Let $f(x)$ be a monic polynomial of degree $1991$ with integer coefficients. Show that the polynomial  $f(x)^{2}-9$ can't have more than $1995$ integer roots .

the numbers 1991, and 1995 seem pretty far fetched.

If f(x)^2 - 9 has more than 1995 roots, say, z_1, ..., z_r, w_1, ..., w_s with r + s > 1995 and f(z_i) = 3 and f(w_j) = -3 for every z_i, w_j,  then

f(x) - 3 = product(x - z_i) * h and
f(x) + 3 = product(x - w_j) * g

for some h,g.  z_i and w_j are distinct, pairwise.  h,g are in Z[x].

Then for any z_i,  f(z_i) = 3 so  6 = product(z_i - w_j) * g(z_i).

so |z_i - w_j| <= 6 for every i,j.  Since r + s > 1995, and z_i and w_j are pairwise distinct integers, this clearly cannot be true.


calculus
integration
trigonometry
calculus computations

$ \int \sin ^ 4 x\cos ^3 x\ dx\equal{}\int \sin ^ 4 x\cos ^ 2 x\cdot \cos x\ dx$

$ \equal{}\int \sin ^ 4 x (1\minus{}\sin ^ 2 x)\cdot \cos x\ dx$

$ \equal{}\int (\sin ^ 4 x\minus{}\sin ^ 6 x)(\sin x)'\ dx$

$ \equal{}\frac{1}{5}\sin ^ 5 x\minus{}\frac{1}{7}\sin ^ 7 x \plus{}Const.$

[quote="Identity"]Correct kunny, thanks. I'm not quite sure what you did there, pervez, but nice try.[/quote]

Kunny was integrating by the substitution method, you can regard $ \int (\sin ^ 4 x \minus{} \sin ^ 6 x)(\sin x)'\ dx$ as $ \int (u^4 \minus{} u^6) du$. You simply integrate this and then substitute $ u \equal{} \sin x$. :P


search
ratio
inequalities
combinatorics proposed
combinatorics

Two mathematicians, lost in Berlin, arrived on the corner of Barbarossa street with Martin Luther street and need to arrive on the corner of Meininger street with Martin Luther street. Unfortunately they don't know which direction to go along Martin Luther Street to reach Meininger Street nor how far it is, so they must go fowards and backwards along Martin Luther street until they arrive on the desired corner. What is the smallest value for a positive integer $k$ so that they can be sure that if there are $N$ blocks between Barbarossa street and Meininger street then they can arrive at their destination by walking no more than $kN$ blocks (no matter what $N$ turns out to be)?

A historic note: this problem is embased in a real history! Carlos Gustavo Tamm de Araujo Moreira and Nicolau Saldanha really lost in Berlin... And then, after some time created this problem!

In the original solution (of temself!) the mathematicians do not separate each other (or, in other words, consider only one lost mathematician :P).

This problem has been solved long before Nicolau Saldanha had been born...

The Mathematics literature contains hundreds of papers on variants and
extensions of this problem. In Computer Science, this problem is usually
called the "Cow-Path" problem.

Can anyone please post the solution? I got a totally weird answer that must be wrong. :(

i thlnk  the smallest value of k=3
It is given that they have to travel along martin luther street to get barb. street to meininger street.now they dont know which dirn. to go.let us assume they go along forrward dirn. crossing N blocks but they dont find the reqd.street ,noow they will be sure that they will reach their destination in backward dirn.since they can travel only along forward or backward dirn.Now total blocks covered --N  BLOCKS IN FORWARD DIRN.+N BLOCKS WHILE RETURNING +N BLOCKS IN BACKWARD DIRN.=3N BLOCKS where k=3 for any N .Similarly assume for starting in backward dirn.

But in the original statement, the value of $N$ is not known, am I right?

Pierre.

The minimum value of $k$ is 9.

We describe a strategy to achieve $k=9$ as follows.  We can represent the blocks on the street as integers, with 0 representing the current position.  A search will be represented by a sequence $(a_{1},-a_{2}, \ldots, (-1)^{n+1}a_{n}, \ldots)$, which indicates that you first travel to the integer $a_{0}$, then backwards to $-a_{1}$, then forwards to $a_{2}$, etc. (Of course one could change directions before travelling to new territory, but that would never be optimal.)  Set $a_{n}= 2^{n-1}$.  Let's say we find the correct block on our way to $a_{n}$;  then the block is some distance $2^{n-3}+i$, where $i \leq 2^{n-1}-2^{n-3}$.  The total distance we have travelled is

$2*1+2*2+\ldots+2*2^{n-2}+2^{n-3}+i = 2^{n}-2+2^{n-3}+i$

and the ratio is

$\frac{2^{n}-2+2^{n-3}+i}{2^{n-3}+i}\leq 9$.

For the opposite direction, assume that $k < 9$.  As above, suppose that we find the correct block on the way to $a_{n}$, then the ratio is

$\frac{2*a_{0}+2*a_{1}+\ldots+2*a_{n-1}+a_{n-2}+i}{a_{n-2}+i}$

For this to be less than or equal to $k < 9$, we need $\frac{a_{n-1}}{a_{n-2}}< \frac{k-3}{2}+e$ for sufficiently large $n$, for any $e > 0$.  So $\limsup_{n \in \bb{N}}\frac{a_{n+1}}{a_{n}}$ exists and equals some number $r$ with $1 \leq r \leq \frac{k-3}{2}$.  This means that, for any $\epsilon > 0$ there exists an $N > 0$ such that for all $n > N$, $\frac{a_{n+1}}{a_{n}}< r+\epsilon$.  So for $n > N+x+2$,

$\frac{2*a_{0}+2*a_{1}+\ldots+2*a_{n-1}+a_{n-2}+1}{a_{n-2}+1}$
$> \frac{\frac{2}{(r+\epsilon)^{x}}+\frac{2}{(r+\epsilon)^{x-1}}+\ldots+2+\frac{2*a_{n-1}}{a_{n-2}}+1+\frac{1}{a_{n-2}}}{1+\frac{1}{a_{n-2}}}$
$= \frac{\frac{2-\frac{2}{(r+\epsilon)^{x+1}}}{1-\frac{1}{r+\epsilon}}+\frac{2*a_{n-1}}{a_{n-2}}+1+\frac{1}{a_{n-2}}}{1+\frac{1}{a_{n-2}}}$
$< k$

As $n \to \infty$, $a_{n-2}\to \infty$, so $\frac{1}{a_{n-2}}$ can be made as small as we desire.  Also, $\epsilon$ and $x$ were arbitrary, so we have

$\frac{2}{1-\frac{1}{r}}+\frac{2*a_{n-1}}{a_{n-2}}+1 \leq k$
$\frac{a_{n-1}}{a_{n-2}}\leq \frac{k-3}{2}-\frac{1}{r-1}$  (1)

We claim that

$\frac{k-3}{2}-\frac{1}{r-1}< r$

After some algebraic manipulation,  we get

$r^{2}-\frac{k-1}{2}r+\frac{k-1}{2}> 0$

The discriminant is

$(\frac{k-1}{2})^{2}-4 \frac{k-1}{2}= \frac{(k-1)(k-9)}{4}< 0$,

so the above inequality holds for all $r$.

But (1) holds for all sufficiently large $n$, which means that the $\limsup_{n \in \bb{N}}\frac{a_{n-1}}{a_{n-2}}$ is bounded below $r$, a contradiction.  So $k = 9$ is the minimum value.

Hmm... what is wrong with this technique? Of course it's not time-efficient in reality, but in the asked problem (where only distance matters) I think it makes $k\le8$...

I will assume every block has the same length, but that doesn't really matter. A walks to the left, B walks to the right. At each corner they reach, both mathematicians wait sufficiently long (enough time for the other mathematician to catch them up if they'd have found the correct block). If the other doesn't come they walk on to the next block. So when one of the mathematician has the correct street (they both have walked $N$), then the mathematician who has found it returns in the other direction -- to find the other mathematician still waiting for him. THen they walk back together. So together they have walked $(1+1)+2+(2+2)=8$ times $N$ blocks, so $k\le 8$...

Bah... I had assumed, from "by walking no more than $kN$ blocks", that they would be walking together.  But now that I think about it, talking about two mathematicians probably implies that they can separate.  In that case, we need clarification on what "by walking no more than $kN$ blocks" means - is it the sum of what the two mathematicians travel, or the maximum?

I agree that if we assume it is the sum, then the minimum is $k=8$, although I'm not seeing a proof yet.  If it's the maximum, $k=5$ looks pretty good as well.


When a satellite is a distance R from the center of Earth, the force due to gravity on the satellite is F. What is the force due to gravity on the satellite when its distance from the center of Earth is 3R?

(1) F/9 (3) F
(2) F/3 (4) 9F

The gravitational force between two masses is inversely proportional to the square of the distance between them. So if the distance rises to the triple, the intensity of the force must be lower by a factor of 1/9. So the correct answer is (1), F/9.

:whistling: [hide="CLICK ME NOW"]
$F_{grav}= G\frac{m_{1}m_{2}}{r^{2}}$
Because $G$, $m_{1}$, and $m_{2}$ are all constants....
$F_{grav}= \frac{1}{r^{2}}$
so at 3x the distance; $F_{grav}= \frac{1}{3^{2}}= \frac{1}{9}$

Answer is $A)$ $\frac{F_{1}}{9}$[/hide]


calculus
derivative
function
real analysis
real analysis unsolved

If $ \displaystyle f: [a, b]\to\mathbb{R}$ is a twice differentiable function such that $ \displaystyle f^{{\prime}{\prime}}(a)\equal{}f^{{\prime}{\prime}}(b)$ then there exists $ c\in (a, b)$ such that
\[ \displaystyle f(c)\minus{}f(a)\equal{}(c\minus{}a)f^{{\prime}}(c)\minus{}\frac{(c\minus{}a)^{2}}{2}f^{{\prime}{\prime}}(c).\]

$ f''(a)\equal{}f''(b)\ \Longrightarrow \ \displaystyle\exists_{u\in]a,b[}\  (u\minus{}a)f''(u)\equal{}f'(u)\minus{}f'(a)$ (Flett's theorem)
Let's define  $ g(x)\equal{}(x\minus{}a)f'(x)\minus{}2f(x)\plus{}xf'(a)$ . It's easy to see that $ g'(x)\equal{}(x\minus{}a)f''(x)\minus{}f'(x)\plus{}f'(a)$ and $ g'(u)\equal{}g'(a)\equal{}0$
So (still  Flett's theorem), there is   $ c\in]a,u[\ \subset \ ]a,b[$ such that $ (c\minus{}a)g'(c)\equal{}g(c)\minus{}g(a)$ and we can write
$ (c\minus{}a)g'_{(c)}\equal{}g_{(c)}\minus{}g_{(a)} \Longrightarrow (c\minus{}a)^2f''_{(c)}\minus{}(c\minus{}a)(f'_{(c)}\minus{}f'_{(a)})\equal{}$$ (c\minus{}a)f'_{(c)}\minus{}2f_{(c)}\plus{}cf'_{(a)}\plus{}2f_{(a)}\minus{}af'_{(a)}\Longrightarrow f_{(c)}\minus{}f_{(a)}\equal{}(c\minus{}a)f'_{(c)}\minus{}\frac{(c\minus{}a)^2}2f''_{(c)}$
 :cool:


I got the vaccine a few weeks ago. It is exactly like the usual seasonal flu vaccine, aside from distribution/supply issues. It may be rationed based on need; if so, you'll have to demonstrate increased risk for some reason in order to get it.

As for what you're risking: that's also just like the usual seasonal flu, except that more people than usual are getting sick because nobody has resistance from previous flu seasons.

[quote="lovejenete"]Hi All,
I read a lot of articles about Swine Flu. I work in contact with people every day.
I am getting worried about the Swine Flu and I am not sure if I want the vaccine?
Has anybody here taken the vaccine?
Any problems from it?[/quote] There's been no recorded problems from the vaccine that have been verified. However, there is the ever-constant possibility of getting the swine flu. So basically, get the shot.

I recently got it a few months ago (early because I'm at high risk, due to being an asthmatic.) It was completely fine, I had a sore arm for a few hours and that was it. I haven't contracted any flu and I've been completely healthy ever since.

I got the one-dose nasal vaccination a couple weeks ago, and my baby brother got a two-dose nasal vaccination. I had no problems with that, but I got three other shots for other things, which may have distracted me.

I live in a sparsely populated area, so it is easier to get the vaccination, but my parents, for example, still haven't had the chance to get it. I think they will give it to babies first, then children, and then adults. Unless you have a special case where you are in great risk of getting it, I think you will have to wait your turn. Be sure to know when to go so they don't run out.

Our school district gave out the swine flu shots for free to any student that wanted it, so I just got mine last week.  The only ill effect I felt was that my right arm was sore for around 2 days afterwards.  So I agree with the above posts, its a potential gain for no potential loss.  (Of course, there was also that provision that you shouldn't get the shot if you're allergic to eggs... )

Agreed. The vaccine really is a good idea.

If you're worried about those people saying it causes autism, that's just laughable. 99.9% of the scientific community agrees the potential ills of the vaccine are very negligible, and furthermore, by not taking the vaccine you put the lives of others around you in jeopardy.

So, yeah. Vaccines are good.

[quote="infinite101"]The usual side effects are swollen arms, but it may be risky to take it.

It bypassed the 6 month safety guarantee, and in 1978, when the same vaccine was issued, more people were killed or harmed than those safe.[/quote]Correct me if I'm wrong, but I'm pretty sure there wasn't any major swine flu outbreak in 1978. Also, it isn't the same vaccine. The vaccines are changed or started from scratch ever year because the virus continues to evolve.

[url=http://en.wikipedia.org/wiki/1976_swine_flu_outbreak]1976[/url], not 1978. And the outbreak never left the grounds of Fort Dix - that particular strain seems to not have been very contagious. And Ihatepie is right, it wasn't the same vaccine.

