Let $ a, b, c$ be positive real numbers satisfying $ a \plus{} b \plus{} c \equal{} 3$.
Prove that: $ ab \plus{} bc \plus{} ca \plus{} \frac {1}{abc} \ge 3 \plus{} abc$
:)

\[ ab \plus{} ac \plus{} bc \plus{} {1 \over {abc}} \ge 3\sqrt {abc}  \plus{} {1 \over {abc}}\]
\[ 3\sqrt {abc}  \plus{} {1 \over {abc}} \minus{} 3 \minus{} abc \equal{} {{(1 \minus{} \sqrt {abc} )(1 \plus{} \sqrt {abc} {{(\sqrt {abc}  \minus{} 1)}^2})} \over {abc}} \ge 0\]

With $ ab+bc+ca=3s $ result $ 0<s\le 1 $
$ (ab+bc+ca)^2\ge 3abc(a+b+c)\Leftrightarrow abc\le s^2 $
We have:
$ LHS\ge 3s+\frac{1}{s^2} $ and $ RHS\le 3+s^2 $
It is sufficient that:
$ 3+s^2\le 3s+\frac{1}{s^2}\Leftrightarrow s^4-3s^3+3s^2-1\le 0 $
$ \Leftrightarrow (s-1)(s^3-2s^2+s+1)\le 0\Leftrightarrow (s-1)[s(s-1)^2+1]\le 0 $
which is obviously true!
_________
Sandu Marin

[quote=nguoivn]Let $ a, b, c$ be positive real numbers satisfying $ a \plus{} b \plus{} c \equal{} 3$.
Prove that: $$ ab \plus{} bc \plus{} ca \plus{} \frac {1}{abc} \ge 3 \plus{} abc$$
:)[/quote]
Solution of [b]Zhangyanzong[/b]:
$$ab+bc+ca\geq \sqrt{3abc(a+b+c)}\iff a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c)$$
$$ab+bc+ca + \frac{1}{abc} \geq \frac{1}{9}(ab+bc+ca) (a+b+c)+ \frac{2}{3}\sqrt{3abc(a+b+c)}+ \frac{1}{abc} $$$$
= abc+ 2\sqrt{abc}+ \frac{1}{abc} \ge abc + 3$$

 [url=https://artofproblemsolving.com/community/c6h311130p1678666]maths.vn [/url]
Let $ a, b, c$ be positive real numbers satisfying $ a(a+1)+b(b+1)+c(c+1)\le 18.$  Prove that  $$  \frac{1}{{1 + a + b}} + \frac{1}{{1 + b + c}} + \frac{1}{{1 + c + a}}\geq \frac{3}{5}$$

Let $ a, b, c$ be positive real numbers satisfying $ a+b+c=3.$ Prove that$$ab+bc+ca + \frac{1}{a+bc} \ge abc + \frac{1}{3}$$
$$ab+bc+ca + \frac{1}{a^2+bc} \ge abc + \frac{1}{9}$$
Let $ a, b, c$ be positive real numbers satisfying $ a+2bc=3.$ Prove that$$ab+bc+ca + \frac{1}{abc} \ge abc + 3$$
Let $ a, b, c$ be positive real numbers satisfying $ a^2+2bc=3.$ Prove that$$ab+bc+ca + \frac{1}{abc} \ge abc + 3$$

[quote=sqing]Let $ a, b, c$ be positive real numbers satisfying $ a+2bc=3.$ Prove that$$ab+bc+ca + \frac{1}{abc} \ge abc + 3$$
Let $ a, b, c$ be positive real numbers satisfying $ a^2+2bc=3.$ Prove that$$ab+bc+ca + \frac{1}{abc} \ge abc + 3$$[/quote]


$ \triangle ABC$,and a point $ P$.$ A_1B_1C_1$ is Cevian triangle of $ P$.$ A_2,B_2,C_2$ are midpoints of $ B_1C_1,A_1C_1,A_1B_1$.$ A_2P,B_2P,C_2P$ cut $ BC,CA,AB$ at $ A_3,B_3,C_3$.
Prove that $ AA_3,BB_3,CC_3$ are concurrent.

[b]Generalization:[/b]
$ A_2, B_2, C_2$ lie on $ B_1C_1, C_1A_1, A_1B_1$ such that $ A_1A_2, B_1B_2, C_1C_2$ concur at $ P'$. Then $ AA_3, BB_3, CC_3$ concur at $ J$ which lies on $ PP'$.
[b]Solution:[/b]
Let $ A_4, B_4, C_4$ be the intersections of $ B_1C_1$ and $ BC, C_1A_1$ and $ AC, A_1B_1$ and $ AB$, respectively then $ A_4, B_4, C_4$ are collinear.
$ P'A_2\cap (A_4B_4C_4) \equal{} \{L\}, AL\cap B_1C_1 \equal{} \{V\}, AP\cap B_1C_1 \equal{} \{W\}$.
We have $ (AA_3VL) \equal{} (APWA_1) \equal{} \minus{} 1$ then $ VW, A_3P, LA_1$ concur at $ A_2$.
Denote $ J \equal{} AL\cap PP'$.
Applying Menelaus's theorem we obtain: $ \frac {JP}{JP'} \equal{} \frac {LA_1}{LP'}.\frac {AP}{AA_1}$.
Denote $ J' \equal{} BB_3\cap PP', M \equal{} BB_3\cap B_4C_4$.
We will show that $ \frac {J'P}{J'P'} \equal{} \frac {JP}{JP'} \Leftrightarrow \frac {LA_1}{LP'}.\frac {AP}{AA_1} \equal{} \frac {MB_1}{MP'}.\frac {BP}{BB_1}$
$ \Leftrightarrow \frac {AP}{AA_1} \equal{} \frac {MB_1}{MP'}.\frac {BP}{BB_1}.\frac {AA_1}{AP} \equal{} \frac {BP}{BB_1}.\frac {C_4B_1}{C_4A_1}$
$ \Leftrightarrow \frac {AA_1}{AP}.\frac {BP}{BB_1}.\frac {C_4B_1}{C_4A_1} \equal{} 1$ (it's right from Menelaus's theorem)
Therefore $ J'\equiv J$. Similarly we are done.

Based on the same idea, as in the solution of the general problem by [b][size=100]livetolove212[/size][/b], we can have another approach applying the [b][size=100]Desarques theorem[/size][/b] and the [b][size=100]Pappos theorem[/size][/b]. It is a good chance for a proof without metrical relations.

At first, let us to restate the general problem as follows :

[b][size=100][color=DarkBlue]GENERAL PROBLEM. - A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup DEF,$ the Cevian triangle of an arbitrary point $ P,$ inwardly to it. Let $ D',\ E',\ F'$ be, three arbitray points on $ EF,\ DF,\ DE$ respectively, such that $ DD'\cap EE'\cap FF'\equiv P'$. We denote as $ A',\ B',\ C',$ the points of intersection of $ BC,\ AC,\ AB$ respectively, from the line segments $ PD',\ PE',\ RF'$. Prove that the line segments $ AA',\ BB',\ CC'$ are concurrent at one point so be it $ Q,$ lying on the line segment $ PP'$.[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] - It is enough to prove that the points of intersection of the line segments $ AA',\ BB',\ CC',$ taken per two of them, lie on the line segment $ PP'$.

So, let be the point $ Q\equiv AA'\cap BB'$ and we will prove that the points $ Q,\ P,\ P',$ are collinear.

$ \bullet$ We denote the points $ X\equiv BC\cap PE'$ and $ Y\equiv AC\cap PD'$ and based on the [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup XDE',\ \bigtriangleup YED'$ are perspective, because of the collinearity of the points $ C\equiv XD\cap YE$ and $ P\equiv XE'\cap ED'$ and $ F\equiv DE'\cap ED'.$

So, we have that the line segments $ XY,\ DE,\ E'D',$ are concurrent at one point so be it $ Z.$

Because of now, the collinearity of the points $ X\equiv PE'\cap SE$ and $ Y\equiv PD'\cap SD$ and $ Z\equiv D'E'\cap DE,$ where $ S\equiv DY\cap EX,$ we conclude that the triangles $ \bigtriangleup PD'E',\ \bigtriangleup SDE$ are perspective and then, we conclude that the line segments $ PS,\ DD',\ EE'$ are concurrent at point $ P'.$ That is, the points $ P,\ P',\ S,$ are collinear.

$ \bullet$ We consider now, the triads of points $ A,\ Y,\ E$ and $ B,\ X,\ D,$ on $ AC,\ AB$ respectively and applying the [b][size=100]Pappos theorem[/size][/b], we have that the points $ S,\ P,\ R\equiv AX\cap BY,$ are collinear.

Similarly, from the triads $ A,\ B',\ Y$ and $ B,\ A',\ X,$ applying again the [b][size=100]Pappos theorem[/size][/b], we have that the points $ P,\ R,\ Q,$ are also collinear.

Hence, becuase of the line segments $ PP'S,\ SPR,\ PRQ,$ taken per two of them, have two common points, we conclude that the points $ Q,\ R,\ P,\ P',\ S,$ are collinear.

That is the point $ Q\equiv AA'\cap BB'$ lies on the line segment $ PP'.$

$ \bullet$ By the same way we can prove that the points $ Q'\equiv BB'\cap CC'$ and $ Q''\equiv AA'\cap CC',$ also lie on $ PP'.$

Hence, we conclude that $ Q\equiv Q'\equiv Q''\equiv AA'\cap BB'\cap CC'$ and the proof is completed.

Kostas Vittas.

[quote="k.l.l4ever"]A simple solution for the general one:
Denote $ \{ X \} \equal{} CC_1 \cap AA_3,\{ Y \} \equal{} CC_1 \cap A_1B_1,\{ Z \} \equal{} CC_1 \cap A_1A_2$
$ \{ R \} \equal{} CC_3 \cap AA_3,\{ R' \} \equal{} PQ \cap AA_3$.Therefore:
$ (A,X,R,A_3) \equal{} C(A,C_1,C_3,B) \equal{} P(A_1,Y,C_2,B_1) \equal{} C_1(A_1,Z,Q,A_2) \equal{} P(A,X,R',A_3)$
$ \rightarrow R' \equiv R$.
From that,easy to conclude that $ {R} \equal{} AA_3 \cap BB_3 \cap CC_3$ and $ R \in PQ$.[/quote]
Thank you very much dear [b][size=100]l.l.l4ever[/size][/b], for a nice solution based on the [b][size=100]Double Ratio[/size][/b] ( = [b][size=100]Cross Ratio[/size][/b] ) theory.

I would like to preserve this proof in the future with your name and if it is possible, tell me please about it.

Thank you in advance, Kostas Vittas.


If $ 10^{\log_{10}9} \equal{} 8x \plus{} 5$ then $ x$ equals:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {5}{8} \qquad \textbf{(D)}\ \frac {9}{8} \qquad \textbf{(E)}\ \frac {2\log_{10}3 \minus{} 5}{8}$

$ 10^{\log_{10}9}\equal{}9\equal{}8x\plus{}5\implies x\equal{}\boxed{\textbf{(B)}\ \frac12}$.


From Hungary 2000 :
If $1 \leq m \leq n$, prove that $m$ divides
\[n(C_n^0 - C_n^1 + C_n^2 - \cdots +(-1)^{m-1}C_n^{m-1})\]

Pierre.

Is it me or has the level of the intermediate level problems gone up substatially after the merge?  I don't think that olympiad level problems should be in the intermediate forum.

ah?
Even if they are quite easy?
Ok, so moderators please move it... and sorry for the misposting.

Pierre.

Yes,this is quite easy.We have:
[tex] n(C_n^0-C_n^1+...+(1)^{m-1}C_n^{m-1})=(-1)^{m-1}C_{n-1}^{m-1} \cdot n=(-1)^{m-1}C_n^m \cdot m [/tex] is divided by [tex] m [/tex]


I'll give the problem and answer, but I need to know how to get the equation. Is it through linear regression or other means?

  At 4.80 per bushel, the annual supply for soybeans in the Midwest is 1.9 billion bushels and the annual demand is 2.0 billion bushels. When the price increases to 5.10 per bushel, the annual supply increases to 2.1 billion bushels and the annual demand decreases to 1.8 billion bushels.

The supply equation is p=1.5x+1.95
The demand equation is p=-1.5x+7.8

Thanks! :)

Hi Inequality.

$ p$ is the price per bushel and $ x$ is the amount in billions of bushels.

For the supply equation, we have two points $ (x, p)$ on a straight line: $ (x, p) \equal{} (1.9, 4.8)$ and $ (2.1, 5.10)$.

The gradient, or slope, of this line is $ \frac {5.10\minus{}4.80}{2.1\minus{}1.9} \equal{} \frac{3}{2} \equal{} 1.5$,

so the supply equation is $ p\equal{}1.5x\plus{}c$,

where we can find $ c$ by substituting the coordinates of either of the two points on the line.

Substituting $ (x, p) \equal{} (1.9, 4.80)$, we get $ 4.80 \equal{} 1.5 \times 1.9 \plus{} c$

$ 4.80\equal{} 2.85 \plus{} c$

$ c\equal{}4.80\minus{}2.85\equal{}1.95$

Therefore the supply equation is $ p\equal{}1.5x\plus{}1.95$

In the same way, for the demand equation we get a gradient of $ \frac{5.10 \minus{} 4.80}{1.8\minus{}2.0} \equal{} \minus{}1.5$

and the equation is $ p\equal{}\minus{}1.5x \plus{}k$.

Substituting $ (x,  p) \equal{} (2.0, 4.80)$, we find that $ k\equal{}7.8$

and so the demand equation is $ p\equal{}\minus{}1.5x \plus{} 7.8$.

I don't know anything about economics, but I hope this helps you.


algebra
polynomial
search
superior algebra
superior algebra theorems

let $ R$ be a ring with $ 1\neq 0$
given a polynomial $ r_nx^n\plus{}\dots\plus{}r_0$ prove that this poly is invertible iff $ r_0$ is invertible and $ r_1,\dots,r_n$ are nilpotent.

we had this several times, use the search function.


I need help.I have a lot of question if someone help about this problem i will solve another problem i need example :(
thanks about all help...

