superior algebra
superior algebra unsolved

give a characterization of all solvable extensions of degree $p$ over $\mathbb{Q}$, where $p$ is a prime.

as a matter of completeness... by saying $L/K$ is solvable i mean that there's an extension $M/L$ such that $M/K$ is galois and $\text{Gal}(M/K)$ is solvable.

no one?
i found this problem pretty interesting...

by the way, could some moderator edit the previous post? of course i meant an extension of [i]degree[/i] $p$, not of [i]order[/i] $p$...

[moderator: as you wish...]

Suppose $E/\mathbb{Q}$ is a solvable extension. Is it necessarily separable? Actually, now that I think about it, I don't think I will even need that.


trigonometry
calculus
calculus computations

Let $ x$ be real and suppose that $ x/2^k$, $ k\geq 0$, is not a multiple of $ \pi/2$. Then find
\[ \prod_{k\equal{}0}^{\infty}\left(\frac{2^k}{x}\tan\left(\frac{x}{2^k}\right)\right)^{2^k}\]

Simple manipulation shows that

$ \prod_{k\equal{}0}^{N} \left[ \frac{2^k}{x} \tan \left( \frac{x}{2^k} \right) \right]^{2^k} \equal{} \frac{2x}{\sin 2x} \left[ \frac{2^{N}}{x} \sin \left( \frac{x}{2^{N}} \right) \right]^{2^{N\plus{}1}}$.

Then by taking $ N \to \infty$ to both sides, we have

$ \prod_{k\equal{}0}^{\infty} \left[ \frac{2^k}{x} \tan \left( \frac{x}{2^k} \right) \right]^{2^k} \equal{} \frac{2x}{\sin 2x}$.

[quote="sos440"]Simple manipulation shows that

$ \prod_{k \equal{} 0}^{N} \left[ \frac {2^k}{x} \tan \left( \frac {x}{2^k} \right) \right]^{2^k} \equal{} \frac {2x}{\sin 2x} \left[ \frac {2^{N}}{x} \sin \left( \frac {x}{2^{N}} \right) \right]^{2^{N \plus{} 1}}$.[/quote]
Would you please elaborate any further?

[quote="Croata"]Would you please elaborate any further?[/quote]
How do you mean to elaborate?

I mean if you can show me how did you get that.

Ah, now I understood what you have meant.

My first approach was direct and intuitive, but somewhat messy. I used the double-angle formula for sine and rearranged the product to obtain the equality.

[hide="Here is a line-by-line flow."]Let $ P_n$ be the $ n$-th partial product. Then

$ \begin{eqnarray*}
P_N & = & \prod_{k=0}^{N} \left[ \frac{2^k}{x} \tan ( x/2^k) \right]^{2^k} \\
& = & \prod_{k=0}^{N} \left[ \frac{2^k}{x} \frac{\sin (x/2^k)}{\cos (x/2^k)} \right]^{2^k} \\
& = & \prod_{k=0}^{N} \left[ \frac{2^{k-1}}{x} \frac{2 \sin (x/2^k) \cos (x/2^k)}{\cos^2 (x/2^k)} \right]^{2^k} \\
& = & \prod_{k=0}^{N} \left[ \frac{2^{k-1}}{x} \frac{\sin (x/2^{k-1})}{\cos^2 (x/2^k)} \right]^{2^k} \\
& = & \left\{ \prod_{k=0}^{N} \left[ 2^{k-1} \sin (x/2^{k-1}) \right]^{2^k} \right\} \left\{ \prod_{k=0}^{N} \left( \frac{1}{x} \right)^{2^k} \right\} \left\{ \prod_{k=0}^{N} \left[ \frac{1}{\cos^2 (x/2^k)} \right]^{2^k} \right\} \\
& = & \frac{\sin 2x}{2x} (\cos (x/2^N))^{-2^{N+1}} \left \{ \prod_{k=0}^{N-1} \left[ 2^{k} \sin (x/2^{k}) \right]^{2^k} \right\}^{2} \left\{ \prod_{k=0}^{N-1} \left( \frac{1}{x} \right)^{2^k} \right\}^2 \left\{ \prod_{k=0}^{N-1} \left[ \frac{1}{\cos (x/2^k)} \right]^{2^k} \right\}^2 \\
& = & \frac{\sin 2x}{2x} (\cos (x/2^N))^{-2^{N+1}} P_{N-1}^2,
\end{eqnarray*}.$

Now divide both sides by $ P_{N-1}$ and arrange it to obtain the formula for $ P_{N-1}$. Replacing $ N-1$ by $ n$, we have exactly what we want.[/hide]

Then I realized that we can obtain that by multiplying $ \sin 2x$ and applying double-angle forumla recursively. Just try it by yourself!


Find all positive integers that prove the equation :

\[ x^6  \plus{} x^3 y^3  \plus{} y^6  \equal{} 3^n
\]

Examining $ \bmod9$ (where $ x^3,y^3\in\{0,\pm1\}$) and remembering that $ 3^n\equiv3\pmod9$ if $ n \equal{} 1$ and $ 3^n\equiv0\pmod9$ if $ n > 1$, we find that if $ \{x^3,y^3\}\pmod9$ equals (order doesn't matter),

$ \{ \minus{} 1, \minus{} 1\}\pmod9$, then $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv3\pmod9$,
$ \{0, \minus{} 1\}\pmod9$, then $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv1\pmod9$,
$ \{1, \minus{} 1\}\pmod9$, then $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv1\pmod9$,
$ \{0,0\}\pmod9$, then $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv0\pmod9$,
$ \{1,0\}\pmod9$, then $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv1\pmod9$,
$ \{1,1\}\pmod9$, then $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv3\pmod9$,

so the only possibilities are $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv3\pmod9\implies(x,y,n) \equal{} (1,1,1)$, and $ x^6 \plus{} x^3y^3 \plus{} y^6\equiv0\pmod9\implies(x,y) \equal{} (3a,3b)$ for positive integers $ a,b$.

Therefore, $ (3a)^6 \plus{} (3a)^3(3b)^3 \plus{} (3b)^6 \equal{} 3^n\implies a^6 \plus{} a^3b^3 \plus{} b^6 \equal{} 3^{n \minus{} 6}$, from which we can conclude that $ \boxed{(x,y,n) \equal{} (3^k,3^k,6k \plus{} 1)}$ where $ k\in\mathbb Z\geq0$ are the only solutions.


There is NO substitute for visiting a college campus.  Sometimes, however, you just can't do that, for whatever reason.  [url=http://campustours.com/]Campus Tours[/url] is an excellent website for online tours of a college campus.

yeah, i'm visiting brown this month, i don't think u even need to sign up or notify them of the visit :o


ratio
combinatorics unsolved
combinatorics

Here is the  problem:
 It is given set $S_n=\{ 1, 3, ..., 2n-1 \}$. Determine maximal integer $k$ such that exist subset $T$ of $S_n$ and for every two $x, y \in T$ $x$ doesn't divide $y$ and $y$ doesn't divide $x$.

Let $K$ be the number of elements of $S_n$, not divisible by 3, i.e. $K=n-[(n+1)/3]$. Clearly, each element of $S_n$ belongs to one of $K$ subsets, in each of which the ratio of any twoelements is a power of 3 (for any element $a$, not divisible by 3, consider a subset $\{a,3a,9a,\dots\}$). So, $k\le K$. But if we consider all elements of $S_n$ except first $[(n+1)/3]$, then the least left element equals $2[(n+1)/3]+1$, and so the ratio of any two lest elements is less then 3. Hence it may be not integer.


Okay so first off ill type out the question for everyone ,

Person A has 8000 dollars and would like to earn 500 dollars from the money, how much would see invest in a stock that has been getting a 10% annual return and how much should she invest in savings bonds that pay 0.4% annual interest? model this situation with a linear system.

Okay so far this is the work ive done

S be the amount in Stocls
B be the amount in bonds

1. 0.1s+0.4b=500
2. s+b=8000---> S=8000-b

0.1(8000-b)+0.4b=500
800-0.1b+0.4b=500

heres where i get stucked i check the answers for it and i should have gotten 0.06 but instead i ended up with 0.05 and sometimes 0.03 can anyone PLEASE PLEASE explain this to me. Thanks!

[quote="Alpha1234"]1. 0.1s+0.4b=500[/quote]

This is incorrect.  $4$% should be $0.04$, so your equation should be

$0.1 s+0.04 b = 500$

This is the wrong forum to ask about homework.  Classroom Math would be more appropriate.  This is the [b]Olympiad[/b] Algebra section, not just the Algebra section.


function
combinatorics unsolved
combinatorics

find the number of solutions of: 2a+3b+5c=1000; where a,b,c are non-negative integers. :oops:

Yes, 16384.
bobbym, did you find this answer by a computer, or not?

Hi Kirill;

You have a typo it is 16834 solutions. I used 2 generating functions which are ideally suited to this type of problem, they are;

$ \sum_{k \equal{} 0}^{500}x^{2k} \cdot \sum_{k \equal{} 0}^{334}x^{3k}\cdot \sum_{k \equal{} 0}^{200}x^{5k} \equal{} 1 \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} 2x^5 \plus{} 2x^6 \plus{} 2x^7 \plus{} 3x^8\plus{}3x^9\plus{}4x^{10}\ldots$

and

$ \frac {1}{(1 \minus{} x^2)(1 \minus{} x^3)(1 \minus{} x^5)} \equal{} 1 \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} 2x^5 \plus{} 2x^6 \plus{} 2x^7 \plus{} 3x^8\plus{}3x^9\plus{}4x^{10}\ldots$

These are essentially the same, in as much as they give the same answer. But they  are expanded in different ways. Now all that was necessary was to find the coefficient of x^1000 in the expansion of either one of them. I played with both trying to get the coefficient by binomial tricks or residues. I then used the second form to generate a recurrence relation:

a(n)=a(n-10)-a(n-8)-a(n-7)+a(n-3)+a(n-2)

with the initial conditions being the coefficients in the above expansions.
and ran it with a computer to a(1000).

HI;

I tried hard to develope an answer that wasn't computer dependent. But:

2a+3b+5c=1000, that one thousand is what makes this intractable to all my hand methods. Another way of phrasing this is asking for the partitions of 1000 from the multiset { 2,2,2,2,2,2,2,2,... , 3,3,3,3,3,3,3,3,3,3,3,3... , 5,5,5,5,5,5,5,5,5,5,5,5,5...}. A really big problem!

I am not saying it can't be done just because I can't but some problems even in math require a computer. In this case the recurrence I provided could be run in any language and even a handheld calculator  or spreadsheet could do it.

Also a simple program (in any language) that ignores the ideas above and just consisted of 3 loops and a counter could be used.


geometry
rectangle
combinatorics unsolved
combinatorics

For a rectangle nxm points
1)
How many ways that an ant move from a point at the "top left - down" (sorry ,i don't know how to mention the place of this point .... ) to the point at the "top right -up"?
An ant only can move up or right

2)How many ways that an ant move from a point at the "top left - down"  to the point at the "top right -up"?
An ant can move up ,down , left ,right but it can't remove to the point that it was occupied before .

I just can solve the 1st proplem ... help me ..

To prove the first one, among the $m+n$ steps, we can freely choose the $n$ vertical ones (and they determine the whole path).

Where did you find the problems (the second one  ;) )?

[quote="Myth"]First problem is well known. Answer is $C_{m+n}^{n}$.

Second problem is quite strange... I have some doubts that it is possible to give an answer for it.[/quote]
^_^
it is nxm [b]points[/b] . So ... the answer is $C_{m+n-2}^{n-1}$.

Do you mean we need $(n-1)$ vertical and $(m-1)$ horizonal moves? If yes, then, indeed, answer is $C_{(m-1)+(n-1)}^{n-1}$.

ofcourse we need ...
^_^
there are only nxm points ..
it is not a hard problem ...
but the second is .... i can't plan any thing ...


linear algebra
matrix
linear algebra unsolved

Let $ N$ be a complex matrix $ 2 \times 2$ such that $ N^2 \equal{} 0$. Show that either $ N \equal{} 0$ or $ N$ is similar, over $ C$, to
$ \begin{pmatrix}
0&0 \\ 1&0
\end{pmatrix}$

If you don't know the Jordan form, I'd suggest doing this in three steps:

* Show that $ N$ is similar to a matrix with zeroes in the right column. (hint: eigenvalues, choice of basis)
* Show that $ N$ is similar to a matrix with all but the lower left entry zero. (hint: eigenvalues)
* Finish the problem. (hint: choice of basis)

If $ N$ is not zero, there is some $ v$ such that $ w \equal{} Nv$ is not zero. We then have $ Nw \equal{} 0$.
With respect to the basis $ \{v,w\}$, $ N$ has the matrix $ \begin{bmatrix}0 & 0 \\
1 & 0\end{bmatrix}$.

This construction, generalized to larger matrices, is an important part of how Jordan form is proved. We can get a block structure with diagonal blocks of the form $ \lambda I \plus{} N$ easily from the existence of eigenvalues, and then this construction lets us pin down the nilpotent parts.