Google combinations of words involving "1976 swine flu Fort Dix" and you'll get a good variety of sources.


A packet of 25 one-dollar bills weighs $ \frac{9}{10}$ of an ounce. Find the number of pounds that a million dollar bills would weigh.

since $ \frac{1,000,000}{25} \equal{} 40,000$, we can do

$ 40,000(\frac{9}{10}oz.) \equal{} 36,000 oz.$.

but we still have to convert it to pounds. so we can do

$ 36,000 oz. (\frac{1 lb.}{16 oz.})$.

this is essentially the same as multiplying by 1. the $ oz.$ factor out and we are left with

$ \frac{36,000}{16} lbs. \equal{} 2,250 lbs.$


Find all integers solution to the system:
$ a+b+c=24$
$ a^{2}+b^{2}+c^{2}=210$
$ abc=440$

This may not be the best way, but...
[hide]
$ abc = 440$
From this we know that either all variables are positive, or one is positive and two other is negative. If the latter is true, then
$ (-a)+(-b)+c=24$
$ c=24+a+b$
$ a^{2}+b^{2}+(24+a+b)^{2}>210$ since $ 24^{2}$ by itself is greater than 210. So all the variables are positive.

$ 440 = 2^{3}\cdot5\cdot11$
This tells us that whichever variable contain the factor $ 11$, is $ 11$ (that is, it doesnt contain any other factors of $ 2$ or $ 5$ which would make it $ 2\cdot11=22$ or $ 5\cdot11=55$). This holds because $ a+b+c = 24$, and if one of the variable is $ 22$ or greater, then $ a+b+c>24$.

WLOG let $ c=11$ Therefore $ a+b = 13$. We know $ 2^{2}\cdot5=20>13$, so either two cases exist: $ (2^{2}, 2\cdot5)$ or $ (2^{3}, 5)$. The first case would make $ a+b=14$ and the second case $ a+b=13$, so our solutions are $ \boxed{(5,8,11)}$ and their permutations.
[/hide]

[quote="riddler"]wow.. red_dog that quick? what did you use there?[/quote]
He used [url=http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas]this[/url].  :wink:


Fermat cutoffs are online.

[url=http://www.cemc.uwaterloo.ca/english/contests/past_result/2005/pcf_avg.shtml]www.cemc.uwaterloo.ca/english/contests/past_result/2005/pcf_avg.shtml[/url]

Does anyone know last year's cutoffs? It seems quite high, but I'm not sure.

No worries Jenn, they can't not invite us!  Honestly!  We're a legend now.   :D

[quote="billzhao"]I find those stats rather meaningless. The only thing that matters to you, Jenn, is whether someone in the Ontario West region got higher than you.[/quote]

That's also true Yufei, but somehow I get the impression that this was easier than other years. I just wanted to confirm that.. I thought maybe I could determine whether I could go to Seminar on the score. I'm fairly certain i'm not the highest scorer because Dr. Rice would have mentioned that to me then. He says all scores are in, and they're getting ready to post the results book. So ya. Stupid quadratic, I have no idea why I made that mistake.. *cries*

[quote="Sunny"]No worries Jenn, they can't not invite us! Honestly! We're a legend now. Mr. Green[/quote]

Somehow I find that very comforting (darn, why don't they have a sarcastic smiley here??), rooming next to Carley two years in a row, and being on students-to-watch-for list makes us legends eh? It'll be a great year, provided that we all make it back!!

Speak of the devil, the results are posted: http://cemc.uwaterloo.ca/english/contests/fermat.shtml

Congrats to gold medallists:
Yiru Chen
Lin Fei
Chris Hennick
Viktoriya Krakovna
Henry Ding Liu
David Rhee
Daniel Ye
Yufei Zhao

Does anyone know who (I assume) left a Part B question blank to score 146?!

Also, does anyone know who Daniel Ye from NB is?  I see he's listed under USAMO qualifiers as a grade nine student, yet on Fermat results he's 16 years old and in grade 11.  Or are there two people from the same school with the same name, who both happen to be really good at math?

Which question did you miss, Jenn?

I am not sure if I am right, but:
I presume I know who Daniel Ye (gr. 11) is.  He used to be in Sir Winston Churchill Secondary School in Vancouver in grade 10, and i was there too, (gr. 9).  I heard of poeple talking about him, but i never know him personally.  Then people told me that as I moved to Toronto, He moved to New Brunswick.

I missed either #22 or #23, the one with square in it..
That was a mistake, but it doesn't make a difference now..  :blush:

Hi, everyone. I am totally a neophyte here. I have never heard of a Waterloo seminar. Can anyone care to explain how do you qualify for it? COMC/Euclid/Fermat?Which one you need to write? And generally how many people are invited?


function
algebra
polynomial
integration
calculus
limit
complex analysis

Let $f$ be an entire function such that $lim_{z\to \infty} \frac{f(z)}{z^2}=0$. Prove that $f$ is a linear function.

[quote="The Captain"][quote="eugene"]It is obvious that if $f$ is entire and $f(z)=a_0+a_1z+a_2z^2+...$ and $lim_{z \to \infty} \frac{f(z)}{z^2}=0$ then $a_i=0$ for all $i \geq 2$.[/quote]
But we must prove that $f$ is a polynomial, eugene.[/quote]

No, i never meant that $f$ mst be a polynomial just under the hypothesis of being entire fucntion. It is just a well-know result in complex analysis that if $f$ is entire it is either a polynomial or transedent, i.e. we can write it as a [b]series[/b] $f(z)= \sum_i a_iz^i$. By the way what are you assuming of being entire what is the definition you are used to?

For every $z$ and every $R$,

$f''(z)=\frac2{2\pi i}\int_{|\zeta-z|=R}\frac{f(\zeta)}{(\zeta-z)^3}\,d\zeta$

where by that I mean the integral around the circle once counterclockwise.

It follows from $\lim_{|z|\to\infty}\frac{f(z)}{z^2}=0$ that given any $\epsilon>0$ and any fixed $z$, there exists $R$ such that if $|\zeta-z|=R,$ then $\left|\frac{f(\zeta)}{(\zeta-z)^2}\right|<\epsilon.$

Then, estimating in the integral above,

$|f''(z)|\le\frac{2\epsilon}{2\pi}\int_{|\zeta-z|=R}\frac{|d\zeta|}{|\zeta-z|}$

$=\frac{2\epsilon}{2\pi}\cdot\frac{2\pi R}{R}=2\epsilon.$

Since this is true for any $\epsilon>0,$ we have that $f''(z)=0$ for all $z.$ The result follows.


calculus integration trigonometry function symmetry Support limit

Hello,

I have seen that kunny is posting integrals pretty often just to see if someone is able to solve it. I think that is great, but then I thought that we could [i]steal[/i] Silverfalcon's idea in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33505]Problem Solving Marathon 2[/url] and use one single thread to post Integrals and series only. Then, if everybody agree, the rules should be the same (very simple)

(1) If there is a problem open (nobody has solve it correctly yet) and you can solve it, you can post the solution ([i]a complete solution, not just the answer[/i]) and give another problem ([i]series and integrals only, the integrals may be definite or indefinite, you can ask to prove that some series is convergent or divergent instead of asking for the exact value of convergence[/i]).
(2) You [b]cannot[/b] solve a problem you proposed yourself, even when it is spected that you know the solution to your problem (let's not give integrals that cannot be expressed in terms of common functions).
(3) If you are posting a solution to a wrongly solved problem, you have to explain why the first solution is right.
(4) Let's keep the level at a standar college calculus level. Let's try to avoid fancy stuff (l[i]ike integrals or sums that requires Fourier Analysis or Complex Analysis techniques to be solved, just because those would leave out A LOT OF PEOPLE[/i]). But of course, you can post challenging problems that require not too fancy stuff, like the one below.

[color=red][size=150][b]EDIT[/b][/size][/color]
(5) Please, hide your solutions, and number the problem (one problem at a time!)

If everybody agrees, here I propose the first problem:

[color=blue][size=200][b]Problem #1[/b][/size][/color]
Find the exact value of convergence of $\sum_{n=0}^{\infty}\arctan\left(\frac{1}{n^2+n+1}\right)$

[hide]$\tan[\arctan 1/n-\arctan 1/(n+1)]=1/(n^2+n+1)$ gives the result.[/hide]

edited to hide

Here are some of my favorites (yes, they probably have been posted before)

Prove that $\int^{2\pi}_0 \frac {dx}{1+\tan^4x}=\pi$

Evaluate $\int^{\pi}_0 \frac {x\sin x dx}{1+\cos^2x}$

Prove that $\int^{\pi}_0 \frac {dx}{\alpha-\cos x}=\frac {\pi}{\sqrt{\alpha^2-1}}$ for $\alpha>1$

In the interests of honesty, I will state that the only obvious way I know to do the third integral is to use the residue theorem.  But it came from a Schaum's outline which has nothing on complex analysis.  So knock your socks off, guys, and remember to think outside the box!

[quote="blahblahblah"]Here are some of my favorites (yes, they probably have been posted before)

Prove that $\int^{2\pi}_0 \frac {dx}{1+\tan^4x}=\pi$[/quote]

Let [tex]x=\arctan(u)[/tex] then use partial fractions
[quote]

Evaluate $\int^{\pi}_0 \frac {x\sin x dx}{1+\cos^2x}$
[/quote]

Prove that $\int^{\pi}_0 \frac {dx}{\alpha-\cos x}=\frac {\pi}{\sqrt{\alpha^2-1}}$ for $\alpha>1$

In the interests of honesty, I will state that the only obvious way I know to do the third integral is to use the residue theorem.  But it came from a Schaum's outline which has nothing on complex analysis.  So knock your socks off, guys, and remember to think outside the box![/quote]

Yes, but if I replace $\tan^4x$ with $\tan^{9827523897.4187}x$, the answer doesn't change...

Interesting observation but it wont hold for all n

For example if n=5, then the function is undefined at [tex]\frac{3\pi}{4}[/tex]

So are you saying as long as the function is well-defined on [tex][0, 2\pi][/tex] the result is true??

Remarkable

And we show this with elementary calculus? (ie no residue theorem)

Sorry, you're of course correct.  I should have stated that:

$\int^{\frac {\pi}{2}}_0 \frac {dx}{1+\tan^nx}=\frac {\pi}{4}$, which eliminates the behavior problems and is seen to be the same elsewhere.

[hide]Hint: Don't go for the antiderivative, look for symmetry[/hide]

[quote="blahblahblah"]$\tan[\arctan 1/n-\arctan 1/(n+1)]=1/(n^2+n+1)$ gives the result.[/quote]

(a) Your term is not defined for $n=0$, even when that is not hard to solve.
(b) The idea is to give a [b]complete solution[/b] to the problem (not a good hint), and after that, to propose [b]one[/b] problem at a time.

Come on David, that's as close to a full solution as it gets.  Besides, for the first part, you just separate the first term.  We end up getting $\arctan(1)+\sum^{\infty}_{i=1}\arctan\left(\frac{1}{i^2+i+1}\right)$ which gives us $\frac {\pi}{4}+\frac{\pi}{4}$ by direct computation of the first term plus the telescoping series.

As for the second part, I pulled them out of a Schaum's outline, which I didn't want to have to dig up again a week from now.  There's no pedagogical value for me to solve these problems, but I think I still have some good problems to share.

There is no pedagogical value for me either (I teach this stuff, I am not learning it!) but it is for everyone  ;)

It is just like give time to the problems to be solved (at least that was my idea) and being proposing one integral after another, or series!

Also, dheardie, I suppose it would be your turn to propose, say, problem #3! :P

[hide="By the way"]Also $\tan[\arctan (n+1)-\arctan n]=1/(n^2+n+1)$ and therefore there is no indefinition ;)[/hide]

come on, some one should post some new problem.... ;)

eh ok
[b]Problem 3[/b]

Find
[tex]\int_{0}^{ \pi/2}\frac{1}{1+\sqrt{\tan(x)}}\,dx[/tex]

[quote="fedja"]Since nobody answered for more than a week (probably because posting the same problem twice puzzled everybody), I'll post just to support the idea  :)

From the identity $\frac 1{1+y^\alpha}+\frac 1{1+(1/y)^\alpha}=1$, we see that for any $\alpha\in\mathbb R$ the function $f(x)=\frac 1{1+\tan(x)^\alpha}$ satisfies $f(x)+f(\pi/2-x)=1$ from where it follows that the integral is $\frac\pi 4$

Now something really different:
[b]Problem 4[/b]
Find all $x\in[0,2\pi)$ for which the series $\sum_{n>1}\frac {(-1)^n}{\sqrt[3]n-\cos nx}$ diverges.[/quote]

fedja that is the most beautiful solution i have ever seen for that problem!

Here is where the problem was from you can see my (horrible) solution for it -  http://www.sosmath.com/CBB/viewtopic.php?t=10759&start=75 :)

[hide]

Answer: $x=\pi/2,\pi,3\pi/2$. Right?

Lemma 1. Partial sums $S_n=\sum_{n=1}^N (-1)^n \cos nx$ are bounded unless $x=\pi$.

Follows from the fact that $S_n$ is a real part of geometric progression $\sum (-1)^ne^{inx}$.

Lemma 2. Partial sums $S_n=\sum_{n=1}^N (-1)^n \cos^2 nx$ are bounded unless $x=\pi/2,3\pi/2$.

Follows from formula $\cos^2 t=(1+\cos 2t)/2$ and then analogously to lemma 1.

We use Abel (or Dirichlet?) criterion. If $a_n$ decreases to 0 and oartial sums of $\sum b_n$ are bounded, than $\sum a_nb_n$ converge.

From here, series $\frac{(-1)^n}{n^{1/3}}$ converges.  Subtract from initial series. We have to investigate series $\sum \frac{(-1)^n\cos nx}{n^{1/3}(n^{1/3}-\cos nx)}$. For $x=\pi$, it behaves like $n^{-2/3}$ and so diverges.

