pigeonhole principle
algebra
polynomial
system of equations
combinatorics unsolved
combinatorics

[quote="hanschinaman"]Prove: Given 2n-1 integers, we can find n integers of them such that the sum can be written in the form kn, where k is a integer.[/quote]
Isn't it somehow too weak? :maybe: Say $ x_{1}, x_{2}, \ldots, x_{n}$ are $ n$ integers, where $ n \in \mathbb{N}$. Then set $ s_{1}= x_{1}$, $ s_{2}= x_{1}+x_{2}$, $ \ldots$, $ s_{n}= x_{1}+x_{2}+\ldots+x_{n}$. If $ n \mid s_{i}$, for any $ i = 1, 2, \ldots, n$, then we're done. Otherwise, by the pigeonhole principle, there are $ i, j = 1, 2, \ldots, n$, with $ i < j$, such that $ s_{j}\equiv s_{i}\mod n$. Then $ n \mid (x_{i+1}+\ldots+x_{j})$, and the proof is complete.

your solution is wrong,the problem asks to find exactly $ n$ numbers which their sum is divisible by $ n$...
but as [b]s.tringali[/b] said this is a classic problem,and I`ll post it`s solution if no one gave you a link of it...

I think it would be better if I paid more attention when reading of a problem... :oops: But then it's just the Erdos-Ginzberg-Ziv theorem. I'll try to find a link, I'm pretty sure it has been already discussed many and many times on the forum.

just note that if the problem is true for $ n=a$ and $ n=b$ then the problem is true for $ n=ab$...
so it remains to prove this problem for $ n=p$ where $ p$ is a prime number...
now the rest has two solutions,which I`ll give it in the hidden section below,because of whom which want to think on the case where $ n$ is prime...
[hide="solution"]
[b]if $ p$ is a prime number and $ 2p-1$ natural numbers $ a_{1},...,a_{2p-1}$ are given,show that the sum of $ p$ numbers of them is divisible by $ p$[/b]
$ \boxed{\text{first solution}}$
we will prove this using contradiction,so assume that the sum of every $ p$ numbers of them isnt divisible by $ p$ so according to fermat`s theorem we get that:
$ A^{p-1}\equiv 1 (\bmod p)$
.
.
.
$ B^{p-1}\equiv 1 (\bmod p)$
where $ A=a_{1}+...+a_{p},...,B=a_{p}+...+a_{2p-1}$ (we have $ binom{2p-1}{p}$ equations)
now add the sides of the above equations to get that:
$ \boxed{A^{p-1}+...+B^{p-1}\equiv \binom{2p-1}{p}(*)}$
but we have:
$ A^{p-1}+...+B^{p-1}=(a_{1}+...+a_{p})^{p-1}+...+(a_{p}+...+a_{2p-1})^{p-1}$
but the coefficient of every term in the last part of the above equation is of the form $ \binom{2p-i-1}{p-i}$ which is divisible by $ p$,so the $ LHS$ of $ (*)$ is equivalent to $ 0$ modula $ p$,so we must have:
$ \binom{2p-1}{p}\equiv 0 (\bmod p)$
which is a contradiction...

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

$ \boxed{\text{second solution}}$
the second solution is based on [b]chevalley's theorem[/b]...
denote polynomials $ f_{1}(x_{1},...,x_{2p-1}),f_{2}(x_{1},...,x_{2p-1})$ as:

$ f_{1}(x_{1},...,x_{2p-1})=a_{1}x_{1}^{p-1}+...+a_{2p-1}x_{2p-1}^{p-1}\equiv 0 (\bmod p)$

$ f_{2}(x_{1},...,x_{2p-1})=x_{1}^{p-1}+...+x_{2p-1}^{p-1}\equiv 0 (\bmod p)$

now $ x_{1}=...=x_{2p-1}=0$ is a solution for the above system of equations,so according to chevalley's theorem,this
system has at least one more solution,but $ x_{i}^{p-1}\equiv 0,1 (\bmod p)$ so we have two cases:

$ i)$all of the $ x_{i}^{p-1}$s are equal to $ 0$

$ ii)$or $ p$ numbers of numbers $ x_{i}^{p-1}$ are equal to $ 1$ modula $ p$,so the second equation of the above system
has only two solutions:

$ x_{i}^{p-1}\equiv 0 (\bmod p)$

and the case where $ p$ numbers of the numbers $ x_{i}^{p-1}$ where equal to $ 1$ modula $ p$...

so the other answer to the above system is the following:

[b]$ p$ numbers equal to $ 1$ and the $ p-1$ other equal to$ 0$ modula $ p$...[/b]

so if we put this case in the first equation of the system we get that:"

$ f_{1}(x_{1},...,x_{2p-1})=a_{1}x_{1}^{p-1}+...+a_{2p-1}x_{2p-1}^{p-1}$

$ \equiv a_{i_{1}}+...+a_{i_{p}}+0+0+...+0 \equiv 0 (\bmod p)$

hence we have:

$ p \mid a_{i_{1}}+...+a_{i_{p}}$ and we are done...

[/hide]


Bill has $ \frac 45$ as much money as Bonnie. Bonnie has $ \frac 56$ as much money as Fran. Bonnie has $ \$480$. What is the total amount of money, in dollars, that Bill and Fran have together?

Since Bill has $ \frac{4}{5}$ the amount Bonnie has, he has $ \$480\times\frac{4}{5}\equal{}\$384.$ Since Bonnie has $ \frac{5}{6}$ the amount Fran has, Fran has $ \$480\times\frac{6}{5}\equal{}\$576.$ Finally, the total amount is $ \$384\plus{}\$576\equal{}\boxed{\$960}.$

Useful: Fran has 6/5 as much money as Bonnie, so added up, Bill and Fran have twice the money as Bonnie, or 960 dollars.


logarithms
function
calculus
integration
Putnam
geometry
real analysis

Let $ F(x)\equal{}x\log^{1}x\log^{2}(x)...\log^{m}(x)$ where $ m$ is the greatest integer (possibly zero) such that the mth iterate of the (natural) logarithm function is greater than or equal to one. Determine whether $ \sum_{n\equal{}1}^{\infty}\frac{1}{F(n)}$ converges.

The integral test tells us that it is enough to answer the question whether $ \int_{0}^{\infty}\frac{dx}{F(x)}<\plus{}\infty$. Now, $ F(e^{t})\equal{}e^{t}F(t)$ for $ t>0$ by definition (the empty product is understood to be $ 1$, as usual). Making the change of variable $ x\equal{}e^{t}$, we see that
\[ \int_{0}^{\infty}\frac{dx}{F(x)}\equal{}\int_{0}^{1}\frac{dx}{F(x)}\plus{}\int_{1}^{\infty}\frac{dx}{F(x)}\equal{} 1\plus{}\int_{0}^{\infty}\frac{e^{t}dt}{F(e^{t})}\equal{} 1\plus{}\int_{0}^{\infty}\frac{dx}{F(x)}\,.\]
Hence the integral equals $ \plus{}\infty$. What makes the problem attractive is that the base of the logarithm matters a lot here and $ e$ gives the borderline case.

That is really neat :lol: My method was essentially identical in substance but I wrote it somewhat more clumsily - using the integral test then splitting the integral up in such a manner that there were infinitely many intervals on which it exceeds 1.
[b]edit[/b]; I like that Putnam question jmerry - if I remember correctly from a previous thread it is your favourite example of a "borderline" series. The one I posted above is mine.

Putnam 2002 A6 was a discrete version of this (with integer base for the logarithms):
Fix an integer $ b\ge 2$. Let $ f(1)\equal{}1,f(2)\equal{}2$, and for each $ n\ge 3$, define $ f(n)\equal{}nf(d)$, where $ d$ is the number of digits of base-$ b$ digits of $ n$.
For which values of $ b$ does $ \sum_{n\equal{}1}^{\infty}\frac{1}{f(n)}$ converge?

This was the solution I handed out to my class after the 2002 Putnam was over. That being several years ago, I can't remember whether it was my work or jmerry's or some blend.

[hide="Solution to Putnam 2002 A6"][b]A6:[/b] For the integer $ b\ge 3,$ we have the sequence $ f(n)$ defined by $ f(1) = 1, f(2) = 2,$ and after that, $ f(n) = nf(d),$ where $ d$ is the number of base-$ b$ digits of $ n.$ Then  $ \sum_{n = 1}^{\infty}\frac{1}{f(n)}$  converges for $ b = 2$ and diverges for $ b\ge 3.$ As a lemma in our proof, we will need to estimate $ \sum_{n = b^{k-1}}^{b^{k}-1}\frac{1}{n}.$ On the one hand, picturing this sum as the sum of the areas of boxes shows that $ \sum_{n = b^{k-1}}^{b^{k}-1}\frac{1}{n}>\int_{b^{k-1}}^{b^{k}}\frac{1}{x}\, dx .$
Hence, $ \sum_{n = b^{k-1}}^{b^{k}-1}\frac{1}{n}>\ln b,$ and for $ b\ge 3,\ \ln b > 1.$ On the other hand, since
 $ \frac{1}{nb}+\frac{1}{nb+1}+\cdots+\frac{1}{nb+b-1}<\frac{1}{n},$  $ \sum_{n = b^{k-1}}^{b^{k}-1}\frac{1}{n}$  is a strictly decreasing function of $ k,$ and, in particular, for $ k\ge 3,\ \sum_{n = 2^{k-1}}^{2^{k}-1}\frac{1}{n}\le\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7} =\frac{319}{420},$ which is less than $ 1.$
Assume that $ b\ge 3.$ Then  $ \sum_{n = 1}^{\infty}\frac{1}{f(n)}=\sum_{k = 1}^{\infty}\sum_{n = b^{k-1}}^{b^{k}-1}\frac{1}{f(n)}=\sum_{k = 1}^{\infty}\frac{1}{f(k)}\sum_{n = b^{k-1}}^{b^{k}-1}\frac{1}{n}> (\ln b)\sum_{k = 1}^{\infty}\frac{1}{f(n)}.$
Thus  $ \sum_{n = 1}^{\infty}\frac{1}{f(n)}> (\ln b)\sum_{n = 1}^{\infty}\frac{1}{f(n)}.$ Since $ \ln b > 1,$ this must mean that the sum diverges.
Now assume that $ b = 2.$
${ \sum_{n = 1}^{\infty}\frac{1}{f(n)}= 1+\frac{1}{2}+\frac{1}{6}+\sum_{n = 4}^{\infty}\frac{1}{f(n)}= 1+\frac{1}{2}+\frac{1}{6}+\sum_{k = 3}^{\infty}\frac{1}{f(k)}\sum_{n = 2^{k-1}}^{2^{k}-1}\frac{1}{f(n)}= 1+\frac{1}{2}+\frac{1}{6}+\sum_{k = 3}^{\infty}\frac{1}{f(k)}}\sum_{n = 2^{k-1}}^{2^{k}-1}\frac{1}{n} < 1+\frac{1}{2}+\frac{1}{6}+\left(\frac{319}{420}\right)\sum_{k = 3}^{\infty}\frac{1}{f(k)}.$
So, ${{ \sum_{n = 3}^{\infty}\frac{1}{f(n)}}<\frac{1}{6}+\left(\frac{319}{420}\right)\sum_{k = 3}^{\infty}\frac{1}{f(k)}},$ or  ${ \left(\frac{101}{420}\right)\sum_{n = 3}^{\infty}\frac{1}{f(n)}}<\frac{1}{6}$   and the sum converges.[/hide]


Hey all!

This is an extra credit question given to us by our Calculus teacher. He said we may use ANY online sources. I could definately use the grade help and a boost in the right direction!

You do not have to solve the problem, just get me started with the right figures. The computation I can do myself. I spent an hour or two looking for the right numbers and testing them, but according to the http://146.142.4.24/cgi-bin/cpicalc.pl, all my numbers have so far been incorrect.

Here is the question :

First make the computation from the CPI what amount in June 2008 equates to June 2000. The amount is $ \$$1000.

After you made the preceding computation, calculate the present value of a dollar at this time for an obligation that matures on June, 2009.  You can use the current treasury bill interest rate for this computation.

Thanks for any help in advance!

I believe that this belongs to Calculus forum, not here. But I have to say asking others to solve this here might be considered the same as asking "real" people to help you (when I say real, I mean through old, real, face-to-face communication). Online source might include only information on internet, not help from people across the globe. If you are unsure of this, you might check with your teacher if this is ok (unless, of course, you really don't care and only wish extra credit).

Just to be sure, you want the value of dollar in 2009 in means of the value in 2000?

No I even asked him infront of the whole class, I said can we use online forums/communities to get help with where to look and he said yes.

What the question is asking is to find what 1000 dollars in June of 2000 is now equivalent to how much in 2008, not 2009.

Bump. Anyone have any idea? Spent awhile this evening after work working on the problem. Still can't seem to find the right data. That or I am not doing the right computation. Hm.

Anyone?

The calculator gives you the answer does it not? What exactly do you (does your teacher) want?

I have to be the calculator by showing my calculation. I have to know HOW TO get that answer.

Actually, the calculator gives a different answer from what you want because it uses the average CPI from each year to calculate its answers. [i]172.2[/i] is the average CPI from the year 2000 and [i]215.303[/i] is the average CPI from the year 2008. What this means, approximately , is that 172.2 dollars from 2000 has the same worth as 215.303 dollars from the year 2008. Therefore, the calculator gives 1250.31 dollars from 2008 as having the same worth as 1000 dollars from the year 2000.

However, from the data here [url]ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt[/url], we can see that the CPI from June 2000 is [b][i]172.4[/i][/b] and that the CPI from June 2008 is [b][i]218.815[/i][/b]. It shouldn't be too hard to see how to get the answer now.


let $ f$ is a function such that:
$ [1;p\infty[===>]0;\sqrt{3}]$
$ x===>\sqrt{x+2}-\sqrt{x-1}$
prove that $ f$ is $ surjective$

$ f(1)\equal{}\sqrt 3$ and $ \lim_{x\to \infty }f(x)\equal{}0$. Because $ f$ is continiosly for all $ y\in (0,\sqrt 3]$ exist x, suth that $ f(x)\equal{}y$.


What is the number of degrees of the angle formed by the minute and hour hands of a clock at 11:10PM?

