questions correct on the amc12. Just make sure you don't make any careless errors.

Hmm it depends you might have problem solving ability and get like a 100 on AMC 12 but mess up on a couple things on SAT and only get say a 780.

umm my only advice for 800 SAT is... don't fall asleep in the middle and waste all your time!! (Personally, it seems like anyone should be able to get an 800 if they just get enough sleep the night before)

sorry if that sounds harsh

I make stupid mistakes a lot, so I will never get an 800 on the SAT math, but can easily get above a 100 on the AMC 12.

Dude the AMCs are much harder. If you don't make retarded mistakes, you could probably get an 800 with your eyes closed.

It is a totally different kind of test... I'd say you cannot really compare because 800 SAT math means a PERFECT score even though almost all the SAT problems are waaaaay easier than AMC problems. You cannot make a single mistake if you want an 800... so it's totally different.

I don't think you need to get every single question right for an 800 on the SAT Math, at least for the IIC test.[/i]

Every genuine sample test for the SAT II Math Level 2 that I have seen allows missing a few questions and still getting a scaled score of 800. In terms of numbers of students, AIME qualifiers are somewhat more common than perfect scorers on the SAT I math section AT THE SAME AGE, but by high school age lots of students perfect the SAT I math section

[url=http://www.collegeboard.com/prod_downloads/about/news_info/cbsenior/yr2006/national-report.pdf]SAT National Report[/url]

and even lots of them perfect the SAT II Math Level 2 test. Qualifying for AIME is not outrageously hard, but getting a [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06-perfect1012AB.html]perfect score on an AMC 10 or 12[/url] is very hard indeed.

Good luck, everyone.

[quote="joml88"]Hm, I think 100 AMC and 800 SAT are about equivalent.  I know someone who got over 120 on the AMC 12 and got a 740 SAT math.[/quote]

Yeah I got a 119.5 AMC 12 in 9th grade and a 720 in SAT I math in 10th grade. :oops:


http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-2-1012-Answers.shtml
shows that the answer for number 9 is 2...

however there is no 2 on the answer choices. so i assume the website has a type-o...

or is my test messed up..

It was A, 7. Solving for $x$ to get $f(5)$ gives you $2$ though.

i dont get it
is it 7 or 2
i got 7..but it's 2?
alright i think i missed that one then because i don't remember putting a 2... argh

$x = 2$, but the answer to the question isn't $x$.

It was likely a typo.  It was asking you for the output value, which is 7.  So the answer A was correct.  Don't worry, none of the answer choices were 2.


probability
combinatorics unsolved
combinatorics

I am reading this example about permutation and can't understand how to go about rationalize it.

Example: Suppose that 2 anthropology, 4 computer science, 3 statistics, 3 biology, and 5 music books are put on a bookshelf with a random arrangement. what is the probability that the books of the same subject are together?

Solution is P(A) = 5! x 2! x 4! x 3! x 3!x 5! / 17!
I understand the bottom is the total numbers of arrangement that you can form with 17 books. How do you explain the numerator part? in detail please. Thanks in advance!

Added two more,
1. A town has six parks. On a Saturday, six classmates, who are unaware of each other's decision, choose a park at random and go there at the same time. what is the probability that at least two of them go to the same park?

2. If we put five math, six biology, eight history, and three literature books on a bookshelf at random, what is the probability that all the math books are together?

Probability is giving me a headache!

[quote="JBL"]These problems are combinatorial; they belong either in one of the high school forums (Intermediate or Pre-Olympiad) or in the Combinatorics section.

My hint for your first question is, "Count the number of ways you can arrange the books that work."

My hint for the second question is, "Count the complement."

My hint for the third question is, "Does it matter if we have separate biology, history and literature books, or can we just call them 'notmath books?'"[/quote]

Thanks! i get it now.. :) It'll takes lots of practice for me to remember.


geometry
trapezoid
circumcircle
cyclic quadrilateral
geometry unsolved

Let $ ABCD$ cyclic quadrilateral with circumcircle $ k$. The line tangent with $ k$ intersects $ A$ meet the extended lines $ BC$ on $ P$ and the line tangent with $ k$ intersects $ B$ meet the extended lines $ AD$ on $ Q$
Prove that if $ AQ\equal{}AD$ and $ BP\equal{}BC$, then $ ABCD$ a trapezoid

So here's how we do it:

QA*AD=QB*QB=QA*2QA so QB=QA*2^0.5
The same way we obtain that AP=PB*2^0.5

We apply Sin Theorem for QBP and ABP and we get:

1) AQ/sin(QBA)=QB/sin(QAB) and 2)PB/sin(PAB)=AP/sin(ABP)

AQ/QB=PB/AP=(2^0.5)/2 so from 1) and 2) we get that sin(QBA)/sin(QAB)=AQ/QB=PB/AP=sin(PAB)/sin(ABP) i.e.
3)sin(QBA)/sin(QAB)=sin(PAB)/sin(ABP)

Now all we have to see is that PAB=ACB=ADB=QBA( from ABCD inscribed in a circle and BQ and AP-tangents)

So sin(PAB)=sin(QBA) and in 3) we get that sin(QAB)=sin(ABP) which means that either QAB=ABP which means that BAD=ABC and DC||AB or
QAB=180-ABP which means that AD||BC. Either way ABCD is a trapezoid.


a^2+b^2=c^2
(a+r)^2+(b+r)^2=(c+r+1)^2
a,b,c,r are all positive integers

I don't get that.
Can you explain it further for me?
I really appreciate it.

I tried  susbstituying: $a=2nm$, $b=n^{2}-m^{2}$, $c=n^{2}+m^{2}$, or $a=n^{2}-m^{2}$, $b=2nm$, $c=n^{2}+m^{2}$, but came up with a larger equation.

$boxedexe$, could you post your solution?


calculus
integration
function
real analysis
real analysis unsolved

Take the integral of the function y = x^2 / (xsinx + cosx)

Why did you put it in National Olympiads?
I will move it to Calculus.

Are you sure this is an elementary function?


Determine all pairs $(x, y)$ of positive integers with $y \vert x^2 +1$ and $x^2 \vert y^3 +1$.

The problem is actually a transformation of problem no. 78 which means,
$y \vert x^2 +1$ => $x^2+1=Ky$ .......(1)   => $x^2=(Ky-1)|(y^3+1)$, where K is positive integer. This is nothing but problem no. 78 that is the question find all ordered pairs (m,n) such that $\frac{n^3+1}{mn-1}$ is integer where m,n are positive integers, which can be solved easily and the solution was given there anyone can see that and from the solution there we get  $(K,y)= (2,2),(1,2),(2,1),(1,3),(3,1),(2,5),(5,2),(3,5),(5,3)$ are the only solutions and (1) gives us when (K,y)=(1,2) or (2,1) we have x=1 and when (K,y)=(2,5) or (5,2) we get x=3. Hence all solutions are (x,y)= (1,1),(1,2),(3,2),(3,5).

which also satisfy the conditions given :D


[quote="math92"]For those of us who live in the affected areas, it’s impossible to ignore the new residents of our neighborhoods…. the cicadas known as Brood X. They are not locusts, but the 1.5-inch insects definitely come in huge numbers. There can be an estimated 1.5 million per acre. What is the equivalent measure of cicadas per square foot if there are 43,560 square feet per acre? Express your answer as a decimal to the nearest tenth.[/quote]

[hide]$\frac{1500000}{43560}\approx34.4$ per sqaure foot.[/hide]

[quote="math92"]For those of us who live in the affected areas, it’s impossible to ignore the new residents of our neighborhoods…. the cicadas known as Brood X. They are not locusts, but the 1.5-inch insects definitely come in huge numbers. There can be an estimated 1.5 million per acre. What is the equivalent measure of cicadas per square foot if there are 43,560 square feet per acre? Express your answer as a decimal to the nearest tenth.[/quote]

[hide]
1500000/43560 is approximately 34.44

Cicadas are cool... ^_^
[/hide]


probability
number theory
greatest common divisor
algebra
function
domain
relatively prime

What is the proof that the probability that two randomly chosen integers are relatively prime is 6/(pi^2)?

Is it
1/2^2 + 1/3^2 + 1/5^2 + 1/7^2 +...?

First let's examine the probability that two positive integers A and B share a common factor.  For a given prime $p$ the probability that A is divisible by $p$ is $\displaystyle \frac{1}{p}$ likewise the probability that B is divisible by $p$ is also $\displaystyle \frac{1}{p}$.  The probability that both A and B are divisible by $p$ is $\displaystyle \frac{1}{p^2}$.  Therefore the probability that A and B don't share these factors is $\displaystyle 1-\frac{1}{p^2}$.  To find the probability that A and B are relatively prime is $P=\displaystyle \prod_{p\in \text{prime}}(1-\frac{1}{p^2})$
\[\displaystyle \frac{1}{P}=\prod_{p\in \text{prime}} \frac{1}{(1-\frac{1}{p^2})}=\sum^{\infty}_{k=1} \frac{1}{k^2}=\frac{\pi^2}{6}\]
Therefore
\[\displaystyle P=\frac{6}{\pi^2}\]

[quote="amd2357"] $\[\displaystyle\prod_{p\in \text{prime}} \frac{1}{(1-\frac{1}{p^2})}=\sum^{\infty}_{k=1} \frac{1}{k^2}$[/quote]Interesting proof.  Can you explain how you did this step in more detail?  :?:

For $x<1$
\[\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+...\]
Let $\displaystyle x=\frac{1}{p^2}$.  Then the LHS of
\[\displaystyle\prod_{p\in \text{prime}} \frac{1}{(1-\frac{1}{p^2})}=\sum^{\infty}_{k=1} \frac{1}{k^2}\]
becomes
\[\displaystyle (1+\frac{1}{2^2}+\frac{1}{2^4}+...)(1+\frac{1}{3^2}+\frac{1}{3^4}+...)(1+\frac{1}{5^2}+\frac{1}{5^4})...\]
If you expand this infinite product you get the sum of all possible reciprocals of squares.  To understand why pick a given integer $k$ write out its prime factorization, then $\frac{1}{k^2}$ equals the unique product of certain terms in each part of the LHS. Example: $k=12=2^23$ so $\frac{1}{k^2}$ would be the product of the terms $\frac{1}{2^4},\frac{1}{3^2},1,1,1,...$.  In general
\[\displaystyle\prod_{p\in \text{prime}} \frac{1}{(1-\frac{1}{p^n})}=\sum^{\infty}_{k=1} \frac{1}{k^n}\]

Another "intutative" way to see this is to consider a nxn lattice where n is very large.
Let $p$ be the probablity that two numbers are coprime.