Again note that for $x\ne \pi$, series $\sum \frac{(-1)^n\cos nx}{n^{2/3}}$ converges. Subtract it, get $\sum \frac{(-1)^n\cos^2 nx}{n^{2/3}(n^{1/3}-\cos nx)}$. Do the same thing once more.
[/hide]

Probelm 5.

$\int\limits_0^{\pi} \frac{\sin 884 x\cdot \sin 1122 x}{2\sin x}  d x$

just notice $\frac{\sin 884 x\cdot \sin 1122 x}{2\sin x}+\frac{\sin 884 (\pi-x)\cdot \sin 1122 (\pi-x)}{2\sin(\pi- x)}=0$
and $\frac{\sin 884 x\cdot \sin 1122 x}{2\sin x}$ is limited.
then the answer will be 0,right?
PROBLEM 6
$\lim_{x \to 0}\frac{f(2x)-f(x)}{x}=a$,and $f(x)$ is continouce at 0.
calculate $f'(0)$(you should to prove f'(0) is exist first.)

[quote="Fedor Petrov"][quote="zhaobin"]just notice $\frac{\sin 884 x\cdot \sin 1122 x}{2\sin x}+\frac{\sin 884 (\pi-x)\cdot \sin 1122 (\pi-x)}{2\sin(\pi- x)}=0$.[/quote]

No.[/quote]
sorry,I had made a mistake. :oops:

[quote="zhaobin"]
$\lim_{x \to 0}\frac{f(2x)-f(x)}{x}=a$,and $f(x)$ is continouce at 0.
calculate $f'(0)$(you should to prove f'(0) is exist first.)[/quote]
For $\epsilon > 0$, we have
$a-\epsilon < \frac{f(2x)-f(x)}{x} < a + \epsilon, \quad 0<|x|<\delta$

Let $x\in (-\delta, +\delta)$ and let $x_n = x / 2^n$ then we have
$a-\epsilon < \frac{f(2x_n)-f(x_n)}{x_n} < a + \epsilon$
and it is
$\frac{a-\epsilon}{2^n} < \frac{f(\frac{x}{2^{n-1}})-f(x_n)} {\frac{x}{2^n}} < \frac{a + \epsilon}{2^n}$

Sum them up, and from the continuity of $f(x)$ at $x=0$, we obtain that
$a - \epsilon \leq \frac{f(x)-f(0)}{x} \leq a+\epsilon$

Thus $f'(0)$ exists and $f'(0) = a$

[b] Moderator edit: Edited for a mistake in $\LaTeX$ (forgotten "}")[/b]


real analysis
number theory
number theory unsolved

Find all integer solutions to the diophantine equation:

$ 2a^2 \minus{} ab \minus{} 3b^2 \equal{} 8$

paramahansa, there are many different forums here so that problems can be posted where they belong, not randomly. Do you know the difference between real analysis and number theory?

yes i do.  if you post reply, please post help - dont be mean

he  was  not mean,  he was asking u to post at the right place.  :D

paramahansa is sorry.  can somebody help me with problem?

$ 2a^2\minus{}ab\minus{}3b^2\equal{}(a\plus{}b)(2a\minus{}3b)\equal{}8$

Let $ x\equal{}a\plus{}b, y\equal{}2a\minus{}3b$ , then
$ \begin{cases} 3x\plus{}y\equal{}5a............(1) \\
2x\plus{}y\equal{}5b.........(2) \end{cases}$

From (1), $ (2,4),(\minus{}2,\minus{}4),(8,1),(\minus{}8,\minus{}1)$ are the only possible values of $ (x,y)$.
After checking $ (2)$, they all give solutions.

So $ (2,0),(\minus{}2,0),(5,3),(\minus{}5,\minus{}3)$ are the only solutions.


trigonometry
calculus
calculus computations

I need to integrate 2dx/(x^2+4) from -inif. to 2. I tried all the tricks but maybe not. I must be missing something but every substitution seems to be failing me. Please help!

NOOOOOOOOOOOOO!!!!!!!!! Its too easy! Thanks alot jmerry. Cant believe I missed that. I was trying all kinds of subs but it was a straight forward integral. Thanks again.

$\arctan x$ is the same as $\tan^{-1}x$, the inverse (not reciprocal) of the tangent function.


[quote="catcurio"]Source: 2005 AMC-10B #23

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$, $E$ as the midpoint of $\overline{BC}$, and $F$ as the midpoint of $\overline{DA}$. The area of $ABEF$ is twice the area of $FECD$. What is $\frac{AB}{DC}$?

$\text{(A)}\ 2 \qquad\text{(B)}\ 3 \qquad\text{(C)}\ 5 \qquad\text{(D)}\ 6 \qquad\text{(E)}\ 8$[/quote]

[hide](C)5

WE know that the midpoint of the given midpoints int he larger trapezoid is 2 times more than the midpoint of the given midpoints of the smaller trapezoid. If we call the larger one x, smaller y,  if you take x=2 y=1, the lengths of the sides are .5 and 2.5..... 2.5/.5=5. the ratio is always the same anyways. [/hide]


Rain falls vertically on the window of a moving car. If a raindrop makes a 45 degree angle with the window when it slants, find the speed of the car in terms of the speed of the falling rain.

This doesn't really work because you don't know the slant of the windshield :P

But I assume you want the window to be vertical, in which case we have:

$ |\tan\theta | \equal{}\frac{\| v_{car}\|}{\| v_{rain}\|}$, and since $ \theta \equal{} 45^{\circ}$,

$ \frac{\| v_{car}\| }{\| v_{rain}\| }\equal{} |\tan 45^{\circ}| \equal{} 1\iff\| v_{car}\| \equal{}\| v_{rain}\|$

If your window makes an angle of $ \phi$ with the vertical, we instead have:

$ \frac{\| v_{car}\| }{\| v_{rain}\| }\equal{} |\tan\left (45^{\circ}\minus{}\phi\right) | \equal{} 1\iff\| v_{car}\| \equal{}\| v_{rain}\| |\tan\left (45^{\circ}\minus{}\phi\right) |$


I think all the MC tests should be of equal difficultly. And thats what the writters would hope to do.

2008-2009 MATHCOUNTS was definitely easier than this year's MATHCOUNTS.

But overall MATHCOUNTS is usually the same difficulty, but the 1990s tests are somewhat easier, I agree.

The most difficult test was 2005, with a countdown cutoff of 30. The easiest in recent memory was 2003 with a cutoff of 40.

[color=red][size=67]AIME15: These are national countdown cutoffs (top 12 written).[/size][/color]

[quote="jksong"]Just out of curiosity, which years in MathCounts have you found to be hard and which ones have you found to be easy, in particular the state and national competitions?

Personally, I think that the mid-upper 1990s states were easy, but starting in the mid 2000s, they have been getting harder. What do you guys think?[/quote]

These old tests were based a lot on computation (which would explain the easiness), but now they are getting more like word problems which require some kind of "out of the box" solution.

In my opinion, the writers alternate between hard-easy over the years.  :lol:

Probably when Mathcounts started in 1983 it was pretty easy, since sprint had 40 problems and just 30 minutes, and you couldn't use a calc for target(and there were 10 total problems I believe)

It seems to be getting harder and harder.

2004 was probably when the mathcounts people really wanted to make things harder.  Case in point for practice tests:
2003 State Sprint Score - 29
2004 State Sprint Score- 21


abstract algebra
Ring Theory
superior algebra
superior algebra theorems

One can prove that an Artinian ring is also a Noetherian ring. However, the converse is false. I find both definitions very symmetric in nature.. Therefore I was wondering if someone could explain me what 'structure' a Noetherian ring lacks to become an Artinian ring? Or can't we describe things like that in terms of 'algebraic structures'?

Maybe this offers some insight? With some nice constraints we get if $ R$ is a commutative Von Neumann regular ring, then it is Noetherian iff it is Artinian.

More loosely we can also get that if $ R$ is a commutative Noetherian ring, then it is Artinian if every $ R\minus{}$module is good.

I think also that every associative Artinian ring with unit is right Noetherian.

In commutative algebra, Artinian rings are "small" and very close to fields: every Artinian ring is a finite direct product of local Artinian rings, and in a local Artinian ring, every element is either a unit or nilpotent.

I think the reason for the asymmetry lies in the fact that it is easy to produce multiples, but hard to find divisors. In any ring, $ (a)\supseteq(a^2)\supseteq(a^3)\supseteq\ldots$ is a decreasing sequence of ideals, but for an increasing sequence, we would have to find divisors of $ a$, and divisors of divisors of $ a$, ... owk


fwiw I got a 43 in 2004; that seems like a long, long time ago now.

That's an excellent score, no matter what your expectations were.  I just found out I got a 35, 4 points lower than my predicted.

[quote="Ethanol"]I'm sure some of the forum-dwellers here have for the past two years been bashing their head against a metal porcupine known as IB.  Results are now out, and many people I know seem to have not lived up to their predicted grades.  Though I do not yet know what I got, I'm curious to see if other people's hopes have been dashed as mine will be tomorrow (I have been told that the highest score at my school was lower than my predicted grade).

-Sad on the east coast[/quote]

Oh yeah, I've been through two years of that too. After I did the exams, I was so happy that I was going to IMO and didn't have to worry about the score, so I didn't bother checking it in the summer (it's not like I could, anyway, because I lost the password). I found out only on the commencement ceremony, and by that time the IB results seemed like something completely irrelevant and otherworldly. The workload was so huge, though... especially at the end of the year, when all the teachers are behind with the material. But that didn't seem to have taught me anything - I went to univ, took a bunch of hard math courses, and had another IB-like year. I think IB addicts you to overachieving  :o .


calculus
integration
inequalities
calculus computations

Who can you give me a solution ? Thanks !


i have a doubt regarding finding the sequence of amino acid in a polypeptide.which is the most efficient method for finding the sequence? is it the overlapping procedure or are there better methods than this? :?:

see this it might be relevant  :D
http://en.wikipedia.org/wiki/Peptide_sequence

[quote="valeriummaximum"]i don understand your question???
do you mean how we find whether a polypeptide is GLYALAGLY!!!!!!
is it what you mean??[/quote]
yes -48 dont cry ranga


Five points on a circle are numbered 1,2,3,4, and 5 in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it will be on point

[asy]
size(80); defaultpen(linewidth(0.7)+fontsize(10)); draw(unitcircle);
for(int i = 0; i < 5; ++i) { pair P = dir(90+i*72); dot(P); label("$"+string(i+1)+"$",P,1.4*P); }[/asy]

$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

[hide]
i believe it repeats after 3 jumps.
5, 1, 2, 4, 1, 2, 4, ... since 1995 is a multiple of 3, the bug will be on 4 after 1995 jumps
D
i wouldnt want to be that bug :)
[/hide]

[hide="Answer"]The bug begins by jumping to point $1$, then point $2$, then point $4$, then point $1$, etc. $1995\equiv 0\bmod{3}$, and the first jump places it on the $1$, so it will end on the $4\Rightarrow \boxed{D}$.[/hide]

Sorry for reviving, but am I missing something? It looks like it will land on 5 (hence, choice E) since 1995 is divisible by 3 and every three jumps it gets back to point 5.

I just checked the original AHSME test and the diagram posted by the OP is incorrect (it doesn't match up with the description given in the problem). The problem specifies CLOCKWISE but, clearly, the diagram is showing COUNTERCLOCKWISE; hence, that's the discrepancy. The solutions given by the previous users stand correct.

Correct diagram (a simple sign change in the code fixes it):
[asy]
size(80); defaultpen(linewidth(0.7)+fontsize(10)); draw(unitcircle);
for(int i = 0; i < 5; ++i) { pair P = dir(90-i*72); dot(P); label("$"+string(i+1)+"$",P,1.4*P); }[/asy]


$ ABCD$ is a cyclic quadrilateral. Prove that the orthocenters of the triangles $ ABC$, $ BCD$, $ CDA$ and $ DAB$ are vertex of a rectangle.


geometry
probability
trigonometry
incenter
perimeter
analytic geometry
algebra

The problems are posted here: http://www.math.sc.edu/contest/1993/exam/PDF/exam1993.pdf

I decided to post all 30 solutions this time since this USC was unusually harder than the others :D

I wasn't able to get #16 and #18

EDIT: I'll post #18 later, i just realized that square roots can be negative or positive :blush:

EDIT: yeah..., for some reason I missed the fact on #16 that it said that M and P are midpoints... :blush:  :blush: I'll post that later too

[hide="1"]Let $a$ be the length and $b$ be the width. We have that $3ab=2b(a+3) \Longrightarrow ab=6$. $a=6, b=1$ works, so $6$ is our answer.[/hide]

[hide="2"]$3+x+3x=23 \Longrightarrow x=5$.[/hide]

[hide="3"]Construct an equilateral triangle whose sidelengths (length $2$) are composed of the radii fo the circles. The area is thus $\sqrt{3}-\pi/2$.[/hide]

[hide="4"]Notice that $(1+i)^{2}=2i$. We then have $(2i)^{50}=-2^{50}$.[/hide]

[hide="5"]Notice that $f(3)=f(2+1)=f(2)+f(1)+1=f(2)+3$. Also, $f(2)=f(1+1)=f(1)+f(1)+1=5$. Thus, $f(3)=3+5=8$.[/hide]

[hide="6"]WLOG, let $p=10$. Thus if $100$ dollars was the original price, after the price reduction, we have $90$ dollars. We need $10$ dollars. Thus, $90(1+x)=100 \Longrightarrow x=\frac{10}{90}$. The percentage thus is choice $e)$.[/hide]