85 degrees, just remember that the hour hand still moves!

A better solution:

Each "number" on the clock is $ 30^\circ$, because there are $ 12$ numbers and $ \tfrac{360}{12} \equal{} 30$.  The hour hand is $ \tfrac{1}{6}$ past the $ 11$th number, so it is $ 30 \cdot 11 \plus{} 30 \cdot \tfrac{1}{6}^\circ \equal{} 330 \plus{} 5^\circ \equal{} 335^\circ$ past vertical.  The minute hand is at the $ 2$nd number, or $ 30 \cdot 2^\circ \equal{} 60^\circ$.  Subtracting yields $ 335 \minus{} 60^\circ \equal{} 275^\circ$.  However, we must subtract from $ 360^\circ$ because the angle measure is greater than $ 180^\circ$.  Thus the answer is $ 360 \minus{} 275^\circ \equal{} \boxed{85^\circ}$.

Bah.
We can use rate bashing methods.
Say we are at 11, which is 330 degrees clockwise from 12. Now, we want to find how much time it takes for the angles to be $ x$, which we want. Equating relative distances and times, we have the equation $ t\equal{}\frac{330\minus{}x}{6\minus{}0.5}$. Since we know $ t\equal{}10$, we substitute to pwn and find that $ x\equal{}275^\circ$. However, we want the acute angle (I'm assuming), so our answer is $ 360^\circ\minus{}275^\circ\equal{}85^\circ$. We can easily prove, using this as a guide, that in general, the angle formed by the HM hands is $ |5.5m\minus{}30h|$, where we subtract this quantity from $ 360^\circ$ if necessary.

Or use the formula $ |5.5m\minus{}30h|\equal{}|55\minus{}330|\equal{}275$, which is the same as a 360-275=85  degree angle.


Does one of the firt $ 10^8+1$ Fibonacci numbers terminate with $4$ zeroes?

Yes.
For any $n$, at least one of the $n^2+1$ first terms (starting at $F_1$ otherwise it is trivial) in the Fibonacci sequence is divisible by $n$.

Let's consider the sequence modulo $n$ :
Then the first $n^2+1$ terms form $n^2$ pairs of consecutive terms, none being $(0,0)$ (otherwise all the terms in the Fibonacci sequence would be divisible by $n$). Thus, at least one pair appears twice, which ensures that the sequence is periodic with period not greater than $n^2-1$ (note that we could define the sequence for negative subscipts by the recursive formula so that the sequence is periodic from the start). It follows that the first pair to appear twice is $(1,1)$ (since $F_1=F_2=1$). Let $a_k$ be the last term before this second occurence : $\cdots , a_k, 1,1, \cdots$
Using the recursive formula of the sequence, it follows that $a_k = 1-1 = 0 \mod [n]$, as desired.

Pierre.


How many numbers is it possible to form with only 3 digits (excluding 0) ?

And what is the total of all these numbers ?

[hide]$9*9*9=729$ possible numbers. Dont know why you would exclude zero though. Not sure how to do the 2nd part.[/hide]

The average is $555$, therefore sum is $555*729=404595$

Modeler, please explain the average.

We could try the same problem, but including zero this time.

Nevermind modeler about the average 555. It's obvious.

Here's another puzzle :

Find a 5-digit number that, if we place a 9 at the beginning to make a new number, is 4 times larger than if we place the 9 at the end instead!


ceiling function
videos
continued fraction

So. Any good movies you watched??

I watched 21 three (?) weeks ago.
I was happy. Except that I (think) on Ben's Birthday Cake, with the fibonnacci sequence, they started with zero... i was gonna say something in the middle of the theater, but that would just make me look like some idiot.
21 is an awesome movie. you should see it. (unless you don't like gambling)

Has anyone seen AVP: Reqium? I heard it was good, but I haven't seen it yet.

I know nothing about the movie 21, but isn't the 0th term of the Fibonacci sequence defined to be 0?

[quote="kimmystar94"]i was gonna say something in the middle of the theater, but that would just make me look like some idiot[/quote]

Well, I was sort of the same.  And I said similar things to my friends anyway.  Like "Switch Doors.... this is the Monty Hall Problem" or "Do you see what he did to solve that continued fraction [in the background]"

I didn't catch any real mistakes, and I have to agree with leoxnlin... $ F_0\equal{}0$.

hmm. i thought $ F_0 \equal{} 1$ but w/e

yes i saw 21--cool movie

ummm.... i give it $ \frac {\lceil{e \plus{} \pi \plus{} i^4}\rceil}{10}$

I saw 21 last week with some relatives of mine.  I liked it.

oh. and i saw the switching doors thing. i was telling my parents.. i remember it now!

and i still think the first term of fibonnacci is 1.

[quote="perfect628"]$ F_0 \equal{} F_1 \equal{} 1$. Yay.[/quote]

But then it doesn't satisfy Binet's formula...

and $ F_{0}\equal{}0$, however, this is the 0th term, not the first.

well, I say it's the 0th

Sorry I know it's off topic but

Ugh I always do that...When I was first taught about Fib I was told there was no 0, and just haven't gotten it out of my head since...

The fibonacci sequence is defined as $ F_n\plus{}F_{n\plus{}1}\equal{}F{n\plus{}2}$, with $ F_1\equal{}F_2\equal{}1$. Thus, $ F_0\equal{}0$.

Okay guys, some authors prefer $ F_1 \equal{} F_2 \equal{} 1$ and some prefer $ F_0 \equal{} F_1 \equal{} 1$. They're essentially the same thing and both are used.

Personally, I prefer $ F_0 \equal{} F_1 \equal{} 1$.

It's much better to use $ F_0 \equal{} 0$, $ F_1 \equal{} 1$, and $ F_2 \equal{} 1$.  Otherwise, some properties become uglier to state.  For example, $ F_{mn}$ is divisible by $ F_m$ and $ F_n$.  Also $ \gcd(F_m, F_n) \equal{} F_{\gcd(m, n)}$.  Those properties are too nice to stick in random plus or minus 1's.

The Fibonacci Quarterly (the journal of Fibonacci numbers) uses this convention, and I recommend sticking with it.

I wonder if this topic is considered spam and will be locked... lol

Movies? Ratatouille was a good one.

[quote="serialk11r"]I saw I am Legend on Friday, it was good :D[/quote]

When I saw that "the last man on Earth" was actually not the last, I felt like being cheated :(
The scriptwriter is too gutless to make the human race extinct :P
imo The Mist is better, especially the ending

I heard I am Legend was amazing on Blu-Ray.

Cloverfield seems like a nice one too.

The Mist was cool

Cloverfield--haven't seen it, heard its disappointing

i don't like scary movies. they freak me out.

Most scary movies nowadays are "the most interesting ways humans die"

I haven't seen anything lately.

[quote="7h3.D3m0n.117"]Cloverfield seems like a nice one too.[/quote]
Hmm... I haven't seen it.  My dad saw it, but he didn't like it.

Personally, I'm not a fan of scary movies.

Yeah, I saw Cloverfield. I didnt like it, it was supposed to be like a home video so the camera moved around a bunch.  I got motion sickness.

Vantage Point was good though

[quote="Quickster94"]Vantage point was a really good movie
It was an interesting and new way to show the same thing
but, the ending was a little bit weird
the terrorist ruins his plans to save a little girl, even when he's already killed so many ppl?[/quote]

before, only police knew. not only would running over the girl stop them; it would attract everyone's attention to them

Do youtube videos count? I have a great one...

Youtube [i]movies[/i]? I thought there was a maximum time limit for video uploads?

Anyone watched The Forbidden Kingdom ?
Want to know how non-Chinese think about it

[youtube]F226lI-lo5M[/youtube]

eh, it's not long, but it's funny.

hahahaha. i recently saw iron man. that movie is pretty awesome. ^^

Okay... Of the movies mentioned so far:

SEEN:
21 (8.5 out of 10, because I say so)
I am Legend (5 out of 10, really corny ending in my opinion)
Iron Man (10+Pi out of 10, really good)

HAVEN'T SEEN:
Cloverfield
The Mist
Vantage Point
The Forbidden Kingdom

I haven't seen Iron Man yet, but I don't really like any movie based on a popular comic (?). Movie makers aren't expressing their creativity, just their demand for money and use of cheap computer graphics. So I'll deduct a point from whatever score I'll give to Iron Man (although it'll probably still be very high around 8-9).

Speaking of cheap computer graphics, has anyone seen the amazing Speed Racer? I'm going to automatically give that zero, no questions, no comments.

haha. speed racer bombed at the box office... bad

they MURDERED speed racer.

at least keep it a graphics movie, not a real-human-acting-out movie.

Apparently Speed Racer pulled in second at the box office (I recall that one of the news anchors said it "stalled in second place"  :rotfl: ) so it must have done somewhat decent. But maybe that's because all the families decided to watch so that bumped up ticket sales to like 4 per view.

InuYasha move: Swords of an Honorable Ruler. (ok so it's kinda old but o well its awesome.)

I want to watch that one. I need to watch it...

[quote="dreambold22"]I've finished all the manga for InuYasha. yay[/quote]

this has [i]what [/i] to do with recent movies?

Iron Man was a Great movie. I'd give it a 9.5 out of 10.

I'm gonna see 21, Indiana Jones, and Hancock.

I AM IRONMAN. BOSS

Don't expect too much from Indian Jones. Next bigun up is CON-Prince Caspian. I will watch that

ironman was funny, tho kinda wierd. we are playing the theme from it in orchestra class. (it does NOT sound right on the violin.)


geometry
ratio
trigonometry
angle bisector
perpendicular bisector
geometry proposed

Let ABC be a triangle. Perpendiculars from B and C meet the bicsector of angle A at P and Q respectively. The line from P, parallel to AB, meets the line from Q, parallel to AC, at R. Prove that AR is the symmedian from A :(

[quote="shoki"]these perpendiculars from $ B$ and $ C$ r perpendicular to
$ AC$,$ AB$ or the bisector of $ \angle A$ ?
for both cases the problem seems correct...[/quote]
Oh, perpendicular to the bicsector of $ \angle A$ :)

For the second case it's easy to show that $ QMP$ is isoceles($ M$ the midpoint of $ BC$) and to prove that $ PM\parallel{}AC,QM\parallel{}AB$ by this we have $ QR \equal{} RP$ and $ RM$ is perpendicular to $ AD$(the angle bisector $ D\in {BC}$).the result directly follows.

For the first one: u can prove that $ H$( the orthocenter of $ \Delta ABC$ ), $ R$ ,lay on the perpendicular bisector of $ QP$.If $ AF$ is the symmedian of $ A$ then by doing SOME computations( :D ) u can find the ratio $ \frac {\sin \angle A'AF}{\sin \angle FAD}$(where $ A'$ is the projection of $ A$ on $ BC$.).after this find the ratio
$ \frac {\sin \angle HAR}{\sin \angle RAD}$(both of these two ratios can be expressed in terms of sinus and cos of the angles($ A,C,B$).prove that this two ratios r equal.the problem is solved, then! :D


Nope. To find the inverse of the function
$y=\frac{3}{5}-\frac{3}{2}\sqrt{1+x^{3}}$,  you have to solve for $x$. Luckily, $x$ appears only once, which makes it possible to solve this equation by moving the terms to the left one by one:
$y-\frac{3}{5}=-\frac{3}{2}\sqrt{1+x^{3}}$,
$-\frac{2}{3}y+\frac{2}{5}={\sqrt{1+x^{3}}}$,
$\left(-\frac{2}{3}y+\frac{2}{5}\right)^{2}={1+x^{3}}$,
etc.

If by "inverse" you meant the reciprocal (as in $1/y$), then please be advised that $1/(a+b)$ is not the same as $1/a+1/b$. Even engineers should know that. :)


I know how to create a basic numbered list using:

[code]\begin{enumerate}
\item
\item
\end{enumerate}[/code]

How would I begin with the number 10 for instance?  I mean, how would I create a list like this?:

10. [info here]

11. [more info here]

...

Put \setcounter{enumi}{10} after \begin{enumerate} but before your first \item.

enumi is a "counter" which LaTeX uses to keep track of item numbers in lists.


plz help me to count this without using a calculator:
tan 10 + tan 70 - tan 50

i know the answer is square root 3 . but how to prove it? :maybe:

Here's a start...I think...

$ \tan 10 \plus{} \tan 70$

$ \equal{} \frac {\sin 10\cos 70 \plus{} \sin 70\cos 10}{\cos 10\cos 70}$

$ \equal{} \csc 20$

thanks but still not the answer. the answer should be tan 60.

We can find $ S$ by some way. Do you want I post it?

yes that's true. But the problem is i get the answer by using calculator. so we should evaluate the problems by not using tan 60.
am i true?