Now total number of lattice points = $n^2$

Number of   points which are coprime are  (approximately since n is very large) = $p n^2$
Number of  points whose HCF is 2  ( Both numbers are even but have no other common factors) = $p n^2 /4$
Number of points whose HCF is 3 = $p n^2/9 $

(Number of points where both  numbers are divisible by 3 are $n^2/9$ out of which $pn^2/9$ will have only 3 as a common factor) ..etc..

And so on  for 4, 5, 6 etc...

Adding up all we should get $n^2$ so

 $p( 1+ 1/4 + 1/9 + 1/16  \dots ) = 1$
which gives $p = \frac 6 {\pi ^2 } $

BTW the problem is sometimes asked as "what is the fraction of lattice points which can be seen from the origin")

[quote="Gyan"]Another interesting (but a little harder)  question is: what is the probablity that two randomly chosen  Gausian  integeres are coprime.[/quote]

what is the definition of GCD for gausian integers?

[quote="seamusoboyle"]Complex numbers that are not the product of two other complex numbers, where all the coefficients are integers. (I think thats it anyway.)[/quote]

I think you're talking about Gaussian primes.
It's a bit more specific than that. 3 is a Gaussian prime, but it's the product of the complex numbers -3 and -1, -3i and i, and some others.
In considering the primality of the Gaussian integer, you have to exclude the four fourth roots of unity from the factorization.
Looking at common factors gets complicated when you have instances such as 13=(2+3i)(2-3i)=(3+2i)(3-2i) because it would appear that 13 has two different prime factorizations.

I read on the net that a number $a+0i$ (or $0+ai$) is a gaussian prime iff $a$ is an integer prime and $a \equiv 3 ( \mod 4 )$.

I have question: how do we factor a number of the form $a+0i$ with $a$ prime, $a \equiv 1 ( \mod 4 )$ ?

And yeah, it seems strange that numbers seem to have more than one factorization :(

[quote="perfect_radio"]
and yeah, it seems strange that numbers seem to have more than one factorization :([/quote]

$1,-1,i,-i$ are "units" here and is "not counted" as more than one factorization. Just in case of integeres, you can multiply a factor  by $-1$  and   sort of get more than one factorization.  $5$ and $-5$ both are factors of $15$  (To avoid confusion, some times, we consider positive  numbers only ) . For Gausian integers if $z$ is a factor, $-z,iz,-iz$  are also factors and are not considered "more than one"

[quote]I read on the net that a number $a+0i$ (or $0+ai$) is a gaussian prime iff $a$ is an integer prime and $a \equiv 3 ( \mod 4 )$.

I have question: how do we factor a number of the form $a+0i$ with $a$ prime, $a \equiv 1 ( \mod 4 )$ ?

([/quote]

Any  prime $p$ which is $1$  ($\mod 4$)  can be written (in non-trivial , unique way) as $a^2+b^2$ and thus can be factored as $(a+ib)(a-ib)$ . Of course this is also true for even prime ($2$) but [b]not[/b]  when $p \equiv 3 \bmod 4$.
To actually  factor a given $p$   ( or find $a,b$ given $p$ such that $a^2+b^2=p$ ),  there are methods  which are better than 'trial and error'  but these methods  are a little more complicated. (Can be seen on net or  math books .. I have also seen it being discussed in AoPs forum too). ..

Hope this helps.

Well, the probabilty for gaussian integers can easily be calculated the sam way:
The probabilty that a gaussian prime $\pi$ divides a number is $N(\pi)$ where $N(x) := |x|^2$ is the norm on the gaussian integers. The probabilty that it divides two numbers is then $N(\pi)^2$. So we get for the probabilty $x$:
$x= \prod_{ \pi} ( 1- \frac{1}{N(\pi)^2} )$, where the product goes over all association classes of gaussian primes.
But since we know all gaussian primes we get:
\[x=(1-\frac{1}{N(1+i)^2})( \prod_{N(\pi) \equiv 1 \mod 4} 1- \frac{1}{N(\pi)^2} ) ( \prod_{p \equiv 3 \mod 4}  1- \frac{1}{{p^2}^2} ) =\\= ( 1- \frac{1}{2^2} )  ( \prod_{p \equiv 1 \mod 4}  1- \frac{1}{p^2} )^2 )  ( \prod_{p \equiv 3 \mod 4}  1- \frac{1}{{p^2}^2} ) \]
By some things already proven in the previous posts (and threads) and some multiplying out we get:
$x=\frac{6}{\pi^2}  ( \prod_{p \equiv 1 \mod 4}  1- \frac{1}{p^2}  )  ( \prod_{p \equiv 3 \mod 4}  1+ \frac{1}{p^2} ) = \frac{6}{\pi^2 \Gamma} $
where $\Gamma = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$ is Catalan's constant.

You can do the same for other domains with unique prime factorisation (but the 'norm' has to be changed since there isnt a euclidean norm everytime), ., e.g. Eisensteinian integers.

how to put $p \equiv 1 ( \mod 4 )$ into the form $a^2+b^2$?

$N$ can be writen in the form $a^2+b^2$ if and only if when we write $N=x^2y$ where y is square free and has no prime factors equivelent to 3 mod(4)

Also note that we just need to prove it for its prime factors, as

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=(ac+bd)^2+(ad-bc)^2$

A prove of the existence of such a solution can be found here: http://www.mathlinks.ro/Forum/viewtopic.php?t=46300

But I think perfect_radio (and when not he: I) also wants to now an effective method for finding the (since there is only one) solution.


When a car's brakes are applied, it travels 5 feet less in each
second than the previous second until it comes to a complete stop.  A car goes 35 feet in the first second after the brakes are
applied.  How many feet does the car travel from the time the
brakes are applied to the time the car stops?

Essentially, we're trying to find $ 35 \plus{} 30 \plus{} 25 \plus{} \dots \plus{} 5 \plus{} 0$. Notice that this is an arithmetic sequence, which means the difference between each term is the same, in this case, 5. To find the sum, we can pair up the terms, with the first and last, second and second last, and so forth. Notice this gives us the same sum: 35 and 5, 30 and 10, etc. There are 16 terms in total, so 8 pairs. Therefore, we have $ 40 \times 8 \equal{} 320$.

um i think ftw only accepts 140 or it just my computer ?

The answer is $140$. Isabella here is what happened:

There are not $8$ pairs of $40$, because $35+0$ is not $40$.

[hide="Solution"]
Match up terms as isabella said, and you get 6, $40$s when ignoring the middle $20$. This means there are 3 pairs: $3\cdot40+20=\boxed{140}$.
[/hide]


Let $P(x)$ be a polynomial with odd degree.
Prove that $P(x^2-1) = (P(x))^2 - 1 \rightarrow P(x) = x$

We could generalise it. It holds both the polynomial is even degree and odd degree!!

Formally differentiating both side gives :

$2 \times x \times P'(x^2-1) = 2 \times P(x) \times P'(x)$

If $n$ is the degree of $P$ then $1+2(n-1) = n+(n-1)$ i.e. $n=1$.

[quote="tµtµ"]

If $n$ is the degree of $P$ then $1+2(n-1) = n+(n-1)$ i.e. $n=1$.[/quote]

Hm... That would be too simple :D $1 + 2(n - 1) = n + (n - 1)$ is an identity.

:D

Should stop funny cigarettes while doing maths  ;)


Here is a problem that I can't solve, though I spent a lot of time:
  Let a,b,c reals with a^2+b^2+c^2=3 . Then a^3(b+c)+b^3(c+a)+c^3(a+b)<=6.

I proved it for the case when \sum a^4 \geq 4 \sum a^2b^2. The other case, \sum a^4 < 4 \sum a^2b^2, is strange and I cannot proceed. Anyone else approach it a similar way?

Here is my approach
We can easily reduce the problem to the case when a, b, c >=0. We can assume that a = min {a, b, c}.
Let f(a, b, c) = a^3(b+c) + a(b^3+c^3) + bc(b2+c2). We will show that
i)f(a, b, c) <= f(a, t, t) where  t = sqrt((b2+c2)/2) >= (b+c)/2 >= a.
ii)f(a, t, t) <= 6 where  a2+ 2t2 = 3.
From these facts we get the result.
Proof: i) f(a, t, t) - f(a, b, c) = a^3(2t-b-c) + a(2t^3-b^3-c^3) + t2(2t2-2bc)
We have b^3+c^3 = (b+c)(b^2+c^2-bc) <= 2t(b2+c2-bc), so 2t^3-b^3-c^3 >= 2t^3 - 2t(b2+c2-bc) = t(2t2 - 2b2 - 2c2 + 2bc) = -t(b-c)2, and
f(a, t, t) - f(a, b, c) >= a^3(2t-b-c) - at(b-c)2 + t2(b-c)2 = a^3(2t-b-c) + t(t-a)(b-c)2 >= 0.
ii) f(a, t, t) = 2a^3t + 2at^3 + 2t4. So, f(a, t, t) <= 6 <=> a^3t + at^3 + t4 <= 3 <=> at(3-t2) + t4 <= 3 <=> a2t2(3-t2)2 <= (3 - t4)2 <=>  (3-2t2)t2(3-t2)2 <= (3-t4)2 <=> 3(t2-1)2(t4-3t2+3) >= 0  which is true.