[hide="7"]If we call the squares $a,b,c,d,e,f$ (in order from left to right), we have:
$b+c+d+e+f=3$
$a+c+d+e+f=8$
$a+b+d+e+f=5$
$a+b+c+e+f=6$
$a+b+c+d+f=2$
$a+b+c+d+e=4$
Adding all the equations gives us $5(a+b+c+d+e)=28 \Longrightarrow a+b+c+d+e=5.6$.[/hide]

[hide="8"]The expression simplifies to $(\frac{x^{6}-1}{x-1})^{6}$. Expanding both the numerator and denominator, we see that the coefficient of the $x^{3}$ term is $\binom{6}{5}+\binom{6}{3}+\binom{6}{6}+\binom{6}{3}=56$.[/hide]

[hide="9"]We have $(y+x)(y-x)=187$. Now, since $187=11 \cdot 17$. Therefore, $y+x=17$ and $y-x=11$. Thus, $x=3, y=14$ is a possible solution and the answer is $42$.[/hide]

[hide="10"]If we construct right triangles for each pair of arguments ($\arcsin, \arccos$ and $\arctan, arccot$), we see that the sum of the angles is $90^\circ+90^\circ=\pi$.[/hide]

[hide="11"]The probability is $\frac{3!+2!+1!}{4!}=\frac{3}{8}$.[/hide]

[hide="12"]WLOG, let $(1)$ have roots $x=1,2$ and $(2)$ have roots $x=1,3$. Thus:
$(1)$ $x^{2}-3x+2=0$
$(2)$ $x^{2}-4x+3=0$

Thus, we know that $(a,b,c,d)=(-3,2,-4,3)$ and our answer coice must equal $3$. The answer is $(a)$.[/hide]

[hide="13"]Expanding $\sin{(x+y)}$ and $\sin{(x-y)}$, we have:
$(1)$ $\sin{x}\cos{y}+\sin{y}\cos{x}=.3$
$(2)$ $\sin{x}\cos{y}-\sin{y}\cos{x}=.5$

$(1)+(2)$ yields $2\sin{x}\cos{y}=.8$ and our answer is $.4$.[/hide]

[hide="14"]For the nine slots, we have to choose two spots for each of our pairs $(1,2), (3,4), (5,6)$. There are $\binom{9}{2}\binom{7}{2}\binom{5}{2}$ ways to do this. We also need to multiply by $3!$ again for the remaining three slots for $7,8,9$. Thus the total number of ways is $\binom{9}{2}\binom{7}{2}\binom{5}{2}3!$, which after simplifying is choice $(a)$.[/hide]

[hide="15"]The fraction will be $\frac{3+5+7+11+13}{3 \cdot 5 \cdot 7 \cdot 11 \cdot 13}=\frac{1}{11 \cdot 7 \cdot 5}$. Our answer is thus $385$.[/hide]

[hide="17"]The sum reduces to $2(1)+4(2)+8(3)+16(4)+32(5)+37(6)=2+8+24+64+160+222=480$.[/hide]

[hide="19"]We look at the number of possibilities by considering each dot. If we start with $A$, there are $4$ distinct colors for it. Then, there are $3!$ ways to color the other three dots since they are all distinct. Thus, the number of ways to color the graph is $3! \cdot 4 \cdot 4=96$. However, $A,B,C,D$ are distinct, or in other words, the graphs must be distinct which can occur one way for each pair of starting dots we choose. Thus, the total number is $96-\binom{4}{2}=84$.[/hide]

[hide="20"]We have $(1+1)(1+2)(1+3)(1+4)(1+5)(1+6)-1$ (The $-1$ since we have one less set). This is $7!-1=5039$.[/hide]

[hide="21"]There are $\binom{8}{2}-12=16$ games.
[/hide]
[hide="22"]Quickly simplifying the factors and cancelling gives us that $A=124/55$ and $B=1001/192$. It is fairly obvious that $C$ will be about $2$. Therefore $B>A>C$.[/hide]

[hide="23"]Any integer that can be created through $M$ can be created through $N$ and vice versa. Thus $M=N$.[/hide]

[hide="24"]Notice that for each $f_{n}(3)$ where $n$ is odd, the value is $1/5$ and for each value of $f_{n}(3)$ where $n$ is even, the value is $3$. It follows that the answer is $1/5$.[/hide]

[hide="25"] Let the circle we are looking for be $(x-h)^{2}+(y-k)^{2}=r^{2}$ where $(h,k)$ is obviously the center. Plugging in points $(6,0)$ and $(0,2)$ gives us that $3k-h=8$. Seeing our answer choices, none of the points work, thus our answer is $e)$.[/hide]

[hide="26"]We only need to worry about the decimal expansion of $\sqrt{n^{2}+1}$ since the $-1$ will do nothing to the decimal digits. We have $(1667^{2}+1)^{1/2}$. Using the extended binomial theorem, we have that $\binom{1/2}{0}1667+\binom{1/2}{1}\frac{1}{1667}+...$. Now we only have to look at the second term since all the following terms will be too small to affect the first nonzero digit of the decimal expansion. We see that $\frac{1}{2 \cdot 1667}=.00029....$. The answer is $2$.[/hide]

[hide="27"]Notice that $P$ is the incenter of the triangle. The incircle has radius $2$. Thus, using $rs=A$, we have $2 \cdot s=24 \Longrightarrow s=12$ and the perimeter is $24$.[/hide]

[hide="28"]First note that $\cos{A}, \cos{B}, \sin{A}, \sin{B}>0$ and $A+B>90$. From these, we deduce that the $x$ value and the $y$ value of the coordinates will always be positive and thus the point can only be in the first quadrant.[/hide]

[hide="29"]Using Heron's Formula and $R=\frac{abc}{4A}$, the answer is $\frac{8}{\sqrt{15}}$.
[/hide]
[hide="30"]Factoring out a $\frac{1}{3}$ telescopes the sum to $\frac{1}{3}(\frac{1}{1 \cdot 2 \cdot 3}-\frac{1}{29 \cdot 30 \cdot 31}=\frac{749}{13485}$.[/hide]

[quote="joml88"]Shobhit, on #6 you cannot say "WLOG".  WLOG = without loss of generality.  By letting p=10 you lose the generality...what you are doing is exploiting the multiple choice aspect of the test and plugging in a value.  If you didn't have the answer choices there you wouldn't be able to solve the problem by just plugging a value in.[/quote]

LMAO.  Wow.  That's Shobhit trying to sound smart.

[quote="joml88"]Shobhit, on #6 you cannot say "WLOG".  WLOG = without loss of generality.  By letting p=10 you lose the generality...what you are doing is exploiting the multiple choice aspect of the test and plugging in a value.  If you didn't have the answer choices there you wouldn't be able to solve the problem by just plugging a value in.[/quote]

yep, exploiting multiple choice is a good thing and yeah i shoulda put in UMCTYA instead of WLOG (using multiple choice to your advantage) :wink:

[quote="mysmartmouth"]
LMAO.  Wow.  That's Shobhit trying to sound smart.[/quote]

I always sound smart fool

[quote="Elemennop"]16 is B and 18 is A, right?[/quote]

u forgot quadrants :wink:

[quote="Elemennop"]Oh yea, that's right, so it's B.

your #14 solution is too tedious. You only need to realize that for all permutations, exactly half will fulfill each of the conditions and half won't... (make 1-1 connection by in each permutation switching 1 and 2, and you'll see it's right). So the number of total perms is $9!$, and you half it three times, so it's $9\cdot 7!$.[/quote]

nice  :)


avval ye ghaziye begim
theorem:
farz konid ke $ P(u,v,w)$ ozvi az $ R[u,v,w]$ bashad.va hamgen va motegharen va az daraje $ 3$ bashad.
in do hokm moadel hastand yani harkodoom oon yeki ro natije mide:
$ (a)$ $ P(1,1,1),P(1,1,0),P(1,0,0)\geq0$
$ (b)$ $ P(x,y,z)\geq0$ baraye hameye $ x,y,z$ na manfi
esbatesh ham vaghean sadas.forme kollisho benevisin bad $ P(1,1,1),P(1,1,0),P(1,0,0)$ ro hesab konid.nahayatan be ye majmooe namanfi miresid(schour baladin? age balad nistin begin ta baratoon benevisamesh :wink: )
Stolarsky's theorem:
farz konid ke $ P(u,v,w)$ ye chandjomleie motegharen haghighi va az daraje 3 bashad.
agar $ P(1,1,1),P(1,1,0),P(2,1,1)\geq0$ angaah baraye har $ a,b,c$ ke toole azlaae mosallasi bashand darim:
$ P(a,b,c)\geq0$
mitoonin baraye esbatesh mese ghabli form benevisin va gharar bedid $ a=u+v$ va $ b=v+w$ va $ c=w+u$ va az ghaziye balayi estefade konid.
agar khodayi nakarde zaboonam laal too neveshtane form moshkeli darin inam az form:(az sigma cyc va sym ham estefade nakardam chon ba latex balad nistam benevisameshoon)
$ P(x,y,z)=A({x^{3}}+{y^{3}}+{z^{3}})+B({x^{2}}y+x{y^{2}}+{y^{2}}z+y{z^{2}}+{z^{2}}x+z{x^{2}})+Cxyz$ va $ A,B,C$ haghighi hastan.(chejoori ba latex ozv boodan ro neshoon bedim? :blush: )

[quote="mehdi"]...chejoori ba latex ozv boodan ro neshoon bedim? :blush: )[/quote]

Example:

$ a\in A$

bebinin namosavie schur ke hatman kheilia mishnasanesh.age kasi nemishnase ine:
$ r\in R$ va $ a,b,c\geq0$ yani $ r$ manfi ham mitoone bashe:
$ {a^{r}}(a-b)(a-c)+{b^{r}}(b-a)(b-c)+{c^{r}}(c-a)(c-b)\geq0$                  $ (1)$
ino haminjoori negash nakonin.age gharaar bedin $ r=0$ be in miresin:
$ {a^{2}}+{b^{2}}+{c^{2}}\geq ab+bc+ca$                        $ (2)$
age gharaar bedin $ r=1$ be in miresin ke binahayat por karborde:
$ a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\geq0$                  $ (3)$
ke age bazesh konin mishe:
$ {a^{3}}+{b^{3}}+{c^{3}}+3abc\geq{a^{2}}b+a{b^{2}}+{b^{2}}c+b{c^{2}}+{c^{2}}a+c{a^{2}}$   $ (4)$
dar zemn mishe oono be in shekl haa ham nevesht:
$ 2(ab+bc+ca)-({a^{2}}+{b^{2}}+{c^{2}})\leq \frac{9abc}{a+b+c}$                             $ (5)$
$ abc\geq(a+b-c)(a+c-b)(b+c-a)$                          $ (6)$
dar zemn age dar in namosavi ye $ a+b+c$ zarb konin in mishe ke khoob chizie:
$ {a^{4}}+{b^{4}}+{c^{4}}+abc(a+b+c)\geq2({a^{2}}{b^{2}}+{b^{2}}{c^{2}}+{c^{2}}{a^{2}})$                        $ (7)$
va dar zemn age ba tarafe raste namosavie 4 inkaro bokonim ye chize jadid va dar zemn mofid giremoon miyad:
bana bar namosavie hesabi hendesi:  $ {a^{2}}b+{b^{2}}a\geq2{(ab)^\frac{3}{2}}$
pas mitoonim benevisim:
$ {a^{3}}+{b^{3}}+{c^{3}}+3abc\geq2({{ab}^\frac{3}{2}}+{{bc}^\frac{3}{2}}+{{ca}^\frac{3}{2}})$         $ (8)$
khob hala rabtesho be in topic migam.
rabtesh ine ke tamaame namosavihayi ke shomaare daran be vasileye ghaziyeye avval ke ghabl az stolarsky neveshtam  mostaghel az baghiye saabet mishan.kheili saade.mitoonin emtehaan konin chon ghaziye shabihe yek algorithm maa ro be hale masale sogh mide.
dar zemn tooye namosavie $ (1)$ age gharaar bedin $ r=-1$ be chize jaaalebi miresin


Let $ a,\ b,\ c$ be real numbers such that $ a<b<c,\ a\plus{}b\plus{}c\equal{}0.$

Prove that $ \frac{1}{2}\leq \frac{a^2\plus{}b^2\plus{}c^2}{(c\minus{}a)^2}<\frac{2}{3}.$

[quote="kunny"]Let $ a,\ b,\ c$ be real numbers such that $ a < b < c,\ a \plus{} b \plus{} c \equal{} 0.$

Prove that $ \frac {1}{2}\leq \frac {a^2 \plus{} b^2 \plus{} c^2}{(c \minus{} a)^2}$[/quote]

It is equivalent to : $ (a\plus{}c)^2 \plus{}2b^2 \geq 0$ ( right )

[quote="kunny"]Let $ a,\ b,\ c$ be real numbers such that $ a < b < c,\ a \plus{} b \plus{} c \equal{} 0.$

Prove that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{(c \minus{} a)^2}< \frac{2}{3}$[/quote]

It is equivalent : $ (a\minus{}b)(c\minus{}b)<0$ ( right )

Done ! :)

Coud you explain the latter part in more deatail? :)

Let $ b\equal{}a\plus{}m,c\equal{}b\plus{}n,m>0,n>0$,so $ a\plus{}b\plus{}c\equal{}3a\plus{}2m\plus{}n\equal{}0$
then the inequality becomes:
\[ \frac{1}{2}\leq \frac{2(m^2\plus{}mn\plus{}n^2)}{3(m\plus{}n)^2}<\frac{2}{3}\]
obvious now.

[quote="kunny"]Coud you expalin the latter part in more deatail? :)[/quote]

The right part is equivalent :

$ a^2\plus{}3b^2\plus{}c^2\plus{}4ca <0$ (*)

And we have : $ a^2\plus{}b^2\plus{}c^2 \equal{}\minus{}2ab\minus{}2bc\minus{}2ca$

:)


Suppose in triangle $ABC$,$D$,$E$ and $F$ are on $BC$,$CA$ and $AB$ that: \[\displaystyle \frac {BD}{DC}=\frac {CE}{EA}=\frac {AF}{FB}=\frac {1- \lambda}{ \lambda}\] that $\lambda$ is real number. We see easily that centroid of $DEF$ is constant as $\lambda$ varies. Find the locus of circumcenter of $DEF$ as $\lambda$ varies.