(Note: This part works over any field. We only need an algebraically closed field for Jordan form so we can guarantee the existence of eigenvalues, and nilpotent matrices have all their eigenvalues equal to zero without any help.)

From this result, I know that we can get the following statement: If $ A$ is a $ 2 \times 2$ matrix of complex entries, then $ A$ is similar, over $ C$ to one matrix of the following kind:
$ \begin{pmatrix}
a&0 \\ 0&b
\end{pmatrix}$
or
$ \begin{pmatrix}
a&0 \\ 1&a
\end{pmatrix}$
but I can't get it, sorry I know this is easy, but it is difficult for me.


[quote="Kevin K."]How can you express the number 100 using only four 9's?
(You can use operations signs, exponents, etc, but not another number.)[/quote]

[hide]$9/9+99=100$[/hide]


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Let A in $M_n(R)$ such that $A^3 + 6A^2 +11A + 6I_n = 0_n$
Prove that there exist

$(a,b_1,...,b_n)$ reals numbers,
$B_1,...,B_n$ matrices
$(B_k)^2 = I_n$ for any k in {1,..,n}

such that

$A = a.I_n + b_1.B_1 + ...+b_n.B_n$

Let $m_a$ be the minimal polynomial of $A$. Then $m_a| x^3+6x^2+11x+6=(x+1)(x+2)(x+3)$. This proves that $m_a$ has no multiple roots, so $A$ is diagonalizable.

Lemma: Let $A$ be a diagonal matrix in $M_n(\mathbb{R})$ whose diagonal entries are ordered increasingly. Let $E$ be the set of eigenvalues of $A$ and let $m=|E|$. Then there exist diagonal matrices $B_1,\ldots,B_m$ with $\pm 1$ entries on the diagonal and $b_1,\ldots,b_m$ reals such as $A=\sum_{k=1}^mb_k\cdot B_k$.

Proof: If $m=2$ then let $B_1=I_n$ and $B_2=diag(I_i,-I_j)$ such that $i+j=n$ and $i,j$ are the multiplicities of $a_1,a_2$, the $2$ eigenvalues of $A$.
Let $b_1=\frac{a_1+a_2}2$ and $b_2=\frac{a_1-a_2}2$. Then $A=b_1B_1+b_2B_2$.

If $m\neq 2$, let $a_1,\ldots,a_m$ be the eigenvalues of $A$ and let $i_1,\ldots,i_m$ be their multiplicities.
Let $B_j=diag(-I_{i_1},-I_{i_2},\ldots,I_{i_j},-I_{i_{j+1}},\ldots,-I_{i_m})$, $j=\overline{1,m}$.

We have to solve $b_1B_1+b_2B_2+\ldots+b_mB_m=A$.
This is equivalent to $(2I_m-UU^t)B=\bar{A}$ [b][*][/b], where $U=(u_{i,1}=1)_{i=\overline{1,m}}$, $B=(b_{i,1}=b_i)_{i=\overline{1,m}}$ and $\bar{A}=(a_{i,1}=a_i)_{i=\overline{1,m}}$.
The characteristic polynomial of $UU^t$ is $X^{m-1}\cdot$ (the characteristic polynomial of $U^tU$) which is $X-m$. So $\det(2I_m-UU^t)=2^{m-1}(2-m)\neq 0$ for $m\neq 2$. So [*] has a solution and this is a solution for the lemma.

Back to the problem: obvious


MIT
college
summer program
Mathcamp
Harvard
AMC
USA(J)MO

MIT sent out their regular decision letters on Friday March 11th.  Well, did anyone get in yet?

[quote="Geehoon"]I got in.

Geehoon Hong
B.N. Cardozo High School, NY
Mathcamp 2004
MIT class of '09[/quote]

Congratulations, and congratulations to all the members who got into the college they wanted.

Best regards,

i really want to go to MIT
how did you get in? what did you do to get in?

our school has two. im SO excited for them.

I got in also...w00t!  (After being pwned in the Early Action round).  No one else from my school got in.

For people who got into MIT, congratulation! I have a vested interest myself. What are you guys' standardized tests score? What types of specials awards or recognition do you have? I am just curious.

[quote="tetrahedr0n"]I'm a junior, but Hannah Zhou, a girl from my school (MMPC silver medalist, SW semifinalist) was accepted.[/quote]

just curious...what does MMPC and SW stand for?

MMPC = Michigan Math Prize Competition, the flagship state competition. Especially nice since it ranks its winners.

SW = Siemens Westinghouse, the research competition.

I just found out I got in. I am very excited. Still waiting on Caltech and Harvard, but that won't be too stressful anymore.

Congratulations Nathan! I should know by tomorrow.  Additional congrats to you and Adam on your 14s on the AIME.

Are the admissions decision available online too?  :?  :?

No.  From what I've read on MIT's website (http://www.mit.edu) they said they will not release admission decisions online or by e-mail

no, a 97 GPA on a scale from 0-100, sometimes referred to as Mastermans'? I'm not sure but I have a pretty good idea that this is what Mariag is talking about

[quote="Mariag"]i am not sure what it's called, my school just gives all grades out of a 100 and the GPA is just the numerical average thereof[/quote]
Are you in the US? in most of the countries where I have been, the averages are given numerically (0-100, 0-20, 0-5, 0-10) and that is why sometimes US schools have problems cathegorizing the real GPA (or what in the US is considered GPA) of forein people. I honestly think that a person with more than 90 GPA in a 0-100 scale probably would have 4.00 GPA in the US!

I've always [b]dreamt of MIT[/b] and want to go there more than anything in the worls (even if  I get Caltech or Princeton)I live in India and somebody told me just 2 students are going to be taken from india this year.Is this true?????????

I've come among the [b]top 40 Young Scientists in India[/b] under the KVPY Scholarship conducted by the [b]IIT's[/b] in India.I've always been top in class and scored 94% in ALL India Board CBSE exams.(regarding very high)
Ive also got 2280 in SAT(new) and am quite confident about SAT II's i'll give next year.I'm also expecting near 100% in TOEFl
I was also adjudged the best All round student in my school in Grade 10.(i had to leave that scholl coz it did'nt have classes beyond grade 10 I go to another school now)

Yet that 2 students taking thing is really a blow to me .Can u guys help me out.This is  a long post but plz rate my chances and what else you think need to be done.

[b]THANK YOU[/b]

djimenez,

actually, i believe 95+/100 is equivalent to 4.0 GPA according to something i read a while ago which I don't quite remember

It depends on the school.

If I got 90 in every class, my GPA would be 4.0
If I got 100 in every class, my GPA would be 4.0

That is how my school does it.

i do go to school in the US. i once saw a chart ranking 97+ to be a 4.0 and so on down. the way frading in my school works is that any grade above a 90 is possible but below 90 grades are only given in steps of 5 and no grade below 45 is awarded. i thought this system was common for the US, but guess some schools do use the 4.0 scale :?

Maria, I think most schools use the 4.0 system.  My school uses a 5.0 weighted scale.  All classes are out of 4 points, but APs are out of 5.  An A is a 4 in a regular class and a 5 in an AP, a B a 3 in a regular class and a 4 in an AP, et cetera.  Minuses and pluses don't have different points.  Plus, because the letter grade translates to the point, your points may be dependent on what teacher you have--our calc teacher had an A- at grades $\ge 85$.

My school, as mentioned earlier is 4.0 system.

Pre-IB weights .02 per semester
IB/AP weights .04 per semester
College courses weight .04 or .08 depending on the class.
I think Honors might weight .02..

[quote="sonar"]Maria, I think most schools use the 4.0 system.[/quote]
My school (Benjamin N. Cardozo HS, Bayside NY) is using 0-100 system, and im not too confient on this, but most public schools in new york do use 0-100 system.

i c, so NY is just different... sorry for the confusion :(

Rushil - Yes MIT is nice, my son  is very happy there. If you send me a pm I can give his e-mail to you if you  want some  more specific  information.

There are many  Indian students in MIT (Hey, they evne  have a  Cricket team, and a  Bhangra group!). Many are graduate students. In undergraduate  classes . natually . most  are  Indian-Americans (whose family is in US)  and  a very few foreign stiudents.   Guess many parents would not send their kids far away for undergraduate studies. Each year they take about 100 foreigh students as undergraduate freshmen (about 8%).  So yes the number is small, but not as low as 2 :)  I would guess that  your chances are good. (But I am not an admisson person!)

Few things to consider, MIT does not give merit scholarships, They do give Financial assistance.  Undergraduate Tution is  about $30,000.  (BTW  Caltech, and some other schools (eg Duke)  have good merit scholarships .)

Check out the web site (mit.edu) write to them etc...  Hope this helps.

It seems that admisson staff is  familier with Indian grading  system etc.. Special awards (IIT d  help,  Also if you have participated in other well kown activities  (eg RMO, INMO, IMO - also physics or chemisty olympiads,  science-talent/national  scholarships etc...) mention it,.- Admisson staff would most likely familier with such things.

 Good luck!

By the way I may have posted it here before but check it out:

 The  video produced by MIT  Indian students
Get videos directly
Small (10 MB): http://web.mit.edu/anoop/www/msur/milesur.html
Big(50 MB): http://web.mit.edu/anoop/www/msur/milesur50.mpg
 -

[quote="djimenez"]last July I heard in a CNN broadcast that there has been a descence in the number of students comming to US Universities, because European, Australian and New Zelandian Univeristies have become very open and attractive for international students. [/quote]

Since 2001 there are many more restrictions on the admission of non-US nationals to US universities.
It has become more difficult to obtain student visas, which is not surprising considering that (as a rough statistical generalization) non-US students at schools below the state university level were generally there for the purpose of obtaining a visa first and a certificate second.   Many professional licenses such as hairdresser, electronics repairman, nurse, biotechnician, etc  are in the USA obtained through professional "colleges" which were, until the new restrictions, basically visa mills for aspiring immigrants and degree mills for academically nonaspiring natives.   The result of the tightened immigration restrictions, however, is seen all the way up to postdocs at MIT, Harvard, etc, and it was only intensive lobbying by the universities and technology industries that prevented the rules from becoming even more draconian.

Meanwhile, the US having become less friendly to foreigners, many schools in Europe and elsewhere have seen a rise in interest from students from e.g. the Middle East, who are able and willing to pay the fees.

[quote="Geehoon"]My school (Benjamin N. Cardozo HS, Bayside NY)[/quote]

Did you, by any chance, have this teacher?

http://sports.espn.go.com/espn/news/story?id=2098847

[quote="blahblahblah"]
Did you, by any chance, have this teacher?

http://sports.espn.go.com/espn/news/story?id=2098847[/quote]
my brother said his global teacher, Mr. K was a professional wrestler :D  It is very possible that it is him, but my brother doesn't have him any more.


inequalities
Cauchy Inequality
inequalities proposed

Let $ a,\ b,\ c$ be real numbers such that $ a,\ b,\ c\in{[\minus{}\sqrt{2},\ \sqrt{2}]}$ and $ abc\neq 0$.

Prove that $ \frac{(2\minus{}a^2)(2\minus{}b^2)(2\minus{}c^2)}{\left(\frac{1}{|a|}\plus{}\frac{1}{|b|}\plus{}\frac{1}{|c|}\right)^6}\leq \frac{1}{729}$.

[color=darkblue]Hello, [b]Kunny[/b] ![/color] [color=darkred][b]Happy new year ![/b][/color]

[quote="kunny"] [color=darkred]Let $ a,\ b,\ c$ be real numbers such that $ a,\ b,\ c\in{[ \minus{} \sqrt {2},\ \sqrt {2}]}$ and $ abc\neq 0$ .

Prove that $ \frac {(2 \minus{} a^2)(2 \minus{} b^2)(2 \minus{} c^2)}{\left(\frac {1}{|a|} \plus{} \frac {1}{|b|} \plus{} \frac {1}{|c|}\right)^6}\leq \frac {1}{729}$. [/color][/quote]
[quote="Virgil Nicula"]$ \{x,y,z\}\subset (0,2]\implies\ 729\cdot(2\minus{}x)(2\minus{}y)(2\minus{}z)\ \le \left(\frac {1}{\sqrt x}\plus{}\frac {1}{\sqrt y}\plus{}\frac {1}{\sqrt z}\right)^6\ .$[/quote]
[color=darkblue][b][u]Proof[/u].[/b] Prove easly that $ t\in (0,2]\ \implies\ \frac {3\minus{}t}{2}\ \le\ \frac {1}{\sqrt t}\ .$ Therefore,

$ (2\minus{}x)\cdot(2\minus{}y)\cdot(2\minus{}z)\cdot 1\cdot 1\cdot 1\ \le\ \left[\frac {(2\minus{}x)\plus{}(2\minus{}y)\plus{}(2\minus{}z)\plus{}1\plus{}1\plus{}1}{6}\right]^6\ \implies$

$ 729(2\minus{}x)(2\minus{}y)$ $ (2\minus{}z)\ \le\ \left(\frac {9\minus{}x\minus{}y\minus{}z}{2}\right)^6\equal{}$ $ \left(\frac {3\minus{}x}{2}\plus{}\frac {3\minus{}y}{2}\plus{}\frac {3\minus{}z}{2}\right)^6\ \le\ \left(\frac {1}{\sqrt x}\plus{}\frac {1}{\sqrt y}\plus{}\frac {1}{\sqrt z}\right)^6\ .$[/color]

Nice Solution, Virgil Nicula.