This approach is not very nice, but very powerfull.

Namdung

Thank you very much, Namdung. Your solution is excellent! However, I will try (in vain, of course) to find another solution to this strong inequality. Congratulations!

Finally, I have found a 3-lines solution:
  All we have to do is to observe that x^4+y^4+4x^2y^2>=3xy(x^2+y^2) and thsi is true because it can be written (x-y)^2(x^2-xy+y^2)>=0. Thus we have a^4+b^4+4b^2a^2>=3ab(a^2+b^2) and the others. Now, sum them up. I maintain my opinion that this is a tough inequality.


Hi all,
My boss passed me 400 keys for 40 rooms (10 keys/room), they come in bunches of ten (1bunch/room) and ask me to find out which bunch belongs to which room, because the stupid supplier labeled the keys all wrongly.

How can I do this rather than the normal way (try and error)? Please help :( Many thanks in advance.

there are 10 keys to each room ? i mean 10 keys that are the same ?

No, all the keys are different from each other and so as the drawers and rooms. (10 drawers/room~10keys/room).

Haha I am sure that trial-and-error is the only way. But guessing usually uses less time :P

if the keys are ALL labelled wrongly, that slightly reduces your options, only look at the derangments.

So of key X is for room(insert random number from 1-40 here) and says that it is for a different room, you don't have to try that different room.


The only rational point on the curve $ y^{2}\equal{}x^{3}\plus{}x$ is $ \left(0,0\right)$. Prove it!

$ (0,0)$ has order $ 2$ in the elliptic curve's group. Check that the only torsion points are $ \mathcal O$ and $ (0,0)$ by Nagel-Lutz' theorem.
Now you may check that $ [E: 2E]\equal{}2$, giving that there is no free part in the group (as $ [E: 2E]\equal{}2^{r\plus{}s}$ with $ r$ the rang of the free part and $ 2^s$ the order of the 2-torsion). To check this, you may unfold the way $ [E: 2E]$ is calculated in Silverman's and Tate's book: find that associated elliptic curve that belongs to the map $ (a,b) \mapsto (2a,b)$ and calculate the images.


Hi ,

Your problem does not belong to this section .

Here is 1 proof :

Just transform your inequality (mosbarzim ) into  $ (\frac{a^{3}}{c^{2}}\plus{}\frac{c^{3}}{a^{2}})\plus{}(\frac{b^{3}}{c^{2}}\plus{}\frac{c^{3}}{b^{2}})\plus{}(\frac{a^{3}}{b^{2}}\plus{}\frac{b^{3}}{a^{2}})\ge 6$

Then  apply AM-GM twice and here we done .

why $ abc\equal{}1$
AM-GM =>$ a^3\plus{}b^3\geq 2(ab)^{\frac{3}{2}}$
then $ S\equal{}\sum\frac{a^3\plus{}b^3}{c^2} \geq \sum \frac{2(ab)^{\frac{3}{2}}}{c^2}$
other AM-GM: $ S \geq 6$

and a smal remarque, Your problem does not belong to this section, you must put it in inequality section.


Sea ABCD un cuadrilatero convexo. Sean P y Q sobre AD y BC respectivamente tales que cumplan:   $BP/PC=AQ/QD=AB/CD$.
Si BA y CD se intersectan en R, demostrar que la bisectriz $l$ del angulo BRC y la recta PQ son paralelas.

Deberia ser $P$ sobre $BC$ y $Q$ sobre $AD$.

Sean $ X,Y$ dos puntos en las diagonales $ BD, CA$ tales que $ \frac {_{BP}}{^{PC}} \equal{} \frac {_{BX}}{^{XD}} \equal{} \frac {_{AY}}{^{YC}} \equal{} \frac {_{AQ}}{^{QD}}.$ De este modo tenemos $ PX \parallel QY \parallel RD,$ $PY \parallel QX \parallel RA$ y por otro lado:

$ \frac {PX}{CD} \equal{} \frac {PB}{BC} \ , \ \frac {PY}{AB} \equal{} \frac {PC}{BC} \Longrightarrow \frac {PX}{PY} \equal{} \frac {CD}{AB} \cdot \frac {PB}{PC} \equal{} \frac {AB}{CD} \cdot \frac {CD}{AB} $

$\Longrightarrow \ PX \equal{} PY  \Longrightarrow \ PXQY$ es un rombo.

Por consiguiente la diagonal $ PQ$ de $ PXQY$ es bisectriz de $ \angle XPY$ y como las rectas $ PX,PY$ son paralelas a $ RC,RB,$ entonces sus bisectrices $ PQ$ y $ l$ también lo son.


Find the length of the longest ladder that can be carried around the corner of a corridor (whose dimensions are indicated in the figure below). Assume that the corner is a right angle and that the ladder can only be carried parallel to the floor.

I'm pretty sure the longest ladder has length

[hide]

equal to the minimum value of a/sin:theta: + b/cos:theta:, over all values of :theta: less than 90 degrees.

of course, there must be better ways of doing this, since this looks a little nasty to find (without calculus and a little bit of pain).

[/hide]

Do you mean the maximum? If so, can you figure out what that is (in terms of a and b alone)?

I meant the minimum though, because [hide]that is the available length of ladder that can squeeze around the corner at some angle :theta:. Therefore, the ladder can be at most as long as the smallest of these values.[/hide]

why can't you fit a ladder of length sqrt(a^2 + b^2)?

You can fit a ladder of that length; you can also fit significantly longer ladders as well -- in fact, you can get a ladder of length up to [hide](a^(2/3) + b^(2/3))^(3/2)[/hide], I think, around the corner.  Not 100% sure on that, though.

ComplexZeta wrote:Oh I see what you're saying. Isn't that just [hide]min{a,b}[/hide] though?

I dont think so... the cos and sin terms are less than 1, so I think the soln lies elsewhere.

Assuming Nukular's characterization is correct, [hide]we can use Holder's inequality to minimize the quantity.  Let p = 2/3 and q = -2.  Their reciprocals add to 1.  (The negative value means that the usual sign in Holder's inequality is reversed.)  Then we get that a/sin(:theta:) + b/cos(:theta:) is at least (a2/3 + b2/3)3/2 times (sin2(:theta:) + cos2(:theta:))-1/2.  The last factor is just 1 by the usual trig identity.  Thus we end up with the same expression that Joel wrote.  The equality case is achieved when tan(:theta:) = (a/b)1/3.

Or we can use calculus to get the same bound.[/hide]

And what if we allow tilting the ladder, and say that the height of the corridor is h?


Which of the following numbers has the largest reciprocal?

\[ \textbf{(A)}\ \frac{1}{3} \qquad
\textbf{(B)}\ \frac{2}{5} \qquad
\textbf{(C)}\ 1 \qquad
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ 1986
\]

The number has to be as small as possible, so A.

[hide="Click for solution"]
Because the numbers are [b]all positive[/b], we want the smallest number, or $ \boxed{\textbf{(A)}}$.
[/hide]

If the number is positive, the answer is the one closest to 0. For example take a look at 1/100000000000. The reciprocal would be very very large.


Suppose that I give an n digit number, is there a way to distinguish if it's a perfect square

[hide]It has to end in a 1, 4, 9, 6, or a 5. If it does prime factorize it. If it has an even number of digits for each power than it is. If not then it isn't.[/hide]

[hide="Another way"]If $n$ is a perfect square, it has to be congruent to $0,1\pmod 4$ or $0,1,4\pmod 8$.[/hide]

[quote="Chocolate milk"]This's my answer. Is it true?
[hide] If it's a perfect square, it will end in 1,4,9,6 or 5.[/hide][/quote]

Yes, but not all numbers that end in $1$, $4$, $5$, $6$, or $9$ are perfect squares.  :)

press 2nd and the $x^{2}$ button and type the number in on your calculator

You guys all forgot $0\pmod{10}$. :roll:

The problem is decidedly unclear.  Do you want a quick computational method?  A quick computer algorithm?  A 100% foolproof positive test or just a negative test?

Edit:  A computational method I might use, if the problem ever arose.

Essentially, we approximate the floor of its square root.  Starting from there, we repeatedly square, taking care to hit only those numbers $\bmod 10$ or $\bmod 16$ or however you want to do it whose squares end in the right digit.  (Obviously we check quickly first to make sure it's a square $\bmod 10, 16, \text{etc.}$ at all.

The initial approximation:  If $n$ is even, take the floor of the square root of the leading digit (or the first three if you have those squares memorized like a good mathlete ;) ).  If $n$ is odd, the first two digits.

[quote="t0rajir0u"]...If $n$ is even, take the floor of the square root of the leading digit...[/quote]
What do you mean by that?

For example, $\sqrt{123476544}> \sqrt{123000000}= \sqrt{123}\cdot 1000 > \sqrt{121}\cdot 1000 = 11000$

You would then start squaring numbers greater than $11000$ whose squares end in $4$ (in other words, $\equiv 2, 8 \bmod 10$ until you find the right number.