This is from a pre-TST? I wouldn't expect an olympiad problem to have a cubic curve as a result...

http://www.mathlinks.ro/Forum/viewtopic.php?t=17597

  darij

Yes , At the exam I obtained this with complex geometry, But I couldn't believe it


Determine the eigenvalues and eigenvectors of the symmetric matrix
\[ \mathsf{A} \equal{} \begin{pmatrix} 3 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{pmatrix}\]Use an identity of the form $ \mathsf{PAP}^{T} \equal{} \mathsf{D}$ where $ \mathsf{D}$ is a diagonal matrix, to find $ \mathsf{A}^{\minus{}1}$

I can find the eigenvalues and eigenvectors, but how does $ \mathsf{P}^T$ come in? And how do we find $ \mathsf{A}^{\minus{}1}$. If $ \mathsf{P}$ isn't orthogonal I don't see how that would help.

point is we can write $ A \equal{} PDP^{ \minus{} 1}$, where $ P$ is the matrix whose columns are $ A$'s normalized eigenvectors ($ P$ *is* orthogonal, so we have $ P^{ \minus{} 1} \equal{} P^{T}$), and $ D$ is the diagonal matrix whose diagonal entries are the eigenvalues of $ A$ (in the same order down the diagonal as the corresponding eigenvectors are as columns of $ P$). once you've written $ A$ in that form (it's called spectral decomposition), then you can easily compute $ A^{ \minus{} 1}$ as $ PD^{ \minus{} 1}P^{ \minus{} 1}$, where $ D^{ \minus{} 1}$ is diagonal with entries the reciprocals of the eigenvalues.

[quote="pleurestique"]point is we can write $ A \equal{} PDP^{ \minus{} 1}$, where $ P$ is the matrix whose columns are $ A$'s normalized eigenvectors ($ P$ *is* orthogonal, so we have $ P^{ \minus{} 1} \equal{} P^{T}$), and $ D$ is the diagonal matrix whose diagonal entries are the eigenvalues of $ A$ (in the same order down the diagonal as the corresponding eigenvectors are as columns of $ P$).[/quote]

In general the normalized eigenvectors may not be orthogonal though? In this case $ A$ happened to be symmetric; what if it wasn't?

I get the second part though, thanks

in general, no, eigenvectors of non-symmetric matrices are not orthogonal.


Find the greatest three-digit number divisible by 6.


geometry
LaTeX
factorial
function
binomial coefficients

im fairly knew to geometry and have come across this fairly hard question
Investigate: If formulae can be used to work out number of intersections
between the crosses and dot (not including ones at dot and cross)
example
[img]http://i231.photobucket.com/albums/ee78/sal_surfer123/327c2c73-1.jpg[/img]

Maybe we can use choose notation.
Like so:
\[ \left(\begin{array}{cc} \mbox{number of originating points} \\
2\end{array}\right)
\]
In your case:
\[ \left(\begin{array}{cc} 4 \\
2\end{array}\right) \equal{} 6
\]

dunno

There are 8 lines in that picture ;)  Although actually, you aren't on such a terrible track.  It's probably a good idea to try a bunch of different pictures -- for example, there's one dot and several crosses, how many intersections?  When there're two dots and (two, three, four, five, six) crosses, how many intersections?  Three dots and (two, three, four, five) crosses?  Then with this data, maybe you can make an improved conjecture and prove it.

A LaTeX suggestions: \binom{n}{k} makes $ \binom{n}{k}$.

Edit: the first sentence of this post refers to an earlier version of the preceding post, which was edited before I posted.

Ohh
I meant the originating points: which were the 4 X's

im confused
could you dumb it down a bit

do you know the "choose notation"
like
\[ _nC_p? \mbox{ or } \binom {n} {p}
\]
Thanks JBL for the LaTeX suggestion
I was using this:
\left(\begin{array}{cc} n \\
p\end{array}\right)?

Teaching the notation isn't very important in comparison with teaching the [i]idea[/i] behind the notation.  But perhaps you should finish solving the problem first?  ;)

$ \binom{n}{r}$ means "the number of ways to choose $ r$ things from a set of size $ n$, where the choices are without replacement and the order of the choices doesn't matter."  So, for example, $ \binom{4}{2} \equal{} 6$ because from the set {a, b, c, d} we can choose the following pairs: ab, ac, ad, bc, bd, cd.  There's a simple formula for the binomial coefficients in terms of the factorial function, and they are very important combinatorial objects that arise in a lot of places.  But that isn't going to help you solve the problem ;)

Probably, you should try what I suggested in my first post.

[hide="fairly large hint"]
Suppose you have some dots and crosses. Take two of those dots and two of those crosses and examine their lines and intersections...[/hide]


[quote="Sporally"]Can someone plz confirm if this is done right or not.

((x^2)-(3x)-4)/((3x^2)+x-2)

and i end with:

(x-4)/(x-0,6667)

(task nr. 316)[/quote]

you should try to use latex .... it isnt hard and it makes ppl easy to see.....

$\frac{x^2-3x-4}{3x^2+x-2}=\frac{(x+1)(x-4)}{(3x-2)(x+1)}=\frac{x-4}{3x-2}$ :)

The first step of using LaTeX is quite easy- just put dollar signs ($\$$) before and after the mathematical expression, and see what happens. At least the exponents will look much nicer :)

[quote="Sporally"]So i got it right but you just don't like using numbers like 0.33, 5,33 and 1,5 so you just put it all in *3? Regarding LaTex: I have promised to take a look at it, but i really haven't had any luck with it so far. Anyway, i don't think i will get too much time to learn it in the very near future, but maybe later i will try and learn it.[/quote]

yeah, it's generally a good idea to keep all factorized stuff in integer form, and writing 1/3 as 0.33 is usually a bad idea if you don't use that bar that signifies repeitition...

I honestly was quite confused when you first wrote $3x-2$, but taking a look at it and 20 secs later i got it :)

WOW, i did just use LaTex :D Just skip the celebrations...


[hide="solution"]We are given
$x+y=2$
$xy=3$
We need to know $\frac 1x + \frac 1y$.
Multiple $\frac 1x$ by $\frac yy$ and $\frac 1y$ by $\frac xx$.
Simplifying, that gives us $\frac{x+y}{xy}$. Substituting, $\boxed{\frac 23}$[/hide]

You have the right idea, but you substituted incorrectly. ;)

Let x and y be the two numbers.

$x+y=2$
$xy=3$

We know that
$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$.

Thus, the problem becomes

$\frac{x+y}{xy}$=$\frac{2}{3}$

Very good trick to know.

Edit: What was wrong with biffanddoc's answer?

He edited. Before, he said that $\frac{x+y}{xy}=\frac{3}{2}$.

[quote="G-UNIT"]The sum of two numbers is 2.  The product of the same two numbers is 3.  What is the sum of their reciprocrals?[/quote]

[hide]
$x+y=2$
$xy=3$
$x=\frac{3}{y}$
$\frac{3}{y}+y=2$
$3+y^2=2y$
$y^2-2y+3=0$
$y=\frac{2\pm\sqrt{4-12}}{2}$
$y=\frac{2\pm2i\sqrt{2}}{2}$
$y=1\pm i\sqrt{2}$
WLOG, $y=1+i\cdot \sqrt{2}$
Thus, $x=3/(1+i\sqrt{2})$.
so now, $1/x+1/y=\frac{1}{1+i\sqrt{2}}+\frac{1+i\sqrt{2}}{3}$
$\frac{(3)+3(1+i\sqrt{2})}{3(1+i\sqrt{2})}$
$\frac{2+i\sqrt{2}}{1+i\sqrt{2}}$
$1+\frac{1}{1+i\sqrt{2}}$
[/hide]

Uhh... let me just take this opportunity to say, O_O.

Seriously, if it was that complicated, it wouldn't be in the MathCounts forum... :roll:

[quote="JesusFreak197"]$\frac{1}{1+i\sqrt{2}}+\frac{1+i\sqrt{2}}{3}=\frac{2}{3}$

You did something wrong adding those two together.[/quote]
Oh, yeah thanks!!
How did I mess up? :oops:
Corrected:

[hide]$\frac{1}{1+i\sqrt{2}}+\frac{1+i\sqrt{2}}{3}=\frac{3+(1+i\sqrt{2})^2}{3(1+i\sqrt{2})}$
$\frac{3+1+2i\sqrt{2}-2}{3(1+i\sqrt{2})}$
$\frac{2(1+i\sqrt{2})}{3(1+i\sqrt{2})}$
$\frac{2}{3}$
As desired.
[/hide]

[quote="IntrepidMath"]Whoops, I did it the brute force way.
[hide="Shinier Solution"]
1/x+1/y=x+y/xy=2/3.
[/hide]
What was I doing??
if anyone is really bored out of their mind, could you see what I did wrong?
Thanks.[/quote]

You also made a bad LaTeX formula. I got the error:
[Unparseable or potentially dangerous LaTeX formula. Error 6.]

And there was one that wouldn't show up on my computer.

[hide]
To get the same denominator, you multiply 1/x and y/y=y/xy and you do
1/y * x/x=x/xy
$\frac{x+y}{xy}=\boxed{\boxed{\boxed{\frac{2}{3}}}}$[/hide]

The LaTeX errors have been fixed now. Please try the Preview button unless you never make mistakes ;) It saves the mods some work :)

[quote="mathnerd314"]It also helps to know the formula for adding fractions:

$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$[/quote]

I don't think a formula is needed.  You should just know to multiply by $\frac{d}{d}$ and $\frac{b}{b}$ to get a common denominator.  This way, you can add thousands of fractions at once (why that would come up in a problem unless there was a shortcut way like a geometric series is beyond me).

[quote="JesusFreak197"]Well, it kind of becomes an intuitive thing. I actually have the $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$ thing memorized, and I can get other ones pretty much instantly unless they're large.[/quote]

My old SAT book calls it the Flying X algorithm. Think about it.

Nice avatar.


This is from another topic, but nevertheless I didn't want to bring back an old topic.

http://www.humanmetrics.com/cgi-win/JTypes2.asp

What is your personality type?

Please say if you the results resemble you.   :lol:

Mine is ENFJ.

Extroverted- 1
Intuitive-12
Feeling- 38
Judging- 11

Yes. I am an extrovert, and I don't always do the right thing.

Your Type is
ENFP
Extroverted Intuitive Feeling Perceiving
Strength of the preferences %
22 12 12 56

My preferences are stronger than yours, so I have a stronger personality! :P

The Portrait of the Champion Idealist (eNFp)

The Champion Idealists are abstract in thought and speech, cooperative in accomplishing their aims, and informative and expressive when relating with others. For Champions, nothing occurs which does not have some deep ethical significance, and this, coupled with their uncanny sense of the motivations of others, gives them a talent for seeing life as an exciting drama, pregnant with possibilities for both good and evil. This type is found in only about 3 percent of the general population, but they have great influence because of their extraordinary impact on others. Champions are inclined to go everywhere and look into everything that has to do with the advance of good and the retreat of evil in the world. They can't bear to miss out on what is going on around them; they must experience, first hand, all the significant social events that affect our lives. And then they are eager to relate the stories they've uncovered, hoping to disclose the "truth" of people and issues, and to advocate causes. This strong drive to unveil current events can make them tireless in conversing with others, like fountains that bubble and splash, spilling over their own words to get it all out.

Champions consider intense emotional experiences as being vital to a full life, although they can never quite shake the feeling that a part of themselves is split off, uninvolved in the experience. Thus, while they strive for emotional congruency, they often see themselves in some danger of losing touch with their real feelings, which Champions possess in a wide range and variety. In the same vein, Champions strive toward a kind of spontaneous personal authenticity, and this intention always to "be themselves" is usually communicated nonverbally to others, who find it quite attractive. All too often, however, Champions fall short in their efforts to be authentic, and they tend to heap coals of fire on themselves, berating themselves for the slightest self-conscious role-playing.

I feel that this test is not accurate at all.
Only giving two choices for these questions is very bad, imo.
For a lot of the questions, I felt like "hmm sometimes yes, sometimes no," "hmm somewhat yes," "to some people yes, to others no," etc.
This should have at least 5 choices("definitely yes,somewhat yes, half half, somewhat no, definitely no" or something like that).
like this questions:
"You are a person somewhat reserved and distant in communication"
If I'm talking to someone I somewhat know, then definitely yes, if I'm talking to someone I know, then definitely no
And some of these questions aren't that good....
"It's difficult to get you excited".... eh what do they mean by "difficult"? how excited?
"Strict observance of the established rules is likely to prevent a good outcome".....eh it depends on the rules and how strict it is, doesn't it?

junggi has right.

but, i manage to reach the end of it and this is what i got:

Your Type is
ISTJ
Introverted Sensing Thinking Judging
Strength of the preferences %
44 12 62 56

Qualitative analysis of your type formula

You are:

-moderately expressed introvert

-slightly expressed sensing personality

-distinctively expressed thinking personality

-moderately expressed judging personality

junggi is right..the test needs an "sometimes" or a "maybe" option as well...but i got...
Your Type is
ENTJ
Extroverted Intuitive Thinking Judging
Strength of the preferences %
22 62 1 22
You are:
slightly expressed extrovert
distinctively expressed intuitive personality
distinctively expressed thinking personality
slightly expressed judging personality

Introverted	Intuitive Thinking	Perceiving
Strength of the preferences %
22	             12	           12            33

INTP

You are:
slightly expressed introvert
slightly expressed intuitive personality
slightly expressed thinking personality
moderately expressed perceiving personality

[quote="mustafa"]I'm an INTP
Introverted:100!	Intuitive:88	Thinking:25	Perceiving:22

Apparently Einstein is my icon or something..[/quote]

I thought you were too introverted..

