inequalities
quadratics
algebra unsolved
algebra

Let $n$ be a positive integer greater than 1. Determine all the collections of real numbers  $x_1,\ x_2,\dots,\ x_n\geq1\mbox{ and }x_{n+1}\leq0$ such that the next two conditions hold:

(i) $x_1^{\frac12}+x_2^{\frac32}+\cdots+x_n^{n-\frac12}= nx_{n+1}^\frac12$

(ii) $\frac{x_1+x_2+\cdots+x_n}{n}=x_{n+1}$

$x_{n+1}$ must be greater than $0$ because all other terms are positive.

Solution:
First of all we have that:
$\sqrt[k] {x_i} \ge \sqrt[l] {x_i} \iff k > l$ equality holds $\iff x_i =1$
$\implies x_i \ge \sqrt{x_i}$
It implies also, that:
$\sqrt{x_{n+1}} = \frac{\sum_{i=1}^{n} \sqrt{x_{i} ^{2i-1}}}{n} \ge \frac{\sum_{i=1}^{n} x_i }{n} = x_{n+1}$
$\implies x_{n+1} = \sqrt{x_{n+1}} = 1$
Now we have that:
$1=\frac{\sum_{i=1}^{n} x_i}{n}\ge \frac{\sum_{i=1}^{n} 1}{n} = 1$
$\implies x_i = 1$ for $1\le i \le n+1$
And it is the only solution.

This post is really old, i didn't remeber it.
For the first question it's just a typo.. The second inequality should be in the other way.

Now for the second question i can't answer, there's a mistake in the way i'm reasoning. Because the problem is that $x^{a}\ge x^{b}$ if $a > b$. So $x_{i}^{\frac{2i-1}{2}}\ge x_{i}\forall i\ge 2$ but  i did not consider the case $x_{1}^{\frac{1}{2}}\le x_{1}$ so there's a mistake. Sorry.

I wrote this post when i was learning to use Tex so there are lots of mistakes.

OK so this does not look at all like the problem that actually came in the contest. The real problem is:

Let $ n > 1$. Find all solutions for $ x_{1},x_{2},...,x_{n \plus{} 1}\geq 1$ such that: $ (1)$ $ x_{1}^{\frac {1}{2}} \plus{} x_{2}^{\frac {1}{3}} \plus{} ... \plus{} x_{n}^{\frac {1}{n \plus{} 1}} \equal{} nx_{n \plus{} 1}^{\frac {1}{2}}$; and $ (2)$ $ \frac {x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n}}{n} \equal{} x_{n \plus{} 1}$

Anyways, the answer to this problem is the following

We want to find all numbers that meet $ x_{1}^{\frac {1}{2}} \plus{} ... \plus{} x_{n}^{\frac {1}{n \plus{} 1}} \equal{} n\sqrt {\frac {x_{1} \plus{} ... \plus{} x_{n}}{n}}$
By the Quadratic-Arithmetic mean, we have that
\[ RHS \equal{} n\sqrt {\frac {(x_{1}^{1/2})^{2} \plus{} (x_{2}^{1/2})^{2} \plus{} ... \plus{} (x_{n}^{1/2})^{2}}{n}}\geq n\left[\frac {x_{1}^{1/2} \plus{} x_{2}^{1/2} \plus{} ... \plus{} x_{n}^{1/2}}{n}\right] \equal{} x_{1}^{1/2} \plus{} x_{2}^{1/2} \plus{} ... \plus{} x_{n}^{1/2}\]
Since $ x_{1},x_{2},...,x_{n}\geq 1$, we have that $ x_{1}^{1/2} \equal{} x_{1}^{1/2}$, $ x_{2}^{1/2}\geq x_{2}^{1/3}$, $ x_{3}^{1/2}\geq x_{3}^{1/4}$, ..., $ x_{n}^{1/2}\geq x_{n}^{1/n\plus{}1}$ with equalities if $ x_{2},x_{3},...,x_{n} \equal{} 1$. Therefore, for the equality $ x_{1}^{\frac {1}{2}} \plus{} ... \plus{} x_{n}^{\frac {1}{n \plus{} 1}} \equal{} n\sqrt {\frac {x_{1} \plus{} ... \plus{} x_{n}}{n}}$ to hold, we need $ x_{2},x_{3},...,x_{n} \equal{} 1$. Also, since we are counting on the quadratc mean holding equality, we also need $ x_{1} \equal{} 1$. Taking this into account, we see that $ x_{n \plus{} 1} \equal{} 1$. Then $ x_{1},...,x_{n \plus{} 1} \equal{} 1$ is the only possible solution.

It would be nice if people who writes the problem, at least they know the problem. I was trying to recopilate problems for a training and now I don´t know the exact problem. Thank you


Fie $A_1 A_2 \ldots A_n$ un poligon convex. Demonstrati ca exista o dreapta in planul poligonului care il imparte in doua parti de arii si perimetre egale.

$n\equiv \ 0 \ (mod \ 2) \ \Rightarrow  A_1A_{\frac{n}{2} +1}$  e dreapta cautata
$n\equiv \ 1 \ (mod \ 2) \ \Rightarrow  M$ mijlocul lui $A_{\frac{n+1}{2}}A_{\frac{n+1}{2} +1}$ si $A_1M$ e dreapta cautata

mi-am dat si eu cu parerea...nu stiu daca e bine

Pentru "spx2": poligonul nu este neaparat regulat!

consideram doua poligoane P is P'
unde P e poligonul dat si P' un poligon intermediar de-al nostru.
laturile poligonului dat P sunt $[l1,...,l_n]$ (ordonate crescator).
P' il facem cu latura cat $l1$ si P' este acum regulat(dupa urmatorul pas nu va mai fi).
dupa care determinam dreapta cum am spus mai sus.
apoi luam urmatoarea latura $l_2$ inlocuim una corespunzatoare din P'
cu aceasta.
cu siguranta noua dreapta va fi determinata de lungimea acelui segment
care a fost inlocuit cu $l_2$.
pasii se repeta pana cand poligonul P' devine P.
(de mentionat este ca laturile inlocuite vor fi inlocuite pe pozitiile pe care
 le au in P)

despre modificarea pozitiei dreptei.
daca dreapta imparte poligonul P' intr-o bucata superioara si una inferioara.
si noul segment inlocuit e in partea superioara atunci partea superioara se va
mari si atunci ori
1) una din extremitatile segmentului de dreapta(din poligon) va trebui sa se mute
putin mai sus pentru a compensa deficitul partii inferioare
2)ori se muta toata dreapta mai sus putin pentru a compensa deficitul partii inferioare

daca noul segment inlocuit e in partea inferioara atunci se intampla invers decat
mentionat mai sus,si tot asa.

oarecum daca dreapta imparte poligonul in 2 parti cu arii egale,aria fiind dependenta
de perimetru,automat se modifica si perimetrele.
si aria cu perimetrul sunt direct proportionale intr-un poligon convex asa ca intuitiv
aceasta separare a poligonului e posibila

[quote="spx2"]si aria cu perimetrul sunt direct proportionale intr-un poligon convex[/quote]

????

Un hint pe care aş putea să ţi-l dau e că problema asta e de analiză.

pai intr-un poligon convex daca se mareste perimetru se mareste si aria nu ?
reciproc daca se mareste aria,se mareste si perimetru nu ?
"solutia" data mai sus e "algoritmica"
(oare ce-am scris are vreo relevanta,sau e doar pur empiric ?)
cu analiza nu prea stiu cum as putea-o aborda.
nu prea ma pricep la analiza,si nu prea stiu cum as putea s-o aplic aici
n-am experienta prea mare in probleme de genul asta

am facut ceva modificari la "algoritm",sper sa fie mai clar
sunt mai mult decat curios cum se rezolva cu analiza!
e o problema frumoasa.dar grea!
eu nu stiu sa o rezolv
("algoritmul" ar putea fi complet daca ar exista o caracterizare
concreta a modificarii pozitiei dreptei in functie de lungimea fiecarui
nou segment inlocuit in P',si astfel el ar da o solutie problemei)

fie $\mathcal{P}$ poligonul si $f: \mathcal{P}\rightarrow\mathbb{R}$ o functie pe care o definim asa :

fie $M$ punct al poligonului. exista $N\in\mathcal{P}$ si e unic, astfel incat dreapta $MN$ imparte poligonul in doua poligoane de acelasi perimetru. acum consideram ca $M$ este pe latura $A_{k}A_{k+1}$ a poligonului (indicii modulo $n$). precizare : $A_{i}$ se afla pe latura $A_{i}A_{i+1}$. atunci $f(M)$ va lua valoarea ariei unui poligon din cele doua poligoane mai mici determinate de dreapta $MN$. precis, acest poligon va fi cel ce il contine pe $A_{k+1}$.
functia $f$ e continua
acuma, fie $B$ "omologul" lui $A_{1}$.
$f(A_{1})$ o sa fie aria poligonului $A_{1}A_{2}...A_{k}B$, iar $f(B)$ o sa fie aria poligonului $BA_{k+1}A_{k+2}...A_{n}A_{1}$. Deoarece poligonul este convex, avem $f(A_{1})+f(B)=S$, unde $S$ e aria poligonului.
daca $f(A_{1})=S/2$, am terminat
daca $f(A_{1}>S/2$, atunci $f(B)<S/2$ si $f$ are darboux, deci exista un punct $M$ pentru care $f(M)=S/2$, analog daca $f(A_{1})<S/2$.

e bine ce zic eu aici ? e cam tarziu si tare imi e frica ca am zis numa prostii  :blush:

Maky , m-ar ajuta mult sa inteleg demonstratia daca
ai raspunde la urmatoarele intrebari:
1) daca e posibil sa definesti notinunea de "omolog"
2)remarca "precis acest poligon va fi cel care-l contine pe $A_{k+1}$" mie
nu mi se pare asa evidenta
3)niste desene ar fi foarte folositoare
4)cand spui $\mathcal{P}$ cred ca te referi la multimea punctelor
   de pe laturile poligonului, e adevarat ?

$\mathcal{P}$ reprezinta, intr`adevar, multimea punctelor de pe laturile poligonului (inclusiv varfurile)
fie $M\in\mathcal{P}$.
numim "omologul" lui $M$, singurul punct $N\in\mathcal{P}$, ales astfel incat dreapta $MN$ imparte poligonul in doua poligoane de perimetre egale
sa presupunem ca $M$ se afla pe latura $A_{k}A_{k+1}$ si $N$ pe latura $A_{l}A_{l+1}$. prin conventie, punctul $A_{i}$ se afla pe latura $A_{i}A_{i+1}$. totii indicii sunt modulo $n$.
atunci, lui $f(M)$ ii dam valoarea ariei poligonlui $MA_{k+1}A_{k+2}...A_{l}N$.
aceasta definitie a lui $f$ este pentru orice punct $M$.
in particular, o vom aplica pentru punctele $A_{1}$ si $B$, unde $B$ este omologul lui $A_{1}$.
presupunem $B$ pe latura $A_{r}A_{r+1}$.
$f(A_{1})$ este aria lui $A_{1}A_{2}...A_{r}B$.
$f(B)$ este aria lui $BA_{r+1}A_{r+2}...A_{1}$.
Deci obtinem imediat $f(A_{1})+f(B)=S$, unde $S$ e aria poligonului.
de aici e simplu, sau sunt amandoua $S/2$ si $A_{1}B$ este dreapta dorita, sau $f$ ia o valoare mai mica decat $S/2$ si o valoare mai mare decat $S/2$, si din continuitate, va lua si valoarea $S/2$ intr`un punct.


algebra
polynomial
function
topology
complex analysis
complex analysis unsolved

If $ f$ is analytic on a simply connected open set $ \Omega$ then for any $ \epsilon > 0$ there exists a polynomial $ g$ such that $ |f(z) \minus{} g(z)| < \epsilon$ for all $ z\in \Omega$.

That's not quite right: consider $ f(z)\equal{}\frac{1}{1\minus{}z}$ on $ \Omega\equal{}\{z\colon |z|<1\}$.

[quote="mlok"]That's not quite right: consider $ f(z) \equal{} \frac {1}{1 \minus{} z}$ on $ \Omega \equal{} \{z\colon |z| < 1\}$.[/quote]
Ah, let me see if I got it right. Uniformly convergent functions that are uniformly continous converge to a uniformly continous function. Since $ f(z)$ is not uniformly continous on $ \Omega$ then the convergence can't be uniform since $ g_n(z)$ are.

Right. Or you could just notice that the limit must be bounded.
This is why the claim should be stated as: for any $ \epsilon > 0$ and any compact set $ K\subset\Omega$ there exists a polynomial $ g$ such that $ |f(z) \minus{} g(z)| < \epsilon$ for all $ z\in K$.

[quote="mlok"]This is why the claim should be stated as: for any $ \epsilon > 0$ and any compact set $ K\subset\Omega$ there exists a polynomial $ g$ such that $ |f(z) \minus{} g(z)| < \epsilon$ for all $ z\in K$.[/quote]
So this is Weierstrass theorem analouge in complex analysis?

A somewhat weak form of it. The ultimate form is [url=http://en.wikipedia.org/wiki/Mergelyan%27s_theorem]Mergelyan's theorem[/url].


Can you find the next two terms in each of the following sequences?