I hope the year 2009 will be good for you!

Hello, kunny!
This is my solution for your problem
$ (2\minus{}a^2)(2\minus{}b^2)(2\minus{}c^2) \le \left(\frac{6\minus{}a^2\minus{}b^2\minus{}c^2}{3}\right)^3$
All we need to do is to prove
$ 3\left(6\minus{}a^2\minus{}b^2\minus{}c^2\right)\le \left(\frac{1}{|a|}\plus{}\frac{1}{|b|}\plus{}\frac{1}{|c|}\right)^2$
We can rewrite
$ \left(\frac{1}{|a|}\plus{}\frac{1}{|b|}\plus{}\frac{1}{|c|}\right)^2\plus{}3\left(a^2\plus{}b^2\plus{}c^2\right)\ge18$
Now, it is time for using Cauchy inequality
$ \left(\frac{1}{|a|}\plus{}\frac{1}{|b|}\plus{}\frac{1}{|c|}\right)^2 \ge \frac{9}{\sqrt[3]{|abc|^2}}$

and $ 3\left(a^2\plus{}b^2\plus{}c^2\right) \ge 9\sqrt[3]{|abc|^2}$

The last step is to prove
$ \frac{9}{\sqrt[3]{|abc|^2}}\plus{}9\sqrt[3]{|abc|^2} \ge 18$ which is true by applying Cauchy inequality for 2 positive real numbers
We are done!

Hello, great math. :)

Now I have just collected 3 solutions. :lol:

I would like to give you another solution of mine, kunny
Using Cauchy inequality, we get
$ \left(\frac {1}{|a|} \plus{} \frac {1}{|b|} \plus{} \frac {1}{|c|}\right)^6 \ge \frac {792}{|abc|^2}$
Therefore, we have $ \frac {(2 \minus{} a^2)(2 \minus{} b^2)(2 \minus{} c^2)}{\left(\frac {1}{|a|} \plus{} \frac {1}{|b|} \plus{} \frac {1}{|c|}\right)^6}\leq \frac{(2 \minus{} a^2)(2 \minus{} b^2)(2 \minus{} c^2)(abc)^2}{729}$
Simultaneously, we still hold the following inequality
$ (2 \minus{} a^2)a^2 \le \left(\frac {2 \minus{} a^2 \plus{} a^2}{2}\right)^2 \equal{} 1$
Likewise, we apply for the other variables b and c.


Let $a+b+c=3.$
Hence, $6/5\geq\frac{ ab + ac}{ a^2 + ( b+c)^2} + \frac{bc+ba}{b^2+(a+c)^2} +\frac{ca+cb}{c^2+(b+a)^2}\Leftrightarrow$
$\Leftrightarrow\sum\frac{a(3-a)}{a^2+(3-a)^2}\leq\frac{6}{5}\Leftrightarrow\sum(\frac{3a-a^2}{2a^2-6a+9}-\frac{9}{25}(a-1)-\frac{2}{5})\leq0\Leftrightarrow$
$\Leftrightarrow\sum\frac{(a-1)^2(2a+1)}{2a^2-6a+9}\geq0.$ :)
By the way, $\sum\frac{ab+ac}{ka^2+(b+c)^2}\leq\frac{6}{k+4},$ where $0.8\leq k\leq8.$ ;)

Arqady what a nice solution!!!
In the most of  solutions  you use a trick like this
[quote="arqady"]
$\Leftrightarrow\sum\frac{a(3-a)}{a^2+(3-a)^2}\leq\frac{6}{5}\Leftrightarrow\sum(\frac{3a-a^2}{2a^2-6a+9}-\frac{9}{25}(a-1)-\frac{2}{5})\leq0$[/quote]

How do you think of it ?  You are on the earth ?

[quote="arqady"][quote="silouan"]
How do you think of it ?  You are on the earth ?[/quote]
Dear, silouan. This is very easy:
If $f(a)=\frac{3a-a^2}{2a^2-6a+9}+k(a-1)-\frac{2}{5}$ then I wish $f'(1)=0$ and obtain $k=-\frac{9}{25}.$ ;)[/quote]

Very good solution RESPECT!
So where can study of using that strategy can you say please :-)

[quote="Davron"]
So where can study of using that strategy can you say please :-)[/quote]
The scope of this strategy is limited enough. But sometimes it works.

[quote="kunny"]The tangent line is the key point, but sometimes.[/quote]

sure i am agree with you!

Do anyone have another solution !!! If you want i can post the original form of the Problem ...

Or everybody have that...?

[quote=wudud]If $ a$, $ b$, $ c$ are nonnegative real numbers, then show that

$ \frac65\geq\frac{ ab+ac}{ a^{2}+\left( b+c\right)^{2}}+\frac{bc+ba}{b^{2}+\left(c+a\right)^{2}}+\frac{ca+cb}{c^{2}+\left(a+b\right)^{2}}$.[/quote]
The following inequality is also true.
Let $a$, $b$ and $c$ be [b]real[/b] numbers such that $\prod_{cyc}(a^2+(b+c)^2)\neq0$. Prove that:
$$ \frac65\geq\frac{ ab+ac}{ a^{2}+\left( b+c\right)^{2}}+\frac{bc+ba}{b^{2}+\left(c+a\right)^{2}}+\frac{ca+cb}{c^{2}+\left(a+b\right)^{2}}$$

[quote=arqady]The following inequality is also true.
Let $a$, $b$ and $c$ be [b]real[/b] numbers such that $\prod_{cyc}(a^2+(b+c)^2)\neq0$. Prove that:
$$ \frac65\geq\frac{ ab+ac}{ a^{2}+\left( b+c\right)^{2}}+\frac{bc+ba}{b^{2}+\left(c+a\right)^{2}}+\frac{ca+cb}{c^{2}+\left(a+b\right)^{2}}$$[/quote]

Because if $x,\,y,\,z$ be real numbers such that $ x + y + z = 1,$ then
$$ \frac {1}{1 + x^2} + \frac {1}{1 + y^2} + \frac {1}{1 + z^2}\leq \frac {27}{10}.$$

[quote=arqady]It seems that he following inequality is also true.
Let $a$, $b$ and $c$ be [b]real[/b] numbers such that $\prod_{cyc}(a^2+(b+c)^2)\neq0$. Prove that:
$$\frac{ ab+ac}{ 16a^{2}+\left( b+c\right)^{2}}+\frac{bc+ba}{16b^{2}+\left(c+a\right)^{2}}+\frac{ca+cb}{16c^{2}+\left(a+b\right)^{2}}\leq \frac{3}{10}$$[/quote]
Yes, it's true!
There is a nice proof for this inequality. ;)

Quite  nice.
Let $x$, $y$ and $z$ be real numbers such that $xy+xz+yz+xyz=4$. Prove that:
$$\frac{x}{4x^2+1}+\frac{y}{4y^2+1}+\frac{z}{4z^2+1}\leq\frac{3}{5}$$


I will be posting another Mock ARML tommorrow at around 4:30 PM...Same rules as the first one and the answers can be submitted to me by Monday night. Thank you :)

I should have several ARML style questions up on Monday or Tuesday to go with Shobhit's problems.


We view the world as particles and discrete units. But De-Broglie says that everything has both a particle and a wave nature. Can we view the world as a continuous landscape by tapping into the wave nature? If not, why? If so, how? Does anyone have any ideas?
   Ashwath :)

Great to see someone from india
MERA BHARAT MAHAN

Let me give you a very practical example for this
Consider a particle located in a region of potential V(x)
Such That

V(x)=    0     if 0<x< a
	     ∞   OTHERVISE
 now the particle is bound between the two limits
x=0 & x=a if it tries to escape it’s pushed by an unbounded force
now the time independent schrondiger wave eqn is
			([u]d^2 ψ[/u])/d^2 x   =   -k^2ψ

k=(√[u]2mE[/u])/h

where ψ(x) =  A sin (kx) (it gives the probability of x) E= total energy
don’t be scared this is just like a wave except that we are writing it for a particle
according to classical theory
total energy  T=K + V
or    T=E
so the energy can have all values
at x=a
now  0 = A sin(ka)
so ka= 0 , +π,+2π,+3π,+4π,+5π
  hence
    k =   (N π)/a  	solving this we get

E=  (N^ 2 π^ 2 h ^ 2)/2ma^2

so the energy is quantized!!!!!!
NOTE:- IF YOU PLOT IT. It’ll LOOK LIKE A WAVE ON A STRING & DIFFERENT VALUE OF n WILL GIVE DIFFERENT MODE.
   now this is true for subatomic particles but for the macroscopic view m is large
   so the energy gap in successive states is small and hence we don’t use QM to describe everything as a wave. Classical physics is an excellent approximation for this
 I HOPE YOU GET THE POINT!!!!!!!!...

Hello,
    There are a couple of minor errors in the above discussion.
 Psi(x), the wave function does not give the probability of the position x of the particle. Infact, Psi(x) is usually a complex function.Hence, it is square of amplitude of Psi(x) which denotes the probability.
 Secondly,the quantization term for energy does not hold for sub atomic particles alone.It holds for any particle in general.
 An addition, the world CAN be viewed as a continuum taking the wave picture in consideration.For instance, a chalk can be 'seen' as a wave of extremely small wavelength.It all depends on one's percepetion.As Born-Oppenheimer approximation states, (I like to put it this way), As mass becomes sizeably larger than those of electronic dimensions, quantum mechanics 'approaches' its classical counterpart.As mass becomes much larger, Physics has to be reorganized as in the case of super massive black holes.The Unification is the word Theoretical Physicists are after now.
Sathej


Find all positive integers a b c s.t.

                               a!b!=a!+b!+c!

Bomb

WLOG suppose that $a\ge b.$
We observe some cases:

(i)$c< a,b$ then it is easy to see that after dividing both sides of $a!\cdot b!=a!+b!+c!$ by $c!,$ the LHS is even RHS is odd,unless $b=c+1,a>c+1,c-even.$ simplifying we get
$a!(c+1)!=a!+(c+1)!+c!\Rightarrow (c+1)!|c!.$ So no solution in this case.

(ii)$c>a,b$ then after dividing both sides of $a!\cdot b!=a!+b!+c!$ by $b!,$ the LHS is even RHS is odd unless $a=b+1$(this case doesn't give solutions) or $a=b\Rightarrow a!=2+c!/a!\Rightarrow c<a+4$ this gives $c=4,a=3.$

(iii)$c=b$ then we have $a!\cdot b!=a!+2b!\Rightarrow a!=a!/b!+2(*)\Rightarrow a!/b!|2$ but this possible iff $(a,b)=(2,1);(1,1)$ but this doesn't work in (*). So this case also doesn't give solution.

(iiii)$c=a$ then $a!|b!$ this leads to contradiction.
So we have the unique solution $(a,b,c)=(3,3,4).$
I think that I didn't miss any case.


[quote="daermon"][hide]$2h+p+s=155\\ h+2p+2s=235\\ 3h+4p=235$  multiplying the first equation by two and subtracting the second from it gives $3h=75$, so $\boxed{h=40}$.  plugging this into the third equation gives $\boxed{p=\frac{160}{4}=40}$.  plugging this into the first equation gives $\boxed{s=155-90=65}$[/hide][/quote]
Mostly correct. You made a mistake with $h.$

:wallbash_red:  :wallbash_red:
oops that was a typo [hide]$\boxed{h=25}$[/hide]

[quote="i_like_pie"]Three people go to a store. One person buys two shirts, one pair of pants and one sweater. The second person buys one shirt, two pairs of pants and two sweaters. The third person buys three shirts and four pairs of pants. Find the price of each item if the first person spent 155 dollars and the other two spent 235 dollars each.[/quote]

[hide="algebra derived answer"]
Let $h$ be shirts
$2h+p+s=155$
$h+2p+2s=235$
$3h+4p=235$
$h=25$
[/hide]


For those of you familiar with the USACO training pages, I'm having trouble with the problem called Contact in section 2.1. It gives a long binary string and asks you to find patterns of a certain length in them. What I do is add a 1 in front of each binary string and convert it into a base 10 number. This allows me to store the number of occurences of each number in an array. However, my code seems to produce incorrect counts (overcounts in varying degrees) and I was wondering if someone could give my code a quick look. All help with this would be most appreciated.

note:I did not include the algorithm I used to display the results because I have tested it and it is accurate.