Well, I'm not sure I understood your last post, but this outline of the solution doesn't need a calculator at all. It's just using addition formulae for trig functions and the known value of $ \tan \frac {\pi}{3}$. If you feel the need to show that $ \tan \frac {\pi}{3} \equal{} \sqrt {3}$, take an equilateral triangle,, use Pythagoras to find the length of the height, divide it with half of the side, and you have your tangent (see why?).

sorry to confuse you.
what i meant is:
the question is:
tan 10 + tan 70  - tan 50 = ?

all angle in degrees. :o

ooo i know the way now.
just add tan 60 and then minus tan60

tan 10 + tan 70 - tan 50 + tan 60 - tan 60

yes u r rite.
 i'm sorry :roll:

[quote="hsiljak"]It equals $ \tan 60^0$, as [b]thanhnam2902[/b] already shown. And as you can use value $ \tan 60^0 \equal{} \sqrt 3$ as a known fact, you have your solution. If you ask could you get the answer without going on $ \tan 60^0e$, the answer is probably not. Every other approach would use the fact $ \tan 60^0 \equal{} \sqrt 3$, although it could be camouflaged.[/quote]But one thing you all need to realize is that [b]thanhnam2902[/b]'s proof assumes that $ S \equal{} \tan10^\circ \plus{} \tan70^\circ \minus{} \tan50^\circ$ evaluates to $ \tan60^\circ \equal{} \sqrt3$. Here is a solution I found by picking up where Carbon57 left off:

(skipping the degree notation)

$ S \equal{} \frac1{\sin20} \minus{} \frac {\sin50}{\cos50}$, $ \tan\theta \equal{} \frac {\sin\theta}{\cos\theta}$

$ \equal{} \frac1{\sin20} \minus{} \frac {\cos40}{\sin40}$, $ \sin\theta \equal{} \cos(90 \minus{} \theta)$

$ \equal{} \frac {\sin40 \minus{} \sin20\cos40}{\sin20\sin40}$

$ \equal{} \frac {\sin20(2\cos20 \minus{} \cos40)}{\sin20\sin40}$, $ \sin2\theta \equal{} 2\sin\theta\cos\theta$

$ \equal{} \frac {2\sin70 \minus{} \cos40}{\sin40}$

$ \equal{} \frac {2\sin30\cos40 \plus{} 2\cos30\sin40 \minus{} \cos40}{\sin40}$, $ \sin(\alpha\plus{}\beta)\equal{}\sin\alpha\cos\beta\plus{}\cos\alpha\sin\beta$, $ \sin30 \equal{} \frac12$

$ \equal{} 2\cos30 \equal{} \sqrt3 \equal{} \tan60$

that's what we're looking for.
it's a cool solution.
Thank you so much!!! :lol:
Please help me again with:

tan 9 - tan 27 - tan 63 + tan 81
all in degrees
 :roll:


geometry
rectangle
vector
inequalities
triangle inequality

[quote="cancer"]Let $A,B,C,D$ be four points on a circle such that $AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}$. What can we tell about quadrilateral $ABCD$?[/quote]

[hide="I am not sure... :maybe: "]The quadrilateral is cyclic and orthogonal.[/hide]
Perhaps someone could disprove or strengthen my statements...

[quote="diophantient"][quote="cosinator"][quote="spiral191"]no, it just tells you the quadrilateral is a parrellogram...[/quote]
and since its in a circle, it must be a rectangle (or a square).[/quote]
We know that the converse of that is true, but what about a quad with sides 3/4/6/8 and diagonals 5 and 10?
It isn't a parallelogram.[/quote]
Yes, that would be a problem, [i]if[/i] the quadrilateral existed... (it doesn't)
[hide="hint"]
Assign vectors to points $A,B,C,D$ and factor the difference $AB^{2}+BC^{2}+CD^{2}+DA^{2}-AC^{2}-BD^{2}$.
[/hide]

[quote="scorpius119"][/quote][quote="diophantient"][quote="cosinator"][quote="spiral191"]no, it just tells you the quadrilateral is a parrellogram...[/quote]
and since its in a circle, it must be a rectangle (or a square).[/quote]
We know that the converse of that is true, but what about a quad with sides 3/4/6/8 and diagonals 5 and 10?
It isn't a parallelogram.[/quote][quote="scorpius119"]
Yes, that would be a problem, [i]if[/i] the quadrilateral existed... (it doesn't)
[hide="hint"]
Assign vectors to points $A,B,C,D$ and factor the difference $AB^{2}+BC^{2}+CD^{2}+DA^{2}-AC^{2}-BD^{2}$.
[/hide][/quote]
Wait, what's wrong with that (sorry for being slow)?

[hide]Let the quadrilateral be $ABCD$, and $AC=10$.  By the triangle inequality, $AB+BC\ge AC$ and $AD+DC\ge AC$.  Since $AB+BC+CD+DA=21$, either $AB+BC=10$ and $AD+DC=11$ or vice-versa.  WLOG assume the scenario given.  Then $B$ is on $\overline{AC}$, so $ABCD$ would be really considered a triangle, not a quadrilateral.  Contradiction.[/hide]

[quote="diophantient"][quote="scorpius119"][quote="diophantient"]a quad with sides 3/4/6/8 and diagonals 5 and 10?[/quote]
Yes, that would be a problem, [i]if[/i] the quadrilateral existed... (it doesn't)[/quote]
Wait, what's wrong with that (sorry for being slow)?[/quote]
Lingomaniac only gets half-credit: in addition to being degenerate, it's over-determined.  A 3-4-6-8 quadrilateral with a diagonal of length 10 must already be degenerate, a 3-8-10 triangle with a fourth "vertex" on the long side, at distance 4 from one vertex.  This determines the length of the other "diagonal" (now a cevian of the triangle) in one of two different ways (we could also have the sides in order 3-6-4-8), giving (e.g., by Stewart's theorem) possible values of $\sqrt 7$ and $\sqrt\frac{218}5$ for the other diagonal.


vector
trigonometry
geometry unsolved
geometry

Prove that the vector $ AM$ , where $ M$ is midpoint of $ BC$ is also hight of the isosceles triangle $ ABC$, where $ AB\equal{}BC$.

For every $ X$, treat $ X$ as the vector from $ A$ to $ X$. Then $ M\equal{}\frac{B\plus{}C}{2}$. So $ (B\minus{}C)\cdot M \equal{} \frac{1}{2}(B\minus{}C)\cdot (B\plus{}C)\equal{}\frac{1}{2}(B\cdot B\minus{}C\cdot C)\equal{}\frac{1}{2}(|B|^2\minus{}|C|^2)\equal{}0$, implying that the vectors $ B\minus{}C$ and $ M$ are orthogonal, i.e. $ BC\perp AM$.

Is this solution correct???

The sum of vectors MB and AM is the vector AB, while the sum of vectors MC and AM is the vector AC. We know that MC=MB,AB=AC.  so from the cosine theorem we obtain that cos CMA= cos BMA and since the sum of the angles CMA and BMA is 180 then both of them are right angles.

can anybody help me and say if my solution is correct or not?? :blush:


inequalities
function
logarithms
calculus
calculus computations

[quote="kunny"]Let $x,\ y$ be real numbers such that $x\geq 1,\ y\geq 1.$
Prove that $\ln (x+e^{y}+1)<\ln x+\frac{1}{x}+y.$[/quote]

Since $e^{z}$ is a strictly increasing function of $z$ applying the exponential function to both sides of the given inequality we have that

\[\ln (x+e^{y}+1)<\ln x+\frac{1}{x}+y\]
\[\Rightarrow x+e^{y}+1<e^{\ln x+\frac{1}{x}+y}= xe^{\frac{1}{x}}e^{y}\]
\[\Rightarrow 1+\frac{1}{x}e^{y}+\frac{1}{x}<e^{\frac{1}{x}}e^{y}\]

since $e^{z}=\sum_{k=0}^{\infty}\frac{z^{k}}{k!}$ it follows that $e^{z}>\sum_{k=0}^{n}\frac{z^{k}}{k!}$ for $z>0$ and $n\in\mathbb{N}$ and hence

\[e^{\frac{1}{x}}e^{y}> \left( 1+\frac{1}{x}+\frac{1}{2x^{2}}\right)e^{y}= \frac{1}{x}e^{y}+\left(1+\frac{1}{2x^{2}}\right)e^{y}\]
\[>\frac{1}{x}e^{y}+\left( 1+\frac{1}{2x^{2}}\right)\left( 1+y\right) =1+\frac{1}{x}e^{y}+\frac{y+1}{2x^{2}}+y\]

so it remains to show that
 \[1+\frac{1}{x}e^{y}+\frac{y+1}{2x^{2}}+y>1+\frac{1}{x}e^{y}+\frac{1}{x}\]
\[\Leftrightarrow \frac{y+1}{2x^{2}}+y>\frac{1}{x}\]
\[\Rightarrow 2x^{2}y+y+1>2x\]
\[\Rightarrow 2x(xy-1)+y+1>0\]

the product of $x\geq 1,\ y\geq 1$ gives $xy-1\geq 0$ multiplying this by $2x\geq 2$ gives $2x(xy-1)\geq 0$ and now adding $y\geq 1$ gives further that $2x(xy-1)+y\geq 1\Rightarrow 2x(xy-1)+y+1\geq 2>0$ which was to be shown.


Chinese Western Mathematical Olympiad 2005 was held on Nov 5-6, 2005 in Chendu, Sichuan.
Here are the problems.

First Day (Nov 5th, 2005, 8AM-12AM)

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=63023

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=63026

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=63029

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=63024

Second Day (Nov 6th, 2005, 8AM-12AM)

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=63027

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=63028

7. http://www.mathlinks.ro/Forum/viewtopic.php?t=59834

8. http://www.mathlinks.ro/Forum/viewtopic.php?t=63025

Here is the PDF file for the problem: [url=http://www.mathlinks.ro/Forum/resources/files/china/63/China-Western_Mathematical_Olympiad-2005-63-37.pdf]click here to download it[/url].

Correction:

I look through the official paper of the problem 5 and find that there are some mistakes in it. Here is the correction:

Circles $C(O_1)$ and $C(O_2)$ intersect at points $A$, $B$. $CD$ passing through point $O_1$ intersects $C(O_1)$ at point $D$ and tangents $C(O_2)$ at point $C$. $AC$ tangents $C(O_1)$ at $A$. Draw $AE \bot CD$, and $AE$ intersects $C(O_1)$ at $E$. Draw $AF \bot DE$, and $AF$ intersects $DE$ at $F$. Prove that $BD$ bisects $AF$.

hello!!!
how about CWMO2006?
where  can I find it?
I'll be very happy, if you post or send me them!!!
thanks a lot!

Please don't post soooooo many times. (Although i'm doin the same)

Find Carefully!


trang WEB về hóa học duy nhất mà tôi biết là hoahocvietnam.
Có ai rành về các trang WEB hóa học của nước ngoài giới thiệu cho tôi với,ở đó cho tải sách thì càng tốt,cảm ơn trước...

Mình cũng không được rành cho lắm, nhưng mình có biết về một Open Couse Ware của trưòng MIT.
Bạn có thể vào đó để xem. Ở đó cho down tài liệu thoải mái.
[url=http://ocw.mit.edu/OcwWeb/Chemical-Engineering/index.htm]MIT Open Couse Ware[/url]

cám ơn nhiều,linkgreencold cũng mới biết một cái này [url=http://www.knovel.com/]Knovel[/url],phải có PRIVATE EMAIL mới tải được,FREE trong 7 ngày...


For those who took the AIME 1, enter in your index. There has been so much discussed about AIME scores and USAMO floors, but not too much about the actual USAMO cutoff index.

Edit: Hmm, sorry, I guess I should have made an option "150 and under", but I can't change the poll now. Oh well.

I haven't got mine back yet but I know the best I could've gotten is a 152.5 :lol: (unless they considered all the doodles and the political messages to be correct answers lol)

You should also edit it to include values such as 239.5 (my index; I put it in the 230-239 option).

Hmm...6 people got 250+ indices??? Who here got those?

argh...smart people!!

i just want a 220+...lol :-)


For Thanksgiving dinner, Mrs. Zappone prepared a huge turkey and had potatoes, carrots, green beans, cranberry salad and dinner rolls. Of course, the problem is that it's hard to fit all the food you want onto your plate-let alone into your stomach! Grandpa Curt takes a little bit of ecerything while Grady sticks to turkey and dinner rolls, How many distinct combinations of food items are possilble at the Zappone family feast, assuming everyone takes some turkey?

AGAIN

MATHCOUNT PEOPLE , try not to answer it

[quote]Either I can't read, or it doesn't say how many people there are o_O .[/quote]

You don't have to know the number of people attending the party...

if you wish to know, it's 8

but it's unneccesary

[quote="shinwoo"]
AGAIN

MATHCOUNT PEOPLE , try not to answer it[/quote]

what if u r in mathcount, but is not very good, and this is your first year, and you don't know if you are even gonna make it to state? can you answer? hehe6

assuming that I can answer this.... :
[hide]if you get some of everything, there's 1 combination, if you get 4 out of 5, there's 5 combinations, and if you get 3 out of 5, there's 10 combinations. 2 out of 5, 10 combinations, and 1 out of everything, there's 5 combinations. and so I think the answer would be 1 + 5 + 10 + 10 + 5 = 31.... [/hide]

but there's the possiblity that I'm not understanding the problem correctly, if that's the case. i"m sorry.

Oh. So most of the information is irrelevant? If so, I misread the problem the first time.

[quote="LuCy4EvA"][quote="KBabe"][hide]7! =5040. is that right?[/hide][/quote]

how did u get 7!? where did the 7 come from? I counted 5 kinds of food[/quote]

"a huge turkey and had potatoes, carrots, green beans, cranberry salad and dinner rolls. "

o oops, it's 6 kinds of food....right?

both, KBabe and LuCy4EvA are wrong

I guess LuCy4EvA,you can answer the question

well,

both of you guys are wrong...

there is a really hard way to solve it, and really easy way to do it...

really hard way,

you have to learn about some stuff that I won't tell until anyone solve this problem...

the other way,

try all the combitnations( it will take you awhile ) :D

hmmm...
so six kinds of food.. you can have on your plate  1; 2; 3; 4; 5; 6 kinds of food on ur plate? right? soo..

6 * 6 * 6 * 6 * 6 * 6 = 46656

soo.. 46656 kinds of combinations of foods....

hmm....did i get it?

I have a question, do you HAVE to have 6 kinds of food on your plate. and wouldn't you use combination since order doesn't matter?

Okay....

since we are supposed to have a turkey, it would actually be 6 sides..

so,

the maximum combinations would be formed by 6 sides..

you forgot the last combinations...

with no sides, just turkey..

that would be counted as 1....

so you were close LuCy4EvA

and KBabe, you're wrong again...
( I'm not being mean )

This whole thing might be resolved if a new clearer problem statement where posted.

For Thanksgiving dinner, Mrs. Zappone prepared:

[b]turkey[/b], [b]potatoes[/b], [b]carrots[/b], [b]green beans[/b], [b]cranberry salad[/b] and [b]dinner rolls[/b].

[size=75]
Grandpa Curt takes a little bit of everything

Grady sticks to turkey and dinner rolls
[/size]
How many distinct combinations of food items are possilble at the Zappone family feast, assuming everyone takes some turkey?

WAIT, DOES THE PART ABOUT GRANDMA AND GRANDPA AFFECTS THE PROBLEM? I'M JUST WONDERING :)

It doesn't. Shinwoo just put that in to screw us all  :P .