THESE ARE PRETTY CHALLENGING

A)
2, 12, 1112, 3112, 132112, ..., ...

B) 1, 1, 1, 2, 3, 7, 23, ..., ...

C) 1, 1, 1, 2, 3, 5, 11, ..., ...

$ \text{HINT:}$ patterns B and C are very similar

[quote="TheBB"][hide]1:

2, 12, 1112, 3112, 132112, ..., ...

Speak each number out loud:

"Two"
"One two"
"One one, one two"
"Three ones, one two"
"One three, two ones, one two"

etc....

2:

Each number is the product of the two preceding ones plus the one preceding those.

3:

Each number is the preceding number plus the product of the two preceding that again.[/hide][/quote]

Oh. Haha.  Those were good ones.  :clap2:


What is the value of $ x$ which satisfies $ \sqrt[3]{x\sqrt{x}}\equal{}7$?

hello,$ \sqrt[3]{x\sqrt{x}}\equal{}\sqrt[3]{x^{\frac{3}{2}}}\equal{}\left(x^{\frac{3}{2}}\right)^{\frac{1}{3}}\equal{}x^{\frac{1}{2}}\equal{}7$,so we have $ x\equal{}7^2\equal{}49$.
Sonnhard.


How many terms does the following sequence have?

\[ 15, 15\plus{}3, 15\plus{}6, 15\plus{}9, \, \ldots \, , 15 \plus{} 39\]

This is just 15+0x3,15+1x3...15+13x3. SO it's the number iof terms from 0-13, or 14.


inequalities
trigonometry
inequalities proposed

[color=darkred]Prove that $ \{x,y,z\}\subset (0,\infty )\implies$ $ \sum\sqrt {\frac {y+z}{x}}\ge 2\cdot\sum\sqrt {\frac {x}{y+z}}\ .$

[b][u]Remark.[/u][/b] A geometrical interpretation : ${ \boxed {\sum\sqrt {\frac {a}{p-a}}\ge 2\cdot\sum\sqrt {\frac {p-a}{a}}}\ge 6\sqrt {\frac rR}}\ .$[/color]

Consider the substitutions $ \sqrt {\frac {x}{y \plus{} z}} \equal{} a$, $ \sqrt {\frac {x}{y \plus{} z}} \equal{} b$, $ \sqrt {\frac {z}{x \plus{} y}}\equal{}c$, respectively. The inequality rewrites as
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \geq 2\left(a \plus{} b \plus{} c\right),
\]
where $ a$, $ b$, $ c$ satisfy $ a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2} \plus{} 2a^{2}b^{2}c^{2} \equal{} 1$.

Now, let $ bc \equal{} \cos{A}$, $ ca \equal{} \cos{B}$, $ ab \equal{} \cos{C}$. The inequality rewrites as
\[ \cos{A} \plus{} \cos{B} \plus{} \cos{C}\geq 2\left(\cos{A}\cos{B} \plus{} \cos{B}\cos{C} \plus{} \cos{C}\cos{A}\right),
\]
which is true since
\[ (\cos{A} \plus{} \cos{B} \plus{} \cos{C})^{2} \geq 3\left(\cos{A}\cos{B} \plus{} \cos{B}\cos{C} \plus{} \cos{C}\cos{A}\right),
\]

\[ \cos{A} \plus{} \cos{B} \plus{} \cos{C} \leq \frac {3}{2}.
\]
[hide="Open problem"]
Let $ a$, $ b$, $ c$ positive real numbers. Then the following inequality holds
\[ \sum_{cyc}{\sqrt {\frac {b \plus{} c}{a}}} \geq 2\left(\sum_{cyc}{\sqrt {\frac {a}{b \plus{} c}}}\right) \cdot \sqrt {1 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 8abc}{4\sum{a(a \plus{} b)(a \plus{} c)}}}.
\]
You are probably wondering what's with the term $ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 8abc}{4\sum{a(a \plus{} b)(a \plus{} c)}}$ under the radical. Well, it can be explained, since an inequality between
\[ \displaystyle\frac {\sum{\sqrt {\frac {b \plus{} c}{a}}}}{2\sum{\sqrt {\frac {a}{b \plus{} c}}}}?\frac {\sqrt {\sum{\frac {b \plus{} c}{a}}}}{2\sqrt {\sum{\frac {a}{b \plus{} c}}}}
\]
isn't quite possible and because of the following older inequality of mine, which seems to hold :D :
\[ \sum_{cyc}{\frac {b \plus{} c}{a}} \geq 4\sum_{cyc}{\frac {a}{b \plus{} c}} \plus{} 1 \minus{} \frac {8abc}{(b \plus{} c)(c \plus{} a)(a \plus{} b)}.
\]
[/hide]

[quote="manlio"]I cannot find a direct algebraic proof for this nice inequality?  :(

Has anyone help me?

Thank you very much[/quote]
Here is my solution. By AM-GM and Cauchy Schwarz inequality we have
\[ \left( {\sum {\sqrt {\frac{{a \plus{} b}}{c}} } } \right)^2  \ge 8\sum {\frac{a}{{b \plus{} c}}}  \plus{} 6
\]
\[ 4\sum {\sqrt {\frac {{ab}}{{\left( {a \plus{} c} \right)\left( {b \plus{} c} \right)}}} } \le 2\sum {\frac {{a \plus{} b}}{{\sqrt {\left( {a \plus{} c} \right)\left( {b \plus{} c} \right)} }}} \le
\]
\[ \le \sum {\left( {\frac {{a \plus{} b}}{{a \plus{} c}} \plus{} \frac {{a \plus{} b}}{{b \plus{} c}}} \right)} \equal{} \sum {\frac {{a \plus{} 2b \plus{} c}}{{a \plus{} c}}} \equal{} 3 \plus{} \sum {\frac {a}{{b \plus{} c}}}
\]
From these above inequality we can get the original, :).

[quote="manlio"]I cannot find a direct algebraic proof for this nice inequality?  :(

Has anyone help me?

Thank you very much[/quote]

Setting $ x\equal{}a,y\equal{}b,z\equal{}c$ we have

$ LHS\minus{}RSH\equal{}\sum(\sqrt{\frac{b\plus{}c}a}\minus{}2\sqrt{\frac a{b\plus{}c}})$
$ \equal{}\sum\frac{(b\minus{}a)\plus{}(c\minus{}a)}{\sqrt{a(b\plus{}c)}}\equal{}\sum (b\minus{}c)[\frac1{\sqrt{c(a\plus{}b)}}\minus{}\frac1{\sqrt{b(c\plus{}a)}}]$
$ \equal{}\sum\frac{a(b\minus{}c)^2}{\sqrt{bc(a\plus{}b)(a\plus{}c)}[\sqrt{b(c\plus{}a)}\plus{}\sqrt{c(a\plus{}b)}]}\ge0$

Thank you very much dduclam and Honey_S for your nice proofs.  :)

It's an old problem.
Square, we have: $ \sum\ \frac {a \plus{} b}{c} \plus{} 2\sum\ \sqrt []{\frac {(a \plus{} b)(a \plus{} c)}{bc}} \geq\ 4\sum\ \frac {a}{b \plus{} c} \plus{} 4\sum\ \sqrt []{\frac {ab}{(a \plus{} b)(a \plus{} c)}}$
Obviously trues because: $ \sum\ \frac {a \plus{} b}{c} \geq\ 4\sum\ \frac {a}{b \plus{} c}$
$ \sum\ \sqrt []{\frac {(a \plus{} b)(a \plus{} c)}{bc}} \geq\ 6$
$ \sum\ \sqrt []{\frac {ab}{(a \plus{} b)(a \plus{} c)}} \leq\ \frac {3}{2}$
 :)


Find the [i]least[/i] possible value of $ a \plus{} b$, where $ a, b$ are positive integers such that $ 11$ divides $ a \plus{} 13b$ and $ 13$ divides $ a \plus{} 11b$.

[quote="mszew"][quote="thomaskurian89"]Find the [i]least[/i] pssible value of $ a \plus{} b$, where $ a, b$ are positive integers such that $ 11$ divides $ a \plus{} 13b$ and $ 13$ divides $ a \plus{} 11b$.[/quote]

$ 11| a \plus{} 13b \iff 11|a \plus{} 2b$ and $ 13| a \plus{} 11b \iff 13|a \minus{} 2b$
Notice that $ a \equal{} 23$ and $ b \equal{} 5$ is a solution then $ a \plus{} b \le 28$
Lets prove that it is the minimum $ a \plus{} b$

If $ b \equal{} 1$ then from $ a \plus{} 2b\equiv 0(11)$ we get $ a \equiv 9 (11)$ however neither $ 9$ nor $ 20$ fullfills the other equation
If $ b \equal{} 2$ then from $ a \plus{} 2b\equiv 0(11)$ we get $ a \equiv 7 (11)$ however neither $ 7$ nor $ 18$ fullfills the other equation
If $ b \equal{} 3$ then from $ a \plus{} 2b\equiv 0(11)$ we get $ a \equiv 5 (11)$ however neither $ 5$ nor $ 16$ fullfills the other equation
If $ b \equal{} 4$ then from $ a \plus{} 2b\equiv 0(11)$ we get $ a \equiv 3 (11)$ however neither $ 3$ nor $ 14$ fullfills the other equation
If $ b \equal{} 5$ then from $ a \plus{} 2b\equiv 0(11)$ we get $ a \equiv 1 (11)$ however neither $ 1$ nor $ 12$ fullfills the other equation
idem for the other $ b\le 28$

Is it true that instead of $ 11$ and $ 13$ we have $ 2n \minus{} 1$ and $ 2n \plus{} 1$ and the $ a$ for a minimum $ a \plus{} b$ is $ 4n \minus{} 1$ ?[/quote]

Thanks.  :)

But is there an easier way than case-bashing b from 1 to 28?

Eg., $ 143|(a \plus{} 2b)(a \minus{} 2b)$
$ 143|a^4 \minus{} 4b^4$

Can we now use Pell equations?


IMO Shortlist
rectangle
matrix
combinatorics
game
invariant

A solitaire game is played on an $m\times n$ rectangular board, using $mn$ markers which are white on one side and black on the other. Initially, each  square of the board contains a marker with its white side up, except for one corner square, which contains a marker with its black side up. In each move, one may take away one marker with its black side up, but must then turn over  all markers which are in squares having an edge in common with the square of the removed marker. Determine all pairs $(m,n)$ of positive integers such that all markers can be removed from the board.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

All markers can be removed if and only if at least one of m, n is odd. Suppose WLOG m is odd and all boards of area less than mn with one of m, n odd satisfy the property. Then we can solve the m*(n-1) board, at which point we have a row of m black markers. We can remove markers 1, 3, 5, ..., m, and then 2, 4, 6, ..., m-1, which are all black now.

Suppose that m, n are even. The perimeter of the entire initial rectangle is 2(m+n). In particular, this is 0 mod 4. When a piece is removed, suppose it shares perimeter k with the entire figure. Then when it is removed, k is lost from the entire perimeter and 4-k is gained, for a net gain of 4-2k, or 2k mod 4. Now all but the original black marker k corresponds to the number of adjacent markers removed (always odd since the removed marker must be black) plus the perimeter the marker in question shares with the original rectangle (odd for edges, even for corners and centers). Hence the perimeter changes by 2 mod 4 each time a center piece or corner (other than the original) is removed. Thus the perimeter changes (n-2)(m-2)+3 times from 2(n+m). This would indicate that the final perimeter is 2 mod 4, but we know it should be 0. Contradiction.

Here is another proof that it doesn't work if [i]m, n[/i] are even:

Suppose that you can remove all [i]mn[/i] counters.  Note that each edge between two counters that have not been removed corresponds to exactly one flip of a counter (as which ever you remove first flips the other).  There are initially [i]2mn - m - n[/i] edges between counters and [i]mn - 1[/i] white counters. Each white counter must be flipped to black at some point so this uses [i]mn - 1[/i] of the [i]2mn - m - n[/i] flips, leaving [i](m-1)(n-1)[/i] spare. The spare must be an even number or else there will be some counter that uses an odd number of the spare flips so will be white, which is not allowed.
Hence [i](m-1)(n-1)[/i] is even so either m or n is odd

[quote=Kayak]What's the motivation between the invariant (as given in Pranav Sriram's book) $$\{ \text{Total number of White markers + Total number of sides common to two markers} \} \mod 2 $$? I find the only sensible way to guess the invariant so try randomly a hell lot of things until it randomly pops up in your brain - is there a better way ? You can guess the answer that it's possible if one of $m,n$ is odd by testing small cases easily, but how you come up with the invariant ?[/quote]

maybe, when you want an example for m,n even, there is not, so you need compare the diferences with the examples the m, n that if they have a solution.
I didn't note this invariance :c, but i like it.

[color=#00f][b]I am presenting a proper and a nice solution to the following problem [/b]-[/color]-

1.) [color=#f00]Graph Theory Conversion - [/color]
Make markers as vertices and color the vertices black or white depending on the top colour of the marker. Join 2 vertices by an edge, if the corresponding markers are placed in adjacent squares. Also note that when we remove any vertex it's incident edges are removed and its neighboring vertices change their color from black to white or vice versa. Now suppose in a move , suppose a vertex $ v$ has degree $d(v)$ ,  and $w$ of the neighboring vertices are white and $b$ of the vertices are black. Note that after removing $v$ no of white vertices changes from $w$ to $d(v)$ - $w$ and change in number of white vertices = $d(v)$-$2w$.

2.) [color=#f00]Finding an Invariant -  [/color]
Now let at any instant $\alpha$ denotes number of white vertices and $\beta$ denotes total number of edges left. Now note that when $\beta$ decreases by $d(v)$ ; $\alpha$ changes by $d(v)$- $w$  , then $\alpha$+$\beta$ decreases by $2w$ , 	$\Rightarrow$  $\alpha$ + $\beta$ $mod 2$ is an [b][color=#00f]invariant[/color][/b] . At end $\alpha$+$\beta$ = 0 $\Rightarrow$ $\alpha$+$\beta$ is always even. At beginning  $\alpha$+$\beta$ = $3mn-m-n-1$ $\equiv$  $mn-m-n-1$ = $(m-1)(n-1)$ $mod 2$
$\Rightarrow$[color=#f00] atleast one of m and n must be odd.[/color]               	$\blacksquare$

[i]A solution without invariance:[/i]
[rule]
The answer is all markers can be removed iff at least one of $(m,n)$ is odd.

First we give a construction. Wlog, $m$ is odd. Below we show the procedure for case $(m,n) = (5,4)$, which easily generalizes (a $\bullet$ means a marker is there but cannot be removed ; a $\color{blue}{\bullet}$ means a marker is there and can also be removed $\color{red}{\times}$ means no marker is there):
\begin{align*}
&\begin{matrix}
\bullet & \bullet & \bullet & \bullet & \bullet \\
\bullet & \bullet & \bullet & \bullet & \bullet \\
\bullet & \bullet & \bullet & \bullet & \bullet \\
\color{blue}{\bullet} & \bullet & \bullet & \bullet & \bullet
\end{matrix} \qquad \longrightarrow \qquad \begin{matrix}
\bullet & \bullet & \bullet & \bullet & \bullet \\
\bullet & \bullet & \bullet & \bullet & \bullet \\
\color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times}
\end{matrix} \qquad \longrightarrow \qquad
\begin{matrix}
\bullet & \bullet & \bullet & \bullet & \bullet \\
\color{blue}{\bullet} & \bullet & \color{blue}{\bullet} & \bullet & \color{blue}{\bullet} \\
\color{red}{\times} & \color{blue}{\bullet} & \color{red}{\times} & \color{blue}{\bullet} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times}
\end{matrix} \\
\longrightarrow \qquad & \begin{matrix}
\bullet & \bullet & \bullet & \bullet & \bullet \\
\color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times}
\end{matrix} \qquad \longrightarrow \qquad \begin{matrix}
\color{blue}{\bullet} & \bullet & \color{blue}{\bullet} & \bullet & \color{blue}{\bullet} \\
\color{red}{\times} & \color{blue}{\bullet} & \color{red}{\times} & \color{blue}{\bullet} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times}
\end{matrix} \qquad \longrightarrow \qquad \begin{matrix}
\color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times}
\end{matrix} \\
\longrightarrow \qquad & \begin{matrix}
\color{red}{\times} & \color{blue}{\bullet} & \color{red}{\times} & \color{blue}{\bullet} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times}
\end{matrix} \qquad \longrightarrow \qquad \begin{matrix}
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} \\
\color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times} & \color{red}{\times}
\end{matrix}
\end{align*}
Now FTSOC assume that both of $m,n$ were even and we can still remove all markers.
We consider a directed graph where vertices correspond to the cells of the board. For two adjacent cells $C_1,C_2$, we draw a directed edge from $C_1 \to C_2$ if $C_1$ was removed before $C_2$. Note that other than one corner cell, all other cells must have an odd number of edges directed towards them. This gives
$$\# ~ \text{edges} \equiv mn-1 \pmod{2}$$
But note that
$$\# ~ \text{edges} = m(n-1) + n(m-1) = 2mn - m - n$$
Thus we obtain
$$2 \mid (2m - m - n) - (mn-1) = (m-1)(n-1)$$
As both of $m-1,n-1$ are odd, we obtain our desired contradiction. $\blacksquare$

We claim that the answer is all $m$ and $n$ such that $m$ and $n$ are not both even. Let $p$ be the number of pairs of adjacent markers and $w$ be the number of white markers. Note that in each move, if a total of $x$ markers are taken out of the board, the number of markers toggled is $x$ and $p$ decreases by $x$. Note that for each marker that is changed, $w$ either increases or decreases by $1$, so $w$ is changed by $w\pmod 2$. In the end, we want $w=p=0$ so in the beginnining we need $w\equiv p\pmod 2$. Indeed, $m(n-1)+n(m-1)\equiv mn-1\pmod 2$. If $m$ and $n$ are both even, we have a contradiction. If $m$ or $n$ is odd, this is not a contradiction.

Now, we'll show that it works as long as one of them is odd. We'll show this by induction on $n$, where $m$ is odd. Then, remove the entire $m\times n-1$ board first. We are left with a single row on the edge of the board, some black and some white. Since each of the markers on the row directly adjacent was marked exactly once, and each removal affects the square of the row that is left that is directly adjacent to it only, all of our markers are currently black. Clearly, we can remove everything.


function
algebra
domain
modular arithmetic
real analysis
real analysis unsolved

Let $f$ be a periodic function on $\mathbb{R}$,meaning $\exists d\in\mathbb{R}$ s.t $f(x+d)=f(x)$ $\forall x\in\mathbb{R}$.  Show that if $f$ is continuous, then $f$ attains its maximum and minimum on $\mathbb{R}$.

My proof for this problem seems a bit too straightforward.  Am I skipping over some key nuance?

Since $\exists d\in\mathbb{R}$ s.t $f(x+d)=f(x)$ where $f$ is a continuous, then $f(x)$ exists $\forall x\in\mathbb{R}$.  So, the interval $[x,x+d]$ is closed and bounded.  By the Heine-Borel Theorem, $[x,x+d]$ is therefore compact.

$\therefore$ by the Extreme Value Theorem, since $[x,x+d]$ is compact and $f$ is continuous, then $\exists a,b\in [x,x+d]$ s.t $f(a)\leq f(c) \leq f(b)$ $\forall x\in [x,x+d]$.  So, $f(a)$ is the minimum and $f(b)$ is the maximum.  QED.

You're not really missing anything here, but just to make sure you have things under control, can you also prove that $f$ is uniformly continuous on $\mathbb{R}?$

Ironically, this problem was listed in my textbook just before the section on uniform continuity.  I'm presuming this problem is designed to shed light on what it means for a function to be uniformly continuous?  I'll need to read the next section, I guess.

The Extreme Value Theorem is essentially a theorem about continuous functions with a compact domain; it's a compactness theorem. It's another big theorem that continuous functions on a compact set are uniformly continuous. The grouping of these topics in adjacent sections of your textbook is quite logical. I would also assume as likely that your text hasn't breathed the word "compact" yet. That's the way I'd handle it - get a collection of compactness-related results like Bolzano-Weierstrass, extreme value, and uniformly continuous first, and only then seek to unite them with the concept of compactness.

Of course $\mathbb{R}$ isn't compact - but then this isn't just any continuous function on $\mathbb{R},$ it's a periodic function. A useful attitude would be to think of $f$ as a function with domain $\mathbb{T}_{d},$ where $\mathbb{T}$ is a circle of circumference $d.$ [hide="Technical language"]It's the quotient group $\mathbb{R}/d\mathbb{Z},$ whose elements are the equivalence classes of the relation $x\equiv y\pmod{d},$ endowed with the appropriate quotient topology.[/hide]


Sound travels at the rate of 1130 feet per second. How far away is a lightning strike if the sound of thunder reaches you 7 seconds after you see the lightning flash? Assume the speed of light is infinite. Round your answer to the nearest tenth of a mile.

What are they saying? :?:

To find how far the lightning is, multiply the speed of sound by the amount of time.

$ 1130*7\equal{}7910$

Now convert to miles-

$ 7910/5280$

This is approximately $ 1.5$ miles.

yes, you are right. I did not know how to do it, but I learned! from you! :lol:

The general rule for that stuff is usually divide the amount of time by 5 and put it in miles.


function
integration
calculus
calculus computations

$ f(x)$ is a $ Real$ $ valued$ function defined by:
$ f(x) \equal{} x^2 \plus{} x^2 \int^1_{ \minus{} 1} t \cdot f(t)dt \plus{} x^3 \int^1_{ \minus{} 1} f(t)dt$
Then,
$ \int^1_{ \minus{} 1} f(x)dx \equal{} ?$

For every $ x\in\mathbb{R}$ holds:

$ f(x) \equal{} x^2 \plus{} x^2\displaystyle\int^1_{ \minus{} 1}t\cdot{f(t)}\,dt \plus{} x^3\int^1_{ \minus{} 1}f(t)\,dt$

For $ x \equal{} \minus{} 1$ we have:

$ f( \minus{} 1) \equal{} 1 \plus{} \displaystyle\int^1_{ \minus{} 1}t\cdot{f(t)}\,dt \minus{} \int^1_{ \minus{} 1}f(t)\,dt$  (1)

For $ x \equal{} 1$ we have:

$ f(1) \equal{} 1 \plus{} \displaystyle\int^1_{ \minus{} 1}t\cdot{f(t)}\,dt \plus{} \int^1_{ \minus{} 1}f(t)\,dt$  (2)

Then (2) minus (1) gives: $ f(1) \minus{} f( \minus{} 1) \equal{} \displaystyle2\int^1_{ \minus{} 1}f(t)\,dt$ $ \Leftrightarrow$ $ \displaystyle\int^1_{ \minus{} 1}f(t)\,dt \equal{} \frac {1}{2}\left({f(1) \minus{} f( \minus{} 1)}\right)$.

Let's denote $ f(x)\equal{}x^2 \plus{}ax^2 \plus{}bx^3,\ b\equal{}\int_{\minus{}1}^1f(t)dt\equal{}\frac 23(1\plus{}a),\ a \equal{} \int_{\minus{}1}^1tf(t)dt \equal{} \frac 25b$
It's easy to find $ a \equal{} \frac 4{11},\ b\equal{}\frac{10}{11}$..  :cool:

[quote="Diogene"]Let's denote $ f(x) \equal{} x^2 \plus{} ax^2 \plus{} bx^3,\ b \equal{} \int_{ \minus{} 1}^1f(t)dt \equal{} \frac 23(1 \plus{} a),\ a \equal{} \int_{ \minus{} 1}^1tf(t)dt \equal{} \frac 25b$
It's easy to find $ a \equal{} \frac 4{11},\ b \equal{} \frac {10}{11}$..  :cool:[/quote]
Nice one....

Thank You!


I have being doing practice AIME exams. I seem to get around 8 on the AIME. My weaknesses seem to be functions, probability/counting, and geometry. Does anyone have any books they recommend to me to help improve my skills???

If you haven't done AoPS v1 or v2, that's a good place to start. ACoPS has really good stuff, too, though it gets more difficult into some olympiad material. Otherwise, I'd suggest hanging around the Intermediate/Pre-Olympiad forums.


percent
AMC
AIME
floor function
ceiling function

[quote="4everwise"]Each of the 2001 students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between 80 percent and 85 percent of the school population, and the number who study French is between 30 percent and 40 percent. Let $m$ be the smallest number of students who could study both languages, and let $M$ be the largest number of students who could study both languages. Find $M-m$.[/quote]
[hide]To find m, note that if 80 % study spanish and 30 % study french there will be a 10 % overlap so m is 201 (we can't have fractional students). To find M, note that the max is when everyone taking French is taking Spanish which would be 40 percent of 2001 which is 800 (we have to round down). So our answer is 599.[/hide]

[quote="usaha"]sorry to bring this back but since im practicing for the AIME, and Im tryin to do problmes, I want to see if Im right.  When I read this thread, i think bpms's solution (which is the only one) is wrong, because the total students for M, accoridng to bpms's way does not add up to 2001
I think the answer is 298?  Can anyone confirm that?
Ill post a solution if thats right[/quote]
You could check on the AMC website...
http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2001-aa/aimeans41001.html
number 2.
You were correct.

solution
[hide]Well, all I did was caculate the amont of Spanish students at 80 percent-1601 and 85 percent-1700 (you cant have fractions of people)
and French students at 30 percent-601 and 40 percent-800 people,

so the fewest amount of people taking both is the fewest percentage for Spanish and french (80 and 30 percents) so using PIE-1601+601-x=2001, x=201
the most is the largets percentage for the two languages (85 and 40), so 1700+800-x=2001, x=499

so 499-201=298[/hide]

Can you explain more in-depth how you apply PIE?  I somehow managed to end up with 299, but I would like to know my mistake.  Thank you.

This is a very simple problem with a small but crucial observation. When considering the maximum cases , we round down , when considering the minimum cases , we round up. This gives $M=499$ and $m=201$ hence consequently our answer is $M-m=298$.

$\square$

Wow, a 5 year bump of a thread that is currently 15 years old. Now that is impressive. Wow, how did you find this friend oh mighty thread reviver? :omighty: What is your secret?  :omighty:

[s]Anyway, this belonged in HSM in 2004, but now, this belongs in MSM lol. I believe this will require a venn diagram. But, I will still give it a go. Let's see, for $M$, we could have $85\%$ Spanish and $40\%$ French, which means that $25\%$ are both

For $m$, we could have Spanish = $80\$$ and French = $30\%$, which means that $m = 10$.

Now for the subtraction of $M - m = 25 - 10 = \boxed{15}$. And as Richard always says in his videos, we are done.[/s]

Nevermind, this does not work, I forgot that they were asking for the population, and not the percentage.

[quote=ProblemPro]What is wrong with saying you have minimum $30 + 80 - 100 = 10$ % both and maximum $40 + 85-100 = 25$% both and just doing $  \lfloor 2001 * .25 \rfloor - \lceil 2001 * .10 \rceil $
Which becomes $500-201 = \boxed{299}$?[/quote]

Hello, can someone please clear up the problem with this method? I did the same thing.

[quote=Akababa][quote="ProblemPro"]Is anyone able to answer my question? Thanks[/quote]

Rounding errors are different if you count them separately vs add them up.[/quote]

I don't understand this at all.

The minimum value is just $2001 \cdot \frac{80}{100}$ rounded up added to $2001 \cdot \frac{30}{100}$ rounded up with $2001$ subtracted, or $201$. The maximum value is just $2001 \cdot \frac{85}{100}$ rounded down added to $2001 \cdot \frac{40}{100}$ rounded down with $2001$ subtracted, or $499$. The answer is $499-201=\boxed{298}$.

It's because you have to consider that you cannot have a non-integer amount of students at any point; if you take 80 percent of the students in one category, you cannot just add that to something else and then round down. You must round down immediately. For example 7.9 + 7.8 with immediately rounding both down is 7+7 = 14. Rounding down after summing gives 15.

$(\lfloor 2001 \cdot .85 \rfloor + \lfloor 2001 \cdot .4 \rfloor) - (\lceil 2001 \cdot .8\rceil + \lceil 2001 \cdot .3 \rceil - 2001) = \boxed{298}.$


[quote="diophantient"][quote="Farnak"]Prove that the number of partitions of a natural number n into unequal (unordered) parts is equal to the number of its partitions into odd (unordered) parts.[/quote]
[hide="hint"]Generating Functions![/hide][/quote]

Yeah, that is totally what you do.

AMP did this exact question in simple fashion.

[quote="Farnak"]Prove that the number of partitions of a natural number n into unequal (unordered) parts is equal to the number of its partitions into odd (unordered) parts.[/quote]

Somebody mentioned via genereting functions.
[hide="Solution"]
Let $ \{ a_{n}\}$ be a sequence of unequal parts. And define $ f(x) \equal{} a_{0}\plus{}a_{1}x\plus{}a_{2}x^{2}\plus{}...$. With some trail and error we find that $ f(x) \equal{} (1\plus{}x)(1\plus{}x^{2})(1\plus{}x^{3})...$. Let $ \{b_{n}\}$ be a sequence of odd parts. And define $ g(x) \equal{} b_{0}\plus{}b_{1}x\plus{}b_{2}x^{2}\plus{}...$. This time $ g(x) \equal{} (1\plus{}x\plus{}x^{2}\plus{}...)(1\plus{}x^{3}\plus{}x^{6}\plus{}...)(1\plus{}x^{5}\plus{}x^{10}\plus{}...)$.

Now, $ f(x) \equal{}\frac{1\minus{}x^{2}}{1\minus{}x}\cdot\frac{1\minus{}x^{4}}{1\minus{}x^{2}}\cdot\frac{1\minus{}x^{6}}{1\minus{}x^{3}}\cdot ... \equal{}\frac{1}{1\minus{}x}\cdot\frac{1}{1\minus{}x^{3}}\cdot ... \equal{} g(x)$.
[/hide]

Thanks for the solution. By the way, who/what is AMP?

It's AwesomeMath [Summer] Program. :wink:  :wink:

As Oleg would say, "Find a buttjection."

This can be generalized. If you still want to find a bijection...

[hide="hint"]

use basis representation

[/hide]


[quote="flip2004"]Consider the sequence  $(x_n)_{n=0}^{\infty}$  defined by  $\begin{array}{|c|} \hline \\ \displaystyle x_{n+2}= \displaystyle \frac{2x_{n+1}^2}{1+x_{n}^2} \\ \\ \hline \end{array}\; \; , \; n\in \{0,1,...\}\; ,\; x_0=a>0 \; ,\; x_1=b>0 .$
[b] Prove that for any pair $(a,b)$ of positive numbers  holds the inequality  [/b]  $\; \; \; \; \; \displaystyle \begin{array}{|c|}\hline \displaystyle x_{2006} < \displaystyle 32\\ \hline \end{array}\; .$[/quote]

I guess you have some sort of a general inequality. What's that?


Let a triangle $ABC$ with $\angle A$ acute,  a point  $D$ on side $AB$ , a point $E$ on side $AC$ , the orthocenter $Z$ of triangle $ADE$ and the orthocenter $H$ of triangle $ABC$. If $K, L$ are the midpoints of segments $BE , CD$, prove that
   $KL \perp ZH$.

[b] Note[/b]

 [i]Posted in the preolympiad section too, but without solvers.EDIT:But now Nicula has sent a nice solution! [/i]

This is equivalent to the fact that the Gauss line of the complete quadrilateral formed by the lines BC, CA, AB, DE is perpendicular to the Aubert line of this complete quadrilateral. And this follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=842 part [b]b)[/b].

  darij

See the my proof at http://www.mathlinks.ro/Forum/viewtopic.php?t=109289

Darij and Virgil,

   thank you both very much. I did not expect that this problem would be so interesting.I tried a little with the metric relation , used by Virgil , but I failed( :( ).

 I had never seen before  the very good result , as it was formulated by Darij. Many thanks again.

[u]Babis[/u]


Here at the AMC, we are experimenting with some ideas for presenting problems for contest preparation.  We have developed a database of problems and indexed them and presented them in various ways.

A good starting place to look at these is
http://www.unl.edu/amc/mathclub/04,1-problems.html

Each problem has the problem statement, a hint, a complete solution, and some indexing and item difficulty notations.

The available problems are indexed in 4 ways:
1.  A listing by year and number
2.  An indexing by NCTM Standard
3.  An indexing by topic using the MathWorld.com heirarchy
4.  A database with descriptions

Problems are presented in 3 formats:
1.  the WS or worksheet format, an HTML page with embedded gifs for the figures and mathematical notation.
2.  PDF
3.  IAP, or "InterActive Problem" which presents the problem with links for the hint and the solution.

We developed this format with teachers and math team coaches as the primary audience.  However, I think the readers of this forum would have some comments to make about the problem format.

Remember, this is experimental and developmental.  We are not sure what is the right way to present all our back catalog for lots of people.  I would appreciate your comments and constructive criticism.

Steve Dunbar
AMC Director

I think that this is a great resource! I will be teaching my school's math club this year, so I plan on using these resources often. :)

You mention that you are in the experimental phase.  Will the link remain the same?  In other words, can AoPS go ahead and use that link in our resources area to point teachers (and students) to that page or will you use a different link later?

That link will remain active for some time, weeks at least, probably months.   Ultimately, it will move but it will probably be next contest year/cycle.  I think you are safe in posting it for now, I don't expect it will change abruptly.

Steve Dunbar
AMC Director

Looks like a great coaching resource, especially the worksheets/quizzes by topic area.  Our math group typically meets once per week for a little over one hour - just about enough time for a 15 minute lesson on a topic, then some time to work problems and discuss solutions.  The quizzes/worksheets will fit nicely with our time constraints.

Well done.

Errata:  The links for the PDF Questions under the High School worksheets point to the solutions instead of the questions.  See link below, click the Q's under High School PDF's.

[url=http://www.unl.edu/amc/mathclub/04,0-quizes.html#progression]Quizzes by Topic Area[/url]

wow great resource.  I'll be using it for a while.  :D

Errata:  The links for the PDF Questions under the High School worksheets point to the solutions instead of the questions.  See link below, click the Q's under High School PDF's.

Thanks for picking that up.  We will get it fixed, but not before next week, our Webmaster is taking a brief but well-deserved few-day vacation after getting all that info up on the web.

Steve Dunbar
AMC Director

wow, this is really well done.  especially the quizzes.

well done, i'll be using this for a while... :)

Hey, just to report another typo, under the elusive formulas chart

http://www.unl.edu/amc/mathclub/resources/05,4a-symboltable.html

the symbols for union and intersection are switched around.  :D

Excellent resource.

One typo: In the number theory quiz, the answer to Question 10 is given as "(B) 160".  Actually the correct answer is "(E) 190".  This typo also appears in the paper copy of the Math Club book.

Yes, this is right, because when you look for the problem by year, it is listed correctly as 190 and a valid solution is included.


I have two question about Chinese Remainder Theorem

1) What is it?

2) Where did it get its name?

http://mathworld.wolfram.com/ChineseRemainderTheorem.html

No idea about the name, but then again, google is your friend.

where it came from (found on Planet Math)... The Chinese remainder theorem is said to have been used to count the size of the ancient Chinese armies (i.e., the soldiers would split into groups of 3, then 5, then 7, etc, and the leftover soldiers from each grouping would be counted).

In Andrew's book "Number Theory", it is said that Sun Tzu investigated certain residue systems and used a version of the theorem. Thus, it was named "chinese" after him.


Let $ a \equal{} x \plus{} y,\ b \equal{} xy,$ we have $ (1 \plus{} 2x)(1 \plus{} 2y) \equal{} 3\Longleftrightarrow a \plus{} 2b \equal{} 1.$

$ \therefore x(1 \minus{} x^3)(1 \plus{} 8y^3) \minus{} y(1 \minus{} y^3)(1 \plus{} 8x^3)$

$ \equal{} (x \minus{} y)\{1 \minus{} 8xy(x \plus{} y) \minus{} (x \plus{} y)(x^2 \plus{} y^2) \minus{} 8x^3y^3\}$

$ \equal{}(x\minus{}y)\{1\minus{}ab\minus{}a(a^2\minus{}2b)\minus{}8b^3\}$

$ \equal{} (y \minus{} x)(a^3 \plus{} 8b^3 \minus{} 1 \plus{} 6ab)$

$ \equal{} (y \minus{} x)(a \plus{} 2b \minus{} 1)(a^2 \minus{} 2ab \plus{} 4b^2 \plus{} 1 \minus{} 2ab \plus{} a \plus{} 2b) \equal{} 0.$ Q.E.D.


USAMTS
email
geometry
3D geometry
analytic geometry
rectangle
trapezoid

No solutions, hints, etc., until Tuesday, Nov 22, just general discussion.

p.s. sorry about this, i clicked on the wrong thread in the moderator control panel... :oops:

When was the latest you guys ever sent out solutions and they were graded?  The Monday after the deadline?  I havent finished yet, and am wondering if they'll still be graded if i send them monday, and if anyone has sent them later than monday.

I emailed Blair Kelly about this and he told me that this year he is being more strict about this; in by Monday 12:00 am EST or they will be considered 'late'.

[i]Date:  Thu, 20 Nov 2003 14:37:25 -0500 (EST)
From:  Blair Kelly III <bfkelly@afterlife.ncsc.mil>
To:  mlin8@hotmail.com
Reply-To:  usamts@nsa.gov
Subject:  Re: [USAMTS] Deadlines

Due to the increased number of participants, we are tightening up on the rules this year. All entries need to be postmarked or faxed before Monday midnight EST or they will be considered 'late' and may not be graded.

Happy Thanksgiving!

---

Blair Kelly
Director, USAMTS
usamts@nsa.gov
http://www.nsa.gov/programs/mepp/usamts.html

> Hi Mr. Kelly,
>
> I've read on a math message board that you will still
> accept entries even if they are a day or two late. Is
> this true?
>
> Thanks for your attention.[/i]

thanks, so that means i cant mail it tomorrow, then.

Sure. You're not the only one rushing them.  :lol:

how are we supposed to send it out if we cant mail it tomorrow? I had a serious issue that impeded my gettting to the post office....

bleep this sux so much..... My mom's business just opened and Ive been doing that day and night.... Arg, I hope they still accept it.

Fax it. By the way, we can start discussing after noon EST Monday (12 hours after deadline).

does anyone know how to fax it, i.e., how do they want it? How are we supposed to keep the pages together when we fax it? Please reply quickly.....

(508) 528-6373.

One page after the other. To avoid confusion, write your name on all pages.

god damn that was a hectic hour. I just had to run 2.5 miles over to my mom's business, get her to sign the cover sheet, then run back..... I swear I must have run faster than i ever have (which is saying something considering i run pretty well... fastest 3-mile is 18 mins).  Then I faxed the solutions off.... not sure if they got them or whatnot, so im also going to visit mr. mailman tomorrow. UGH.....

Whew! No need. Simply check your email tomorrow for a receipt confirmation notice from Blair Kelly.  :)

why the rush?  still like 5 hours left, no?

On top of all of the crap ive been through today with this, I just saw that my type up for number 3 has a typo in it that makes [b]can[/b] become [b]cant[/b]. I sent another copy through to them with the typo fixed, saying what happened, but that really angers me. And Im also goign to send an email to the Director telling him what the deal was with that. Stupid word processor.

Since the deadline has 1.5 hours past, and mlin has kindly cleared this up for us, I believe it's all right to start discussing, no?

These are the ANSWERS I got - these are useless without full solutions anyway so it's not like it matters.

1. 24/27 cubes
2. n=84
3. 34 unique circles
4. 2700 triangles
5. [img]http://www.geocities.com/tasdoom212/math/usamts5.jpg[/img]

4a. 1980 acute
4b. 45 right
4c. 675 obtuse

http://68.38.137.202/round2.pdf

Since Blair Kelly is being strict about it this year... here are my solutions.

I think don't think it is necessary to share solutions very soon after they are due. Even if Mr. Kelly said they are being strict about punctuality of solutions, that doesn't mean that they are. Until someone gets a late solution rejected, we don't know what their criteria are for accepting and rejecting late ones. (Actually, we won't know even then, but we'll have a better idea.)

Do they give u some points even if you got it wrong

You can read what they say about it on their website.  Basically, if they think you moved in the direction of a solution, but didn't get the final answer, you will get partial credit.  In other words, it's the solution, not the answer that counts the most.

[quote]fairyofwind Posted: Mon Nov 24, 2003 1:35 pm    Post subject:

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

tasdoom... 1980 obtuse... 675 acute
 [/quote]

yup, thats what i got. btw, does anyone have a "simple" solution to number 5?

my short solution for nmber 5:

There are at least 9 vertices in the given polyhedron. Every vertex of a polyhedron must be connected to at least three other vertices. If each vertex in the given polyhedron is connected to exactly three other vertices, there would be a total of 9*3/2= 13.5 edges, which is impossible. The number of edges must be a whole number greater than 13.5. The smallest possible value is 14.

dang, 4/5...i missed number 5...does anyone have an actual picture of the polyhedron?

http://mathworld.wolfram.com/HeptahedralGraph.html

Look at the graph in row 2, column 6. It is planar and 3-connected.

There is another nonisomorphic graph which also has 14 edges and 3 pentagons but I am too lazy to look for it.

Mm.  I'm glad I sent mine in on Wednesday...

But bah, I missed #5.  Grr.

Basically.  I found that 14 circles pass through 4 points, accounting for 56 triangles but only 14 circles; since there are 9C3 - 8 = 76 triangles total, the other 20 triangles each have their own circle, for a total of 14+20=34.

How did you prove that there were exactly 14 cyclic quadrilaterals?

I got 14, but my reasoning was more or less equivalent to "Because I said so."

[quote]Did anyone solve it by counting (ways to choose 3 points) - (lines) - 3(cyclic quadrilaterals)?[/quote]

yes, i did, but i didn't have a proof of cyclic quadrilaterals... i just counted them. maybe i should've put why there are no more... eh, whatever, too late now.

oh well, missed no. 3.

can anyone give coordinates of some cyclic quadralaterals which aren't squares?

on that particular problem:

first of all, all rectangles are cyclic as well.

second of all, symmetrical trapezoids are:
for instance: (0,0) ; (1,0) ; (2,1) ; (2,2) is one. there are four of those.

and thats all i counted... i guess thats right then.

I just drew them all, and prayed that I didn't miss any.   :D

Basically, the cyclic quadrilaterals are:
- 4 1x1 squares
- 1 :rt2:x:rt2: square
- 1 2x2 square
- 4 2x1 rectangles (say, (0,0),(2,0),(0,1),(2,1))
- 4 isosceles trapezoids with bases of length 2:rt2: and :rt2: and legs of length 1

total 14.

lol, i cannot believe that i missed number 3 since i considered a straight line a triangle. ah, at least i got 1,2,4,5 right.

[quote="314159265358979323846"]I missed number 3... i mailed in a solution which ended up with the right numerical answer but was incredibly incomplete. Did anyone solve it by counting (ways to choose 3 points) - (lines) - 3(cyclic quadrilaterals)?[/quote]

yeah, i did. i found all cyclic quadrilaterals by the following algorithm: choose any 2 points, and choose another point and reflect it through the perpendicular bisector of the first two points. If it lands on a valid point in the grid, well, there's a cyclic quadrilateral. We don't have any cyclic n-gons, n>4, because we don't have any 2 cyclic quadrilaterals with 3 points in common. BTW, I didn't send in USAMTS solutions, but this is what I came up with (and what I would have sent in if I did)


function
induction
algebra unsolved
algebra

[quote="ll931110"]Find all functions $ f: N \rightarrow N$ satisfying two conditionals below:
$ f(1) > 0$
$ f(m^2 \plus{} n^2) \equal{} f^2(m) \plus{} f^2(n)$ for all non-negative integers m and n

I had found a solution of this problem (Perhaps I may post it next days), but it isn't very natural. Could someone help me to solve it?[/quote]

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=272002[/url]


A triangle $ ABC$ can be divided into two isosceles triangles by a line through $ A$. Given that one of the angles of the triangles is $ 30^{\circ}$, find all possible values of the other two angles.

It is easy to work out some of the answers,but it is difficult to work out all the answers ,I find six triangles,maybe there are more than 6 answers.(T1:30,60,90;T2:15,75,90;T3:15,135,30;T4:37.5,67.5,75;T5:15,45,120;T6:37.5,112.5,30..[T just means triangle.]) :maybe:  :blush:


Prove that for any algebraic number $x$ with minimal polynomial of degree $n$, there is a constant $c$ s.t.
$|x-\frac ab|> \frac c{b^n}$ for any integers $a,b$

Let $q$ be an algebraic number of order $n$.

Let $f(x)=a_0x^n+...+a_n$ is an irreducable polynomial s.t. $f(q)=0$. We have $f(x)=(x-q)g(x)$.
Suppose $\delta>0$; $|g(x)|$ is continuous, so it is bounded on $[q-\delta,q+\delta]$, i.e. $|g(x)|\leq M$ on $[q-\delta,q+\delta]$. Let $c=\min(1/M,\delta)$.
For an arbitrary rational number $a/b$ there are two possibilities.
[b]Case 1.[/b] $\frac{a}{b}\not\in [q-\delta,q+\delta]$. Then $|q-a/b|>\delta\geq c\geq c/b^n$.
[b]Case 2.[/b] $q-\delta\leq a/b\leq q+\delta$, then $|g(a/b)|\leq M$, thus $|f(a/b)|=|q-a/b|\cdot g(a/b)\leq \frac{1}{c}|q-a/b|$.
On the other hand $|f(a/b)|=\frac{|a_0a^n+a_1a^{n-1}b+...+a_nb^n|}{b^n}\neq 0$, moreover the numerator is at least 1, therefore $|f(a/b)|\geq 1/b^n$. It follows that $\frac{1}{c}|q-a/b|\geq \frac{1}{b^n}$. QED

And, additionally, I want to state the following result.

[b]Theorem (Thue-Zigel-Roth).[/b]
Let $q$ is an algebraic number of order $n$. Then for all $\varepsilon>0$ thre are finitely number of rational numbers $a/b$ s.t. $|q-a/b|<\frac{1}{b^{2+\varepsilon}}$.


Ring Theory
algebra
function
domain
superior algebra
superior algebra unsolved

I was wondering if every quotient of PID is PID and in the affirmative case how to prove it

any help would be mostly welcome

A quotient of PID need not to be a domain, e.g. $ \mathbb{Z}/4\mathbb{Z}$

can you please give an example of an ideal of $ \mathbb{Z}/4\mathbb{Z}$ which is not principal

Of course I can't because every ideal in $ \mathbb{Z}/4\mathbb{Z}$ is principial (in fact every ideal in any quotient of any PID is principial). But still, that ring is not a principial ideal [b]domain[/b], since it is not a domain.

A quotient of a principal ideal ring is a principal ideal ring.

This is easy to prove using the projection homomorphism and the fact that if $ I$ is an ideal, then $ \pi^{\minus{}1}(I)$ is an ideal.


Let $x,y,z>0$ s.t. $\sum\frac1x=1$. Prove that
$\sum\sqrt{x+yz}\ge\sqrt{xyz}+\sum\sqrt x$

This is an easy ineq
 $ \sqrt{xyz} = \sqrt{\frac{xy}{z}} + \sqrt{\frac{yz}{x}} + \sqrt{\frac{xz}{y}} $
 So , we prove : $ \sqrt{z+xy} \geq \sqrt{z} + \sqrt{\frac{xy}{z}} $ (1)
 $ (1) <==> (\sqrt{x}-\sqrt{y})^2 \geq 0 $


algebra
polynomial
function
Ring Theory
superior algebra
superior algebra unsolved

I confused :?  , but I need more help.

Let [b]R[/b] be a ring [b]R[x][/b] of polynomial with real coefficients.
Look at  Principal ideal [b]<x^2+1> [/b]in [b]R.[/b].
Show that every element in factor ring [b]R/<x^2+1> [/b]has one unambiguous define representative of that subset [b]E[/b] of [b]R[/b], which consist of those polynomials of [b]R [/b] which has degree maximum 1.

This cause of bijective function [i]f[/i]: E --> R/<x^2+1>.
Explaine why?

I cannot :wallbash_red:

Best regards
Tanja

[quote="julien_santini"]Use the euclidean division.[/quote]

How?  :?   :blush:

Best regards
Tanja


Given that A and B are nxn matrix. How do we prove that $ \parallel{}AB\parallel{} < \equal{} \parallel{}A\parallel{}.\parallel{}B\parallel{}$ where ||.|| is the sup norm or matrix norm.

By definition, $ \|A\|$ is the smallest number such that $ \|Ax\|\le \|A\|\|x\|$ for all $ x$. Since $ \|ABx\|\le \|A\|\|Bx\|\le \|A\|\|B\|\|x\|$, the claim follows.


analytic geometry
graphing lines
slope
quadratics

[quote="Farcus"]Edit - Also a problem I can't solve.
"Find all values of k such that the equation of X^2+kX+38=0"

I seriously dont know where to begin this even with the quadratic equation.[/quote]
Do you mean:"Find all values of k such that X^2+kX+38=0 has a real solution."?
in that case:
[hide]for it to have a real solution, the discriminant must be nonnegative.
So $b^{2}-4ac=k^{2}-152\geq0\rightarrow k^{2}\geq152\rightarrow k\geq2\sqrt{38}\ or\ k\leq-2\sqrt{38}$[/hide]

Well, yes, that problem seems as though that's not the end of the sentence.  If it's $\text{Find all the roots of }x^{2}+kx+38=0$, the answer would be:
[hide]$\frac{-k\pm\sqrt{k^{2}-152}}{2}$[/hide]

[quote="Farcus"][quote="ch1n353ch3s54a1l"][quote="Farcus"]Find the point p on the graph y=1/3x such that the slope of (1,1/3) and p is -3/4.

is p (39/31,31/117)?

if not please tell me why and how to solve this problem.

Edit - Also a problem I can't solve.
"Find all values of k such that the equation of X^2+kX+38=0"

I seriously dont know where to begin this even with the quadratic equation.[/quote]

well by substituting the point back in you see it doesn't work.  :)
[hide="1"]Call p $(x, y)$. then we know $y=\frac{1}{3x}$(correct? because your original point satisfied this) and $\frac{y-\frac{1}{3}}{x-1}=\frac{-3}{4}$ and solving we find $x,y=(1,1/3) or (4/9,3/4)$. but x can't be 1 because then the slope is infinite so we have (4/9,3/4), which is our solution.[/hide]

don't you need restrictions or something on x or k? because there should be infinite...if you're saying all values of k because the roots x, y have product 38 and sum -k...but you have infinitely many x, y that satisfy $xy=38$.[/quote]

How did you get that answer? I know you have to substitude y for 1/3x and solve for X but i get bogged down in the algebra for some reason[/quote]

You have $y=\frac{1}{3x}$ and $\frac{y-\frac{1}{3}}{x-1}=\frac{-3}{4}$.

Substitute $y=\frac{1}{3x}$ into the second equation to get $\frac{\frac{1}{3x}-\frac{1}{3}}{x-1}=\frac{-3}{4}$ and then you have $\frac{\frac{1-x}{3x}}{x-1}=-3/4$, then $\frac{1-x}{3x}=\frac{3-3x}{4}$ then you get the quadratic $9x-9x^{2}=4-4x$ and then $9x^{2}-13x+4=0$, $(9x-4)(x-1)$ and...yeah.


There are 4 couples. What is the probability that when they are arranged in a row that no couple will be touching each other? How about if they are arranged in a circle?

i don't really get it if they line up so the just don't touch wouldn't there be an infinite number of ways?

Touching just means the couple aren't standing next to each other.

OH brother... you people just complicate everything. This is just an order sequence type of question - you know like how can you arrange A, A1, B, B1, C, C1, D, D1 so that A does not border A1, B does not border B1, ...etc.

You're being far too literal -- the question is about lining them up and having two people not be adjacent.  Does that make sense?

Not two people JBL - there are four couples that cant be adjacent. That is eight people, which makes the problem harder.

Just a clue[hide]Lable each couple as A, A1, B, B1, and so on... Then you find the probability that only one couple of each of the four possible couples touches, then the probability that any two of the different combinations of two couples are touching, and so on... Then you subtract this from one. Use an inclusive - exclusive equation to solve.[/hide]

i suppose you could do A B C D E F G H then A2 B2 C2 D2 E2 F2 G2 H2 for a start. Then ACBDEFGH, and A2C2B2D2E2F2 so I  guess if nCr could work here it would be something like 16C8. Does that help????


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Given a field $K$ with characteristic $0$ such that every polynomial of degree $3$ in $K[X]$ has a root in $K$, and an irreducible polynomial $f$ of degree $4$ whose discriminant is a square in $K$, find $Gal_{K}(f)$.

We have $S_4 \supset A_4 \supset V_4 \supset C_2 \supset \{e\}$.
When solving the equation, we see that the normal series of the Galois group $G$ we look for fulfills that the step induced by $A_4 \supset V_4$ is trivial (has degree $3$). But also the step $S_4 \supset A_4$ is trivial since the discriminant is a square. Thus since $G$ must have degree at least $4$, we have that $G=V_4$.


if a,b,c>0 then prove that a^3+b^3+c^3>=a^2(2c-b)+b^2(2a-c)+c^2(2b-a)

cheers!

[i]unknown source[/i]

We have ab + b \geq 2ab, bc + c \geq 2bc, ca + a \geq 2ca.
Summing gives a + b + c + ab + bc + ca \geq 2(ab + bc + ca).
Hence a + b + c \geq a(2c - b) + b(2a - c) + c(2b - a).


although SAT does have hard problems they seem to fit well as basic here. I know the solution and the answer but this answer is suppose to have an easier solution than mine.

Question:

If x^3 = y^2, then x^6 - [sqrt]x^6 =

the multiple choices are expressed in y something and there is one 0
[/list][/code]

[hide]$x^{6}-\sqrt{x^{6}}=(x^{3})^{2}-x^{3}=y^{4}-y^{2}$.  So the answer on the multiple choice should be some form of that.  [/hide]

[quote="8parks11"]i found that solution to be the easiest (the answer is y^2) but how is

is this theory true?

x^n -  [sqart]x^n   = X^ (n/2)[/quote]
You mean...
$x^{n}-\sqrt{x^{n}}= x^{n/2}$?

Well if I plug 2 and 4 in...
$2^{4}-2^{2}=12=2\sqrt{3}$
not
$2^{2}$ as your theory says

its not my theory.. its from some mathematician i believe..and i saw it recnetly..as a new one but i can't find it anymore.

i think it just wrote it wrong

Well, it's true that

$\sqrt{ x^{n}}= x^{ \frac{n}{2}}$

From this observation, we can conclude that

$x^{n}-\sqrt{ x^{n}}= x^{ \frac{n}{2}}\left( x^{ \frac{n}{2}}-1 \right)$


There is this one problem I sort of having trouble on. First, I must mention that I do have an answer that I proposed for this except I'm not sure whether it's not right or not. Second, this isn't too tough question and I feel somewhat sad that I'm not so sure about it. Anyway, here it is:

$2x [(x-5)^{-3}] - 4x^2 [(x-5)^{-4}]$

Okay... That's equation.. And here are the steps:

$2x \frac {1}{(x-5)^3} - 4x^2 \frac {1}{(x-5)^4} $

$\frac {2x}{(x-5)^3} (1 - \frac {2x}{x-5})$

For the question, maybe I didn't need big bracket [] but I just put there to make it neat. :)

So please help me on this. I'm pretty sure and thinking I'm write but you can't be always so sure.

I did it the long way by doing everything out and got

\[-2x\frac{x+5}{(x-5)^4}\]

(this is my first time with the tex, so I hope everything look right)

But I studied your steps and they are correct.  ;)

[quote="Cathy Jane"]oops, sorry, I don't think I understand how to do the tex.  :([/quote]

Don't worry I fixed it for you.  The only problem was that you had [ \tex ] and [  \spoiler ] at the end.

[quote="Silverfalcon"]So please help me on this. I'm pretty sure and thinking I'm write but you can't be always so sure.[/quote]

There is a pretty sure way to check.  Plug in a few values.  It's easier than typing all of that :).


Denote by $u(k)$ the greatest odd divisor of $k \in \mathbb{N}.$ Prove that $\forall n \in \mathbb{N}$ we have: \[\dfrac{1}{2^n} \cdot \sum^{2^n}_{k=1} \dfrac{u(k)}{k} > \dfrac{2}{3}.\]

Hi,

My attempt at the solution.

Solution :

$\forall k \in N$, $k$ can be written as $k = 2^t * s$, where $t \geq 0$, $s \geq 1$ and $\gcd(s,2) = 1$.

Thus, the greatest odd divisor of $k \in N$,
\[u(k) = s = \frac{k}{2^t}  \Rightarrow  \frac{u(k)}{k} = \frac{1}{2^t}. \]

Thus, essentially we have to sum over $\frac{1}{2^t}$ for all $t$'s possible.

Now, one can see that number of integers contributing $\frac{1}{2^t}$ is $2^{n-t-1}$ for $t = 0, 1, ... , (n-1)$.
For $t = n$, only $2^n$ contributes.

So,
         \[ \frac{1}{2^n} \sum_{k=1}^{2^n} \frac{u(k)}{k}
      = \frac{1}{2^n} \left \{ \sum_{t=0}^{n-1} \frac{1}{2^t} * 2^{n-t-1} + \frac{1}{2^n} \right \} \]
      \[ = \frac{1}{2^n} * \left \{ 2^{n-1} * \sum_{t=0}^{n-1} \frac{1}{2^{2t}}  +  \frac{1}{2^n} \right \} = \frac{1}{2^n} * \left \{ 2^{n-1} * \frac{2^{2n} - 1}{2^2 - 1}  +  \frac{1}{2^n} \right \} \]
      \[ = \frac{\frac{2^{2n} - 1}{2} + \frac{3}{2^{2n}}}{3} \]

The numerator $= 9/4 > 2$ for $n = 1$.
For $n > 1$, one part in the numerator $\frac{2^{2n} - 1}{2}$ is greater than $2$ itself while the other part is positive. Thus, the numerator is greater than $2$ always.

Thus, $\frac{1}{2^n} \sum_{k=1}^{2^n} \frac{u(k)}{k}  > \frac{2}{3}$

****************************************************************************************
Sambit

Let $\displaystyle S_n = \frac 1 {2^n} \sum_{k=1}^{2^n} \frac {u(k)} k$.
Then, $\displaystyle 2^n\cdot S_n = \sum_{k=1}^{2^n} \frac {u(k)} k$.

Now observe that $\frac {u(2k)} {2k} = \frac 1 2 \cdot \frac {u(k)} k$. Then :
$\displaystyle 2^n \cdot S_n = \sum_{k odd} \frac {u(k)} k + \sum_{k=1}^{2^{n-1}} \frac {u(2k)} {2k} = 2^{n-1} + 2^{n-2}S_{n-1}$
Thus :
$S_n = \frac 1 2 + \frac {S_{n-1}} 4$. (1)

The desired result now follows easily by induction. But, we may go further :
Let $S_n = U_n + \frac 2 3$. It is easy from (1) to see that the sequence $(U_n)$ is geometrical with ratio $\frac 1 4$. Since $S_0 = 1$ and $U_0 = \frac 1 3$, it follws that $S_n = \frac 2 3 + \frac 1 3 \cdot (\frac 1 4)^n$.

Pierre.


What are the rightmost three digits of $ 5^{1993}$?

If you calculate the first few powers of 5, you notice that after 5^3, the last three digits alternate between 125 and 625:

5^3=125
5^4=625
5^5=3125
5^6=15625
5^7=78125
etc.

So 5^1993 ends in 125, because 1993 is odd.


Hi,

So can students around the world take it when it's administered (for example, American Universities)?

And if so, do we take it on our campus, since it's open in format? And how do we register? Do the problems post themselves online?

Info would be much appreciated.

It seems to me this is an national-level competition.

http://en.wikipedia.org/wiki/Mikl%C3%B3s_Schweitzer_Competition
http://www.math.u-szeged.hu/~mmaroti/schweitzer/


articles calculus integration Support videos

What is your opinion of the new OS Windows Vista?

I'm going to tell everybody who asks me what I think about it what a peice of garbage it is. If you're gonna move to a new OS, move to Linux, and you'll be a bit happier. Certainly you won't be using your old programs and you might have a little trouble with multimedia (might), but you won't be using terribly many of your old programs on Vista either, you'll have DRM lockin under Vista, and you'll have Vista phoning home and telling M$\$$ all about your computer.

I was quite impressed: the Atlanta Journal-Constitution had a front-page article on whether Vista was worth it and they actually mentioned Linux as an alternative if you didn't want to move up to Vista. It was, of course, under the "stick with what you've got" and the "OS X rocks" paragraphs, but the Linux paragraph was longer than either of the preceding paragraphs. :)

I don't know, but I've been told . . .

I know nothing myself, but the descriptions from some of the beta testers seems to be that this, aside from the pricing and publicity, is just a major upgrade of XP.  One that other companies would have done most long ago and as upgrades, not as a new version to make more money upon.

I hear no major new bugs and some old major ones are less frequent or have been fixed.

Yea, M$ is really mature. And Linux is nice for some people, but many others don't like to move to an OS where half their programs don't run or run at a crippled pace.

Let's compare Windows and Linux.

Windows: Has DRM integration, which restricts rights.
Linux: Has no DRM integration, which allows piracy.
I choose: I don't really care, I haven't had any DRM issues.

Windows: Insecure, allowing a hacker to attack your computer.
Linux: Secure, but it's easy to accidentally damage your own computer.
I choose: Windows, I've never been hacked, but I have [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119925]deleted my own files in Linux[/url].

Windows: Costs money.
Linux: Free.
I choose: Linux, obviously.

Windows: Easy to use, easy to learn.
Linux: Easy to use, but more difficult to learn.
I choose: Windows. I suppose that means I'm not a true geek.

Windows: Makes it hard to change advanced settings.
Linux: Advanced settings are easy to change, but only if you know the commands.
I choose: Windows, because I like the shiny graphical interfaces.

Windows: Supports most commonly used software, but not a lot of free software.
Linux: Supports most free software, but not a lot of commonly used software.
I choose: Linux, because free software is, well, free, and often just as good or better than commercial software.

Windows: Has Clippit.
Linux: Does not have Clippit.
I choose: Do I really need to say?*

Not seeing a lot of pros on either side here. I'm not trying to insult either side or give a straw man argument, so if something I said is wrong please don't get mad. It's probably just ignorance.

*For the record, I actually like Clippit. *dodges tomatoes* But he's just so easy to make fun of.

It depends on what you want.

I am a pure video editor, shooter, and other video stuff. Therefore, I'd stay with Windows, with my Adobe Premiere Pro 2.0 and Avid Pro. However, if you are pestered about linux, I'd go with gentoo (lightweight) or SuSE (better, but longer to download).

DO NOT GET REDHAT!!! I JUST INSERT TEH DISC, CLICK EXIT ON THE FIRST SCREEN...THE TERMS OF AGREEMENT, AND MY WINDOWS PARTITION IS DELETED. WHAT IS THIS REDHAT GARBAGE?

Vista doesn't support some of my video editing software.  :o

I have a disc and product key for Beta2, but when I go ahead to install it, it says, "Delete Adobe Premiere Pro 2.0, Delete Sonic MyDVD, Delete Avid XPress Pro, etc etc etc."

Keep XP for video editing and multimedia, but I personally like SuSE because of its great interface and stuff, but I'm like sort of forced to stay with XP.  :(

Read [url=http://www.suseforums.net/index.php?showtopic=27137&st=0&gopid=144938&#entry144938]this[/url] post I posted on the SUSE FOrums for help on getting Adobe Premiere for Linux, so I am going to keep Windows, after 2 days with Linux.

[quote="jmadsen"]Macs are also good for video editing (or so I hear, I don't use them). And for video work in Linux you can use [url=http://en.wikipedia.org/wiki/Cinelerra]Cinelerra[/url]. I think it's supposed to be the equivalent of Adobe Premiere.[/quote]

Already tried.  :wink: Macs stink, in my opinion, in the field of Video Editing. Cinerella doesn't have good transistitions, effects, color correction, etc. Adobe Premiere Pro 2.0 is much better than FCP HD, no matter how much you praise FCP HD. FIrstly, it supports HDV native editing, available in both. However, if you are using APP2 on a bad PC, you're going to suffer crashes. I use an XPS Gen 600, so I'll use APP2. FCP HD works well only on good macs. How much are you going to pay to get a mac with these specs:

Intel Pentium D Extreme Edition 3.2 GHZ Processor
3X300 GB Hard Drives Internal
2X500 GB Hard Drives Internal
1X600 GB Hard Drive Internal
1X320 GB Hard Drive External
4GB Dual Channel DDR2 SDRAM at 667MHz - 4 DIMMs
nVidea SL4 Video Processor
48x Combo + 16x DVD+/-RW w/ dbl layer write capable
24 inch UltraSharp™ 2407FPW Widescreen Digital Flat Panel (they cost a ton)
512MB nVidia GeForce 7900 GTX
AGEIA® PhysX® physics accelerator

and more...

All for just $\$$4999.  :)

Try to get a mac with stuff like that...

See how it all adds up.

I downloaded it but couldn't run it on my PC so I took it to my friend. He installed it. Looks cool, otherwise disappointing. Now he hasn't registered it for fourtenn days and it has expired now.

The games are REALLY stupid! One is about making cakes and all ... :rotfl:

I just have windows XP, and it works for me!

eh I'm gonna wait for Microsoft to fix out the bugs and then buy it. Halo 2 on PC!


If $x,y,z>0$  with  $x+y+z=1$ then prove the following ineq.

$\sqrt{\frac{1-2x}{x}}+\sqrt{\frac{1-2y}{y}}+\sqrt{\frac{1-2z}{z}}\geq 3$

1.$x,y,z \leq 1/2$ :D
2.If we take $a=\sqrt{(1-2x)/x},b=\sqrt{(1-2y)/y},c=\sqrt{(1-2y)/y}$we have this ineq.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64636]www.mathlinks.ro/Forum/viewtopic.php?t=64636[/url] :)


topology
function
advanced fields
advanced fields unsolved

I've had very little experience with category theory, so this is a really basic question. Let $\left(\times_{\gamma}X_\gamma, \Pi_\gamma T_\gamma\right)$ be the product topology and $\{\pi_\gamma\}$ be the relative projection mappings for some indexing set $\Gamma$. Show that these satisfy the universal property.

I know that I have to assume that I have a topological space $\left(Y, S \right)$ such that there are continuous mappings $p_\gamma : Y \to X_\gamma$. Then I need a unique $h: Y \to \times_\gamma X_\gamma$ that makes the mapping diagram commute. I have no idea how to define $h$ though. I think it has to be something like $h(y)=\pi_\gamma^{-1}\circ p_\gamma (y)$, but this doesn't seem to be well-defined. Defining this function is all I would like help on, I want to work out the details myself (I'll post it so as not to let people down  :lol: ).

The $\gamma$ coefficient of $h(y)$ is $p_\gamma(y)$. That tuple is the only element of the intersection of the $\pi_\gamma^{-1}(p_\gamma(y))$.

Maybe I'm not understanding due to arbitrary sized products. For the case of two indices we have $(X_{1}\times X_{2}, T_{1}\Pi T_{2})$ a topological space with projections $\pi_{1}$ and $\pi_{2}$. Suppose there exists a topological space $(Y, S)$ such that there are continuous mappings $p_{1}: Y \to X_{1}$ and $p_{2}: Y \to X_{2}$, then there is a unique continuous map $h: Y\to X_{1}\times X_{2}$ given by $h(y)=\pi_{1}^{-1}(p_{1}(y))=\pi_{2}^{-1}(p_{2}(y))$, so when defining it arbitrarily $h(y)=\pi_\gamma^{-1}(p_\gamma(y))$ doesn't depend on choice of $\gamma$ or is it the case that the mapping $\pi_\gamma^{-1}(p_\gamma)$ only makes sense for a single $\gamma$ once a $y$ is chosen in $Y$ or neither of these is what you meant?  :blush:

The inverse images of points under projections are large. You don't want to use them to define an element, but you do want an element of the intersection. To define an element of the product, specify each of its coordinates.

I see it now, but it seems kind of awkward notationally. Here is what I think $h(y)=\bigcap_{\gamma\in\Gamma}\left\{ \pi_\gamma^{-1}(p_\gamma(y))\right\}$.

Let $\times U_\gamma$ be a an open set in the product space. Then $h^{-1}(U_\gamma)=\bigcup_{\gamma}\left\{p_\gamma^{-1}(\pi_\gamma(U_\gamma))\right\}$ but since projections are open, each of those are open sets, and since $p_\gamma$ are continuous the preimage of those opens sets are open, and thus the union is open. (Briefly sketching).

So that is the existence part. I think uniqueness comes from the fact that any continuous function making the diagram commute is determined through a composition of the functions, so it must be that function. ? Maybe?


calculus
integration
number theory unsolved
number theory

Find all possible(or general) integral solutions of the equation

$x+x^{3}+2 = 2^{y}$

This is equivalent to $(x+1)(x^{2}-x+2)=2^{y}$, which leads to $x+1 = 2^{a}$ and $x^{2}-x+2 = 2^{b}$ for some non-negative integers $a,b$.
Substituing, we deduce that $2^{2a}-3\cdot 2^{a}+4 = 2^{b}$ which forces $a \leq 3$ or $b \leq 3$.
Direct checking leads to $(x,y) = (0,1),(1,2),(3,5).$

Pierre.


Hey do you guys know how you get on the Indiana ARML team?

I know its early, but I want to find out as early as possible.  Thanks.

Also, is there a facebook group of Indiana ARML?
Are there webpages (the old one only goes up to 2006) and email links?

Also, if you are going, would you please post your name and school...I dont know anybody from my school that is going, so I might have to ask one of you guys.

Thanks

I'm probably going this year. I don't think you need to worry about ARML until April... i don't know though, but if you have questions, you should email Dr. Rickert.

for more information, look here, [url]http://www.rose-hulman.edu/~rickert/ARML/IN/[/url]

Anirudh Prabhu, West Lafayette High School.

Who are you?

yea like I said, we will probably post more about it early April.

Let me find a link for last year.

[quote="undefined117"][quote="mwpl11"]gold team :

phil
sachin
jeffrey
richard
lyndon
mihika
lauren
rj
mason (?)
ashley
shawn
justin
joseph
anirudh
youkow[/quote][/quote]

This was last years gold team (from this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199790]link[/url]).  Sorry, but I could not find who was on the Silver Team.
Im pretty sure most of them are going again.


If a triangle is randomly selected
from the triangles pictured, what
is the probability that at least
part of the interior of that triangle
will be shaded? Express your
answer as a common fraction.

[asy]size(100); import patterns; add("hatch",hatch(.5mm,E,linewidth(.5))); draw((0,0)--(1,0)--(0,.34)--cycle); draw((.5,0)--(.5,.5*.34)--(0,0)); fill((.5,0)--(.5,.5*.34)--(1,0)--cycle,pattern("hatch"));[/asy]

after a little time staring at the picture, we can conclude that there are 5 different triangles to choose from (do you see them?)

out of those 5, three of them have a region shaded, so our answer is $ \frac{3}{5}$

You already have $3$ from the smallest triangles.
Then the bottom forms another triangle.
Then the big triangle itself.

That gives us $5$ triangles.

[quote="thkim1011"]I don't get it. I can see only 4 if you separate the large triangle.[/quote]

The large triangle is supposed to be included.  :)


Do you know how can I use LaTeX on forum(to write a^3 or sth)? :oops:

Put dollar signs $\$$ around your text. Then a^3 becomes $a^{3}$.

Put dollar signs around the material you want to be latexed.

put dollar signs around it ($\$$) to turn it into latex

a^3 looks like $a^{3}$ in latex

more info at http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

but when you're in the classroom you have to put a semicolon before the first dollar sign.


(a) Positive  integers  $p,q,r,a$  satisfy  $pq=ra^2$, where  $r$  is prime  and $p,q$ are  relatively  prime. Prove  that one  of the  numbers  $p,q$ is a  perfect  square.
(b) Examine if there  exists  a  prime  $p$  such that  $p(2^{p+1}-1)$ is a  perfect square.

a) $r$ is prime, so, WLOG, suppose that $r|p$ and that $p = rx$.
so, $xq = a^2$, with $(x,q) = 1$. ($x$ and $q$ are perfect squares).

b)if $p = 2$ is trivial.
if $p|x^2$, we have that $p^2|x^2$.
so, $p|2^{p+1} -1$.
by fermat little theorem, we have that $p|2^{p-1} - 1$
so, $4 \equiv 1 (mod.p) \Longrightarrow p = 3$.
for $p = 3$, is trivial

a)
$ \ pq=ra^{2} $=$ \ pq=r^{2\alpha+1}b^{2} $ where $ \alpha $ may be zero.

r must divide atleast one of p,q. wlog assume  r|p as (p,q)=1 ;so $ \r^{2\alpha+1}|p $ so this can be rewritten as $ \ p'q=b^{2} $ where (p',q)=1.

so both p' and q are perfect squares.
hence a)

b) if it have to be a perfect square then $\ p|2^{p+1}-1$ i.e. $ \ p|2^{p}-2+2^{p}+2-1 $
as $ \ p|2^{p}-2 $ by fermat's little theorem.
so $\ p|2^{p}+1 $
so $\ p|(2^{p}+1 )-(2^{p}-2)=3 $
so only possibility is p=3
but it is not a perfect square then.

so no such p exists.


algebra
polynomial
invariant
Ring Theory
logarithms
inequalities
Galois Theory

Say we have algebraic numbers $\alpha,\beta$ for which

\[
[\mathbf{Q}(\alpha):\mathbf{Q}] = n, \,\, [\mathbf{Q}(\beta):\mathbf{Q}] = m \quad \mbox{ with } \quad (n,m)=1.
\]
Does it follow that

\[
\mathbf{Q}(\alpha+\beta) = \mathbf{Q}(\alpha,\beta)?
\]

[i]Posed by Bhargav Bhatt[/i]

Let $m=[\mathbb{Q}(\alpha):\mathbb{Q}]$ and $n=[\mathbb{Q}(\beta):\mathbb{Q}]$. Since $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$ does not exceed $mn$ and is divisible by both $m$ and $n$, it equals to $mn$.  Since $\mathbb{Q}(\alpha+\beta)\subseteq \mathbb{Q}(\alpha,\beta)$ and $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=mn$, it suffices to show that $[\mathbb{Q}(\alpha+\beta):\mathbb{Q}]=mn$. Suppose $\gcd([\mathbb{Q}(\alpha+\beta):\mathbb{Q}],m)<m$. Then $[\mathbb{Q}(\alpha+\beta,\beta):\mathbb{Q}]$ divides $n[\mathbb{Q}(\alpha+\beta):\mathbb{Q}]$ and so cannot be equal to $mn$. However, $\mathbb{Q}(\alpha+\beta,\beta)=\mathbb{Q}(\alpha,\beta)$ which has degree $mn$ over $\mathbb{Q}$. The contradiction shows that $\gcd([\mathbb{Q}(\alpha+\beta):\mathbb{Q}],m)=m$. Similarly, $\gcd([\mathbb{Q}(\alpha+\beta):\mathbb{Q}],n)=n$. In other words, $[\mathbb{Q}(\alpha+\beta):\mathbb{Q}]$ is divisible by both $m$ and $n$. QED.

Boris

[quote="bbukh"]Suppose $\gcd([\mathbb{Q}(\alpha+\beta):\mathbb{Q}],m)<m$. Then $[\mathbb{Q}(\alpha+\beta,\beta):\mathbb{Q}]$ divides $n[\mathbb{Q}(\alpha+\beta):\mathbb{Q}]$
Boris[/quote]

WHY?

I apologize for an incomplete proof.

Let $f$ be the minimal polynomial for $\beta$ over $\mathbb{Q}(\alpha+\beta)$. Let $g$ be the minimal polynomial for $\beta$ over $\mathbb{Q}$.  Clearly, $\sigma f\mid \sigma g=g$ for every $\sigma\in Gal(\mathbb{Q}(\alpha+\beta)/\mathbb{Q})$. The polynomial $\sigma f$ is irreducible because $f$ is irreducible. Consider $f^*:= \prod_{\sigma\in Gal(\mathbb{Q}(\alpha+\beta)/\mathbb{Q})} \sigma f$. I would like to say that $f^*$ divides $g$, but I can't quite do that since $\mathbb{Q}(\alpha+\beta)[X]$ does not have to be an UFD. However, $\sigma_1 f$ and $\sigma_2 f$ do not have common roots if $\sigma_1\neq \sigma_2$ because if they shared a root $\tau$, then neither would be the minimal polynomial for $\tau$ contradicting their irreduciblity. Each root of $\sigma f$ is a root of $g$. Therefore, we can conclude that $f^*$ divides $g$. However, $f^*$ is invariant under the action of $Gal(\mathbb{Q}(\alpha+\beta)/\mathbb{Q})$ and so $f^*\in\mathbb{Q}[X]$. Since $g$ is irreducible, we conclude $f^*=g$. Hence $\deg f=\deg g/\lvert Gal(\mathbb{Q}(\alpha+\beta)/\mathbb{Q})\rvert$ divides $\deg g$. Since $\deg f=[\mathbb{Q}(\alpha+\beta,\beta):\mathbb{Q}(\alpha+\beta)]$, and $[\mathbb{Q}(\alpha+\beta,\beta):\mathbb{Q}]= [\mathbb{Q}(\alpha+\beta,\beta):\mathbb{Q}(\alpha+\beta)][\mathbb{Q}(\alpha+\beta):\mathbb{Q}]$ we have the desired result.

Boris

Boris,

I am slightly confused. Are you assuming Q(a+b) is Galois over Q?

If so, I think your proof works. As it happens, k[X] is a UFD for any field k, so Q(a+b)[X] is also a UFD.

If not, does Gal(Q(a+b)/Q) refer to the Galois group of its normal closure? If so, then I am not sure how to interpret the polynomial f*. Which field does it have coefficients in? Over what field does it divide g?

I posted this problem to sci.math.research where, after a few months of no luck, Lenstra pointed Buzzard to a paper of Isaacs which answers this question in the affirmative, and in much greater generality. It can be viewed at http://mathforum.org/kb/thread.jspa?threadID=568606&messageID=1696268#1696268

(I also wrote up my rather trivial results on this problem at http://bhargav.com/bhargav/neumann.pdf where, among other things, there is a proof (different from yours) for the case that Q(a+b) is Galois over Q)

Cheers,
Bhargav

You are right. My argument is flawed in the case $\mathbb{Q}(a+b)/\mathbb{Q}$ is not a Galois extention.

Boris

There is a short elementary proof in the paper reviewed below.

MR0432581 (55 #5568) 10L05
Browkin, J.; Divis, B.; Schinzel, A.
Addition of sequences in general fields.
Monatsh. Math. 82 (1976), no. 4, 261--268.
http://www.digizeitschriften.de/home/services/pdfterms/?ID=502041
http://www.springerlink.com/index/N318706143115X62.pdf

The authors prove that if $ L$ is a field of characteristic $ p$,
$ X\subset L$, $ Y\subset L$, $ \text{card} X=m$, $ \text{card} Y=n$ and
either $ p=0$ or $ p > \min\{2^{m+n-2},\,m^{n-1},n^{m-1}\},$
then there exists a $ z\in L$ represented uniquely as $ x+y$ with $ x\in X$, $ y\in Y$.
(This result is connected with field extensions $ K(\alpha,\beta)$ such that $ K(\alpha,\beta)=K(\alpha+\beta)$.)
Furthermore, they prove that if $ L$ is a field of characteristic $ p$, $ X\subset L$, $ Y=±X$,
$ \text{card} X=n$ and $ p>2^{n-1}$, then there exists a $ z\in L$
represented uniquely as $ x+y$, $ x\in X$, $ y\in Y$. Let $ f(p)$ denote
the greatest number such that in any set of residues
$ {a_1,\cdots,a_n}$ modulo $ p$ ($ p$ prime) with $ n\leq f(p)$ one of the
differences $ a_i-a_j$ occurs only once. E. G. Straus proved that
$ f(p)\geq 1+log(p-1)/\log 4$ [J. Number Theory 8 (1976), 40--42;
MR0392876 (52 #13689)]. The latter result of the present paper stated
above improves on this estimate since it yields that $ f(p)\geq\log
p/\log 2$.
On the other hand, Schinzel proved that $ f(p)\leq 2+o(1)/\log 3\log p$
(``An inequality for determinants with real entries'', to appear in Colloq. Math.);
this fact shows that the authors' results are not far from the best possible.

Reviewed by Andras Sarkozi


function
number theory solved
number theory

f is a real-valued function on the reals such that | f(x) | <= 1 and f(x + 13/42) + f(x) = f(x + 1/6) + f(x + 1/7) for all x. Show that there is a real number c > 0 such that f(x + c) = f(x) for all x.

a solution at:

http://www.kalva.demon.co.uk/short/soln/sh967.html


Kimler oldugunu bilmiyorum, ama sizler icin iyiyi haber var :
Turkiye takimi 193 punala JBMO( Junior Balkan Math Olympiad) ucuncusu oldu. Ve full ceken ogrencide vardi, yani 40 puan.

Saygilarimla
Davron Latipov

3. olmak bizim için çok iyi bir haber olmasa gerek ama full çeken tek kişinin turkiye den olmasi guzel bir haber.


function
real analysis
real analysis unsolved

Let $\displaystyle f : \mathbb R \to \mathbb R$ be a continuous function. For all $\displaystyle a,b \in \mathbb{R}$, $\displaystyle a<b$, there is $\displaystyle c_1,c_2 \in \left[ a,b \right]$, $\displaystyle c_1 \leq c_2$, s.t. $\displaystyle f \left( c_1 \right) = \min_{x \in \left[ a,b \right]} f \left( x \right)$ and $\displaystyle f \left( c_2 \right) = \max_{x \in \left[ a,b \right]} f \left( x \right)$.

Prove that $\displaystyle f$ is increasing.

If not,
there exist $x_1<x_2$,such that $f(x_1)>f(x_2)$
Let $f(\xi)=M=\max_{x \in [x_1,x_2]}f(x)$ then $\xi<x_2$
let $a=\xi,b=x_2$
we find find a contradiction.
I hope it is ok :)

If $f([a,b])=[f(c_1),f(c_2)] \forall a<b$ can we say that $f$is injective?

No, we can't. This function can be constant on intervals.

The appropriate correction to zhaobin's solution: Let $\xi=\max\{x: f(x)=M,x_1\le x\le x_2\}$. The set $f^{-1}(M)\cap [x_1,x_2]$ is closed (hence compact), so it has a maximum.


a cube has a diagonal of (3*3^0.5). find the length of a side of the cube.

Do you mean space diagonal?

[hide]$\frac{3\sqrt3}{\sqrt3}=3$ [/hide]

yes, i mean space diagonal. i got 3 as my answer.
[hide]
assume the side of the cube is 3. then, the diagonal of each face of the cube would be (3*2^0.5). the space diagonal would then be the square root of 3^2+(3*2^0/5)^2=27. the square root of 27 is (3*3^0.5)

please correct me if i'm wrong[/hide]

i misread the problem, you're right  :blush:
I have edited the post. Do you know latex?

[quote="h78reg"]no i don't know latex. sorry if that made the problem look unclear. :blush:[/quote]

It was just my eyes from being on AoPS for 236! hours. As you get into higher levels of forums, latex will more necessary. You might want to learn it sometime (there's a Latex link on the toolbar).


Find the no. of ways in which [b]511[/b] could be expressed as a product of 3 factors.

[quote="M@re"]I think that it is ${11+3-1\choose 3}={13\choose 3}=286$

Am I correct??[/quote]

I don't have the correct ans. with me right now..In fact I also took an approach similar to urs..but then realised that this approach would have repititions..e.g. (1,1,511),(1,511,1),(511,1,1) are all the same..

so the actual no. of ways would be much less

Well, i'll have to think a little more about this, then i will anwser, but for now i don't hink that I am wrong  :D  :D

As you said I was wrong. This is new solution. Answer $15$. Let $(x\geq y\geq z)$
denote powers of $5$ in that expression so this are
possibilities:\[(11,0,0), (10,1,0), (9,2,0), (9,1,1), (8,3,0), (8,2,1), (7,4,0)\]
\[(7,3,1), (7,2,2), (6,5,0), (6,4,1), (6,3,2), (5,5,1), (5,4,2), (5,3,3), (4,4,3)\]

I think that is all :D :D :D

Now if you want, you can try same problem for $5^n$, I will do that as soon as i
disconnect, and I will post it here too.

[quote="M@re"]As you said I was wrong. This is new solution. Answer $15$. Let $(x\geq y\geq z)$
denote powers of $5$ in that expression so this are
possibilities:\[(11,0,0), (10,1,0), (9,2,0), (9,1,1), (8,3,0), (8,2,1), (7,4,0)\]
\[(7,3,1), (7,2,2), (6,5,0), (6,4,1), (6,3,2), (5,5,1), (5,4,2), (5,3,3)\]

I think that is all :D :D :D

Now if you want, you can try same problem for $5^n$, I will do that as soon as i
disconnect, and I will post it here too.[/quote]

Could there be any pattern cause if we wrote down the number of ways which can start with 11 to 1 , which i predict might be

$1,1,2,2,3,3,3,4,4,4,4$ . Adding up to get $32$ and divide by 2 since it repeats .  :?  :?

I've heard of this way though -

Find how many way you can put two lines between one of two fives in here -

 5 5 5 5 5 5 5 5 5 5 5

For example, you could have  ||5 5 5 5 5 5 5 5 5 5 5, or  5|5 5 5 5 5|5 5 5 5 5.  Would that work for this problem?

No it can't work, I and someone beside me said it before, in that case you have repetition

[quote="K81o7"]Yeah, then this can be solved via generating sequence...[/quote]

I relle don't know generating functions and recurent equations but I don't think that this can be solved via that because look here http://www.mathlinks.ro/Forum/viewtopic.php?t=48438 general formula is just to wierd   ;)  ;)

Well, if permutation mattered, it would be simple, but otherwise, there's just one more step which needs to be added.  Essentially, you're trying to find the number of triplets of numbers whose sum is 11, using integers from 0 to 11.  But suppose you just try to find the coefficient of $x^{11}$ in $(1+x+x^2+x^3+...)^3$?  That's the idea of generating sequences.  In problems where permutation does mater, it makes things a whole lot simpler: in this case, the answer would just be the 12th triangular number, 78.  Then, with permutation, you just have to find the number which are two different numbers and subtract that number twice (because each triplet of the form (a, b, b) has 3 permutations).  There are 6 or those (we have to find the number of possible b: you can have 0, 1, 2, 3, 4, 5).  So you know that there are 18 which are actually just 6, each repeated 3 times.  Then, you know the remaining ones are of the form (a, b, c) with three different numbers.  There are 60 more, and each is counted 6 times.  Therefore, you get 10 in the end.  Thus, the final answer is 6+10=16.  Pretty simple method without tedious calculations.  Something like this would be very powerful if you needed to find, say, quadruplet factors of something like $5^{96}$...

H_s_potter's way is also a very good way for which permutation matters.  That's a very good method for a problem like the following...
There is a jelly bean company.  They produce jelly bean jars which hold 1000 jelly beans each, and each jelly bean is one of five colors.  They claim that "No jar of jelly beans is the same".  How many possible jars can be made?
You can just think of 1004 slots, and 4 pegs.  You have to place the 4 pegs into four slots, and the five regions created represent the different colors, and each slot represents a bean...

[quote="santosh_blr"][quote="M@re"]As you said I was wrong. This is new solution. Answer $15$. Let $(x\geq y\geq z)$
denote powers of $5$ in that expression so this are
possibilities:\[(11,0,0), (10,1,0), (9,2,0), (9,1,1), (8,3,0), (8,2,1), (7,4,0)\]
\[(7,3,1), (7,2,2), (6,5,0), (6,4,1), (6,3,2), (5,5,1), (5,4,2), (5,3,3)\]

I think that is all :D :D :D

Now if you want, you can try same problem for $5^n$, I will do that as soon as i
disconnect, and I will post it here too.[/quote]

Man..that's perfect except the fact that u missed one triplet (4,4,3)  :idea:[/quote]

They already said that to me  :D  :D  :D  OK i will edit it now  :roll:  :roll:

[quote="K81o7"]  Pretty simple method without tedious calculations.  Something like this would be very powerful if you needed to find, say, quadruplet factors of something like $5^{96}$...
[/quote]

I already posted the solution for $5^n$ in topic "triplets factors... generalized$ so if you interested go there, you have link in some of my earlyer posts in this topic


Dar mosallase haddato zavieye  ABC  ke G markaze seghl an ast noghteye P ra tori biabid ke       "  AP.AG + BP.BG + CP.CG = Min

$\sum \overrightarrow {GA}=\overrightarrow 0\ ;\ \sum PA\cdot GA\ge \sum \overrightarrow {PA}\cdot \overrightarrow {GA}\ .$
$P(x,y,z)\ ,\ x+y+z=1\Longrightarrow x\overrightarrow {PA}+y\overrightarrow {PB}+z\overrightarrow {PC}=\overrightarrow 0\ .$ $\overrightarrow {PA}\cdot \overrightarrow {GA}=\left(\overrightarrow {PM}+\overrightarrow {MA}\right)\cdot \overrightarrow {GA}=\overrightarrow {PM}\cdot \overrightarrow {GA}+\frac 23m^2_a\ : \ \ (y+z)\overrightarrow {PM}=y\overrightarrow {PB}+z\overrightarrow {PC}$ a.s.o.

wow, mr virjil nicula az koja farsi baladeh bekhooneh?

means:how can mr virgil nicula read persian.???

[quote="Virgil Nicula"]The mathematics is a universal Language !
\[ \mathrm {U.M.L.\ \equiv\ Universal\ Mathematical\ Language}  \][/quote]

ofcourse it is...

but the writer of this problem wrote  persian...

i wonderd because of that. ;)

Ashegh joon Mr virgil faghat $AP.AG + BP.BG + CP.CG = \text{Min}$  fahmide.

hamin ham vase fahmidan soal kafie. ;)

areh valla....

akhe in dafeye avvale ishoon nist. chon to community e china ham ye hamchin kari kardan. :P


In our number system the base is ten. If the base were changed to four you would count as follows: $ 1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be:

$ \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$

Obviously the answer is not 20 (choice A). In base 4 the digits are only 0,1,2,3, so B,C, and D are wrong. Therefore by elimination the answer is E.

[quote="grn_trtle"][hide="...without elimination  :D "]$ 20_{10}\equiv 4\pmod{16}\equiv 0\pmod{4}$, so it must be $ 110_4\implies\boxed{\textbf{(E)}}$.

Sorry if this is unclear, it's kind of weird to describe my method for converting bases  :|
Yes, I realize I have weird modular arithmetic notation.
[/hide][/quote]We could say that the expanded form of 20 is
$ 20\equal{}1 \cdot 4^2 \plus{} 1 \cdot 4^1 \plus{} 0 \cdot 4^0$ and that the numbers multiplied by the powers of the base $ (4$ in this case$ )$ in such an expansion are the digits of the number in that base. It does not sound much clearer, though.

OK, I'll attempt to explain it more...
Basically, try to find the largest power of $ 4$ that's less than $ 20$, which is $ 2$.
Then find the remainder, which is $ 20\minus{}16\equal{}4$.
Because of that $ 4^2$, the number must have a $ 1$ in the $ 4^2$ place.
Similarly, we find that $ 4^1\equal{}4$, so there's a $ 1$ in the $ 4^1$ place.
We've broken down $ 20$ as much as possible now, so the number is $ 110$.