/*
PROG:contact
ID:pplatyp1
*/

using namespace std;

#include <iostream>
#include <fstream>

ifstream fin("contact.in");
ofstream fout("contact.out");

//ps is the the count of each pattern
int ps[8192];
//each digit is temporarily stored here
int pattern[14];
char tmp;

int  a,b,c;
int coount=0;

void sort (char j)
{
count++;

if (j=='0')
{

pattern[12]=pattern[11];
pattern[11]=pattern[10];
pattern[10]=pattern[9];
pattern[9]=pattern[8];
pattern[8]=pattern[7];
pattern[7]=pattern[6];
pattern[6]=pattern[5];
pattern[5]=pattern[4];
pattern[4]=pattern[3];
pattern[3]=pattern[2];
pattern[2]=pattern[1];
pattern[1]=0;
}

if (j=='1')
{

pattern[12]=pattern[11];
pattern[11]=pattern[10];
pattern[10]=pattern[9];
pattern[9]=pattern[8];
pattern[8]=pattern[7];
pattern[7]=pattern[6];
pattern[6]=pattern[5];
pattern[5]=pattern[4];
pattern[4]=pattern[3];
pattern[3]=pattern[2];
pattern[2]=pattern[1];
pattern[1]=1;
}

int y;
int num;
int l;
int p,z;

//tries all pattern lengths
for(y=a;y<=b;y++){
num=0;

if(y<count){
for(l=y; l>=1; l--){
p=1;
if (pattern[l]==1){
for(z=l; z>1; z--)
p=p+p;
num=num+p;
}
}
p=1;
for(z=1; z<=y; z++)
p=p+p;
num=num+p;
ps[num]++;

}
}

}

int main()
{
char t;

//a is shortest length, b is longest length, c isn't important for now

fin>>a>>b>>c;

while(fin.get(t)){

sort(t);

}

Hi

Well, I think you've got the main thing right, but really it is really tough to make out exactly where the problem might be because usually, the bugs creep in when you put separate things together. It may stem from things such as array indices etc... So, it would probably be easier if you could post  all the code.  Also if you could post a small case for which it doesnt work.

But really, the best thing you can do is to try to do it yourself(If you havent already tried a lot). This would give you more experience in debugging your own programs and help you in future.

In general, for debugging, try and look for cases for which it doesnt work and print values at different points in the program(if you're not using a debugger). Dry run small cases.

Hope this helped...

Yes, I've had this basic code for weeks, and my debugging efforts have only allowed me to isolate the problem with this part of the code. I was hoping a fresh set of eyes could help me with the problem.

The way my program overcounts is bizarre.

in the following test case
2 4 10
01010010010001000111101100001010011001111000010010011110010000000

my program overcounts all patterns with a 0 by 1

in
1 12 20
11111111111(480 1's)

it overcounts all patterns by 6

in larger and more complicated tests my program overcounts erratically.

here's my display algorithm if you think it will help (but I know the code in the first part overcounts.)

int ca,cb,cc,cd,ce,cf,cg;
int max;

note: the array asd holds the powers of 2
note:the array bps holds the set of frequencies about to be displayed

//each iteration displays one frequency
for(ca=c;ca>0;ca--){

max=0;

//finds the proper frequency to display
for(cb=asd[a];cb<asd[b+1];cb++){

if(ps[cb]==max){
bps[cd]=cb;
cd++;
}

if (ps[cb]>max){
max=ps[cb];

for(cc=1;cc<=4096;cc++)
bps[cc]=0;

cd=2;
bps[1]=cb;
}

}

//if no more values to display, terminate early
if(max==0)
break;

fout<<max<<endl;

//display each pattern
for(cg=1;cg<cd;cg++){
ps[bps[cg]]=0;

//subtract off the leading 1
for (ce=1;;ce++){
if (bps[cg]>=asd[ce] && bps[cg]<asd[ce+1])
break;
}

bps[cg]=bps[cg]-asd[ce];

//display the number
for(cf=ce-1; cf>=0; cf--){

if(bps[cg]<asd[cf])
fout<<"0";

if(bps[cg]>=asd[cf]){
fout<<"1";
bps[cg]=bps[cg]-asd[cf];
}
}

if(cg!=cd-1)
fout<<" ";
}

fout<<endl;
}

return 0;
}

I didn't really get a chance to look through the whole program, but the following part of the program seemed incorrect.

if(y<count){
for(l=y; l>=1; l--){
p=1;
if (pattern[l]==1){
for(z=l; z>1; z--)
p=p+p;
num=num+p;
}
}
p=1;
for(z=1; z<=y; z++)
p=p+p;
num=num+p;
ps[num]++;

  First, should the if condition be y <= count instead of y < count?  Second, the end brace for that if statement seems too early; shouldn't it cover this whole section of code, including the part where you calculate the leading '1'?

  There are lots of comments I could make about the coding style, but I don't have time.

  I would recommend that you try your program on all trivial test cases, like input strings of length 1, 2, and 3, before you try more complicated cases.

Ok, thanks for the advice. I believe you're right on that condition. However, I think the end brace for that statement does include the part where I calculate the leading 1.

Hmm...you're also right about the trivial test case checking.

the test case of
1 4 10
1

gives two 1's two 01's one 0 and one 001.

weird.
Perhaps there is something fundamentally wrong with how I'm handling the input. is the fin.get function the right one to use?

If you ever do have time to comment on my coding style it would be most appreciated.

After you read a, b, and c, isn't the input cursor still at the end of the first line?  So when you call fin.get(t), won't the first one or two characters you read be the carriage return and/or the line feed from the first line (depending on your operating system)?

If I had to solve the problem, I would have approached it quite differently.  Instead of an array as the main data structure, I would have used a map (or hash_map) from (C++) strings to integers.  Then the whole program would have been about 15 lines of code.  The STL library is quite powerful.


geometry
integration
analytic geometry
calculus
rotation
algebra
polynomial

Basically I got to

$x=12(y^{2}-y^{3}) from 0\to 1$

what is the volume of the solid created when this is revolved over the y=axis.

Thanks for any help in advance. I remember doing these last semester no problem but now in a review section I can only start but never get anywhere. Thank you again.

(writing in my sleep)

The basic form for the surface area of revolution is $\int 2\pi r\sqrt{dx^{2}+dy^{2}}$, where $r$ is the distance from a point $(x,y)$ to the axis of rotation. The term $\sqrt{dx^{2}+dy^{2}}$ is a useful mnemonic device: you can transform it by pulling out $dx$ or $dy$ from the square root. In this case $r=x$ (distance to the $y$-axis is measured by the $x$-coordinate, and vice versa). The variable of integration should be $y$; otherwise you'd have a hard time solving for $y$ in terms of $x$. So, $SA=\int_{0}^{1}2\pi 12(y^{2}-y^{3})\sqrt{(dx/dy)^{2}+1}dy$. Since $dx/dy=12(2y-3y^{2})$, we have $SA=24\pi \int_{0}^{1}(y^{2}-y^{3})\sqrt{144(2y-3y^{2})^{2}+1}dy$, a nightmarish integral. Time to wake up. :surrender:

When mlok wakes up, he might see that vasion asked for the volume, not the surface area.

Slice this whichever way is convenient - in this case into strips parallel to the $x$ axis. Each strip is located at height $y,$ has width $dy$ and extends from $x=0$ (which is the $y$ axis) to $x=12(y^{2}-y^{3}).$

Now rotate this strip about the $y$ axis. It becomes a thin circular disk centered on the $y$ axis, of thickness $dy$ and radius $12(y^{2}-y^{3}).$ This has volume equal to $\pi r^{2}dy$ or

$dV=\pi(144)(y^{2}-y^{3})^{2}\,dy$ so

$V=144\pi\int_{0}^{1}(y^{2}-y^{3})^{2}\,dy.$

This is a polynomial integral, not the "nightmarish" integral for the surface area.


Let $x,y,z \in [0,1]$. Prove that $x+y+z-xy-yz-zx \leq 1$.

The LHS is an affine expression in each of the variable so it reaches its maximum at the endpoints of [0,1], that is when x,y,z are in {0;1}.
Direct computation shows that if (x,y,z) = (0,0,0) or (1,1,1) then the lhs is 0, and if (x,y,z) is a permutation of (1,0,0) or (1,1,0) then the lhs is 1.

Pierre.


Let $n$ be an integer between 1 and 9 (both included) Consider all n-digit numbers formed by using the digits 1 through $n$ once and only once in each number formed. What is the sum of all of these $n$-digit numbers?

Hardly Pre-Olympiad.

[hide="Solution"] Each digit $k$ appears in each place $(n-1)!$ times.  The sum is

$\frac{n(n+1)}{2}(n-1)! \left(1+10+...+10^{n-1}\right)$ [/hide]


What scores are you and your team getting on national mathcounts problems?

(only write if you are in nats)

I'm getting around averages of 28s on nats tests and our team has an average of 8 on team rounds.

I made nats but I haven't done any practice national tests yet. Where do you get them?

You can buy them (but they're really expensive) from the MC store (only like 1999-2007 though). The only other option is to ask people to e-mail or pm them to you (which most people will call "illegal") or ask your coach or other coaches for them.

lol i know. it's fun talking like that.

And for your scores, your team should average perfect scores on the team rounds (if they want to place) and get at least 35s. It might be better to aim for the higher 30s, as the test is probably going to be pretty beastly.

...
Guys could you please stop spamming the MC forums just so you can increase your posts?
And yeah, you wanna average more than 35 on most Nats tests.

Meh, I always get in the uber-high thirties to low forties for individuals, but then again, I've done 1998-2004 like so many times, and though I don't remember the answers, the questions are pretty familiar, which is annoying and unfair, I admit.  As for the recent ones, I can get in the mid-30's for 2005 cuz it's harder than most of the previous tests (in my opinion) and in the semi-high-almost-competent-nearly-acceptable 30's for 2006.  I have the 2007 test, but I don't wanna retake it until like weeks before the competition.  The last time I saw those questions was in May last year at Nats, so I'm saving these relatively new, unrecycled problems for later.

As for team scores, I practice by myself and can get 7's to 10's depending on difficulty.  There was the rare 4 when I totally had no time to finish the 2005 one, but the rest were pretty easy. :D

PS: Sorry for the immodesty if there was any.... :blush:   Then again, my scores are nothing compared to other people's on AoPS, so I'm fine, I think. :maybe:

just so you all know, all you need to make is countdown is somewhere from 31-35 depending on the year difficulty. Probably the reason we aren't doing that good as a team is that we have all never been to nats  (I'm new to mathcounts, I only learned about it in october) so we just have to get used to these problems. Well our scores have been going up:
7,7,10, so I think we will do better as a team.

Edit: thanks peijin-simple mistake.

I'm pretty sure you got the meaning of veteran wrong because a veteran is someone who is experienced and already has been through the event before.
But anyways, Team Round counts a lot in MC, it basically counts as 8 points that you guys score.

meh, I keep on missing something for a stupid reason

old nats:  41-44

after 2000 nats: 33-41

If I were going back to Nationals I would probably not waste my time worrying how other people are preparing, and preparing yourself.

You should probably try it.

I guess he just wants to know what scores he needs to get to be successful, or is just bored of studying...
Either way, I think you [b]should[/b] be trying to do some old Nats tests instead of coming on AoPS and posting questions you already know how to solve here.

Gotta agree with Catalyst on this one.  If you knew you only needed a 38 wouldyou leave 6 blank since you didn't NEED them?  Areyou going to try HARDER if you think you need a 39?  Go do some problems, write some problems, prove some solutions, understand what you did wrong on the last one.

Assume you need to get them all right and go out and try to get them all right.  Or, assume you have no chance and just enjoy the challenge and what ever will be will be.

Good luck to everyone, and relax!!!!!

[quote="frost13"]Or, assume you have no chance and just enjoy the challenge and what ever will be will be.  [/quote]
THAT'S HOW I MADE NATS IN SEVENTH, THEN NOT IN EIGHTH.

It works--I thought I woud fail chapter in seventh, then wound up getting second at states. Gain that mentality, then don't lose it like I did.

Learn how to approach problems. Be persistent.  Look for creative solutions to problems you've done before.  Check problems over using different solutions, and listen to Carl for once.

At wiseidiot: You don't think that setting your goal as "beat Lyndon Ji" would help you? (Sorry Lyndon, you're just a veteran whose name I know. I don't want you not to win again, I'm just using you as a benchmark. :))

In general, my philosophy on national competitions (and competitions in general) is: go to win, but don't let winning become more important than having fun. (I went to Nat MC, nat vocabulary competitions 4 times, and national science fair twice so far, if that lends my words any weight. And I've done well in some small subset of those, lest you think having fun has gotten in the way of success.)

Who's this "Lyndon Ji" you're speaking of?  I must beat him this year!! I will utterly destroy him! Wait a second...

My practice tests range from 0 to 46. Oh. That's how good I am.

i 100% agree. i mean, why do you (or anyone) do mathcounts? $ For Fun$.

what kind of place does cough cough low 30s cough cough get you at nats? My friend who is awful at math is wondering...

In the old ages (1990's), that used to be considered average.
Nowadays (2002-present), that is an amazing score to get.

Nationals seem to be much harder nowadays from what I can tell. Also, the problems are less "formulaic approach" and more creative.

Last year the national written winner, Justin Ahmann, got 25+16 = 41. Most years that woulda put him in like 8th place.

well, considering last year's scores, 40-46's would probably get top 3...

[quote="mathcrazed"]Does your friend happen to have the username dynamo729 on AoPS =P?

What place do you think 40-46s would get you... My friend (who is good at making "smart mistakes" =P and is horrible at math) wants to know. =D[/quote]

Maybe. I'll say undefined.