Your Type is
INTJ
Introverted	Intuitive	Thinking	Judging
100	25	1	56

I've done these tests quite a few times, everytime I'm INTJ, but the percentages change slightly (perhaps depending on my mood)

I seriously doubt the accuracy of these tests. I mean, how are they going to cover your whole personality in seventy questions? Me, for example, lean toward one side in one thing and the other in another.

However, here's my results.

Your Type is
ISTJ
Introverted Sensing Thinking Judging
Strength of the preferences %
78 12 1 56

Qualitative analysis of your type formula

 You are:very expressed introvert

slightly expressed sensing personality

slightly expressed thinking personality

moderately expressed judging personality

Introvert: 22
Intuitive: 62
Thinking: 1
Percieving: 56

We took this in class for career shadowing...

ENFP

Extrovert: 50%
Intuitive: ~66%
Feeling: ~23%
Perceiving: ~13%

People were s-h-o-c-k-e-d that I wasn't thinking lol. But i'm bipolar in that aspect so =/  :lol:

[quote]Introvert: 56
Intuitive: 12
Thinking: 12
Percieving: 44

You are:
moderately expressed introvert

slightly expressed intuitive personality

slightly expressed thinking personality

moderately expressed judging personality[/quote]

what does this mean?

-jorian

Your Type is
ENTP
Extroverted Intuitive Thinking Perceiving
Strength of the preferences %
44 25 25 33

Qualitative analysis of your type formula

You are:

* moderately expressed extrovert
* moderately expressed intuitive personality
* moderately expressed thinking personality
* moderately expressed perceiving personality

from

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=519143&highlight=intp#519143

There is a poll here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=519143

Over 5 months later, I take it again and get:

Your Type is
ENTJ
Extroverted	Intuitive	Thinking	Judging
Strength of the preferences %
44	12	1	1

Qualitative analysis of your type formula

You are:

    * moderately expressed extrovert
    * slightly expressed intuitive personality
    * slightly expressed thinking personality
    * slightly expressed judging personality

It appears I've gone 13% from moderate Intuitive to slight Intuitive, 24% from moderate Thinking to right at the border, and 34% from Moderate Perceiving to crossing the border to Judging.

[quote="tokenadult"]These tests are completely bogus, and have never been properly validated.[/quote]

nevertheless, they show the vast amount of different personality types that you could be. One of the descriptions probably fits you.

Your Type is
ESTJ
Extroverted	Sensing	Thinking	Judging
Strength of the preferences %
11	50	38	44

but it's weird. I'm never 'always' on those choices

[quote="tokenadult"][quote="now a ranger"]One of the descriptions probably fits you.[/quote]

You will gain insight into the objections to personality testing by learning about the [url=http://en.wikipedia.org/wiki/Forer_effect]fallacy of personal validation[/url]. Most of the descriptions could fit most people, because they are vague, flattering, and ambiguous, and that is the same problem newspaper horoscopes have.[/quote]

well it is allocated toward people who want to know what they are most like or have a certain curiousity. There are 24 possible kinds. So it isn't common for two people to have the same Personality.


Why is it I can do things like $2^2$, but when I attempt to do $2^13$ it gets messed up?


geometry parallelogram vector invariant trapezoid

[quote="myc"]Can anyone explain question?  Which side of parellelogram is AA' directly above?  AD or AB?
This is from the 1997 AHSME #25.
Let ABCD be a parallelogram and let (ray arrow sign)$AA'$, (ray arrow sign) $BB'$, (ray arrow sign) $CC'$, and (ray arrow sign) $DD'$ be parallel rays in space on the same side of plane determined by ABCD.  If AA' = 10, BB' =8, CC'=18, DD'=22, and M and N are the midpoints of line A'C' and line B'D', respectively, then MN= ___?

A.0        B.1         C.2        D. 3          E.4

Thanks![/quote]

Basically, the parallelogram determines a plane, and all four vectors are parallel to one another; it doesn't specify the angle that the vectors make with the plane though. Usually when a piece of information is not given such as that and no other restrictions are given that determine a [u]unique[/u] situation, you can use the handy problem-solving rule:

[b]Make life easy for yourself![/b]

This distance $MN$ is most likely invariant between different angles of the vectors with the plane. Just use right angles, and worry only if your answer is not one of the given choices.

But make sure that your situation fits [u]all[/u] of the restrictions; oftentimes a unique situation is determined by some sneaky restrictions or relations of the diagram or numbers.

[hide="For this problem:"]
Heck, why not make the parallelogram a square? Nothing is saying we can't. So make it a square with vertices $A(0,0,0), \ B(2,0,0), \ C(2,2,0), \text{ and }D(0,2,0)$. Make the vectors right angles in the positive Z direction. Then we have the points $A'(0,0,10), \ B'=(2,0,8), \ C'(2,2,18), \ D'(0,2,22)$. Find the midpoint of the two lines; we get that $M$ is at $(1,1,14)$, and $N$ is at $(1,1,15)$. The distance, then, is simply $\boxed{1}$.[/hide]

Note that if you wanted to be rigorous, you would need to prove that this distance remains invariant under changes both in the shape of the parallelogram and in the angle between the vectors and the plane.

uhh...the line through M, N passes through the center of the parallelogram, $O$, because diagonals bisect each other, and M, N are midpoints, etc, SAS similarity...so we just need to subtract the heights... OM, and ON, which are the averages of the other heights [some trapezoids here]...the answer is just $\frac{1}{2}|(10+18)-(8+22)|=1$


Find $ c_{max}$ that for any $ x,y,z$, we have $ 1 \plus{} |x \plus{} y \plus{} z| \plus{} |xy \plus{} yz \plus{} zx| \plus{} |xyz| > c(|x| \plus{} |y| \plus{} |z|)$

If $ x,y,z$ be [b]real[/b] numbers, then

$ 1 \plus{} |x \plus{} y \plus{} z| \plus{} |yz \plus{} zx \plus{} xy| \plus{} |xyz|$

$ \geq\left(\frac {3\sqrt [3]{207 \plus{} 48\sqrt {3}}}{32} \plus{} \frac {99}{32\sqrt [3]{207 \plus{} 48\sqrt {3}}} \minus{} \frac {3}{32}\right)\left(|x| \plus{} |y| \plus{} |z|\right),$

with equality if $ x \equal{} \frac {1}{2} \minus{} \frac {\sqrt [3]{7 \plus{} 4\sqrt {3}}}{2} \minus{} \frac {1}{2\sqrt [3]{7 \plus{} 4\sqrt {3}}} ,y \equal{} z \equal{} \frac {\sqrt [3]{7 \plus{} 4\sqrt {3}}}{4} \plus{} \frac {1}{4\sqrt [3]{7 \plus{} 4\sqrt {3}}} \minus{} \frac {1}{4}.$

[hide="Remark"]$ \frac {1}{2} \minus{} \frac {\sqrt [3]{7 \plus{} 4\sqrt {3}}}{2} \minus{} \frac {1}{2\sqrt [3]{7 \plus{} 4\sqrt {3}}} \equal{} \minus{} 0.91082\cdots$ is the unique real root of $ 2a^3 \minus{} 3a^2 \plus{} 4;$

$ \frac {\sqrt [3]{7 \plus{} 4\sqrt {3}}}{4} \plus{} \frac {1}{4\sqrt [3]{7 \plus{} 4\sqrt {3}}} \minus{} \frac {1}{4} \equal{} 0.45541\cdots$ is the unique real root of $ 4b^3 \plus{} 3b^2 \minus{} 1;$

$ \frac {3\sqrt [3]{207 \plus{} 48\sqrt {3}}}{32} \plus{} \frac {99}{32\sqrt [3]{207 \plus{} 48\sqrt {3}}} \minus{} \frac {3}{32} \equal{} 0.99421\cdots$ is the unique real root of $ 64c^3 \plus{} 18c^2 \minus{} 54c \minus{} 27.$[/hide]

I conjecture the following nicer inequality holds for any [b]complex[/b] numbers $ x,y,z$ :

$ 1 \plus{} |x \plus{} y \plus{} z| \plus{} |yz \plus{} zx \plus{} xy| \plus{} |xyz|\geq\frac {\left(|x| \plus{} |y| \plus{} |z|\right)}{\sqrt [3]{4}},$

with equality if $ x \equal{} \minus{} \frac {1}{\sqrt [3]{2}},y \equal{} \frac {e^{\frac {\pi}{3}i}}{\sqrt [3]{2}},z \equal{} \frac {e^{ \minus{} \frac {\pi}{3}i}}{\sqrt [3]{2}}.$

[b]Remark[/b] $ \frac {1}{\sqrt [3]{4}} \equal{} 0.62996\cdots.$


a free- floating astronaut 'A' pushes another free- floating astronaut 'B' in space . mass of A>mass of B. Force exerted by A on B would be

1 greater than force by B on A
2 equal to "...above......................"
3 less than ".....above................"
4 equal to 0

According to Newton's 3rd law, the forces would be equal.

thats what i thought , guess what the answer given is......... equal to ZERO !!!!

why, i cant understand....

Even i agree with equal ..but goin for guesses...if u push the astronaut,he woul keep goin and never stop,right?so no acceleration i guess...i know its a dumb explaination,just tryin to think hoq did they get 0!!!

Don't spoil your understanding of Physics just for answering an ill'framed question. AIEEE etc. local Indian exams have A LOT of mistakes in questions. Just don't forget the correct rules just for getting the answer to a question written by an examiner not at all interested. The Answer IS Equal

Even I the dumbest at physics say agree with you all!

Part of the confusion is in the fact there are two time segments in this case:

(1) The time during which one astronaut imparts force upon both, and accelerates them.

(2) The time after contact is broken, in which no acceleration *afterward* is possible.

Both qualitatively different portions of the event require time.  Only the first converts muscle energy into translational motion, and according to properly formulated laws of conservation of momentum (Newtonian), both astronauts get an equal amount of momentum, but end up with different velocities since they have different masses.

I am Extremely Sorry guys if i have been posting stupid t my questions , but thats what my @##@$%##@@ SOurce says !! I apologise to bus for saying his soln was wrong of my other question cos the soln was correct , only there wasnt ne option like that .........

damn i am throwing away that stupid source ...........

You don't need to apologize for anything really :) .

Yo eminem,could ya tell me that @@$$###!** source?? :D

forget it Mr.shady , i thought u had enough sources of your own and didnt need nemore ,  :rotfl:  :rotfl:  :rotfl: ahhahahahahhahahha

hey i'll tell ne one who's intereasted xcept u and .....ummmm.......perhaps RIDDLER???

HAHAHAahahahahahahhaahahaah :lol:  :lol:  :lol: ,

sorry guys if you consider this spam, a little jokin never hurts .......


calculus
integration
limit
function
real analysis
real analysis unsolved

Assume that $ \lim_{t\to\infty}\int_{\minus{}t}^{t\plus{}k}f$ exists for each $ k\in\mathbb R.$ Provide an example where $ \int_{\mathbb R}f$ diverges.

Note that with $ k\equal{}0,$ this is the usual definition of principal value for an integral on the line. Hence the hypothesis is slightly - but only slightly - stronger than the existence of a principal value.

How about this: let $ f(x)\equal{}\text{sgn}\,x,$ the function that is $ 1$ for positive $ x$ and $ \minus{}1$ for negative $ x.$

$ \int_{\minus{}t}^{t\plus{}k}\text{sgn}\,x\,dx\equal{}k,$ for all $ t>0,$ so the limit equals $ k.$

The function is clearly not integrable on $ \mathbb{R}.$

Other examples I thought of included $ f(x)\equal{}\tanh(x)$ (same idea as the signum, only continuous) and $ f(x)\equal{}\frac{x}{1\plus{}x^2}.$ These also work.

Okay, thanks a lot for your answer but, if we consider $ f(x)\equal{}\frac1x$ ? Can we first work with the integral and then take the limit? Since obviously $ \int_{\mathbb R}f$ diverges.

The problem is that you need $ \int_{ \minus{} t}^{k \plus{} t}f(x)\,dx$ to exist in the first place, before you take any limits, which it doesn't for $ f(x) \equal{} \frac1x.$

You want to create trouble as $ |x|\to\infty,$ but you don't want local trouble. You can consider the example of $ f(x) \equal{} \frac {x}{1 \plus{} x^2}$ to be the same as $ \frac1x$ at infinity, but without the local trouble.

Here is a slightly trickier one. Suppose that the limit $ \lim_{t\to\plus{}\infty}\int_{\minus{}t}^{t\plus{}k}f$ exists for all $ k\in[2009,2010]$. Does it follow that $ \lim_{t\to\plus{}\infty}\int_{\minus{}t}^{t}f$ exists?

In my first five minutes of "that must be false, let me concoct a counterexample," none of my alleged counterexamples actually work. Maybe it's going to take more than five minutes.

He says trickier, so it could be false either true! :)


I recently heard about the "Chicken McNugget Theorem." For those of you who don't know about it, it states that the greatest number which cannot be formed by adding any multiples of the numbers $A$ and $B$ can be found with $AB-A-B$. I thought this was really neat, but is there any way to apply it to more than 2 numbers?

There is not a way that I know of to use this with more than two numbers.  Also, please use the search function; this topic comes up fairly often.   :)

$AB-A-B=(A-1)(B-1)-1$

So my best guess for three numbers $A,B,C$ is:

$(A-1)(B-1)(C-1)-1=ABC-AB-AC-BC+A+B+C-2$

This is probably wrong though.