WHEN I FIRST SAW THE PROBLEM, I THOUGHT THAT THE ANSWER WOULD BE X-2, WHEN X IS 32..... I DONT KNOW WHY I THOUGHT OF THAT, BUT I THOUGHT YOU HAVE TO SUBTRACT GRANDPA AND GRANDMA

wouldnt the answer be 720.
probably its wrong i m too young to get it rite.
LOL

no one is EVER too young to answer a problem..... but you should probably try and say how you got 720.... :D  :D  :D  :D

i still don't get how you can solve this problem by using combination....

wouldnt it be something like
7C1/7! coz there is 7 foods and u are sure that everyone takes turkey.
and ya misregard the 720 i made a mistake in calculating

Lucy, it's ok not to know everything in the world..

you'll learn those eventually later on..

and, the reason I posted on the Basics section is because I wanted to know how my age people would solve this problem...
 :D

i no how to do permutations and combinations but i still dont get how u got 5! and 2!.
Where did u guys get it from? Can someone explain to me PLZ! lol i m going dumb

and ya i m the same person as mathnerd89. that was my bro's SN so i was using it. that mite be a good clue that tells u that i m STUPID

yea, same here.... I don't get where the numbers 2 and 5 came from.... haha, feeling stupid because this topic has been going on for 4 pages and I still don't get it.... can someone please explain  :?  :?  :?  :?  and to shinwoo, I know I don't have to know everything, but I hate it when I see a problem and never figure out how to do it..... or you can say i like challenges....  :D  so please tell me where you got the 5 and 2 :D  :D  :D


inequalities quadratics algebra function

Let $ a,b,c\geq 0$ such that $ a\plus{}b\plus{}c\equal{}1$. Prove that:

$ a^2\plus{}b^2\plus{}c^2\geq 4(ab\plus{}bc\plus{}ca)\minus{}1$

Well,

$ (a\plus{}b\plus{}c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab\plus{}bc\plus{}ca) \equal{} 1 \rightarrow
2(ab\plus{}bc\plus{}ca) \equal{} 1 \minus{} (a^2\plus{}b^2\plus{}c^2) \rightarrow
4(ab\plus{}bc\plus{}ca) \equal{} 2 \minus{} 2(a^2\plus{}b^2\plus{}c^2)$

So we have to prove

$ a^2\plus{}b^2\plus{}c^2 \geq 2 \minus{} 2(a^2\plus{}b^2\plus{}c^2) \minus{} 1 \rightarrow
3(a^2\plus{}b^2\plus{}c^2) \geq 1 \rightarrow
a^2\plus{}b^2\plus{}c^2 \geq \frac{1}{3}$

But for QM-AM inequality we have

$ \sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{3}}\geq \frac{a\plus{}b\plus{}c}{3} \rightarrow
\frac{a^2\plus{}b^2\plus{}c^2}{3} \geq \left(\frac{1}{3}\right)^2 \rightarrow
{a^2\plus{}b^2\plus{}c^2} \geq \frac{1}{3}$ as wanted.  :)

[hide="Solution"][color=darkblue]We can write it so that all the terms are quadratic by using $ (a\plus{}b\plus{}c)^2\equal{}1$:

$ a^2 \plus{} b^2 \plus{} c^2 \geq 4(ab \plus{} bc \plus{} ca) \minus{} (a \plus{} b \plus{} c)^2$

Expand and rearrange: $ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \minus{} 2ab \minus{} 2bc \minus{} 2ca \geq 0$

Factor left side: $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \geq 0$[/color][/hide]

[b]Edit:[/b] Fixed the typo pointed out by [b]mathwizarddude[/b]. Almost five years later :)

[quote="DreamTeam"]Let $ a,b,c\geq 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that:

$ a^2 \plus{} b^2 \plus{} c^2\geq 4(ab \plus{} bc \plus{} ca) \minus{} 1$[/quote]

$ a^2 \plus{} b^2 \plus{} c^2\geq 4(ab \plus{} bc \plus{} ca) \minus{} 1$

$ \leftrightarrow (a\plus{}b\plus{}c)^2 \minus{}2(ab\plus{}bc\plus{}ca) \ge\ 4(ab \plus{} bc \plus{} ca) \minus{} 1$

$ \leftrightarrow 1 \ge\ 3(ab\plus{}bc\plus{}ca)$ which is obvious  :wink:

We have done!  :)

It's easy to see that: $ a^2\plus{}b^2\plus{}c^2\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\geq 9abc$([color=blue]AM-GM inequality[/color]). We can prove a stronger inequality:

\[ 9abc\geq 4(ab\plus{}bc\plus{}ca)\minus{}1.\]

After normalization the inequality becomes:
\[ \frac{9abc}{a\plus{}b\plus{}c}\geq 4(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2\]
\[ \Longleftrightarrow (a\plus{}b\plus{}c)^3\plus{}9abc\geq 4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\]
\[ \Longleftrightarrow a^3\plus{}b^3\plus{}c^3\plus{}3abc\geq ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)\]
\[ \Longleftrightarrow\displaystyle\sum_{cyc}a(a\minus{}b)(a\minus{}c)\geq 0\]

which is [color=blue]Schur's Inequality[/color] for $ \boxed{r\equal{}1}.$

By Power Mean Inequality: $ \left(\frac{a^2\plus{}b^2\plus{}c^2}3\right)^{\frac12}\ge\frac{a\plus{}b\plus{}c}3\equal{}\frac13$, and hence $ 3(a^2\plus{}b^2\plus{}c^2)\ge1$. So $ a^2\plus{}b^2\plus{}c^2\ge 2\minus{}2(a^2\plus{}b^2\plus{}c^2)\minus{}1\equal{}2(1\minus{}a^2\plus{}b^2\plus{}c^2)\minus{}1\equal{}2((a\plus{}b\plus{}c)^2\minus{}a^2\plus{}b^2\plus{}c^2)\minus{}1\equal{}2(2ab\plus{}2bc\plus{}2ca)\minus{}1\equal{}4(ab\plus{}bc\plus{}ca)\minus{}1$. Q.E.D.

Another way(Using Mixing varialble)
Let $ f\left( {a,b,c} \right) \equal{} a^2  \plus{} b^2  \plus{} c^2  \minus{} 4(ab \plus{} bc \plus{} ac) \plus{} 1$
so $ f\left( {a,\frac{{b \plus{} c}}{2},\frac{{b \plus{} c}}{2}} \right) \equal{} a^2  \plus{} (\frac{{b \plus{} c}}{2})^2  \plus{} (\frac{{b \plus{} c}}{2})^2  \minus{} 4(ab \plus{} (\frac{{b \plus{} c}}{2})^2  \plus{} ac) \plus{} 1$
And $ f(a,b,c) \minus{} f\left( {a,\frac{{b \plus{} c}}{2},\frac{{b \plus{} c}}{2}} \right) \ge 0$
So we just have to prove that $ f\left( {a,\frac{{b \plus{} c}}{2},\frac{{b \plus{} c}}{2}} \right) \ge 0$ which is obvious.

$ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$
by rearrangement inequality/AM-GM/Anything really
so
$ 2(a^2\plus{}b^2\plus{}c^2) \plus{} 2(ab\plus{}bc\plus{}ca)\geq 4(ab\plus{}bc\plus{}ca)$
$ a^2\plus{}b^2\plus{}c^2 \plus{} (a\plus{}b\plus{}c)^2 \equal{} a^2\plus{}b^2\plus{}c^2 \plus{}1 \geq 4(ab\plus{}bc\plus{}ca)$
$ a^2\plus{}b^2\plus{}c^2 \geq 4(ab\plus{}bc\plus{}ca)\minus{}1$

[quote="TZF"][color=darkblue]
Expand and rearrange: $ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \minus{} 2a \minus{} 2b \minus{} 2c \geq 0$[/color][/quote]
I think it should have been $ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \minus{} 2ab \minus{} 2bc \minus{} 2ac$, but anyways, nice solution.

let $ a^2\plus{}b^2\plus{}c^2 \geq 4(ab\plus{}bc\plus{}ca) \minus{} 1$ be denoted by (P).
Then, $ a\plus{}b\plus{}c\equal{}1 \Rightarrow (a\plus{}b\plus{}c)^2\equal{}1;$ hence:
    $ (P) \Leftrightarrow a^2\plus{}b^2\plus{}c^2 \geq 4(ab\plus{}bc\plus{}ca) \minus{} a^2 \minus{} b^2 \minus{} c^2 \minus{} 2(ab\plus{}bc\plus{}ca)$
          $ \Leftrightarrow a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ca$, which is true by AM-GM. Hence the result. :)

[quote="DreamTeam"]Let $ a,b,c\geq 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that:

$ a^2 \plus{} b^2 \plus{} c^2\geq 4(ab \plus{} bc \plus{} ca) \minus{} 1$[/quote]
[hide="This may be an alternative sol."]
AM_GM
$ a^2\plus{}b^2\ge 2ab ,b^2\plus{}c^2\ge 2bc, b^2\plus{}c^2\ge 2bc$
$ \implies  a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$
$ \implies  2(a^2\plus{}b^2\plus{}c^2)\ge 2ab\plus{}2bc\plus{}2ca$
$ \implies  a^2\plus{}b^2\plus{}c^2\plus{}(a^2\plus{}b^2\plus{}c^2\plus{} 2ab\plus{}2bc\plus{}2ca)\ge 4ab\plus{}4bc\plus{}4ca$
$ \implies  a^2\plus{}b^2\plus{}c^2\plus{}(a\plus{}b\plus{}c)^2\ge 4ab\plus{}4bc\plus{}4ca$
$ \implies  a^2\plus{}b^2\plus{}c^2\plus{}1\ge 4ab\plus{}4bc\plus{}4ca$
$ \implies  a^2\plus{}b^2\plus{}c^2\ge 4ab\plus{}4bc\plus{}4ca\minus{}1$
 :roll:[/hide]

[quote="DreamTeam"]Let $ a,b,c\geq 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that:

$ a^2 \plus{} b^2 \plus{} c^2\geq 4(ab \plus{} bc \plus{} ca) \minus{} 1$[/quote]

[hide]
With the method of Lagrange Multipliers, we see

$ \lambda\equal{}2a\minus{}4b\minus{}4c\equal{}\equal{}2c\minus{}4a\minus{}4b\equal{}2b\minus{}4a\minus{}4c$

Looking at the first two:

$ 6a\equal{}6c \implies a\equal{}c$

With the others we see $ b\equal{}c$ and $ a\equal{}b$

So the critical point is at $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$

We also have to check boundary points (when one of a,b,c=0)

WLOG let a=0

$ b\plus{}c\equal{}1$

We must show $ b^2\plus{}c^2 \geq 4bc\minus{}1$

Note that $ \frac{1}{2}\equal{}\frac{b\plus{}c}{2} \geq \sqrt{bc} implies bc \leq \frac14$

Thus the maximum the RHS can possibly be is 0. Clearly, $ b^2\plus{}c^2 \geq 0 \geq 4bc\minus{}1$

As the domain is compact, the relative max can only occur at a critical point or boundary point. However, at the critical point, equality is clearly achieved, so the inequality is always satisfied/[/hide]

[quote="Mewto55555"][quote="DreamTeam"]Let $ a,b,c\geq 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that:

$ a^2 \plus{} b^2 \plus{} c^2\geq 4(ab \plus{} bc \plus{} ca) \minus{} 1$[/quote]

[hide]
With the method of Lagrange Multipliers, we see

$ \lambda \equal{} 2a \minus{} 4b \minus{} 4c \equal{} \equal{} 2c \minus{} 4a \minus{} 4b \equal{} 2b \minus{} 4a \minus{} 4c$

Looking at the first two:

$ 6a \equal{} 6c \implies a \equal{} c$

With the others we see $ b \equal{} c$ and $ a \equal{} b$

So the critical point is at $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$

We also have to check boundary points (when one of a,b,c=0)

WLOG let a=0

$ b \plus{} c \equal{} 1$

We must show $ b^2 \plus{} c^2 \geq 4bc \minus{} 1$

Note that $ \frac {1}{2} \equal{} \frac {b \plus{} c}{2} \geq \sqrt {bc} implies bc \leq \frac14$

Thus the maximum the RHS can possibly be is 0. Clearly, $ b^2 \plus{} c^2 \geq 0 \geq 4bc \minus{} 1$

As the domain is compact, the relative max can only occur at a critical point or boundary point. However, at the critical point, equality is clearly achieved, so the inequality is always satisfied/[/hide][/quote]
dont u think this is overkill metwo5555? i mean it is very simple then why go for a method like langrange multipliers..

Well since we have the constraint of a+b+c=1, then this just means all the partials are equal, and its really easy to find the critical points.

Since $(a+b+c)^2=1$,
$a^2+b^2+c^2 \geq 4(ab+bc+ca)-1
\Leftrightarrow
a^2+b^2+c^2 \geq ab+bc+ca$

This is just Rearrangement inequality :)


Knowing that $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$,prove the ineq:
$ (1\plus{}a)(1\plus{}b)(1\plus{}c) \geq (1\minus{}a^2)^2\plus{}(1\minus{}b^2)^2\plus{}(1\minus{}c^2)^2$

[quote="SUPERMAN2"]Knowing that $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$,prove the ineq:
$ (1 \plus{} a)(1 \plus{} b)(1 \plus{} c) \geq (1 \minus{} a^2)^2 \plus{} (1 \minus{} b^2)^2 \plus{} (1 \minus{} c^2)^2(*)$[/quote]

We have :

$ (*) \Leftrightarrow ab\plus{}bc\plus{}ca \geq 2(ab\plus{}bc\plus{}ca)^2\plus{}3abc$

$ \Leftrightarrow ( \frac{ab\plus{}bc\plus{}ca}{3}\minus{}3abc)\plus{}2(ab\plus{}bc\plus{}ca)(\frac{1}{3}\minus{}ab\minus{}bc\minus{}ca) \geq 0$ ( true )

Done ! :)


what is meant by polynomials sign variation ? what does sign variation mean ? different signs ?
what is the sign variation in this poly...   $p(x)=x^{2}-3x+2$?

Signs variations, by the Descartes Rules of Signs, gives boundaries to possible roots of the polynomial.

thanks , but i know that , i dont get what variation sign is, i mean how to find how many variation of sign a polynomial has, do you count the number of + and - before the coeffiecients? or ? what is the variation of signs in this $P(x)=x^{2}-3x+2$

Think of the coefficients of the polynomial.
For this one, we have
+1, -3, +2
Look for instances where it changes over...
So in this one, we go from
+ to -

at

1 --> 3

and

- to +

at 3 --> 2

That's the end of the polynomial, so it has two variations in sign.