This problem is made easy by the fact that $ a$, $ b$, $ c$, and $ d$ appear in no more than two equations.  We have
$ 3a\equal{}a\plus{}4$
$ 3b\equal{}6\plus{}a\plus{}b$
$ 3c\equal{}\minus{}1\plus{}c\plus{}d$
$ 3d\equal{}2d\plus{}3$
Solving the first equation yields $ a\equal{}2$.  Plug this into the second to get $ b\equal{}4$.  Solve the last equation to get $ d\equal{}3$, and plug this into the third to get $ c\equal{}1$.  Then the original matrix is
$ \begin{pmatrix}
2&4\\
1&3
\end{pmatrix}$


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Any help with this one?

Let D be the set consisting of 0 and the zero divisors of a commutative ring R (with 1).  Prove D contains at least one prime ideal.

Thanks!

[quote="Soarer"]$ D$ is actually the union of all the associated primes of $ R$, so of course it contains a prime ideal.[/quote]

Be careful with the associated primes in the case when $ R$ is not noetherian ring.
For instance one can give an example of ring $ R$ for which $ \text{Ass}(R) \equal{} \varnothing,$ so the statement above is not true.

But it is true that $ D$ is a sum of prime ideals.
By Zorn lemma for every $ a\in D$ there exists maximal element of family:
$ \{I\triangleleft R \ : \ (a)\subset I \subset D\}$
and, by maximality in this family, such ideal is prime.

[quote="ifk"][quote="Soarer"]$ D$ is actually the union of all the associated primes of $ R$, so of course it contains a prime ideal.[/quote]

Be careful with the associated primes in the case when $ R$ is not noetherian ring.
For instance one can give an example of ring $ R$ for which $ \text{\minus{}}(R) \equal{} \varnothing,$ so the statement above is not true.

But it is true that $ D$ is a sum of prime ideals.
By Zorn lemma for every $ a\in D$ there exists maximal element of family:
$ \{I\triangleleft R \ : \ (a)\subset I \subset D\}$
and, by maximality in this family, such ideal is prime.[/quote]

thanks.  could you explain why maximality in that family implies it is prime?

Denote by $ I_{0}$ ideal which is maximal in that family.
Suppose that $ I_{0}$ is not prime.
Then $ \exists_{s,t\in R\setminus I_{0}} \ st \in I_{0}.$
Since $ (s) \plus{} I_{0}, \ (t) \plus{} I_{0}\supsetneq I_{0}$ and by maximality of $ I_{0}$ there exists:
$ x\in ((s) \plus{} I_{0})\setminus D, \ y \in ((t) \plus{} I_{0})\setminus D.$
$ xy \in ((s) \plus{} I_{0})((t) \plus{} I_{0}) \subset I_{0},$ because $ st \in I_{0}.$
But this is contradiction with the fact that $ R\setminus D$ is closed under multiplication.

Thanks ifk for the reminder.

A more conceptual way of doing it, I think, would be the following. Let $ S$ be the multiplicative set of nonzero divisors. Then $ S^{\minus{}1}A$ is nonzero, so there is a prime ideal. Pull back, we will get a prime ideal $ p$ not intersecting $ S$, this means that $ p \subset D$.


Prove that polynom \[ p(x) = (x-a_1)(x-a_2) ... (x-a_n) - 1  \] is ireducibile.

[quote="omjfv"]Prove that polynom \[ p(x) = \left((x-a_1)(x-a_2) ... (x-a_n)\right)^2 + 1  \] is ireducibile.[/quote]
Thats the one I know and that was posted before... (at least I think so)

I think that the true version of problem is:                                                  [img]http://www.mathlinks.ro/Forum/latexrender/pictures/a319742aef18fdd9783b801e099c277a.gif[/img]
are integer  numbers such that every two of them bieng unequal(n>1) prove that[img]http://www.mathlinks.ro/Forum/latexrender/pictures/ade409d2934c79aa3da282b3e87c1d7b.gif[/img] is ireducibile.(in Z[X])


Consider a sequence $\{a_{n}\}_{1}^{\infty}$ of positive reals verifying :
a) there exist $c \in \mathbb R$ such that $a_i \leq c$ for $i=1,2,...$
b) $|a_i-a_j| \geq \frac{1}{i+j}$ for all $i \neq j$
Prove that $c \geq 1$

Yes $Badr$ you are right, its from ISL 2002.
Hope we will see different solutions :?


Let $z_1, \ldots, z_n$ be complex numbers.  Prove that there is a set $S \subseteq \{ 1, \ldots, n \}$ such that
\[
\big|\sum_{j \in S} z_j\big| \geq \frac{1}{\pi} \sum_{j=1}^n |z_j| \, .
\]

For partial credit:  Prove the result with "$\pi$'' replaced by a larger constant.

Source: I once saw this result cited in the endnotes of Rudin's book on [i]Real and Complex Numbers[/i], in the chapter on complex measures.  Unfortunately, I don't remember who to credit the result to.  (If anyone knows, please let me know.)

i thought this a pretty interesting result, so i took the liberty of posting it in the inequalities forum for more people to see.  it already received a response.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=205886#p205886[/url]


function
LaTeX
calculus
calculus computations

Suppose that f(x) tends to L as x tends to infinity, and let g(x)=f(1/x). Prove that g(x) tends to L as x tends to zero+ (tend to zero from above)
If you have any idea, your help would be valuable.

What have you tried so far? What happens when you try to directly apply the definition of limit of a function at a point?

What i cannot do is to argue about the limit as x tends to zero..I don't know how to arrive to that fact from the epsilon-delta definition of the limit of a function, which involves the limit as x tends to infinity.

There aren't any hints? I am really stuck with this problem, so any ideas would be really useful.

What does "$ f(x)$ tends to $ L$ as $ x$ tends to infinity" mean? Can you write this down explicitly, something like "for every <something> there is <something else> such that <something happens>"?

If you can, you should be well on your way - write this down in this thread and we'll take the next step.

If you can't, you have to go back to your notes or textbook and figure it out since otherwise there's no way you can get started on this problem.

it means that given any e>0 there exists a d>0 such that |f(x)-L|<=e whenever |x|>d

In $ \text{\LaTeX}$:

1. Given any $ \epsilon > 0$ there is a $ d$ such that $ |f(x) - L| \leq \epsilon$ whenever $ x > d$.

Since $ x \to \infty$ you actually know that $ x$ is large, not that $ |x|$ is large. You don't know anything about $ f(x)$ when $ x$ is negative.

This is the data. What you [i]want [/i]to show is this:

2. Given any $ \epsilon > 0$ there is a $ \delta$ such that $ |f(\frac {1}{z}) - L| \leq \epsilon$ whenever $ |z| < \delta$.

I changed the variable to $ z$ to make it easier to see what's going on. I did not change the $ \epsilon$ because you can, and should, rely on the $ \epsilon$ in statement 1 to prove statement 2.

Your task is to define this $ \delta$, given $ \epsilon$. It's unlikely you can somehow define $ \delta$ directly from $ \epsilon$, but it is very likely that you should make use of statement 1., which gives you a $ d$. Can you guess how $ \delta$ should be defined as a function of $ d$? Can you make it so that if $ |x| < \delta$, you know something about $ 1/x$ that enables you to conclude what you need about $ f(\frac {1}{x}) - L$?


can anyone tell me why the sum of the angles of a triangle is 180?

Draw a line going through a vertex parallel to the opposite side of the vertex.  Then, you see that angles of the triangle can all lie on the line, thus $180^{\circ}$.

here's a drawing. line MN  is parallel to side EF. so, angle MTF is congruent to angle EFT and angle FET is congruent to NTE. since angle MTN is 180, the triangle's angles must also sum to 180.

[quote="joseph14i"]then, how do you know that angle ftm is congruent to angle tfe? what about angle ETN and angleTEF? if you are trying to say that it's because MN is paralell to FE, can you prove that in a logical way?[/quote]
Since the two lines are parallel, the two angles are congruent by alternate interior angles.

For the sake of contradiction, assume that the angles you mentioned are not congruent.  If not, then draw a line (i.e. Tx) such that $\widehat{ETN}$ and $\widehat{TEF}$ are congruent. Then $Tx\parallel FE$. In this case, we would have two parallels drawn from a single point to another line, but that's not possible according to Euclid's Parallel Postulate.


integration
logarithms
real analysis
real analysis unsolved

More generally, let $ k$ be a natural number. Then

$ \sum_{m_{1},m_{2},\cdots, m_{n}\geq 1}\frac{1}{m_{1}\cdots m_{n}(m_{1}\plus{}m_{2}\plus{}\cdots\plus{}m_{n}\plus{}k)}\equal{}$

$ \equal{}(\minus{}1)^{n}\int_{0}^{1}x^{k\minus{}1}\ln^{n}(1\minus{}x)dx$.

We have that

$ S\equal{}\sum\frac{1}{m_{1}\cdots m_{n}}\int_{0}^{1}x^{m_{1}\plus{}\cdots\plus{}m_{n}\plus{}k\minus{}1}dx$

$ \equal{}\int_{0}^{1}x^{k\minus{}1}\sum_{m_{1}}\frac{x^{m_{1}}}{m_{1}}\cdots\sum_{m_{n}}\frac{x^{m_{n}}}{m_{n}}dx$

$ \equal{}\int_{0}^{1}x^{k\minus{}1}\left(\ln\frac{1}{1\minus{}x}\right)^{n}dx$

$ \equal{}\int_{0}^{1}(\minus{}1)^{n}(1\minus{}x)^{k\minus{}1}\ln^{n}(x)dx$.

When $ n\equal{}2$ and $ k\equal{}2$ we get that the sum equals

$ \int_{0}^{1}(1\minus{}x)\ln^{2}(x)dx\equal{}\frac{7}{4}$.


function
modular arithmetic
algebra
functional equation
algebra unsolved

Find all functions with real values, defined for integers and such that $ f(1)+f(ab)+f(a+b)= f(a)+f(b)+f(ab+1)$

Hi,

We can see quite quickly that :
1) $f(n) = xn^{2}+yn+z$ is a family of solutions
2) if $f$ and $g$ are solutions, $x f+y g$ is solution
3) $f(0)$ can be any value
4) the knowledge of $f$ in 1, 2, 3, 4, 5 and 6 determines totally $f$ (except $f(0)$, of course, which can take any value)

Well, but, from here to the identification of all the solutions, there's still a big step...

mathmanman has stated that there's linearity (property 2), so it would be enough to find all the independent solutions (from linearity, we can get all the others).
So far, he has given some of them:
$f(x)=z^{2}$,
$f(x)=z$,
$f(x)=1$,

If $f(0)$ is completely independent of the other values, then $f(x)=\delta (x)$ (defined as usual, $\delta (0)=1$, and $\delta (x)=0$ if $x\neq 0$) should be another independent solution.
But it's not.
Taking $a+b=0$,
$f(1)+f(-a^{2})+f(0)=f(a)+f(-a)+f(1-a^{2})$,
which doesn't hold in general for $f(x)=\delta (x)$, because
$\delta (0)=1$,
$\delta (1)=\delta (a)=\delta (-a)=\delta (-a^{2})=\delta (1-a^{2})=0$ if $a\notin\{-1,0,1\}$.

Hi,

First, in my last post, in the 4th point, the "5" shouldn't be here.

Second, now I have the solution, but in a very unsatisfying way. I'm interested in any direct and elegant solution.

$P(a,b) : f(a+b)+f(1)+f(ab) = f(a)+f(b)+f(1+ab)$
And we thus have $P(a, b)$ true for all $a, b$ integers.

1) the set of the solutions is a $\mathbb{R}$-vectorial space
================================
obvious : $f(x)$ solution $\implies xf$ is solution
$f(x)$ and $g(x)$ solutions $\implies f(x)+g(x)$ is solution

2) the dimension of this space is at most 6
==============================
$P(a,2) \implies E_{1}: f(a+2)+f(1)+f(2a) = f(a)+f(2)+f(2a+1)$
$P(a,4) \implies E_{2}: f(a+4)+f(1)+f(4a) = f(a)+f(4)+f(4a+1)$
$P(2a,2) \implies E_{3}: f(2a+2)+f(1)+f(4a) = f(2a)+f(2)+f(4a+1)$

$E_{1}: f(2a+1) = f(2a)+f(a+2)-f(a)+f(1)-f(2)$
$E_{3}-E_{2}\implies E_{4}: f(2a+2) = f(2a)+f(a+4)-f(a)+f(2)-f(4)$

Then let us assume $f$ fixed for 0, 1, 2, 3, 4 and 6 with the values $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{6}$.

$E_{1}$ with $a=2$ then allows us to determine $f(5) = 2x_{4}-2x_{2}+x_{1}$, $E_{1}$ and $E_{4}$ then allow us to determine all the other values.
Warning, I am not saying that $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{6}$ can be any values. I say that if $f$ is solution, the knowledge of these values determines entirely $f$.

This shows well that the dimension of the vectorial space is at most 6. It can possibly be less if we cannot choose $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{6}$ independently. I am going to show that this dimension is exactly 6 by finding a basis.

3) we can find 6 solutions linearly independent.
====================================
$S_{1}: f_{1}(0) = 1 and f_{1}(n)=0$ for all $x > 0$
$S_{2}: f_{2}(n) = 1$
$S_{3}: f_{3}(n) = n$
$S_{4}: f_{4}(n) = n^{2}$
$S_{5}: f_{5}(n) =$
36 if $n=0 \pmod 6$
11 if $n=1 \pmod 6$
20 if $n=2 \pmod 6$
27 if $n=3 \pmod 6$
20 if $n=4 \pmod 6$
11 if $n=5 \pmod 6$
Checking that $f_{5}$ is solution is done easily considering the 36 values of the couple $(x \mod 6, y \mod 6)$
$S_{6}: f_{6}(n) =$
12 if $n=0 \pmod 6$
5 if $n=1 \pmod 6$
8 if $n=2 \pmod 6$
9 if $n=3 \pmod 6$
8 if $n=4 \pmod 6$
5 if $n=5 \pmod 6$
Checking that $f_{6}$ is solution is done as for $f_{5}$.

The linear independence of these 6 solutions is quite easy to see :
Let $x_{1}f_{1}(x)+x_{2}f_{2}(x)+x_{3}f_{3}(x)+x_{4}f_{4}(x)+x_{5}f_{5}(x)+x_{6}f_{6}(x) = 0$.
$f_{1}, f_{5}$ and $f_{6}$ being bounded, this equality implies (let $x \to+\infty$) that $x_{4}$, then $x_{3}$ are zero.
Then letting $a=0, 1, 2$ and 3, it comes :
$x_{1}+x_{2}+36x_{5}+12x_{6}= 0$
$x_{2}+11x_{5}+5x_{6}= 0$
$x_{2}+20x_{5}+8x_{6}= 0$
$x_{2}+27x_{5}+9x_{6}= 0$
and $x_{1}= x_{2}= x_{3}= x_{4}= x_{5}= x_{6}= 0$

4) the set of the solutions is thus a $\mathbb{R}$-vectorial space of dimension 6 of which we know a basis. We thus have the set of the solutions.

We indeed find back the binomials (combinations of $f_{2}, f_{3}$ and $f_{4}$).
We find back the independence of $f(0) (f_{1})$
We add to it two strange periodical solutions : $f_{5}$ and $f_{6}$.

That's it.

Wow, mathmanman :!: Astonishing work!

I still have lots of doubts about your solution, though. :(

Don't worry, most of them are because of my lack of knowledge about vectorial spaces :( , so I don't quite understand the way you proved the linear independence of the solutions. But I understand it could be easily checked with advanced algebra and/or a computer, so I'm not specially worried about that.

Anyway, maybe you could help me with these two easier questions:

There's some typo in your definition of $f_{1}$.
Did you mean $f_{1}(0)=1$ and $f_{1}(n)=0$ for all $n\neq 0$?
I still feel that this is not a solution. Taking $a=t$, $b=-t$, where $t\in\mathbb{Z}\setminus\{-1,0,1\}$, the functional equation doesn't hold!

Did you find $f_{5}$ and $f_{6}$ by computer? They are hard to check, they must have been harder to find. Maybe some other linear combination would have been more meaningful.

I'm still stuck with this problem.
It seems that mathmanman didn't consider the negative integers.
Still, if I'm not wrong, with his equations $E_{1}$ and $E_{4}$ it can indeed be proved that the dimension of the vectorial space of solutions is $6$.
On the other hand, using $E_{1}$ and $E_{4}$, $f(0)$ has an influence on the values of $f(n)$ for negative values of $n$. If $f_{1}$ isn't a solution of the functional equation, we should find a sixth independent solution.

Sorry for the late reply, but I was on a vacation last week.

[quote="lordWings"]Wow, mathmanman :!: Astonishing work!
[/quote]
Thanks! :D

[quote="lordWings"]Did you mean $f_{1}(0)=1$ and $f_{1}(n)=0$ for all $n\neq 0$?
I still feel that this is not a solution. Taking $a=t$, $b=-t$, where $t\in\mathbb{Z}\setminus\{-1,0,1\}$, the functional equation doesn't hold!
[/quote]
You are right. But for me "integers" means "natural integers".
So I did not take into account the aspects "relative integers". :(

[quote="lordWings"]
Did you find $f_{5}$ and $f_{6}$ by computer? They are hard to check, they must have been harder to find. Maybe some other linear combination would have been more meaningful.[/quote]
Yes, in my previous example, the solutions $f_{5}$ and $f_{6}$ rely on two sextuplets (36, 11, 20, 27, 20, 11) and (12, 5, 8, 9, 8, 5) that I have found experimentally.

If we replace $f_{5}$ and $f_{6}$ with :
$f_{5b}= 4f_{2}+f_{5}/3-4f_{6}/3$
$f_{6b}=-f_{5}/4+3f_{6}/4$
We have again a basis of the vectorial space and $f_{5b}$ and $f_{6b}$ are nicer since they rely on the following sextuplets :
for $f_{5b}$ : (0, 1, 0, 1, 0, 1)
for $f_{6b}$ : (0, 1, 1, 0, 1, 1)

And so :
$f_{5b}(x) = \mod{(x,2)}= x-[x/2]$
$f_{6b}(x) = \mod{(x^{2}, 3)}= x^{2}-[x^{2}/3]$

We can make a last change :
$f_{5c}(x) = f_{3}(x)-f_{5b}(x) = [x/2]$
$f_{6c}(x) = f_{4}(x)-f_{6b}(x) = [x^{2}/3]$

Which gives a general form much more exploitable than in my previous message :

$x = 0 \implies f(0) = c_{0}$
$x > 0 \implies f(x) = c_{1}x^{2}+c_{2}x+c_{3}+c_{4}[x/2]+c_{5}[x^{2}/3]$

with $[y]$ denoting the integer part of $y$.

Will note that over all integers, the dimension of the solution set is only $5$.


abstract algebra
linear algebra
matrix
vector
algebra
polynomial
modular arithmetic

Let $R$ be a ring. (This means a commutative ring with $1$ here.) Let $n$ and $m$ be two positive integers.

[b]a)[/b] Show that any module homomorphism $f$ from the module $R^{n}$ to $R^{m}$ uniquely corresponds to a $n\times m$ matrix (i. e. a matrix with $n$ columns and $m$ rows; I am not sure whether this notation is the common one) $A$ such that $Av=f\left(v\right)$ for every vector $v\in R^{n}$.

[b]b)[/b] Let $X_{1}$, $X_{2}$, ..., $X_{u}$, with $u=\binom{m}{n}$, be all $n\times n$ minors of the matrix $A$. Show that $f$ is injective if and only if there exists no nonzero element $t$ of $R$ such that $t\cdot\det X_{i}=0$ for every $i\in\left\{1;\;2;\;...;\;u\right\}$. [Particularly, if $n>m$, then the matrix $A$ has no $n\times n$ minors at all, so there exists such a nonzero element $t$ of $R$ (namely, every nonzero element of $R$ can be choosen as $t$), and thus $f$ cannot be injective.]

[b]c)[/b] Let $Y_{1}$, $Y_{2}$, ..., $Y_{v}$, with $v=\binom{n}{m}$, be all $m\times m$ minors of the matrix $A$. Show that $f$ is surjective if and only if there exists an element $\lambda_{i}$ of $R$ for every $i\in\left\{1;\;2;\;...;\;v\right\}$ such that $\sum_{i=1}^{v}\lambda_{i}\cdot\det Y_{i}=1$. [Particularly, if $n<m$, then the matrix $A$ has no $m\times m$ minors at all, so the sum $\sum_{i=1}^{v}\lambda_{i}\cdot\det Y_{i}$ is empty and thus cannot be $1$, and therefore $f$ cannot be surjective.]

[b]d)[/b] Assume that $n=m$. Show that if $f$ is surjective, then $f$ is injective.

This is a proposed problem, i. e. I have a solution (though it is most likely not the shortest one). I am interested in others's solutions, generalizations and applications.

  Darij

Nice problem, Darij! The last one can also be proved using Nakayama's lemma. I am in a hurry right now, so I cannot post more details, but tomorow I will try to post some more comments. PS: sorry for the useless message.

[quote="harazi"]Nice problem, Darij! The last one can also be proved using Nakayama's lemma.[/quote]

Little hint: It can be without as well. ;)

In fact, Peter used Nakayama, too, and I think he even proved one of [b]b)[/b] and [b]c)[/b] with it. Probably I have the advantage of not knowing Nakayama (more precisely, not knowing how to apply it in the interesting cases) which really helps in this problem... None of the parts is really hard actually, but I am interested in the Nakayama proof.

By the way, I used [b]b)[/b] to solve problem 2 [b]iii)[/b] in Chapter 1 of Atiyah-Macdonald. The problem asks to show that if $f=a_{0}+a_{1}x+...+a_{n}x^{n}\in R\left[x\right]$ is a polynomial over $R$, then $f$ is a zero-divisor in $R\left[x\right]$ if and only if there exists an element $t$ of $R$ such that $tf=0$. (As a matter of fact, this problem is still not trivial using [b]b)[/b]: you have to induct properly.) At the moment I am trying to apply [b]c)[/b] to problem 2 [b]iv)[/b], which states that if $f\in R\left[x\right]$ and $g\in R\left[x\right]$ are two polynomials over $R$, then the polynomial $fg$ is primitive if and only if the polynomials $f$ and $g$ are primitive. Here, a polynomial $a_{0}+a_{1}x+...+a_{n}x^{n}\in R\left[x\right]$ is called [i]primitive[/i] if the ideal generated by $a_{0}$, $a_{1}$, ..., $a_{n}$ is $R$.

  Darij

I'm also very far from being familiar with this commutative algebra stuff, so I don't see how to solve $ii)$ or $iii)$ with Nakayama. Anyway, here are some ideas. Of course, I won't insist on $i)$, which is common sense. I will note $R=A$  :D (french notation) and $M$ the matrix associated to this homomorphism. First, indeed, $iv)$ is very easy without Nakayama or even without all the other parts. Indeed, assume that $f$ is surjective. Then if $e_{1},...,e_{n}$ is the canonical basis of $A^{n}$, there are vectors $x_{1},...,x_{n}$ such that $Mx_{i}=e_{i}$. By taking the matrix $N$ with columns $x_{1},...,x_{n}$ we get $MN=I_{n}$. This shows that the determinant of $M$ is invertible in $A$ and so $f$ is injective: if $Mx=0$ we multiply by $Com(M)$ to get $det(M)x=0$ and by invertibility we are done.
    Here's a proof for $iv)$ using this Nakayama lemma: if $M$ is an $A$-module of finite type (I guess it's finitely generated in english) and $I$ is an ideal in $A$ such that $M=IM$ then one can find $a\in I$ such that $(1+a)M=0$. The proof is very easy, I will skip it. Now, suppose that $M$ is an $A$-module of finite type (note that we do not assume it is free, so this is stronger than your $iv)$) and $f$ is a surjective homomorphism.  You know that $M$ becomes an $A[X]$ module by $P.m=P(f)(m)$. Of course, it is of finite type too. But the surjectivity of $f$ implies that $M=f(M)=X.M$ and so $M=IM$ for $I=(X)$. By Nakayama's lemma there is $Q\in A[X]$ such that $(1+XQ(X)).M=0$, from where the inverse of $f$ can be explicited as $-Q(f)$.
    Ok, I must confess that my proofs for $ii)$ and $iii)$ are not probably correct, since I never checked my computations in multilinear algebra. I just hate it! Here's my idea. Let's work with ii). One part is not difficult, so let's suppose that there is some nonzero $t$ which annulates all minors of order $n$. Let $r$ be the maximal integer for which one can find a minor of order $r$ which is not annulated by $t$. Then $r<n$ and there are $i_{1},...,i_{r}$ and $j_{1},...,j_{r}$ such that the minor with lines $i_{1},...,i_{r}$ and columns $j_{1},...,j_{r}$ is not annulated by $t$. Now, take some $j$ different from all $j_{1},...,j_{r}$ (possible since $r<n$) and take the elements $x_{j_{k}}=(-1)^{k}\cdot{Thing}$, where $Thing$ is the minor whose lines are $i_{1},...,i_{r}$ and whose columns are $j,j_{1},...,j_{k-1},j_{k+1},...,j_{r}$ (this works for $x_{j}$ as a definition, too). All the others $x_{s}$ are zero. Thus, at least one of the $tx_{s}$ is not zero and also you can check that $Mx$ is just the vector whose components are the minors of lines $i,i_{1},...,i_{r}$ and $j,j_{1},...,j_{r}$ for all $i$ and this is $0$ by definition of $r$. This shows that $f$ is not injective. Note that sometimes I made the confusion between minor and its determinant...
   Finally, at least one part of the third question can be done with the help of the previous defined $x_{i}$, by taking $r=n$ and making a sum of such minors, but I will let someone else explain this, I just hate these minors.

Grobber posted a proof of b) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=114519]here[/url], which is quite similar to harazi's.

I think I can prove the $\Longleftarrow$ part of c) in a similar way. For any $J \in \overline{1,n}, |J| = m$, we can find a matrix $B_{J}$ with $n$ rows and $m$ columns (and with the last $n-m$ rows filled with $0$'s) such that $A B_{J}= \left( \det A_{J}\right) \cdot I_{m}$. Now consider the matrix $B = \sum_{J}\lambda_{J}A_{J}$. We have that $A B = I_{m}$, from where surjectivity is pretty clear.
As for the other direction, I have no idea :blush:

d) If f is surjective, the inhomogeneous linear system of equations Ax=b has a solution for every b \in R^n.
w.l.o.g. the matrix A is in linestairsform Z.
If Z had less than n stairs, the system of equations Zx=e_n would have no solution, so if you went back with the elementary line transformation from Z to A, the vector e_n would change to a vector b and this system of equations Ax=b would have no solution, too. That’s a contradiction to the condition that f is surjectif. So the linestairsform Z has to be an upper triangular matrix
The system of equations Zx=0 has the only solution x=0 (because it is an upper triangular matrix) This solution is the same solution as the solution of Ax=0, so Ax=0 hast the only solution x=0, too. So the homogeneous system of equations has only the solution x=0 which means that f is injective.

(i hope you can understand my mathemathic english, i had to look up many words ;) )

[quote="Kent Merryfield"]I'm going to guess that "linestairsform" would, in usual mathematical English, be "row echelon form."

There's a distinction between row echelon form, in which there is nothing below and to the left of a stairstep boundary, and reduced row echelon form, in which all of the entries above a pivot elements are also zero, and the pivot elements themselves are one. From the context of your post, I think you mean the first concept.[/quote]
No, it's the upper triangular matrix that is a row echelon form, how you described it.
a "linestairsform" is nearly the same but the "diagonal" (i.e. the first a_i,j which is not 0 in the i-th line) doesn't has to be every a_i,j where i=j, it may also be that the first a_i,j, which is not 0, is in a column in which i<j.

edit: If someone's interested, i can post here a link to our (German) script, where these forms are explained (i think page 7 and 8), perhaps it's enough to look at the matrices to understand what i mean....:
http://www1.am.uni-erlangen.de/~prechtel/LinAlgI.pdf
"linestairsform":Zeilenstufenform, "uppper triangular matrix": Diagonalsystem)

Let me provide a complete proof, though most of it is already done.

[b]a)[/b] is basic Linear Algebra, where $ f: R^n\to R^m$ corresponds to the $ m\times n$  :!: matrix $ A$,
which has as columns the images of the standard basis of $ R^n$ under $ f$.

[b]d)[/b] follows immediately from [b]b)[/b] and [b]c)[/b].

[b]b)[/b]
"$ \Leftarrow$"
Assume $ f$ is not injective and take $ v\in R^n,v\not =0$ with $ Av=f(v)=0$.
Let $ t\not =0$ be some non-zero entry of $ v$ and $ P$ some $ n\times n$ permutation matrix
with $ Pv=\left(\begin{array}{c}t\\ \vdots\end{array}\right)$ and set $ Q: =\left(\begin{array}{cccc}t &&&\\ \vdots &1&&\\ \vdots &&\ddots &\\  \vdots &&&1\end{array}\right)$.
Then $ T: =P^{-1}Q$ has first column $ v$ and $ \det T=\pm \det Q=\pm t$.
And now $ X_iT$ has first column $ X_iv=0$ for all $ i=1,\ldots,u$
and therefore $ 0=\det X_iT=\pm t\cdot \det X_i$ for all $ i$.
"$ \Rightarrow$"
Take $ d\le \min\{m,n\}$ minimal with $ t\cdot \det X=0$ for all $ d\times d$ minors $ X$ of $ A$.
Take $ Y$ the $ (d-1)\times d$ minor of $ A$ from the first $ d-1$ rows and $ d$ columns.
Let $ Y_i$ be the $ (d-1)\times (d-1)$ minor of $ Y$, which you get by canceling the $ i$-the column for $ i=1,\ldots ,d$.
After permuting the rows and columns of $ A$ we may assume $ t\cdot \det Y_1\not =0$. ($ \det Y_1: =1$ for $ d=1$)
With $ A=(a_{ij})_{i=1,\ldots,m\atop j=1,\ldots ,n}$ we set $ Z_i: =\left({\textit{\Large{Y}}\atop \begin{array}{ccc}a_{i1}&\cdots & a_{id}\end{array}}\right)$ for $ i=1,\ldots ,m$.
$ Z_i$ is a $ d\times d$ minor of $ A$ for $ i\ge d$ and otherwise has a multiple row.
So by [url=http://en.wikipedia.org/wiki/Laplace_expansion]Laplace's formula[/url] we get
$ 0=t\cdot \det Z_i=t(-1)^{d+1}a_{i1}\det Y_1+\cdots+t(-1)^{d+d}a_{id}\det Y_d$
for all $ i=1,\ldots ,m$. (also for $ d=1$)
This gives $ f(v)=Av=0$ for $ v: =
\left(\begin{array}{c}(-1)^{d+1}t\cdot \det Y_1\\
\vdots \\
(-1)^{d+d}t\cdot \det Y_d\\
0\\
\vdots \\
0\end{array}\right)\in R^n\setminus\{0\}.$
And now for $ m<n$ we can add a zero-row to $ A$ and get a $ (m+1)\times n$ matrix $ \tilde{A}$,
in which every $ (m+1)\times (m+1)$ minor has a zero-row, and therefore determinant $ 0$.
The above argument then gives $ v\in R^n\setminus\{0\}$ with $ \tilde{A}v=0$, which gives $ f(v)=Av=0$.
So either way we get $ f$ not injective.

[b]c)[/b]
"$ \Rightarrow$"
If $ f$ is surjective we find $ v_1,\ldots,v_m\in R^n$, such that $ f(v_1),\ldots,f(v_m)$ is the standard basis of $ R^m$.
Then the $ n\times m$ matrix $ B: =\left(\begin{array}{ccc}v_1&\cdots&v_m\end{array} \right)$ satisfies $ AB=I_m$ and $ \det AB=1$,
and the result follows from the [url=http://en.wikipedia.org/wiki/Cauchy-Binet_formula]Cauchy-Binet formula[/url], including $ m\le n$.
"$ \Leftarrow$"
For $ J\subseteq \{1,\ldots,n\},|J|=m$ let $ A_J$ be the $ \{1,\ldots,m\}\times J$ minor of $ A$ and $ \text{adj}(A_J)$ be its [url=http://en.wikipedia.org/wiki/Adjugate]adjugate[/url].
Let $ B_J$ be the $ n\times m$ matrix, which has $ \text{adj}(A_J)$ as $ J\times \{1,\ldots,m\}$ minor, and otherwise zero-entries.
Then for $ \sum_{J\subseteq \{1,\ldots,n\},|J|=m}\lambda_J\det A_J=1$ and $ B: =\sum_{J\subseteq \{1,\ldots,n\},|J|=m}\lambda_JB_J$
we get
\begin{align*}AB&=\sum_{J\subseteq \{1,\ldots,n\},|J|=m}\lambda_JAB_J\\
&=\sum_{J\subseteq \{1,\ldots,n\},|J|=m}\lambda_JA_J\text{adj}(A_J)\\
&=\sum_{J\subseteq \{1,\ldots,n\},|J|=m}\lambda_J\det A_J\cdot I_m=I_m\end{align*}
and so for all $ v\in R^m$ we have $ f(Bv)=ABv=v$, which gives $ f$ surjective.
[b]q.e.d.[/b]

[quote="darij grinberg"]
At the moment I am trying to apply [b]c)[/b] to problem 2 [b]iv)[/b], which states that if $ f\in R\left[x\right]$ and $ g\in R\left[x\right]$ are two polynomials over $ R$, then the polynomial $ fg$ is primitive if and only if the polynomials $ f$ and $ g$ are primitive. Here, a polynomial $ a_{0} + a_{1}x + ... + a_{n}x^{n}\in R\left[x\right]$ is called [i]primitive[/i] if the ideal generated by $ a_{0}$, $ a_{1}$, ..., $ a_{n}$ is $ R$.[/quote]
This is trivial.
$ f\in R[x]$ primitive $ \Leftrightarrow$ the coefficients of $ f$ don't lie all in some maximal ideal $ I\subsetneq R$
$ \Leftrightarrow$ $ f\not\equiv 0\pmod{I}$ for all maximal $ I$.
And since $ R/I$ is a field for all $ I$ maximal, we get
$ fg\not\equiv 0\pmod{I}\Leftrightarrow f\not\equiv 0\pmod{I}$ and $ g\not\equiv 0\pmod{I}$
for all $ f,g\in R[x]$ and $ I$ maximal.

[quote="darij grinberg"]
By the way, I used [b]b)[/b] to solve problem 2 [b]iii)[/b] in Chapter 1 of Atiyah-Macdonald. The problem asks to show that if $ f = a_{0} + a_{1}x + ... + a_{n}x^{n}\in R\left[x\right]$ is a polynomial over $ R$, then $ f$ is a zero-divisor in $ R\left[x\right]$ if and only if there exists an element $ t$ of $ R$ such that $ tf = 0$.[/quote]
Could you show your proof?
I only found the following awkward induction.
[hide="Proof"]
Write $ f(x)=\sum_{i\ge 0}\alpha_ix^i$ and $ g(x)=\sum_{i\ge 0}\beta_ix^i\in R[x]\setminus\{0\}$ with $ f(x)g(x)=0$.
Then $ \sum_{i,j\ge 0,i+j=n}\alpha_i\beta_j=0$ for all $ n\ge 0$.

[b]1)[/b] Assume we have $ t\in R$ and $ n\ge 0$ with $ tg(x)\not =0$
and $ t\alpha_ig(x)=0$ for all $ 0\le i<n$.
Then  $ t\alpha_n^{k+1}\beta_k=0$ for all $ k\ge 0$ by induction.
[b]Proof:[/b]
Assume $ k\ge 0$ and $ t\alpha_n^{l+1}\beta_l=0$ for all $ 0\le l<k$.
Then $ t\alpha_n^{k+1}\beta_k=-t\alpha_n^k\sum_{0\le i\le n+k\atop i\not=n}\alpha_i\beta_{k+n-i}$.
By assumption for $ i<n$ we have $ t\alpha_i\beta_{k+n-i}=0$ as coefficient of $ t\alpha_ig(x)=0$
and for $ i>n$ we have $ l: =k+n-i<k$ and $ l+1\le k$, and therefore $ t\alpha_n^k\beta_{k+n-i}=0$.
This gives $ t\alpha_n^{k+1}\beta_k=0$, which completes the induction.
[b]q.e.d.[/b]

[b]2)[/b] For all $ n\ge 0$ we find some $ t\in R$ with $ tg(x)\not =0$
and $ t\alpha_ig(x)=0$ for all $ 0\le i\le n$ by induction.
[b]Proof:[/b]
Assume we already have some $ t\in R$ with $ tg(x)\not =0$
and $ t\alpha_ig(x)=0$ for all $ 0\le i< n$. (Take $ t=1$ for $ n=0$)
Then for $ d: =\deg g+1$ we get $ t\alpha_n^d\beta_k=0$ for all $ k\ge 0$ using [b]1)[/b].
So $ t\alpha_n^dg(x)=0$ and we find $ k\ge 0$ maximal with $ t\alpha_n^{k}g(x)\not =0$.
So we can take $ \tilde{t}: =t\alpha_n^{k}\in R$ with $ \tilde{t}g(x)\not =0$
and $ \tilde{t}\alpha_ig(x)=0$ for all $ 0\le i\le n$.
[b]q.e.d.[/b]

And now using [b]2)[/b] for $ n=\deg f$ we get $ t\in R$ with $ tg(x)\not =0$
and $ t\alpha g(x)=0$ for all coefficients $ \alpha$ of $ f(x)$.
So we find some coefficient $ \beta$ of $ g(x)$ with $ t\beta\not=0$ and $ t\beta f(x)=0$.
This completes the proof.[/hide]
In fact this shows, we can find $ t$ as a product of coefficients of $ f$
each with multiplicity $ \le \deg g$ and only one coefficient of $ g$.

The last problem seems to have been discussed [url=http://www.mathlinks.ro/viewtopic.php?search_id=1074977430&t=180120]here[/url].

[quote="darij grinberg"]This paragraph is totally useless now.[/quote]
Thank you. This is much better. :D

[quote="darij grinberg"]
I knew this proof. Let's say it this way, if you find a straightforward argument to make it constructive, I would be (positively) surprised.[/quote]
Now I get it! You want to avoid Zorn's lemma. (though, I'm not sure why)
Anyway if $ f$ is not primitive, and $ I\subsetneq R$ the ideal generated by the coefficients of $ f$,
then $ fg\equiv 0\pmod{I}$, so the coefficients of $ fg$ also lie in $ I$ and $ fg$ is not primitive.
Conversely if $ f,g$ are primitive and not $ fg$, with $ I\subsetneq R$ the ideal generated by the coefficients of $ fg$,
then $ f,g\not\equiv 0\pmod{I}$ and $ fg\equiv 0\pmod{I}$, so $ f,g$ are zero-divisors over $ R/I$,
and the other exercise gives some $ t\in R\setminus I$ with $ tf\equiv 0\pmod{I}$.
But $ f$ is primitive, so the ideal generated by the coefficients of $ tf$ is $ tR\subseteq I$. Contradiction!

[quote="darij grinberg"]I have not read your induction yet, but it definitely does not look more awkward than mine.[/quote]
But it is awkward.
[url=http://www.mathlinks.ro/viewtopic.php?search_id=1074977430&t=180120]Here[/url] you find a three-line proof from -oo-. (thanks for that, perfect_radio)

[quote="olorin"]Now I get it! You want to avoid Zorn's lemma. (though, I'm not sure why)[/quote]

Not only. I try to have proofs as constructive as possible. I like your new proof quite much, but I [i]think[/i] we can do better. There are different levels of constructivity here: Think of proofs as algorithms. We are given two primitive polynomials $ f$ and $ g$ - that is, we know the coefficients $ f_i$ and $ g_i$ of these polynomials [i]and[/i] we know coefficients $ \lambda_i$ and $ \mu_i$ such that $ \sum_i\lambda_i f_i = 1$ and $ \sum_i\mu_i g_i = 1$. We want an algorithm to "show" that $ fg$ is primitive, and by that I mean: to actually compute coefficients $ \nu_i$ such that $ \sum_i\nu_i h_i = 1$, where $ h_i = \sum_{k + l = i}f_kg_l$ are the coefficients of $ fg$.

Your first proof gives such an algorithm [b]if[/b] we have some oracle that takes as input a finitely generated ideal $ I$ (given by its generators) and gives as output a maximal ideal containing $ I$ [b]or[/b] a proof that $ I$ is the whole ring (and by proof, I mean again coefficients for a linear combination of the generators that equals $ 1$). Actually, we don't need the output ideal to be maximal; prime ideal is enough, I think. Anyway, we do have such black boxes in many cases (e. g. if our ring is $ \mathbb{Z}$). But we can always wonder whether we can do better.

Your new proof provides an algorithm which, at least in the version you have written it, requires an oracle that takes as input a ring element and an ideal (again given by its generators, so we are talking about finitely generated ideals), and tells you whether the element lies in the ideal or not. (In other words, whether the equivalence class of the element in the ring modulo this ideal is zero.) You need this because you apply part [b]b)[/b] of my problem, which obviously needs an oracle that tells you whether a given element of the ring is zero or not (how else could you obtain a [b]nonzero[/b] element as a result?) - and you apply this to $ R\slash I$ as the ring here, so actually you want to know whether an element lies in your ideal $ I$ or not.

The question is what kind of oracles we can avoid - for instance, we will most likely need oracles that can add, subtract and multiply ring elements. But in the case of this problem, I don't see an a priori reason that we need anything more than this - in fact, for given "small" degrees of the polynomials $ f$ and $ g$, we can write the required coefficients for $ fg$ by explicit formulas:
$ \lambda_1f_1 + \lambda_0f_0 = 1$; $ \mu_1g_1 + \mu_0g_0 = 1$ $ \Longrightarrow$
$ \left(\lambda_1\mu_1^2g_1 + 2\lambda_1\mu_0\mu_1g_0 - \lambda_0^2\mu_1^2g_0\right)\cdot f_1g_1 + \left(\lambda_1\mu_0^2g_0 + \lambda_0\mu_1^2g_1\right)\cdot\left(f_1g_0 + f_0g_1\right)$
$ + \left(\lambda_0\mu_0^2g_0 + 2\lambda_0\mu_0\mu_1g_1 - \lambda_1\mu_0^2g_1\right)\cdot f_0g_0 = 1$.
I computed this by hand using my proof, not by some random guessing. The proof doesn't really yield formulas for the coefficients, but an algorithm to find such formulas given the degrees of the polynomials. But to say the least, the proof is not really fun. If I were to write it up, it would be some 8 pages long even after problem [b]c)[/b] above (which is used in the proof) is solved.

Here is a hint on how I proceed:

To any integer $ a$ and any polynomial $ u\left(X\right) = \sum_{i = 0}^au_iX^i$ (here, we do not have to exclude the case $ u_a = 0$, and we should not because we don't want to use oracles telling us whether something is zero) corresponds a linear mapping $ M_a\left(u\right) = R^{a + 1}\to R$ given by the matrix $ M_a\left(u\right) = \left(\begin{array}{ccccc}u_a & u_{a - 1} & ... & u_1 & u_0\end{array}\right)$ (here I identify linear mappings with their corresponding matrices). It is clear that the polynomial $ u$ is primitive if and only if this matrix $ M_a\left(u\right)$ (or, if you want, the corresponding linear mapping) is surjective.

Let $ f\left(X\right) = \sum_{i = 0}^af_iX^i$ and $ g\left(X\right) = \sum_{j = 0}^bg_jX^j$. Then, $ \left(fg\right)\left(X\right) = \sum_{k = 0}^{a + b}h_kX^k$, where the coefficients $ h_k = \sum_{i + j = k}f_ig_j$ satisfy

$ M_{a + b}\left(fg\right) = M_a\left(f\right)N\left(g\right)$,

where $ N\left(g\right)$ is the linear mapping $ R^{a + b + 1}\to R^{a + 1}$ given by the matrix

$ N\left(g\right) = \left(g_{a + i - j}\right)_{0\leq i\leq a;\; 0\leq j\leq a + b}$.

Here, as usual, $ g_k$ is considered to mean $ 0$ if $ k < 0$ or $ k > b$. Here is how the matrix $ N\left(g\right)$ looks like:

$ N\left(g\right) = \left(\begin{array}{cccccccc}g_b & g_{b - 1} & ... & g_1 & g_0 & & & \\
& g_b & g_{b - 1} & ... & g_1 & g_0 & & \\
& & \ddots & \ddots & \ddots & \ddots & \ddots & \\
& & & g_b & g_{b - 1} & ... & g_1 & g_0\end{array}\right)$,

where empty spaces signify zeroes. This matrix has $ a + 1$ rows and $ a + b + 1$ columns.

We know that $ M_a\left(f\right)$ is surjective (since $ f$ is primitive) and that $ g$ is primitive. We want to show that $ fg$ is primitive, i. e. that $ M_{a + b}\left(fg\right)$ is surjective. In view of $ M_{a + b}\left(fg\right) = M_a\left(f\right)N\left(g\right)$, this will be clear as soon as we have shown that $ N\left(g\right)$ is surjective. This is a simplification of the problem, because we don't need to care for $ f$ anymore (though $ a$ is still important). But here the ugly part starts: of course, I use problem [b]c)[/b] above to show the surjectivity, and I even show more (namely, I prove that for arbitrary ring elements $ g_0$, $ g_1$, ..., $ g_n$, the ideal generated by the determinants of the $ \left(a + 1\right)\times\left(a + 1\right)$ minors of the matrix $ N\left(g\right)$ is the ideal $ \left\langle g_0,g_1,...,g_n\right\rangle^{a + 1}$; note that this looks like a typical result of late 19th century invariant theory). But this still requires an induction or maybe even a double induction. I can give some more details if you ask, but don't expect me to ever give a complete description of the proof, unless I succeed to find a serious simplification.

  darij

[quote="darij grinberg"] I am interested in others's solutions, generalizations and applications.
[/quote]it is a consequence of basic statements about Fitting ideals (I remembered them today thinking of some lecture of the Son of God [he used them to show that some coherent sheaf is a vector bundle <- this is a in the broader sense related application]). They can also be used for statements about the minimal number of generators of a module (here, this is $ 0$).


Let [tex] a,b,c [/tex] be three positive numbers. Prove that:
[tex] \frac{a^3+b^3+c^3}{4abc}+\frac{1}{4} \geq (\frac{a^2+b^2+c^2}{ab+ac+bc})^2 [/tex]

It's equivalent to
$(\sum ab)^2(\sum a^3 + abc) \ge 4abc(\sum a^2)^2$
Minus $4abc(\sum ab)^2$ on both sides
$(\sum ab)^2 (\sum a^3-3abc) \ge 4abc((\sum a^2)^2 -(\sum ab)^2)$
$(\sum ab)^2 (a+b+c) \ge 4abc(\sum a^2+\sum ab)$ (Cancelled $a^2+b^2+c^2-ab-bc-ca$)
$\sum_{sym} (a^3b^2) + \frac{5}{2} \sum_{sym} (a^2b^2c) + \sum_{sym} a^3bc \ge 2 \sum_{sym} a^3bc + 2 \sum_{sym} a^2b^2c$
This is a strict inequality true by muirhead ($(3,2,0) \succ (3,1,1)$)

So it's true with equality holds when $a=b=c$

[quote="Anh Cuong"]Umm... the last inequality you wrote, it is strange, RHS is 4 and LHS is ....tatata .
 :roll:[/quote]

but I have canceled $a^2+b^2+c^2-ab-bc-ca$ in the previous step. So if you regroup the initial inequality you get something like $A(a^2+b^2+c^2-ab-bc-ca) \ge 0$ where A is L.H.S. - R.H.S. in my last inequality.  :lol:


DEATH
[u]+   LIFE[/u]
 EREFI

D+E+A+T+H+L+I+F+R=IH

what digits are represented by each letter?

p.s. each letter is a different positive digit, and IH is a integer, not I*H

Are all the letters different digits? or are some of them the same? Also, for $\ D+E+A+T+H+L+I+F+R=IH$, does $\ IH$ mean $\ I*H$ or the integer $\ IH$?

Just looking at the Problem...I think the Letter stand for digits so that I=I E=E etc. IH is supposed to be 1 2 digit number like um.....eg.23, 34,25, 12...etc/

Why is this starting to get ignored?

I think it's the date that pulls it far back on the list

Perhaps it's too hard for the forum.  I'll move it up a level.


Numbers of the form $ F_{n} \equal{} 2^{2^{n}}\plus{}1$ are called fermat numbers.Without actually calculating $ F_{5}$,show that $ F_{5}$ is divisible by $ 641$.

I'm quite sure that you mean
$ 641 | F_{5}\equal{}{2^2}^5 \plus{} 1$ since a fermat number has the form $ F_{n}\equal{}{2^2}^n\plus{}1$

Yes sorry I meant $ F_{5}\plus{}1$

I have edited my post.

Since $ 5 \cdot 2^7 \equiv_{641} \minus{}1$ taking the fourth powers of the congruence gives us $ 5^4 \cdot 2^{28} \equiv_{641} 1$ and  $ 2^{32}\plus{}1\equal{}F_{5} \equiv_{641} 0$ since $ 5^4 \equiv_{641} \minus{}16\equal{}\minus{}2^4$


Let $x,y,z$ be positive real numbers such that $x+y+z=3$.Prove that:
$\frac{\sqrt{x+y^{2}z^{2}}}{yz}+\frac{\sqrt{y+z^{2}x^{2}}}{zx}+\frac{\sqrt{z+x^{2}y^{2}}}{xy}\geq 3\sqrt{2}$

[quote]How to prove it?[/quote]
two times Minkowsky


calculus
integration
trigonometry
real analysis
real analysis unsolved

If $ch\, t$ means $\cosh t=\frac12(e^t+e^{-t})$, the answer is $\pi$. Indeed, it is easy to see that, for every $\delta>0$, the integral $\int_\delta^\infty$ doesn't matter and near $0$ the denominator is equivalent to $\frac 12(t^2+a^2)$, so $a$ times the integral from $0$ to $\delta$ is almost the same as $\int_0^\delta\frac{2a\,dt}{t^2+a^2}=2\arctan(\delta/a)\to \pi$ as $a\to 0+$.

[quote="alekk"]$\int_0^{\infty} \frac{dt}{ch(t) - cos(a)}$.
[/quote]

the integral is $\frac{\pi-a}{\sin a}$

the require limit is $\pi$


calculus
integration
limit
real analysis
real analysis unsolved

f:[0;1]->R , continuous such that
$\any x \in ]0,1], f(x) \neq 0$
f(0)=0, derivable at 0 with
$ f'(0)=a\neq 0$
Let $k>1$
Compute $\lim_{x->0}\int_{x}^{kx}\frac{1}{f(t)}dt$

Because of the mean-theorem, we can write :
Sum(x =< t =< k*x){(t/f(t))*(1/t)} = (u(x)/f(u(x)))* Sum(x =< t =< k*x){1/t} = (u(x)/f(u(x)))*ln(k) , where x < u(x) < k*x
Manifestly if x --> 0 then  u(x)--> 0 , so (u(x)/f(u(x))) --> 1/f(0) = 1/a , and finally the initial limit is   ln(k)/a
 :cool:


[quote="duj12358"]a,b,c are positive real,and abc=1 Prove:
1/a(a+b)+1/b(b+c)+1/c(c+a)>=3/2
thanks.[/quote]
do you mean $ \frac{1}{a(a\plus{}b)}\plus{}\frac{1}{b(b\plus{}c)}\plus{}\frac{1}{c(c\plus{}a)}\ge \frac{3}{2}$?


limit
trigonometry
function
logarithms
absolute value
real analysis
real analysis unsolved

This one is very hard, I think. You can admit here that $\pi$ is badly approximable with rational numbers (Mahler or Mignotte or Hata's theorem). Then study the convergence of $\sum_{n\geq 1}{\frac{(-1)^n\cdot |sinn|}{n}}$.

The only thing that comes into my mind is the well-known limit:

                                         \[ \lim_{n\to\infty} \frac{|\sin 1|+|\sin 2|+\ldots+|\sin n|}{n}=\frac{2}{\pi}. \]

 Harazi, is this the key of solving this problem? :?

Cezar, I say I consider it very difficult, much much more difficult than that limit. Sorry, but you have to do many more steps in order to solve this problem, it's not an ordinary one. At least, I think so.

I know a proof of this one (which is not mine of course :rotfl:). Indeed this is very difficult, to say the least.
I can give hints or directions if someone is interested.

Of course, you can give a hint or something.

Maybe that way i'll stop looking at this problem like the calf at the new gate. (I know, it sounds horrible in English)

Please post a full solution for this serie, please. :)  I tried to solve it , but no result...... :(

[quote="cezar lupu"]Please post a full solution for this serie, please. :) [/quote]
OK, here goes a sketch :).
First, decompose the function $|\sin x|$ into its Fourier series. You'll get
\[ |\sin x|=\sum_{m=0}^\infty c_m\cos mx \]
with some $c_m$. All we need to know about $c_m$ is that they decay like $m^{-2}$.
Now let $N_1<N_2$ be two very large numbers. Our aim is to show that $\sum_{n=N_1}^{N_2} \frac{(-1)^n|\sin n|}n$ is small. Using the Fourier expansion, rewrite it as
\[ \sum_{m=0}^\infty c_m \sum_{n=N_1}^{N_2}\frac1n\cos n(m+\pi)\,. \]
Note now that, since $\pi$ is irrational, each series $\sum_{n=1}^{\infty}\frac1n\cos n(m+\pi)$ converges (Abel-Dirichlet test).  It remains to estimate the partial sums of the $m$-th series. According to Hata's theorem, the distance from $m+\pi$ to the nearest integer multiple of $2\pi$ is at least $cm^{-10}$ (the exact power doesn't matter). Thus, the partial sums of the series $\sum_{n=1}^{\infty}\cos n(m+\pi)$ are bounded by $Cm^{10}$ for large $m$. Now, to bound the partial sums of the series $\sum_{n=1}^{\infty}\frac1n\cos n(m+\pi)$, note that $\left|\sum_{n=1}^{N}\frac1n\cos n(m+\pi)\right|\le C\log m$ for $N<m^{10}$ (this is just a crude absolute value estimate) and $\left|\sum_{n=m^{10}}^{N}\frac1n\cos n(m+\pi)\right|\le C$ (this is Abel-Dirichlet). Therefore, the partial sums of the $m$-th series are bounded by $C\log m$. But $\sum_m |c_m|\log m$ converges and, therefore, we can first choose $M$ such that $\sum_{m>M}|c_m|\log m$ is very small and then choose $N$ such that the sums from $N_1$ to $N_2$ of every of the first $M$ series are small if only $N_2>N_1>N$. This finishes the proof of convergence.

Beautiful! In our homework, the indications were to use the Hardy-Littlewood tauberian theorem and approximations of $\pi$, but this is much more interesting.

Here are the steps, I will let you complete the proof:

   1) Use the fact that $\pi$ is badly approximable with rationals to prove that $\frac{ ln|cosn|}{lnn}$ is bounded and that for all $k>1$ the series $\sum {ln|cos n|}{n^k}$ is convergent.

  2) Put $h(x)=\sum {\frac{(-1)^n|sinn|x^n}{n}}$. Express $h(x)$ in terms of $\sum_{p\geq 1}{\frac{1}{4p^2-1}\cdot\sum_{n\geq 1}{\frac{(-1)^{n+1}cos(2pn)x^n}{n}}=g(x)}$ where $g$ is defined on $(-1,1)$

  3) Use 1) to prove that $g$ is defined in 1.

  4) Prove that $h$ has a limit in 1.

  5) Conclusion: use Hardy-Littlewood tauberian theorem: if $a_n=O(1/n)$ and there exists $lim_{x goes to 1}{\sum{a_n x^n}}$ then $\sum{a_n}$ converges.
   The last "exercise" is in fact a very serious theorem, I would be glad to see a simple approach, since the only solution I know for it is quite long and complicated.

The connection between cezar lupu's post (#2) and fedja's post (#5): cezar has essentially given the constant term, $c_0,$ of fedja's Fourier series.

Is it a oral exam or written exam of Ecole Centrale ?
Wich year ?

I think not even examinators at ENS can give such a problem in oral examination. It is written exam, but I don't know the year.

The convergence of the series and some other related results are also in "Diophantine Approximations and Convergence of Alternating Series", by N.V.Rao in Proceedings of the American Math. Soc.  Vol.93, No.3 1985

Another approach that also uses the irrationality measure of pi is given here by me: https://math.stackexchange.com/questions/2202138/determine-whether-sum-n-1-infty-frac-1n-sinnn-converges/2207289#2207289

Main ingredients are
1. Koksma inequality
2. Erdos -Turan inequality.

wow , just proving the convergence is that hard ,  finding the limit   is a whole another story   :maybe:  there is a lemma that I think  can help with both if I don't make any nonsense !
for $f : \mathbb{R} \to \mathbb{C}$  continuous and $2\pi$ periodic ,  let $\alpha \in \mathbb{R}-\mathbb{\pi\mathbb{Q}}$  for a positive integer $n\geq 1$
$I_{n}(f)=\sum_{k=1}^{n} \frac{f(k\alpha)}{k}$   then $(\frac{I_{n}(f)}{\ln(n)})$ converge to $\mu_{f}$  The average of $f$ .

As harazi said it was problem exam Centrale 1980 MP , analysis

See also
[url=https://math.stackexchange.com/questions/735283/convergence-of-an-alternating-series-sum-n-geq-1-frac-1n-sin-nn?noredirect=1&lq=1] see here [/url]

https://math.stackexchange.com/questions/735283/convergence-of-an-alternating-series-sum-n-geq-1-frac-1n-sin-nn?noredirect=1&lq=1

Here a generalization
N.V.Rao diophantine approximation and convergence of alternating series
Proceeding of the American Mathematical Society ,volume 93, number 3,March 1985
page 420-422

[url=https://www.ams.org/journals/proc/1985-093-03/S0002-9939-1985-0773994-4/S0002-9939-1985-0773994-4.pdf] see here[/url]
https://www.ams.org/journals/proc/1985-093-03/S0002-9939-1985-0773994-4/S0002-9939-1985-0773994-4.pdf

The problem exam Centrale 1980 M can be find in
[url=https://concours-maths-cpge.fr/] see here [/url]

https://concours-maths-cpge.fr/

année—>1980
concours—> Centrale-Supelec
filière —> MP (ex M)
matière—> mathématiques

and choose "rechercher " , you will see

Centrale MP math 1 and choose  "details"  you will find

énoncé means the problem exams
corrigé means solution


[color=green][b] If $a,b\in {\mathbb R}^n \; ,\; a=(a_1,a_2,...,a_n)\; ,\; b=(b_1,b_2,...,b_n),$ then  prove the inequality
\[ \begin{array}{|c|} \hline \\ \left[n\sum\limits_{k=1}^{n}b_k^2 -\left(\sum\limits_{k=1}^{n}b_k\right)^2\right]\left[ \left(\sum\limits_{k=1}^{n}a_k^2\right) \left(\sum\limits_{k=1}^{n}b_k^2\right)-\left(\sum\limits_{k=1}^{n}a_kb_k\right)^2\right]\ge \\ \\ \ge \left[\left(\sum\limits_{k=1}^{n}a_k\right)\left(\sum\limits_{k=1}^{n}b_k^2\right)- \left(\sum\limits_{k=1}^{n}b_k\right) \left(\sum\limits_{k=1}^{n}a_kb_k\right)\right]^2\\ \\ \hline \end{array}\; .  \]
[/b][/color]

Wow,I find this nice inequality is not hard:
just notice:
$n\sum\limits_{k=1}^{n}b_k^2 -(\sum\limits_{k=1}^{n}b_k)^2= \sum_{i> j} (b_i-b_j)^2$
and
$(\sum\limits_{k=1}^{n}a_k^2) (\sum\limits_{k=1}^{n}b_k^2)-(\sum\limits_{k=1}^{n}a_kb_k)^2=\sum_{i > j} (a_jb_i-a_ib_j)^2$
and
$(\sum\limits_{k=1}^{n}a_k)(\sum\limits_{k=1}^{n}b_k^2)- (\sum\limits_{k=1}^{n}b_k) (\sum\limits_{k=1}^{n}a_kb_k)=\sum_{i > j} (b_i-b_j)(a_jb_i-a_ib_j)$
then use Cauchy-Schwarz.
Edited a small mistake

[quote="flip2004"][quote="zhaobin"]Wow,I find this nice inequality is not hard: just notice ...... [/quote]
OK, but please post a complete solution !
However, it's clear that the posted inequality implies Cauchy- Schwarz inequality.
Can you post all equality cases ? Thanking for interest,[/quote]
Do you mean when does the equality holds?
It is a problem :( .Let me think about it.Now I must to go to sleep first :P

[quote="flip2004"][quote="zhaobin"][quote]  Can you post [u][b]all equality cases[/b][/u] ? [/quote] Do you mean when does the equality holds?  It is a problem ...:[/quote]
[size=75][color=blue][b]I appreciate that my posted inequality is not equivalent with Cauchy-Schwarz inequality ! [/b][/color][/size]
This may be observed by investigating all equality cases (in posted inequality) , then to compare them with equality case
 in Cauchy-Schwarz inequality.[/quote]
Yes ,But I find it is not easy.Have you find all equality cases?

[quote="zhaobin"]... Can you post  [u][b]all equality cases[/b][/u] ? ...  Have you find all equality cases?[/quote]
Assuming that $n$ is a fixed positive integer, then [b]equality case holds if and only if there exist real numbers $\alpha,\beta,\gamma$
such that   $\alpha a_k+\beta b_k +\gamma =0$ for all $k\in \{1,2,...,n\}.$ [/b]
...There are similar (but different) conditions  for equality when  $n$ is supposed to be variable in  ${\mathbb N}^* .$([size=75][color=blue][b]Try to find it ! [/b][/color][/size] ).


Does anyone know any "popular" NJ math contests? :maybe:

Don't forget CML! And for high schoolers, PUMaC.


trigonometry
function
calculus
calculus computations

Let
f(x)=(a+sinx)(4+sinx)/(1+sinx)
where a>1 , a and x are real numbers.
As x varies, find the minimum value of f(x).

$ f$ is periodic with period $ 2\pi$, so we need only consider values of $ x$ in the interval $ [0, 2\pi]$.

Setting $ z \equal{} 1 \plus{} \sin x$ may simplify the function somewhat.

In other words, consider a new function $ g(z) \equal{} (a \minus{} 1 \plus{} z)(3 \plus{} z)/z$.  You could find the minimum of $ g$ on an appropriate interval using standard techniques, and then you will have found the minimum of $ f$.

Can we use high school maths to solve this problem??


hello, put $ n\equal{}4$ in your equation.
Sonnhard.

Hello
[hide="Hint (by integration)"]
u know $ (1\plus{}x)^n\equal{}C_n^0 \plus{} xC_n^1\plus{}x^2C_n^2\plus{}...........\plus{}x^{n}C_n^n$         ...........$ equation(1)$
Transfer x by $ \frac{1}{x}$ and then integrate the equation.
On RHS of ur new equation u will get $ C_n^0 (logx)\plus{}C_n^1x\plus{}C_n^2\frac{x^2}{2}..........\plus{}C_n^n\frac{x^{n}}{n}$
multiply this equation with the following equation
$ (1\plus{}x)^n\equal{}C_n^0x^n\plus{}C_n^1x^{n\minus{}1}\plus{}............C_n^n$
compare the coefficient of $ x^n$ and then put x=1.
u will get the value of n.
[/hide]
Thank u.

hello, for $ n\equal{}4$ we get $ 16\plus{}18\plus{}\frac{16}{3}\plus{}\frac{1}{4}\equal{}\frac{475}{12}$, so your equation is fulfilled.
Sonnhard.


Hi

Prove that:

a/b+b/c+c/a>=(a+b)/(a+c)+(b+c)/(b+a)+(a+c)/(b+c)  with (a,b,c>0)

bye

This one was (1) of the thread

  http://www.mathlinks.ro/viewtopic.php?t=5263

And it is really hard, see the first solution in the [url=http://www.mathlinks.ro/Forum/contest/1st%20Edition/1st%20round%20sol.pdf]proposed solutions[/url].

[[b]EDIT:[/b] Links updated.]

  Darij

strange, both of those links refuse to open here :?


Does anyone know where I can find some AIME-level problems?

Thanks.

Harvard MIT guts round is good, and so is the canadian open mathematics challenge

some of the last  problems (probably only#25) of the Cayley competition (canadian) might help... took me a while to set up.

other than that, you probably have heard of the other ones... ARML individual, HMMT, stanfordmathmeet (although in recent years ive heard that they have dropped in the quality and difficulty).


If x and y are each positive integers, how many distinct ordered pairs (x,y) exist such that (x,y) lies on one of the bisectors of the angles formed by the lines x-2y+5=0 and 2x-y-3=0

This is not a pre-olympiad level problem at all.  It sounds like Analytic Geometry textbook.

Find the locus of points equidistant from both. That's your bisector.


I wasn't sure where to put this but I'm going nuts that I can't figure out the logic behind this.
Here's the webpage that "explains" it.
http://en.wikipedia.org/wiki/Monty_hall_problem

How am I more likely to win by switching than if I stayed on the same door? To me, there's a 50/50 chance. I don't know which door the car is behind, so switching or staying has the same chance of winning me the car.

The way I understand it is you have a 66% chance of picking a goat, which means switching would give you the car.
And you have a 33% chance of picking the car, which means switching will give you a goat.

So the chances that you picked the goat are greater than if you picked the car, which means switching is usually a better idea.

Statements about probability are in essence statements about statistics.  The statement [i]"The outcome of experiment $ E$ is $ A$ with probability $ p$"[/i], is essentially the same as saying [i]"If experiment $ E$ were performed $ n$ times, we should expect $ A$ to be the outcome approximately $ np$ times."[/i]  This is a rather simplified view of probability, but 'typically' this is a good way of thinking about it.

So now let's let $ E_{\text{switch}}$ be the Monty Hall experiment in which you always choose to switch doors, and $ A$ be the outcome that you win the car.  If $ E$ is performed $ n$ times, we expect by symmetry that the door [i]originally[/i] chosen will have a car behind it about $ \frac {1}{3}n$ times.  Therefore, switching to the other door will give you a car about $ \frac {2}{3}n$ times.  Thus, if we accept the interpretation of probability as given in the first paragraph, we must conclude that the probability of event $ A$ is $ \frac {2}{3}$.

I don't see why switching would give you a 2 in 3 chance but staying doesn't give you a 2 in 3 chance.

It can be easily seen from the picture in link you wrote, but I'll write it here:

1) You firstly have chosen the right door $ \longrightarrow$ if you change, you lose
2) You firstly have chosen the goat door $ \longrightarrow$ as you see the another goat door opened, if you change, you win

Probability that 1) case will appear is $ \frac{1}{3}$ and 2) case will apear in $ \frac{2}{3}$ cases because there are $ 2$ goats and $ 1$ car from $ 3$ objects

[quote="Bugi"]It can be easily seen from the picture in link you wrote, but I'll write it here:

1) You firstly have chosen the right door $ \longrightarrow$ if you change, you lose
2) You firstly have chosen the goat door $ \longrightarrow$ as you see the another goat door opened, if you change, you win

Probability that 1) case will appear is $ \frac {1}{3}$ and 2) case will apear in $ \frac {2}{3}$ cases because there are $ 2$ goats and $ 1$ car from $ 3$ objects[/quote]
I think I understand it now. It's just that there are so many explanations that make it sound a LOT more difficult than it really is.

Hmm what if the game show host did not have to open a door with a goat in it?

Then the probability is 1/2.  (This is true, for example, of the game show "Deal or No Deal.")  This restriction (that Monty is guaranteed to open a door hiding a goat) is crucial, and it's very important when you state the Monty Hall problem that you make it clear.

what if the game show host did not know what was behind each of the doors, and randomly opens one of the other two?

but then what if there was a car behind the other door?
then the game would be pointless

[quote="Bugi"]violin321, enough useless questions. I could have also asked some questions like: What if we can open all 3 doors? :rotfl:[/quote]

This is hardly a useless question.  If Monty opens a door at random and reveals a goat, the answer is not the same as if Monty opens a door which he knows does not have the car behind it.

In the Wikipedia article, the figure under the Solution section clearly illustrates why its a statistically better choice to change your option when one option is removed. The key to why it can be confusing lies in that only 3 doors are considered, if you read 'Increasing the number of doors' under the section 'Aids to Understanding' this is easily seen.

Incidentally, I remember this problem being mentioned on a NUMB3RS episode! Nice show...  :D


Shake hands with them as often as possible. And give good speeches.

Another one:
You have 3 seconds to answer the question, so if you're pretty fast at it, and youve just about got it, then just raise you hand, and then in those 3 seconds, finish working the question, and answer.

AND THE BIG TIP:
[u][b]PRACTICE!!![/b][/u] you can ask for past countdown rounds from MATHCOUNTS, and the Elisa Saabs web page has hundreds of them.

Practice also helps... lol.  Lack of that is probably why I failed countdown when it counted :(.

[quote="ln(dx/dy)"][quote="dwx314"]Another one:
You have 3 seconds to answer the question, so if you're pretty fast at it, and youve just about got it, then just raise you hand, and then in those 3 seconds, finish working the question, and answer. [/quote]

Actually, you have about 6 seconds.  First they announce your name and school, then you get 3 seconds, then you can waste 1 second by starting your answer as "umm".

Another one~~If there is 10 seconds left and you can't get the answer, and you can see your opponent is just about to get it, buzz in, then pretend like you're thinking a lot until they ask for the answer. Then give them a huge answer that takes about 5 seconds to say. ;) Your opponent will run out of time, and no one will win.(that question)[/quote]
Now that's just cruel! But hey, whatever it takes, huh? :D My advice is to not think of how much it counts-just focus on the problem. Disregard that if you're like me and thrive under pressure, though. Adrenaline helps me think.

You definitely will not get 6 seconds at Nationals. They'll say your first name and expect an answer.

3 seconds is a reasonable expectation.

I suggest practicing without pencil and paper fairly regularly.  Learn to do things like multiply, factor integers, reduce fractions, and convert between decimal/fraction in your head.  Winning in the count down round has a lot to do with mental arithmetic abilities -- particularly at the state and national levels where many students know how to work nearly every problem.

I'll add just two:

1 -  Answer the question before your opponent  does.
2.    Make sure that the answer is correct.

Good ones Gyan. Also, RIng the buzzer and don't raise ur hand(surprising how many people do.). Bring a pen also. Writes faster.

[quote="SirErnest"]I won Countdown in the Indiana State competition, but, unfortunately, it doesn't count here.  I just scan the question really fast and mainly look for the numbers.  Then I read the last line and see what they are asking for.  It also helps that I am on the Science Bowl team.  My Science Bowl team made it to Nationals in Colorado last year.[/quote]

I remember that...state CountDown I mean. Another hint that I picked up at state, when teh answer says to answer as a percent, answer as a percent not a fraction. And when the guy next to you answers as a fraction, quickly convert it to a percent and buzz in adn get the correct answer. :)

[quote="sponge008"][quote="ln(dx/dy)"]
[quote="dwx314"]
Another one:
You have 3 seconds to answer the question, so if you're pretty fast at it, and youve just about got it, then just raise you hand, and then in those 3 seconds, finish working the question, and answer.
[/quote]

Actually, you have about 6 seconds. First they announce your name and school, then you get 3 seconds, then you can waste 1 second by starting your answer as "umm".

Another one~~If there is 10 seconds left and you can't get the answer, and you can see your opponent is just about to get it, buzz in, then pretend like you're thinking a lot until they ask for the answer. Then give them a huge answer that takes about 5 seconds to say.  Your opponent will run out of time, and no one will win.(that question)
[/quote]

Now that's just cruel! But hey, whatever it takes, huh?  My advice is to not think of how much it counts-just focus on the problem. Disregard that if you're like me and thrive under pressure, though. Adrenaline helps me think.[/quote]You have to think about how much it counts. If it is in sudden death, sometimes you have to break the rules a little if you want any chance of winning. Once last year in 6th grade, when we were doing a practice cd round in school, I had a coughing fit when I buzzed in, which gave me a lot of time to finish the problem. The answer was accepted, but I got it wrong anyway. :D  :D  :D  :D  :D  :D  :D

Countdown is unofficial in my chapter and state  :D  (this is NY)

[quote=Ultroid999OCPN][b][s]12.5 year bumps are lit[/s][/b]
.........Really?
@GrilledBrain: Please don't bump threads.
Personally, I feel that if you freak out, you will lose, so staying calm is important.[/quote]

Please follow your own advice
If you really want to tell them not to bump threads, just PM them

It really depends what level you're on.
If it's a school or chapter round, then countdown is probably unofficial, so you don't need to worry about it too much. Some chapters/schools have buzzers; some do not.
If it's state, it almost certainly has buzzers, and in some states, countdown is official, while in others, it's not. Unlike school and chapter, people will generally be much more strict about the [b]three seconds[/b] you have to answer.
If it's nationals, then you also have the pressure of being on live TV in front of 200+ people, besides facing some of the best mathletes in the country. Besides that, it's mostly the same as state.

All in all, just know your strengths and ability, and play accordingly.

[quote=MathMan5000]our chapter does official countdown. also, u have 45 seconds to answer questions in countdown....[/quote]

They mean after you hit the buzzer.

Don't do what I did this year on the Chapter Countdown Final for this year. I buzzed in and when I said the correct answer, the time for speaking ran out. And then the other person buzzed in and immediately said the same answer... Got it right. This happened twice, cuz i hesitated before answering. Somehow i still made a comeback.......  :10:

And also, it helps to be relaxed during the countdown as much as possible. Just don't stress out.

[quote=ln(dx/dy)]Do you have any tips you can share for doing well in the Count Down round :?:[/quote]

when you're "working out" the problems on paper, try writing really noisily to distract and/or scare your opponent...

no joke it works

Our countdowns just if you get the answer right, you move on. Make a silly mistake, well your out.

The most important thing about countdown that you need to remember is to not hesitate. Remember that it's a fast-paced round, and that if you take too long, you'll be beat to the question.

[quote=e_power_pi_times_i]It is very hard to "keep your left hand at the buzzer and right hand for writing" when you are writing with your left hand. Thunk.[/quote]

[quote=GameMaster402]then switch it lol[/quote]

as a lefty individual i can confirm that this strategy is very effective

My state and chapter both use countdown officially, and this is pretty basic, but read the problem as FAST as you can and understand it quickly, as it can give you a split second advantage which could be what decides who wins that question.

um so try skipping to the end of the question to see what they want, then pick off the important info

I wouldn't recommend this strategy. Read the problem normally.

[quote=KevinTheFirst][quote=thedoge]It really depends what level you're on.
If it's a school or chapter round, then countdown is probably unofficial, so you don't need to worry about it too much. Some chapters/schools have buzzers; some do not. [/quote]
Yeah lol our chapter doesn't even do countdown
And why is this thread from 2005?[/quote]

No KevinTheFirst, our chapter does do countdown, it's just that it's not official.

More than likely, if you read the first sentence and the last sentence, you will already know what the question will be about.

EDIT:300 MSM posts!!!!!!!!!!!!!!!!!!!!!!!!!!!


Two circles whose radii are in the ratio $4 : 1$ touch each other externally at $M$ an lie inside a rectangle $ABCD$ such that the larger circle touches sides $AD,BC$ and $CD$, and the smaller circle touches the sides $AB$ and $AD$. The common tangent at $M$ to the circles meets sides $AD$ and $AB$ at $P$ and $Q$.
Find the ratios $AP/PD$ and $AQ/QB$.

Is it 1/3 and 3/5 ?

My circle in triangle geometry's kinda sketchy...

Are you sure?
Can you post a complete solution?

I'm getting $\frac{AP}{PD}=\frac{1}{2}$ and $\frac{AQ}{QB}=1$.  Here's my solution...
Let the center of the larger circle be $O$, and that of the smaller circle be $I$.  Draw the radius of the larger circle, $OE$ so that $E$ is on $AD$.  Now let the smaller circle touch $AB$ and $AD$ at $X$ and $Y$ respectively.  Let the circles touch at $L$.  Drop a perpendicular from $I$ to $OE$: call the intersection point $N$.  Let $IN$ intersect $PQ$ at $M$.  Let the radius of the larger circle be 4, that of the smaller 1.
Now that those are all defined, the solution...
Clearly, $AB=8$.  $OL=5, ON=4-1=3 \implies IN=4 \implies NX=5 \implies AD=9$.  Note the following about the following triangles.  $ION~INL~QNX~QPA$.  Thus, $QPA$ is a 3-4-5 right triangle.  Let $AX=AY=1$, $PY=PL=n$, $QL=QX=x$.  Then $(1+n) : (1+x) : (n+x)=3:4:5 \implies n=2,x=3$.
$\therefore$ $\frac{AP}{PD}=\frac{n+1}{8-n}=\frac{1}{2}$, and $\frac{AQ}{QB}=\frac{x+1}{7-x}=1$.


quadratics
number theory proposed
number theory

Prove for all positive integer $n$: $2^n+1$ has no prime divisor in form $8k-1$.

It's from Vietnam 2004 TST. The proof is simple for those who know quadratic reciprocity and practically impossible to find otherwise.


limit
trigonometry
calculus
calculus computations

[quote="gotztahbeazn"]Prove that:

$\displaystyle\lim_{x\rightarrow 0} \frac{1-\cos x}{x} = 0$[/quote]

$= \mathop {Lt}\limits_{x \to 0} \frac{{1 - \cos ^2 x}}{{x(1 + \cos x)}} = \mathop {Lt}\limits_{x \to 0} \frac{{\sin ^2 x}}{{x^2 (1 + \cos x)}}.x = \mathop {Lt}\limits_{x \to 0} \frac{{\sin ^2 x}}{{x^2 }}\frac{1}{{(1 + \cos x)}}.x = 0$

L'Hopital rule works here.
\[\displaystyle\lim_{x\rightarrow 0} \frac{1-\cos x}{x}=\displaystyle\lim_{x\rightarrow 0} \frac{\sin x}{1}=0\]

Let $f(x)=\cos x$, we have $f'(x)=-\sin x$.

$\lim_{x\rightarrow 0} \frac{1-\cos x}{x}=-\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=-f'(0)=0$


ARML's in less than a month, so I was wondering who all from AoPS is going to ARML at Iowa in 07.

I'll be there with Indiana Gold.

hmm I'll probably go on bronze, if I'm lucky I"ll get on silver...(from IN)

Well, I'll be there...

...from Iowa City.

I think I'll walk to the test site!

[quote="minsoens"][quote="Altheman"][quote="jackb1115"]I'll be there from Chicago, hopefully Chi A.

JB[/quote]

me too[/quote]ill be there with the Chicago also[/quote]

I aswell, but I dont think I have much of a chance at A team

[quote="pkothari13"][quote="Klebian"][quote="sapphyre571"]I'll be there from TX[/quote]
same
texas might be interested in balling against indiana[/quote]

yayuhhh thats what i'm talkin' 'bout. we'll show you how we do it up in h-town and d-town like them rockets and mavs... oh wait.  :( . i meant.. san antonio...[/quote]

well the pacers didn't even make the playoffs, much less win 60+ in the regular season

i'm going from IN. hopfully on Blue.

Nat Mc, are you "EG"?

I hope I'll be there with Wisconsin RED (top team)

Probably going to the lecture also, unless my team holds a Mao tournament or something :D

[quote="pianoforte"]Go Missouri!  :laugh:

Hopefully the A team.[/quote]

about sums it up

I'll be schleppin' along with the MN team, spending Saturday in the grading room, praying that there are no incorrect solutions this year  :maybe:

Also, for those MN kids on here, how'd the last two practices go? (We should use the cobweb-bedecked MN forum, come to think.)

[quote="nat mc"]if anyone wants to play basketball on friday, some ppl from indaina will probably be there playin so we could meet up[/quote]

basketball? where?

Wait, good relays?!  That's a new one.  :P
Good luck, guys.  I'll be looking for you in the top 7 this year.

I'll be there with the Minnesota Maroon. Fun, fun.

[quote="Danbert"]Well, why not add to the multiple off-topic converstions?

[quote="generating"]Also, for those MN kids on here, how'd the last two practices go?[/quote]

Generally speaking both practices went well - lots of 10s on team events, good relay scores, etc.  In particular, the most recent practice (i.e. the one I most clearly remember) featured what was probably the best-done team event I've ever seen - everybody worked together well, all the problems were done and checked, team dynamics were great - and on our practice NYSML, the Gold team lost to NYC A by either 3 or 6 points, which is very little considering the potential total score and the fact the we were missing Nick and Rohan for a large part of the day.  So, we're all excited, hoping to finally do really well again.  This is a good year for us.[/quote]

I agree with this assessment of the practices, it also seems like both teams do pretty well on power comparing our scores and NYC A's.  The maroon team also beat, I believe, the top teams in both divisions on the individual round questions 5 and 6.

I'll be going on either Indiana's A team or B team, not sure yet.

Yeah, the Missouri people will finalize A/B teams the Friday before the competition.

Procrastinators of the world, unite! Tomorrow.

just to be clear,

[b]ON FRIDAY, MEET BY THE EULER SPOILERS LECTURE[/b]

[i think that is good at least]

Will be there at Penn State :)

Argh, didn't see that there was a post for each location :blush:

I'm going from Indiana. Probably B team, though. What day do the actual competitions start?

The actual competitions start on Saturday, but ARML wants everyone to be there on Friday night.


consider :D  A 4 by 8 rectangle.with 8 vertical and 4 horizontal lines if a king stays at the bottomless left in how many ways can he get to his castle just at the bottomless right in exactly n-moves(lets say n=7)if all movements are allowed(vertical,horizontal&diagonally)

Do you mean to say that the king is on a 4 square by 8 square board? And also, the word is just "bottom" [i]("bottomless" can be one of several things, none of which you probably intend to mean :P)[/i]


without calculators  find the measure of acute angle x
$\cos x=\frac{1}{1+(2+\sqrt{3}-\sqrt{2}-\sqrt{6})^{2}}$

Could you please post how you found that?

We have:
$\cos^{2}x=\frac{1}{1+\left(2+\sqrt 3-\sqrt 2-\sqrt 6\right)^{2}}=\frac{1}{1+\left[\left(\sqrt 3-\sqrt 2\right)\left(\sqrt 2-1\right)\right]^{2}}\\ =\frac{2\sqrt 6+2\sqrt 2+8}{16}=\frac{\sqrt 6+\sqrt 2+4}{8}$
$\Longrightarrow \cos 2x=2\cos^{2}x-1=\frac{\sqrt 6+\sqrt 2}{4}=\cos\frac{\pi}{6}\cos\frac{\pi}{4}+\sin\frac{\pi}{6}\sin\frac{\pi}{4}=\cos\frac{\pi}{12}$
$\Longrightarrow x=\pm\frac{\pi}{24}+k\pi\quad k\in \mathbb{Z}$
On the otherhand $x<\frac{\pi}{2}\Longrightarrow x=\frac{\pi}{24}$


I'm taking AMC-10 next year on Feb. 16. I hope I can make it to AIME but it's kind of nervous you know. ;) My average score is always above 100 but just little bit below than 120.  :( I hope I can increase my skill better... During the winter vacation.

Tell which AMC you taking!

I'm taking the AMC 10B on February too. But I'm not too confident. My average practice test score is 95...  :(  I'm really worried. I've been studying more during this vacation, and I hope to improve. Anyone with last minute tips?

AMC 12...i didn't qualify with the AMC 10 8th grade year....with the AMC 12 9th grade year i qualified comfortably! :lol:

12A and 10B is my plan, I'm not really worried about AMC because I'm a 9th grader and all I need is a 7 or 8 on AIME(very difficult!)...

Probably AMC 10A only. I'll try 12B if I can handle old AMC 12 tests though. I got 112 on the one AMC 10 test I've taken so far (it was the 2002 test or something, so blank answers were 2 points instead of 2.5) so I'm not sure if that's, for me, unusually low, unusually high, or average. So for the AIME, I'm borderline at best and don't have a chance at worst. I'm glad to know qualifying isn't entirely impossible, though.

12 A for sure, and, if I can arrange things with another school 12 B as well.

I'm doing 10A and 12B. I have just qualified on all the practice tests I've taken, so I think I had better take both of them.

Personally, i dont like AMC10, last year i scored higher in AMC12 than AMC10.  Basically, you really have to be aware of computational mistakes in AMC10, and DON'T guess!!!! (this is the way i messed up last year) Leave them blank!!!!

AMC12, not sure which... and i think my school will only order 1 set for everybody

my school doesn't even give AMC10s, so during my 9th and 10th grade years i only had the option of taking the AMC12 =|

let's hope I qualify for the aime  :?

[quote="Bictor717"]I'm taking both AMC 12's. I've never taken two AMCs the same year before, is it strenuous?[/quote]

I wouldn't say so; you need only study once, although you may need to review a few times.


In the equilateral triangle $ ABC$ is chosen a point $ O$. Its symmetric point with respect to the sides $ BC$, $ CA$ and $ AB$ are denoted respectively with $ A_1$, $ B_1$ and $ C_1$. Prove that the lines $ AA_1$, $ BB_1$ and $ CC_1$ intersects at a common point.

[quote="Luis González"]$\triangle ABC$ becomes a Napoleon triangle (either internal or external) of $\triangle  A_1B_1C_1,$ so  the lines $ AA_1, BB_1,CC_1$ concur at a Napoleon point of $\triangle A_1B_1C_1.$[/quote]
1st of all, what's the proof of this? I mean $ ABC$ may be just homothetic to the Napoleon's triangle of $ A_1B_1C_1$....

And the concurrency here is not so well know as far as I know, (and I also think that this problem is true for not for the reflected points  but also for the glide reflected points)

Dear it's the celebrated Jacobi's theorem: In a $\triangle ABC,$ points $X,Y,Z$ are constructed, all outside or all inside, such that $\angle BCX=\angle ACY,$ $\angle CAY=\angle BAZ$ and $\angle ABZ=\angle CBX.$ Then the lines $AX,BY,CZ$ concur.

For a proof try a search, it has been posted before.


If $ x \plus{} y \equal{} 1$, then the largest value of $ xy$ is:

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 0.5\qquad
\textbf{(C)}\ \text{an irrational number about }{0.4}\qquad
\textbf{(D)}\ 0.25\qquad
\textbf{(E)}\ 0$

[hide="Solution"]$ x \plus{} y \equal{} 1\geq 2\sqrt {xy}$ (AM-GM)

$ 1\geq 4xy$

$ xy\leq\frac {1}{4}$.

Equality when $ x \equal{} y \equal{} 1/2$.

Answer: $ \boxed{\textbf{(D)}\ 0.25}$.[/hide]

Is this correct? It depends on $ x,y$ being positive which is not given, but if either was negative their product would be too small. (And both can't be negative).

Yeah, it's correct. You also also view this geometrically by graphing $ y \equal{} \minus{}x \plus{} 1$ and finding the point at which the product $ xy$ is maximized.

or you could substitute y=1-x and find the derivative of x*(1-x)=x-x^2. The derivative is 1-2x, and the maximum is found when 1-2x=0. so x=1/2, and y=1/2, to give 1/4 or 0.25. BAM!
ok, i realize calculus is overkill, but it makes me feel better to actually use it once in a while.


Solve this equation in $ \mathbb{Z}$ :

$ x\equal{}\sqrt{y^2\minus{}\sqrt{y^2\plus{}x}}$

Let $ k\equal{}\sqrt {y^2 \plus{} x}$.

The equation
$ \Leftrightarrow \begin{cases} x \ge 0 , k \ge \sqrt {x} \\
x^2 \equal{} k^2 \minus{} k \minus{} x \Rightarrow [k \minus{} (x \plus{} 1)](k \plus{} x) \equal{} 0 \Rightarrow k \equal{} x \plus{} 1 or k \equal{} x \equal{} 0 \end{cases}$

$ \Leftrightarrow (x \plus{} 1)^2 \equal{} y^2 \plus{} x$ or $ x \equal{} y \equal{} 0$

So $ (x,y) \equal{} (k,\pm \sqrt {m^2 \plus{} m \plus{} 1})$ for all non-negative $ m$s and $ (x,y) \equal{} (0,0)$ are the solutions.


how many subsets of the set {1,2,3,4,5} contain the number 5?

can anyone tell me how to work with subsets along with this problem? :D

[hide]I don't know much about subsets, but here's what I got:
Subsets w/5 numbers: $\binom{4}{4}=1$
Subsets w/ 4 numbers: $\binom{4}{3}=4$
Subsets w/3 numbers: $\binom{4}{2}=6$
Subsets w/2 numbers: $\binom{4}{1}=4$
Subsets w/1 numner: $\binom{4}{0}=1$
So the answer is $16$.[/hide]

[hide]what about all five numbers? that makes $\boxed{16}$[/hide]

[quote="SplashD"]what about all five numbers?[/quote]
I guess he thought $S\notin\left\{\text{subsets of }S\right\}$  :)
You could also [hide]just begin with {5} and add subsets of {1,2,3,4}. You'll get all the subsets you wanted, so we just have to count the number of subsets of {1,2,3,4}, which equals $2^{4}= 16$.[/hide]

This reminds me of a simple aime problem... we can pair up each subset with a 5 to each subset without a 5 (by ignoring the element 5, obviously) and since every subset either has a 5 or doesn't, the number of subsets with a 5 is half the number of subsets. How many subsets are there? We have 5 choices: include a 1, or not? include a 2, or not? etc. so there are 2^5 subsets. Therefore, there are 16 subsets with a five. Btw, I know this is overkill for an easy problem like this but it generalizes...


Hi
please give me just a hint to solve this problem

There are $ 6$ different points in the plane. Let $ D$ be the greatest distance between them and $ d$ the smallest one. Prove that $ \frac{D}{d}>\sqrt{3}$.

 :(

[hide="Hint, even though I can't do it myself"]Consider the two points that are the farthest away from each other. Where must the other 4 points lie?[/hide]

Adding on to what Brut3Forc3 said...

Consider the points that are furthest apart. The distance between these points is $ D$. The other 4 points have to be in the intersection of the 2 circles with radius $ D$ with centers at the 2 points furthest points. But there's still a more restricted area...since in that intersection, the furthest possible distance of 2 points in that intersection is $ D\sqrt {3}$. Not sure where to go from here...

Carbon57, try a little harder, you are close to the idea :) Redraw the diagrams and look closely at where those 4 points must be.

I think I have an idea. Would the maximum length of $ d$ be when the points are... [hide]in a regular hexagon?[/hide]
If so, how is it proven?

That would be the length of a smaller diagonal, the larger diagonal will have twice the length of a side of the hexagon, so a hexagon wouldn't be ideal for this problem.

EDIT 10PM: Actually, equality can be achieved for $ \frac{D}{d}\geq\sqrt3$. Try to find this configuration. It's easier than you think :)


A small, fully loaded passenger plane tried to approach the runway during a violent snowstorm. The control tower regretfully informed the pilot that due to the inclement weather, the runways were closed to all air traffic. Furthermore, all airports within a three-hundred mile radius were also closed. Upon hearing this, the pilot immediately informed the passengers of the news while turning the plane around and heading back from where they had just come. Incredibly enough, within a half hour all the passengers were safely inside an airport terminal building.
How was this possible?

[hide="Because..."]It hadn't taken off yet.[/hide]

[hide]The plane was grounded, and never took off in the first place[/hide]


Let ABC be a triangle and P a point in its interior, not lying on any of the medians of ABC. Let $A_1$, $B_1$, $C_1$ be the intersections of PA with BC, PB with CA, PC with AB, respectively, and let $A_2$, $B_2$, $C_2$ be the intersections of $B_1C_1$ with BC, $C_1A_1$ with CA, $A_1B_1$ with AB, respectively. Prove that if some two of the circles with diameters $A_1A_2$, $B_1B_2$, $C_1C_2$ intersect, then they are coaxal.

In http://www.mathlinks.ro/Forum/viewtopic.php?t=21701 , I showed that the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ are the diagonals of a certain complete quadrilateral. Now, by the Bodenmiller theorem, the circles whose diameters are diagonals of a complete quadrilateral are coaxal.

The Bodenmiller theorem is proven at http://cut-the-knot.com/Curriculum/Geometry/CircleOnCevian.shtml#GB ("Corollary").

  Darij

What about http://cut-the-knot.com/ ? I did not see any new articles recently.


Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a commutative ring with unit, $ \mathfrak{m}$ its maximal ideal, and $ q$ an ideal of $ R$ such that $ \mathfrak{m}^k\subset q$ for some $ k\geq 1$. Prove that $ q$ is primary.

a general fact is that if the radical of an ideal is maximal ideal $ m$, then it is a $ m$-primary ideal.


geometry
rectangle
similar triangles
Pythagorean Theorem

In the diagram, the two circles are tangent, and $ PB$ and $ PD$ are common tangents to the circles. If PA=AB=6, what is the area of the larger circle?

[quote="hohoho11"]In the diagram, the two circles are tangent, and $ \line{PB}$ and $ \line{PD}$ are common tangents to the circles. If PA=AB=6, what is the area of the larger circle?[/quote]

Is what he meant to say.

[hide="Hint"]Draw a line from the center of the circle to the point of contact for each circles. Now, make a rectangle connecting the center of the smaller circle to the radii of the bigger circle. I believe it should be simple from there.[/hide]

[hide="Can somebody find a simpler solution please?"]We can create a system of equations. We can draw lines from the center of each circle to its respective points of tangency with the lines $ PD$ and $ PB$. A pair of similar right triangles will be created. Denoting the radius of the smaller circle to be $ r_1$ and that of the bigger circle to be $ r_2$, we have:
\[ \frac {r_2}{12} = \frac {r_1}{6} \Rightarrow r_2 = 2r_1
\]
Furthermore, we can find two ways to calculate the hypotenuse of the triangle with legs $ 12$ and $ r_2$. Setting them equal, we have:
\[ \sqrt {12^2 + r^2_2} = r_2 + r_1 + \sqrt {6^2 + r^2_1}
\]
Then, replacing $ r_2$ with $ 2r_1$, and rearranging, we have:
\begin{align*} & \sqrt {12^2 + {\left(2r_1\right)}^2} = 2r_1 + r_1 + \sqrt {6^2 + r^2_1} \\
& \Rightarrow \sqrt {144 + 4r^2_1} - \sqrt {36 + r^2_1} = 3r_1 \\
& \Rightarrow 144 + 4r^2_1 + 36 + r^2_1 - 2\sqrt {\left(144 + 4r^2_1\right)\left(36 + r^2_1\right)} = 9r^2_1 \\
& \Rightarrow - 2\sqrt {\left(144 + 4r^2_1\right)\left(36 + r^2_1\right)} = 4r^2_1 - 180 \\
& \Rightarrow \sqrt {\left(144 + 4r^2_1\right)\left(36 + r^2_1\right)} = 90 - 2r^2_1 \\
& \Rightarrow 5184 + 288r^2_1 + 4r^4_1 = 4r^4_1 - 360r^2_1 + 8100 \\
& \Rightarrow 5184 + 288r^2_1 + 4r^4_1 = r^2_1 - 45 \\
& \Rightarrow 648r^2_1 = 2916 \\
& \Rightarrow r_1 = \frac {3\sqrt {2}}{2} \\
& \Rightarrow r_2 = 2r_1 = 3\sqrt {2} \end{align*}The area is then $ \pi\left(3\sqrt{2}\right)^2=\boxed{18\pi}$.
[/hide]

[hide="Solution."]Call the distance from $ P$ to the first circle $ x$.  That means that the distance from the first circle to $ P$ is $ r \plus{} x$, with $ r$ being its radius.  By similar triangles, the radius of the second circle is $ 2r$ and its distance from $ P$ is $ 4r \plus{} x$.  Using the Pythagorean Theorem on both of these triangles:

$ (2r)^2 \plus{} 12^2 \equal{} (4r \plus{} x)^2$
$ (r)^2 \plus{} 6^2 \equal{} (r \plus{} x)^2$

Simplifying and multiplying the latter by 4:

$ 4r^2 \plus{} 144 \equal{} 16r^2 \plus{} 8rx \plus{} x^2$
$ 4r^2 \plus{} 144 \equal{} 4r^2 \plus{} 8rx \plus{} 4x^2$

Therefore:

$ 12r^2 \equal{} 3x^2$
$ x \equal{} 2r$

Plugging it back into the original equation:

$ r^2 \plus{} 36 \equal{} (3r)^2$
$ 8r^2 \equal{} 36$

You're interested in finding $ \pi(2r)^2$, so divide by $ 2$ and multiply by $ \pi$.  The final answer is $ \boxed{18\pi}$.[/hide]

We have:
+) $ IA \equal{} 2IB$
+) $ OA \plus{} IB \equal{} OI \equal{} IP$
+) $ IP^2 \equal{} IB^2 \plus{} BP^2$
$ \leftrightarrow$ $ (OA \plus{} IB)^2 \equal{} IB^2 \plus{} 36$
$ \leftrightarrow$ $ (2IB \plus{} IB)^2 \equal{} IB^2 \plus{} 36$
$ \leftrightarrow$ $ 9IB^2 \equal{} IB^2 \plus{} 36$
$ \leftrightarrow$ $ IB^2 \equal{} \frac {9}{2}$ $ \leftrightarrow$ $ IB \equal{} \frac {3}{\sqrt {2}}$ $ \rightarrow$ $ OA \equal{} 3 \sqrt {2}$
$ \rightarrow$ $ S_{(O;OA)} \equal{} \pi . OA^2 \equal{} 18.\pi$.


[b][size=150]A box contains 3 red balls, 2 blue balls and 5 green balls. Two balls are drawn from the box, one after the other and not replaced.

1.) What is the probability that they are both blue?

2.) What is the probability that they are both the same colour?[/size][/b]

1)[hide]

There are 10 marbles in the box, so there is a 2/10 prob that the first marble is blue and a 1/9 prob that the second is blue, so the overall probability is $\frac{1}{5}*\frac{1}{9}=\frac{1}{45}$

[/hide]

2)[hide]

There is a $\frac{1}{45}$ prob that they are both blue, a $\frac{1}{15}$ prob that they are both red, and a $\frac{2}{9}$ prob that they are both green.  The total is $\frac{1}{45}+\frac{1}{15}+\frac{2}{9}=\frac{14}{45}$[/hide]

This isn't quite Intermediate level. :?