[quote="The QuattoMaster 6000"][quote="ckck"][quote="Interval"]Polygon ABCD is congruent to polygon A'B'C'D'.
If angle A corresponds to angle A',
the value of angle A' is 5x - 65 and
the value of angle A is 2x - 5,
find angle A.[/quote]
[hide]That means angles A and A' are congruent. So $2x-5=5x-65$. So solving for x which is 20. Then plug that back in to find the measure of angle A or $\boxed{95}$.[/hide][/quote]
Everything up to x=20 is fine.  But 5x-65=35, not 95.[/quote]
Wait. What? Aren't we looking for A not A'?

[quote="The QuattoMaster 6000"]Still the same thing: 2x-5=2(20)-5=40-5=35.[/quote]
Oh. Whoops. I plugged the x in for the first one and then subtracted 5 from the second one. :blush:


inequalities
symmetry
inequalities proposed

x,y,z are all nonnegative，and x+y+z=1

prove：x*x*y+y*y*z+z*z*x+x*y*z<=4/27

there is a weird solution to it ,but I want something that is easy to understand,esp one by Schur's inequality. :)

[quote="toetoe"]

$(x+y+z)^3\geq 3(x^2y+y^2z+z^2x)(3)$

quote]
It' s wrong. :)

[quote="mysteryfigure"]x,y,z are all nonnegative，and $x+y+z=1$.

prove：$x^2y+y^2z+z^2x+xyz\le \frac{4}{27}$

there is a weird solution to it ,but I want something that is easy to understand,esp one by Schur's inequality. :)[/quote]
I think this inequality ist'n true.
inequality: $x+y+z=1 ,x^2y+y^2z+z^2x\le \frac{4}{27}$true .

Yes,I known but it's this inequality: $x^2y+y^2z+z^2x\le \frac{4}{27}$.
If $x^2y+y^2z+z^2x+xyz\le \frac{4}{27}$ is true then I don't know it and want to see your proof. :)

[quote="Albanian Eagle"]the solution has been posted many times (also to the generalisation)
We can't use schur because equality happens at (2\3 1\3 0)  ;)[/quote]
But it does hold at(1/3,1/3,1/3)

[quote="mysteryfigure"][quote="zhaobin"]let $f(x,y,z)=x^2y+y^2z+z^2x+xyz$
we may assume $x \ge y \ge z$(it is easy we do this assume although it is only cyclic)
we have:$f(x+z,y,0)-f(x,y,z)=(x+z)^2y-(x^2y+y^2z+z^2x+xyz)=xyz+z^2y-y^2z-z^2x=z(x-y)(y-z) \ge 0$
so we need to prove:$f(x+z,y,0) \le \frac 4 {27}$
the rest is easy.
the solution maybe not easy to acceptible.
But I think it is right.
the equality hold when $x+z=\frac 2 3,y=\frac 1 3,$ and $x=y,y=z,or, z=0$
that means $x=y=z=1/3$ or $x=\frac 2 3,y=\frac 1 3,z=0$[/quote]
I am afraid your assume is partially right because the original term is not symmetry,but the partial answer doesn't seem to have errs,indeed.
Would you find out a way to complete the answer?[/quote]
if $x \ge z \ge y$
we have:$f(x,z,y) \ge f(x,y,z)$


Positive integers are written in the squares of an infinite square grid, such that if a number $ a$ is written in one square, the sum of the number in the square to its right and the square below it is $ 2a\plus{}1$. Prove that on each diagonal parallel to the line $ x\equal{}y$, all numbers are different.


Let $ x$, $ y$ be two integers with $ 2\le x, y\le 100$. Prove that $ x^{2^n} \plus{} y^{2^n}$ is not a prime for some positive integer $ n$.

Proposed solution.

We prove that for given $ x,y$ and $ x\neq y$ there exists positive integer $ n$ such that $ x^{2^n} \plus{} y^{2^n}$ is multiple of $ 257 \equal{} 2^8 \plus{} 1$.

Let $ z \equal{} x/y, z\neq 0,1$ is the element of field $ \mathbb{F}_{257}$. The order of multiplication group $ G \equal{} \mathbb{F}^{*}_{257}$ is $ 256 \equal{} 2^8$. Thus $ ord_{G}(z) \equal{} 2^s, s\geq 1$ and $ z^{2^{s \minus{} 1}} \equal{} \minus{} 1$. Then $ x^{2^{s \minus{} 1}} \equal{} \minus{} y^{2^{s \minus{} 1}}$ and $ x^{2^{s \minus{} 1}} \plus{} y^{2^{s \minus{} 1}} \equal{} 0$.

Upd.: Then $ x^{2^{s \minus{} 1}} \plus{} y^{2^{s \minus{} 1}}$ is multiple of 257. Equation $ u^2\plus{}v^2\equal{}257$ has two solutions $ (1,16)$ and $ (16,1)$ but $ x,y\geq2$.

And finally $ z\equal{}x/y\neq\minus{}1$ in field $ \mathbb{F}_{257}$ since $ x,y\leq100\le257/2$

Assume that $ x$ and $ y$ are different -if they are equal it is trivial. Let this assumption to be wrong. Since $ 257 \equal{} 2^8 \plus{} 1$ is a prime number and $ x$ and $ y$ are coprime with $ 257$, $ x^{2^8} \minus{} y^{2^8} \equal{} (x^{2^7} \plus{} y^{2^7})...(x \plus{} y)(x \minus{} y)$ is divisible by $ 257$ (Little Fermat). Moreover $ 257$ can't divide $ x \minus{} y$ so $ 257$ must divide $ x^{2^k} \plus{} y^{2^k}$ for some $ k$, hence from the claim it must be equal to $ 257$. Due to $ x, y < \frac {257}{2}$, $ 257$ must be the sum of two squares. It is easy to check that only solutions of this equation is $ (1, 16)$ (and negatives, permutations). This gives that one of the numbers is $ 1$ which contradicts our claim.

Edited because of my absence of knowladge about 129 is not a prime :D So sorry about this.

$ 129 \equal{} 3 \cdot 43$? In general $ 2^k \plus{} 1$ prime requires $ k$ to be a power of $ 2$.

:blush:  :lol: Yes, I know this of course. I can't believe myself, I solved the problem in our team's preparation exam, wrote the solution there,  one more time here and i haven't realized that 129 is not prime until I saw mellowmellon's warning. Anyways, the solution still works with a little fixing.

I considered $ 2^{16}\plus{}1\equal{}65537$.

Firstly $ 65537$ is a prime.
Since $ 2\le x,y \le 100$, there exists an integer $ r$ such that $ r \equiv \frac{x}{y} (mod 65537)$.
Take $ d$ as the index of $ r$ $ (mod 65537)$.
According to Fermat's Theorem, $ d|\phi (65537) \equal{}65536\equal{}2^{16}$, so $ d$ is a power of $ 2$ and $ d \ge 4$ because $ r$ cannot be $ 1$ or $ \minus{}1$.
Denote $ d\equal{}2^k$ then $ k \ge 2$ and $ k$ is an integer
Then $ r^{\frac {d}{2} \equiv \minus{}1 (mod 65537)}$, which means $ 65537|x^{2^{k\minus{}1}}\plus{}y^{2^{k\minus{}1}}$.
If $ x^{2^{k\minus{}1}}\plus{}y^{2^{k\minus{}1}}\not \equal{} 65537$, then $ x^{2^{k\minus{}1}}\plus{}y^{2^{k\minus{}1}}$ is not a prime.
If $ x^{2^{k\minus{}1}}\plus{}y^{2^{k\minus{}1}} \equal{} 65537$, since every prime of the form $ 4k\plus{}1$ can be expressed as the sum of two perfect square and that express is unique, we deduce that $ \{ x^{2^{k\minus{}1}}, y^{2^{k\minus{}1}} \} \equal{} \{ 65536, 1 \}$
Without loss of generality we assume y=1, then $ x^{2^k}\plus{}y^{2^k}\equal{}2^{32}\plus{}1\equal{}641*6400417$ is not a prime, too.
Q.E.D

This is a nice problem but way too easy for P8. Here is my solution.

Since $x^{256}\equiv y^{256}\pmod{257}$ but $x\not\equiv y\pmod{257}$, we take the maximal $k$ such that $x^{2^k}\not\equiv y^{2^k}\pmod{257}$. By maximality,
$$257\mid x^{2^{k+1}} - y^{2^{k+1}} = (x^{2^k}+y^{2^k})(x^{2^k}-y^{2^k})$$
thus $257\mid x^{2^k}+y^{2^k}$. Manually check that $x^{2^k}+y^{2^k}\ne 257$, so we are done.

[quote=Swad][quote=linboll]since every prime of the form $ 4k+1$ can be expressed as the sum of two perfect square and that express is unique,[/quote]

Didn't know it's unique too. Maybe there's some easy proof of this fact that I'm missing. Can you give me a link to it's proof please? Thanks in advance :)[/quote]

Assume on the contrary that
$$ p = a^2 + b^2 = c^2 +d^2$$
with $p$ prime ; $a,b,c,d \in \mathbb Z_{\ge 0}$ ; $\{a,b\} \ne \{c,d\}$. Then,
$$  (ad + bc)^2 + (ac - bd)^2 = (ac +bd)^2 + (ad - bc)^2 = (a^2+b^2)(c^2 + d^2) = p^2   $$
But we also have
$$ (ac+bd)(ad+bc) = cd(a^2 + b^2) + ab(c^2 + d^2) = p(ab+cd) $$
So $p$ divides at least one of $ac+bd,ad+bc$. WLOG $p \mid ac+bd$. Recall
$$p^2  = (ac+bd)^2 + (ad-bc)^2$$
We obtain $ad - bc = 0$, i.e. $ad = bc$. But also
$$ a^2 + b^2 = c^2 + d^2 $$
This forces $a=c$ and $b=d$, which is a contradiction.

[quote=Schur-Schwartz][quote]
This forces $a=c$ and $b=d$, which is a contradiction.
[/quote]
Wait, you didn't use p=4k+1 at all. Is this lemma correct for every p?[/quote]

If $p = 4k + 3$, by $\pmod{4}$, there are no solutions to $a^2 + b^2 = p$.


search
MATHCOUNTS
complementary counting
number theory

Well, 1-99 will have it 9+1 times, 100-199 will have it 9+9+1 times, 200-299 will have it 9+1 times, and it's that way to 500. Then it's just a simple sum.

Oh yeah, just add another 8 to each hundreds span, sorry, my mistake. I only counted 11...

i think i figured something out

can you solve it by counting the amount of ones in each place:

like, in the units place, we know that 1 will appear 50 times (because it appears once in every ten)

and in the tens place, doesn't the one appear 10 times every hundred, and so it appears 50 times

and in the hundred's place, we know the 1 appears 100 times

so its 100+50+50= 200

right?

Yes, but I would think it easier to to it by range.[/u]

[quote="Altheman"]Here is a slightly different problem:

How many pages have a $1$ in the number of a 500 page book numbered $1$ to $500$?

[hide="hint"]We can apply the principle of inclusion-exclusion. Let $A$, $B$, and $C$ be the sets of all integers $1$ to $500$ inclusive such that the hundreds, tens, and ones digit is $1$...[/hide][/quote]

Or you can do complementary counting and get a 1 step answer of $176$ :D

[quote="Anaxerzia"][quote="kohjhsd"][quote="mr. math"]A book has 500 pages numbered 1,2,3 and so on. How many times does the digit 1 appear in the page numbers?[/quote]

[hide]
001, 011, ..., 491: 50
010, 011, ..., 419: 50
100, 101, ..., 199: 100

Therefore 50+50+100=200.[/hide][/quote]

Arent you over counting a lot of them? o_o[/quote]

I am just counting how many 1's there are...

[quote="kohjhsd"][/quote][quote="Anaxerzia"][quote="kohjhsd"][quote="mr. math"]A book has 500 pages numbered 1,2,3 and so on. How many times does the digit 1 appear in the page numbers?[/quote]

[hide]
001, 011, ..., 491: 50
010, 011, ..., 419: 50
100, 101, ..., 199: 100

Therefore 50+50+100=200.[/hide][/quote]

Arent you over counting a lot of them? o_o[/quote][quote="kohjhsd"]

I am just counting how many 1's there are...[/quote]

Woops  :blush:. I misread the question and thought the original was asking how many pages.

[quote="Altheman"]Here is a slightly different problem:

How many pages have a $1$ in the number of a 500 page book numbered $1$ to $500$?

[hide="hint"]We can apply the principle of inclusion-exclusion. Let $A$, $B$, and $C$ be the sets of all integers $1$ to $500$ inclusive such that the hundreds, tens, and ones digit is $1$...[/hide][/quote]

Can't you use Number Theory to solve this problem by changing the base of 500 ad subtracting? If so, how?[/hide]

[quote="Anaxerzia"][/quote][quote="kohjhsd"][/quote][quote="Anaxerzia"][quote="kohjhsd"][quote="mr. math"]A book has 500 pages numbered 1,2,3 and so on. How many times does the digit 1 appear in the page numbers?[/quote]

[hide]
001, 011, ..., 491: 50
010, 011, ..., 419: 50
100, 101, ..., 199: 100

Therefore 50+50+100=200.[/hide][/quote]

Arent you over counting a lot of them? o_o[/quote][quote="kohjhsd"]

I am just counting how many 1's there are...[/quote]

Woops  :blush:. I misread the question and thought the original was asking how many pages.

[quote="Altheman"]Here is a slightly different problem:

How many pages have a $1$ in the number of a 500 page book numbered $1$ to $500$?

[hide="hint"]We can apply the principle of inclusion-exclusion. Let $A$, $B$, and $C$ be the sets of all integers $1$ to $500$ inclusive such that the hundreds, tens, and ones digit is $1$...[/hide][/quote][quote="Anaxerzia"]

Can't you use Number Theory to solve this problem by changing the base of 500 ad subtracting? If so, how?[/hide][/quote]
No, besides. its a lot easier to use complementary counting..it takes 1 step doing that.