The problems, as far as I can tell, are getting less and less about arithmetic and more and more about actual problem solving. Many people say that this is making them more difficult, but it just depends on the way your mind thinks. For example, I thought this year's problems were much easier than usual, while all of my friends and teammates told me it was much harder. I really dislike the formula/arithmetic types of problems, and like where they are going with this. It's making competitions much easier for me  :D .

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191755

there might be some states left?

edit your post for difficulty please.

i know, its hard  :wink:

[quote="1=2"]I get into the 30's. I will not win nationals. :([/quote]

well it all depends. For example, a 34 or so will get you into countdown at nationals. However, most people who do well at practice end up getting very nervous at the competition and do very badly.

About Turkishgenius, the years in general get harder, for example, at nationals, 2005 was considered the hardest year, yet 2006 and 2007 were much harder than the tests before 05.


As avea o intrebare :
Daca \[\displaystyle ABCD \] e un patrulater inscriptibil si \[\displaystyle M,N,P,Q\]mijloacele laturilor\[AB,BC,CD,AD\] si luam in exteriorul patrulaterului \[\displaystyle M_1,N_1,P_1,Q_1 \ a.i. \ MM_1=\frac{AB}{2} \ , \ NN_1=\frac{BC}{2}\ , \ PP_1=\frac{DC}{2}\ , \ QQ_1=\frac{AD}{2} \.\]Putem demonstra ca \[\displaystyle M_1N_1P_1Q_1\] inscriptibil ?

Sau, cu alte cuvinte, construim patrate pe laturile patrulaterului, in exteriorul lui, si $M_1,$etc. sunt centrele patratelor, nu?

Am facut un desen pe calculator, in Euklides si imi pare rau, dar nu pare sa fie adevarat in general.

daca construiesti patrate in afara patrulaterului pe laturile sale atunci $MM_1$,etc. vor fi perpendiculare pe laturi.caz particular.

ce se *poate* demonstra este ca diagonalele acestui patrulater, $ M_1N_1P_1Q_1 $, sint perpendiculare si au lungimi egale.

Daca $X,\ Y,\ Z,\ T$ sunt centrele patratelor construite in exteriorul patrulaterului convex $ABCD$ pe laturile $[AB],\ [BC],\ [CD],\ [DA]$ respectiv, atunci :
[b]1).[/b] $XZ\perp YT;\ XZ=YT;$
[b]2).[/b] Mijloacele diagonalelor $[AC],\ [BD]$ si mijloacele segmentelor $XZ,\ YT$ sunt varfurile unui patrat.

[u]Observatie.[/u] Aceasta clasica problema apare frecvent in ilustrarea metodei de  rezolvare vectoriala sau a metodei de utilizare a numerelor complexe.


algebra
polynomial
limit
search
calculus
calculus computations

Suppose $ f(x)$ is a polynomial of degree $ n$ such that $ f(x)\geq  0$ for each $ x\geq 0$  Prove that $ f(x)\plus{}f'(x)\plus{}f'(x)\plus{}...\plus{}f^{(n)}(x)\geq 0$
for all $ x\geq 0$

Let $ g(x) \equal{}\left( f^{(0)}(x)\plus{}\cdot\plus{}f^{(n)}(x)\right) e^{\minus{}x}$. Then $ g'(x) \equal{}\minus{}f(x)e^{\minus{}x}\leq 0$ and $ g$ is decreasing on $ x\geq 0$.

Also, $ \lim_{x\to\infty}g(x) \equal{} 0$, implying $ g(x)\geq 0$ for all $ x\geq 0$.

Therefore, $ f^{(0)}(x)\plus{}\cdot\plus{}f^{(n)}(x)\; \equal{}\; g(x)e^{x}\;\geq\; 0$.

Actually, mdk has posted this before (and quite recently I might add): http://www.mathlinks.ro/viewtopic.php?search_id=1657165616&t=159238

That makes me wonder: do you try to solve the problems you post or even bother reading the solutions posted?

[quote="perfect_radio"]
That makes me wonder: do you try to solve the problems you post or even bother reading the solutions posted?[/quote]
No, this is the 3rd time I see mdk do this. And I wonder why he does this. Ask JBL he can probably come up with many more such instances.

Since sylow_theory mentioned me: mdk posted six questions yesterday.  Of these, I have seen five appear on the forum.  This one and one other have definitely been posted by mdk before.  As far as I can tell, mdk solves only a small fraction of the problems s/he posts and reads only a small fraction of the threads s/he starts, and never (or hardly ever) posts on a thread that s/he didn't start.


Hi. Does anyone know the top team scores for last year?

Also, if you are:

1. Mathcounts age
2. Living in NJ
3. Interested in being on a Mathcounts team
4. Homeschooled

Please contact me ASAP.

Thanks.

I happen to know the top team score from last year.

Thomas R. Grover Middle School

46,35,34,31 Team:20 Total: 56.50

Anybody know top individual scores for Bergen Chapter?

I don't remember exact numbers, but I know the #1 guy got 46/46, and there were largely a bunch of high 40s.


Hi ,
prove that if  |ax² + bx + c| <= 1 for every x belonging to [-1 1] we then have  |cx² + bx +a| <= 1 for every x belonging to [-1 1] .

Hi again , can someone just give some idea or a plausible way to proceed , thanks in advance. :roll:

Inequality is not true,be cause $ |\minus{}2x^2\plus{}1|\leq 1 ,x\in [\minus{}1,1]$ but $ |x^2\minus{}2|\nleq 1,x\in[\minus{}1,1]$

  Correct is :$ |ax^2\plus{}bx\plus{}c|\leq 1$ for any $ x\in [\minus{}1,1]$, then $ |cx^2\plus{}bx\plus{}a|\leq2$ for any $ x\in [\minus{}1,1]$

 Indication:    $ cx^2\plus{}bx\plus{}a\equal{}c(x^2\minus{}1)\plus{}(a\plus{}b\plus{}c)\frac {1\plus{}x}{2}\plus{}(a\minus{}b\plus{}c)\frac{1\minus{}x}{2}$

A similar problem was proposed in the 1996 Spanish National Olympiad:

Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, $ g(x) \equal{} cx^2 \plus{} bx \plus{} a$.  If $ |f( \minus{} 1)|,|f(0)|,|f(1)|\leq1$, then show that $ |f(x)|\leq\frac {5}{4}$, $ |g(x)|\leq2$ for all $ x\in[ \minus{} 1,1]$.

[quote="Marius Mainea"]Inequality is not true,be cause $ | \minus{} 2x^2 \plus{} 1|\leq 1 ,x\in [ \minus{} 1,1]$ but $ |x^2 \minus{} 2|\nleq 1,x\in[ \minus{} 1,1]$

  Correct is :$ |ax^2 \plus{} bx \plus{} c|\leq 1$ for any $ x\in [ \minus{} 1,1]$, then $ |cx^2 \plus{} bx \plus{} a|\leq2$ for any $ x\in [ \minus{} 1,1]$

 Indication:    $ cx^2 \plus{} bx \plus{} a \equal{} c(x^2 \minus{} 1) \plus{} (a \plus{} b \plus{} c)\frac {1 \plus{} x}{2} \plus{} (a \minus{} b \plus{} c)\frac {1 \minus{} x}{2}$[/quote]

Thanks vm Marius Mainea for this counterexample and for the pretty decomposition as well . i think that it's ok now . What a pity our school manuals still contain such errors . :maybe:

thanks Daniel73 , i'll try to work on this one.

A sort of related problem from the 2008 Canadian Math Olympiad Repechage Problem Set:
9. Suppose that a, b, c are real numbers with $ \minus{} 1 \le ax^2 \plus{} bx \plus{} c \le 1$
for all real numbers x where $ \minus{} 1 \le x \le 1$. Prove that $ \minus{} 2 \le a \le 2$.

Similar problem from Japan, here.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=145147&search_id=1371640373[/url]


Suppose that a river is the set $ \{(x,y),x\in\mathbb{R},0\leq y\leq l\}$. The water flows in the middle of the river at the speed $ v_1$, in the shore at speed $ v_2\leq v_1$ and the speed decreases linearly as the distance to the middle of the river increases. Swimmer is on the shore $ (y \equal{} 0)$ and wants to swim to the point $ (0,l)$ using  as little energy as possible. At calm water the swimmer swims at the speed $ v_3$. Where the swimmer must start his swimming trip and how long does the swimming trip takes?

Note that the swimmer does not need to swim always to the same direction. He may turn his direction continuously.

Could you define what you mean by energy here, a bit better?

Energy means the energy that a swimmer have to use to manage his swimming trip. I don't know enough physiology to compute how much it takes energy to swim in a flow.

Or do you mean energy as in kinetic energy?

Okay, this is what I make of the problem. Assuming the river is flowing from right to left, let the swimmer start at the point $ (d,0)$ on the shore and swim forwards with speed $ v_3$. If he swims at that constant forward speed all the way, he will reach the opposite shore in a time of $ \frac {l}{v_3}$. In the meantime, he lets the current carry him horizontally. He must therefore calculate his starting point in such a way that the current will carry him the exact horizontal distance $ d$ in the time $ \frac {l}{v_3}$.

The speed $ v$ of the current at distance $ y$ from the shore is easy to calculate.

$ v\ = \ \{\begin{array}{ll}v_2 + \frac {2y}{l}\,(v_1 - v_2) & 0\leq y\leq\frac {l}{2} \\
 \\
v_1 - \frac {2}{l}\,(y - \frac {l}{2})(v_1 - v_2) & \frac {l}{2}\leq y\leq l \end{array}$

But since the forward speed is constant, $ y=v_3t$.

Hence

$ d\ = \ \int_0^\frac {l}{2v_3}{[v_2 + \frac {2v_3t}{l}\,(v_1 - v_2)]}\,\mathrm{d}t + \int_{\frac {l}{2v_3}}^\frac {l}{v_3}{[v_1 - (\frac {2v_3t}{l} - 1)(v_1 - v_2)]}\,\mathrm{d}t$

And if my calculation is correct, this should turn out to be $ d = \frac {l}{2v_3}\,(v_1 + v_2)$.

I don’t know if this is the most energy-efficient solution but I sure hope it makes sense. :roll:

Well, that would be good, but you didnt solve the problem.
The component of the velocity $ \vec v_3$ perpendicular to the velocity of the river is not at all constant. since the direction of $ v_3$ can always change.
Maybe this is the most energy efficient, way if that is constant, but I dont see a proof of that, which clearly is the lone difficulty of the problem.
Again, Puuhikki, could you specify by what you mean, energy, here?
kinetic energy, or potential energy, or the work done by the swimmer...

[quote="yeahnigga"]Again, Puuhikki, could you specify by what you mean, energy, here?
kinetic energy, or potential energy, or the work done by the swimmer...[/quote]
The work done by the swimmer.

the swimer is in y=0 and x=?
and y is refer (droit qui paralle avec la longueur de revier)? im sorry i can write all that with english

I guess that the swimmer should not turn when swimming.


[quote="Magic Man"]Im having some trouble at the moment with a certain questions in log functions could some one explain them to me or help me out with them please.

If Log(a^2) = x and log(1/b) = y, find an expression in x and y for the following:

1. log a/b
2. log a^3
3. log (ab)^1/2
4. log (5ab)

Their all to the common base of 10
Thank you for any help
Magic Man[/quote]

First make them easier to deal with

[hide="Like this"]
$ log(a^2)\equal{}2log(a)\equal{}x$ and $ log(\frac{1}{b})\equal{}log(1)\minus{}log(b)\equal{}\minus{}log(b)\equal{}y$
[/hide]

Then you can start to solve the problems

[hide="1"]
$ log(\frac{a}{b}) \equal{} log(a)\minus{}log(b)\equal{}\frac{x}{2}\plus{}y$
[/hide]

[hide="2"]
$ log(a^3)\equal{}3log(a) \equal{} \frac{3}{2}x$
[/hide]

[hide="3"]
$ log((ab)^{\frac{1}{2}}) \equal{} \frac{1}{2}log(ab)\equal{}\frac{1}{2}(log(a)\plus{}log(b))\equal{}\frac{1}{2}(\frac{x}{2}\minus{}y)$
[/hide]

[hide="4"]
$ log(5ab)\equal{}log(5)\plus{}log(a)\plus{}log(b)\equal{}log(5)\plus{}\frac{x}{2}\minus{}y$
[/hide]

[quote="Magic Man"]i kinda understand what you are doing but where do you get the 2 from at the end each equation?[/quote]

Do you mean the $ \frac{x}{2}$

That is because $ x\equal{}log(a^2)$. We know this is the same as $ 2.log(a)$ and so everytime I need to reference $ log(a)$ I have to use $ \frac{x}{2}$ (since this is $ \frac{2log(a)}{2}$

I hope thats what you meant?

Oh alright i see that u put the 2 on the bottom because so then the to two's cancel each other out okay! thank you very much for your help i get it now

No problems, I think most people have trouble with logs when they first see them (well, not so much with the rules, but anyway)

One last question Number three isent acctually written like that i acctually rearanged the formula because i dont have square root on ma computer
the original question was:
Log square root of (ab)
will it give me the same answer as u gave me for 3

Yes, thats how powers work $ x^{\frac{1}{2}=\pm \sqrt{x}}$ remember, the square root symbol only means the positive root. Also, you can use LaTeX to put the square roots in if you need to[/u]

For future reference, [code]\log[/code] gives $ \log$, which looks better than $ log$. Text is awful in latex.