[quote="Farenhajt"]Then again, why he doesn't take a long square root and see if it's integer? Do kids these days still know how to take square root by pen and paper and no calculator?[/quote]

Maybe he wants a really quick, one-glance way to do it.  :maybe:

[quote="life=tennis+math"][quote="Farenhajt"]Then again, why he doesn't take a long square root and see if it's integer? Do kids these days still know how to take square root by pen and paper and no calculator?[/quote]
I certainly wasn't taught this, but would appreciate if you could show how (as in, I know like the first 37-38 but I don't know how I could find the square root of say 26803 without a calculator). So yeah, can you show this method?

In addition, perfect squares ending with $5$ should actually end in $25$, and ones ending with $0$ should end with an even amount of $0$s, except for $0$ itself.[/quote]

Arrgh. I knew how to do that. but i forgot.  :blush:

See [url]http://mathforum.org/library/drmath/sets/select/dm_sqrt_nocalc.html[/url], especially [url=http://mathforum.org/library/drmath/view/52610.html]Longhand Square Roots[/url]


conics
ellipse
geometry proposed
geometry

Given an ellipse e with focii F and F' and a 2n-gon $A_1A_2...A_{2n}$ whose sides $A_1A_2$, $A_2A_3$, ..., $A_{2n-1}A_{2n}$, $A_{2n}A_1$ touch the ellipse e. Prove that $\measuredangle A_1FA_2+\measuredangle A_3FA_4+...+\measuredangle A_{2n-1}FA_{2n}=180^{\circ}$.

  Darij

What we need to prove is the following thing:

Given two fixed lines $d,d'$ tangent to the ellipse $\mathcal E$ with foci $F,F'$ in $T,T'$ respectively, and a line $\ell$ tangent to $\mathcal E$ in $P$ which cuts $d,d'$ in $A,A'$ respectively, as long as $\ell$ remains on the same side of the ellipse (the ellipse divides the angle formed by $d,d'$ in two regions which have $T,T'$ in common), the angle $\angle AFA'$ is constant $(*)$.

We can move the sides of the $2n$-gon one by one with small movements until the sides $A_1A_2,\ldots,A_{2n-1}A_{2n}$ become the sides $A_2A_3,\ldots,A_{2n}A_1$ respectiely. $(*)$ then proves that $\sum\angle A_{2k-1}FA_{2k}=\sum A_{2k}A_{2k+1}$ (the indices are modulo $2n$), and since the two sums add up to $2\pi$, we get the result.

In order to prove $(*)$ we can use the fact (discussed before and proven on the forum, I think; I'll prove it if someone wants me to :)) that $\angle TFA=\angle PFA,\ \angle T'FA'=\angle PFA'$, so $\angle AFA'=\angle PFA+\angle PFA'=\frac 12\cdot\angle TFT'$, which is fixed (but it has to be measured in the right direction; that's why it's important that $\ell$ stays on the same side of the ellipse).


geometry
parallelogram
trigonometry
algebra
polynomial
geometry solved

Consider an octagon with equal angles and with rational sides. Prove that it has a center of symmetry.

Take octogon $A_1A_2A_3A_4A_5A_6A_7A_8$ such that $\overline{A_1A_2}$ is parallel to the $x$-axis. Let $d_n$ denote the length of the edge connecting vertices $A_n$ and $A_{n+1}$ where $A_{9} = A_1$. Clearly, $d_1 + d_2 \cdot \frac{1}{\sqrt{2}} + d_3 \cdot 0 - d_4 \cdot \frac{1}{\sqrt{2}} - d_5 - d_6 \cdot \frac{1}{\sqrt{2}} + d_7 \cdot 0 + d_8 \cdot \frac{1}{\sqrt{2}}$
$= (d_1 - d_5) + \frac{1}{\sqrt{2}} \cdot \left(d_2 - d_4 - d_6 + d_8\right) = 0$. Since the $d_i$ are rational, it is clear that $d_1 = d_5$. Then we have that $A_1A_2A_5A_6$ is a parallelogram, thus, its diagonals bisect each other. Simiarly, all the diagonals are current at a point $P$ where they bisect each other. $P$ is the desired point of symmetry.

Again, an algebraic approach: let $a_{1},\ldots,a_{8}$ be the side lengths of the equilateral octogon. Then, if
\[
\alpha=\cos\frac{2\pi}{8}+i\sin\frac{2\pi}{8}=\cos\frac{\pi}{4}+i\sin\frac {\pi}{4}
\]
then
\[
a_{1}+a_{2}\alpha+\ldots+a_{8}\alpha^{7}=0.
\]
Since $\alpha^{4}=-1,$ we obtain
\[
\left(  a_{1}-a_{5}\right)  +\left(  a_{2}-a_{6}\right)  \alpha+\left(a_{3}-a_{7}\right)  \alpha^{2}+\left(  a_{4}-a_{8}\right)  \alpha^{3}=0,
\]
therefore $\alpha$ is a common root of the polynomials
\[
f\left(  x\right)  =x^{4}+1
\]
and
\[
g\left(  x\right)  =\left(  a_{1}-a_{5}\right)  +\left(  a_{2}-a_{6}\right) x+\left(  a_{3}-a_{7}\right)  x^{2}+\left(  a_{4}-a_{8}\right)  x^{3}.
\]
If $\deg g>0,$ then the $\gcd\left(  f,g\right)  $ is a nonconstant polynomial with rational coefficients, implying that $f$ is reducible in $\mathbb{Q}%
\left[  X\right]  ,$ a contradiction. It results that the opposite sides of the octogon are congruent and the conclusion follows easily.


Its given a triangle $ ABC$.Penpendicular line $ l$ to $ A\minus{}median$ through $ A$ intersects $ B\minus{}altitude$ and $ C\minus{}altitude$ in $ P,N$ , respectively.Prove that $ AN\equal{}AP$.

Let $ C',B'$ be the projections of $ C,B$ on $ AB,AC$.to prove the equality we must prove:
$ \frac{AC'}{\cos \angle C'AN}\equal{}\frac{AB'}{\cos \angle B'AP}$
$ \frac{AC'}{\sin \angle BAM}\equal{}\frac{AB'}{\sin \angle CAM}$
$ \frac{AC'}{AB'}\equal{}\frac{\sin \angle BAM}{\sin \angle CAM}$,which is obvious.


function
real analysis
real analysis unsolved

Give an example of measurable functions $F \to \mathbb{R}$,$G \to \mathbb{R}$ such that their composition isn't measurable.

I believe the [url=http://mathworld.wolfram.com/CantorFunction.html]Cantor's function[/url] is somehow involved.

(I'll edit that post if I find something/erase it if I'm totally wrong)

How about this:

Every function from $\mathbb R$ to itself is a composition of two measurable functions.

You can modify the proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=measurably&t=115441]here[/url] (if it's correct, that is :)).


trigonometry
limit
calculus
calculus computations

compute this [b]infinite [/b]product?

(i managed to do it in 'bout 45+ mins...  :surf:  )

$\hspace* \;\;\;\;\;\quad \;\;\;\;\; \prod_{k=2}^{\infty}\;\frac{k^{4}-1}{k^{4}+1}$

[hide]
$Ans: \;\;\;\boxed{\frac{\pi\;\cdot\;\sinh{\pi}}{\cosh{( \pi\;\cdot\;\sqrt{2})}-\cos{( \pi\;\cdot\;\sqrt{2})}}}$
[/hide]

[quote="misan"]
where,
$\prod_{k=2}^{\infty}\;\frac{k^{4}-1}{k^{4}}\;=\;\frac{1}{4}\;\cdot\;\left( \frac{\sinh\pi}{\pi}\right)$

$\prod_{k=1}^{\infty}\;\frac{k^{4}+1}{k^{4}}\;=\;\frac{1}{2}\;\cdot\;\left( \frac{\cosh{(\pi\;\cdot\;\sqrt{2})}-\cos{(\pi\;\cdot\;\sqrt{2})}}{\pi^{2}}\right)$[/quote]

But how did you know these identities? Did you have them memorized, look them up, or derive them?


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_0^{\frac {\pi}{2}} \frac{dx}{4sinx\plus{}7cosx\minus{}1}$

hello, this will help you
http://www.mathpropress.com/stan/bibliography/trigSubstitution.pdf
Sonnhard.

It's obvious ( :rotfl: ) that $ \minus{} \frac {1}{4}\left(\mbox{arctanh}\left(\frac {1}{4}\left(2 \minus{} 3\tan\left(\frac {x}{2}\right)\right)\right)\right)$ is a primitive of $ \frac {1}{4\sin x \plus{} 7\cos x \plus{} 1}$ (well, it's true, you can check it for yourself).

So the answer is $ \minus{} \frac {1}{4}\left(\mbox{arctanh}\left( \minus{} \frac {1}{4}\right) \minus{} \mbox{arctanh}\left(\frac {1}{2}\right)\right)$, which happens to be equal to $ \frac {\log 5}{8}$

hello, i think you have a typo, the integrand must be $ \frac{1}{4\sin(x)\plus{}7\cos(x)\minus{}1}$.
Sonnhard.

Thanks, you're right. Unfortunately, the typo appeared in the very beginning of my calculation. Then it gets easier: the antiderivative is $ \minus{} \frac {1}{4}\left(\mbox{arctanh}\left(\frac {1}{2}\tan\left(\frac {x}{2}\right)\right)\right)$, thus the answer is $ \frac {1}{2}\mbox{arctanh}\frac {1}{2}$

2 Sonnhard
Danke shon :) , your file helped me to solve this integral


The vertices of a regular $ 2n$-gon (with $ n>2$ an integer) are labelled with numbers $ 1,2,...,2n$ in some order. Assume that the sum of the labels at any two adjacent vertices equals the sum of the labels at the two diametrically opposite vertices. Prove that this is possible if and only if $ n$ is odd.