I just finished taking biology Honors (not AP), should I take the SAT II biology?  What are the SATII tests for ?  Should I take one everytime I finish a course?

Do you have to win states at sports or music or something to get into really good colleges, say for ex. MIT?  My grades and stuff are good, but I'm not extrodinarly good at something.  I'm participating in the math club, speech and debate, chinese club.  I've just started playing pingpong.  but I haven't won any awards or anything.  I'm going to be a sophomore this fall.

thanks

SAT II test scores are required as part of the application process at some elite colleges. The largest number of required SAT II tests I have heard of is three. At most colleges, you choose which SAT II scores to submit, but a few colleges specify one or two of the SAT II tests you should take, and and some others specify that they should come from broad categories, for example one science SAT II, one math SAT II, and one SAT II from among the other SAT II tests. You can find the current details about required or suggested SAT II tests for applicants on the admission pages of each college's Web site.

I do not think it is generally expected that you would already have done something as fancy as winning a state award by the end of your Freshman year (Tim Abott comes to mind as a rather extreme example). In my opinion, do the things you love to do, try your best, and everything else will fall in to place.

Think about it this way -- there are only 52 states to win a state award in...and I do not know of any college with a freshman class size of 52 students or smaller (okay, so there are multiple forms of competitions to accrue awards in, but I think you see my point).


In isosceles triangle $ ABC(AC\equal{}BC)$  the point $ M$ is in the segment $ AB$ such that $ AM\equal{}2MB ,F$ is the midpoint of $ BC$ and $ H$ is the orthogonal projection of $ M$ in $ AF.$ Prove that $ \angle BHF\equal{}\angle ABC$


Given an integer $a_1$($a_1 \neq -1$), find a real number sequence $\{ a_n \}$($a_i \neq 0, i=1,2,\cdots,5$) such that $x_1,x_2,\cdots,x_5$ and $y_1,y_2,\cdots,y_5$ satisfy $b_{i1}x_1+b_{i2}x_2+\cdots +b_{i5}x_5=2y_i$, $i=1,2,3,4,5$, then $x_1y_1+x_2y_2+\cdots+x_5y_5=0$, where $b_{ij}=\prod_{1 \leq k \leq i} (1+ja_k)$.


trigonometry
induction
inequalities proposed
inequalities

You are right I lost the condition $x_{i}>0 i=\overline{1,n}$ thus
Let $0<x_{1},...,x_{n}\in R$ such that $x_{1}+x_{2}+..+x_{n}=\pi$ prove that
$\sin x_{1}^{2}+\sin x_{2}^{2}+...+\sin x_{n}^{2}\le \frac{9}{4}$

[quote="N.T.TUAN"]If so, we only need prove for $n=3$. But then, it is assertion follow

"Let triangle $ABC$ , prove that $\sum\sin^{2}{A}\leq\frac{9}{4}$"[/quote]

I think for $n=3$ we have to prove $\sum\sin{A^{2}}\leq\frac{9}{4}$, not $\sum\sin^{2}{A}\leq\frac{9}{4}$.

ok it must be
Let $0<x_{1},...,x_{n}\in R$ such that $x_{1}+x_{2}+..+x_{n}=\pi$ prove that
$\sin^{2}x_{1}+\sin^{2}x_{2}+...+\sin^{2}x_{n}\le \frac{9}{4}$
But I don't understand your ideal
[quote="N.T.TUAN"]If so, we only need prove for $n=3$. But then, it is assertion follow
"Let triangle $ABC$ , prove that $\sum\sin^{2}{A}\leq\frac{9}{4}$"[/quote]
How do you do?

[quote="N.T.TUAN"][quote="skywalker"]Why do you think?
$\sin^{2}x_{n-1}+\sin^{2}x_{n}\le \sin^{2}(x_{n-1}+x_{n})$
I belive it is wrong![/quote]
WLOG choose $x_{n-1},x_{n}$ agree![/quote]

I don't understand anything you said, unless you can reply decently, you shouldn't reply!

[quote="N.T.TUAN"]If $x_{n-1}+x_{n}\leq \frac{\pi}{2}$ then
$\sin^{2}x_{n-1}+\sin^{2}x_{n}\le \sin^{2}(x_{n-1}+x_{n})$

Now, do you understand?[/quote]
I don't understand, why  $x_{n-1}+x_{n}\leq \frac{\pi}{2}$ while $x_{1}+..+x_{n}=\pi$ and moreover
why $x_{n-1}+x_{n}\leq \frac{\pi}{2}$ then $\sin^{2}x_{n-1}+\sin^{2}x_{n}\le \sin^{2}(x_{n-1}+x_{n})$
You only know to say, you don't explain anything.

[quote="N.T.TUAN"][quote="skywalker"]
I don't understand, why  $x_{n-1}+x_{n}\leq \frac{\pi}{2}$ while $x_{1}+..+x_{n}=\pi$ [/quote]
Because $(x_{1}+x_{2})+(x_{n-1}+x_{n})\leq \pi(n>3)$ therefore $x_{1}+x_{2}$ or $x_{n-1}+x_{n}$ must $\leq\frac{\pi}{2}$.

[quote]
why $x_{n-1}+x_{n}\leq \frac{\pi}{2}$ then $\sin^{2}x_{n-1}+\sin^{2}x_{n}\le \sin^{2}(x_{n-1}+x_{n})$[/quote]
In triangle $ABC$, if $C\geq\frac{\pi}{2}$ then $c^{2}\geq a^{2}+b^{2}$ by Cosin law, also $\sin^{2}{C}\geq \sin^{2}{A}+\sin^{2}{B}$ by sin law. Now, note that $x_{n-1},x_{n},\pi-(x_{n-1}+x_{n})$ are angles of a triangle and $\pi-({x_{n-1}+x_{n}})\geq\frac{\pi}{2}$.
[quote]
You only know to say, you don't explain anything.[/quote]
Sorry, Now,  do you understand?[/quote]
ok all are right I understood, but you won't be able to do with inductive
[quote="N.T.TUAN"]You can use induction on $n$.
When does we have $\sin^{2}{x_{1}}+...+\sin^{2}{x_{n}}\leq \sin^{2}{x_{1}}+...+\sin^{2}{x_{n-2}}+\sin^{2}{(x_{n-1}+x_{n})}$ ? That is all!.[/quote]
You had $x_{n-1}+x_{n}\leq \frac{\pi}{2}$ thus $x_{1}+...+x_{n-2}\ge \frac{\pi}{2}$ how do you do?

Assume assertion is true with $n=1,2,...,k(k\geq 3)$. Now for $n=k+1$ , WLOG assume $x_{k}+x_{k+1}\leq\frac{\pi}{2}$, then $\sum_{i=1}^{k+1}\sin^{2}{x_{i}}\leq \sin^{2}{x_{1}}+...+\sin^{2}{x_{k-1}}+\sin^{2}{(x_{k}+x_{k+1})}\leq\frac{9}{4}$ by hypothesis induction.

ok that's right good solution! My solution is base on contruction the polygon but maybe it will be longer than this solution, I won't post it, thanks for the solution! :)


Lemon Ice sold 3.9 million CD's at $\$14.00 each.  Lemon Ice received a royalty of 6% for each CD sold.  How many dollars did Lemon Ice receive?

i think it is a profit. For each CD Lemon Ice got 84 cents.

[hide da answer]
yea it's 3,276,000[/hide]


6-th Chinese Girls' Mathematical Olympiad was held on August 13-14, in Wuhan, China. Here are the problems:

Thanks a lot Zhaoli. If you have free time ,Can you please translate Chinese TST 2007 problems (not solutions)to everybody?

Thanks Zhaoly. Problems from 2007 Chinese Girls' Mathematical Olympiad are interesting :P


AMCDirector,

I would assume that you work for AMC (based on your user id?)

You mentioned one time, parents cannot register AMC test for their own kids. But can they register it and then find a free or paid third party (non-affiliate) proctor to administer the exam in a public building (similiar to home schoo kids)?  Of course the test material shall be mailed to proctor by AMC directly.

If the answer is yes, then it should help a lot of students who happen to not have a supportive school. Assuming that the parent is willing to pay all the cost.

If the answer is no, then, tell us why ?

I would point you to [url=http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-CU-list.shtml]this list[/url] of colleges that offer the B date, but the only Texas location listed is in Dallas.

Thanks, but I am not asking the question for myself. I heard people asking this kind of questions all the time from regions that no math programs at all. I have heard in some schools, students could not even find a teacher who was willing to do the proctoring ! I felt bad for those kids. I believe that parents should be able to register their kids just like home schools.


I wonder if it would be possible to enabled some HTML - in particular, the <SUP> and <SUB> tags. Then we won't have to deal with subscripts like a_2...

(TeX support would be too much to ask.  :P )

Tex support is futuristic at this point...lol...especially knowing how much stuff is involved...

Dreams beyond dreams, I say, because it's almost impossible. Again, look at the source code by right clicking anywhere in the forum and click on "View Source", then bet :)

I think superscripts and subscripts are now supported (add this AM).  The subscripts can get a little hard to read if they're the last line of the post.

Thanks so much! This'll make typing up math equations much easier.

Yay!

So is there any BBCode which makes the text go superscript and subscript? Copy and paste isn't working on my computer, it just shows up as a normal text. Any tips? Oh I get it now, thanks!

use < sup > and < / sup > (without the spaces in between) to raise a block of text to superscript; sub subscript.

You guys should really enable super and sub scripts. It's annoying typing ^ all the time. It gets very confusing after awhile.

They did.  You just have to uncheck "disable HTML in this post"

Yup, or change your setting for all your posts in your profile (Click on Yes at "Always allow HTML:")

by the way, why is the default set to "disable HTML" in the first place?

Default is set into all Nos mainly because they can get in way of normal posting
Examples: "Suppose you have the following:(2-x)(y-5)"
"Inversion:Diversion?"

I doubt it'll become a big problem but just in case I guess.

That's why I usually space my words/terms/signs...


trigonometry
continued fraction
complex numbers
algebra unsolved
algebra

Prove that the solution of the given equation is$x = 2cos \frac{\pi}{n+2}$
Equation is $\frac{1}{x} = x - \frac{1}{x - \frac{1}{x - \frac{1}{x - ... n times}}}$

here there are $n$  $'x's$ in RHS

[hide="only a little hint"]take the inverse of all the sides ($^{-1}$ to understand :P ) and note that you can stop everywhere on the continued fraction obteining another element of the serie: so try sum and subtraction formulas for sin and cosine after defining the serie; otherwise, you can also help with regular polygons and complex numbers or probabilities polynoms and fibonacci[/hide]

i am unable to solve the recurring series????
plz help


[quote="zaya_yc"]$ x,y,z\geq 0$ and $ xy \plus{} xz \plus{} yz \equal{} 1$  .

Find min:

$ x(1 \minus{} y^{2})(1 \minus{} z^{2}) \plus{} y(1 \minus{} z^{2})(1 \minus{} x^{2}) \plus{} z(1 \minus{} x^{2})(1 \minus{} y^{2})$[/quote]

we have

$ \sum {x\left( {1 \minus{} y^2  \minus{} z^2  \plus{} y^2 z^2 } \right)}  \equal{} \sum x  \plus{} xyz \minus{} \sum {xy\left( {x \plus{} y} \right)}  \equal{} 4xyz \ge 0$


function
limit
integration
real analysis
real analysis unsolved

$V:R^n\rightarrow R$ function $C^1$
(S) $x_k''(t)=f_k(x(t))$ for all $1\leq k \leq n$
where $f_k=-\frac{\partial V}{\partial x_k}$

Let $x(t)=(x_1(t),...,x_n(t))$ a solution of the system  (S) on the finite interval
$a<t<b$

1/Prove that $H(t)=\frac{1}{2}||x'(t)||^2+V(x(t))$ is constant on $]a,b[$

here $||z||^2=(z,z)$ scalar product

2/Assuming that $V(x)\geq M >-\infty$ for all $x\in R^n$
Prove that $x(t),x'(t),x\"(t)$ are bounded on $]a,b[$
and prove that $\lim_{t\rightarrow b}x(t),\lim_{t\rightarrow b}x'(t),\lim_{t\rightarrow b}x\"(t)$ exist

[quote="Moubinool"]$V:R^n\rightarrow R$ function $C^1$
(S) $''_k(t)=f_k(x(t))$ for all $1\leq k \leq n$
where $f_k=-\frac{\partial V}{\partial x_k}$
[/quote]
Is it $x''_k(t)=\cdots$?

1/ $H'(t)=(x''(t);x'(t))+\sum_{k=1}^{n}x_{k}'(t)\frac{\partial V}{\partial x_k }$ with the
statement you have $H'(t)=0$ so $H$ is constant

Try it for 2/


search function parameterization \/closed

I noticed that when members post, they have something on the
top of their 5 star thingies and their avatar. some say Riemann Hypothesis, and navier-strokes equations, and I'm new member.
What is this all about????????

These rankings directly correspond with the number of stars you have. In general, the more posts you have, the higher the ranking. As for the names, they're the Millenium Problems - how fitting for a site like AoPS to use those :P

What's the highest ranking?
Do you know anyone who has it?

The highest one is administrator  :wink:
And you could have used the search function to find a lot of topics about this.

Above the forum topics or posts is a search button. Click on it, and you'll see a new window which allows you to set search parameters and then search what you want.


geometry
perimeter
function
calculus
derivative
calculus computations

I need help with this problem, i don't think it is that hard but i just don't know how to do it.

Problem: How fast is the perimeter of an equilateral triangle increasing when the area is 36 sq.cm and increasing 3 sq. cm/sec.?

please help-thanks

don't you come all at once, you might flood the site, you will get your chance to get the question answered.
 :(

This is what's called a "related rates" problem.  The idea is that you have two things, related via some function, and so the change in one thing is related to the change in the other, via the derivative of said function.

So, to do this, write down the perimeter as a function of the area.  If you still don't see what to do then you should talk to your teacher because these sorts of exercises are given in beginning calculus courses and understanding them is fairly important.

I know it is a related rate, but I am having trouble finding a relationship between the perimeter and the area.  My teacher is not that helpful.  A=1/2bh and perimeter would equal to B+2 *(some other variable)  If I can find the relatioship then it would be easy to solve.  I was thinking mabe the equalateral triange can be divided into two right triangles and then the area would be cut into half.OMG i think i just found a way on how to do it!  Let me try this on paper and come back.

i think i had it but nope, somebody help me please....
 :)


1) Prove that for any integer k>1 and any positive integer n, nk is the sum of n consecutive odd numbers.

2) Let n be a positive integer and m be any integer with the same parity as n. Prove that the product mn is equal to the sum of n consecutive odd integers.

The first follows from the second one, so let's prove that:

$mn=n(n+m-n)=n(a+n-1)$, where $a=m-n+1$ is an odd number. This means that $mn=na+2\frac{(n-1)n}2=a+(a+2)+\ldots+a+2(n-1)$.

I think this should be in the "Number Theory" section if you want it in the "Advanced" section, but it should go in the "Intermediate" forum.


O=circuncenter
I=incenter
Xa= incircle touch with BC

we have $ r$ and we have $ OI^2 \equal{} R^2 \minus{} 2rR$ then we have $ R\equal{} r \plus{} \sqrt{OI^2\plus{}r^2}$. Then the circle with center O and radius R meet the perpendicolar line from Xa to IXa on B and C and the tangent from B to the incircle ($ \neq BC$) meet the tangent from C to the incircle ($ \neq BC$) on A.


algebra
polynomial
modular arithmetic
number theory
prime numbers
number theory proposed

Find all integers $ n\ge 2$ having the following property: for any $ k$ integers $ a_{1},a_{2},\cdots,a_{k}$ which aren't congruent to each other (modulo $ n$), there exists an integer polynomial $ f(x)$ such that congruence equation $ f(x)\equiv 0 (mod n)$ exactly has $ k$ roots $ x\equiv a_{1},a_{2},\cdots,a_{k} (mod n).$

I have a small question : k can take all values in the set ${ \{1,..,n}$ or it a fixed number ? If it takes all values in the set $ \{1,..,n\}$ then only  $ n=1,2,4,p$ satisfy condition .

Can you answer my question ,Fang-jh ?
To prevent spamming , I will post my solution in the case $ k$ can take all values in the set $ \{1,..,n\}$ .
The first ,we will show that $ n$ must be a power of prime . Suppose the contrary , then $ n$ can be written in the form $ pq$ where $ p,q$ is relative prime and greater than 1 . Consider $ n\minus{}1$ numbers $ {1,..,n\minus{}1}$ . By the condition of problem ,exist a polynomial f such that the equation $ f(x)\equiv 0(\mod n)$ has exactly  n-1 solution 1,..,n-1 . We will show that 0 is also a solution of this equation .
Because $ p,q>1$ so $ p,q<n\minus{}1$ , by the fundamental property of polynomial , we have $ p|f(p)\minus{}f(0)$ and $ q|f(q)\minus{}f(0)$ ,and because $ p|f(p),q|f(q)$ so $ pq|f(0)$ ,and this gives contrary .
Therefore n must be a power of prime . Let $ n\equal{}p^a$ ,we have two cases :
1) $ p$ is an odd integer (or $ p>2$ ) . We will show that $ a\equal{}1$ ,if not consider two numbers $ 0,p^{a\minus{}1}$ .By the condition of the problem ,exist f such that equation $ f\equiv 0 (\mod p^a)$ has exactly  two solution 0 and $ p^{a\minus{}1}$ (obvious solutions are considered in $ F_n$) . We can write f in the form : $ f(x)\equal{}(x\minus{}0)(x\minus{}p^{a\minus{}1})G(x)\plus{}ax\plus{}b$ . Because $ p^a|f(0),f(p^{a\minus{}1})$ so $ p^a|b$ and $ p|a$.Consider $ x\equal{}2p^{a\minus{}1}$ , it is not very hard to show that it is a solution of equation $ f\equal{}0 (\mod p^a)$ and it gives contrary .
Hence in this case only $ a\equal{}1$ ,or $ n\equal{}p$ is a prime number satisfies condition .
2) p=2 . By the similar trick ,we have $ a\leq 3$ (by consider two numbers $ 0,2^{a\minus{}2}$ and and other number $ 2^{a\minus{}1}$ ) . If $ a\equal{}3$ we only need to consider two number 0,6 and 4 .
Hence $ a\leq 2$ and then  $ n\in \{1,2,4\}$ .
Thus all numbers satisfy condition are 1,2,4,p.

The official solution is $ n\equal{}4$ or $ n$ is prime.

[u]Another solution.[/u]

Let $f(x)=a_nx^n+...+a_0x^0=\sum_{k=0}^{n} a_kx^k$
Suppose we can represent $n$ as $n=pq$, Where $1<p\le q$. If $p=q=2$, then, obviously, the statement holds. Otherwise assume $q<p+q<pq$ $(1)$

Notice that expanding binom, we get $(p+q)^k=p^k+q^k$ for any $k>0$. So, if we set $f$ to have $0, p, q$ as its roots (or $0, p$ if $p=q$) then we have:

$f(p+q) \equiv f(p)+f(q) \equiv  0 \pmod{n}$

yielding that $p+q$ is a root too, which contradicts $(1)$

The only cases left (when we can't represent $n$ as $pq$ where $1<p \le q$) are $2,p$ and also $n=4$ when $p=q=2$. For prime numbers and $n=4$ the statement, obviously, holds.

Hurray!

I am assuming that $k$ is a fixed integer.Clearly any $n \le k$ is a solution since the condition we will be nothing then.From here we will assume that $n >k$.First I will show that $n$ is a prime power.Assume for country that $m$ has at least two prime divisors write $m=pq$ where $gcd(p,q)=1$ then greedily take all residues mod $q$ and mod $p$ that weren't taken before(I mean we haven't taken a $a$ so that $a$ produces the same residue mod$p$ or mod $q$) if there isn't any take one that has a residue mod $p$(or $q$ if $q>p$) not taken before if there is non take one arbitary then it is easy to see that  there is a residue $r$ that $r$ is not in the set but $r \pmod q$ and $r \pmod p$ are in the set which by CRT we will get a contradiction.Now assume that $n=p^{\alpha }$ where $p$ is an odd prime then let $0,p^{\alpha -1}$ be in our set but $2.p^{\alpha -1}$ not to be there then we will get a contradiction using the fact that $f(a+kp^{\alpha -1}) \equiv f(a)+kp^{\alpha -1}f'(a)$ and so $\alpha =1$.In this case Since we have a fild we can use lagrange interpolation formula.Next assume that $n=2^{\alpha }$.If $\alpha \ge 3$ then take $0,2^{\alpha -1},2$ in the set but not $2+2^{\alpha -1}$ in the set then again using hensel's lemma and the fact that $f'(0) \equiv f'(2) \pmod 2$  we get a contradiction.So $\alpha \le 2$ meaning that all the solutions are:

$\{x| x \le k \} \cup \{p|p is a prime \} \cup \{ 1,4 \}$


Points $A, B, C, D$ lie on a line $l$, in that order. Find the locus of points $P$ in the plane for which $\angle{APB}=\angle{CPD}$.


trigonometry
inequalities
inequalities proposed

[quote="mateivld"]In any triangle we have $\frac{1}{2}\sum \sin^2 {2A}\leq\sum \cos^2 A+2\sum \cos^2 A\cos^2 B\leq \frac{3}{2}\sum cos^2 2A$[/quote]

Second inequality is wrong. Let $A=B=\frac{\pi}{6}$ and $C=\frac{2\pi}{3}$, then $LHS=\frac{29}{8}$ and $RHS=\frac{9}{8}$

And here is first:

It is equivalent to:

$\sum\sin^22A-2\cos^2A\leq4\sum\cos^2A\cos^2B$

$\sum\sin^22A-2\cos^2A=\sum4\sin^2A\cos^2A-2\cos^2A=\sum 2\cos^2A(2\sin^2{A-1}) = \sum -2\cos^2A\cos2A$

Now we have inequality:

$\sum2\cos^2A\cos^2B+\cos^2A\cos2A\geq0$

$\sum\cos^2A(2\cos^2B+\cos2A)=\sum\cos^2A(2\cos^2A+2\cos^2B-1)=2\sum\cos^4A+2\sum\cos^2A\cos^2B -\sum\cos^2A=\left(\sum\cos^2A\right)^2+\sum\cos^4A-\sum\cos^2A$

Left to prove:

$\left(\sum\cos^2A\right)^2+\sum\cos^4A\geq\sum\cos^2A$

By Cauchy-Schwarz:

$\left(\sum\cos^2A\right)^2+\sum\cos^4A\geq\frac{4}{3}\left(\sum\cos^2A\right)^2$

So we get inequality:

$\sum\cos^2A\geq\frac{3}{4}$

Using $\sin^2A+\cos^2A=1$ we get:

$\sum\sin^2A\leq\frac{9}{4}$

And finally it is true by Jensen:

$\sum\sin^2A\leq3\sin^2\left(\frac{A+B+C}{3}\right)=3\sin^2\left(\frac{\pi}{3}\right)=\frac{9}{4}$

I managed to prooved this one just now. And accidentally, I use the totally same method.


We can cancel $ 12!$ from both sides.  Thus, we have $ 15 \cdot 14 \cdot 13 \equal{} n$, and $ n \equal{} \boxed{2730}$.


linear algebra
matrix
floor function
Olimpiada de matematicas

[quote]Problema 3
Se tiene un tablero de $n\times n$ con los números del $1$ al $n^{2}$ en sus casillas. Un rey se encuentra en la casilla con el $n^{2}$ y de ahi se mueve a la casilla con el número $1$ y en su camino solo pasa de una casilla a alguna otra casilla adyacente usando siempre el camino mas corto entre ambos números.
Luego viaja a la casilla con el número $2$ de igual manera y sigue asi hasta volver a la casilla $n^{2}$ de
nuevo. Sea $N$ el número de pasos que el rey da en su recorrido, es decir saltos de una casilla a otra
adyacente. Halle el menor valor posible para $N$ y el mayor valor posible para $N$.[/quote]

Bueno, esto va a estar mal y es un chorizón, pero por probar no creo que pase nada.
a) El rey se desplaza de una casilla a otra $n^{2}$ veces y en cada uno de dichoss desplazamientos efectúa por lo un paso: por lo tanto $N\geq n^{2}$ y se podrá cumplir sólo en caso de que exista una numeración del tablero en la que números consecutivos (modulo $n^{2}$) se hayen siempre asignados a casillas adyacentes. Si $n$ es par entonces es fácil ver que un recorrido de ese tipo para el rey existe; puede empezar en la casilla $(1;n-1)$ (en el borde superior), por ejemplo, y barrer con con un camino en forma de $s$ todo el rectángulo a la izquierda de la columna $n$, acabando en la casilla $(n;n-1)$; de aquí pasa a $(n;n)$ y sube luego hasta $(1;n)$ por la columna $n$, hasta la posición de partida. Para $n$ impar el mínimo $N=n^{2}$ no se puede cumplir, pues si coloreamos el casillero como un tablero de ajedrez, casillas adyacentes serán siempre de distinto color, algo irrealizable siendo la longitud de nuestro ciclo, $n^{2}$, un número impar. Sin embargo se demuestra por inducción que para todo $n=2m+1$ existe un recorrido para el que $N=n^{2}+1$ y en el que $(1;n)$ y $(2;n)$ son consecutivas. Para $3$ lo cumple:

$\begin{bmatrix}2&1&9\\3&5&8\\4&6&7\end{bmatrix}$.

Ahora consideremos dentro del tablero de $(n+2)\times(n+2)$ el cuadrado de $n$ de lado a la izquierda de la columna $n-1$ y por encima de la fila $n-1$; apliquémosle la hipótesis de inducción y extendamos el camino que de ella resulta para que recorra por entero el tablero; entre $(2;n)$ y $(2;n+1)$ añadimos la secuencia (para menor pesadez dada una secuencia en línea recta decasillas se mencionarán sólo los extremos):
$\mapsto (2;n)\mapsto (2;n+1)\mapsto (n+1;n+1)\mapsto (n+1;1)\mapsto$
$\mapsto (n+2;1)\mapsto (n+2;n+2) \mapsto (1;n+2) \mapsto (1;n)$;

como no hemos añadido desplazamientos que no sean entre casillas adyacentes hemos acabado.

b) Cada vez que el rey da un paso lo hace, sin tener en cuenta el sentido, entre una columna o fila $i$ y una $i+1$ con $1\leq i<n$. Ahora bien, cada par de elementos de ese tipo determina la división del tablero en dos secciones de tal manera que cada uno de los pasos antes mencionados ocurre en el contexto durante un desplazamiento entre dos casillas pertenecientes a secciones distintas. Por lo tanto, para cada dicho par de elementos se efectuarán como mucho tantos pasos como el doble (uno para llegar y otro para salir) de casillas haya en la sección con mínimo número de ellas de las dos por ellos determinada, es decir $min \{ ni,\;n(n-i) \}$. Haciendo la suma para todo $i$ tanto por filas como por columnas se obtiene la cantidad $S(n)=4\sum_{i=1}^{n-1}(min \{ ni,\;n(n-i) \})$  tal que: $N\leq S(n)$; la igualdad se cumplirá si y sólo si por cada desplazamiento $(i,j)\mapsto (i',j')$, $|i-i'|,|j-j'|>\lfloor\frac{n}{2}\rfloor$, pues sólo en ese caso el número de saltos entre cada par de "elementos consecutivos" (f. o c.) será máximo.
Por ello para $n=2m$ la condición no se puede cumplir. En efecto, si dividimos el tablero en cuatro cuadrados disjuntos: de lado $m$, los desplazamientos que cumplen la propiedad anterior tienen lugar sólo entre casillas en cuadrados diagonalmente opuestos; por lo tanto de ser $N\geq S(n)-1$, tendría que haber como mucho $1$ desplazamiento entre casillas de una y otra parejas de cuadrados, pero dado que en total se tiene que entrar  y salir por lo menos una vez de cada una ,  concluimos $N\leq S(n)-2$.
Ahora sólo hace falta ver que la igualdad puede cumplirse en ambos casos; basta con comprobar que la cantidad $S(n)-N$ es igual a $0$ para todo $n$ impar y $2$ para todo $n$ par. Procedamos por inducción.
Utilizaremos como hipótesis suplementaria que en el recorrido del rey son consecutivos dos pares de esquinas opuestas, pongamos $a=(1;1)\mapsto b=(n;n)$ y $c=(1;n)\mapsto d=(n;1)$.
Para $n=3$ se cumple con la distribución:
$\begin{bmatrix}7&4&2\\3&9&6\\1&5&8\ \end{bmatrix}$.

