limit
real analysis
real analysis solved

a) Let $(x_n)_{n\ge0}$ and $(y_n)_{n\ge0}$ be two sequences of real numbers, with $y_n=\displaystyle\frac1{2^n}\sum_{k=0}^nC_n^kx_k,\forall n\in\mathbb N$. Prove that $(x_n)_{n\ge0}$ is convergent [b]iff[/b] $(y_n)_{n\ge0}$ is convergent.
b)Compute $\displaystyle\lim_{n\to\infty}\frac n{2^n}\sum_{k=1}^nC_{n-1}^{k-1}\left\{\sqrt{k^2+2k+2}\right\}$ where $\{x\}$ denotes the fractional part of $x$.

[i]Octavian Stroe, Paul Valceanu[/i]

a/ if $x_n$ then $y_n$  converge it is an application of Cesaro theorem

If $y_n$  converge this not give $x_n$ converge

Consider $x_k=(-1)^k$

$y_n=0$ converge but $x_n=(-1)^n$ does not converge

ohh... I'm sorry... I'm dumb  :blush:
if $x_n$ is convergent then $y_n$ is convergent and their limits are equal.

For a/ just $x_n \;converge \Rightarrow y_n \; converge$ is true

there exist $n_0$ s.t. for any $n\geq n_0$  $ |x_n-x|\leq \epsilon$

$|y_n-x|\leq \frac{\sum_{k=0}^{n_0}|x_k-x|C_n^k}{2^n}+\frac{\sum_{k=n_0+1}^{n}|x_k-x|C_n^k}{2^n}$

The fisrt term on RHS $\leq \frac{max_{0\leq k \leq n_0}|x_k-x|}{2^n}(1+n+n^2/2!+...+n^{n_0}/n_0!)$ wich tends to $0$

The second term on RHS $\leq \epsilon$

$y_n$ tend to $x$

Cerinta corecta a fost; "Aratati ca daca sirul Xn este convergent atunci si sirul Yn este convergent...."

[quote="Kuba"]
b)Compute $\displaystyle\lim_{n\to\infty}\frac n{2^n}\sum_{k=1}^nC_{n-1}^{k-1}\left\{\sqrt{k^2+2k+2}\right\}$ where $\{x\}$ denotes the fractional part of $x$.

[i]Octavian Stroe, Paul Valceanu[/i][/quote]

Use $kC_n^k=nC_{n-1}^{k-1}$ you get

$x_k=k\left(\sqrt{k^2+2k+2}-E(\sqrt{k^2+2k+2}\right)$

Now what is  $\lim x_k$ ?


Η ακολουθία ορίζεται ως εξής :
         α(0)=0, α(1)=1, α(2)=2, α(3)=6 και
    α(n+4)=2α(n+3)+α(n+2)-2α(n+1)-α(n) , n>=0.
   Να δειξετε οτι το n διαιρει το α(n) για καθε n>0.

kali Nick... :wink:

hint
[hide]eukola pernume oti $a_{x}=\frac{x}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^{x}-\frac{x}{\sqrt{5}}(\frac{1-\sqrt{5}}{2})^{x}$
to opio ine akereo polaplasio tu $x$[/hide]

 :D

πολυ ωραια Ράλλη! ειναι δηλ α(n)=nf(n) opoy f(n) ο νιοστος αριθμος fibonacci!
  Βρηκα κ μια ωραια θεωρια αριθμων χθες .
  Να βρειτε ολους τους πρώτους αριθμούς p, q ωστε το pq να διαιρει το
  2^p +2^q.

Νίκο γιατί δε βάζεις λέω αυτές τις ωραίες ασκήσεις με τις λύσεις τους στα αντίστοιχα τόπικς του βιβλίου? :wink:

Δεν έχω παρακολουθήσει από την αρχή αυτή την ωραία προσπάθεια που γίνεται και δεν σκέφτηκα να τη βάλω εκεί .
Την επόμενη θα τη βάλω εκεί .

[quote="nikos"]
  Βρηκα κ μια ωραια θεωρια αριθμων χθες .
  Να βρειτε ολους τους πρώτους αριθμούς p, q ωστε το pq να διαιρει το
  2^p +2^q.[/quote]
Elpizw na einai safes auto pou akolouthei(k swsto :D )

An $q=2$ tote $2p/ 4+2^{p}$ dld $p/4+2^{p}$.  $2^{p}\equiv2 \mod p$. Epomenws $p/6$ kai $p=2,3$. Elegxontas pairnoume ta zeugh $(2,2)$ kai $(3,2)$.
Estw $(p,2)=1$ kai $(q,2)=1$.

An $q=3$ pairnoume $3p/8+2^{p}$ 'h $3/8+2^{p}$ pou tha simaine $p$ artios, atopo. Ara $q\neq 3$ k profanws k $p\neq 3$.

$2^{p}\equiv 2\mod p$ kai $2^{q}\equiv 2 \mod q$ kai etsi eukola katalhgoume stis $2^{p-1}\equiv-1\mod q$ kai $2^{q-1}\equiv-1\mod p$.

Estw $r,z$ oi elaxistoi thetikoi arithmoi tetoioi wste $2^{r}\equiv 1 \mod q$, $2^{z}\equiv-1 \mod q$. Ws gnwston, an $2^{f}\equiv 1 \mod q$, tote $r/f$. Parathroume oti $2^{2z}\equiv 1 \mod q$ k epomenws $r/2z$ pou shmainei $2z\geq r$. Estw $p-1=rk+l$, $0\leq l< r$. Tote

an $l=0$ exoume $-1\equiv 2^{p-1}\equiv 2^{rk-1}\equiv 2^{r(k-1)+r-1}\equiv 2^{r-1}\mod q$ kai epomenws afou $2^{r}\equiv 1 \mod q$ pairnoume $3\equiv 0 \mod q$, dld $q=3$ pou exoume apokleisei.

ara $l\neq 0$ kai epomenws $-1\equiv2^{rk+l-1}\equiv 2^{l-1}\mod q$ kai epomenws $z\leq l-1<r-1$. Epomenws $z+1<r\leq2z$ pou se sunduasmo me $r/2z$ dinei $r=2z$.

Twra eimaste etoimoi na deiksoume oti an $2^{m}\equiv-1 \mod q$ tote $z/m$. Pros touto thewroume oti $m=dz+j$ me $0\leq j<z$. Tote $-1\equiv 2^{m}\equiv 2^{dz+j}\equiv (-1)^{d}\cdot 2^{j}\mod q$.  Omws $j<z<r$ kai epomenws h teleutaia dn ginetai na isxuei ektos an $j=0$ k $d$ perittos.

Xrhsimopoiwntas to parapanw gia $m=p-1$ pairnoume $p-1=z(2c+1)$(1).

An orisoume omoia ta $r_{1},z_{1}$ $\mod p$ (antistoixa pros ta $r,z$) tote epanalambanontas tin diadikasia pairnoume $q-1=z_{1}(2v+1)$(2) me $r_{1}=2z_{1}$(3). Shmeiwteon oti $r=2z$(4) k epeidh $2^{p-1}\equiv 1 \mod p$, $r_{1}/p-1$(5) kai omoia $r/q-1$(6).

$2/p-1$, apo thn (1) $2/z$, apo thn (4) $4/r$, apo thn (6) $4/q-1$, apo thn (2) $4/z_{1}$, apo thn (3) $8/r_{1}$, apo thn (5) $8/p-1$. Sunexizontas auth th diadikasia mporoume na briskoume osodhpote megales dunameis tou $2$ pou na diairoun to $p-1$(k to $q-1$). Atopo.

Telika oi mones luseis einai oi $(2,2), (2,3), (3,2)$. :ninja:

Και σαφης και εννοειται σωστη η λυση σου Galeo!


I need to prove that any 3 points on the curve f(t)=(t,t^2,t^3) form a plane, rather than a line. I can't find a good way to start. Any help would be appreciated!