[hide="Solution"]First we show that it's possible for $ n$ odd.  For every odd integer $ k$ with $ 0
<k<2n$ we label the $ k$th vertex with the number $ k$.  Note that if $ k$ is odd then $ k\pm n$ is even, so for each vertex we already labelled, the vertex diametrically opposite is still unlabelled.  Then we put every even number $ 2m$ diametrically opposite from $ 2m\minus{}1$.  Now every choice of two adjacent vertices gives us one with odd label, which is one less than the label of the vertex diametrically opposite it, and one with even label, which is one more than the label of the vertex diametrically opposite it.  Therefore the equal sum property holds.

Now we show that it's impossible for $ n$ even.  Suppose there were a working labelling.  Then consider the first $ n$ vertices.  We can split these into adjacent pairs since $ n$ is even, and the sum of the labels in all these pairs is equal to the sum of the labels of the diametrically opposite pairs.  This means that the sum of the labels of the first $ n$ vertices, and similarly any set of $ n$ adjacent vertices, is half of $ 1\plus{}2\plus{}\cdots \plus{}2n$.  But then the sum of the first $ n$ labels is equal to the sum of the second through $ n\plus{}1$st lable, which means that the first and $ n\plus{}1$st vertices are labelled with the same number.  Contradiction.[/hide]


complex numbers
superior algebra
superior algebra unsolved

Let $ S$ be a set of $ m$ pairwise different complex numbers $ z_1, \ldots z_m$ such that if $ \sum_{ 1\leq i \leq n \atop z_i \in S} z_i \equal{}\sum_{1\leq j \leq n  \atop  {z}_j' \in S} {z}_j'$ then there exists $ \sigma \in \mathfrak{S}_m$, $ \forall i,j,$ $ {z'}_j \equal{} z_{\sigma(i)}$.
How many different complex numbers $ \sum_{1\leq i \leq n \atop z_i \in S} z_i$ are there ?

Can you do the case $ m\equal{}1$?  How about $ m\equal{}2$ and $ m\equal{}3$?  Do you see a pattern yet?

$ m\equal{}1$ ... You have a set $ S$ made up of of one (that's [b]1[/b] only!) complex number.  How many different answers can you get if you add together $ n$ terms, all chosen from the set $ S$?
And your answer is $ n$?  That's not my answer.

Of course, typo, I corrected. The rest should be OK.


Dear everymathlinker
I was noticing that myth is absent for a long time and harazi is leaving .
Myth is by far better than a lot mathematician i have ever met( i am more than forty  years old):i noticed that he is great in global viewing thing and very sympathetic to have him around you.
Harazi also is just like myth and great in seeing all the details wich also a kind of global viewing thing.
I propose for every math linker to send a message to wyth and harazi to beg them to come back.
They are for us like sun to the earth:we were moving aroud them to get math-light and math-warmness.

People come, people go. Most likely they have moved on to busier life, especially college/university level of math research requires substantial time. I started in MathLinks as well and I remember Myth, Lagrangia, harazi, grobber, and others, were the most frequent participants of ML. This is a moment I'll never forget.

After the AoPS and ML communities merge, things have changed. Some people thought it was bad, but some liked it. In any case, it depends on the willingness to further enjoy solving problems in ML. For example, like me, I started with Peter VDD in the Basics section and I remember solving problems with him. Now I see Peter VDD has moved on to college and I sometimes return to check out the olympiad section to solve problems for fun.

As for the old timers like Myth, harazi, Lagrangia,... the question whether they would return should be left intact. I see senouy wants every ML'ers to send them a message to bring them back but that's not appropriate. Bombarding them with tons of messages is not the solution. Let it as it be, sooner or later, whenever they feel like solving problems again, they will come back to greet us

3X.lich

Yes, you are right. I think it is not a good idea to beg them to return. Personally, I won't do that. I apreciate them, but we have to go further. Things like this happened at "big houses", but....they went further. :D
In conclusion, I have to say that I learned lots of things from Harazi (Myth had already left the forum when I joined in), but like i said: we have to pass that. ;)

I see,yes you're right.But i think it would be nicer if they left the forum in beauty not because of misunderstanding or  dispute.It seemed to me that myth left in angriness. Myth : a great talented mathematician.But let the bitterness make life sweater for them.

[quote="romano"]I saw Myth on ML yesterday . It's a good sign ![/quote]Don't worry. As a matter of fact Harazi enters frecquently on the College Playground now. I know that he is very busy with his upcoming examinations for Ecole Normale Superioure, so he appears less often on the site now ;)

Yes, they are great problem solvers and I enjoy reading their proofs. I mean I learned so much from them so I miss them and their solutions...I hope they'll be back :)

[quote="3X.lich"]Bombarding them with tons of messages is not the solution.[/quote]

I agree with you, 3X.lich!!!!!.


How many distinct positive integers can be represented as the
difference of two numbers in the set $ \{1, 3, 5, 7, 9, 11, 13\}$?

the first thing we can do is find all possible two digits sets
but after a little thought, we see that if we consider the sets $ (1,3), (1,5), (1,7), (1,9), (1,11), (1,13)$, we'll have considered all the options

now we see that we only have to consider the positive integers
therefore, in each of these sets, there is only one difference

with that said, we see that there are $ \boxed{6}$ distinct positive integers


Eight rooks are placed on an 8 x 8 chessboard, so that there is just one rook in each row and column. Show that we can find four rooks, A, B, C, D, so that the distance between the centers of the squares containing A and B equals the distance between the centers of the squares containing C and D.

do the two pairs $A, B$ and $C, D$ have to be disjoint, or just distinct? if just distinct, then for each $i$, $i = 1, ..., 8$ let $x_i$ and $y_i$ denote the column and row, respectively, of the $i$-th rook. there are $\binom{8}{2} = 28$ pairs of rooks, and for each pair $i, j$, $i\neq j$, the pair $\{|x_i-x_j|,|y_i-y_j|\}$ of values has its elements from $\{1, ..., 7\}$ (because no two rooks are in the same row or column). there are $\frac{7\cdot7-7}{2}+7 = 27$ such pairs of values, and 28 pairs of rooks, so there must be 2 pairs of rooks,$(i, j)$ and $(k, l)$, $(i, j) \neq (k, l)$, for which $\{|x_i-x_j|,|y_i-y_j|\} = \{|x_k-x_l|,|y_k-y_l|\}$. the distance from $i$ to $j$ is then the same as the distance from $k$ to $l$.

[quote]...there are$\frac{7\times 7-7}{2}+7=27$  such pairs of values...[/quote]

why?

and $\frac{7\times 7-7}{2}+7=28\not =27$


Let be $ a,b,c,d\in \mathbb{R}$ such that $ (1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)(1\plus{}d^2)\equal{}16$ . Prove that :

$ \minus{}3\le ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd\minus{}abcd\le 5$

When does the equality occurs ?

[quote="alex2008"]Let be $ a,b,c,d\in \mathbb{R}$ such that $ (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2)(1 \plus{} d^2) \equal{} 16$ . Prove that :

$ \minus{} 3\le ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd \minus{} abcd\le 5$

When does the equality occurs ?[/quote]
We have:
$ (ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd\plus{}ac\plus{}bd\minus{}abcd\minus{}1)^2 \equal{}((1\minus{}ab)(cd\minus{}1)\plus{}(a\plus{}b)(c\plus{}d))^2 \le ((1\minus{}ab)^2\plus{}(a\plus{}b)^2)((1\minus{}cd)^2\plus{}(c\plus{}d)^2) \equal{}16$
Then,we have Q.E.D

[quote="arqady"][quote="alex2008"]Let be $ a,b,c,d\in \mathbb{R}$ such that $ (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2)(1 \plus{} d^2) \equal{} 16$ . Prove that :

$ \minus{} 3\le ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd \minus{} abcd\le 5$
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78246[/quote]

The same question : When does the equality holds ?

this is a trigonometric ineq which equivalent to : $ \minus{}1 \le Cos \ \alpha \le 1$


Show that $ (a \plus{} b)(a \plus{} c)(b \plus{} c)$ is equivalent to $ (ab \plus{} ac \plus{} bc)(a \plus{} b \plus{} c) \minus{} abc$. Can a similar generalization be made about $ (a \minus{} b)(a \minus{} c)(b \minus{} c)$? How about $ (a \plus{} b)(a \minus{} c)(b \minus{} c)$?

[hide="Solution"]Suppose $ a,b,c$ are the roots of the cubic polynomial $ x^3\plus{}dx^2\plus{}ex\plus{}f$.

By Vieta's, $ ab\plus{}ac\plus{}bc\equal{}d$, $ a\plus{}b\plus{}c\equal{}\minus{}e$, and $ abc\equal{}\minus{}f$

Then $ (ab\plus{}ac\plus{}bc)(a\plus{}b\plus{}c)\minus{}abc\equal{}f\minus{}de$

We can write $ (a\plus{}b)(a\plus{}c)(b\plus{}c)\equal{}(\minus{}e\minus{}c)(\minus{}e\minus{}b)(\minus{}e\minus{}a)\equal{}\minus{}(e\plus{}c)(e\plus{}b)(e\plus{}a)$. We'll treat this as a polynomial in $ e$. We can see pretty easily that if we pick an $ e$ from one of the factors, we must necessarily choose two of $ a,b,c$ from the other two to get an $ e$ term, so the coefficient of $ e$ is $ \minus{}ab\minus{}ac\minus{}bc\equal{}\minus{}d$. Obviously the term that doesn't include $ e$ is $ \minus{}abc\equal{}f$. Putting it all together, we get $ f\minus{}de$, which proves the equality.[/hide]

Meh, there might be a better way, but at least I didn't expand it all out.