Para $n=2$:
$\begin{bmatrix}1&4\\3&2\ \end{bmatrix}$.

Ahora, dado un el tablero de lado $n+2$ extendamos el recorrido que nos proporciona la hipótesis de indución para el cuadrado central de lado $n-2$ de la siguiente manera. Dividimos las $(n+2)^{2}-n^{2}=4n+4$ casillas de borde en $2$; por un lado todas las de los cuadrantes inferior derecho y superior izquierdo por otro todas las demás; en el caso en que $n$ sea impar los puntos medios de los lados se distribuyen uno a cada cuarto de borde. Por el motivo ya mencionado no hace falta contar cual es el número de pasos nuevos realizados (espero), ni siquiera hace falta especificar del todo la secuencia de casillas; es suficiente (tomando como ejemplo el primer conjunto de los dos de la partición, el otro es análogo) con que de $a$ se pase a una casilla del tramo inferior y a partir de ese momento se alterne entre arriba y abajo hasta acabar en una casilla superior desde la que pasar a $b$.

Piedad.  :stretcher:

Edit.: lo he puesto espero que un poco más conciso.

Bueno sólo para msotrar mi solkución que creo es mas ingeniosa que la que recibió el premio!!! (aunque esto es sólo una opinion personal)

Para la cota superior simplemente, de resto es igual. Denotemos por $S(i,j)$ la suma de distancias de cada casilla a la casilla $(i,j)$. Entonces:

$N\leq 2S(i,j)$

y esto se prueba pues en cada paso de $N$ tomamos el camino mas corto, sin embargo podríamos tomar uno mas largo llendo primero a la casilla $(i,j)$ y luego a la siguiente.

Tomando $(i,j)$ como la casilla central si $n$ es impar y alguna de las cuatro centrales si $n$ es par, hallamos precisamente que:

$N\leq n^{3}-n$ si $n$ es impar.

$N\leq n^{3}-2$ si $n$ es par.

Además la igualdad se tiene si los pasos consecutivos tienen uncamino corto que pasa por el centro, esto nos da la idea de dividir el tablero en cuadriculas y alternar pasos entre ellas para la construcción. Esta parte se la dejo a ustedes...

you can consider that the shortest distance between two consecutives number is the  number of squares of the "L" shape,I solved easily the problem in that manner.


Let ABC be a surface of centroid G. It is possible to divide this surface in  two surfaces $S_1, S_2$ such that
$\frac{S_1}{S_2}=\frac{m}{n}$,  $m,n\in$$N^*$, using only a ruler?

[quote="Marius Damian"]Fiind date un triunghi si centrul sau de greutate, sa se precizeze daca poate fi partitionata suprafata triunghiulara in doua suprafete aflate in raportul m/n (m,n naturale, nenule, prime intre ele), folosind doar RIGLA NEGRADATA.[/quote]
Can someone translate this into English? :D


What happened to the download forum? I can't find it.

Edit-Ok, thanx

[quote="filletwho"]What happened to the download forum? I can't find it.[/quote]It's in the [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Section[/url] just at the end (because all of the materials were Olympiad ones anyway :) ).


trigonometry
real analysis
real analysis unsolved

Show that $ \sum_{n\equal{}1}^{\infty} \frac{1}{n^{2\plus{}\cos{n}}}$ diverges

I read that a limit points of cos n is -1. In particular...


[color=darkred]Determine all functions $ f : [0; \plus{} \infty) \rightarrow [0; \plus{} \infty)$, such that
\[ f(x \plus{} y \minus{} z) \plus{} f(2\sqrt {xz}) \plus{} f(2\sqrt {yz}) \equal{} f(x \plus{} y \plus{} z)\]
for all $ x,y,z \in [0; \plus{} \infty)$, for which $ x \plus{} y\ge z$.[/color]

An old problem  :wink: ,see here : http://www.mathlinks.ro/viewtopic.php?p=1130155#1130155

As far as I understood, you should qualify for IMO and Balkan MO, Ahiles. Congrats.


Why do you believe it's 3, and why does your friend believe it's 4?

Also, should this be moved to the AMC forums?

There was no AIME in 1977, so I doubt it's and AMC problem.

I assume that initials means two.

so, there are 26x26=676 possibilities for initials.

2029/676 is a little greater than 3, making the answer 4

[quote="BlueMouse"]My friend and I have a rather large argument over the following problem:

There is a city of exactly 2029 people. If x= The least numer of people sharing the same initials, what is the least value of x?

Disregarding vacuously true answers like -12, I say 3. My friend says 4.

Who's right?[/quote]

First of all Are you sure this is an 1977 AIME problem?

i don't think there was AIME back then. Maybe you mean AHSME?

Anyway, 0.

There are $ 26^3\equal{}17576$ possible initials.

Therefore, you can have them all having different initials.

Two clarifications:

1) It's from ARML.
2) Two initials.
3) I think it's four because least. Least case in the optimal scenario is 3. 4 is the greatest.

so now both u and ur friend think it's four?  along with me : )

O_o oops. Probably lack of sleep. I meant three.

As Quickster94 pointed out, there are 26x26=676 possibilities for initials. If there were no more then 3 people sharing the same initials, there could be no more than 3*676 = 2028 people in the city. That extra person, person #2029, must force there to be a group of 4 people sharing the same initials. So the answer is 4.

BlueMouse, for a generalization of this type of problem (and others!), see [url=http://en.wikipedia.org/wiki/Pigeonhole_principle]Dirichlet's box principle[/url].

[quote="Twiz"][url=http://en.wikipedia.org/wiki/Pigeonhole_principle]Dirichlet's box principle[/url][/quote]
Commonly known as the pigeonhole principle. :wink:

Well,

$ 26^2=676$, so we want

$ \lceil ing\frac{2029}{676}\rceil ing$

Which is $ 4$.

[quote="Twiz"]I commonly call it Dirichlet's principle. :P[/quote]
If you want people to know what you're talking about, say "pigeonhole".


Computing Combinations Part 4

We compute several examples of the identity C(n,r) = C(n,n-r), and then we prove the identity both with algebra and with a counting argument.

[hide="My solution's"]
a. $\binom{11}{2}+\binom{11}{3}$ compared to $\binom{12}{3}$

$\binom{11}{2}=\frac{11!}{2!9!}\implies \frac{11*10}{2}\implies \frac{11*\cancel{10}}{\cancel{2}}\implies 55$
$\binom{11}{3}=\frac{11!}{3!8!}\implies \frac{11*10*9}{3*2*1}\implies \frac{11*\cancel{10*9}}{\cancel{3*2}}\implies 3*5*11=165$
$\binom{12}{3}=\frac{12!}{3!9!}\implies \frac{12*11*10}{3*2*1}\implies \frac{\cancel{12}*11*\cancel{10}}{\cancel{3*2}}\implies 220$

${\boxed{165+55=\boxed{220}} \text{or} \binom{11}{2}+\binom{11}{3}=\boxed{\binom{12}{3}}}$

b. $\binom{14}{12}+\binom{14}{13}$ compared with $\binom{15}{13}$
$\binom{14}{12}=\binom{14}{2}$ because $\binom{n}{r}=\frac{n!}{n-r!*r!}$, so we have for example: $\binom{8}{3}=\frac{8!}{8-3!*3!}\implies \frac{8!}{5!3!}=\binom{8}{5}=\frac{8!}{8-5!*5!}\implies \frac{8!}{3!5!}$

$\binom{14}{12}=\binom{14}{2}$
$\binom{14}{13}=\binom{14}{1}$
$\binom{15}{13}=\binom{15}{2}$

$\binom{14}{2}=\frac{14!}{12!2!}\implies \frac{14*13}{2}\implies \frac{\cancel{14}*13}{\cancel{2}}\implies 7*13=91$

$\binom{14}{1}=\frac{14!}{13!1!}\implies \frac{14!}{13!}\implies \frac{14*\cancel{13*12*11*10*9*8*7*6*5*4*3*2}}{\cancel{13*12*11*10*9*8*7*6*5*4*3*2}}=14$

$\binom{15}{2}=\frac{15!}{13!2!}\implies \frac{15*14}{2}\implies \frac{15*\cancel{14}}{\cancel{2}}\implies 15*7=105$

\[\boxed{14+91=\boxed{105}}\text{or} \boxed{\binom{14}{2}+\binom{14}{1}=\boxed{\binom{15}{2}}}\]

We have just identified a new identity!  The hockey stick identification!
Note how we have two equations here:
$\binom{14}{1}+\binom{14}{2}=\binom{15}{2}$
$\binom{11}{2}+\binom{11}{3}=\binom{12}{3}$

Let's look at the numerators of the binomials.  The first numerator is going to be called $n$
We would have $\binom{n}{x}+\binom{n}{y}+\binom{n+1}{z}$ for integers x,y,z that we are going to solve right now.
Let's call the first denominator of the binomials $r$
\[\boxed{\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}}\]
That is known as the hockey sticky identification, and it is very useful for solving problems involving $\binom{n}{r}+\binom{r+1}=x$ or pascal's triangle problems.

Note how in the first equation $\binom{14}{1}+\binom{14}{2}=\binom{15}{2}$
\[n=14\text{ and } r=1\]
and in our second equation $\binom{11}{2}+\binom{11}{3}=\binom{12}{3}$
\[n=11\text{ and } r=2\] [/hide]


Putnam
number theory unsolved
number theory

Let $ N$ be the positive integer with $ 2007$ decimal digits, all of them $ 1$; that is,
$ N \equal{} 1111\cdots 11$.
Find the thousandth digit after the decimal point of$ \sqrt {N}$.

$ A_n\equal{}1111...1\equal{}\frac{10^n\minus{}1}{9}$ (n times). Therefore $ \sqrt{A_n}\equal{}\frac{\sqrt{10^n\minus{}1}}{3}.$
When n is even we can easy calculate $ \sqrt{A_n}\equal{}\frac{10^{n/2}}{3}\sqrt{1\minus{}10^{\minus{}n}}\equal{}\frac{10^{n/2}}{3}(1\minus{}\frac{1}{2}10^{\minus{}n}\minus{}\frac 18 10^{\minus{}2n}\minus{}\frac{1}{16}10^{\minus{}3n}\minus{}...)$
But when n is odd we need $ \sqrt{10}$ with 1000 digits.

yeah.. i believe $ n \equal{} 1998$ was a putnam problem though that putnam doesn't appear to be on this forum.


How many different positive integers can be represented as a
difference of two distinct members of the set $ \{1, 2, 3, 4, 5, 6 \}$?

First, we will find the number of ways to make a sum.

The largest possible integer we can make is $ 6 \plus{} 5$ or $ 11$. The smallest possible integer is $ 1 \plus{} 2$ or $ 3$. $ 11 \minus{} 3 \equal{} 8$.

But this is not our final answer. We want to find a POSITIVE integer, and it's a difference, so we must divide by $ 2$ to get $ 4$. We're counting, and to find the number of integers between $ a$ and $ b$, we do $ a \minus{} b \plus{} 1$, so we add $ 1$ to $ 4$ to get $ \boxed{5}$.

wow, that is complicated....
i just do...
since they're asking for difference, then, all the possibilities are
$ 6\minus{}5, 6\minus{}4, 6\minus{}3, 6\minus{}2, 6\minus{}1$

anything else would have the same answer as one of those
therefore, there are 5 different positive integers

Oh, oops.

Was my method correct, or was I just lucky and got the right answer?

idk
it might've been right, i didn't really look into it


limit
function
calculus
real analysis
real analysis solved

In the last week I had some Internet problems .. couldn't stay on the net for more than 5 minutes and get disconected.. I hope I will "move" to cable internet :D :D
wanted to make some comments on some problems but couldn't ... wait till I have cable :D:D

here is the main problem:

I found a problem in a book given at a contest in Romania ..
Build a series a_n of pozitive numbers with the following properties:
1)  \lim (n-->oo) (a_(n+1)-a_n)=oo;
2)  \lim (n-->oo) (sqrt(a_(n+1))-sqrt(a_n)=0.

okay .. I even have the answer a_n=n*ln (n) and they prove that this series verifies the conditions... that's quite simple!! :D

BUT HOW DID THEY COME UP WITH THIS ONE!!?? :?

please.. really need help! :D :D

My opinion is that, intuitively, we would like a sequence that does eventually grow (in the "Cauchy" sense of condition (i)) but not way too fast (as dictated by condition (ii)). Hence, from experience the ominpresent log function is fatally bound to appear (sounds stupid?). The rest is more of a "trail-and-error" thing.

Cherrs,
Thanos

this is calculus and should go to maths for college :D

they also say that n*sqrt(n) verifies these conditions.. oh my... intuition :D :D :D

cheers!

OKay ... here is another one! :(

[b]1)[/b]Build up a sequence (xn)n>=1 with xn in (0,1) so that the series (xn^n)n>=1 has no limit! :D

cheers!!

Here is another one to keep you up! :D :D

[b]2)[/b] Build up a sequence (xn)n>=1 with xn in (0,1) so that the series (n*xn^n)n>=1 has no limit!

cheers! :D :D

for the first one just consider:

x_(2n+1)=1/2 and x_(2n)=1/(4n^2) and this is it! :D

the computations are easy! :D

cheers! :D

But doesn't this sequence have limit 0? Here's what I thought: x_(2k-1)=1/2 and x_(2k)=1-1/2k. Maybe I just misunderstood the sequence you wrote.

Oh, maybe by series you mean S_n=x_1+..+x_n?