Let $ x,y,z \geq 0$ such that $ \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx} \plus{} \sqrt {xyz} \equal{} 4$
Prove that 1)$ \sum \sqrt x \geq \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx}$
                2)$ x^2 \plus{} y^2 \plus{} z^2\geq \sum \sqrt [4] {x^3y ^3}$

1)  Let  $ \sqrt {x} \equal{} \frac {2a}{b \plus{} c}$ , $ \sqrt {y} \equal{} \frac {2b}{c \plus{} a}$  ,  $ \sqrt {z} \equal{} \frac {2c}{a \plus{} b}$

 and 1)  reduces to
\[ \sum {\frac {2a}{b \plus{} c}}\ge \sum {\frac {4ab}{(b \plus{} c)(c \plus{} a)}}
\]
or
\[ \sum {a(a \plus{} b)(a \plus{} c)}\ge 2\sum {ab(a \plus{} b)}
\]
or
\[ \sum_{cyc}{a^3} \plus{} 3abc\ge \sum_{sym}{a^2b}
\]
wich is Schur Inequality.

 2)
$ LHS\ge \frac {1}{3}(\sqrt [4]{x^6} \plus{} \sqrt [4]{y^6} \plus{} \sqrt [4]{z^6})(\sqrt [4]{x^2} \plus{} \sqrt [4]{y^2} \plus{} \sqrt [4]{z^2})\ge \frac {1}{3}(\sum {\sqrt [4]{x^3y^3}})(\sum {\frac {2a}{b \plus{} c}})\ge RHS$


number theory
greatest common divisor
least common multiple

Determine all pairs $ (a,b)$ in positive integers which satisfy next equation:

$ LCM(a,b)\plus{}GCD(a,b)\plus{}a\plus{}b\equal{}ab$, $ a\ge b$.

where $ LCM$ means the Least Common Multiple,and $ GCD$ does the Greatest Common Divisor of $ a,b$

Let $ (a,b) \equal{} g, a \equal{} gA,b \equal{} gB$. Then $ [a,b] \equal{} gAB$ and our equation becomes
\[ gAB \plus{} g \plus{} gA \plus{} gB \equal{} g^2AB
\]
which is equivalent to
\[ (A \plus{} 1)(B \plus{} 1) \equal{} gAB
\]
Hence $ AB\mid (A \plus{} 1)(B \plus{} 1)$. But since $ (A,A \plus{} 1) \equal{} 1$ we conclude that $ A\mid B \plus{} 1$ and $ B\mid A \plus{} 1$. Let $ B \equal{} kA \minus{} 1$. Then $ kA \minus{} 1\mid A \plus{} 1$ implies $ kA \minus{} 1\le A \plus{} 1\Leftrightarrow (k \minus{} 1)A\le 2$. Therefore $ (k,A) \equal{} (3,1),(2,2),(2,1),(1,2)$. Hence $ (k,A,B)\equal{}(1,2,1),(3,1,2),(2,2,3),(2,1,1)$. Thus we get the pairs $ (a,b) \equal{} (3,6),(4,6),(4,4)$.

Remark: I worked with $ A\mid B \plus{} 1$ so I got the pairs where $ a\le b$. For $ a\ge b$, we can simply reverse each pair.


For any positive real numbers $a,\ b$ and $c$.
\[ \frac{c}{a(b+c)}+\frac{a}{b(c+a)}+\frac{b}{c(a+b)} \ge \frac{9}{2(a+b+c)}  \]

Cauchy + AM-GM?

$((b+c)+(c+a)+(a+b))\left( \frac{c}{a(b+c)}+\frac{a}{b(c+a)}+\frac{b}{c(a+b)} \right ) \geq \left(\sqrt{\frac ca} + \sqrt{\frac ab} + \sqrt{\frac bc} \right)^2 \geq 9$


I don't know how to solve this without a computer program...

I'm horrible at manipulating sums - maybe someone can give me a hint?

What is $\displaystyle\sum_{k=1}^{360} \frac{1}{k\sqrt{k+1} + (k+1)\sqrt{k}}$ in lowest terms?

it telescopes.  the denominator can be written as $\sqrt{k}\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})$.  then go from there.

how come that contest doesnt seem to be too popular?  geehoon told me about it, but we couldn't find anything on it.  i'd think that since anyone can take it a lot of people on this site would have known about it.

Ah, thank you.

I figured it out now.

It's [hide]$\frac{18}{19}$[/hide].

It's so nice how all the terms cancel...

Hmm... I still don't get it, can someone please post a full solution.

[hide]
As suggested, the summand simplifies to 1/ (:sqrt: k :sqrt: (k+1) ( :sqrt: k +  :sqrt: (k+1)).  If you rationalize the denominator, it simplifies further to 1/  :sqrt: k - 1/ :sqrt: (k+1) which telescopes.[/hide]


Each of the integers 1, 2, 3, ..., 16 is written on a separate slip of paper and these slips are placed in a pile. Jillian will randomly draw slips from the pile without replacement and will continue drawing until two of the numbers she has drawn from the pile have a product that is a perfect square. What is the maximum number of slips that Jillian can draw without obtaining a product that is a perfect square?

Beastly hard problem.

Well just think...  what makes a number square?

9=3^2
16=2^2 x 2^2
36=2^2 x 3^2

Now if we split all the numbers 1-16 to their prime factorization, which numbers absolutely cannot be together? Which number "pose a threat" to other numbers? If you think about it logically, I think the answer is 11?

Darn, I cant explain it well. Would someone else like to try?

The numbers with exactly 1 prime factor, exactly 2 [u]different[/u] prime factors, and 1 are the only numbers that can be drawn.  This is because if you multiply 1 to anything, it'll be the same, when you have a number with exactly one prime factor, it is prime and thus isn't a square, and when you multiply a number with exactly 1 prime factor with a number that has exactly 2, it still cannot be a square.  For example, 3*15=45, which isn't a square.

The primes between 1 to 16 are 2,3,5,7,11,13; a total of 6 numbers.

The numbers with exactly 2 distinct prime factors between 1-16 are 2*3,2*5,2*7,3*5; for a total of 4 numbers.  Any other product of the previous primes we showed is greater than 16.

Lastly, we have to include 1. (Which is, of course, 1 number).  Thus, the greatest numbers of slips that can be drawn is $ 6 \plus{} 4 \plus{} 1 \equal{} \boxed{11}$.

I failed this one at state because I misread the question and thought it was the numbers from 1-12.  :(  :wallbash_red:  :wallbash:

so you are saying after you pick the primes, one, and the products of the primes less than 16, the rest of the numbers products are going to be perfect squares?

I am a noob.

wait no, that is not teh problem.
you want to find out how many slips she has to pick so that none of the pairs of products out of the ones that she picked are perfect squares?


Two players play the following game: first player writes down one digit, after his move second player write one digit to the left or right to the previously written digit, after his move first player write one digit to the left or write to previously written number, etc. Prove that the first player can play in that way such that after each of the moves of the second player, the obtained number is not the exact square.

I think you may have confused the word "write" with the word "right"
Please edit your post :)

Daniel


The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:

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

[hide="Click for solution"]
Let $ P_1$ and $ P_2$ be the perimeters of the smaller and larger triangles, respectively.  Let the radius of the circle be $ r$ and the side length of the large triangle be $ s$.  Then $ r\equal{}\frac{s\sqrt{3}}{2}$ and $ s\equal{}\frac{2r}{\sqrt{3}}$, so $ P_1\equal{}2r\sqrt{3}$ and $ P_2\equal{}3r\sqrt{3}$, so the desired ratio is $ 2: 3$, or $ \boxed{\textbf{(E)}}$.
[/hide]

It may be faster to just note that the ratio of any two corresponding lengths of the two triangles must be equal to the ratio of the altitudes, which by the properties of medians and centroids is always $ 2: 3$.


guys, i dont know how you ppl remember all the reagents and their uses but i think its a real pain in the ass. besides there are too many reagents and there are no proper sources to search for them. so why dont we start posting all the reagents we know( at least the important ones) over here. we could organise it later. i know it is boring to type all the reagents but it would be useful for everyone especially for those trying for jee n the olympiads

sorry posted it twice by mistake. (at least will get attention)


Hi,

I read over the parts about function transformation in the Intro to Algebra book, but I am still not completely sure why the transformation of something like $ f(x \plus{} 3)$ moves the graph of $ f(x)$ three to the left, and something like $ f(x \minus{} 3)$ moves the graph of $ f(x)$ three to the right.

Thank you for the help.

Ahh...never mind. I seem to found a subtle explanation for why this works convincing myself. Other perspectives are certainly welcomed.

Here's how I am trying to picture it:

Let there be a function such that $ f(x) \equal{} x\plus{}1.$ Let us now transform this function as follows:
$ f(x\plus{}3) \equal{} (x\plus{}3)\plus{}1.$ A horizontal shift of 3 units in this case corresponds to the decrease of the x value by three. We see, that in order for the first function to acheive $ f(0),$ we merely have to have $ x\equal{}0.$ However, in the second function, $ x\equal{}\minus{}3$ in order for $ f(x\plus{}3) \equal{} f(0),$ therefore this corresponds to the x-value being three lower than in the $ f(x).$ Thus, it is a shift 3 units to the left.

The book perhaps had the same/similar explanation that I failed to understand?  :maybe:

The way I think about it when you have a function, let's say $ f(x)\equal{}x$ and $ f(x)\equal{}x\plus{}1$, in order for $ y\equal{}0$, you must have $ x\equal{}0$ for the first and $ x\equal{}\minus{}1$ for the second.  When you would plot the points, the second one would be to the left of the first.  Just think about like that.

$ f(x)\equal{}x\plus{}a$
$ x$ is translated $ \minus{}a$ units
If $ \minus{}a>0$, Then $ a$ moves right.
If $ \minus{}a<0$, Then $ a$ moves left.

Or if you are adding a quantity to $ x$, then your graph will be translated to the left by that quantity.
The opposite holds true when you subtract.

Hope this helps, even though you basically said the same thing in the last post.

Also we can take a look at the concept of roots. $ f(x)\equal{}x^2$ is 'centered' (if you will) at the origin, but $ f(x\minus{}3)$ is centered at $ x\equal{}3$. Applying this to a more complicated function, we have a set of roots at $ x\minus{}r_i$. Rather than showing the function, observe the resulting roots when we apply $ f(x\minus{}s)\equal{}x\minus{}r_i\minus{}s\Rightarrow x\equal{}r_i\plus{}s$.

Clearly, the functions roots shift in the direction of $ \frac{s}{|s|}$, as does the function itself.

I would assume the $ i$th root of $ f(x).$

As RoFlLoLcOpT said, $ f(x)$ has zeros at $ x\equal{}r_i$


algebra
function
domain
logarithms
geometry
search
complex analysis

Could anyone help me with these problems?

1. Map 1-1, conformally the upper half of the unit disk onto the upper half-plane, $ \{z\in\mathbb{C}|Im(z)>0\}$

2. Map 1-1, conformally the circular sector $ 0<arg(z)<\pi/3,0<|z|<1$ onto the unit disk.

3. Find a Mobius transformation which maps the unbounded domain in the extended plane $ \mathbb{C}_{\infty}$ whose boundary consists of $ |z\minus{}5|\equal{}4$ and $ |z\plus{}5|\equal{}4$ onto the annulus $ \{w|1<|w|<R\}$. Evaluate R.

4. Map that portion of the z-plane which lies outside the circles $ |z\minus{}1|\equal{}1$ and $ |z\plus{}1|\equal{}1$, 1-1, conformally onto the exterior of the unit disk so that the points at infinity correspond to one another.

Much appreciated,
ftr

1. A logarithm map might help you get started. (After that, I would rescale the strip and exponentiate again to get the punctured disc, which I can Mobius transform in a suitable way, but there may well be a more direct approach.)

2. Again, it looks like a logarithm might help.

3. It's a theorem in olympiad geometry that any two nonintersecting circles can, under a suitable inversion, be made to be concentric. This is really just that geometry problem and it might help to ignore the complex analysis trappings. (the question of why R is uniquely determined is harder; search this forum for "extremal length" and you'll get a post where I tried to give a solution)

4. Is there an obvious point about which to invert?

Thank you for the hints.

However, I'm still stuck with 3 and 4. I'm looking at choosing 3 distinct points to form the transformations and intersections of domains.


let $n$ is natural number and $a,b$ are natural numbers and relative togther are prime prove that        $ \frac{1}{a}\ + \frac{1}{a+b}\ + \frac{1}{a+2b}\ + ... + \frac{1}{a+nb}\ $   is'nt natural number

Postet at least one time before (don't know if solved...).
But please tell me: what is your book¿

This problem was one of the hardest problem , that has been discussed in iran olympiad and lots of our team members couldent solved , but a nice , elegant solution by[b]Hesame Mahdavifar[/b](IMO's $2002,2003$ silver medalist) was given in one of the magezines published by iran olympiad instructores, Algebra you can go to that magazine (i think number 16, the artcile with the name :a hard problem from primery number theory), if anyone is interested in that solution , i can copy that in  the forum.

well if you got time, lomos_lupin, please do :D
just make sure that it is not an entire copy of it because they might be copyrighted... :D

ye, palytpxine, i must be carefull for that :lol:  but to speak the truth we dont have brute rull of copyright here.
 :|
I have found it as [b]zetax[/b] said. there is a solution by [b]ali[/b] in this.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37417[/url]

but if you guys what a real peice of [b]Art of problemsolving[/b] , you must see the
solution of [b]hesam mahdavifar[/b]

excellent piece of solution by ali, but i still want to see the solution of hesam mahdavifar :D

See here for a solution:
https://artofproblemsolving.com/community/u257113h1331945p8387502


calculus
derivative
trigonometry
function
calculus computations

Hello everyone,

This is my first post. I would like to know how to solve the following problem quickly, beside obtaining the explicit expression for the $13th$ derivatives and then substituting $x=0$:

$\frac{d^{13}}{dx^{13}}\arcsin{x}|_{x=0}$.

Thank you to everyone in advance.

Do you know the power series for arcsin?

Sorry, I don't know how to obtain the power series of arcsin

$f(x)=\sin^{-1}x$

$f'(x)=\frac{1}{\sqrt{1-x^{2}}}=(1-x^{2})^{\frac{-1}{2}}\\=1+\frac{1}{2}x^{2}+\frac{\frac{-1}{2}\frac{-3}{2}}{2!}(-x^{2})^{2}+\cdots+\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2^{k}\cdot k!}x^{2k}+\cdots$
(using binomial expansion .

so we can write

$f'(x)=1+\sum^{\infty}_{k=1}\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2^{k}\cdot k!}x^{2k}$

so if we differentiate $f(x)$  , $r$ times , we get

$f^{(r)}(x)=\sum^{\infty}_{k=1}\frac{(2k)(2k-1)(2k-2)\cdots (2k-r+2)\cdot 1\cdot 3\cdot 5\cdots (2k-1)}{2^{k}\cdot k!}x^{2k-r+1}$

Then in this case , let $r=13$ and notice that that the coefficient of $x$ power 1 or above will not affect since we want to find $f^{(13)}(0)$ . So we only need to consider when $r=13$ and $2k-r+1=0$ so $k=6$ , for this value you can find

$f^{(13)}(0)=\frac{12!11!!}{2^{6}6!}$

where $11!!=1.3.5.7.9.11$


Your paycheck each week is  500.00 dollars . If you want to save 2,000.00 dollars in one year, how much should you set aside from your paycheck each week?

you would set aside 2000/52 which, when rounded, equals $38.46 each week.

no problem. happy birthday by the way. :)

Hello to all the members of the Forum. This is Smith. I am here to provide you information on Jobs and job searches. Are you tired of in finding a new jobs. Then dont worry we are here to provide you information on Jobs and job searches.
______
Smith

[url=http://jobs.bizoppjunction.com]Job Search[/url]

[quote="cyao15"]isn't this a MATH forum not a JOB forum or and ADVERTISING forum

:play_ball:[/quote]

Sometimes I think that there should be a test to become a member (easy problems), and after 25 productive posts or so, they can access all forums.  There could be a beginners forum.  Then again, not many advertisers are here...

[quote="cyao15"]you know what?
that's an awesome idea
i agree, there should be a sort of test thing
maybe like old moems questions or whatever.[/quote]

But some people need to ask questions and aren't good enough to answer the MOEMS.

You guys, drop it.  If you want to propose something, ask it in the Questions, [b]Suggestions[/b], and Announcements forum.


$ A$ is the set of positive multiples of 6 less than 600, and $ B$ is the set of positive multiples of 8 less than 800. How many elements are in $ A\cap B$?

We need to find the number of numbers that are in both sets.  First, we do $ LCM(6, 8)\equal{}24$.  Then, we need to find how many multiple of $ 24$'s there are.  We use set $ A$, since there are less numbers here than in set $ B$.  Therefore, we have $ \lfloor600\div24\rfloor\equal{}\boxed{25}$.  But wait!  The restriction in set $ A$ says the multiples of $ 6$ [i]less than[/i] $ 600$, so $ 600$ itself wouldn't work.  So we have $ 25\minus{}1\equal{}\boxed{\boxed{24}}$


For positive real numbers $ a,b,c$ prove that \[ \frac{a}{(b\plus{}c)^4}\plus{}\frac{b}{(c\plus{}a)^4}\plus{}\frac{c}{(a\plus{}b)^4}\ge \frac{3}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)}\]

[cyclic sums]

It is easy to prove :
\[ \frac { (a)(a + b)(a + c) }{ (b + c)^3 } + \frac { (b)(b + a)(b + c) }{ (a + c)^3 } - \frac {(a + b)^2}{ (a + c)(b + c) } \\
\[= \frac {(a + b)(a - b)^2}{(c + a)^3(c + b)^3}*(a+b+2c)(a^2+b^2+c^2+2ac+2bc+ab)\ge 0
\]

Hence after clearing the denominator on the LHS,
\[ \sum \frac {a(a + b)(a + c)}{(b + c)^3}\ge \frac {1}{2}*\sum \frac {(b + c)^2}{(a + c)(b + c)}\ge \frac {3}{2}
\]
where the last step is am-gm.

[quote="nayel"]Interesting. Very nice proof Alex! :)  I've tried to create stronger inequalities of the same type but your proof seems to kill them all. :(  :D  Now finally I have the following. Have fun! :wink:

For $ a,b,c > 0$ prove that
\[ \frac {a}{(b \plus{} c)^4} \plus{} \frac {b}{(c \plus{} a)^4} \plus{} \frac {c}{(a \plus{} b)^4}\ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)(a \plus{} b)(b \plus{} c)(c \plus{} a)}
\]
[/quote]
By Holder inequality, we have:
$ (a(b\plus{}c)\plus{}b(c\plus{}a)\plus{}c(a\plus{}b))^4(\frac {a}{(b \plus{} c)^4} \plus{} \frac {b}{(c \plus{} a)^4} \plus{} \frac {c}{(a \plus{} b)^4}) \ge (a\plus{}b\plus{}c)^5$
Or:
$ 2(ab\plus{}bc\plus{}ca)(\frac {a}{(b \plus{} c)^4} \plus{} \frac {b}{(c \plus{} a)^4} \plus{} \frac {c}{(a \plus{} b)^4}) \ge \frac{(a\plus{}b\plus{}c)^5}{2^3(ab\plus{}bc\plus{}ca)^3}$
We will prove that:
$ (a\plus{}b\plus{}c)^3(a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 2^3(ab\plus{}bc\plus{}ca)$
We also have (a famous problem!):
$ \sqrt[3]{\frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{8}} \ge \sqrt{\frac{ab\plus{}bc\plus{}ca}{3}}$
And: $ a\plus{}b\plus{}c \ge \sqrt{3(ab\plus{}bc\plus{}ca)}$
and problem is solved!

I made a different solution .

By the rearrengement inequality

$ 3(\sum\frac{a}{(b\plus{}c)^4})\geq (\sum\frac{a}{b\plus{}c})(\frac{1}{(b\plus{}c)^3})$

Now use Nesbitt inequality for the first term and AM-GM for the second to obtain the result .

Yes silouan that was my solution too. And note that it also solves my second proposed problem in this thread. :)

[quote="NguyenDungTN"]We also have (a famous problem!):
$ \sqrt [3]{\frac {(a + b)(b + c)(c + a)}{8}} \ge \sqrt {\frac {ab + bc + ca}{3}}$
[/quote]

Well, I never saw this before and thought I'd give it a try:
[hide]Let $ x=a+b,y=b+c,z=c+a$. Then notice that \[ 4(ab+bc+ca)=2(xy+yz+zx)-(x^2+y^2+z^2)\]
So we need to show that \[ \sqrt[3]{xyz}\ge\sqrt{\frac{2(xy+yz+zx)-(x^2+y^2+z^2)}{3}}\]
Which is equivalent to
\[ x^2+y^2+z^2+3(xyz)^{\frac 23}\ge 2(xy+yz+zx)\]
This follows from Schur and AM-GM:
\begin{eqnarray*}(x^{\frac 23})^3+(y^{\frac 23})^3+(z^{\frac 23})^3+3(xyz)^{\frac 23} &\ge & \sum_{\text{sym}}\left(\left(x^{\frac 23}\right)^2y^{\frac 23}+x^{\frac 23}\left(y^{\frac 23}\right)^2\right)\\\
&\ge& \sum_{\text{sym}}2xy \end{eqnarray*}
[/hide]
Are there other solutions to this one? Just wondering.


abstract algebra
superior algebra
superior algebra unsolved

Given any finite group $ G$ show there exists fields $ F\leq E$ with $ E$ normal over $ F$ such that $ \mbox{Gal}(E/F)\simeq G$.

What about if $ F\equal{}\mathbb{Q}$, it is my understanding that this an unsolved problem in mathematics?

[quote="Abel"]When F=Q, it seems to be easy only when G is abelian.[/quote]
That the Hilbert's 12th Problem. [url=http://en.wikipedia.org/wiki/Kronecker%E2%80%93Weber_theorem]Here[/url]. (In same form or another).

But isn't Hilbert's problem for fields other than the rationals? I meant the simple case when F=Q (i.e. the rationals) and G is abelian (and finite).


quadratics
modular arithmetic
number theory unsolved
number theory

Show that if $ x$ is not divisible by $ 3$ then $ 4x^2+3$ has at least one prime factor of the form $ 12n+7$. Deduce that there are infinitely many primes of this sort.

[hide="What I have so far:"]
Let $ p = 12n+7$.

Since $ p \neq 3$ and $ 3$ and $ p$ are both odd, the quadratic reciprocity law gives
$ \begin {eqnarray*}\left(\frac{-3}{p}\right) &=& \left(\frac{-1}{p}\right) \left(\frac{3}{p}\right) \\
&=& (-1)^{\frac{p-1}{2}}\left(\frac{12n+7}{3}\right)(-1)^{\frac{3-1}{2}*\frac{p-1}{2}} \\
&=&  (-1)^{6n+3}\left(\frac{1}{3}\right)(-1)^{6n+3} \\
&=& (-1)(1)(-1) \\
&=& 1$

Hence, $ y^2 \equiv -3 \pmod p$ has a solution.

From here, how do I get $ 4x^2 \equiv -3 \pmod p$ also has a solution? How would I know letting $ y = 2x$ where $ x$ is not divisible $ 3$ will guarantee a solution? Thanks.

For the infinite prime part of the problem, I guess you can just assume finitely many primes of the form $ 12n+7$, denoted by $ p_1$, $ p_2$, ... , $ p_r$

Consider the number $ 4(p_1p_2...p_r)^2 + 3$. From above, there is a prime factor of the form $ 12n+7$ that divides this number, but is not equal to any of the $ p_1, p_2, ... , p_r$. Hence, the contradiction shows infinitely many primes of the form $ 12n+7$.[/hide]

[hide]You want to go the other way instead; assume $ p$ is a prime dividing $ 4x^2 \plus{} 3$, then $ \minus{} 3$ is a square mod p. Then

$ \left(\frac { \minus{} 3}{p}\right) \equal{} \left(\frac { \minus{} 1}{p}\right)\left(\frac {3}{p}\right) \equal{} ( \minus{} 1)^{\frac {p \minus{} 1}{2}}\left(\frac {p}{3}\right)( \minus{} 1)^{\frac {p \minus{} 1}{2}}( \minus{} 1)^{\frac {3 \minus{} 1}{2}} \equal{} \left(\frac {p}{3}\right)$.

Since $ \left(\frac {p}{3}\right) \equal{} \left(\frac { \minus{} 3}{p}\right) \equal{} 1$, p must be a square mod 3, so $ p\equiv 1\mod 3$.

You also know that there must be at least one prime dividing $ 4x^2 \plus{} 3$ that is $ 3\mod 4$ (otherwise $ 4x^2 \plus{} 3$ would be $ 1\mod 4$). Take this prime $ p$, and since $ p\equiv 1\mod 3$ and $ p\equiv 3\mod 4$, you know that $ p\equiv 7\mod 12$.[/hide]


topology
superior algebra
superior algebra solved

HELLO can yuo help me?
Everey opensubset of a locally path space is itself locally path connected.

Being a locally path connect space is a local property which means that it transmits to any open subset.

The general statement is:
Let $L$ be a property of topological spaces such that a topological space $X$ has $L$ iff it has a base formed of open subsets each of which has $L$.
Let $U$ be an open subset of $X$. Then $U$ has $L$.

Proof: The open subsets of $U$ which are also in the base for $X$ form a base for $U$. The conclusion follows.


1. A man leaves for work at a random time b/w 7:00AM and 8:00Am. His paper arrives at his house at a random time b/w 6:30 AM and 7:30 AM. What is the probability that he has his newspaper before he leaves for work?

I draw 2 axes, one horizontal and one vertical, and label the horizontal one 6:30, 7:00, and 7:30 for the newspaper, and the vertical one 7:00, 7:30, and 8:00. What I'm confused about is what to shade in for the region where he gets the newspaper before he leaves for work. Could anyone please explain this one to me?

Thanks.

You want the graph $x < y$ (since the newspaper is supposed to arrive before he leaves) except remember the origin is a little shifted.

You want to shade above $x < y$ since as long as the man leaves at the same time or later (higher y) than the newspaper arrives, he will get it before he's gone.

If you're even more clever, another nice way to look at this problem is as follows:

The probability of the man leaving between 7 and 7:30 is 1/2
The probability of the paper arriving between 7 and 7:30 is 1/2

Together there is 1/4 chance this will happen.

If this happens, there is a 1/2 chance the paper will after before the man leaves.

Hence there's a 1/8 chance the paper will arrive after the man leaves, so a 7/8 chance the man will get his paper.


function
combinatorics proposed
combinatorics

Suppose we have some proteins that each protein is a sequence of 7 "AMINO-ACIDS" $A,\ B,\ C,\ H,\ F,\ N$. For example $AFHNNNHAFFC$ is a protein. There are some steps that in each step an amino-acid will change to another one. For example with the step $NA\rightarrow N$ the protein $BANANA$ will cahnge to $BANNA$("in Persian means workman"). We have a set of allowed steps that each protein can change with these steps. For example with the
set of steps:
$\\ 1)\ AA\longrightarrow A\\ 2)\ AB\longrightarrow BA\\ 3)\ A\longrightarrow \mbox{null}$
Protein $ABBAABA$ will change like this:
$\\ ABB\underline{AA}BA\\ \underline{AB}BABA\\ B\underline{AB}ABA\\ BB\underline{AA}BA\\ BB\underline{AB}A\\ BBB\underline{AA}\\ BBB\underline{A}\\ BBB$
You see after finite steps this protein will finish it steps.
Set of allowed steps that for them there exist a protein that may have infinitely many steps is dangerous. Which of the following allowed sets are dangerous?
a) $NO\longrightarrow OONN$
b) $\left\{\begin{array}{c}HHCC\longrightarrow HCCH\\ CC\longrightarrow CH\end{array}\right.$
c) Design a set of allowed steps that change $\underbrace{AA\dots A}_{n}\longrightarrow\underbrace{BB\dots B}_{2^{n}}$
d) Design a set of allowed steps that change $\underbrace{A\dots A}_{n}\underbrace{B\dots B}_{m}\longrightarrow\underbrace{CC\dots C}_{mn}$
You see from $c$ and $d$ that we acn calculate the functions $F(n)=2^{n}$ and $G(M,N)=mn$ with these steps. Find some other calculatable functions with these steps. (It has some extra mark.)

[hide=a)] This set is dangerous. Consider the word $NOO$, we have
$$ NOO \longrightarrow OONNO \longrightarrow OONOONN.$$
Since $NOO$ appears again in the right word, we can repeat this process ad infinitum.[/hide]
[hide=b)] Not dangerous. Suppose that we are given a word $w$ in $C$ and $H$ (if there are other letters we could view them as space and apply our argument to each word in $C$ and $H$). Since$w$ contains a finite number of $C$ we can use operation $1$ only finitely many times. Thus, if $w$ had infinitely many steps, there would exist a word $w'$ to which operation $2$ is applied infinitely many times. However, this operation reduces the number of pairs of letters $(X,Y)$, where $X=H \, Y=C$ and $X$ appears before $Y.$ Since there are only finitely many such pairs in $w'$, we get a contradiction.
[/hide]
Does anybody have an idea for c) and d) ?


summer program
MathPath
function
vector
inequalities

Source: MathPath

Prove that for all a, b, c, d, e, and f (very exciting variable names, huh?), that

$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{e^2+f^2}\ge\sqrt{a^2+c^2+e^2}+\sqrt{b^2+d^2+f^2}$

[quote="mathnerd314"]Yes! There is a deceptively simple solution to it all.

Hint: Think of a certain Greek guy.[/quote]
I know, Im thinking pythagorous, but Im also trying to learn basic music theory from google. I will do it later.  :(

Well ,hmm , this one is  advanced and its really over kill but  :D :
See this link for any defination you want :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975
WLOG Assume that $b \le a,c,d,e,f,$
Function $f(x)=\sqrt{x}$ is concave for $x \le 0$ .
we should prove $f(a^2+b^2)+f(c^2+d^2)+f(e^2+f^2) \ge f(a^2+c^2+e^2)+f(b^2+d^2+f^2)$
so according to Karamata ineqaulity we should prove :

                      $(a^2+b^2,c^2+d^2,e^2+f^2)  \prec  (a^2+c^2+e^2,b^2+d^2+f^2,0)$

which is equal to $b^2 \le c^2+e^2$ which is true .Problem solved.

[quote="mathnerd314"]there is a much easier way than the[b] incomprehensible calculations[/b] that lomos_lupin [color=red]just[/color][b] demonstrated[/b][/quote]
Which part is incomprehensible ?
I shall provide any needed explanations,if you want  (Yes i agree its a not easy ,but the links i gave can provide precise details)
I have already mentioned that this technic is not appropriate for this problem but ,I posted it to introduce this strong tool of solving ineqaulities .
This techin help you to solve the stronger problems like :

$\sqrt[2n]{a^2+b^2}+\sqrt[2n]{c^2+d^2}+\sqrt[2n]{e^2+f^2}\ge\sqrt[2n]{a^2+c^2+e^2}+\sqrt[2n]{b^2+d^2+f^2} ,n \in \mathbb{Z}$

and about [color=red] just demostrated[/color]?
This is not demonstration,there is no need of demonstartion,I really wonder why do you think in such a way?

and about this ( Two Greek guys) , well clearly

$\sqrt{a^2+b^2}=\text{Chord of a triangle ABC such that C=90,BC=a,CA=b}$

and similar for the two other ones,but by this way you will reach this :

$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{e^2+f^2}\ge\sqrt{(a+c+e)^2+(b+d+f)^2} $

i wonder how did you work with the Right side,Would you provide a complete formal solution?

[hide]$a^2+b^2=c^2$ the right side of the equation being $a^2+b^2$.squaring both terms makes it $a^2+b^2+c^2+d^2+e^2+f^2$ on the right side of the equation. when u square the whole "term"  on the left side of the equation, there are middle terms when you square multiple terms in one "term" so that the left side HAS to be larger than the right side. i think..[/hide]
do all proofs have to be two columned? if theyre not do you just write out your explanation step by step?

[quote="lomos_lupin"][quote="mathnerd314"]there is a much easier way than the[b] incomprehensible calculations[/b] that lomos_lupin [color=red]just[/color][b] demonstrated[/b][/quote]
Which part is incomprehensible?
[/quote]

Not really incomprehensible. I just couldn't think of another word at the moment. Hard to understand by the average or accelerated junior high student? Would that be more descriptive?

What's the $\prec$ symbol? Is it something about being concave? And what do you mean by concave? From which side?

Sorry, and I didn't see the link.

MAthnerd,i really cant understand the pythagoras hint of yours.How can you just take $a^2+b^2=c^2$ or something like that?That'll just be an assumption...is it appropriate to make suvh an assumption?

[quote]What's the $\prec$ symbol? Is it something about being concave?.[/quote]

A $\prec$ B means that A precedes B

Ah. I get the \prec code now. But what does precede mean in this context?

[quote="mathnerd314"]a, b, c, d, e, and f are sides of three right triangles. a, c, and e are collinear, and b, d, and f are collinear and perpendicular to the three others. Remember that the shortest path between two points is a straight line. Go from there.[/quote]

How come this condition was not mentioned in the question?

It's a creative way of thinking of the problem; it doesn't have to be included in the actual problem.  I dismissed it at first but i guess it's of use.

Yeah so the answer is (with your Pythagorean hint):

If you consider the vectors (a,b), (c,d), (e,f)

The sum of the vectors are (a+c+e,b+d+f).

The distance from this to the origin is the right side, and equality is reached when the vectors are collinear, and in the same direction.

[quote="mathnerd314"]Sorry for reviving this topic, but there is a much easier way than the incomprehensible calculations that lomos_lupin just demonstrated ;). I forgot to post this when I should have (i.e. a week ago) :blush:

[hide="hint"]
[hide="are you sure?"]
Think of the variables as sides of right triangles. Two Greek guys really, both dealing with the same subject.
[/hide]
[/hide][/quote]
First, Proof by Lomos_lupin  was excellent , certainly ,IMO, anything but incomprhensible but more to the point,   this and few  later posts ,perhaps  are forgetting that the RHS  of the problem given above is

$\sqrt{a^2+c^2+e^2}+\sqrt{b^2+d^2+f^2}$ and is  NOT:

$\sqrt{(a+c+e)^2+(b+d+f)^2}$  as pointed out  before.
[quote="lomos_lupin"]

[b]and similar for the two other ones,but by this way [color=red][size=117]you will reach this[/size][/color][/b] :

$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{e^2+f^2}\ge\sqrt{(a+c+e)^2+(b+d+f)^2} $

i wonder how did you work with the Right side,Would you provide a complete formal solution?[/quote]

One still has some work to do, even after using something like "Remember that the shortest path between two points is a straight line" ..

Regards.

This kind of inequality proof has no place in the Getting Started forum.  Please read the posting guidelines before posting any more problems in that forum.

As you wish Mr.Mcrawford.

This proof was posted due to the request of :
[quote="PenguinIntegral"]Is there any other way than squaring all of it?[/quote]

To introduce a new idea.but as you said its not appropriate for getting started.

lomos, it's the entire thread that was inappropriate for the particular forum, not any one part.

the shape of question is like karamarata inequality.

i wish its possible to solve it by karamarata ;) .


factorial
floor function
number theory unsolved
number theory

Can be found the number of digits of  1999! ?

Well, yes, since we have this formula for the number of digits of x (x > 0) :
$d(x) = \lfloor log_{10}(x)\rfloor + 1$ ;)


To whoever posted 2005=5*401 on the important things to know... thx, it really really saved alot of time.

It will probably be useful on the aime too (but not so much)

I wonder what the factors of 2006 are...
dividing by 2 gives 1003...

and seeing as we all know our multiples of seventeen,

2006 = 2*17*59

Thats the Voyage 200 answer anyway... :-)

[quote="mathfanatic"]Don't most calcs have a factor function?  (TI-89 definitely has one; if you have a TI-83+, you should definitely have one of the factoring/prime programs).[/quote]

But that would involve reaching for my calc, pressing the one button, pressing those numbers, and waiting for the answer. Frankly, I'm just too lazy.

I wonder if there's any coorelation between average AMC-12 scores from year to year and the difficulty of factoring that year's number?


f:R -> N* satisfy:
  Every x is belong to Q; every y is belong to I if   $ \frac{\minus{}1}{2^n} < x\minus{}y < \frac{1}{2^n}$
   With every n is belong to N*, exist k is belong to Z satisfy $ f(x)f(y)\equal{}2^nk$

  :P My english is weak

In the future, please post problems of this difficulty in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=217]Olympiad[/url] forums.


function
logarithms
geometric series
real analysis
real analysis theorems

Hi, I'm new here

I'm having difficulty understanding how, once having expanded a function via maclaurin's one can say what values of x it is valid for.

for instance

for the expansion of f(x)=ln(1+x)

f(x)=x-(x^2)/2+(x^3)/3-(x^4)4 etc....

my textbook says that this is valid for |x|<1 but I can't see how this conclusion is arrived at!

any help or links to help would be appreciated

Note that the series diverges outside that interval, so it can't work for larger $x$. It does converge conditionally when $x=1$ to $\ln 2$, but that's it.

The standard error estimates from Taylor's theorem will tell you that the series converges to the function when $|x|<\frac12$, or maybe even $-\frac12<x<1$. To get the last piece of the interval, we have to use something else.

The geometric series $1-x+x^2-x^3+\cdots$ converges uniformly to $\frac1{1+x}$ on the interval $[-a,a]$ for any $a<1$ (The maximum error of the partial sum $\sum_{j=0}^n (-x)^j$  is $\sum_{j=n+1}^\infty (a)^j=\frac{a^{n+1}}{1-a}$ on this interval, which tends to zero)
Integrating term by term (we can do that by uniform convergence), the series $x-\frac x2+\frac{x^2}3-\cdots$ converges uniformly to $\ln(1+x)$ on the interval $[-1,1]$.

thanks jmerry,

however the mathematics is further than what I'm familiar with I'm afraid (which I guess explains why my textbook didn't explain this item).

hopefully i'll get the time to go into the more advanced maths this summer!


group theory
abstract algebra
superior algebra
superior algebra solved

Let G be a group such that it has at most two nontrivial subgroups.
Prove that G is cyclic

let's say a,  of order n, is an element in the group.
If n has more than one prime factor then there exists two nontrivial subgroups. So n has only one prime factor.
If there is another element b, not in <a>, we see that its order, say m, has also one prime factor only.
Here we see that ab is not in <a> or <b>. So the cyclic group of ab is another nontrivial subgroup.
So there can't be another element.