Hello!!!
[hide="generalization for 2"]
$ (a \plus{} b)(a \minus{} c)(b \minus{} c)$
let $ c \equal{} \minus{} c$
$ (ab \minus{} ac \minus{} bc)(a \plus{} b \minus{} c) \plus{} abc$
[/hide]
Also, expanding to get the solution isn't that complicated
:P


i found this information and thought you may be interested. here is what you have to do.

Finding the cube root of a number without calculator
For finding the cube root of any number, first put a comma after 3 digits from right.
e.g. Finding the cube root of 9261.
Put a comma after three digits from right. i.e. 9,261.
Now check the last digit of the number, here 1. And put the number whose cube has the last digit as the last digit of the number whose cube root is to be found. This will be the RHS of the answer.
Now, check the number on the left of the comma.
Check which number has its cube less than the number left of the comma. Over here, it will be 2 as cube of 2 = 8 and 8 is less than 9 and the cube of 3 will be more than 9. This will be the LHS of the answer.
So the answer will be 21.

for more comments on this topic visit http://www.sosmath.com/CBB/viewtopic.php?t=10267&highlight=

Thank You.

Basically you know the last number as each number mod 10 will cube to give a different number. Then, you cross out the last three digits, and take the floor of the cuberoot of what's left. This only works for perfect cubes of course.

$1728$. $8$ gives $2$, and $1$ gives $1$. so is $12^3$.
$262144$. $4$ gives $4$, $6^3 = 216 \leq 262$ ($7^3 = 343 > 262$). so is $64^3$.

And so on.

I think I get it...

So for 148877, 3^3 is 27 so the last digit is 3, and 148 yields 5 as 5^3 is 125 and 6^3 is 216, so - Voila! - 53.

Well, from what I've seen, it only works with perfect cubes. I'll try and prove it in a sec.

[hide="Okay. Here goes nothing"]

Assume x is a perfect cube. Let's call the cube root of x, "y." Notice that every one digit number (including zero) when cubed has a different units digit. (Try this, it's true!) Therefore, whatever digit is the units digit of x, there is a distinct units digit for y; whichever one-digit number when cubed has the units digit that x has. If x<1000, we can stop here, since 10^3=1000 so all the smaller cubes are under 1000.

Now, assume x is such that 1000<=x<1,000,000. x will be a two-digit number, since 100^3=1,000,000 and 10^3=1000. A two-digit multiple of ten cubed is always 1000 times the tens digit alone cubed. (A0^3=[A*10]^3=A^3*1000.) Therefore, taking the cube root of the part to the left of the comma gives the tens digit. But wait! It may not be a perfect cube!  :o Yes, the extra numberage added on by a units digit greater than zero will probably add something to this part to the left of the comma. However, it will always be less than the next multiple of ten cubed, so take the cube root of the greatest one-digit cube [i]less[/i] than this part to get the tens digit.

If x has a millions part, repeat with the hundreds digit of y. A billions part, the thousands digit, etc.

[/hide]

Wow, I think I did it! :D

[quote="09husbri"]
Notice that every one digit number (including zero) when cubed has a different units digit. (Try this, it's true!)
[/quote]
Doesn't

0^3=0
1^3=1
4^3=64
5^3=125
6^3=216
9^3=729

???

Good proof though

I think there was a bit of a miscommunication. What I meant was that the ten cubes of the one-digit numbers have all different unit digits. What biffanddoc thought I meant was that the cubes of the one-digit numbers all have different units digits than their roots, which is indeed false. Sorry if I wasn't clear enough.

[b][u][color=red]I'm definitly lost I wish I could understand you[/color] :wacko:  :wacko: [/u][/b]


For a positive integer $ A$, define $ a_k \equal{} A^{k \plus{} 2} \plus{} (A \plus{} 1)^{2k \plus{} 1}$ for all nonnegative $ k$.
Find, in terms of $ A$, the largest positive integer $ n$ such that $ n|a_k$ for all nonnegative $ k$.

[hide="Solution"]$ a_0=A^2+A+1$, so $ n\le A^2+A+1$. Taking the definition $ \pmod {A^2+A+1}$, we have:
\begin{eqnarray*}
a_n&=&A^2(A)^k+(A+1)(A^2+2A+1)^k\\
&\equiv&A^2(A)^k+(A+1)(A)^k\pmod {A^2+A+1}\\
&\equiv&A^k(A^2+A+1)\pmod {A^2+A+1}\\
&\equiv&0\pmod {A^2+A+1}
\end{eqnarray*}So equality can be achieved and thus the maximum possible $ n$ is $ n=A^2+A+1$.[/hide]


Farmer John stores grain in a large silo located at the edge of his farm. The cylinder shaped silo has one flat, rectangular face that rests against the side of his barn. The height of the silo is 30 feet and the face resting against the barn is 10 feet wide. If the diameter of the silo is 14 feet, determine the capacity of Farmer John's silo in cubic feet of grain.

lol  :lol:  How are you on right now? I’m at school  :D

School was over one week and two days ago
do you bring a laptop to school?

what? you're lucky! we were in the computer lab and I was doing math (COP round 2 [b]#6 + #7[/b]  :mad: to be specific :P) instead of my work.  :D I also go to the library at lunch and stuff sometimes too

[quote="nat mc"]$\frac{49\pi(180-\sin^{-1}{\frac57})}{6}+300\sqrt6=1470\pi+300\sqrt6-\frac{49\pi\sin^{-1}{\frac57}}6$[/quote]
that is the most complicated and unsimplified answer i've seen on this forum

yeah, it's kinda convoluted. However, type it into a calculator and the answer is correct.


I will admit, i got 87. that's right is stink :(

What stinks more is that i took the packet home and realized all the problems were pretty easy. I think I just got really nervous.

I got a 123 with 2 REALLY stupid errors. For #9 I forgot to divide by 2 to find the average and for #13 I took the sum of the 2008 terms instead of the 2008th term :( . And I felt stupid when after the competition I was easily able to do the cylinder/water problem (#19).

[quote="tenniskidperson3"]Ha! I got 144, missing number 25.[/quote]

same...............

[quote="the future"][quote="mround"]for #23, I got three integer pairs, but I forgot b > a, so I chose (C)  :mad:

for #19, I do not even want to talk about why I got wrong.

But I am satisfied with 123, at least I have another chance at AIME (I hope) this year.

Anyone is going to Univ of Iowa for ARML on May 29-31? I made into the Chicago team.

Good luck to you all.[/quote]

yea i am...going to ARML
i think anyone can make Indiana teams...
For the gold, im guessing you have to make USAMO though since im guessing atleast 15 people from Indiana will make USAMO this year.[/quote]

i'll see you there, shawn.

130.5!!!!!!!!

That is a miracle for me, especially considering that I got less than 120 on the 10A.   :) :lol:

I got a 133.5.
At least I did better than on the 10a (114).

I got 138 (same as last year) but screwed up on #2 :blush: ...

I think I might get perfect next year(if i don't screw up).

I was sick when I was taking the test. But I still passed.


inequalities
induction
algebra unsolved
algebra

prove that
$ (1998n)! \le \left( \frac{3995n\plus{}1}{2} \cdot \frac{3993n\plus{}1}{2} \cdot \frac{3991n\plus{}1}{2} \cdot \dots \cdot \frac{n\plus{}1}{2} \right)^n$
for $ n \in N$

trivial. prove by induction more general ineq:

$ (kn)! \le \left( \frac {(2k \minus{} 1)n \plus{} 1}{2} \cdot \frac {(2k \minus{} 3)n \plus{} 1}{2}\cdot \dots \cdot \frac {n \plus{} 1}{2} \right)^n$
for $ n,k \in N$
(use AM-GM)


Dear Friends,
                  Can we not adopt,or say,assigned a  "Numerical
Value Points" in every math forum members contribution?Probably,our forum member will be more creative,and original
everytime they make their "post messaging" in this math forum.
Example,instead of just "RANKING," or making easy "RATINGS,"
we can designate a permanent "numerical value" in every post
message-"DEPENDING" on the "QUALITY" of the posted message.
Why not give it a "TRY." I believed that this is my best way of
making the "Art Of Problem Solving" more "COMPETITIVE."

                                          Sincerly,
                                        Mikhail S.

How would you come up with a system to determine the value of what's in each post?  Administrators and Moderators do not have enough time to read every post and give it a value based on its content.  While this may be a good idea, it is not very practical.

I think most people "know" whether certain users post good quality posts or not as good of quality posts.

Maybe we should have a Karma system. Any user can +1 or -1 any other user (once every two or three days for the same user-to-user interaction). Your "Karma" count will show up under your username and everyone starts at zero.

Actually.... nah.


In the equation:
                            x + y + z + w = 17
x, y, z, w, are all positive integers.

How many ordered quadruples (x,y,z,w) are there that satisfy the equation?

Let 17 dots  * * * ... * * represent the 17 in the equation. We can separate them with 3 dividers to create 4 regions that represent the values of the 4 variables. There are 16 spaces available to place the dividers, so there are $ \binom{16}{3} \equal{} \boxed{560}$ solutions.

You had the right reasoning/intuition about it, but you made a mistake about how to think of the string of 17 dots. Remember, if you have the 3 dividers there, they add to the amount of "dots" in the string.

Take a look at the 17 dots you have. As you said, there should be 3 dividers. So in your string of 17 dots, add 3 more dots, but make them look different, and they'll denote the dividers. So in total, your string now contains 20 characters. Now look at the repeating characters that you have. This just becomes a factorial problem.

Ok, but the dividers can not be at the ends and they can not be next to each other, so it is $ 16 * 14 * 12 \equal{} 2688$?

According to what you're saying it should be $ \frac{20!}{17!3!} \equal{} 1140$ ?

Neil30z you are wrong because x, y, z, and w are positive integers.  If they were nonnegative integers then your reasoning would hold.  modularmarc101's initial answer is correct.

You could just do 20 choose 3 and get 1140, because first you're adding the 3 dividers to the string, and then you could just choose the different positions of the dividers in that string.

Ihatepie is right, the answer is $ 560$.
[hide="why?"]
The reason is that $ x$, $ y$, $ z$, and $ w$ are all positive integers, not whole numbers.
To change the problem to whole numbers, we let $ a\equal{}x\minus{}1, b\equal{}y\minus{}1, c\equal{}z\minus{}1, d\equal{}w\minus{}1$.
Now we have $ a\plus{}b\plus{}c\plus{}d\equal{}13$, but $ a, b, c, d$ are whole numbers.

We can use balls and urns now to get $ \binom{13\plus{}3}{3}\equal{}\binom{16}{3}\equal{}\boxed{560}$
[/hide]

[quote="Ihatepie"]For anyone confused, modular had the right answer the first time, but for the wrong reasons.[/quote]
Why the wrong reasons?

[quote="modularmarc101"]But in that case, the arraw of dots has 14 spaces in total.. not 16 ??[/quote]
That's not how it works. Basically, you have 13 dots, and then you put in the 3 dividers. So you have 16 things in all. Then you have to choose three of those spots to be dividers. So it is $ \binom{16}{3}$

[quote="Neil30z"][quote="modularmarc101"]

According to what you're saying it should be $ \frac {20!}{17!3!} \equal{} 1140$ ?[/quote]

yes, now this answer is correct.[/quote]

Actually, this is the correct answer, not 560, because w, x, y, and z form an ORDERED quadruple, hence the four dividers are distinguishable.

No, the dividers aren't distinguishable. The spot to the left of the first divider is x, in between 1st and 2nd is y, etc.

Sorry everybody, but I just realized that the problem said positive, therefore modularmarc's first answer is correct. My answer (1140) is for nonnegative.


So i qualified for AIME for the first time, and I just heard about AoPS.  I heard AoPS 2 is good for AIME.  Should I read AoPS 1 first or can I start off with AoPS 2.  I don't think I'll be able to read both by next month when we take AIME.

If you made AIME, AoPS 1 will probably serve as mostly review for you.
That said, if you barely made AIME, you would probably want to go over the basics more as well.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1409478#1409478

similar thing.


conics parabola quadratics algebra polynomial

Source: MAML 2006 States!!!

Find the sum of solutions to $\sqrt{1-\sqrt{1-x}}=x$

Oh yea, do it in 3-4 minutes.

[hide]i get the answers 0 and 1, so the answer would be 1. are there any other solutions?[/hide]

[hide="Guess"]Since x can't be negative (sqrt on the LHS) and it can't be greater than 1 (nonreal LHS) that pretty much leaves 0 and 1, and anything in between. 0 works, 1 works, and my guess is that nothing in between works. So my guess is [b]1[/b]. Also, I think there are only two possible x because (I think) there are two intercepts of $y=\sqrt{1-\sqrt{1-x}}$ and $y=x$ since the first is a quartic (pretty much a parabola, no $y^3$ or $y$ term.)[/hide]

wow, yours is a lot more complicated. :)

Well, I just guessed 0 and 1 like you did, then I tried to see whether or not there were any others.

[hide]Um...
$\frac{-1+\sqrt{5}}{2}$? So then the sum is $\frac{1+\sqrt{5}}{2}$[/hide]

[quote="amcavoy"]Factor it and use the quadratic formula.  You get three solutions

$x=1,\frac{-1\pm\sqrt{5}}{2}$

So the sum is $0$.[/quote]

$\frac{-1-\sqrt{5}}{2}$ isn't a solution. x can't be negative because the right hand side has x and the left hand side has square root, which is already positive.

Oh, right.  I was looking at the polynomial I had after squaring everything...

EDIT:
[hide]Doing $\sqrt{1-x}=y$ we have,

$\{\begin{array}{cc}\sqrt{1-x}=y \\ \sqrt{1-y}=x\end{array} \Longleftrightarrow \{\begin{array}{cc}1-x=y^2 \\ 1-y=x^2\end{array}\Longleftrightarrow (y-x)(y+x)=(y-x) \implies x=y$ or $(x+y)=1$

the equation $(x+y)=1$ yields to $\sqrt{1-x}=1-x \implies x=0,1$ and the equation $x=y$ yields to $x^2+x-1 \implies x=\frac{\sqrt{5}-1}{2}$[/hide]

Yes you are right Farenhajt I type it faster and do not look for the other condition.

Ach, well, no more random guessings for me. Maybe I should have worked the problem  :P

[hide]
By inspection, if $\sqrt{1-r}=r$, then $r$ will be a solution of $\sqrt{1-\sqrt{1-x}}=x$.

Since square roots are nonnegative, $r$ must be nonnegative if not extraneous. Squaring both sides produces $r^2+r-1=0$ with solutions $\frac{-1\pm \sqrt{5}}{2}$, of which only $\frac{-1+\sqrt{5}}{2}$ is nonnegative.

Now convert $\sqrt{1-\sqrt{1-x}}=x$ to a polynomial equation; dividing by $x^2+x-1$ will give a quadratic. $\sqrt{1-\sqrt{1-x}}=x\Rightarrow -\sqrt{1-x}=x^2-1\Rightarrow 0=(x^2-1)^2+x-1$.

It turns out that there are obvious factors other than $x^2+x-1$, because the polynomial becomes $(x-1)([x+1][x^2-1]+1)=(x-1)([x^3+x^2-x-1]+1)=x(x-1)(x^2+x-1)$. So the other two roots are 0 and 1, both of which work.

The sum is therefore $0+1+\frac{-1+\sqrt{5}}{2}=\boxed{\frac{1+\sqrt{5}}{2}}$.
[/hide]


integration
calculus
calculus computations

\[ \sum_{n\equal{}1}^\infty \frac{1}{n^{1\plus{}2\epsilon}} \sim \int_1^\infty \frac{dx}{x^{1\plus{}2\epsilon}}\]

I can not understand it . why ?

thanks

$ \sum_{n\equal{}1}^{N}\frac{1}{n}\sim\ln(N)$ for large $ N$.


logarithms
induction
inequalities
real analysis
real analysis solved

I don't know this one! God, how ugly it is. But how interesting it seems also.

It is great this problem! Unfortunately, I can solve only the easy part, by observing that $\sqrt[k]{a}+1\geq\sqrt[k]{a+1}$. By induction, it follows that our sequence is greater than $\sqrt[p_1 p_2...p_{n-1}]{n-1}$ from where one part of the problem.

By the way, in your problem you should put $p_i$ greater than or equal to 1. If not, the other part of the problem is trivial.  ;) . So, can I assume that and to claim that I solved it? :rotfl:

I said that I cannot prove the second part of the problem. The best I could to is to majorize the sequence with $(lnn)^k$ for a suitable constant k.

Sorry. Probably I don't understand your previous reply :blush:

Oh, yes, in fact I asked if I can assume that $p_n>1$ for all n... ;) . In this case I can solve it.  :P . I really cannot obtain better bounds than a certain power of $lnn$ for this damn sequence.  :blush:

Hmm... But fedja wrote $p1,p2,...>1$. What is the question?

Exactly, but in this case $b_n$ is certainly bounded (not to speak about $a_n$ which is smaller than 2 in this case). So, I want to say that it must be replaced by $p_n\geq 1$, otherwise the problem is silly. And if we replace by $p_n\geq 1$, then I still do not have solution for the main part of the problem.

[quote="harazi"]Exactly, but in this case $b_n$ is certainly bounded (not to speak about $a_n$ which is smaller than 2 in this case).[/quote]
???
What about $p_n=1+1/2^n$?

So, $p_n$ are real numbers???? Oh, my god, I completely misunderstood the problem. Seeing they are orders of radicals, I thought they are natural numbers... Sorry. Anyway, the first part of the solution remains true.

[quote="harazi"]Anyway, the first part of the solution remains true.[/quote]
That's no question ;)

[quote="fedja"]Seems harazi knows Leningrad problems very well  :lol: . How about this one:

Let $p_1,p_2,\dots>1$. Let $a_n=\sqrt[p_1]{1+\sqrt[p_2]{1+\sqrt[p_3]{1+\dots+\sqrt[p_n]{1}}}}$.
Prove that $a_n$ converges if and only if the sequence $b_n=\frac{\ln n}{p_1p_2\dots p_n}$ is bounded.[/quote]

Next time please give the year and stage.

$p_n >1$
$u_n>0$

In your book does they say conditions for convergence of

$b_n=\sqrt[p_1]{u_1+\sqrt[p_2]{u_2+\sqrt[p_3]{u_3+\dots+\sqrt[p_n]{u_n}}}}$

What do you think about
 $b_n$ converge iff $\frac{\ln (u_1+u_2+...+u_n)}{p_1...p_n}$ is bounded ?

[quote="Moubinool"]

$b_n=\sqrt[p_1]{u_1+\sqrt[p_2]{u_2+\sqrt[p_3]{u_3+\dots+\sqrt[p_n]{u_n}}}}$

What do you think about
 $b_n$ converge iff $\frac{\ln u_n}{p_1...p_n}$ is bounded ?[/quote]
It is not that simple: when $u_n=1$ for all $n$ the condition you suggest does not agree with the answer for the original problem. I'll think of it though...

[quote="fedja"][quote="orl"]
Next time please give the year and stage.[/quote]
This one was invented specially for the book. It never appeared anywhere else before officially. For other problems it will be too hard for me now to trace them back to their origins; the authors of SPRA initially had an intention to include references but gave the idea up. Don't ask too much of poor me :(. Isn't it enough that I tell where I take the problems from?[/quote]

No, it is not. There are so many SPb problems and just giving the source it is really hard to look up the problem. If you use the SPb book 1961-1993 by Dmitrii Fomin the years are clearly indicated and it should be not a problem to provide that as well. If there is a problem that did not appear on a contest but you found it in a book you can give the book name and the page.  :)

I agree with Fedja (even if in the last days he killed me with these problems), first we should try to solve the problems and then worry about their origin.

HERE THE CASE IF ALL $p_n > 2$

 $a_n=\sqrt[p_1]{1+\sqrt[p_2]{1+\sqrt[p_3]{1+\dots+\sqrt[p_n]{1}}}}\leq w_n =\sqrt{1+\sqrt{1+\sqrt{1+\dots+\sqrt{1}}}}$

$w_n$ is convergent, bounded imply $a_n$ is bounded

$a_n$ is increasing bounded, convergent

The best result I managed to prove is that the sequence is convergent if $\sum_{n>1}{\frac{1}{np_1p_2...p_n}}$ converges. Unfortunately, it is not the case of the problem, but I find it already a nice result.  :(

I think I got it  :?
Please, wait, I will try to write it down.

Deceived... No, there is no problem, I made enormous efforts and I could not prove more...  :( . fedja, you won! But I must confess that even painfull, it was a nice death!  :(

Seems it's time for a
[hide="hint"]
Let $C p_1\dots p_n\geq \ln n$ for all $n$. Divide $a_n$ by $e^{2C}$....
[/hide]

[quote="Moubinool"]
$a_n=\sqrt[p_1]{1+\sqrt[p_2]{1+\sqrt[p_3]{1+\dots+\sqrt[p_n]{1}}}} \leq
K\sqrt[p_1]{\frac{1}{n}+\sqrt[p_2]{\frac{1}{n}+\sqrt[p_3]{\frac{1}{n}+\dots+\sqrt[p_n]{\frac{1}{n}}}}}$
...[/quote]
Actually it gives $a_n=\sqrt[p_1]{1+\sqrt[p_2]{1+\sqrt[p_3]{1+\dots+\sqrt[p_n]{1}}}} \leq
K\sqrt[p_1]{\frac{1}{1}+\sqrt[p_2]{\frac{1}{2}+\sqrt[p_3]{\frac{1}{3}+\dots+\sqrt[p_n]{\frac{1}{n}}}}}$   , which is very close to the solution but still not there ;)

$c_n=\sqrt[p_1]{\frac{1}{1}+\sqrt[p_2]{\frac{1}{2}+\sqrt[p_3]{\frac{1}{3}+\dots+\sqrt[p_n]{\frac{1}{n}}}}}$

maybe we should prove that $c_n$ is bounded to finish

Almost there Myth! :lol:  The only problem is that your very last inequality is wrong (you can remove the roots only if the numbers under the roots are at least $1$). Can someone now solve the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28979]general version[/url] by Moubinool?


Is there any really fast way to average out the numbers 70, 75, 80, 80, 80, 85, 90, 90, 95, 100 without actually adding and dividing by 10? (Yes this was a question from somewhere)

Shame on you, dragon.

Let's see how far they are from...85.

-15-10-5-5-5+5+5+10+15

Conveniently everything cancels but a -5, so the AVERAGE will be -5/10 + 85 = 84.5.

That helps. How about anything besides guassing and checking and then correcting for the error?

What do you mean by "Shame on you"?

It isn't a guess and check. For example, you could start with 42. You get:

28+33+38+38+38+43+48+48+53+58 = 425

Dividing by 10 gives 42.5 and adding that to 42 equals 84.5, which is our average. Obviously, it is a lot easier to pick an integer in the middle of all the data instead of a number very far away. This method works no matter what the numbers are.

If you pick 42, though, the math you have to do is probably harder than simply calculating the average, but when the numbers are close together like this problem, it is much faster using this method if you pick a number in the middle of the data.

[quote="PowerOfPi"]28+33+38+38+38+43+48+48+53+58 = 425[/quote]

Where did you get those numbers from?


FTW
probability
ratio
geometry
3D geometry
prism

There are a few problems on FTW I don't understand how to solve - can someone please explain them to me? Thanks in advance!

1. Three players each have a red card, blue card, and green card. The players will play a game that consists of three rounds. In each of the three rounds, each player randomly turns over one of his/her cards without replacement. What is the probability that, at the end of each game, one card of each color, was turned over in each of three rounds? Express your answer as a common fraction.

2. If all angles are measured in degrees, the ratio of three times the measure $ \angle A$ to four times the complement of $ \angle A$ to half the measure of the supplement of $ \angle A$ is $ 3 : 14: 4$. What is the measure of the complement of $ \angle A$?

3. When the width of a rectangular prism is increased by 2 cm, and the length is increased by 1 cm, the volume of the resulting cube exceeds the original volume by $ 96 \text{cm}^3$. Find the number of centimeters in the sum of the height, length, and width of the original box.

4. What is the smallest positive integer greater than $ \frac{3}{2} \plus{} \frac{3}{4} \plus{} \frac{3}{8} \plus{} \frac{3}{16} \plus{} \frac{3}{32} \plus{} \dots \plus{} \frac{3}{2^n}$, for a finite value of $ n$?

5. On each tile in a set of dominoes, each integer 0 through 12 is paired exactly once with each integer 0 through 12. One number of the pair is written on one half of the domino, and the other number is written on the other half. If one domino is randomly selected from all dominoes in the set, what is the probability that the same number will be on both halves? Express your answer as a common fraction.

Answers:

[hide="Number 1"]
$ \frac{1}{18}$
[/hide]

[hide="Number 2"]
$ 70$
[/hide]

[hide="Number 3"]
$ 15$
[/hide]

[hide="Number 4"]
$ 3$
[/hide]

[hide="Number 5"]
$ \frac{1}{7}$
[/hide]

Oh dang #1 is nats all over again. Those are days I don't want to relive...

4. That's $ 3(\frac{1}{2}\plus{}\frac{1}{4}\plus{}...\plus{}\frac{1}{2^n})<3(2)\equal{}6$ so 6.

Actually, for #4, $ (\frac{1}{2} \plus{} \frac{1}{4} \plus{} \frac{1}{8} \plus{} ...)$ is equal to $ 1$, so the answer is $ 3 \cdot 1$ or $ \boxed{3}$.

[b]#2[/b]:
$ 3A: 4(90\minus{}A): \frac{180\minus{}A}{2}\equal{}3: 14: 4$
Complement of A: 90-A
Supplement of A: 180*A
[b]#3[/b]:
$ a,b,c$-sides
$ abc\plus{}96\equal{}(a\plus{}1)(b\plus{}2)c\equal{}abc\plus{}bc\plus{}2ac\plus{}2c \equal{}>96\equal{}c(2\plus{}2a\plus{}b)$
Also $ a\plus{}1\equal{}b\plus{}2\equal{}c$, so $ a\equal{}c\minus{}1, b\equal{}c\minus{}2\equal{}> 96\equal{}c(3c\minus{}2)$
$ c\equal{}6, b\equal{}4, a\equal{}5$
$ a\plus{}b\plus{}c\equal{}15$
[b]#4[/b]:
Ernie is right
[b]#5[/b]:
Dominoes: $ \frac{13\cdot 14}{2}\equal{}91$
Pairs: 0-0,1-1,...,12-12- 13 pairs
Probability: $ \frac{13}{91}\equal{}\frac{1}{7}$

Hmmm... for #1, I got something different.

First player chooses a card.  Doesn't matter.  He chooses color A.

Second player must choose color B or color C.  He/she chooses Color B. (prob. $ \frac{2}{3}$)

Third player must choose color C.  He chooses color C. (prob. $ \frac{1}{3}$)

Thus, the total probability of this scenario in [b]one[/b] round is $ \frac{2}{9}$.  However, for it to be in $ 3$ rounds, we must cube $ \frac{2}{9}$ to get $ \boxed{\frac{8}{729}}$.

Wow, my answer is waaaaay different.

Ernie, you didn't account for the fact that after the first round player A only has colors B and C left.

[quote="mewto55555"]Oh dang #1 is nats all over again. Those are days I don't want to relive...

4. That's $ 3(\frac {1}{2} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{2^n}) < 3(2) \equal{} 6$ so 6.[/quote]

How is $ \frac {1}{2} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{2^n}\equal{}2$? Thats not even over 1. The infinite series is equal to 1, but since its finite, the number is a little under 3(1). SO the greatest number greater than that is 3.

I know I made a mental booboo. enrnie already called me on that one.

3. For simplicity sake lets just call angle a=x

$ 3x: 4(90\minus{}x): \frac{180\minus{}x}{2}\equal{}3: 14: 4$

Therefore (taking the first 2 only)

$ 42x\equal{}12(90\minus{}x)$
$ 42x\equal{}1080\minus{}12x$
$ 54x\equal{}1080$
$ x\equal{}20$

Thus the complement is 90-x=70

[quote name="Mewto55555" url="https://artofproblemsolving.com/community/p1267617"]
Whoops! Good catch ernie!

1-RGB
2-RGB
3-RGB

Round 1: 1 has 3 ways,2 has 2 ways,3 has 1 - 6 ways

Say 1 picked red, 2 picked green 3 blue,
1-GB
2-RB
3-RG

Round 2- A has 2, B and C have 1

Thus Round 3 has 1 way (one card left for all) So we have 6x2=12 ways

$ \frac{12}{3^3\cdot2^3}\equal{}\frac{1}{18}$
[/quote]

Great solution! The problem was not actually that hard :P I guess its easy to say that after you have looked at the solution.