I said:

Build up a sequence (xn)n>=1 with xn in (0,1)  so that the sequence ((x_n)^n)n>=1 has no limit!

seems correct! :D

cheers!

and for the 2nd x_(2n+1)=1/2 and x_(2n)=1/(2n)

cheers! :D

Yeah, but (1/2)^(2n+1)->0 and (1/(4n^2))^2n-> 0, so the sequence you posted for the first one has limit 0. I think I'm doing this wrong somehow?

And doesn't the sequence you posted for the 2'nd one also->0? For the second I think this might work: take x_(2k-1)=1-1/(2k-1) and x_2k= 1/(2k)^(1/2k).

Could you please explain to me what I'm doing wrong abt the sequences you posted? Maybe I didn't understand because of some brackets or sth like that. :D


Euler
geometry
geometric transformation
reflection
circumcircle
geometry proposed

Let $ ABC$ be an acute triangle with $ AA_1,BB_1,CC_1$ is the altitudes wrt $ \triangle ABC$. $ A_1C_1\cap BB_1\equiv \{B_0\}$, $ CC_1\cap A_1B_1\equiv \{C_0\}$. Let $ A_2$ be the projection of $ A$ onto $ B_0C_0$. Define the same for $ B_2,C_2$. Prove that: $ AA_2,BB_2,CC_2$ are concurrent.
You can see at this: http://www.mathlinks.ro/viewtopic.php?t=276546

[quote="jayme"]Dear Mathlinkers,
A0B0C0 and ABC are orthologic... and we are done
Sincerely
Jean-Louis[/quote]
Dear [b]Jayme[/b], can you point out the point that $ AA_2,BB_2,CC_2$ meet at? :P

[quote="jayme"]Dear Mathlinkers,
Ithink that your point is on the Euler's line because it is after Sondat's theorem perpendiculaire to the orthic axis.
Sincerely
Jean-Louis[/quote]
Dear [b]Jayme[/b], according to my proof, it is exactly the [b]$ Euler$ circle center[/b]. Very intersting!! I am surprised by myself :P

See at the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51168]AOa perp DE[/url], of [b][size=100]jhaussmann5[/size][/b].

The line segment $ AA_{2}\perp B_{o}C_{o}$ here, passes through the point $ O_{a},$ as the reflection of the circumcenter $ O$ of the given triangle $ \bigtriangleup ABC,$ with respect to the side-segment $ BC$ and so, because of $ AH\parallel \equal{} OO_{a},$ where $ H$ is the orthocenter of $ \bigtriangleup ABC,$ we conclude that this line bisects the segment $ OH$ and similarly, for the line segments $ BB_{2}\perp A_{o}C_{o}$ and $ CC_{2}\perp A_{o}B_{o}.$

Hence, the line segments $ AA_{2},\ BB_{2},\ CC_{2},$ are concurrent at [b][size=100]nine point center[/size][/b] of $ \bigtriangleup ABC$ and the proof is completed.

Kostas Vittas.


x,y,z>0
prove that:
$ 3(x\plus{}y)(y\plus{}z)(z\plus{}x)(xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x))\geq 16xyz(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2)$

[quote="tuandokim"]x,y,z>0
prove that:
$ 3(x \plus{} y)(y \plus{} z)(z \plus{} x)(xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} zx(z \plus{} x))\geq 16xyz(x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2)$[/quote]
This is not true, for counter-example, $ x\equal{}1,\, y\equal{}1,\, z\equal{}4.$


hello, i have found $ x\equal{}\minus{}\frac{1}{2}\plus{}\frac{\sqrt{3}}{2}$ $ y\equal{}\minus{}\frac{1}{2}\minus{}\frac{\sqrt{3}}{2}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have found $ x \equal{} \minus{} \frac {1}{2} \plus{} \frac {\sqrt {3}}{2}$ $ y \equal{} \minus{} \frac {1}{2} \minus{} \frac {\sqrt {3}}{2}$
Sonnhard.[/quote]
Can you post your solution,please ?

thank you ...  :oops: Can you solve it with Inequalities , please ? i think this problem has an easier solution !! :maybe:

Add it up and we get

$ (x^2\plus{}x\minus{}\frac{1}{2})^2\plus{}(y^2\plus{}y\minus{}\frac{1}{2})^2\equal{}0$

[quote="cosinator"]Add it up and we get

$ (x^2 \plus{} x \minus{} \frac {1}{2})^2 \plus{} (y^2 \plus{} y \minus{} \frac {1}{2})^2 \equal{} 0$[/quote]
thank you !!


Given a rectangular board MxN divided into MxN cells

How many subset of cells are there such that no two cells in the subset are adjacent to each other (two cells are adjacent if they share a common edge)?

(Own problem)

Could anyone help me solve this problem? Thanks

Equivalently, we are asked for the number of independent sets in a grid graph.  For $ m \equal{} n$ we get sequence [url=http://www.research.att.com/~njas/sequences/A006506]A006506[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url], while the whole table is [url=http://www.research.att.com/~njas/sequences/A089934]A089934[/url]; probably there is no nice formula in general.  For $ m \equal{} 1$ we get the Fibonacci numbers; for $ m$ constant we always get some sequence that satisfies a linear recursion relation.  For $ m \leq 4$ it's not so bad to compute what this relation should be, and for $ m \leq 7$ you can actually find this sequence in the OEIS (follow the links from A089934).


[quote="guile"]Problem 1 (Refer to Figure 1)
ABCD circumscribes the circle
AB = 8, BC = 10, CD=21. Find AD.
Problem 2 (Refer to Figure 2)
BC and DC are tangents of the circle.
BC= 15, AC = 17, FE=EH. What is FH + FC + CH?[/quote]
nice problems guile. :)
[hide="1"]Quite simple. There are many tangents, so let
E, F, G, H be the points (respectively) on AD, AB, BC, and CD such that these points are tangent to the circle. so we have,
FB=BG=a
CG=CH=b
ED=DH=c and
AE=AF=d
Thus, AD= c+d
a+b=BC=10, so
AB+CD=c+d+a+b=c+d+10=29
therefore, c+d=29-10=19=AD=19 :)
[/hide]
[hide="2"]This is very easy. I did this in my head too, so i'll write a summary of my thoughts: :D
-triangle BCA is congruent to triangle DAC
FE=EH, by power of point C,
BC=DC
BF=EF=DH=EH
so, obviously 15+15=30 :) [/hide]
Nice solution fireforce. :lol:

[hide="Lame-o method for #1"]
The sum of the opposite sides equal each other
$ \text{AB}+\text{CD}= \text{BC}+\text{AD}\\ 8+21 = 10+X \\ X = 19$
There is a proper way, but I am just too lazy
[/hide]

thats not lame, its ingenious! :rotfl:
Mines a bit ingenious, and a bit common sense (mostly common sense). :D

I don't think it has a name because it is the same as saying that tangents from a point are equal; fireforce's proof is the same idea as inspired by nature's proof.

[quote="fireforce"][quote="guile"][quote="fireforce"][hide="Lame-o method for #1"]
The sum of the opposite sides equal each other
$\text{AB}+\text{CD}= \text{BC}+\text{AD}\\ 8+21 = 10+X \\ X = 19$
There is a proper way, but I am just too lazy
[/hide][/quote]

Gee, I never knew that  :blush:. (Im poor in geometry) Do you know what theorem states that the sum of the length of the opposite sides of the circumscribed quadrilateral are equal? I mean, does the theorem has a name?[/quote]

A) Don't worry about it I learned it when I was at a math comp. too and when I was competing in geometry. I learned it the hard way (getting it wrong on some test then memorizing the solution into oblivion so I would never do that again)

B) It is only lame-o because my Geometry teacher taught me how to do it the proper way, waaaay later and it works based on Algebra. Its longer too, but it will work regardless of the shape of the object (in this case it was a quad.).  :wink:

C)Basically you use the Ice Cream Cone Theorem and Segment Addition Th. like 4 times. There is no actual theorem name its more like using a combination of techniques.

[hide="The proper way to do #1"]
1) Label each point of tangency (Going clockwise starting with AB E,F,G,H)
2) Using the Ice Cream Cone Th. AE = AH, BE = BF, CF = CG, and DG = DH
3) We know that AE + EB = AB using Segment Addition Th.
4) Use that with Th. with the other 3 sides.
5) LET'S START MAKING EQUATIONS!  :lol: :
    A) Label AE = X
    B) That makes EB = 8-X or -X + 8
    C) Using the conclusions in Step 2; BF = -X + 8
    D) That makes FC = X + 2 (Plug in BF from Step 5c into the equations you made in Step 4)
    E) (Using techniques in steps C and D) this makes   DG and DH = -X + 19
    F) We know using the Segment Addition Th. that AH + DH = AD
    G) We know AH = X (From Step 2)
    H) Plug everything in: X + -X + 19 = $19$ [/hide]

On a side note:
1)Never assume that a theorem does not have a name (I learned this while studying for a history of math test at a math competition)
2) Never take a history of math test if optional at competion (unless you are better at history, then at math)[/quote]

Is there really an ice cream cone theorem?

[quote="guile"]Is there really an ice cream cone theorem?[/quote]

Yes there is an ice cream cone theorem, though this term is really a layman way of naming the theorem:

If two line segments that originate from a single point and are both tangent to a given circle, then they are congruent. (if you draw this out it looks like a ice cream cone)

Also, if anyone who is reading this topic is having trouble with circle geometry it may be a good idea to memorize and understand this theorem, and Power of a Point. These two theorems are often used in the geometry part of competions and major tests.

[quote="guile"][quote="lotrgreengrapes7926"]The "Ice Cream Cone Theorem" is a specific case of Power of a Point.[/quote]

Can you site a specific website where I can learn this. I have read the Power of a Point Theorem from Cut-the-Knot, but it I  cannot use it in solving problems.  :blush:[/quote]
There are many different forms of power of the point theorem, each one very useful.
http://en.wikipedia.org/wiki/Power_of_a_point
http://illuminations.nctm.org/ActivityDetail.aspx?ID=122
Power of the point is also talked about in vol.2. :)

[/quote]
There are many different forms of power of the point theorem, each one very useful.
http://en.wikipedia.org/wiki/Power_of_a_point
http://illuminations.nctm.org/ActivityDetail.aspx?ID=122
Power of the point is also talked about in vol.2. :)[/quote]

Thanks


function
modular arithmetic
real analysis
real analysis unsolved

here an interesting pb:
let $p$ be a prime number and $v_p(n)=max \{k \geq 0 s.t. p^k | n \}$ for $|n|>0$.
For a rational $r=\frac{a}{b}$, $v_p(r)=v_p(a) - v_p(b)$.

1/prove that $d_p(r_1, r_2)=p^{-v_p(r_1-r_2)}$ is a distance (with $d(a,a)=0$)
2/let $p1,p2,..,p_n$ be n distinct prime numbers. We will define a distance on $Q^{n+1}$.
  For $x=(x_0,..,x_n)$ and $y=(y_0,..,y_n)$ in $Q^{n+1}$,
  define their distance by: $d(x,y)=|x_0-y_0| + \sum_{1}^{n}d_{p_i}(x_i, y_i)$

  The diagonal function from $Q$ in $Q^{n+1}$ is: $f(r)=(r,r,r..r)$.
  Prove that $f(Q)$ is dense in $Q^{n+1}$.

To give a quick sketch, if we are given a rational $\frac uv$, all one need to do in order to find $\frac mn$ close to $\frac uv$ $p$-adically, is fix $m,n$ modulo some power of $p$ (maybe a different power for each of $m,n$). We do this for all the primes $p_1,\ldots,p_n$, and we find that the rational we are looking for, which we want close to a list of $n+1$ rationals in the $n+1$ metrics respectively, has the form $\frac{aM+b}{cN+d}$, for fixed $M,N,b,d$ (i.e. its denominator and its numerator are fixed modulo two integers). Of course, rationals of that form are dense in $\mathbb Q$ with the usual metric, so we can find such a number which is close to a given rational in the usual metric as well.

To illustrate this for a prime $p$, assume we're given $\frac uv$ in lowest terms, and we're looking for $\frac mn$ in lowest terms, close to it $p$-adically. If $p\not|v$, we also take $n\not|p$ (take $n\equiv 1\pmod p$, for instance), and then all we need to do is take $m\equiv unv^{-1}\pmod {p^t}$ for some large $t$. On the other hand, if $p|v$, then take $n$ s.t. its $p$-adic norm is equal to that of $v$ (i.e. the same maximal power of $p$ divides both), simplify $\frac{mv-un}{nv}$ with this power of $p$, and we are left with $mv'-un'$ at the numerator, with $p\not|v',n'$, so, again, we can make $mv'-un'$ divisible by powers of $p$ as large as we want by fixing $m\pmod{p^t}$ for large $t$.


Hi I want to learn Algebraic topology, could you please recommend me some good book?
thanks


I feel it is better to provide more detail in a solution.
Along with the computation, a section called "think about it..." or something could provide other ways to think about the problem.

For example, in this, you might say:

"in how many ways can I choose 1 item from a collection of 1293 items?" While this is what is said symbolically, it may help a student to read a text description.

Or you could have a "simpler problem approach" section, or a link to a "simpler problem hint" giving steps to a simpler problem, for example:

numeric simpler:  3 choose 1

context simpler:
In how many ways can you choose 1 cookie from a plate containing a chocolate chip, oatmeal, and a peanut butter cookie.
You could choose the oatmeal
You could choose the peanut butter
or You could choose the chocolate chip

So there are 3 ways you can choose 1 cookie from 3,

3 choose 1.

and maybe even a link to derivations and proofs of formulas that might be used.

If the formula n!/(n!(n-r)!) is used in a solution, then a link to an proof or a lesson on why that formula works might be helpful.

No one posted the answer  :rotfl:  It's 1293.

[hide=Solution]Since $\binom{n}{1} = n$, we have that $$\binom{1293}{1} = \boxed{1293}$$[/hide]
[hide=Answer]$$\boxed{1293}$$[/hide]

[quote=AIME15][quote="ernie"]To sum it up, $ \binom{n}{1} \equal{} n$.[/quote]

How do you know?

Use the definition of $ \binom{n}{r}$.  Then we have $ \binom{n}{1}\equal{}\frac{n!}{1!(n\minus{}1)!}\equal{}\frac{n!}{(n\minus{}1)!}\equal{}n$.[/quote]

How about 0?