No you cant change bases (im betting your thinking of base 9 cahnge but that doesnt work because its not skipping the 1's its only counting the ones)

refer to this
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=630240869&t=118820

see the difference in problems?

[quote="Anaxerzia"]Oh I see now. How would you do this with complimentary counting then?[/quote]

Just try to use constructive counting, and never use the digit $1$...

[quote="Anaxerzia"]Oh I see now. How would you do this with complimentary counting then?[/quote]
Instesd of counting the number of pages with at least one $1$ or using PIE, just count the # of pages[i] without [/i]any $1$s.  Then subtract from 500.  That is,
[hide]First digit: 0,2,3, or 4
Second digit: 0,2,3,4,5,6,7,8,9
Third digit:  0,2,3,4,5,6,7,8,9
Notice how we haven't counted $500$, but we have counted $000$ so that makes up for it.  We have
$4\cdot9\cdot9=324$.
Now we subtract that from $500$ to get $176$.[/hide]

Haha yea...thats correct. I sometimes forget my elementary counting techniques :P I would get owned at mathcounts haha.


Find all positive integers $x$ and $y$ such that
\[ax+xy+ay = (2000+a) \]
for a given integer $a$, say 6.

[quote="t0rajir0u"]$(x+a)(y+a) = 2000+a-a^{2}$

The RHS looks irreducible to me.[/quote]

But this problem is suppose to be solvable. (without a computer)

[quote="ZetaX"]What t0rajir0u said already solves the problem ;)[/quote]

Well for an arbitrary $a$, but not 6 or say 7 ?

Umm, what the problem in setting $a=6$¿
General solution implies special case solution.

Easy divisor $d|(2000+a+a^{2})$ give one solution $x=d-a,y=\frac{2000+a+a^{2}}{d}-a$. In this way we find all solutions.


Consider the function $ f(x)\equal{}|2x\minus{}1|$ defined in $ 0\leq x\leq 1$.

Find the number of the roots such that $ f(f(f(x)))\equal{}x$.

[hide="Solution"]
We note that 1/2 gets mapped to 0, and 0 and 1 get mapped to 1.

This, the graph of f(x) looks like V, the graph of f(f(x)) looks like VV, and the graph of f(f(f(x))) looks like VVVV, where the whole thing is bounded inside the unit square.

If we now draw the graph of y=x, we get one intersection between y=f(f(f(x))) and y=x for each of the two branches of each V, for a total of 8 roots.[/hide]

[hide="answer"]
The 8 lines of the graph of $ f(f(f(x)))$ have equations of $ y\equal{}8x\minus{}c$ and $ y\equal{}\minus{}8x\plus{}c$
for $ c\equal{}1,3,5,7$. The intersection of these 8 lines with $ y\equal{}x$ occur at $ x\equal{}\frac 17,\frac 37,\frac 57,\frac 77,\frac 19,\frac 39,\frac 59,$ and $ \frac 79$.
[/hide]


At a party, each man danced with exactly three women and each
woman danced with exactly two men. Twelve men attended the party.
How many women attended the party?

There were exactly $ 12\cdot3\equal{}36$ dances total. Since each woman danced with exactly two men, we must divide by $ 2$ to get $ \boxed{18}$ women.

[hide=Logic]
If you draw it out, you see that for every $2$ men you need $3$ women to satisfy the conditions of the party. There are $12$ men, so there are $3\times6=18$ women that attended the party.
[/hide]

[quote=ElectricFox]What? I am really confused. What if only 3 women came but each man danced with each 3? Or what if 36 women came and each man danced with three? Or what if 100 women came and each man danced with three? The women would still dance with two men. I do not understand. Someone help please![/quote]

But there are only 12 men

[quote=jmiao][quote=pinkpig]the women's should be loyal to their husband and not dance with too many other men[/quote]

ilp is taking an effect on you[/quote]

what do you mean?

[quote=ZETA_in_olympiad][quote=ElectricFox]Each man could dance with each of the 3 women or
Each man could dance with 3 of the 36 women or
Each man could dance with a bunch of the 100 women.[/quote]

Absurd since each woman would dance with [b]exactly[/b] 2 men.[/quote]

Oh  :oops_sign: , I did not notice it said exactly.  :wallbash:

[hide=solution]if you draw it out,you'll see it takes 3 women for 2 men
the answer is 3*6=18[/hide]


I was reading the solution on Kalva, and the solution used a particular theorem that I have never seen before.  I was wondering if anybody could tell me the name of that theorem.

I initially thought it might be Menelaus's Thm., but when you draw a diagram, it seems to be slightly different.  ???

 :huuh:

Thanks.


function
limit
calculus
calculus computations

[quote="jagdish"]
[b](1) Let f and g are two function for which
f(x-y)=f(x)g(y)-f(y)g(x) and
g(x-y)=g(x)f(y)+f(x)g(y) x,y belong to R.
If d/dx{f(0+)} exists, then prove that f(x) is differentiable at x=0, also find
d/dx{g(0)}.  [/b]

[/quote]

Let, $ x \equal{} y$
$ g(0) \equal{} 2f(x)g(x) \\
g'(0) \equal{} 2[f(x)g'(x) \plus{} f'(x)g(x)]$

I think if $ y \to x^ \minus{}$ we get $ \lim _{x \to 0 \plus{} } f(x) \equal{} 0$ and if $ y \to x^ \plus{}$ we get $ \lim _{x\to 0^ \minus{} } f(x) \equal{} 0$ and $ x \equal{} y, \; f(0) \equal{} 0$, hence differentiable.

Put $ x\equal{}y$ in $ f(x\minus{}y)$ to obtain $ f(0)\equal{}0$

Put $ y\equal{}0$ in $ f(x\minus{}y)$ to get $ f(x)\equal{}f(x)g(0)$.Either $ f(x) \equal{}0$  in which case  whatever has been asked to prove is obvious.However if    $ f(x)\neq 0$,

then $ g(0)\equal{}1$ and we proceed as follows:

Put $ x\equal{}0$ in $ f(x\minus{}y)$  to get $ f(\minus{}y)\equal{}\minus{}f(y)$.Similarly,it can be proved that $ g(x)\equal{}g(\minus{}x)$.

$ f'_{\minus{}}(0)\equal{}\lim_{h\to 0}\frac{f(0\minus{}h)\minus{}f(0)}{\minus{}h}\equal{}\lim_{h\to 0}\frac{f(\minus{}h)}{\minus{}h}\equal{}\lim_{h\to 0}\frac{f(0\plus{}h)\minus{}f(0)}{h}\equal{}f'_{\plus{}}(0)$

Therefore,$ f(x)$ is differentiable at $ x\equal{}0$.

$ g'_{\minus{}}(0)\equal{}\lim_{h\to 0}\frac{g(0\minus{}h)\minus{}g(0)}{\minus{}h}\equal{}\lim_{h\to 0}\frac{g(\minus{}h)\minus{}f(0)}{\minus{}h}$

$ \equal{}\minus{}\lim_{h\to 0}\frac{g(h)\minus{}f(0)}{h}\equal{}\minus{}\lim_{h\to 0}\frac{g(0\plus{}h)\minus{}g(0)}{h}\equal{}\minus{}g'_{\plus{}}(0)$

If $ g(x)$ is differentiable at $ x\equal{}0$ then $ g'_{\minus{}}(0)\equal{}\minus{}g'_{\plus{}}(0)$ in which case $ g'(0)\equal{}0$.But then it should be given that $ g(x)$ is differentiable at $ x\equal{}0$

There is something very strange with this question.

If I put $ y\equal{}0$ in $ g(x\minus{}y)\equal{}g(x)f(y)\plus{}f(x)g(y)$ i.e. $ g(x\minus{}0)\equal{}g(x)f(0)\plus{}f(x)g(0)$ I get $ g(x) \equal{}f(x)$

Will you tell me the source of this question.It appears to be manufactured in some coaching institute for IITJEE.They take some original question and try to twist it and in the process the validity of the question goes for a toss

[quote="pankajsinha"]
Will you tell me the source of this question.It appears to be manufactured in some coaching institute for IITJEE.They take some original question and try to twist it and in the process the validity of the question goes for a toss

[/quote]

Must be FIITJEE.......!

 :rotfl:


induction
complex numbers
inequalities proposed
inequalities

I had a very nice idea for the following problem, but I feel that it is trivial. Anyway, I will post it, since my idea is too nice not to be shared:
    Prove that for any complex numbers $ a_1,...,a_n,b_1,...,b_n $ we can find three indices $ 1\leq i_1<i_2\leq n, 1\leq j\leq n $ so that $ (n-1)| a_{i_1}\cdot a_{j_1}+b_{i_1}\cdot b_{j_1}|\geq | a_j^2+b_j^2| $.

I am sorry harazi. I'm sure I misunderstood something.
can'te we just delete a_n,b_n and use induction?.
Then we can find i1,j1,j s.t.
(n-2)|a_i1*a_j1+b_i1*b_j1|>=|a_j^2+b_j^2|
Then of course
(n-1)|a_i1*a_j+b_i1b_j|>=|a_j^2+b_j^2|

Well, that's it, another mathematical gem of Harazi! The most idiot problem ever to be discussed. Oh, boy, how stupid can I be. And it was such a nice idea. And I was so idiot... Too bad  :blush:  :blush:  :blush:


probability
combinatorics unsolved
combinatorics

let $ X$ be a set with $ n$ elements,and let $ F$  be the family of subsets from $ X$ such that,for all $ A,B$ $ \subset$$ F$ ,$ A$ $ \cap$ $ B$ $ \neq$ $ \phi$,prove that,if there is not a subset $ C$ such that $ C$ $ \not<$ $ F$ ,and for all  $ A$ $ \subset$ $ F$ we have $ A$ $ \cap$ $ C$ $ \neq$ $ \phi$ then, $ Card(F)=2^{n-1}$

Let $ X$ be a set with $ n$ elements and $ F$ a family of subsets of $ X$ such that for all $ A, B \in F$, $ A \cap B \neq \emptyset$.  Prove that if there is no subset $ C \subset X$ such that $ C \not \in F$ and for all $ A \in F$ we have $ A \cap C \neq \emptyset$ then $ \textrm{card}(F) = 2^{n-1}$.

In other words, "A maximal family of subsets of an $ n$-set with the probability that every two have a point of intersection has size $ 2^{n-1}$."  It has appeared several times before, most recently here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160677


Find the sum to $ n$ terms of the series when $ T_n\ \equal{} \ \frac{1}{n^4\plus{}n^2\plus{}1}$

Here's what I could do-:
[hide] The above simplifies to $ T_n\equal{} \frac{1}{2}. \left(\frac{1}{n^2\minus{}n\plus{}1} \ \minus{} \ \frac{1}{n^2\plus{}n\plus{}1} \right)$
How do I further this by getting a form such that $ \Sigma n^a$can be applied, $ a$ being 1,2,3,4..etc.
[/hide]

Thx

You have the right manipulation but you're looking for the wrong trick.  Try using your new form and testing out the first few values of the sequence.

Ahhh... got it. $ (n\plus{}1)^2\minus{}(n\plus{}1)\plus{}1\equal{}n^2\plus{}n\plus{}1$
Thx a lot JBL  :)


linear algebra
matrix
algebra
polynomial

Let $ A,B \in \mathcal{M}_3 \left( \mathbb{R} \right)$ such that $ \det \left(A \right)\equal{}\det \left(B \right)\equal{}\det \left(A\plus{}B \right)\equal{}\det \left( A\minus{}B\right)\equal{}0$. Prove that \[ \det \left(xA\plus{}yB \right)\equal{}0, \forall x,y \in \mathbb{R}.\]

The cubic projective polynomial $ P(X, Y) \equal{} \det(AX \plus{} BY) \equal{} 0$ has four roots, hence is identically zero.  (Projective polynomials, like regular polynomials, have at most $ n$ roots if they are of degree $ n$.)


Can someone tell me how to find the sum of a infinite geometric series that has first term 3 and common ratio 1/2?

Thanks, please write a clear solution that I can understand.

meaning...$ 3 \plus{} \frac32 \plus{} \frac34 \plus{} \frac38 \plus{} ...$?

if so,
[hide="solution"]

first. lets factor out a $ 3$. this gives:

$ 3(1 \plus{} \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} ...)$?

its a pretty well known fact that $ (1 \plus{} \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} ...) \equal{} 2$, but supposing you didn't know that, lets solve. forget about the three for now, lets try to figure out what $ (1 \plus{} \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} ...)$ equals. let $ (1 \plus{} \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} ...) \equal{} x$

factor out $ \frac12$. we get:

$ \frac12(2 \plus{} 1 \plus{} \frac12 \plus{} \frac14...) \equal{} x$

$ \frac12(2 \plus{} x) \equal{} x$

$ 1 \plus{} \frac {x}{2} \equal{} x$

$ \frac {x}{2} \equal{} 1$

$ x \equal{} 2$

going back to the original question,

$ 3 \plus{} \frac32 \plus{} \frac34 \plus{} \frac38 \plus{} \cdots \equal{} 3(1 \plus{} \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} \cdots) \equal{} 3(2) \equal{} \boxed{6}$

it looks a lot longer than it really is...

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hope that helps. :)

Thanks so much. It really helps. Do you know any textbooks I can buy that would teach these kinds of concepts?

Thanks.

heh

AoPS books teach this. I think in the introduction to algebra and intermediate algebra

The ones available right here on the site. (See "Bookstore".) Specifically, Introduction to Algebra and AoPS Volume 1.

You can also check AoPS volume 1!

the basic formula for this type of question is $ \frac{a}{1 \minus{} r}$

Although it's good to know how to prove/construct this formula
which is taught in AoPS volume 1

...like I just said.  :wink:

[quote="isabella2296"]The ones available right here on the site. (See "Bookstore".) Specifically, Introduction to Algebra and [b]AoPS Volume 1.[/b][/quote]

:blush: Sorry about that! I didn't read carefully! :blush: