geometry
trapezoid
complex numbers
geometry proposed

$ ABCD$ is an isosceles trapezoid with $ AB \parallel CD, AB>CD$. Denote by $ O$ the intersection point between the diagonals of the trapezoid and by $ M,N,P$ the midpoints of $ OC,OB$ and $ AD$. If $ PM\equal{}MN$ then find the angle values of $ \triangle PMN$.

Take C' on (AC so that CC' = AO, and C between A and C'. Obviously, AM = C'M and triangle ADC' is isosceles, having AD and C'D the double of MN and MP respectively. From AO = CC' it follows that CD = DO, but DO = CO and DOC is an equilateral triangle.
Then see that <PMA = <DC'C and <C'DC = <ADO, but <ADO = <OCB and <OCB = <AMN, that is,
<NMP = <PMA + <AMN = <DC'C + <CDC' = <DCO = 60 degs, so MNP is an equilateral triangle.

Best regards,
sunken rock

[color=darkblue][b][u]Proof[/u].[/b] $ PM \equal{} MN$ $ \Longleftrightarrow$ $ 2\cdot PM \equal{} AD$ $ \Longleftrightarrow$ $ MD\perp MA$ $ \Longleftrightarrow$ $ DO \equal{} DC$ $ \Longleftrightarrow$ $ ODB$ is equilateral $ \Longleftrightarrow$ $ m(\angle PMN) \equal{} 60^{\circ}$ because $ m(\angle PMN) \equal{}$ $ m(\angle PMA) \plus{} m(\angle NMA) \equal{}$ $ m(\angle OAD) \plus{} m(\angle OCB) \equal{}$ $ m(\angle OBC) \plus{} m(\angle OCB) \equal{}$ $ m(\angle DOC) \equal{} 60^{\circ}$ , i.e. the triangle $ MNP$ is equilateral.

[b][u]Remark[/u].[/b] The proof with complex numbers is much more shorterly.[/color]


linear algebra
matrix
algebra
polynomial

Suppose $A\in\mathcal M_n(C)$,$rank(A)=p$ is it true that $A$ has $n-p$ null eigenvalues? How do you prove it?

Remember that the algebraic multiplicity of an eigenvalue can be larger than its geometric multiplicity. Saying the that rank of $A$ is $p,$ where $p<n,$ is saying that zero is an eigenvalue of geometric multiplicity $n-p.$ But the algebraic multiplicity of zero as an eigenvalue could be larger than $n-p.$

Example: let $A=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.$

Then the rank of $A$ is $2$ and zero is an eigenvalue of geometric multiplicity $1.$ But the characteristic polynomial is $x^3$ and zero is an eigenvalue of algebraic multiplicity $3.$

$\textrm{rank} A = p$ means that all minors of order bigger than $p$ are null. Thus, by Cayley-Hamilton, the last $p$ coefficients are equal to $0$. That's what I meant when I said I could only get an inequality.

Yes but it still works ... becauese you can choose $q$ to be the number of null eigenvalues...

I don't understand what you're trying to say. Can you explain?


Candy bought a video system for a total price of 1,280 dollars. she agreed to make a down payment of 280 dollars. and finance the balance at 15 percent. Her monthly payment will be 62 dollars. What will the interest be for the second month of the loan?

[hide]Since she paid $280$, the remaining balance is $1000$ for the first monthly payment.  The monthly interest percentage is $\frac{.15}{12}= .0125$.  Then the amount of the monthly payment that is applied to the balance is $62-(.0125)(1000) = 49.5$. Then the balance for the second month is  $950.50$. Then the interest for the second month is $(.0125)(950.50) = 11.88$.[/hide]


How much 90 percent vinegar do you have to add to a gallon of 5 percent vinegar to get 20 percent vinegar?   ____gallons

i believe this is right

[hide]
we have 1 gallon of .05 vinegar mix, making .05 vinegar
we have x gallons of .9 vinegar mix, making .9x vineger
mixing those together,
1+x gallons .2 vinegar mix, making .05+.9x vineger

.2(1+x)=.05+.9x
solve
[/hide]


geometry
vector
geometric transformation
homothety
parallelogram
geometry unsolved

I have three geo problems which I have tried hard but can't solve.[Don't use co-ordinate geo in ur soln plz]

1. Given three non-intersecting circles,draw the intersection of the external tangents to each pair of circles. Show that these three points are collinear.

2. Let ABCD be a convex quadrilateral and O is the intersection point of AC and BD. Let $ m,n\in AB$such that $ AM\equal{}MN\equal{}NB$ and let $ P,Q\in CD$ such that $ CP\equal{}PQ\equal{}QD$. Prove that $ \triangle MOP$ and $ \triangle NOQ$ have the same area.

3. Let ABCDE be a convex pentagon and F,G,H,I,J be the respective midpoints of line segments CD, DE, EA, AB, BC.  If AB, BG, CH, DI pass through a common point prove that EJ also passes through that point.

I have tried these hard. Plz help me solve these...

The first one is very easy:
Let $ \zeta_{1}(O_{1},R_{1}),\zeta_{2}(O_{2},R_{2}),\zeta(O_{3},R_{3})$ be the circles and $ P_{1},P_{2},P_{3}$ be the intersection points of external tangent lines to $ \{\zeta_{2},\zeta_{3}\},\{\zeta_{1},\zeta_{3}\}\{\zeta_{2},\zeta_{1}\}$,respectively.It is obvious that $ \{P_{1},O_{2},O_{3}\},\{P_{2},O_{1},O_{3}\},\{P_{3},O_{2},O_{1}\},$ are collinear.Now using Menelaus theorem for $ \triangle O_{1}O_{2}O_{3}$ and transversal $ \overline{P_{1}P_{2}P_{3}}$ we get the result.(Use that $ \frac{P_{3}O_{1}}{P_{3}O_{2}}\equal{}\frac{R_{1}}{R_{2}}$.) :wink:

Third Problem is from Balkan Math Olympiad - 1996. It is proposed by Former Ugoslavia / Serbia.

3. Let ABCDE be a convex pentagon, and let M;N; P; Q;R be the
midpoints of sides AB;BC;CD;DE;EA, respectively. If the segments
AP;BQ;CR;DM have a common point, show that this point
also lies on EN.

Solution: Let T be the common point, which we take as the origin
of a vector system. Then AXP = 0, or equivalently A X (C+D) = 0,
which we may write A X C = D X A. Similarly, we have B X D = E X B,
C X E = A X C, D X A = B X D. Putting these equalities
together gives E X B = C X E, or E X (B + C) = 0, which means
the line EN also passes through the origin T.

P.S It is copied from a book... I believe there is a simpler solution. Maybe with similarities, homothethy, trugonometry or Menelaus/Cheva...I think you may try to find in math excalibur, but I'm not sure...

second problem:

Call E the midpoint of CD, F the midpoint of AB and U the symmetric of O wrt E, V the simmetryc of O wrt F.
Call S the centroid of OCD, R the centroid of DCU, K the centroid of ABO and J the centroid of ABV.
$ Z: AC\cap UV$, $ X: JR\cap MP$, $ Y: QN\cap SK$, $ W: UV\cap DB$, $ G: QN\cap NP$ and SK meet AC in I and BD in H.
Clearly with a homothety with center E and factor 3 we obtain the parallelogram OCUD from SPRQ
and with a homothety with center F and factor 3 we obtain the parallelogram OBVA from KNJM.
So for a composition of homotheties Z, X and O are on the same line and W,Y and O are on the same line
so $ X : AC\cap PM$ and $ Y: BD\cap QN$ and equally for proove that $ I: QN\cap AO$ and $ H: MP\cap DB$
But so for thales' theorem in XOY we have that $ XI\equal{}IO$ and $ YH\equal{}HO$ so G is the centroid of XOY, then

$ [NOQ] \equal{} [IOY]\plus{}[YOQ]\plus{}[ION] \equal{} [IYX]\plus{}[HOP]\plus{}[XOM] \equal{} [MOP]$

Do you know what is the source of the second problem? I like it most.

"U the symmetric of O with respect to  E, V the symmetric of O with respect to F. "

Gabriel- plz explain what do u mean by two points being symmetric with resptect to other point.
I know abt two points being symmetric with respect to a straight line or a circle.

And many many thanks to Erken and borislav_mirchev for helping me in prob 1 and 3.

The 2nd prob(triangle with the same area) is attributed to Razvan Gelca who is a mathematician an is associated with MOP.[/quote]

Let two points are given - A and M. To find point B - symmetric to A by M means that B is on line AM, and AM=BM. What is MOP? Mathematical Olympiad of Poland?


When Betty substitutes x=1 into the expression $ax^3-2x+c$, its value is -5. When she substitutes x=4, then the expression equals 52. Find one value that makes expression equal to zero?

A) 2
B) 5/2
C) 3
D) 7/2
E) 4

[hide]Two equations are; 64a-8+c=52
                                         a-2+c=-5
When you subtract, you get:
                                      63a-6   =57; therefore, 63a=63, and a=1.
Now we have $x^3-2x+c$
I will use x=1 again; 1-2+c=-5
Therefore, c=-4, so;
$x^3-2x-4=0$
x=2
A)
[/hide]


Let $ {(I_{n})}$ be a sequence of intervals such that the infinite series of their diameters converges. Does there then exist an $ x$ such that all integer multiples of $ x$ lie outside of these intervals.

I think that if $ L>\sum |I_{n}|$, then such an $ x$ can be found within the interval $ [L,2L]$. Try it.


algebra
polynomial
calculus
integration
function
domain
linear algebra

At http://en.wikipedia.org/wiki/Discriminant_of_a_polynomial , the discriminant of a monic polynomial is defined as the determinant of a $(2n-1) \times (2n-1)$ matrix. How do we show that equals $\displaystyle \prod_{i<j} (z_i-z_j)^2$, where $z_i$ are the roots of the polynomial?

The main goal of this algebraic digression is to present some properties of the discriminant of a polynomial. We will mainly use it to prove that a polynomial with coefficients in an arbitrary field $k$ has a multiple root in an algebraic closure of $k$ if and only if the discriminant of the polynomial is 0. We will also find some formulas for the discriminant of a polynomial.

Let $A$ be an integral domain, let $k$ be its field of fractions and let $\Omega$ be an algebraic closure for $k$. Let $f\in A[X]$ be a monic polynomial. If $k$was$\mathbb{R}$ or $\mathbb{C}$ then we had the notion of a derivate for $f$ and in both cases we had the formula  $f'(X)=na_nX^{n-1}+(n-1)a_{n-1}X^{n-2}+\ldots+a_1$if$f(X)$was$a_nX^n+a_{n-1}X^{n-1}+\ldots+a_1X+a_0$. \begin{defn} If$f\in A[X]$,$f(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots+a_1X+a_0$then define the formal derivate of$f$by $f'(X)=na_nX^{n-1}+(n-1)a_{n-1}X^{n-2}+\ldots+a_1$.\end{defn}

The following properties of the formal derivate are easy exercises:

\begin{enumerate}
\item $(f+g)'=f'+g'$ for all $f,g\in k[X]$.
\item $(a\cdot f)'=a\cdot f'$ for all $f\in k[X]$ and $a\in k$.
\item $(fg)'=f'\cdot g+f\cdot g'$ for all $f,g\in k[X]$.
\item If $k$ is a field of characteristic 0, then $f'=0\Leftrightarrow f$ is a constant polynomial. If $k$ is of characteristic $p$, then $f'=0\Leftrightarrow \exists g\in k[X]$ such that $f(X)=g(X^p)$.
\end{enumerate}

\begin{defn} Let $f,g\in A[X]$,$f(X)=a_nX^n+\ldots+a_0$ and $g(X)=b_mX^m+\ldots+b_0$. The resultant of $f$ and $g$, is defined as the determinant of the $(n+m)\times(n+m)$ matrix:
\[ Res(f,g)=\det\left(\begin{array}{cccccc}a_n&a_{n-1}&\ldots&a_0&0&\ldots\\ 0&a_n&\ldots&a_1&a_0&0\\ 0&0&a_n&\ldots&a_1&a_0\\ \ &\ &\ &\ddots\\b_m&b_{m-1}&\ldots&b_0&0&\ldots\\ 0&b_m&\ldots&b_1&b_0&0\\ 0&0&b_m&\ldots&b_1&b_0\\ \ &\ &\ &\ddots\end{array}\right).  \]
The first $m$ rows are obtained by permuting circularly the $1\times(n+m)$ vector $(a_n,\ldots,a_n,0,\ldots,0)$ and the last $n$ rows are obtained
by permuting circularly the $1\times(n+m)$ vector $(b_m,\ldots,b_0,0,\ldots,0)$.\end{defn}

\begin{defn} By definition, the discriminant of a polynomial is \[ \Delta_f=(-1)^{n(n-1)/2}\cdot\frac1{a_n}\cdot Res(f,f').  \] \end{defn}

The bare definition of the resultant of two polynomials may look a bit scary, but we will see how it appears naturally in trying to solve the following problem: Given two polynomials with coefficients in $A$, when can we say that they have a common factor? Simple ring theory proves that it is enough to consider the problem in $k$, the quotient field of $A$. The following proposition proves that the resultant of these polynomials plays a decisive role in solving this problem. The proof of the proposition sheds some light on how the definition of the resultant first came to life.

\begin{prop}\label{resdisc:p1} If $f,g\in A[X]$, then $f$ and $g$ have a nonconstant common divisor if and only if $Res(f,g)=0$.\end{prop}
\par\textbf{Proof:} It is enough to consider the problem in $k$, the quotient field of $A$. Let $deg(f)=n$ and $deg(g)=m$. First notice that $f$ and $g$ have a nonconstant common factor if and only if there exists a nonzero polynomial $h\in k[X]$ such that $f|h$,$g|h$ and $deg(h)\leq m+n-1$. If $f,g$ have a common factor $d$, then $f=d\cdot f'$ and $g=d\cdot g'$ for some polynomials $f',g'\in k[X]$. For $h=df'g'$, we have $f|h$,$g|h$ and
$deg(h)=deg(d)+deg(f')+deg(g')=n+m-deg(d)\leq n+m-1$. Conversely let $h=f\cdot f'$ and $h=g\cdot g'$. If $f$ and $g$ have no common
factor then $f|h=gg'\Rightarrow f|g'$ which is impossible because $deg(g')=deg(h)-deg(g)\leq n+m-1-m=n-1$. Therefore $f$ and $g$
have a common factor.

Let $V_f$ be the $k-$vector space of polynomials of degree least of equal to $n+m-1$ divisible by $f$. It is easy to see that a basis for $V_f$ is given by $\{f,Xf,X^2f,\ldots,X^{m-1}f\}$. Similarly define $V_g$. It has the basis $\{g,Xg,X^2g,\ldots,X^{n-1}g\}$. $f$ and $g$ have a nonconstant
common factor if and only if there exists a nonzero polynomial of degree least or equal to $m+n-1$ divisible both by $f$ and $g$. This is equivalent to $V_f\cap V_g\neq 0$ i.e $V_f$ and $V_g$ are not in direct sum in the vector space $V$ of polynomials of degree least or equal to $n+m-1$. A basis for $V$ is given by $\{1,X,X^2,\ldots,X^{n+m-1}\}$. Now $V_f$ and $V_g$ are not in direct sum if and only if $\{X^{m-1}f,\ldots,Xf,f,X^{n-1}g,X^{n-2}g,\ldots,Xg,g\}$ are linearly dependent over $k$ in $V$. By writing these elements of $V$ in the basis $\{X^{n+m-1},\ldots,X,1\}$, the condition that they are linearly dependent is equivalent to that the determinant of the resulting matrix of coefficients is zero. The matrix of coefficients is the same as the matrix of $Res(f,g)$.

\begin{cor}\label{resdisc:c1} If $f\in A[X]$, then $f$ has a multiple root in $\Omega$ if and only if $\Delta_f=0$.\end{cor}

\par\textbf{Proof:} $f$ has a multiple root in $\Omega$ if and only if $f$ and $f'$ have a nonconstant common factor in $k[X]$.\eop\\

We now give some formulas for the resultant and for the discriminant of a polynomial.
\begin{prop} If $f,g\in A[X]$,$f(X)=a_n(X-\alpha_1)\ldots (X-\alpha_n)$ and $g(X)=b_m(X-\beta_1)\ldots (X-\beta_m)$ with $\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m\in\Omega$, then $Res(f,g)=a_n^mb_m^n\prod_{i=1}^n\prod_{j=1}^m(\alpha_i-\beta_j)$.\end{prop}

\par\textbf{Proof:} Consider $F,G\in A[X_1,\ldots,X_n,Y_1,\ldots,Y_m][X]$ defined by \[ F(X_1,\ldots,X_n,Y_1,\ldots,Y_m)(X)=a_n(X-X_1)\ldots(X-X_n),  \]
\[ G(X_1,\ldots,X_n,Y_1,\ldots,Y_m)(X)=b_m(X-Y_1)\ldots(X-Y_m).  \]
Clearly $f=F(\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m)$ and $g=G(\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m)$ in $\Omega[X_1,\ldots,X_n,Y_1,\ldots,Y_m]$.

$\tilde A=A[X_1,\ldots,X_n,Y_1,\ldots,Y_m]$ is an integral domain, so $Res(F,G)$ is a well defined element of $\tilde A$ and it is clear from the definitions that \[ Res(F,G)(\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m)=Res(f,g).  \] Hence it suffices to prove $Res(F,G)=a_n^mb_m^n\prod_{i=1}^n\prod_{j=1}^m(X_i-Y_j)$. The formalism of a complete proof is tedious, but the idea is simple.
For all $i=\overline{1,n}$,$j=\overline{1,m}$ we prove that $X_i-Y_j|Res(F,G)$. Let $\tilde A_j$ be the polynomial ring obtained from $\tilde A$ by removing the variable $Y_j$. Let $F_{ij}$,$G_{ij}$ and $R(F,G)_{ij}$ be the polynomials in $\tilde A_j[X]$ obtained from $F$,$G$ and $R(F,G)$ respectively by replacing $Y_j$ with $X_i$. We have $X_i-Y_j|Res(F,G)\Leftrightarrow Res(F,G)_{ij}=0$. But clearly in $\tilde A_j$,$Res(F,G)_{ij}=Res(F_{ij},G_{ij})$ and the last is 0 because $F_{ij}$ and $G_{ij}$ have an obvious common factor, namely $X-X_i$. These can be formalized rigourously to prove $\prod_{i=1}^n\prod_{j=1}^m(X_i-Y_j)|Res(F,G)$.

Let $V=\prod_{i=1}^n\prod_{j=1}^m(X_i-Y_j)\in\tilde A$. By expanding $F$ and $G$ we see that the variable $X_i$ appears in the matrix of $Res(F,G)$ only on the first $n$ lines and on these lines it appears in degree 1, hence the degree of $Res(F,G)$ in $X_i$ is $n$ for all $i=\overline{1,n}$. Similarly we prove that the degree of $Res(F,G)$ in each $Y_j$ is $m$. From these we conclude that $Res(F,G)=\tau\cdot V$ with $\tau\in A$.

Notice that $\tau=Res(F,G)(1,\ldots,1,0\ldots,0)=Res(a_n(X-1)^n,b_mX^m)$. Computing $Res(a_n(X-1)^n,b_mX^m)=a_n^mb_m^n$ is a good
exercise.

\begin{cor} $Res(f,g)=a_n^m\cdot\prod_{i=1}^ng(\alpha_i)=(-1)^{n+m}b_m^n\cdot\prod_{j=1}^mf(\beta_j)$.\end{cor}
\begin{prop} If $f\in A[X]$,$f(X)=a_n(X-\alpha_1)\ldots (X-\alpha_n)$ and $\alpha_i\in\Omega\ \forall i=\overline{1,n}$,
then \[ \Delta_f=a_n^{2n-2}\left(\prod_{1\leq i<j\leq n}(\alpha_i-\alpha_j)\right)^2.  \]\end{prop}

\par\textbf{Proof:} Use $f'(\alpha_i)=\prod_{j\neq i} (\alpha_i-\alpha_j)$.\eop\\

These formulas are often useful for computations and could have just as well been used to define both the resultant and the discriminant.
However they are not quite useful in proving more theoretical results. For example don't make clear why $Res(f,g)$ and $\Delta_f$ are not just elements of $\Omega$ but also of $A$.

The following is a very important property of the resultant of two polynomials.
\begin{thm}\label{resdisc:t1} If $f,g\in A[X]$, then $R(f,g)\in (f,g)$ i.e. $R(f,g)$ is in the ideal generated by $f$ and $g$ i.e. there exist
polynomials $u,v\in A[X]$ such that $uf+vg=Res(f,g)$.\end{thm}

\par\textbf{Proof:} Consider $R(f,g)$ as the determinant of a matrix with entries in $A[X]$. For all $i=\overline{1,n+m-1}$, add the column
$i$ in $Res(f,g)$ multiplied by $X^{n+m-i}$ to the last column, $m+n$. The only difference between the matrix of $Res(f,g)$ and
the new matrix, $M$, is the last column, which is $^{\top}(fX^{m-1},\ldots,fX,f,gX^{n-1},\ldots,gX,g)$. The determinant of $M$ is $Res(f,g)$. Expanding $\det M$ by the last column gives $Res(f,g)\in (f,g)$.\eop\\

\begin{rem} Let $f(X)=X^3+aX^2+bX+c$ be a polynomial with coefficients in an arbitrary field $k$. We have:
\[ \Delta_f=-4a^3c+a^2b^2+18abc-4b^3-27c^2.  \]\end{rem}


I was thinking about organizing tag-team chess games, where the teams would post their move on the forum using the longer (but clearer) [url=http://en.wikipedia.org/wiki/Algebraic_chess_notation#Long_algebraic_notation]algebraic notation[/url] (e.g. d2-d4, Qg4xc8+, etc.) and it would just be the first person to post on the thread who determines the team's move for that turn.

I think we should have like 4 to 6 people in each team, at least for now. We could get more games going if there's enough of a response.

Teams could strategize via PM.  :P

Anyone up for this?  :)

Methinks we'll get a lot of "unfair"s, but me as well.

I'm interested but the time for each move should be long...maybe 1 week...so the team can discuss moves.

I am interested and I agree that there should be some time to discuss what move would be best, but I think 1 week would be too much. A game would take several months or maybe even a year (if it got really stuck) with a 1 week discussing period. Maybe 2 or 3 days per move would be enough? Either way, I definitely support this and I would like to participate (I may be away for 2 weeks sometime soon though).

this sounds fun, but im not very good at chess. i think 1 - 3 days per move would be fine.

count me in too? :maybe:

actually i was thinking about a similar idea for some time...

10 people = 5 v 5.

Let's split into teams and start the game in a new topic?

[quote="AnonAmoz"]I want teams of three too... when is the sign-up thread coming up?, because I want to join.[/quote]
After the first game is over...

For the first tournament, we'll have teams of 3 people. That's official, so no more arguing over it. If it goes well, we'll do 3 people. If not, we might have to change it.

I sign up as captain for the next available game.

If you really were interested, maybe you would read at least the most recent posts?

And besides, in a team of 2 or 3 people, a captain is not needed, and actually destroys the purpose of having more than one person on a side. Team members collaborate and come up with the move they think would be best, but having a captain gives them more power and therefore defeats the purpose. Teams can elect captains if they really want to, but you can't sign up as captain.

LOL, [b]i_like_pie[/b], not that there's really any purpose in arguing with you, but what gives you the authority to say <<...we'll have teams of 3 people. That's official.>> ?

Anyways, obviously I'm interested in playing in the tournament. I'm not sure whether teams should be chosen randomly or not. (If not, there will probably be a mess of PMs flying back and forth, people feeling cheated and betrayed, and so on.  LOL.)

Eh, what do you guys think?

[quote="i_like_pie"][quote="cincodemayo5590"]LOL, [b]i_like_pie[/b], not that there's really any purpose in arguing with you, but what gives you the authority to say <<...we'll have teams of 3 people. That's official.>> ?[/quote]
Nothing. :P It's just that more people wanted 3 people per team... And why do you say there's no point in arguing? I'm not even sure I want 3 people per team myself. :rotfl:

From the first game, we can basically keep our teams the same (almost), then we can have some people who already know who they want to be with make their own teams, and the rest will be randomized. But we have to make sure that people will actually play, not like what happened in game 1 (a bunch of people quit, and now it's 4v4).[/quote]

Actually, three people voted for 3, and three for 2.

Plus, [b]definitely[/b] no captains allowed.

When I said captain, I meant the best player of course.  :D
And how does me signing up violate teams of 3? :maybe:

i think if u sign up, we'll have more than a multiple of 3, and wont divide evenly

oh, and whats yr uscf rating? i think im better :D  :D  :D

How about this: organize yourself into groups of 2-3. The people that don't will be randomized.

what's uscf?
us chess fed?
what's your fide rating? :rotfl:

i can be in a group of two

USCF is the us chess federation.  (http://www.uschess.org)

To get an Fide rating requires you to get a certain amount of points out of the first so and so many games (I forgot the details) in FIDE rated events (which I think has to be the open section) against already rated FIDE players, which should be at least masters or close to it.  So if somebody has one that is FIDE rated that is on team, that team probably will win.

Five years ago I got close to a FIDE rating with a win against one in a rated tournament. If I had pushed on and won two more, I'd have got a rating. But around that point I left chess.  :( Then the listed rating went down to 1800, making it even easier to get a rating. Many of my friends got it. :(

Anybody who wants to can sign up for the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164632]Team Chess Tournament[/url]!

Note that this will be taking place in the Math and Strategy Games forum (under Other Problem Solving Topics).

You can sign up by post, PM, or email. Please read all the rules and stuff before signing up so there aren't any disputes later. Sign-ups on this topic will not be accepted as I don't want to have to look all over AoPS for them (this includes posts prior to this one because they were made before reading the official rules). If you prefer being on a team with your friends, let me know and I will set up your team. If not, the rest of the teams will be randomized once we have enough people.

Thank you and good luck to all that participate! :)


p, q, and r are three non-zero integers such that p + q + r = 26 and
 (1/p) + (1/q) + (1/r) + (360/pqr) = 1
Compute pqr.


geometry
3D geometry
number theory proposed
number theory

Show that no integer of the form $ xyxy$ in base $ 10$ can be a perfect cube. Find the smallest base $ b>1$ for which there is a perfect cube of the form $ xyxy$ in base $ b$.

[quote="moldovan"]Show that no integer of the form $ xyxy$ in base $ 10$ can be a perfect cube. Find the smallest base $ b > 1$ for which there is a perfect cube of the form $ xyxy$ in base $ b$.[/quote]

$ \overline{xyxy}\equal{}(10x\plus{}y)(101)$. Assume that $ \overline{xyxy}$ is perfect cube. Then as $ 101$ is prime we have

$ 101^3|\overline{xyxy}<101^3$, contradiction. So no integer of the form $ xyxy$ in base $ 10$ can be a perfect cube.

$ \overline{xyxy}$ in base $ b$ $ \equal{}xb^3\plus{}yb^2\plus{}xb\plus{}y\equal{}(xb\plus{}y)(b^2\plus{}1)$

[b]Case 1:[/b] $ b^2\plus{}1$ is square-free then we have $ (b^2\plus{}1)^2 |xb\plus{}y$ to make $ (xb\plus{}y)(b^2\plus{}1)$ a cube.
But $ xb\plus{}y<2b^2<(b^2\plus{}1)^2$, contradiction.

[b]Case 2:[/b] $ b^2\plus{}1$ is not square-free. It's easy to check that the smallest $ b$ to make $ b^2\plus{}1$ not square-free is $ 7$.

When $ b\equal{}7$, it's easy to find that $ (7 \times 2\plus{}6)(7^2\plus{}1)\equal{}1000$ is a cube satisfies the condition.

So the smallest base $ b$ is $ 7$


Its true by muirhead, which suggests an AMGM solution.

$a^4 + a^4 + b^4 + c^4 \ge 4a^2bc$ and summing gives us the required.

[quote="varun"]Can some one please prove the following identity:
a^4 + b^4 + c^4 > abc(a+b+c)
I have tried a lot and I think there is something in it I don't know.
Please give your views on this identity.[/b][/quote]

We have Generalized AM-GM inequality
If a, b, c,...k are 'n' positive quantities, p, q, r ... are positive integers then
$\frac{{a^{p + q + r....}  + b^{p + q + r....}  + c^{p + q + r....} ..... + k^{p + q + r....} }}{n} > \frac{{a^p  + b^p  + c^p ..... + k^p }}{n}.\frac{{a^q  + b^q  + c^q ... + k^q }}{n}.\frac{{a^r  + b^r  + c^r ... + k^r }}{n}...$

Using this for a, b, c with p=3, q=1 we get
$\frac{{a^4  + b^4  + c^4 }}{3} > \frac{{a^3  + b^3  + c^3 }}{3}.\frac{{a + b + c}}{3} > abc\frac{{a + b + c}}{3}$

Hence the answer.

[quote="xxreddevilzxx"]rearrangement. WLOG, assume that $a>b>c$. Then the LHS is ordered in the same way, whereas the RHS is oppositely sorted.[/quote]

First of all, it's wrong, but I think that that would be more apparent if you had bothered to actually write it out.  Posting full 'solutions', even if they're wrong, is always more helpful.

[quote="Peter VDD"][quote="Singular"]Its true by muirhead, which suggests an AMGM solution.

$a^4 + a^4 + b^4 + c^4 \ge a^2bc$ and summing gives us the required.[/quote]
To be exact, it should be $a^4 + a^4 + b^4 + c^4 \ge 4 a^2bc$. This is indeed the best way to solve it... short and simple.[/quote]

I did not understand this message.  Will you please clarify?

[quote="vidyamanohar"]I did not understand this message.  Will you please clarify?[/quote]

Ok, by AM-GM, applied for the 4 numbers $a^4$, $a^4$, $b^4$ and $c^4$ we have

\[\frac{a^4 + a^4 + b^4 + c^4}{4} \geq \sqrt[4]{a^4 \cdot a^4 \cdot b^4 \cdot c^4}\] or \[2a^4 + b^4 + c^4 \geq 4a^2bc.\] Similarly we get \[2b^4 + c^4 + a^4 \geq 4ab^2c\] and \[2c^4 + a^4 + b^4 \geq 4abc^2.\] When we add these inequalities we get \[4\left(a^4 + b^4 + c^4\right) \geq 4\left(a^2bc + ab^2c + abc^2\right)\] and hence we are done.


induction
analytic geometry
Pascal\u0027s Triangle
combinatorics proposed
combinatorics

Finding a nice solution for a hard problem of combinatorics

Prove that 2 following equalities

1) $ \sum_{i \equal{} 2}^{n}C^{2}_{i}\equal{} C^{3}_{n\plus{}1}$
2) $ \sum_{i \equal{} 3}^{n}C^{3}_{i}\equal{} C^{4}_{n\plus{}1}$

And the generalization:

$ \sum_{i \equal{} m}^{n}C^{m}_{i}\equal{} C^{m\plus{}1}_{n\plus{}1}$, with $ m,n$ are integer numbers, $ m < n$

Is it true?

We will prove the general result using mathematical induction on $ n\minus{}m$.
[b]Base Step:[/b] If $ n\minus{}m \equal{} 0$ then $ \sum_{i \equal{} 1}^{m}C_{i}^{m}\equal{} C_{m\plus{}1}^{m\plus{}1}$ which is true, so the result is true for $ n\minus{}m \equal{} 1$.
If $ n\minus{}m \equal{} 1$ then $ \sum_{i \equal{} 1}^{i}C_{i}^{m}\equal{} C_{m}^{m}\plus{}C_{m\plus{}1}^{m}\equal{} C_{m\plus{}1}^{m\plus{}1}\plus{}C_{m\plus{}1}^{m}\equal{} C_{m\plus{}2}^{m\plus{}1}$ which is true by Pascal's formula, hence the result is true for $ n\minus{}m \equal{} 1$.
[b]Induction Step:[/b] Assume the result is true for $ n\minus{}m \equal{} k$. then, $ \sum_{i \equal{} 1}^{k}C_{i}^{m}\equal{} C_{k\plus{}1}^{m\plus{}1}$.
Then, for $ n\minus{}m \equal{} k\plus{}1$ have $ \sum_{i \equal{} 1}^{k\plus{}1}C_{i}^{m}\equal{} C_{k\plus{}1}^{m\plus{}1}\plus{}C_{k\plus{}1}^{m}\equal{} C_{k\plus{}2}^{m\plus{}1}$ by Pascal's formula. Therefore, $ \sum_{i \equal{} 1}^{k\plus{}1}C_{i}^{m}\equal{} C_{k\plus{}2}^{m\plus{}1}$ and teh result is therefore true for $ n\minus{}m \equal{} k\plus{}1$.
The result follows by mathematical induction.

Here is a nice solution to this problem using double-counting.
[b]Lemma 1:[/b] The number of paths from $ (0,0)$ to $ (m,n)$ that consist only of unit steps along the lines parallel to the coordinate axes and that go to the right or up is $ C_{m\plus{}n}^{m}$.
[b]Proof:[/b] Every such path will contain $ m\plus{}n$ steps with $ m$ horizontal and $ n$ vertical steps, and any order of $ m$ horizontal and $ n$ vertical steps will uniquely determine the path with $ m\plus{}n$ steps as every vertical step is a unit step traveling up, and every horizontal step is a unit step traveling to the right. Therefore, the number of paths is equal to the number of ways to choose these paths with $ m\plus{}n$ steps with $ m$ vertical and $ n$ horizontal steps, which is $ C_{m\plus{}n}^{m}$. Then lemma 1 is proved.

Now, consider all the paths from $ (0,0)$ to $ (m\plus{}1,n\minus{}m)$ that consist only of unit steps along the lines parallel to the coordinate axes and that go to the right or up. For each path, there is a unique point $ (m,r)$ (where $ r\in\{0,1,2,...,n\minus{}m\}$) that the path includes until a move to the right is made, call this the special point. For every $ i \equal{} 0,1,2,...,n\minus{}m$ let $ f(i)$ be the number of paths with the special point $ (m,i)$. Then because after passing through the special point $ (m,i)$ the path goes to $ (m\plus{}1,i)$ and then as there are no moves to the left or down, it can only then go vertically all the way to point $ (m\plus{}1,n\minus{}m)$. So, the total number of paths from $ (0,0)$ to $ (m\plus{}1,n\minus{}m)$ is $ \sum_{i \equal{} 0}^{n\minus{}m}f(i)$.
But for $ i \equal{} 0,1,2,...,n\minus{}m$, $ f(i)$ is equal to the number of paths from $ (0,0)$ to $ (m,i)$ which is $ C_{m\plus{}i}^{m}$ by lemma 1.
Then, the number of paths from $ (0,0)$ to $ (m\plus{}1,n\minus{}m)$ is:
$ \sum_{i \equal{} 0}^{n\minus{}m}f(i) \equal{}\sum_{i \equal{} 0}^{n\minus{}m}C_{m\plus{}i}^{m}\equal{}\sum_{i \equal{} m}^{n}C_{i}^{m}$.
But by lemma 1, the number of paths from $ (0,0)$ to $ (m\plus{}1,n\minus{}m)$ is $ C_{m\plus{}1\plus{}n\minus{}m}^{m\plus{}1}\equal{} C_{n\plus{}1}^{m\plus{}1}$.
So, we get $ \sum_{i \equal{} m}^{n}C_{i}^{m}\equal{} C_{n\plus{}1}^{m\plus{}1}$ QED.

A similar way would be to notice that $ \binom{n}{r}$ is the $ r$th number in the $ n$th row of Pascal's triangle (zero-based indices) by considering the coefficients of $ (1 \plus{} x)^n \equal{} (1 \plus{} x)(1 \plus{} x)^{n \minus{} 1}$ then the general result follows immediately.


calculus integration probability invariant symmetry function

Is it possible to have 2 six sided dice with positive integral values on each face (the dice may be distinct) such that the probability of getting any value from 2-12 is identical to getting it with a pair of normal six sided dice?

Wouldn't a pair of normal 6-sided dice work?
Is that allowed or am I reading the question wrong.

Yes.  I suppose that would.   And you read the question right.  But you read the intent of the question wrong.  :)

Well, clearly the sum of the numbers on both dies must be 42 (it's an invariant)

We must have exactly 1 on each die and a 6 on each die. (1)

We must also have at least one 2 and one 5 (2)

From (1) and (2), it's clear that we have at least one 2 and one 5 on each die.  Now our dies look like this:

1 2 X X 5 6
1 2 X X 5 6

Now look at the number of ways that we can make 4 from the two dies (there are normally 3 ways).

There are a few possibilities.  One is the usual die, we'll ignore this for now.

The others are:

Case 1:
1 2 3 3 5 6
1 2 X X 5 6

Case 2:
1 2 2 X 5 6
1 2 3 X 5 6

Case 1: By symmetry we must have
1 2 3 3 5 6
1 2 4 4 5 6

But now the probability of 7s is higher than it was before.

Case 2: The probability of 3s is higher than it was before.

So the answer is no.  I'll prune my casework later, it's a bit unnecessary right now.

Didn't read your solution because it's wrong  :)

The answer is yes.

Yeah, assuming that there was 2 6s was my mistake.  I'll fiddle around some more.

This problem angers me.  In any event, the answer is yes, and:

1 3 4 5 6 8
1 2 2 3 3 4

will do the trick.

There ya go.  This problem would probably be far less frustrating with generating functions.

Besides-you know that result rocks. :lol:

How would you do it with generating functions?

Refactor (x+x^2+x^3+x^4+x^5+x^6)^2 into two different factors, both with 6 terms (counting 2x^2 as 2 terms)

That's a pretty cool factorization.  Did you find it using the usual difference of squares/cubes rigamarole?

They are actually called Sicherman Dice.


inequalities
function
inequalities unsolved

If $ a,b,c,x,y,z>0$ show that :

$ xyz\left(\frac{1}{ax\plus{}by\plus{}cz}\plus{}\frac{1}{bx\plus{}cy\plus{}az}\plus{}\frac{1}{cx\plus{}ay\plus{}bz}\right)\le \frac{xy\plus{}yz\plus{}zx}{a\plus{}b\plus{}c}$

[quote="alex2008"]If $ a,b,c,x,y,z > 0$ show that :

$ xyz\left(\frac {1}{ax \plus{} by \plus{} cz} \plus{} \frac {1}{bx \plus{} cy \plus{} az} \plus{} \frac {1}{cx \plus{} ay \plus{} bz}\right)\le \frac {xy \plus{} yz \plus{} zx}{a \plus{} b \plus{} c}$[/quote]
It's obvious because $ f(x) \equal{} \frac {1}{x}$ is convex function when $ x>0.$  :wink:

Maybe this requires an explanation. Jensen's inequality gives

$ \frac{a}{a \plus{} b \plus{} c}\frac{1}{x} \plus{} \frac{b}{a \plus{} b \plus{} c}\frac{1}{y} \plus{} \frac{c}{a \plus{} b \plus{} c}\frac{1}{z} \ge \frac{a \plus{} b \plus{} c}{ax \plus{} by \plus{} cz}$

Adding the inequalities obtained when we change $ a, b$ and $ c$ cyclically, this gives

$ \frac{xy \plus{} yz \plus{} zx}{xyz} \equal{} \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \ge (a \plus{} b \plus{} c)*$ what you want.


ceiling function
analytic geometry
graphing lines
slope
function
algebra
system of equations

Positive integers $ a$, $ b$, and $ c$ are chosen so that $ a<b<c$, and the system of equations
\[ 2x\plus{}y\equal{}2003\text{  and  }y\equal{}|x\minus{}a|\plus{}|x\minus{}b|\plus{}|x\minus{}c|
\]has exactly one solution. What is the minimum value of $ c$?

$ \textbf{(A)}\ 668 \qquad
\textbf{(B)}\ 669 \qquad
\textbf{(C)}\ 1002 \qquad
\textbf{(D)}\ 2003 \qquad
\textbf{(E)}\ 2004$

[hide]
Is it suppose to be $ a\le b\le c$ or $ a<b<c$?
If it is the first case then to minimuse $ c$ let $ a\equal{}b\equal{}c$
So we have the equation

$ y\equal{}2003\minus{}2x\equal{}3|x\minus{}c|$

Now, note that the graph of $ y\equal{}2003\minus{}2x$ has a gradient of $ \minus{}2$ but the graph $ 3|x\minus{}c|$ has a gradient of $ \pm 3$, and has a turning point at $ x\equal{}c$. So if we sketch it we can easily see that in order for the equation to have only one solution, $ c$ must be greater than the $ x\minus{}$axis intercept of $ y\equal{}2003\minus{}2x$.

Now since $ c$ is an intiger $ c\equal{}\left\lceil \frac{2003}{2} \right\rceil \equal{} 1002$

NOTE: if a<b<c then the answer is the same but we origionally set $ b\equal{}a\plus{}1, c\equal{}a\plus{}2$ and the answer is $ 1004$ but this is not an option :P [/hide]

[hide]Let's look at the slope of the absolute value equation as a function of $ x$:
$ x < a \implies y' \equal{} \minus{}3$
$ a < x < b \implies y' \equal{} \minus{}1$
$ b < x < c \implies y' \equal{} 1$
$ c < x \implies y' \equal{} 3$

The slope of the other equation is $ \minus{}2$, so the only way there can be exactly one solution is if the solution occurs at $ x\equal{}a$.

The $ y$ value of the abs-value equation here is $ |a\minus{}b| \plus{} |a\minus{}c| \equal{} (b\minus{}a)\plus{}(c\minus{}a) \equal{} b\plus{}c\minus{}2a$ and the $ y$ value of the linear equation is $ 2003\minus{}2a$.

Thus, $ 2003 \minus{} 2a \equal{} b\plus{}c\minus{}2a \implies b\plus{}c\equal{}2003$.

Since $ b < c$, the minimum $ c$ is produced by $ (b,c) \equal{} (1002,1003)$[/hide]

[quote="cdymdcool"][quote="TZF"][hide]Let's look at the slope of the absolute value equation as a function of $ x$:
$ x < a \implies y' \equal{} \minus{} 3$
$ a < x < b \implies y' \equal{} \minus{} 1$
$ b < x < c \implies y' \equal{} 1$
$ c < x \implies y' \equal{} 3$

The slope of the other equation is $ \minus{} 2$, so the only way there can be exactly one solution is if the solution occurs at $ x \equal{} a$.

The $ y$ value of the abs-value equation here is $ |a \minus{} b| \plus{} |a \minus{} c| \equal{} (b \minus{} a) \plus{} (c \minus{} a) \equal{} b \plus{} c \minus{} 2a$ and the $ y$ value of the linear equation is $ 2003 \minus{} 2a$.

Thus, $ 2003 \minus{} 2a \equal{} b \plus{} c \minus{} 2a \implies b \plus{} c \equal{} 2003$.

Since $ b < c$, the minimum $ c$ is produced by $ (b,c) \equal{} (1002,1003)$[/hide][/quote]
How can we conclude that there can be exactly on solution at x=a when we know that the slope of the line is -2?[/quote]

[hide]Because in order for there to be a unique solution, the line with slope -2 needs to cross all those 4 lines of the absolute value equation at only one point. Hence, it has to be that at this intersection,

$ a \plus{} b \plus{} c \minus{} 3x \equal{} y$
$ b \plus{} c \minus{} a \minus{} x \equal{} y$
$ c \minus{} b \minus{} a \plus{} x \equal{} y$
$ 3x \minus{} a \minus{} b \minus{} c \equal{} y$ must intersect also.

Plugging in $ x \equal{} a$ satisfies this.[/hide]

Here's an intuitive way of thinking about it:

The absolute value equation decreases with slope -3 until x=a,
then decreases with slope -1 until x=b,
then increases with slope 1 until x=c,
then increases with slope 3 after that.

Now consider a line that goes through (a,y(a)) with slope -2. It touches the graph at that point, but to the left and to the right, it is below the graph of the absolute value function.

On your picture, draw cases in which a line with the same slope (-2) is either higher or lower than the line I proposed. In the first case, there will be exactly two solutions. In the second case, there will be no solutions.

great explanation TZF that made a lot of sense to me.

Oops sorry I just realized what I said doesn't even work for the other 2 lines.  :blush: TZF, thanks for the explanation.

Here's a slightly less handwavy/more explicit way of saying what I said before.
[hide="More explicit"]
$ 2x \plus{} y \equal{} 2003\text{ and }y \equal{} |x \minus{} a| \plus{} |x \minus{} b| \plus{} |x \minus{} c|$

Substituting for $ y$, we get:

$ 2003 \equal{} |x \minus{} a| \plus{} |x \minus{} b| \plus{} |x \minus{} c| \plus{} 2x$

Now, the RHS is:
decreasing with slope -1 until x=a
increasing with slope 1 until x=b
increasing with slope 3 until x=c
increasing with slope 5 after x=c.

The important information is that the RHS is:
infinity at x=-infinity
decreasing until x=a
increasing after x=a
infinity at x=infinity

Note that if the RHS is equal to 2003 for some x < a, then it must continue decreasing until x=a, and then increase back to 2003 at some x > a. In other words, each candidate solution for x < a must come paired with a candidate solution for x > a, and vice versa. The only way there can be exactly one solution is if it happens at x=a.

So, if the RHS is to equal 2003 exactly once, it must do so at x=a.[/hide]

Since it hasn't been a month yet, and this one-line solution is brilliant...

[hide=" :) "]

$ b \plus{} c \equal{} 2003$. Therefore, the minimum value of $ c$ is $ \displaystyle\left\lceil\frac{2003}2\right\rceil \equal{} 1002 \Rightarrow \boxed{\textbf{(C)}}$.

I guess a proof will be necessary...

Substitute $ 2y \plus{} x \equal{} 2003$ into the second equation:

$ 2003 \equal{} |x\minus{}a| \plus{} |x\minus{}b| \plus{} |x\minus{}c| \plus{} 2x$.

$ c \equal{} 2003 \minus{} b$, so:

$ 2003 \equal{} |x\minus{}a| \plus{} |x\minus{}b| \plus{} |x\minus{}2003\plus{}b| \plus{} 2x$.

Now plug in $ x\equal{}a$ (since that's where they intersect):

$ 2003 \equal{} |a\minus{}a| \plus{} |a\minus{}b| \plus{} |a\minus{}2003\plus{}b| \plus{} 2a$.

Note that $ a \equal{} a$, $ a \leq b$, and $ a\plus{}b \leq 2003$ because $ b\plus{}c\equal{}2003$:

$ 2003 \equal{} 0 \plus{} b\minus{}a \plus{} 2003 \minus{}a \minus{} b \plus{}2a$.

Which reduces to $ 2003 \equal{} 2003$.

 :yup:

[/hide]

[quote="TZF"][hide]Let's look at the slope of the absolute value equation as a function of $ x$:
$ x < a \implies y' \equal{} \minus{} 3$
$ a < x < b \implies y' \equal{} \minus{} 1$
$ b < x < c \implies y' \equal{} 1$
$ c < x \implies y' \equal{} 3$

The slope of the other equation is $ \minus{} 2$, so the only way there can be exactly one solution is if the solution occurs at $ x \equal{} a$.

The $ y$ value of the abs-value equation here is $ |a \minus{} b| \plus{} |a \minus{} c| \equal{} (b \minus{} a) \plus{} (c \minus{} a) \equal{} b \plus{} c \minus{} 2a$ and the $ y$ value of the linear equation is $ 2003 \minus{} 2a$.

Thus, $ 2003 \minus{} 2a \equal{} b \plus{} c \minus{} 2a \implies b \plus{} c \equal{} 2003$.

Since $ b < c$, the minimum $ c$ is produced by $ (b,c) \equal{} (1002,1003)$[/hide][/quote]

brilliant solution, i think you mean $ (b,c) \equal{} (1001,1002)$ though?

$\text{AoPS's solution:}$
Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$.

When $x<a$, the slope is $-3$.

When $a<x<b$, the slope is $-1$.

When $b<x<c$, the slope is $1$.

When $c<x$, the slope is $3$.

Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$, so $(b,c-a)$ is a point on $f(x)$. In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$. Thus, graphing this will produce a figure that looks like a cup: [asy] import graph; size(100); draw((0,6)--(3,0)); xaxis(0,8.5); yaxis(0,10); real f(real x) { return -3(x-2)+5; } real f2(real x) { return -1(x-2)+5; } real f3(real x) { return 1(x-4)+3; } real f4(real x) { return 3(x-7)+6; } draw(graph(f,0,2)); draw(graph(f2,2,4)); draw(graph(f3,4,7)); draw(graph(f4,7,8.5)); draw((2,-0.25)--(2,0.25)); label("a",(2,0),N); draw((4,-0.25)--(4,0.25)); label("b",(4,0),N); draw((7,-0.25)--(7,0.25)); label("c",(7,0),N); dot((2,5)); label("P",(1.9,5.2),E); [/asy] From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$. Since the line where $a<x<b$ has slope $-1$, the positive difference in $y$-coordinates from $x=a$ to $x=b$ must be $b-a$. Together with the fact that $(b,c-a)$ is on $f(x)$, we see that $P=(a,c-a+b-a)$. Since this point is on $x=a$, the only intersection point with $2x+y=2003$, we have $2 \cdot a+(b+c-2a)=2003 \implies b+c=2003$. As $c>b$, the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$. This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$, and thus the answer is $\boxed{\mathrm{(C)}\ 1002}$.


inequalities
trigonometry
trig identities
Law of Cosines
inequalities solved

If a, b, c are the sides of a triangle, and s = (a+b+c)/2 is its semiperimeter, then prove

$\left( s-a\right) \sqrt{a}+\left( s-b\right) \sqrt{b}+\left( s-c\right)
\sqrt{c}\leq \frac{3}{2}\sqrt{abc}$

  Darij

Hmm... applied to which variables? I'm sorry for the question, I have never applied Schur before...

  Darij

Just take $a=x^2, b=y^2, c=z^2 $ and apply Schur for x,y,z.

Oh yeah, this was really clear... and I wonder why I was so dumb not to note it for myself.

My proof of the inequality used the substitution x = (b+c-a)/2, y = (c+a-b)/2 and z = (a+b-c)/2, so that a = y+z, b = z+x and c = x+y, and the inequality takes the form

[tex]\frac{x}{\sqrt{\left( x+y\right) \left( x+z\right) }}+\frac{y}{\sqrt{\left(y+z\right) \left( y+x\right) }}+\frac{z}{\sqrt{\left( z+x\right) \left(z+y\right) }}\leq \frac{3}{2}[/tex],

which was a (quite simple) problem from the German TST 2001... But now I see this was not necessary - thanks!

  Darij

[quote="darij grinberg"]
[tex]\frac{x}{\sqrt{\left( x+y\right) \left( x+z\right) }}+\frac{y}{\sqrt{\left(y+z\right) \left( y+x\right) }}+\frac{z}{\sqrt{\left( z+x\right) \left(z+y\right) }}\leq \frac{3}{2}[/tex],
[/quote]

I have a direct algebraic proof of this: just applying Cauchy-Schwarz to the arrays $(\sqrt{x}, \sqrt{y}, \sqrt{z})$ and $(\sqrt{x(y+z)}, \sqrt{y(z+x)}, \sqrt{z(x+y)})$, it can be reduced into $8(x+y+z)(xy+yz+zx)\leq 9(x+y)(y+z)(z+x)$ which is well-known.

But I found the following pretty amazing. Use another substitution $a=\sqrt{y+z}, b=\sqrt{x+y}, c=\sqrt{y+z}$ (it's the reverse of who's substitution, darij?), then a, b,c are sides of an acute triangle and the German TST 2001 inequality is equivalent to:

(1) $\displaystyle \frac{b^2+c^2-a^2}{2bc} + \frac{c^2+a^2-b^2}{2ca} + \frac{a^2+b^2-c^2}{2ab} \leq \frac{3}{2}$, which is (by the Law of Cosines):

(2) $\displaystyle \cos{A}+\cos{B}+\cos{C} \leq \frac{3}{2}$ (another well-known fact).

Wait a minute, if we clear the denominators in (1) and move negative terms over to the right hand side, then (1) is equivalent to:

(3) $a(b^2+c^2) + b(c^2+a^2) + c(a^2+b^2) \leq a^3+b^3+c^3 + 3abc$ (Guess what this is!!)

Yes, everything is equivalent to Schur here  :D

Here is how the proposed proof of

[tex]\frac{x}{\sqrt{\left( x+y\right) \left( x+z\right) }}+\frac{y}{\sqrt{\left(y+z\right) \left( y+x\right) }}+\frac{z}{\sqrt{\left( z+x\right) \left(z+y\right) }}\leq \frac{3}{2}[/tex]

for any three positive numbers x, y, z goes:

By GM-AM, we have

[tex]\frac{x}{\sqrt{\left( x+y\right) \left( x+z\right) }} = x\cdot \sqrt{\frac{1}{x+y}\cdot\frac{1}{x+z}} \leq x\cdot \frac12 \cdot \left(\frac{1}{x+y}+ \frac{1}{x+z}\right)[/tex]
[tex]= \frac12 \cdot \left( \frac{x}{x+y} + \frac{x}{z+x}\right)[/tex].

Now write down two analogous inequalities for the other two terms of the sum and add them together; this completes the proof.

  Darij


Find all polynomials f(x) (with real coefficients) satisfying for each x from real:

f(x) x f(1-x) + x^3 +100 >= 0


Suppose that $S$ is a set of positive integers, $Z^+$. For $n\in S$ we consider the selection of spaces $\{I_n\}$ where

$I_n$= $( n-1/n , n+1/n)$

Does the selection $\{I_n\}$ cover $S$?

If yes, can we choose a finite number of spaces from $\{I_n\}$ in a way that it covers $S$ ?

[quote="ICSD"]Does the selection $\{I_n\}$ cover $S$?[/quote]
$n \in I_n$  :roll:

[quote="ICSD"]If yes, can we choose a finite number of spaces from $\{I_n\}$ in a way that it covers $S$ ?[/quote]
$S = \cup I_{n_i}$, a finite union, $max (n_i) + 1 \notin \cup I_{n_i}$

Thus $\mathbb{Z}^+$ is not compact (which is not really a big surprise ) ;)


find the smallest positive integer k such that

You should check your problem : If it is $ 33|(33\cdot 26^2\minus{}k)$, then $ 33|k$ and so the smallest positive integer possible is $ k\equal{}33$


JBL mentioned a study cited in Peter Schrag's book FINAL TEST:

"In a major study of students in Tennessee public schools ... student performance
... was charted in [kindergarten through high school], and it was noted that some teachers were "good teachers" and some were "bad teachers," by which I mean that student performance in the classes of some teachers went up but in other teachers classes it didn't. This was a sufficiently large study that this was controlled for income, race, etc. of the students and education of the teachers, and there wasn't any of those factors that explained the difference. [b]So, in other words, some of the teachers were just better. [/b]More interesting, the change was permanent -- a student who got a good teacher, then a bad teacher, then a good teacher would have her performance go up, then stagnate, then go up. But, in the second year (when having a bad teacher), her performance would still be better than that of students who had had a bad teacher the year before."

I don't think one needs to resort to notions of "better teacher" to
explain the above.   It could be, for example, that there are a set
of skills that promote performance and that particular ones were
inculcated by the practices of specific teachers.   For example,
an algebra teacher who requires students to write out clear, line-by-line
derivations of their reasoning, may or may not be a good teacher
of algebra, but if the systematic work habit is imparted then it
would presumably have a lasting effect in later courses.

A lot of performance, or survival, in a school context is the
sum of a lot of meta-learning of this kind over and above particular
subject matter learned in classes.  Maybe teachers who are
"better" in the usual (hazy, informal) senses of that term are also
more effective in conveying the meta-knowledge, but it's not
clear to me that the correlation would be all that strong.

Okay, I'm going to be honest -- I have no idea what your point is.  Would you be happy if I had written "Some teachers are just more effective than others," instead of what I did write?  I don't think that my sentence implied that the "better performance" was necessarily a result of innate superiority, although the study did show no correlation between being "good" and having received any higher degrees than a bachelors.  (I don't have the book on-hand, so I can't look and see if it went into any more detail than that, and I've never actually looked at the study itself, just Schrag's presentation of it.  I think there may also have been some correlation between the race of the teacher and the affect of the teacher on the racial gap between students, but I wouldn't swear on it.)
I personally think it's unquestionable that some people are better teachers than others, in an innate sense, and also true that like most innate advantages, such an advantage probably could be overcome by a correct collection of knowledge, skills and practice.  I really don't know what else to say in response to your post -- am I addressing the point you raised?

JBL,

I was not deconstructing the notion of "better teacher", but saying
that this interesting study needs some care in the interpretation.

Specifically, Schrag and/or you appear to interpret the results
as attributes of the teachers (apparently not very replicable ones,
in light of the noncorrelation with education) rather than as attributes of
what they taught  (e.g. did a math teacher impart writing habits
that had an effect in later courses).

The permanence of the effect in later courses could be from
education being cumulative (i.e. greater effectiveness in imparting
this year's algebra will help next year's geometry), or it could also
be that different teachers cover the same curriculum in essentially
the same way but adding different non-curricular meta-knowledge
that still applies in later classes (note-taking, writing, library, etc  skills).

Also, he points out that good teaching is more effective earlier in a child's academic career.

[quote] (JBL wrote:)
the study did show no correlation between being "good" and having received any higher degrees than a bachelors. (I don't have the book on-hand, so I can't look and see if it went into any more detail than that, [/quote]

I looked briefly at the book.  No disrespect intended, but the studies and Schrag's conclusions are more limited than what was described here.

Schrag cites three studies, and briefly discusses two of them. The one you remembered was the TVAAS, which tested some Tennessee students over a 3-year period (not K-12 or anything like that).  The TVAAS exams are for grade 2-8; the study followed the progress of some/all students from grade 2 in grades 3,4,5. The study designers claimed amazing 50 per cent (statistically huge) differences in outcome attributable to having the "good" teachers.

None of the studies claimed that there is no correlation with higher education, and there was nothing whatsoever about controlling for this or any other pertinent
variables such as income, race, etc. (It wouldn't be a major surprise if such issues were dealt with in analyzing the results, but "controlling" for it from the beginning
would be much harder and would lend more weight to the results.)

Schrag does cite the author of one of the other studies,
based on the Texas state exams, as saying that teachers' education beyond
a first degree doesn't make a big difference.  The study author
didn't necessarily claim that this was a finding from the study, as opposed
to his opinion as an education researcher.

Schrag also notes that the Tennessee study was never published except
as a progress report.  He gives a WWW link to the report, but the link
is dead.

[quote]No disrespect intended ...[/quote]
None taken -- you're right, and I apologize.  It appears I conflated several different things into one uber-study.  Some of my descriptions were wrong and gave the study more weight than it deserved.  However, the key point -- that good teachers have a large and measurable effect -- was supported by the TVAAS.  I don't know what the implications of it not being published are -- perhaps research is continuing, or perhaps it was an in-house study, never designed for publication, or perhaps it doesn't meet the rigor of peer review.

With respect to controlling for race, I meant in analyzing the results of the study, not in conducting the study.  This may have been grabbed from somewhere else entirely.  Oops  :blush:

[quote]Schrag does cite the author of one of the other studies,
based on the Texas state exams, as saying that teachers' education beyond
a first degree doesn't make a big difference. The study author
didn't necessarily claim that this was a finding from the study, as opposed
to his opinion as an education researcher.[/quote]

I doubt that someone who has had access to as large a supply of data as Texas state test results would claim something as fact that was not supported in the data, whether or not it was in the study.

Okay, now on to something where I no longer have to rely on mis-remembered studies :) :
[quote]...you appear to interpret the results
as attributes of the teachers (apparently not very replicable ones,
in light of the noncorrelation with education) rather than as attributes of
what they taught[/quote]

I think perhaps that you are choosing to make a very literal interpretation of my use of the words "good" and "better."  Two years ago, when I as a high school student refered to a teacher as "good," I knew what that meant: a good teacher is one who ensured that I learned the material and learned it cold.  I've had good teachers who were good because they were intensely terrifying and strict and I've had teachers who were good because they managed to include all thirty-odd students in the class in a way that drew us in.  Those are two very different types of teachers.  I doubt they would get along well with each other and I doubt even more that their methods are transferable -- there is no way Ms. Thoms could have tought a class like Mr. Geller, and vice-versa.  There are many, many things that go into making a good teacher, far beyond the scope of the curriculum or even the meta-curriculum.  What skills a teacher teaches are irrelevant if students don't incorporate those skills into their lives sufficiently much that they can reproduce them in class, on exams, in future classes, and for the rest of their lives.  And the ability to make students absorb those skills and that knowledge is the ability that is "good teaching."  It manifests itself in the two ways I mentioned, but also in dozens of others -- very rarely do two different teachers teach in exactly the same way -- and it is grounded in the different personality traits and temperments of each teacher.  You can give a teacher as many lists of knowledge and meta-knowledge as you want for them to impart, but some will do a better job than others.  And you can probably help those who do less well with specific suggestions on how to teach specific things, but that will only work for a teacher who is able to actually take that knowledge and put it to good use.

Finally: do you accept that there are students who, in the same learning environment, will preform statistically-measurably better on any kind of educational measurement than their peers?  Do you also accept that, while that gap could be closed somewhat by giving the lower-performing students extra tutoring, smarter teachers and better materials, there exist pairs of students whose abilities are such that, regardless of the relative quality of education they receive above what is offered at a regular public school, one will always outperform the other?  (Sorry that sentence is so long.)  If so, why should we believe anything different about teachers (or, for that matter, people doing anything other than studying)?

[quote="Valentin Vornicu"][quote="JBL"]Finally: do you accept that [...][/quote]It's incredible how you manage to close up all your arguments with this phrase :D :D Please don't mind my intrusion in this discussion, which is otherwise very serious.  :blush:[/quote]

I used to do a form of debate in which the audience would get an opportunity to sometimes make small speeches or to question the speakers.  I found that when one wasn't able to get an opportunity to speak, the best way to make a point was to start a question with "Do you accept that ..." or "Is it not true that ..." and then talk until the moderator of the debate started to give you nasty looks :):)

[quote="JBL"][quote][cited] the author of one of the other studies, based on the Texas state exams, as saying that teachers' education beyond a first degree doesn't make a big difference. The study author didn't necessarily claim that this was a finding from the study, as opposed to his opinion as an education researcher.[/quote]
I doubt that someone who has had access to as large a supply of data as Texas state test results would claim something as fact that was not supported in the data, whether or not it was in the study. [/quote]

Schrag's source in Texas seemed to be saying both more and less.  On the one hand, that in his professional capacity (i.e. taking into account other people's work as well as his own) he'd never heard of much link between postbaccalaureate education and teacher performance.  On the other hand, he didn't seem to be saying that his own study had anything significant to say for or against such a link, either.  Recall that it was the assessments in Tennessee, not Texas, that were geared to discerning the effects of particular teachers.

Also, the irrelevance of post-bac education would support Schrag's point and it's hard to imagine that in interviewing the Texan he wouldn't have inquired further about that point.
In other words, if the Texas guy HAD said anything more specifically supporting this idea in connection with the huge mass of statewide data, Schrag would probably have quoted it.
So it may be there simply wasn't much for him to say.

I didn't check the endnotes about the Texas study for
any additional information, but that was my impression from reading in the bookstore.

I'll try to answer the rest of your posting later.

[quote="BHorseMath"]Which Texas State exam did he work with??? It changed 2 years ago and about 6 years before that it was different.  TExas can't get its act together with state exams so they're are not really a good thing to look at.[/quote]

I think the educationist interviewed was Ronald Ferguson and the Texas exams were the "accountability" exams introduced by Bush the Lesser.   Web-searching for him finds, for instance,

http://www.cfequity.org/Fergus.html
http://educationright.tripod.com/id102.htm

Quoting from the latter:
"In an analysis of 900 Texas school districts, Ronald Ferguson found that teachers’ expertise—as measured by scores on a licensing examination, masters degrees, and experience—accounted for about 40% of the measured variance in students’ reading and mathematics achievement at grades 1 through 11, more than any other single factor."

This seems to contradict the irrelevance-of-postbac-education quotation in Schrag, so it could be that that quotation came from the author of the other Texas study that he cites (though I think I'm correctly remembering Ferguson as the source! I will check.)


1 is obvious. take circle C1, it contains 2 pts. Take circle  C2 not intersecting or touching C1 it has 2 different pts from A. Draw a circle throgh 3 of the 4 pts we have, and get a contradiction to the requirement os the problem. the answer is NO.

[quote="pavel25"]1) Is there a group of points $A$ in a plane such that for each circle in the plane, there are exactly 2 circle's points that belong to group $A$?
2) Is there a group of points $B$ in a plane such that for each circle with radius 1 in the plane, there are exactly 2 circle's points that belong to group $B$?[/quote]

For the A is no, since you can take circumference with ray $\to 0$ and calling d_m the least distance between two of the points, thake the circumference with ray=0 [size=59]-n^2[/size] and center on the midpoint of d_m

For B, i say yes, and it's a group of parallel lines each of the ones have distance 1 from the previous and the following ones ;


inequalities
logarithms
inequalities unsolved

Prove that $(1+8cos^2A)(1+8cos^2B)(1+8cos^2C)\geq 64sin^2Asin^2Bsin^2C$, where $A+B+C=\pi$.

I have a proove but it`s reallllllllly ugly. please show me the shortest proove

I think it can be proven by jensen .

rewrite it as

$\prod (cosec^2A+8\cot^2 A)\geq 64$ $\longleftrightarrow$ $\prod (9\cot^2A+1)\geq 64$

Then using the convexity of  $f(x)=\ln (9\cot^2x+1)$  (if my calculation is correct , it is convex) , then the result follows .


logarithms
inequalities proposed
inequalities

In fact $f(x) = \ln \left( 1+\frac{1}{x^{3}}\right)$ is convex, so

$\ln A = \sum f(a) \ge 3 f \left( 2 \right) = 3 \ln \frac{9}{8}\implies$
$A \ge \frac{9^{3}}{8^{3}}$

As expected.


function
complex numbers
complex analysis

Find the maximum of $ |e^z|$ on $ |z| \leq 1$.

Find the maximum of $ |cos z|$ on $ [0,2\pi]$ X $ [0,2\pi]$.

Find the maximum of $ |e^{z^2}|$ on the unit disk.

note that the first and the last questions are the same as $ z \mapsto z^2$ maps the unit disc to itself

the modulus of the exponential function is the exponential of the real part, and the real part clearly attains a maximum of $ 1$, so the answers to these are $ e^1 \equal{} e$.

for the last, you can bring in the maximum principle to very slightly simplify the argument and find that the maximum of $ \cosh (2\pi)$ will be taken along the upper edge


calculus
integration
calculus computations

[quote="borislav_mirchev"]Calculate:
1. $ \int_{}^{ } \frac {1}{x \plus{} tgx}dx$
2. $ \int_{}^{ } \frac {dx}{x^2(1 \plus{} e^x)}$[/quote]

Could these actually be definite integrals? I am relying on an online integrator for asking this question  :huh:

hello, Maple 12 says
$ \int\frac{1}{x\plus{}tan(x)}dx\equal{}ln(x\minus{}i)\plus{}\int\minus{}\frac{2i}{\left(x\left(e^{(ix)}\right)^2\plus{}x\minus{}i\left(e^{(xi)}\right)^2\plus{}i\right)\left(x\minus{}i\right)}dx$.
Sonnhard.

They were given at the site as an indefinite ones - if there is a mistake - they may.


vector
abstract algebra
symmetry
linear algebra
linear algebra unsolved

Let $ V$ be a vector space over $ F$ where $ char(F) \not\equal{} 2$ and $ \alpha, \beta$ be [i] projections [/i] on $ V$, i.e. $ \alpha^2 \equal{} \alpha, \beta^2 \equal{} \beta$. Show that (a) $ \alpha\plus{}\beta$ is also a projection on $ V$ $ \iff$ (b) $ \alpha \beta \equal{} \beta \alpha \equal{} 0$.

As proposed, my question is "Could this be done using algebraic manipulation only?"

I'm reasonably sure it can't, and if I'm right it's because this result isn't true in a general ring.  One can construct the (non-commutative) algebra $ F < \alpha, \beta > /(\alpha^2 \minus{} \alpha, \beta^2 \minus{} \beta, \alpha \beta \plus{} \beta \alpha)$ in which the left result is true without requiring the right.

I believe the truth of this result rests on finite-dimensionality (it's not true if $ V$ is infinite-dimensional by the above example), so there is something going on beyond here besides algebraic manipulation.

There's a purely algebraic proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=241826]here[/url]. The only ring properties needed are that $ 1$ exists and $ 2$ is not a zero-divisor.

In the example t0rajir0u gave, $ \alpha\beta \equal{} \alpha^2\beta \equal{} \alpha(\alpha\beta) \equal{} \alpha( \minus{} \beta\alpha) \equal{} ( \minus{} \alpha\beta)\alpha \equal{} (\beta\alpha)\alpha \equal{} \beta\alpha^2 \equal{} \beta\alpha$, and the algebra is actually commutative.

How about proof utilizing the Linear Transformations/Operators properties: image, kernel, etc. ? But anyway, [b] Carcul [/b]'s proof is really nice.

$ \alpha\beta \plus{} \beta \alpha \equal{} 0 \Rightarrow \alpha^2 \beta \plus{} \alpha\beta\alpha \equal{} 0 \Rightarrow \alpha\beta \plus{} \alpha\beta\alpha \equal{} 0 \Rightarrow \beta\alpha \equal{} \alpha\beta\alpha$ or $ (\alpha\minus{}\text{id}_V) \beta\alpha \equal{} 0 \Rightarrow Im(\beta\alpha) \subseteq \ker(\alpha \minus{} \text{id}_V)$.

By symmetry, $ \alpha\beta \equal{} \beta\alpha\beta$ and $ Im(\alpha\beta) \subseteq \ker(\beta \minus{} \text{id}_V)$.

Note that $ Im(\beta\alpha) \equal{} Im(\alpha\beta) \Rightarrow Im(\alpha\beta) \equal{} Im(\beta\alpha) \subseteq \ker(\beta \minus{} \text{id}_V) \cap \ker(\alpha \minus{} \text{id}_V) \equal{} \{ 0_V \}$. (for any $ x \in \ker(\beta \minus{} \text{id}_V) \cap \ker(\alpha \minus{} \text{id}_V)$, $ \beta(x) \equal{} \alpha(x) \equal{} x$. Then $ \alpha\beta(x) \plus{} \beta\alpha(x) \equal{} \alpha(x) \plus{} \beta(x) \equal{} x \plus{} x \equal{} 2x \equal{} 0 \iff x \equal{} 0_V$ due to $ char(F) \not\equal{} 2$).

So $ \alpha\beta \equal{} \beta\alpha \equal{} 0$.


limit
trigonometry
algebra
polynomial
calculus
calculus computations

1)

$ n \in \mathbb{N*}$ , find :

$ \lim_{x\rightarrow \frac {\pi}{2n}}\frac {\sqrt {\frac {\sin(2nx)}{1 \plus{} \cos(nx)}}}{4(nx)^2 \minus{} \pi^2}$

2)

find :

$ \lim_{x\rightarrow 0}\frac {\sin(\pi\sqrt {x \plus{} 1} \: )}{x}$

L'hopital's rule is the laziest I can say.

The second has a polynomial for denominator. The first can be simplified before applying L'hopitals rule.

I stopped seeking elegant solutions to such problems a while ago.  :D

thank's

can you help me for the first i tried  L'hopital's rule but there is a lot of calculs

$ \alpha)$ For $ \sin(2nx)\geq0$, we  must remain in $ x\leq{\frac {\pi}{2n}}$ and left limit.
Then $ \displaystyle\mathop{\lim}\limits_{x\to{\frac {\pi}{2n}}^{\minus{}}}\frac {\sqrt {\frac {\sin(2nx)}{1 \plus{} \cos(nx)}}}{4(nx)^2 \minus{} \pi^2}\equal{}\mathop{\lim}\limits_{x\to{\frac {\pi}{2n}}^{\minus{}}}\frac {\sqrt {\sin(2nx)}}{\minus{}(\pi\minus{}2nx)\sqrt{1\plus{}\cos(nx)}(\pi\plus{}2nx)}\equal{}$
$ \minus{}\displaystyle\mathop{\lim}\limits_{x\to{\frac {\pi}{2n}}^{\minus{}}}\left[ {\frac {\sqrt {\sin(2nx)}}{\sqrt{\pi\minus{}2nx}}\frac {1}{\sqrt{\pi\minus{}2nx}}\frac {1}{\sqrt{1\plus{}\cos(nx)}(\pi\plus{}2nx)}}\right] \equal{}$
$ \minus{}\displaystyle\mathop{\lim}\limits_{x\to{\frac {\pi}{2n}}^{\minus{}}}\left[ {\sqrt {\frac {\sin(\pi\minus{}2nx)}{\pi\minus{}2nx}}\frac {1}{\sqrt{\pi\minus{}2nx}}\frac {1}{\sqrt{1\plus{}\cos(nx)}(\pi\plus{}2nx)}}\right] \equal{}$
$ \minus{}\displaystyle\sqrt {\mathop{\lim}\limits_{x\to{\frac {\pi}{2n}}^{\minus{}}}\frac {\sin(\pi\minus{}2nx)}{\pi\minus{}2nx}}\displaystyle\mathop{\lim}\limits_{x\to{\frac {\pi}{2n}}^{\minus{}}}\frac {1}{\sqrt{\pi\minus{}2nx}}\displaystyle\mathop{\lim}\limits_{x\to{\frac {\pi}{2n}}^{\minus{}}}\frac {1}{\sqrt{1\plus{}\cos(nx)}(\pi\plus{}2nx)} \equal{}$
$ \minus{}\sqrt {1}\cdot(\plus{}\infty)\cdot\frac{1}{2\pi}\equal{}\minus{}\infty$.
$ \beta)$ If we admit $ \sin(2nx)\leq0$, then we can talk about right limit which, by similar method, is $ i\cdot(\plus{}\infty)$. $ \square$

:) v  thank's a lot , your methode is the best , because lhospital's rule is not alod in our systeme


geometry
geometric transformation
rotation
superior algebra
superior algebra unsolved

Let $ X \equal{} \{1,2,\cdots, p\}$ with $ p$ prime, and  $ Y \equal{} \{1,2,\cdots,t,i_1,i_2,\cdots,i_{p \minus{} t}\}$ with $ 1\leq t\leq p \minus{} 1$ and $ i_j\not\in X$ for $ 1\leq j\leq p \minus{} t$. Supose that $ a$ and $ b$ are both   $ p$-rotation in $ X$ and $ Y$, respectively. Whether or not $ p$ don't divide the order of $ ab$?

Sorry, but what exactly do you mean by "$ p$-rotation"?

That is the order of $ a$ is $ p$ such as $ (123\cdots p)$.

Do you mean that $ a$ and $ b$ are permutations on $ X$ and $ Y$, respectively?

I have found a counterexample as  follows.

$ a\equal{}(123)$ and $ b\equal{}(142)$.

I revised my question.

Let $ X \equal{} \{1,2,\cdots, p\}$ with $ p$ prime, and  $ Y \equal{} \{1,2,\cdots,t,i_1,i_2,\cdots,i_{p \minus{} t}\}$ with $ 1\leq t\leq p \minus{} 1$ and $ X\cap Y\equal{}\{1,2,\cdots, t\}$. Supose that $ a\equal{}(12\cdots p)$ and $ b$ is a   $ p$-cycle in $ Y$.  Whether or not there exist an element $ x\in <b>$ such that $ p$ don't divide the order of $ ax$?


geometry
incenter
algebra proposed
algebra

Three congruent circles have a common point $ O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incentre and the circumcentre of the triangle and the common point $ O$ are collinear.

This topic must be in Geometry forum.
See here: http://www.mathlinks.ro/viewtopic.php?t=291966


Prove that  the equation $ x^3\plus{}3y^3\equal{}5$ has infinitely many rational solutions

LHP highshool has a student like you?shameful  :D
To mod: Pls lock this topic because it's on Mathematical and Youth magazine


pigeonhole principle
induction
combinatorics proposed
combinatorics

Let $ n\geq3$ be an integer. Prove that the set $ X$ = {$ 1,2,3,\cdots,n^2 \minus{} n$} can be expressed as the disjoint union of two non-empty subsets such that each doesn't contain $ n$ elememts $ a_{1},a_{2},\cdots,a_{n}, a_{1} < a_{2} < \cdots < a_{n},$ satisfying $ a_{k}\leq\frac {a_{k \minus{} 1} \plus{} a_{k \plus{} 1}}{2},k \equal{} 2,\cdots,n \minus{} 1.$

I was able to solve this problem during the competition, which fulfills my goal as 'getting a positive score in CMO' :lol:
The key step is to let $ A\equal{}\cup_{k\equal{}1}^{n\minus{}1}\{k^2\minus{}k\plus{}1,k^2\minus{}k\plus{}2,...,k^2\}$ and $ B\equal{}\cup_{k\equal{}1}^{n\minus{}1}\{k^2\plus{}1,k^2\plus{}2,...,k^2\plus{}k\}$, then show $ A,B$ satisfy the problem.

The proof is not really hard:
[hide]Notice that for any $ n$ elements from a set, say $ A$, at least two consecutive terms, say $ a_i,a_{i\plus{}1}$ are in the same set $ A_k$ for some $ i,k$. If $ i>1$ and $ a_{i\minus{}1}\in A_{k'}$, $ k'<k$ then one may check $ a_i \leq \frac{a_{i\minus{}1}\plus{}a_{i\plus{}1}}{2}$, and we can prove that such $ i,k$ are always exist by pigeonhole principle and extreme principle.[/hide]

We can also use induction:
Consider the same sets. Define inductively as $ A_{3} \equal{} {1,3,4}$, $ B_{3} \equal{} {2,5,6}$, and $ A_{n \plus{} 1} \equal{} A_{n} \cup (n^2 \minus{} n \plus{} 1, n^2 \minus{} n \plus{} 2 ... n^2), B_{n \plus{} 1} \equal{} B_{n} \cup (n^2 \plus{} 1, n^2 \plus{} 2 ... n^2 \plus{} n)$. We wish to prove that $ A_{n \plus{} 1}$ and $ B_{n \plus{} 1}$ have sequences $ a_{1} < a_{2} < ... < a_{k}$, where $ a_{i} \leq \frac {a_{i \minus{} 1} \plus{} a_{i \plus{} 1}}{2}$, of length only $ k \leq n$. Let's use induction to do this:
Assume that $ A_{n}$ doesn't contain such sequences, of length longer than $ n \minus{} 1$. Suppose that $ A_{n \plus{} 1}$ contains such a sequence of length $ n \plus{} 1$ (if it doesn't, we are done). Since $ A_{n \plus{} 1} \equal{} A_{n} \cup (n^2 \minus{} n \plus{} 1 ... n^2)$, since at most $ n \minus{} 1$ elements of $ A_{n}$ can contain such a sequence, at least $ 2$ members of such a sequence must be from the set $ (n^2 \minus{} n \plus{} 1 ... n^2)$. Let the largest element of the sequence from $ A_{n}$ be $ k$, and the two smallest elements from this sequence, in the set $ (n^2 \minus{} n \plus{} 1 ... n^2)$ be $ l$ and $ m$. Clearly we have $ k \leq (n \minus{} 1)^2$, $ l \geq n^2 \minus{} n \plus{} 1$, $ l \leq m \leq n^2$, so $ l \minus{} k \geq n$, but $ m \minus{} l \leq n \minus{} 1$, so $ l \minus{} k \geq m \minus{} l$, so $ \frac {m \plus{} k}{2} < l$, which is a contradiction, since $ k, l, m$ are consecutive elements of set $ (a_{1} ... a_{k})$. So $ A_{n \plus{} 1}$ cannot contain such sequences of length $ n \plus{} 1$, and by the same method, neither can $ B_{n \plus{} 1}$, so we are done.


In triangle $ABC$, $AB > BC > CA$, $AB=6$, $\angle{B}-\angle{C}=90^o$. The incircle touches $BC$ at $E$ and $EF$ is a diameter of the incircle. Radical $AF$ intersect $BC$ at $D$. $DE$ equals to the circumradius of $\triangle{ABC}$. Find $BC$ and $AC$.

i cant solve but i think it is not too hard

i think there is a mistake about this problem
$ \angle{B}\equal{}\angle{C}\plus{}90^{o}$ so AC is the greatest side
can somebody correct it!

It is very difficult but my results are:
BC=7-7^.5
AC=8-2*7^.5


inequalities
function
Cauchy Inequality
3-variable inequality
Iran

here's the solution as i promised.it's by Marcin Kuczma (his name appears many times in the imo authors list on the forum)
so here we go
let F be the expression on the left hand side of the inequality.
Assume without loss of generality x  \geq  y  \geq  z  \geq  0, with y > 0
A=(2x+2y-z)(x-z)(y-z)+z(x+y) 2
B=1/4*[z(x+y-2z)(11x+11y+2z)]
C=(x+y)(x+z)(y+z)
D=(x+y+z)(x+y-2z)+x(y-z)+y(x-z)+(x-y) 2
E=1/4*[z(x+y)(x+y-2z) 2 (x+y+2z) 2 ]
now verify that (is ne1 willing :D )
C 2 (4F-9)=(x-y) 2 [(x+y)(A+B+C)+(x+y)(y+z)*D/2]+E

working out when the equality hold is the easy part

I have just found another proof of the above inequality which seems to be a bit less ugly than the familiar ones.

[color=blue][b]Iran 1996 inequality:[/b] If a, b, c are positive reals, then

$\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}+\frac{1}{\left(a+b\right)^2}\geq\frac{9}{4\left(bc+ca+ab\right)}$.[/color]

[i]Proof.[/i] We first prove a lemma:

[color=blue][b]Lemma 1.[/b] If a, b, c, x, y, z are six nonnegative reals such that $a\geq b\geq c$ and $x\leq y\leq z$, then

$x\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+y\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+z\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$
$\geq 0$.[/color]

[i]Proof of Lemma 1.[/i] Since $a\geq b$, we have $ab\geq b^2$, and since $b\geq c$, we have $bc\geq c^2$. Thus, the terms $3ca+ab+bc-b^2$ and $3ab+bc+ca-c^2$ must be nonnegative. The important question is whether the term $3bc+ca+ab-a^2$ is nonnegative or not. If it is, then we have nothing to prove, since the whole sum

$x\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+y\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+z\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$

is trivially nonnegative, as a sum of nonnegative expressions. So we will only consider the case when it is not; i. e., we will consider the case when $3bc+ca+ab-a^2<0$. Then, since $\left(b-c\right)^2\geq 0$, we get $\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)\leq 0$, and this, together with $x\leq y$, implies that $x\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)\geq y\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)$. On the other hand, since $3ab+bc+ca-c^2\geq 0$ and $\left(a-b\right)^2\geq 0$, we have $\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)\geq 0$, what, combined with $y\leq z$, yields $z\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)\geq y\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$. Hence,

$x\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+y\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+z\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$
$\geq y\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+y\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+y\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$
$=y\left(\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)\right)$.

But

$\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$
$\geq\left(b-c\right)^2\left(-bc+ca+ab-a^2\right)+\left(c-a\right)^2\left(-ca+ab+bc-b^2\right)+\left(a-b\right)^2\left(-ab+bc+ca-c^2\right)$
                             (since squares of real numbers are always nonnegative)
$=\left(b-c\right)^2\left(c-a\right)\left(a-b\right)+\left(c-a\right)^2\left(a-b\right)\left(b-c\right)+\left(a-b\right)^2\left(b-c\right)\left(c-a\right)$
$=\left(b-c\right)\left(c-a\right)\left(a-b\right)\left(\underbrace{\left(b-c\right)+\left(c-a\right)+\left(a-b\right)}_{=0}\right)=0$;

thus,

$x\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+y\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+z\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$
$\geq 0$,

and Lemma 1 is proven.

Now to the proof of the Iran 1996 inequality:

We first rewrite the inequality using the $\sum$ notation as follows:

$\sum\frac{1}{\left(b+c\right)^2}\geq\frac{9}{4\left(bc+ca+ab\right)}$.

Upon multiplication with 4 (bc + ca + ab), this becomes

$\sum\frac{4\left(bc+ca+ab\right)}{\left(b+c\right)^2}\geq 9$.

Subtracting 9, we get

$\sum\frac{4\left(bc+ca+ab\right)}{\left(b+c\right)^2}-9\geq 0$,

what is equivalent to

$\sum\left(\frac{4\left(bc+ca+ab\right)}{\left(b+c\right)^2}-3\right)\geq 0$.

But $\frac{4\left(bc+ca+ab\right)}{\left(b+c\right)^2}-3=\frac{\left(3b+c\right)\left(a-b\right)}{\left(b+c\right)^2}-\frac{\left(3c+b\right)\left(c-a\right)}{\left(b+c\right)^2}$.

Hence, it remains to prove

$\sum\left(\frac{\left(3b+c\right)\left(a-b\right)}{\left(b+c\right)^2}-\frac{\left(3c+b\right)\left(c-a\right)}{\left(b+c\right)^2}\right)\geq 0$.

But

$\sum\left(\frac{\left(3b+c\right)\left(a-b\right)}{\left(b+c\right)^2}-\frac{\left(3c+b\right)\left(c-a\right)}{\left(b+c\right)^2}\right)=\sum\frac{\left(3b+c\right)\left(a-b\right)}{\left(b+c\right)^2}-\sum\frac{\left(3c+b\right)\left(c-a\right)}{\left(b+c\right)^2}$
$=\sum\frac{\left(3b+c\right)\left(a-b\right)}{\left(b+c\right)^2}-\sum\frac{\left(3a+c\right)\left(a-b\right)}{\left(c+a\right)^2}=\sum\left(\frac{\left(3b+c\right)\left(a-b\right)}{\left(b+c\right)^2}-\frac{\left(3a+c\right)\left(a-b\right)}{\left(c+a\right)^2}\right)$
$=\sum\frac{\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)}{\left(b+c\right)^2\left(c+a\right)^2}$.

Thus, the inequality in question is equivalent to

$\sum\frac{\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)}{\left(b+c\right)^2\left(c+a\right)^2}\geq 0$.

Upon multiplication with $\left(b+c\right)^2\left(c+a\right)^2\left(a+b\right)^2$, this becomes

$\sum\left(a+b\right)^2\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)\geq 0$.

In other words, we have to prove the inequality

$\left(b+c\right)^2\left(b-c\right)^2\left(3bc+ca+ab-a^2\right)+\left(c+a\right)^2\left(c-a\right)^2\left(3ca+ab+bc-b^2\right)+\left(a+b\right)^2\left(a-b\right)^2\left(3ab+bc+ca-c^2\right)$
$\geq 0$.

But now it's clear how we prove this - we WLOG assume that $a\geq b\geq c$, and define $x=\left(b+c\right)^2$, $y=\left(c+a\right)^2$ and $z=\left(a+b\right)^2$; then, the required inequality follows from Lemma 1 after showing that $x\leq y\leq z$ (what is really easy: since $a\geq b\geq c$, we have $\left(a+b+c\right)-a\leq \left(a+b+c\right)-b\leq \left(a+b+c\right)-c$, what rewrites as $b+c\leq c+a\leq a+b$, and thus $\left(b+c\right)^2\leq\left(c+a\right)^2\leq\left(a+b\right)^2$, or, in other words, $x\leq y\leq z$).

This completes the proof of the Iran 1996 inequality. Feel free to comment or to look for mistakes (you know, chances are not too low that applying a new method one can make a number of mistakes)

 Darij

Fixing LaTeX in payman_pm's post:

[quote="payman_pm"]$(xy+yz+zx)(\frac{1}{(x+y)^2}+\frac {1}{(y+z)^2}+\frac {1}{(z+x)^2}=$
$=(xy+yz+xz)(\frac{(x+y)^2(y+z)^2+(y+z)^2(z+x)^2+(z+x)^2(x+y)^2}{(x+y)^2(y+z)^2(z+x)^2}$
but we have
$(x+y)^2(y+z)^2+(y+z)^2(z+x)^2+(z+x)^2(x+y)^2=$
$\sum (x^5y+2x^4y^2+\frac{5}{2}x^4yz+13x^3y^2z+4x^2y^2z^2)$
and
$(x+y)^2(y+z)^2(z+x)^2=\sum(x^4y^2+x^4yz+x^3y^3+6x^3y^2z+\frac{5}{3}x^2y^2z^2)$
and by some algebra
$\sum (4x^5y-x^4y^2-3x^3y^3+x^4yz-2x^3y^2z+x^2y^2z^2)\ge 0 $
and by using $schur$ inequality we have
$\sum(x^3-2x^2y+xyz)\ge0 $
and if multiply this inequality to $xyz$ :
$\sum(x^4yz-2x^3y^2z+x^2y^2z^2)\ge 0$     $(I)$
and by using $AM-GM$ inequality we have
$\sum ((x^5-x^4y^2)+3(x^5y-x^3y^3))\ge0$   $(II)$
and by using $I$ ,$II$ the problem is solve.[/quote]

This is actually the canonical proof of the Iran 1996 inequality. I don't like it because of the lots of calculation involved. Note that there are two typos in the above:

$(x+y)^2(y+z)^2+(y+z)^2(z+x)^2+(z+x)^2(x+y)^2=
\sum (x^5y+2x^4y^2+\frac{5}{2}x^4yz+13x^3y^2z+4x^2y^2z^2)$

should be

$\left(xy+yz+xz\right)\left(\left(x+y\right)^2\left(y+z\right)^2+\left(y+z\right)^2\left(z+x\right)^2+\left(z+x\right)^2\left(x+y\right)^2\right)$
$=\sum \left(x^5y+2x^4y^2+\frac52 x^4yz+13x^3y^2z+4x^2y^2z^2\right)$,

and

$\sum ((x^5-x^4y^2)+3(x^5y-x^3y^3))\ge0$   $(II)$

should be

$\sum \left(\left(x^5y-x^4y^2\right)+3\left(x^5y-x^3y^3\right)\right)\ge 0$   $(II)$.

  Darij

Have you a nice proof for the generalization below?

If $ a,b,c$ and $ x,y,z$ are nonnegative real numbers, then

$ \frac 2{(a\plus{}b)(x\plus{}y)}\plus{}\frac 2{(b\plus{}c)(y\plus{}z)}\plus{}\frac 2{(c\plus{}a)(z\plus{}x)}
\\ \ge \frac 9{(b\plus{}c)x\plus{}(c\plus{}a)y\plus{}(a\plus{}b)z}.$

[quote="Vasc"]Have you a nice proof for the generalization below?

If $ a,b,c$ and $ x,y,z$ are nonnegative real numbers, then

$ \frac 2{(a \plus{} b)(x \plus{} y)} \plus{} \frac 2{(b \plus{} c)(y \plus{} z)} \plus{} \frac 2{(c \plus{} a)(z \plus{} x)} \\
\ge \frac 9{(b \plus{} c)x \plus{} (c \plus{} a)y \plus{} (a \plus{} b)z}.$[/quote]
This generalization is very nice, Vasc. However, I think it is a direct consequence of the Iran Inequality (although from it, we can deduce the Iran one). Let me explain wy:

Because $ (b \plus{} c)x \plus{} (c \plus{} a)y \plus{} (a \plus{} b)z \equal{} a(y \plus{} z) \plus{} (b \plus{} c)x \plus{} cy \plus{} bz,$ one can write the desired inequality as
$ \sum \frac {2[a(y \plus{} z) \plus{} x(b \plus{} c) \plus{} cy \plus{} bz]}{(b \plus{} c)(y \plus{} z)} \ge 9,$
which is equivalent to
$ \sum \frac {2a}{b \plus{} c} \plus{} \sum \frac {2x}{y \plus{} z} \ge 9 \minus{} \sum \frac {2(cy \plus{} bz)}{(b \plus{} c)(y \plus{} z)}.$
Now, since
$ 3 \minus{} \sum \frac {2(cy \plus{} bz)}{(b \plus{} c)(y \plus{} z)} \equal{} \sum \frac {(b \minus{} c)(y \minus{} z)}{(b \plus{} c)(y \plus{} z)} \le \frac {1}{2}\sum \frac {(b \minus{} c)^2}{(b \plus{} c)^2} \plus{} \frac {1}{2}\sum \frac {(y \minus{} z)^2}{(y \plus{} z)^2},$
it suffices to prove that
$ \sum \frac {2a}{b \plus{} c} \plus{} \sum \frac {2x}{y \plus{} z} \ge 6 \plus{} \frac {1}{2}\sum \frac {(b \minus{} c)^2}{(b \plus{} c)^2} \plus{} \frac {1}{2}\sum \frac {(y \minus{} z)^2}{(y \plus{} z)^2}.$
This is equivalent to
$ (ab \plus{} bc \plus{} ca)\left[ \sum \frac {1}{(b \plus{} c)^2}\right] \plus{} (xy \plus{} yz \plus{} zx)\left[ \sum \frac {1}{(y \plus{} z)^2}\right] \ge \frac {9}{2},$
which is true according to the Iran Inequality.

Very nice your solution, Can_hang. :lol:
My solution is as follows.
Due to homogeneity in $ a,b,c$ and in $ x,y,z$, we may assume that $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z\equal{}1$. Thus, the inequality becomes
$ \sum\frac 2{(1\minus{}a)(1\minus{}x)}\ge \frac 9{1\minus{}ax\minus{}by\minus{}cz}$,
or
$ \sum\frac 2{1\minus{}(a\plus{}x)\plus{}ax}\ge \frac 9{1\minus{}ax\minus{}by\minus{}cz}$.
Let us denote $ p\equal{}\frac {a\plus{}x}{2}$, $ q\equal{}\frac {b\plus{}y}{2}$, $ r\equal{}\frac{c\plus{}z}{2}$, where $ p\plus{}q\plus{}r$=1. Since $ ax\le p^2$, $ by\le q^2$, $ cz\le r^2$, it suffices to show that
$ \sum\frac 2{1\minus{}2p\plus{}p^2} \ge \frac 9{1\minus{}p^2\minus{}q^2\minus{}r^2}$.
This inequality can be rewritten as
$ \sum \frac 1{(p\plus{}q)^2} \ge \frac 9{4(pq\plus{}qr\plus{}rp)}$,
which is Iran Inequality.

[quote="Ji Chen"]If $x,y,z$ are positive real numbers, then
\[(yz+zx+xy)\left(\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}+\frac{1}{(x+y)^{2}}\right) \geq \frac{9}{4}. \]
CRUX Mathematicorum, 1994, No.4, p.108, Problem 1940.[/quote]
 CRUX Mathematicorum, 2005, No.5, p.335, Problem 3062:
If $x,y,z$ are positive real numbers such that $x+y+z=1$, then
\[(yz+zx+xy)\left(\frac{x}{y+y^{2}}+\frac{y}{z+z^{2}}+\frac{z}{x+x^{2}}\right) \geq \frac{3}{4}. \]

[quote=sqing]
If $x,y,z$ are positive real numbers such that $x+y+z=1$, then
\[(yz+zx+xy)\left(\frac{x}{y+y^{2}}+\frac{y}{z+z^{2}}+\frac{z}{x+x^{2}}\right) \geq \frac{3}{4}. \][/quote]
See here:
http://www.artofproblemsolving.com/community/c6h154962

[quote=manlio]For $x,y,z$ positive real numbers prove that

$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]

We can suppose that $ab+bc+ca=1$ and we'll prove that $f(a,b,c)\ge\frac{9}{4}$.
To prove that one we'll prove that $f(a,b,c)\ge f(t,t,c)$ where $t$ satisfy the condition above, it means that $t^2+2tc=1$.
Assume that $c=\text{min}${$a,b,c$}.
Easy to prove that $f(a,b,c)\ge f(t,t,c)$ so finally we only to show that $f(t,t,c)\ge\frac{9}{4}$.
Since $c=\frac{(1-t^2)}{2t}$, transform the previous inequality to: $f(t,t,c)-\frac{9}{4}=\frac{(1-t^2)(1-3t^2)^2}{4t^2(1+t^2)}\ge0$, which is true.

Nice Solution Re1gnover! However, I don't understand how f(a,b,c) $\geq$ f(t,t,c).

I tried using casework for t (t $\geq \frac{a+b}{2}$ and t $\leq \frac{a+b}{2}$), but only the first case worked out.

How would you go about proving f(a,b,c) $\geq$ f(t,t,c)?

[quote=Elcin_bayramli]it is my post"s end[/quote]

It is wrong....

[quote=Elcin_bayramli]$a^2+b^2+c^2 \ge ab+bc+ac$
$\frac{9}{2(a^2+b^2+c^2)+2(ab+bc+ac)} \ge \frac{9}{4(ab+bc+ac)}$[/quote]

It's wrong because $\sum a^2\ge\sum ab$

[quote=Elcin_bayramli]$a^2+b^2+c^2 \ge ab+bc+ac$
$\frac{9}{2(a^2+b^2+c^2)+2(ab+bc+ac)} \ge \frac{9}{4(ab+bc+ac)}$[/quote]

Reverse sign.

[quote=xzlbq][quote=manlio]For $x,y,z$ positive real numbers prove that

$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]

For $x,y,z$ positive real numbers prove that

\[(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{5}{4}\sum{{\frac { \left( x+y+z \right)  \left( y+z \right) }{2\,x \left( x+y+z \right) + \left( y+z \right) ^{2}}}} \geq \frac{9}{4}\][/quote]

${\it \sum} \left( {\frac { \left( x+y+z \right)  \left( y+z \right) }
{2\, \left( x+y+z \right) x+ \left( y+z \right) ^{2}}} \right) -\frac{9}{5}$
$=\frac{1}{20}\,{\frac {{\it \sum} \left(  \left(  \left( y+z-x \right) ^{2}
 \left( 8\,{y}^{2}+6\,xy+15\,zx \right) +yz \left( 30\,{z}^{2}+48\,yz+
5\,xy \right) +5\, \left( y-x \right) ^{2}yz \right)  \left( y-z
 \right) ^{2} \right) +16\,xyz{\it \sum} \left(  \left( 2\,x-y-z
 \right) ^{2}z \right) }{{\it \prod} \left( 2\,xy+{y}^{2}+2\,yz+2\,zx+{
z}^{2}+2\,{x}^{2} \right) }}$

[quote=manlio]For $x,y,z$ positive real numbers prove that

$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]

For $x,y,z$ positive real numbers prove that

\[\left(xy+yz+zx-{\frac {20}{3}}\,{\frac { \left( y-z \right) ^{2} \left( z-x \right) ^
{2} \left( x-y \right) ^{2}}{ \left( x+y+z \right) ^{4}}}
\right)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}\]

[quote=manlio]For $x,y,z$ positive real numbers prove that

$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]

and stronger:

For $x,y,z$ positive real numbers prove that

\[(xy+yz+zx)\sum{\frac{1}{(x+y)^2}}+15\sum{\frac{x}{x+y}}\sum{\frac{y}{x+y}} \geq 36\]

[quote=xzlbq][quote=manlio]For $x,y,z$ positive real numbers prove that

$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]

and stronger:

For $x,y,z$ positive real numbers prove that

\[M=(xy+yz+zx)\sum{\frac{1}{(x+y)^2}}+15\sum{\frac{x}{x+y}}\sum{\frac{y}{x+y}}-36\geq 0\][/quote]

\[M=\frac{1}{9}\,{\frac {8\,{\it \sum} \left(  \left( 2\,{x}^{2}-{y}^{2}-{z}^{2}
-xy-zx+2\,yz \right) ^{2}yz \right) +{\it \sum} \left(  \left( 4\,{x}^
{2}-2\,xy-2\,zx-2\,yz+{y}^{2}+{z}^{2} \right) ^{2}yz \right) }{
 \left( y+z \right) ^{2} \left( z+x \right) ^{2} \left( x+y \right) ^{
2}}}\geq 0\]

[quote=xzlbq]

and stronger:

For $x,y,z$ positive real numbers prove that

\[(xy+yz+zx)\sum{\frac{1}{(x+y)^2}}+15\sum{\frac{x}{x+y}}\sum{\frac{y}{x+y}} \geq 36\][/quote]

So nice and stronger,if have others method do it?

[quote=xzlbq]

For $x,y,z$ positive real numbers prove that

\[(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{5}{4}\sum{{\frac { \left( x+y+z \right)  \left( y+z \right) }{2\,x \left( x+y+z \right) + \left( y+z \right) ^{2}}}} \geq \frac{9}{4}\][/quote]

others method do this stronger,please?

[quote=Mildorf]What about substituting $x = \cot(A)$, etc.? This gives the equivalent

$\sum_{cyc} \left(\frac{\sin(A)\sin(B)}{\sin(C)}\right)^2 \ge \frac{9}{4}$.

for acute triangles $ABC$, which may have an easy solution I am just missing. It is easily seen from the above that Cauchy gives an elegant proof of the weaker inequality

$(xy+yz+zx)\left(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}\right) \ge 2 + \frac{2xyz}{(x+y)(y+z)(z+x)}$.[/quote]

In acute triangle,prove

\[\sum_{cyc} \left(\frac{\sin(A)\sin(B)}{\sin(C)}\right)^2 \ge \frac{9}{4}\]

[quote=xzlbq][quote=xzlbq]
In acute triangle,prove

\[\sum_{cyc} \left(\frac{\sin(A)\sin(B)}{\sin(C)}\right)^2 \ge \frac{9}{4}\][/quote]

method 3:

<=>
In any triangle,prove

\[\sum{\frac{\cos^2{\frac{B}{2}}\cos^2{\frac{C}{2}}}{\cos^2{\frac{C}{2}}}}\geq \frac{9}{4}\]

<=>

\[M={\frac { \left( x+y+z \right) yz}{ \left( y+z \right) ^{2}x}}+{\frac {
 \left( x+y+z \right) zx}{ \left( z+x \right) ^{2}y}}+{\frac { \left(
x+y+z \right) xy}{ \left( x+y \right) ^{2}z}}-\frac{9}{4}\geq 0\]

But

\[M=\frac{1}{4}\,{\frac {2\,xyz{\it \sum} \left( yz \left( yz-xy \right)
 \left( yz-zx \right)  \right) +{\it \sum} \left(  \left( 4\,{x}^{2}+7
\,xy+4\,{y}^{2} \right)  \left( x-y \right) ^{2}{z}^{5} \right) }{
 \left( y+z \right) ^{2}x \left( z+x \right) ^{2}y \left( x+y \right)
^{2}z}}\geq 0\][/quote]

用这个刻度可以将$M$取开：

\[{\frac { \left( x+y+z \right) yz}{ \left( y+z \right) ^{2}x}}+{\frac {
 \left( x+y+z \right) zx}{ \left( z+x \right) ^{2}y}}+{\frac { \left(
x+y+z \right) xy}{ \left( x+y \right) ^{2}z}}-\frac{9}{4}\geq 15\left( \frac{9}{4}-
 \left( {\frac {x}{x+y}}+{\frac {y}{y+z}}+{\frac {z}{z+x}} \right)
 \left( {\frac {y}{x+y}}+{\frac {z}{y+z}}+{\frac {x}{z+x}} \right)
 \right) \]
<=>

\[{\frac { \left( x+y+z \right) yz}{ \left( y+z \right) ^{2}x}}+{\frac {
 \left( x+y+z \right) zx}{ \left( z+x \right) ^{2}y}}+{\frac { \left(
x+y+z \right) xy}{ \left( x+y \right) ^{2}z}}+15\, \left( {\frac {x}{x
+y}}+{\frac {y}{y+z}}+{\frac {z}{z+x}} \right)  \left( {\frac {y}{x+y}
}+{\frac {z}{y+z}}+{\frac {x}{z+x}} \right) \geq 36\]

[quote=xzlbq]

用这个刻度可以将$M$取开：

\[{\frac { \left( x+y+z \right) yz}{ \left( y+z \right) ^{2}x}}+{\frac {
 \left( x+y+z \right) zx}{ \left( z+x \right) ^{2}y}}+{\frac { \left(
x+y+z \right) xy}{ \left( x+y \right) ^{2}z}}-\frac{9}{4}\geq 15\left( \frac{9}{4}-
 \left( {\frac {x}{x+y}}+{\frac {y}{y+z}}+{\frac {z}{z+x}} \right)
 \left( {\frac {y}{x+y}}+{\frac {z}{y+z}}+{\frac {x}{z+x}} \right)
 \right) \]
<=>

\[{\frac { \left( x+y+z \right) yz}{ \left( y+z \right) ^{2}x}}+{\frac {
 \left( x+y+z \right) zx}{ \left( z+x \right) ^{2}y}}+{\frac { \left(
x+y+z \right) xy}{ \left( x+y \right) ^{2}z}}+15\, \left( {\frac {x}{x
+y}}+{\frac {y}{y+z}}+{\frac {z}{z+x}} \right)  \left( {\frac {y}{x+y}
}+{\frac {z}{y+z}}+{\frac {x}{z+x}} \right) \geq 36\][/quote]
<=>
\[8\,{\it \sum} \left( yz \left( yz-xy \right)  \left( yz-zx \right)
 \right) xyz+ \left( y-z \right) ^{4}{x}^{5}+ \left( x-y \right) ^{4}{
z}^{5}+ \left( z-x \right) ^{4}{y}^{5}\geq 0
\]

[quote=manlio]For $x,y,z$ positive real numbers prove that

$$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$$[/quote]
For $x,y,z$ positive real numbers prove that
$$(xy+yz+zx)(\frac{1}{(x+\sqrt{yz})^2} + \frac{1}{(y+\sqrt{zx})^2} + \frac{1}{(z+\sqrt{xy})^2}) \geq \frac{9}{4}$$
Let $a,b,c>0$,[url=https://artofproblemsolving.com/community/c6h1590915p9860040]Prove that[/url]$$(ab+bc+ca)(\frac{1}{(a+b)^2}+\frac{1}{(a+c)^2}+\frac{16}{(b+c)^2})\geq 6$$
[url=https://artofproblemsolving.com/community/c6h585834p3467155]here[/url]
[url=https://artofproblemsolving.com/community/c6h1134212p5287568]here[/url]
[url=https://artofproblemsolving.com/community/c6h1045p3250]here[/url]
[url=https://artofproblemsolving.com/community/c6h605910p3599304]here[/url]

[quote=manlio]For $x,y,z$ positive real numbers prove that

$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]

$LHS- RHS =\frac{ 2xyz\left( x^3+y^3+z^3+3xyz-x^2(y+z)-y^2(z+x)-z^2(x+y) \right) + \sum xy\left(4(x+y)^2-xy\right)(x-y)^2  }{4(x+y)^2(y+z)^2(z+x)^2} \ge 0 $

[quote=manlio]For $x,y,z$ positive real numbers prove that

$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]

It's very easy by Buffalo Way! Without loss of generality, assume that $x=\min\{a,b,c\}$.
Let $x = a, y = a + u,z = a +v$ so $u,v \geq 0$. Then:
$$VT - VP \equiv \frac{f(a;a+u,a+v)}{4(2a+u)^2 (2a+u+v)^2 (2a+v)^2}$$
Where: $f(a;a+u;a+v) = 34a^4 (u^2 -uv+v^2) +32a^3 (2u^3 -u^2 v -uv^2 +2v^3) +8a^2 (5u^4 +2u^3 v -9u^2 v^2 +2uv^3 +5v^4) + 4a(2u^5 +5u^4 v -6u^3 v^2 -6u^2 v^3 +5uv^4 +2v^5) +uv(u-v)^2 (4u^2 +7uv+4v^2) \geq 0$
qed.

[quote=Nguyenhuyen_AG][quote=manlio]For $x,y,z$ positive real numbers prove that
$(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{9}{4}$[/quote]
In fact
\[(xy+yz+zx)\left [ \frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2} \right ] - \frac{9}{4} \leqslant \frac{x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)}{(x+y)(y+z)(z+x)},\]
for all $x,\,y,\,z$ are real numbers.[/quote]

Very nice inequality, how did you find it? And how did you find the SOS form of it: [url=https://nguyenhuyenag.wordpress.com/2017/01/06/iran-tst-1996]Iran TTS 1996 - Nguyenhuyenag[/url]?
Thank you so much, I really hope for your answer.

[quote=Nguyenhuyen_AG][quote=mudok]
$LHS-RHS=-\frac{(a-b)^2(b-c)^2(c-a)^2}{4(a+b)^2(b+c)^2(c+a)^2}\le 0$[/quote]
 :first:
[quote=SBM]
Very nice inequality, how did you find it? And how did you find the SOS form of it: [url=https://nguyenhuyenag.wordpress.com/2017/01/06/iran-tst-1996]Iran TTS 1996 - Nguyenhuyenag[/url]?
Thank you so much, I really hope for your answer.[/quote]
I use computer  :D[/quote]

OMG, impossible! I believe that you was using a very powerful algorithm and algorithm to write that software.
P/s: Em là tthnew bên VMF nè.

[quote=xzlbq]

For $x,y,z$ positive real numbers prove that

\[(xy+yz+zx)(\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2}) \geq \frac{5}{4}\sum{{\frac { \left( x+y+z \right)  \left( y+z \right) }{2\,x \left( x+y+z \right) + \left( y+z \right) ^{2}}}} \geq \frac{9}{4}\][/quote]
The right inequality easy to prove by $uvw$.
The left inequality is not so trivial, but it's true.


blogs
floor function
induction
number theory
prime numbers
number theory unsolved

prove or disprove that for every natural $ m$ , $ H(m)=2(\frac{2m+1}{2})^{[{\frac{2m+1}{(2m)!+1}[{\frac{(2m)!+1}{2m+1}}]]}}$ is a prime number.
and it also generate all prime numbers.

I have seen this here too:  http://www.primenumbersformula.com/default.htm
this link has complete information about this formula :)

Also theres another formula for prime numbers and that is Ghandi's Formula which i have posted about it long ago in my Blog : http://www.mathlinks.ro/weblog_entry.php?t=131537

Sorry, my friends, but the problem is pretty much a low-level olympiad problem (maybe suitable as an IMO problem 1 or 4) and the formula is more or less a restatement of Wilson's theorem with converse ($ \left(t-1\right)!\equiv-1\mod t$ holds if and only if $ t$ is prime or $ t=1$ or $ t=4$). I really want to know how anyone can claim proving the twin primes conjecture with the help of such a formula...

  Darij

[quote="darij grinberg"]I really want to know how anyone can claim proving the twin primes conjecture with the help of such a formula...

  Darij[/quote]
They cannot.  This formula is not useful in finding large prime numbers, it appears to just be, as you said, strange, but also an interesting formula.   :)

[quote="Amir.S"][quote="darij grinberg"]Sorry, my friends, but the problem is pretty much a low-level olympiad problem (maybe suitable as an IMO problem 1 or 4)[/quote]

can you prove it?[/quote]

Not now, since I am preparing for a seminar which starts tomorrow, but here is the main idea:

[b]1.[/b] For any positive real $ a$, we have
$ \left\lfloor\frac{\left\lfloor a\right\rfloor}{a}\right\rfloor=\left\{\begin{array}{c}1,\text{ if }a\text{ is integer;}\\ 0,\text{ if not}\end{array}\right.$.

[b]2.[/b] For any positive integer $ m$, the number $ \frac{\left(2m\right)!+1}{2m+1}$ is integer if and only if $ 2m+1$ is prime.

  Darij

A similar formula was given by [i]W. Sierpinski[/i]. For completeness, I post the reasoning behind it:

Denote $ p_{n}$ the sequence of prime numbers. $ p_{1}=2$.
[b]1.[/b] $ p_{n}\leq2^{2^{n-1}}$. [b][i]Proof:[/i][/b] This is proved easily by induction using $ p_{n+1}\leq p_{1}p_{2}\dots p_{n}+1$.

[b]2.[/b] $ \sum_{k=1}^{\infty}\frac{p_{k}}{10^{2^{k}}}$ converges. [b][i]Proof:[/i][/b] Follows from 1.

So, denote $ a=\sum_{k=1}^{\infty}\frac{p_{k}}{10^{2^{k}}}$.

Then $ p_{n}=\left[10^{2^{n}}\cdot a\right]-10^{2^{n-1}}\left[10^{2^{n-1}}\cdot a\right]$ ([i]W. Sierpinski[/i]).
The proof is simple and straightforward.

I guess similar formulas can be found for various sequences for which statements like [b]1[/b] and [b]2[/b] hold.
Of course the formula isn't quite useful, yet it's interesting and beautiful.


There are $ 4$ ways to choose a bowl. From there, you can also choose any of the $ 4$ glasses to go with each bowl. So there are $ \boxed{16}$ possibilities.

When you do stuff like this, you always multiply.

[ quote = " Ernie "] ALS JE Dingen te DOEN ALS zegt , je Altijd te vermenigvuldigen . [/ Quote]

Waarom niet 4! , die niet kan ?

german people multiply when they do it as well.

[quote="isabella2296"]To translate, SCP asked why you can't do 4!. (It was Dutch.)[/quote]

Yes, it was Dutch-translated (but bad) and asked why the answer was $4*4$ and not $4!$, but how I read it, it looks like for each thing we have $4$ chooses, and than I get $4^4$, no?
So, I didn't understand and ask how we get the answer?

I hope this isn't translated automatically or so.

[quote="GameBot"]Chandra has four bowls.  Each one is a different color (red, blue, yellow, green).  She also has exactly one glass the same color as each bowl.  If she chooses a bowl and a glass from the cupboard, how many pairings are possible?  One such pairing is a blue bowl and a yellow glass.[/quote]
I made this one because 2 people couldn't figure this out...i was spectating the game...SMH!