Suppose that the points $ (s, s^2, s^3), (t, t^2, t^3)$ and $ (u, u^2, u^3)$ are colinear.  Then ....

and the same thing works in $ \mathbb{R}^n$: this is a Vandermonde determinant


function
inequalities
real analysis
real analysis unsolved

Hey there,
I am trying to solve the following continuity problem...
--------------------------------------------------------------------------------------
f,g:[a,b]->R
f,g are continous functions such that for each x in [a,b], f(x)<g(x).
Prove that exists c>0 such that for each x in [a,b] f(x)+c < g(x)
--------------------------------------------------------------------------------------
I have tried:
h(x) = g(x) - f(x), and so h(x) is continues and h(x) > 0.
Then probably some variation of the intermediate value theorem should be used, but I am not sure in what way?

Thanks.

I guess that you shoud think about uniform continuity of f and g.

hello. Yes, I've tried with uniform continuity. And here is the proof:

$ \ h(x)$ is uniformly continous. (By Cantor theorem).

so $ \forall \epsilon > 0 , \exists \delta > 0 : |h(x) \minus{} h(y)| < \epsilon \ , \ when \ |x \minus{} y| < \delta , \ \ \ \forall x,y \in [a;b]$

$ \ h(y) \ \minus{} \ \epsilon \ < \ h(x) \ \ \forall \ x,y \in [a;b]$

$ \ h(y) \ \minus{} \ \epsilon \ > \ 0 \ because \ h(x) \ > \ 0$

$ \ (g(x) \ \minus{} \ f(x))$ < $ \ (h(y) \ \minus{} \ \epsilon) \ > \ 0$

so, $ \ \exists \epsilon \ > 0 \ such \ : \ f(x) \ \minus{} \ g(x) \ < \ (\epsilon \ \minus{} \ h(y) ): \equal{} \ \minus{} c \ ; \  c: \equal{} \ h(y) \ \minus{} \epsilon \ , \ \ \ \forall \ x,y\in[a;b]$ . h(y) > 0, and $ \epsilon > 0$ is "as small as you want", so we took such $ \epsilon > 0$ that $ \ c > 0$.

Hm... there might be a few errors in Blymblams post, but the idea is there.Denote $ h(x)\equal{}g(x)\minus{}f(x)$, let's take some $ \varepsilon>0$ we can find $ \delta>0$ that for all $ x,y\in [a,b]$ such that $ |x\minus{}y|<\delta$ we have $ |h(x)\minus{}h(y)|<\varepsilon$. From that we have $ 0<h(y)<h(x)\plus{}\varepsilon$, that is $ 0 <h(y)\minus{}\varepsilon<h(x)\equal{}g(x)\minus{}f(x)$. We can now make a partition of $ [a,b]$, $ a\equal{}x_1<\dots<x_n\equal{}b$ such that $ x_i\minus{}x_{i\minus{}1}<\delta$. Notice that for $ x,y\in [x_{i\minus{}1},x_i]$ we have $ 0<h(y)\minus{}\varepsilon<h(x)\equal{}g(x)\minus{}f(x)$. Now $ h(y)\minus{}\varepsilon$ is continous on $ [x_{i\minus{}1},x_i],  \forall i$ and strictly greater then zero, so it has a minimum ( Weierstrass theorem ) that is greater then zero. On $ [x_{i\minus{}1},x_i]$ we denote that minimum $ c_i$. Now take the smallest of all that $ c_i$ and that is the desired constant.
I hope everything is ok now  :lol: .

hello

what is the purpose of constructing the interval: $ \ [x_{i\minus{}1} \ ; \ x_i]$ ? It's because $ \ h(x) \ > \ h(y) \ \minus{} \ \epsilon$ and $ \ h(y) \ \minus{} \ \epsilon$ wouldnt be continous in $ \ [a;b]$ ? sorry if i dismissing smth :)

[quote="Blymblam"]
what is the purpose of constructing the interval: $ \ [x_{i \minus{} 1} \ ; \ x_i]$ ? It's because $ \ h(x) \ > \ h(y) \ \minus{} \ \epsilon$ [/quote]

Yes, on $ [x_{i\minus{}1},x_i]$ we have the desired inequality. Notice, that this inequality doesn't hold on all $ [a,b]$, only on intervals that have length smaller then $ \delta$. ( that was the mistake in your solution, among some other... ).

[quote="Blymblam"]
and $ h(y)\minus{}\varepsilon$ wouldnt be continous in  ?[/quote]

$ h$ is continous on $ [a,b]$ and adding/subtracting a constant still makes it continous.


:? Can you solve this equation
sin(sin(sin(sin(x))))=cos(cos(cos(cos(x))))
Is it nice??? :(  :blush:  :P  :rotfl:

Interesting, $\sin (\sin \cdots( \sin(x))\cdots ) - \cos (\cos \cdots( \cos(x))\cdots )$ seems to be always negative if the number of iterations is large enough. I wonder how many iterations do they need to have?

[quote="phuchung"]
Is it nice??? [/quote]

Hmmm...I think that the nicest part is the so precise reference you gave... ;)

Btw, you may have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sin%28sin&t=10285

Pierre.


Aqui les dejo dos problemas que pusieron en los entrenamientos de YUCATAN:
el entrenador nos dijo que era una APMO, y tambien les dejo el otro el cual lo saco de un libro de teoria de numeros del cual no se su nombre, en el inciso b del problema 2 todavia no lo eh resuelto a si que si tienen alguna solucion que la pongan para poder verla:

1)determine todos los enteros positivos n para los que la ecuacion:

                                                   xn + (2+x)n + (2-x)n = 0

tiene una solucion entera.
los n despues de los numeros puestos significan que estan elevados a la potencia n.

2)(a)Pruebe que cada elemento del conjunto:

                                                   (n! + 2, n! + 3, ... , n! + n)

es divisible por un primo que no divide a ningún otro elemento del conjunto.

b)Pruebe que para todos los enteros positivos n y k, existe un conjunto de n enteros consecutivos tal que cada uno de los elementos de este conjunto es divisible por k primos distintos ninguno de los cuales divide a los otros elementos del conjunto.

[quote="Tony2006"]
1)determine todos los enteros positivos n para los que la ecuacion:

                                                   xn + (2+x)n + (2-x)n = 0

tiene una solucion entera.
[/quote]

Lo paso: $x^n+(2+x)^n+(2-x)^n=0$

[quote="Tony2006"]si es eso ya lo hiciste? :lol:[/quote]

No. Solamente lo pasaba para que lo entiendan mejor.. alguna ayuda?

para el problema 1 de las soluciones que se la que mas me gusto fue una que fue por congruencias, las otras: por polinomios y otra por el teorema del binomio de newton.
una sugerencia para congruencias es ver que el numero x es de la forma 2 ala "a" para "a" entero no negativo
donde 2 ala "a" divide a 2 ala "n+1"

para el problema 2 inciso a la sugerencia es que factorizen n! + k para k entero entre 2 y n.
y para el inciso b no lo se por que todavia no lo eh hecho si alguien tiene alguna idea que lo comente a qui por favor!! :lol:


Dear Math Enthusiast,

     We at National Assessment & Testing are pleased to announce our 2006-2007 competition schedule (appended).  The five contests we administer cost just fifty dollars each (or you can register for all five and receive the package price of two-hundred dollars) and can be more convenient than local face-to-face contests, requiring only two hours of your coach's time during the week.  Our contests are easy to administer, and frequently a contest and its solutions provide material for two team meetings, reducing your coach's need to create original materials.  Each of our competitions feature problems that range from simple to advanced, to provide both confidence and challenges to all students regardless of their level of expertise.  Our contests can also be rewarding, with high-scoring individuals in multiple divisions receiving plaques and certificates, high-scoring schools receiving awards and recognition through press releases to local newspapers, and the highest scoring individual at each school receiving a certificate.  In addition to the contests we administer, we also produce a competition that your team can administer for local elementary and middle school students as both a service project and fundraiser.

     If these sound like the kind of extracurricular mathematical experiences you might like to experience, ask your coach to look over the appended schedule.  If they have any questions, they can contact me at clymer@natassessment.com or (206) 433-7320.  If your school would like to register for one or more of our competitions, your coach can simply e-mail us with the contests you'd like to register for and your school's registration information.

Thank you,

Tom Clymer
Director of Academic Competitions
National Assessment & Testing
http://www.natassessment.com

2006-2007 Competition Schedule

Thursday, September 28th, 2006:  2006 Fall Startup Event

     This is a fun, fast-paced competition in which individuals have 30 minutes to solve 100 problems.  The questions cover a wide variety of topics, and are all quick and easy for students familiar with the subject.  Analogous to a musician practicing scales, this contest can help students and teams identify areas they want to focus on.

"Excellent time at the start of the year.  Excellent opportunity to interest new ninth-graders in math team - it's the competition that draws them in." - Jane Fenn, Corning-Painted Post West High School

Thursday, November 2nd, 2006:  2006 Team Scramble

     In this exciting collaborative experience, many students work together to answer 100 questions in thirty minutes.  Team leaders who understand the strengths of various team members will be able to divide the work to efficiently answer the most questions, while recruiting many students allows for more double-checking of the team's work.

"Kids went crazy when we gave them the solution key after the test." - Jim Ferguson, Dana Hills High School

Thursday, December 7th, 2006:  2006 Ciphering Time Trials

     This individual competition consists of ten rounds in which students have three minutes to solve three problems.  Students must decide which problems they will be able to answer quickly, and whether to attempt less familiar problems or double-check questions with which they are more comfortable.

"Kids loved the fast pace!" - Patrick Montague, Science Academy of South Texas

Thursday, February 1st, 2007:  2007 Four-by-Four Competition

     This is a team contest with ten rounds in which teams of four students have three minutes to solve four problems.  Students learn to work together and quickly distribute problems to team members with strengths in the corresponding subjects.

"It's fun to have a creative format like this." - Sue Forster, Bismarck High School

Wednesday, March 28th, 2007 - April 4th, 2007:  2007 Collaborative Problem-Solving Contest

     This competition consists of fifteen intricate multi-part questions that your entire school (including staff) can collaborate on for an entire week.  Students get experience using real-world problem-solving skills such as brainstorming, collaboration, research, and application of technology.

"We had a great time with this contest - we can't wait until next year!  With better planning, we hope to make this an even bigger event at our school.  As it was, I had to chase kids home tonight." - Tim Corica, The Peddie School

2006 & 2007 Middlementary Math Bonanzas

     We provide everything your high school math team needs to administer a face-to-face competition for your local elementary and middle school students!  Administering a math contest is a terrific experience for your high school students, a great fundraiser for your team, a good service to provide to your community, and an investment in the students that will make up your team in the future.  You can currently register for the 2006 and 2007 versions of this contest at a cost of two-hundred dollars each.

The Fall Start Up really drew me into math team.

I look forward to Time Trials!


Find the integer numbers $x,y,z,t$ which satisfy $x+y+z=t^2$, $x^2+y^2+z^2=t^3$.

[i]Lucian Dragomir[/i]

by cauchy

$t^3=x^2+y^2+z^2\geq\frac{1}{3}(|x|+|y|+|z|)^2\geq\frac{1}{3}(|x+y+z|)^2=\frac{t^4}{3}$

So $t^3(t-3)\leq 0$ or $0\leq t\leq 3$ .

Now the rest is easy by case work  :)


If the sum of two numbers is 8 and the sum of their cubes is 20, what is the sum of their squares?

You might want to check that third step.

[hide="Solution"] A slight re-shuffling of my hint.

$(a+b)^{3}= 512 = (a^{3}+b^{3})+3ab(a+b) = 20+24ab \implies ab = \frac{41}{2}$

Then

$a^{2}+b^{2}= (a+b)^{2}-2ab = 64-41 = \boxed{ 23 }$. [/hide]


A topological space is called normal if every point in it is a closed set, and further, given disjoint closed sets A and B there are disjoint open sets U and V such that U contains A and V contains B
A topological space is termed metrizable if there exists a metric such that the topology arises as the topology of the metric
[b]Problem:[/b] Prove that any metrizable space is normal.

Do the obvious!  Let $ A$ and $ B$ be closed disjoint sets.  For each $ a \in A$, there exists $ \epsilon_a$ so that $ B(a, \epsilon_a) \cap B \equal{} \emptyset$.  Similarly, for each $ b$ there exists $ \epsilon_b$ so that the corresponding ball misses
$ A$.

Let $ U \equal{} \cup_{a \in A} B(a, \epsilon_a /2)$ and $ V\equal{} \cup_{b \in B} B(b, \epsilon_b /2)$.

Clearly, $ U$ and $ V$ are neighborhoods of $ A$ and $ B$ respectively.  Show that they are disjoint.

Standard proof of that fact  bases on $ f(x)\equal{}\frac{dist(x,A)}{dist(x,A)\plus{}dist(x,B)}$. It is continuous, so $ f^{\minus{}1}(\minus{}\infty,\frac{1}{2})$, $ f^{\minus{}1}(\frac{1}{2},\plus{}\infty)$ are open and seperates given sets $ A$, $ B$.


Let $Squish(n)$ be a discrete function such that $Squish(n) = n$ for all $n$ with no digit of $0$ and for $n$ with one or more digit of $0$, $Squish(n)$ is $n$ with all the digits of $0$ removed.  For example: $Squish(919) = 919$ but $Squish(909) = 99$.  Let $k$ be the smallest possible positive integer so that $Squish(k) | k$ but $Squish(k) \ne k$ and th units digit of $k$ is not $0$.  Compute $k+Squish(k)$.

[hide="Solution"]Obviously, $k=10$. Thus, the answer is $\boxed{11}$.[/hide]

[hide] Clever problem, hopefully I did this right. I'm just going to assume the number only has a single zero in it and since I get a small enough number that's good enough for me.

Write $k$ as $10^{k+1}\cdot a+b$, where $b < 10^{k}$ and doesn't end in a zero. This represents where the zero would be. For example, $123037 = 10^{3}\cdot 123+37$. Then we have

$(10^{k}\cdot a+b) | (10^{k+1}\cdot a+b)$,

or equivalently

$10^{k+1}\cdot a+b = m(10^{k}a+b) \Rightarrow (10-m) \cdot 10^{k}\cdot a = (m-1) \cdot b$.

The best way I can finish it from here is simply to check $m = 2, 3, \ldots, 9$ which gives me $k = 105$ when $m = 7$.[/hide]

So I approached it in a slightly less generic way. Basically I

[hide="like cheese"]
said that $11|1001$, so either:
The answer is $1001+11=1012$

or

There is a 3-digit solution (it clearly can't be 2-digits).

Looking specifically for a 3-digit solution is much easier because you have $\overline{a0b}$ and $\overline{ab}$, and you have $\overline{ab}| \overline{a0b}-\overline{ab}= 90a$... or something to that effect.
[/hide]

A similar question:
Compute the number of positive integers $n < 1000$ such that $Squish(n) | n$ and $Squish(n) \ne n$.

[hide]Case 1: Three-digit number, just one zero in the tens place
We have $10a+b|100a+b$, or $90a\equiv 0\pmod{10a+b}$, for digits $a$ and $b$, $a,b\neq 0$.
Now, it's just counting.  As $a$ goes from 1 to 9, we must count the number of factors of $90a$ with a tens digit of $a$.
$a=1$: 15 and 18
$a=2$: None exist
$a=3$: None exist
$a=4$: 45
$a=5$: None exist
$a=6$: None exist
$a=7$: None exist
$a=8$: None exist
$a=9$: None exist
So there are three.

Case 2: Three-digit number, zero in the ones place
Any number with a zero will work.  $\text{Squish}(n)$ will either be $\frac{n}{10}$ or $\frac{n}{100}$, depending on the tens digit.   So there are 90 possible numbers.

Case 3: Two-digit number, zero in the tens place
Any number will work, as $\text{Squish}(n)$ would be $\frac{n}{10}$.  So there are nine.

There can't be any one-digit solutions, as this would violate the rule that $\text{Squish}(n)\neq n$.

So there are $90+9+3$, or $\boxed{102}$, numbers in all.[/hide]


geometry
geometric transformation
reflection
trapezoid
geometry proposed

[color=darkblue][b][size=150]A very nice construct problem ![/size][/b][/color]
[quote][color=darkred][b]Are given a triangle $ABC$ and a point $D\in (BC)$. Construct the points $E\in (CA)$ and $F\in (AB)$ so that
the middlepoints of the segments $[DE]$ and $[DF]$ are collinearly with the vertex $C$ and
the middlepoints of the segments $[DE]$ and $[EF]$ are collinearly with the vertex $A$.[/b][/color][/quote]
[color=darkblue][b][u]The proper construction.[/u][/b]

[b]1.[/b] Construct the reflection $X$ of the point $D$ w.r.t. the point $C$.
[b]2.[/b] Construct the point $Y\in AB$ for which $DY\parallel AC$.
[b]3.[/b] Construct the point $E\in CA$ so that $YE\parallel BC$.
[b]4.[/b] Construct the point $F\in XE\cap AB$.

[b][u]Remark.[/u][/b] The positions of the points $E\in (CA)$, $F\in (AB)$ are defined by the following relations :
$\boxed{\frac{EA}{EC}=\frac{DC}{DB}}$ and $\boxed{\frac{AF}{AB}=\left(\frac{CD}{CB}\right)^{2}}$. Denote the reflection $Z$ of the point $Y$ w.r.t. the point $A$.
The above second relation applied in the trapezoid $DYZX$, where $AC$ is the middle line, i.e. $\frac{AF}{AB}=\left(\frac{YF}{YB}\right)^{2}$,

shows that [b]the point $F$ is the conjugate of the point $B$ w.r.t. the pair $\{Y,Z\}$[/b], i.e. $FI\parallel AC$, where $I\in XY\cap DZ$.[/color]

Nice solution, Virgil Nicula! :lol:

My solution is even more simple:
[quote]The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$[/quote]
Let $K,L$ and $M$ be the midpoints of $DE, EF$ and $DF$ and let $BM$ intersects $AC$ at point $G.$ Then $BG\parallel EF,$ triangles $BCG$ and $ECF$ are similar and $KC$ is a median of triangle $BCG.$ Therefore $K$ is common midpoint of $DE$ and $BG,$ i.e. $BEGD$ is parallelogram.
[i]Construction:[/i] draw $DG\parallel BC,$ then $GE\parallel AB$ and $EF\parallel BG.$
[i]Proof:[/i]By the construction the intersection point of $BG$ and $DE$ is the midpoint of $DE.$ Since $EF\parallel BG$ the intersection point of $KC$ and $EF$ is the midpoint of $EF.$ Finally, the intersection point of $BG$ and $DF$ is the midpoint of $DF.$
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Post any mathematical paradoxes you know here!
(and explain why if needed)

I know one is: what is 0^0?

we know that 0^n = 0 and n^0 = 1. A sequence is needed for n^n:
3^3=27
2^2=4
1^1=1
.5^.5=.707
.1^.1=.79433
0^0 = ?
-.1^-.1=-1.2589
-.5^-.5=-1.4142
-1^-1=-1
-2^-2=-1/4
-3^-3=-1/27

(these are the values my calculator gave me)

So i don't know. 2 out of 3 imply that 0^0 = 0. Is it 0?

Also, theres another one i'm not sure of, since i don't know everything about  :inf: .
If 0x = 0, and  :inf: x =  :inf: , then what is 0* :inf: ?

Yeah I think [tex]0^0[/tex] is undefined.  For the second paradox, I would think that if 0 and :inf: were pure 0 and :inf:, then 0*:inf: would be undefined.  But if not, then it depends on how fast they approach their respective values.  Of course I have nothing but my own thoughts to back all this up.

[tex]0^0[/tex] is undefined. But sometimes we can regard it according to its context, e.g. since [tex]\lim_{x \to 0} x^x = 1[/tex].

[quote="zscool"]I've always been fond of Berry's Paradox:

consider all of the positive integers that cannot be described in fewer than nineteen syllables.  Let L be the smallest of those integers.[/quote]

Yeah that one is funny, it stumps a lot of people, surprisingly, along with the faulty proof by induction that everyone has the same birthday.

[quote]along with the faulty proof by induction that everyone has the same birthday.[/quote]

Can you post that one, Nukular? I haven't seen it before.

One of my favorites is Zeno's Paradox, which states that movement is impossible, and uses infinite series to "prove" it.

And of course, the timeless classic, 1=2.

Let x=y=1.

x=y
x :^2: = xy
x :^2: - y :^2: =xy - y :^2:
(x+y)(x-y) = y(x-y)
x+y=y
2=1.

0* :inf:  equals all real numbers.

For any constant k, k/ :inf:  = 0.  So, by multiplying both sides by  :inf: , 0* :inf: = k.

[hide]This is all nonsense anyway, since infinity isn't a number, it's a concept.  k/ :inf:  doesn't actually equal zero, but it does approach it (hence the use of a limit when solving it).  If you divide k by something ungodly big (but defined), it will equal something infinitessimally small, but not zero.[/hide]

Here's a paradox for you:

1^2 = 1

1^5 = 1

1^13852359236 = 1

1^ :inf:  is indeterminate

I know it uses the logic that lim(x-> :inf: )  (1 + 1/x)^x doesn't equal 1 (it equals e, in the same line of thinking I put in spoiler above).  But why can't a very simple 1^ :inf:  equal 1!?

all women are blondes:

let P(n) be the statement that for any set of n women, if at least one of them is a blonde then they all are:

base case: P(1) is obviously true by def'n

induction hypothesis: assume for all r < m  that P(r) is true

inductive step: given a set S of m women, one of which is blonde, take two subsets A and B whose intersection is the blonde, and whose union is S.  By the induction hypothesis, A and B must contain only blondes and therefore all m women are blondes so P(m) is true.

Now let W be the set of women on earth.  Since there is at least one woman who is blonde, all women are blondes.

hey jeff, were you guys talking about 0* :inf:  when i was visiting your calculus BC class?

zscool -- that's great!  Not a paradox, exactly, but a fabulous "proof" none the less.  Berry's is pretty good, too, although I'm sure it suffers from a mathematical vs. meta-mathematical fallacy somewhere.  I thought I had one, but it disappeared, so I'll see if I can find it again later.

its not a paradox in the classical sense, its a [i]falsidical paradox[/i]
which establishes a result that not only appears false but actually is false

but yea, its a really great proof that is "hard" to find the flaw in (or at least not obvious)

The inductive process fails when n = 2. ;)


integration
limit
trigonometry
logarithms
calculus
algebra
partial fractions

1. $ \int_{a}^{b}\frac {1}{\sqrt {1 \plus{} x^{2}}}dx$

2. $ \int_{a}^{b}\frac {1}{1 \minus{} x^{2}}dx$, for $ \left\vert a\right\vert$,$ \left\vert b\right\vert < 1$

3. $ \underset{x\rightarrow0^{ \plus{} }}{\lim}x^{x}$=1

for the first two use trig subsitution
the the third, take the log of it and find the limit of that.

1.substitute x=$ {tany}$

2.substitute..x=$ {siny}$

2. Decomposition into Partial Fractions:

$ \frac {1}{1 \minus{} x^2} \equal{} \frac {p}{1 \minus{} x} \plus{} \frac {q}{1 \plus{} x}$.

3. Consider the behaviour of $ y\equal{}x\ln x$.

3.
$ exp$ is continuous
$ \underset{x\rightarrow0^{ \plus{} }}{\lim}x\log x \equal{} \underset{x\rightarrow0^{ \plus{} }}{\lim }\frac { \minus{} \log\frac {1}{x}}{\frac {1}{x}} \equal{} \minus{} \underset{x\rightarrow\infty}{\lim }\frac {\log x}{x} \equal{} \minus{} \underset{x\rightarrow\infty}{\lim}\frac {\frac {1}{x}}{1} \equal{} 0$
$ \underset{x\rightarrow0^ \plus{} }{\lim}x^x \equal{} \underset{x\rightarrow0^ \plus{} }{\lim }e^{x\log x} \equal{} e^{0} \equal{} 1$

For 3 show that $ \mathop {\lim }\limits_{x \to  \plus{} \infty } x^{\frac{1}{x}}  \equal{} 1$. That will be easy assuming you know the sequence version. Then simplify your problem.

2. How about the substitution $ x\equal{}t\minus{}\frac{1}{t}$?

Infact the second is the standard integral:

$ \int \frac {1}{a^{2} \minus{} x^{2}}\,dx \equal{} \frac {1}{2.a}\ln|{\frac {a \plus{} x}{a \minus{} x}}| \plus{} C$

ur third answer is absolutely right ICQ-2008

@chemrock, You should add the sign of absolute value to anti logarithm.

They work, but I wouldn't use the second. The integral of $ \sec$ is less elementary than the integrals of $ \frac1{x\minus{}c}$ that we get by partial fractions. Trigonometric substitution is poorly suited to integrals that don't have square roots in them, except for the $ \frac{1}{(x^2\plus{}a^2)^m}$ case.

Since we assumed that both limits of the integral were in $ (\minus{}1,1)$, we can pick a sign and use $ \frac12\ln(1\plus{}x)\minus{}\frac12\ln(1\minus{}x)$ as our antiderivative.

For the first one let $ u\equal{}x\plus{}\sqrt{x^2\plus{}1}.$

hey Black, can u elucidate in ur method...

But valerium the first is also a standard integral:
$ \int \frac {1}{\sqrt {a^{2} \plus{} x^{2}}}\,dx \equal{} \ln{|x \plus{} \sqrt {x^{2} \plus{} a^{2}}|} \plus{} C$

u derive this standard integral using my substitution...

and yes it is a standard integral..


let M  be in interior of convex polygon A1A2...An.show atleast one of angles,
     MA1A2,MA2A3,...,MAnA1 does not exceed pie times(n-2)by 2n.  :!:
 i got,have you any :idea:

:P  my sol. uses isoperimetric pro. for polygon,do you  :D notice connection of this generalisation with generalised erdos mordell :welcomeani:  :?:


I think this thread should be locked.

The meaning behind this post is not to help people argue correctly, but to further bash ID supportists. This post will not help anyone and will most likely cause more anger and resentment, something we don't need any more on these boards.

No, the article clearly states some of the mistakes made while arguing.

Im going to more the comment to another thread, to make you happy.

[quote="WindSlicer"]
EDIT: So you create ANOTHER thread on ID?! Not only does that solidify my point that you created this thread in a seemilngly veiled attempt to bash ID, but it shows that you are making threads for discussions which we have threads for already.[/quote]
1) To move it out of this thread, because you didn't like it
2)I learned something from the article and wanted to share it
3)there is no thread on the disscussion the main idea of ID, evolution is bad so we need an alternate theory
4)Sorry, Im hyped up on caffeine

[quote="ccy"]I am really puzzled by how some people think.  Well, there ARE threads to discuss ID but the IDers all tend to stray off to the merits of evolution.

The article's pretty good.  I printed it out and read it on my way home.[/quote]
You should check the links at the bottom too.

I agree with Windslicer.  There are already two topics on evolution, and I don't want to see any more.  If there is something you want to say, I would prefer if you did it in the existing topics.

[quote="WindSlicer"][quote="jmadsen"][quote="WindSlicer"]I wasn't commenting on the article. I was commenting on the way you expressed the article, not just mentioning how it may help your arguing skills, but then stating how Id has a "cruddy, Pyrrhonic argument". If you can't see how you are directing this article at a specific group of people at AOPS, which will most likely cause more arguments then I don't quite see why you are a mod.

I'm sorry if I'm being blunt, I'm just tired of this stuff. Keep it in ONE thread, at least.

EDIT: So you create ANOTHER thread on ID?! Not only does that solidify my point that you created this thread in a seemilngly veiled attempt to bash ID, but it shows that you are making threads for discussions which we have threads for already.[/quote]
Is it bad that it causes more argument? Isn't the point of the evolution/ID debate to argue? Argument is healthy because it allows us to exchange opinions. Do you want the post deleted because you realize he is right? Do you want the thread locked while ID still has a chance?

The reason he says ID has a Pyrrhonic argument is because it does. If you disagree with that, then you may post your reasoning.

But I agree that there are too many threads on this topic.[/quote]

What the heck? Did you just contradict yourself?

Anyways, I could care less about what people on the internet have to say about this kind of stuff and when they go creating these type of underlying-bashing posts on the other side, and are moderators, that just crosses the line. Maybe you just didn't see the actual post because he deleted it but it was an obviously veiled ID attack. And there is a difference between arguments that produce good discussion and arguments that just create resentment. And we DON'T need ANY of the latter.

Do you really think that I am trying to get this thread deleted because I want ID to live on? Give me a break, we've already got 3 topics on this stuff, and don't need another, especially one that started off like this one.[/quote]

I love how you reword "Sentence stating his opinion and asking for the opinions of other while referencing a good non-partial article" to "attack". I just love that.

I don't care for the "attack" you are making on me now etheir. You seem to have more problems with me than what I did, and are painting me as someone who doesn't care about other people, I do. I also have friends that aren't zealots on my side, I talk to many different people.

This was not bashing, this was a controled argument. If you didn't like the word cruddy in there, it was not put there to describe ID but rather that its main argument has several flaws. This thread/the one I started because of your demands is for discussion of the philisophical points of ID.

You've made your point about not liking what I did, and I fixed that. When you start yelling at me because I made another thread...

[quote="WindSlicer"]Do you really think that I am trying to get this thread deleted because I want ID to live on? Give me a break, we've already got 3 topics on this stuff, and don't need another, especially one that started off like this one.[/quote]
Sorry, I misunderstood what you were saying. I don't see why we have 3 (at least, I think we're up to 5 now) threads on the same topic either.

Thread has been locked. Here's why, so you don't have to PM me to ask:

1) Clearly this is upsetting a lot of personal beliefs, something we don't need.

2) PenguinIntegral, in the future you should not stereotype all ID believers as looking for a different theory because they don't like evolution. There are plenty who find the evidence in favor of ID - not evolution - quite overwhelming. And there are certainly also those on the other side of the coin. Realizing that indeed evolutionists are the majority, please respect the stance of the minority. If this is too much to ask, then these posts will be altered to contain some fundamental respect that AoPS users should be able to find on this board - or at the very least, an absence of disrespect.

I'm picking on you (but this certainly doesn't make you solely responsible for this) because being a moderator for AoPS - ML should give you an even greater burden to set an example. Please do so.

3) Everyone picking on the fact that there are too many EV/ID threads, this is why there are moderators. Significant differences must exist for multiple threads on one topic for the many different threads all to be left active. However, in this case you guys are correct and this will be now locked. The link will be left intact, but further discussion should continue on another threads.

4) And again, for people debating: get off the personal attacks or all of the threads will be locked.


For a), couldn't we just say that $ \lim_{n\rightarrow\infty}(1\plus{}\frac1n)^n\equal{}e$ by the definition of $ e$?

[quote="brainomega"]For a), couldn't we just say that $ \lim_{n\rightarrow\infty}(1 \plus{} \frac1n)^n \equal{} e$ by the definition of $ e$?[/quote]
We can't use $ e \sim 2.7$.


when is your chapter competition. mine is the 2nd

Mine is the 30th.

btw, note the contradictions here.

I think like the 30th of January or sometime around that.

Umm, it's the Saturday after the last Friday of February Vacation.

surely you're not too lazy to check the calendar...

It's Saturday the 2nd for us.

February 2nd. I won't be competing this year though, I'll be grading.  :P

Feb 16.
from now 'till feb 16, I have to do every problem from 1986-1996 and some tests from the 21st century. altogether, that's about 5,000+ problems. yay!

[quote="abacadaea"]9th graders are allowed to grade?

flood the paper with comments. lol. my friend got his with a solution for almost every problem he missed[/quote]

I don't think you're supposed to get the test papers back, as it says (in the general instructions I believe) that all tests become property of MathCounts afterward (i.e. they get to shred them).

missing school isn't fun...

anyways, the one i'm helping out at in CT is on February 2nd...but I had no clue 9th graders were allowed to grade, i thought we could just proctor...well anyways  :o

Well, we miss like the last block of school. (~45 min)

I'm one of those who are waiting for one week to pass quickly?

February 2nd..

Mine was early last year, too.

My chapter is on Febuary the 5.  AKA Tuesday.  We get to miss school, go to a restuarant of our pick (Fazoli's) and after the compitition, we go to Dairy Quenn (our school tradition).  I can' wait to go (and lose)!!!!!!!!(We miss ALL of school!!!!!!)


It’s well known that if a topological space $X$ is separable, then a subspace $Y$ of $X$ need not be separable. But what if $Y$ is also dense?

If there's a counterexample to this, it can't be from a metric space.


ratio
function
analytic geometry
geometry solved
geometry

ok soo i am desingning skis......and i need HELP ASAP!!!!!

So lets imagine that a ski has a turning radius of 16 m, no draw a circle of radius 16 m, from a radial line draw a line perpendicular to it. so now you have a circle with a tangent line to it. Consider this tangent line as part of the ski. On the left side of the tangent line the ski runs for 84cm, and then a line is drawn perpendicular to the tangent line (L1) until it touches the circumfrence of the circle. On the right side of the tangent line the ski runs for 74 cm, and then a line (L2) perpendicular to it runs down until it too touches the circle.  Now i pose as my question, is there some ratio between L1,L2,84,and 74? L1 is know as the front sidecut of the ski, and L2 is the tail sidecut of the ski.

Here is a known number system - This was figured out on a CAD program (Accuracy is probally not 100%)
84(Tangent length left), L1= 19 cm
74(Tangent length right), L2= 14cm
Radius of the circle is 18m

So what sidecuts are needed to supply a 16m turning radius?

I have a formula that i made that gives the sidecut length (true size, usually veyr close to Tangent length)
and from that you can find the sidecut length and turnign radius....but for only one side of the ski ( front or tail) so i will not poisen anyones mind with that dirty information. PLEEASE HELP! ANSWER AS SOON AS POSSIBLE! and can i make a function out of this so that i can graph the data? or two functiooons that show L1 and L2 depentdant on R? there are also angles phata and beta. Phata is the angle made on the left side, and beta is made on the right side. Please view diagram if you can...it is attached! click on the attachment to see in betetr quality!

THANK YOU!!!!!!
[/b]

From your description and drawing, it follows that you are looking for the distances $L_1, L_2$ of a horizontal tangent to the circle with radius $R = 16\ \text{m} = 1600\ \text{cm}$ centered at the coordinate origin from the points of the upper semicircle with x-coordinates $x_L = 74\ \text{cm}$ and $x_R = -84\ \text{cm}$. The tangent has the equation $y = R$ and the circle has the equation $x^2 + y^2 = R^2$. Hence the upper semicircle has the equation $y = +\sqrt{R^2 - x^2}$ (and the lower one, not needed, $y = -\sqrt{R^2 - x^2}$). The distances $L_1, L_2$ are then

$L_1 = R - \sqrt{R^2 - x_L^2} = 1600 - \sqrt{1600^2 - 74^2} \doteq 1.71\ \text{cm}$

$L_2 = R - \sqrt{R^2 - x_R^2} = 1600 - \sqrt{1600^2 - 84^2} \doteq 2.21\ \text{cm}$

Similarly, for the radius $R = 18\ \text{m} = 1800\ \text{cm}$, you would get

$L_1 = R - \sqrt{R^2 - x_L^2} = 1800 - \sqrt{1800^2 - 74^2} \doteq 1.52\ \text{cm}$

$L_2 = R - \sqrt{R^2 - x_R^2} = 1800 - \sqrt{1800^2 - 84^2} \doteq 1.96\ \text{cm}$

It is quite impossible for these distances to be $L_1 = 14\ \text{cm}$ and $L_2 = 19\ \text{cm}$ or anything close to this.

yyea sorry i ment 19mm and 14mm, THANK YOU SOO MUCH!


function
limit
integration
real analysis
real analysis solved

a very nice one I've just had during an oral examination:
let $A(t),B(t):R^{+}->R$ be 2 functions such that:$lim_{x->+\infty} A(t)=0$,$lim_{x->+\infty} B(t)=1$
Let $f$ be a non-null solution of $y''+A(t)y'+B(t)y=0$
Show that the number of zeros of $f$ between $0$ and $t$ is equivalent to $\frac{t}{\pi}$

Moubinool, could you please post a proof?

Theorem : (1) $y\"+q(t)y=0$, (2) $z\"+wz=0$ with $q\leq w$ on $I$ interval

Between two roots $a,b$ of non-zero $y$ solution of (1), any solution of (2) has at least a root

$y(a)=y(b)=0$ for any $z$ solution of (2) there exist $c$ such that $z(c)=0$

Proof

Wronskian $W=yz'-y'z$ satisfy $W'=yz(q-w)$

let $a<b$ two consecutiv roots of $y$

$y(a)=y(b)=0$ suppose that $y>0$ on $]a,b[$

If $z$ has no zero on $]a,b[$ say $z>0$, $W$ is decreasing $[a,b]$

$W(a)\leq 0$, $W(b)\geq 0$ this mean $W=0$ Wronskien is 0 on $[a,b]$, the solutions $y,z$ are

colinear $z $has a zero on $[a,b]$ contradiction.

Back to our problem by a change variable we work with

(3) $y''+(1+g(t))y=0$ where $g$ function wich tends to 0 at +oo.

With the above theorem any solution of (3) has an infinite zero $r_n$ such that
$r_n\leq r_{n+1}$ and $r_{n+1}-r_n$ tends to $2\pi$
The number of zeros of $y$ is equivalent $\frac{t}{\pi}$

More over you have the following

$q:[a,+\infty[\rightarrow ]0;+\infty[ $, class $C^{1}$ if $\lim_{+\infty}\frac{q'}{q^{3/2}}=0$

Then for any non-zero solution of $y\"+q(x)y=0$, the number $N(x)$ of zero of $y$

on $[a,x]$ verify $N(x)\sim \frac{1}{\pi}\int_{a}^{x}\sqrt{q(t)}dt$

A "Professeur de Mathématiques Spéciales $MP^{*}$
told me there is an article in

Revue de Mathématiques Spéciales
RMS page 99
1985-1986

Equations Differentielles Linéaires d'ordre 2:
comportement global ou asymptotiques des solutions
-------------------------------------------------------------------------

See also the book

Hartman, Ordinary Differential Equations page 348

thanx a lot Moubi, but I can't read this article because there are not available in my CDI  :(  do you know where i can have access to it ?

PS:
another solution:
look at $(y, y')/|(y, y')|$ and use relevement theorem.


If $ a\plus{}b\plus{}c\equal{}0$ ,prove that:

I.    $ \dfrac{a^5\plus{}b^5\plus{}c^5}{5}\equal{}\dfrac{a^3\plus{}b^3\plus{}c^3}{3}* \dfrac{a^2\plus{}b^2\plus{}c^2}{2}$

II.   $ \dfrac{a^7\plus{}b^7\plus{}c^7}{7}\equal{}\dfrac{a^5\plus{}b^5\plus{}c^5}{5}* \dfrac{a^2\plus{}b^2\plus{}c^2}{2}$

[hide="Solution"]
By Vieta's Formulas, the cubic with roots $ a,b,c$ is

$ P(x) \equal{} x^3 \minus{} (a \plus{} b \plus{} c)x^2 \plus{} (ab \plus{} bc \plus{} ca)x \minus{} abc \equal{} x^3 \plus{} (ab \plus{} bc \plus{} ca)x \minus{} abc \equal{} 0$.

Let $ S_n \equal{} a^n \plus{} b^n \plus{} c^n$ so that $ \boxed{S_1 \equal{} 0}$ and $ \boxed{S_2 \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca) \equal{} \minus{} 2(ab \plus{} bc \plus{} ca)}$.

Since $ P(a),P(b),P(c) \equal{} 0$, we have $ P(a) \plus{} P(b) \plus{} P(c) \equal{} 0 \equal{} S_3 \plus{} (ab \plus{} bc \plus{} ca)S_1 \minus{} 3abc \equal{} S_3 \minus{} 3abc\implies\boxed{S_3 \equal{} 3abc}$.

Similarly, $ aP(a) \plus{} bP(b) \plus{} cP(c) \equal{} 0 \equal{} S_4 \plus{} (ab \plus{} bc \plus{} ca)S_2 \minus{} abcS_1$

$ \equal{} S_4 \minus{} 2(ab \plus{} bc \plus{} ca)^2\implies\boxed{S_4 \equal{} 2(ab \plus{} bc \plus{} ca)^2}$,

$ a^2P(a) \plus{} b^2P(b) \plus{} c^2P(c) \equal{} 0 \equal{} S_5 \plus{} (ab \plus{} bc \plus{} ca)S_3 \minus{} abcS_2$

$ \equal{} S_5 \plus{} 3abc(ab \plus{} bc \plus{} ca) \plus{} 2abc(ab \plus{} bc \plus{} ca)\implies\boxed{S_5 \equal{} \minus{} 5abc(ab \plus{} bc \plus{} ca)}$, and

$ a^4P(a) \plus{} b^4P(b) \plus{} c^4P(c) \equal{} 0 \equal{} S_7 \plus{} (ab \plus{} bc \plus{} ca)S_5 \minus{} abcS_4$

$ \equal{} S_7 \minus{} 5abc(ab \plus{} bc \plus{} ca)^2 \minus{} 2abc(ab \plus{} bc \plus{} ca)^2\implies\boxed{S_7 \equal{} 7abc(ab \plus{} bc \plus{} ca)^2}$.

It is easy to see that $ \frac {S_5}5 \equal{} \frac {S_3}3\cdot\frac {S_2}2$ and $ \frac {S_7}7 \equal{} \frac {S_5}5\cdot\frac {S_2}2$.
[/hide]


Suppose you have 12 different flavors and cones are made with 3 different flavors. How many possible ice cream cones are possible? How many with 3 different flavors? Note chocolate on the top is for example different from chocolate on the bottom.

So it is $ 12(11)(1)$ for first one. And $ \binom{12}{3}$ for second one?

Assuming as you said that the permutations of the ice cream flavors are significant, $ \binom{12}{3}$ would not give the correct total number of cones.

The number of permutations, without repeated flavors, would be $ 12*11*10 \equal{} 1320$ possible cones. This is because there are 12 flavors to choose for the first position, 11 for the second, and 10 for the final position.

$ \binom{12}{3}$ would be correct in describing the number of cones where ice cream order did not matter however.

Note that more cones would be possible for the first problem, as the second problem's answer is a subset of the first's.

For your first question, it would be, assuming that a cone has to have at least 1 flavor, $ \binom{12}{1}\cdot1!\plus{}\binom{12}{2}\cdot2!\plus{}\binom{12}{3}\cdot3!\plus{}\binom{12}{4}\cdot4!\plus{}\dots\plus{}\binom{12}{12}\cdot12!$.

[quote="Math Champion"]For your first question, it would be, assuming that a cone has to have at least 1 flavor, $ \binom{12}{1}\cdot1! \plus{} \binom{12}{2}\cdot2! \plus{} \binom{12}{3}\cdot3! \plus{} \binom{12}{4}\cdot4! \plus{} \dots \plus{} \binom{12}{12}\cdot12!$.[/quote]As specified in the problem, cones are made with 3 different flavors. It would be crazy to stack 12 flavors creating a humongous cone! :rotfl:

[quote="ragingbullrocky"]Note chocolate on the top is for example different from chocolate on the bottom.[/quote]If the same flavor can be used on top and bottom (assuming that you can't have same [i]adjacent[/i] flavor), there can be $ 12\cdot11\cdot11$ different cones.
 :icecream:

adjacent flavors are like...?  :(
one on top of each otha?

Say the flavor toppings in order is chocolate, vanilla, chocolate. Vanilla is [i]adjacent[/i] to both chocolate. This combination is allowed, per that note I have quoted.

So for the first one, its $ 12^3$, and for the second one, its $ 12\cdot11\cdot11$?


let N is a point On $ l$ .
$ A_1,A_2,B_1,B_2,C_1,C_2$ are 6 point that :
$ A_1,B_1,C_1$ are symmetry of $ A_2,B_2,C_2$ toward $ l$ in order.
$ (AN,B_1C_1) \equal{} P,(BN,C_1A_1) \equal{} Q,(CN,A_1B_1) \equal{} R$
prove P,Q,R are collinear...


[quote="demji"]Find all ordered triples $(x,y,z)$ that satisfy:

$x+y-z=0$
$zx-xy+yz=27$
$xyz=54$

I can't seem to fit this in with Vieta's formula[/quote]

[hide="Hint"]First sub $w=-z$

Now we have:

$x+y+w=0$
$xy+yw+xw=-27$
$xyw=-54$

Can you finish?

[I think I did it right][/hide]

oh, nice thx, now i make a polynomial and find roots.

[hide="Or just say..."]$(g+x)(g+y)(g-z)=g^3 + (x+y-z)g^2 + (xy-xz-yz)g - xyz = g^3 - 27g - 54$. Then we solve the equation.[/hide]


Let $e_{n}$ and $a_{n}$ be the number of dissections of a convex $n$-gon by nonintersecting diagonals into an even or odd number of regions, respectively. Show that $e_{n}-a_{n}=\left(-1\right)^{n}$.

Prove it by induction on $n$:
For small $n$, it's trivially true.

Order the vertives in the canonical order to $A_1,A_2,...,A_n$.
First, lets look at the number of dissections for that no 'cutting' goes through $A_n$:
Since we can add/remove the edge $A_{n-1}A_1$ then, there are equally many 'odd' end 'even' dissections of that type, thus lets assume that there is at least one 'cutting' through $A_n$.

Now lets go on:
For some dissection, let $k\in\{2,3,...,n-3,n-2\}$ be the smallest one such that $A_k$ that is connected with $A_n$.
Now when $k\geq 3$, we can add/remove $A_1A_k$ again, so there are equally much of both types.
So, we are keft with a cut at $A_nA_2$. Now the number of these left odd dissections is $e_{n-1}$, the number of left even ones is $a_{n-1}$ (just dissect the $(n-1)$-gon where $A_1$ is missing), thus, $e_n-a_n = a_{n-1}-e_{n-1} = - (-1)^{n-1} = (-1)^n$.


[quote="phymax"]obviously,by AM-GM,
$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \geq3$

so its sufficient to prove
$ \frac {3(ab \plus{} bc \plus{} ca)}{a^{2} \plus{} b^{2} \plus{} c^{2}}\geq1$

$ 3(ab \plus{} bc \plus{} ca)\geq\ a^{2} \plus{} b^{2} \plus{} c^{2}$

i guess now now u shall try it[/quote]
The last inequality 's false :wink: .Try $ a\equal{}10,b\equal{}c\equal{}1$

Yes, $ a^2\plus{}b^2\plus{}c^2\geq 3(ab\plus{}bc\plus{}ca)\Longleftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\geq 0$.

:blush: Sorry for my mistake....I make a typo...
Now the problem is not so easy...

[quote="kunny"]Yes, $ a^2 \plus{} b^2 \plus{} c^2\geq 3(ab \plus{} bc \plus{} ca)\Longleftrightarrow (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2\geq 0$.[/quote]
It 's false,kunny :P

that's right my friend.
it is false in every way!

:blush:

That was $ (a\plus{}b\plus{}c)^2\geq 3(ab\plus{}bc\plus{}ca)$.

[quote="phymax"]i wanted Jianxing113725 to realise the fault by self trying the problem and not mentioning it wrong[/quote]

I have edited the problem....So sorry for my mistake....

still check your question  :wink:  :huh: !!

[quote="Jianxing113725"]$ a,b,c$ are positive numbers,prove that
$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq4$

May be posted before...Please give me a hint...[/quote]
[i]
Proof:[/i]

Assume: $ b\equal{}max\{a,b,c\}$

$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq4$

$ < \equal{} > (\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \minus{} 3) \plus{} (\frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \minus{} 1 )\geq 0$

$ < \equal{} > (\frac {(a \minus{} c)^2)}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {(b \minus{} a)(b \minus{} c)}{(a \plus{} b)(c \plus{} a)}) \minus{} \frac {(a \minus{} c)^2 \plus{} (b \minus{} a)(b \minus{} c)}{(a \plus{} b \plus{} c)^2} \ge 0$

$ < \equal{} > \frac {(a \minus{} c)^2(a^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(a \plus{} b \plus{} c)^2} \plus{} \frac {(b \minus{} a)(b \minus{} c)(b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b)(c \plus{} a)(a \plus{} b \plus{} c)^2} \ge 0$

Which is obviously true.  :P

[quote="hophinhan"][quote="Jianxing113725"]$ a,b,c$ are positive numbers,prove that
$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq4$

May be posted before...Please give me a hint...[/quote]
[i]
Proof:[/i]

Assume: $ b \equal{} max\{a,b,c\}$

$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq4$

$ < \equal{} > (\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \minus{} 3) \plus{} (\frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \minus{} 1 )\geq 0$

$ < \equal{} > (\frac {(a \minus{} c)^2)}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {(b \minus{} a)(b \minus{} c)}{(a \plus{} b)(c \plus{} a)}) \minus{} \frac {(a \minus{} c)^2 \plus{} (b \minus{} a)(b \minus{} c)}{(a \plus{} b \plus{} c)^2} \ge 0$

$ < \equal{} > \frac {(a \minus{} c)^2(a^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(a \plus{} b \plus{} c)^2} \plus{} \frac {(b \minus{} a)(b \minus{} c)(b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b)(c \plus{} a)(a \plus{} b \plus{} c)^2} \ge 0$

Which is obviously true.  :P[/quote]

man  $ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \geq\ 3$ and $ \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^{2}}\leq1$

[quote="hophinhan"][quote="Jianxing113725"]$ a,b,c$ are positive numbers,prove that
$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq4$

May be posted before...Please give me a hint...[/quote]
[i]
Proof:[/i]

Assume: $ b \equal{} max\{a,b,c\}$

$ \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\geq4$

$ < \equal{} > (\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b} \minus{} 3) \plus{} (\frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \minus{} 1 )\geq 0$

$ < \equal{} > (\frac {(a \minus{} c)^2)}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {(b \minus{} a)(b \minus{} c)}{(a \plus{} b)(c \plus{} a)}) \minus{} \frac {(a \minus{} c)^2 \plus{} (b \minus{} a)(b \minus{} c)}{(a \plus{} b \plus{} c)^2} \ge 0$

$ < \equal{} > \frac {(a \minus{} c)^2(a^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(a \plus{} b \plus{} c)^2} \plus{} \frac {(b \minus{} a)(b \minus{} c)(b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)}{(a \plus{} b)(c \plus{} a)(a \plus{} b \plus{} c)^2} \ge 0$

Which is obviously true.  :P[/quote]
Very nice solution :)


geometry
incenter
circumcircle
ratio
geometric transformation
homothety
trigonometry

Dear Yetti and Mathlinkers,
Let ABC be a triangle, (Ha) the A-altitude of ABC, I the incenter of ABC, D the foot of the perpendicular to BC passing through I, E the foot of the perpendicular to (Ha) passing through I, 0 the circumcircle of ABC and A' the midpoint of the arc BC opposite to A.
Prouve synthetically that A', D and E are collinear.
Sincerely
Jean-Louis

Denote $ J \equal{} AI\cap BC$. We can easily to see that triangles $ A'CJ$ and $ A'AC$ are similar, which implies that $ \frac {A'C}{A'A} \equal{} \frac {CJ}{AC}$, i.e. $ \frac {A'I}{A'A} \equal{} \frac {CJ}{AC}$ $ (1)$. On the other hand, $ CI$ is the internal angle bisector of the angle $ ACJ$, so $ \frac {CJ}{AC} \equal{} \frac {IJ}{IA} \equal{} \frac {ID}{AE}$ $ (2)$ (because of $ \triangle IJD\sim\triangle AIE$). Combining $ (1)$ and $ (2)$, we deduce that $ \frac {A'I}{A'A} \equal{} \frac {ID}{AE}$. Hence, due to Thales' theorem, we conclude $ A'$, $ D$ and $ E$ are collinear.

[b]P.S.[/b] I have discovered this property moths ago, I used it during solve a problem named that "An unexpected concurrency" on Hyacinthos Group. I also sent this solution to [b]Kostas Vittas[/b] via PM  :D

[quote="Virgil Nicula"] [color=darkred] [size=117][b]Here is a new problem ![/b][/size] Let $ ABC$ be a triangle with the incircle $ C(I,r)$ , the $ A$ - excircle $ C(I_a,r_a)$ and the circumcircle $ w = C(O,R)$ . Denote : the projection $ D$ , $ D'$ of $ I$ , $ I_a$ to $ BC$ respectively. ; the point $ E$ for which $ IE\parallel BC$ and $ AE\perp BC$ ; the point $ F$ for which $ I_aF\parallel BC$ and $ AF\perp BC$ ; the second intersection $ S$ (south) of the line $ \overline {AII_a}$ with the circle $ w$ ; the diameter $ [AL]$ of the circle $ w$ ; the intersections $ X\in SD\cap LI$ , $ Y\in SD'\cap LI_a$ . Prove ([b]synthetically[/b] !) that $ \{E,X\}\subset SD$ and $ \{F,Y\}\subset SD'$ .[/color] [/quote]

[color=darkred]I"ll present my [u]synthetical proof[/u] (in my opinion, [b][u]interesting and very nice ![/u][/b] ).

I dedicate it to [b]Cosmin Pohoata[/b], [b]April[/b], [b]Jean-Louis[/b] and [b]Kostas Vittas[/b].[/color]

[color=darkblue][b][u]Proof.[/u][/b] Observe that $ ID = r$ , $ AL = 2R$ and $ IA\cdot IS = 2Rr$ . Hence $ \{\begin{array}{c} \widehat {DIS}\equiv\widehat {IAL} \\
 \\
\frac {ID}{IS} = \frac {AI}{AL}\end{array}\ \stackrel {s.a.s.}{\implies}\ \triangle DIS\ \sim\ \triangle IAL$ $ \implies$

$ \widehat {ISD}\equiv\widehat {ALI}$ $ \implies$ $ \boxed {\ X\in w\ }$ $ \implies$  $ XA\perp XI$ . Since $ EA\perp EI$ obtain that $ AXEI$ is cyclically, i.e. $ \widehat {IXE}\equiv\widehat {IAE}$ .

From $ \widehat {LXE}\equiv \underline {\widehat {IXE}\equiv \widehat {IAE}}\equiv\widehat {LAS}\equiv\widehat {LXS}$ obtain $ \widehat {LXE}\equiv\widehat {LXS}$ $ \implies$  $ E\in XS$ $ \implies$  $ E\in DS$ , i.e. $ \{E,X\}\subset SD$ .

The second part of the conclusion, i.e. $ \{F,Y\}\subset SD'$ , you can prove analogously. I let it as an interesting and easy dual problem. [/color]

[quote="jayme"]... Let ABC be a triangle, (Ha) the A-altitude of ABC, I the incenter of ABC, D the foot of the perpendicular to BC passing through I, E the foot of the perpendicular to (Ha) passing through I, 0 the circumcircle of ABC and A' the midpoint of the arc BC opposite to A.
Prouve synthetically that A', D and E are collinear...[/quote]

$ H$ is the orthocenter. $ M$ is the midpoint of $ BC,$ $ X \equiv AI \cap BC$ is foot of the internal bisector of the $ \angle A,$ $ D$ is that tangency point of the incircle $ (I)$ with $ BC$ and $ K \equiv AH \cap BC$ is foot of the A-altitude $ h_a.$ Inversion in a circle $ (M)$ with center $ M$ and radius $ MD$ takes the 9-point circle $ (N)$ into the common internal tangent of the incircle $ (I)$ and A-excircle $ (I_a)$ other than $ BC.$ Therefore, $ D$ stays put, $ K$ goes to $ X$ and the other way around. By the basic properties of inversion,

$ \overline {DK} \equal{} \minus{}\overline{DX} \cdot \frac{MD^2}{\overline{MX} \cdot \overline{MD}} \equal{} \frac{\overline{XD} \cdot \overline{DM}}{\overline{XM}}\ \Longrightarrow$

$ \frac{\overline{DK}}{\overline{DM}} \equal{} \frac{\overline{XD}}{\overline{XM}} \equal{} \frac{\overline{XI}}{\overline{XA'}} \equal{} \frac{\overline{DI}}{\overline{MA'}} \equal{} \frac{\overline{KE}}{\overline{MA'}}\ \Longrightarrow\ \frac{\overline{MA'}}{\overline{MD}} \equal{} \frac{\overline{KE}}{\overline{KD}}$

which means the right $ \triangle DMA' \sim \triangle DKE$ are centrally similar with similarity center $ D$ and $ A', D, E$ are collinear.

[color=darkblue][b][u]Here are three similar well-known dual properties.[/u][/b][/color]

[color=darkred][b]V.N.[/b][/color] [color=darkblue]and[/color] [color=darkred][b]Jayme[/b][/color] [color=darkblue]wish, if it is possibly, three synthetical proofs, without cross ratio ![/color]

[color=darkred]Let $ \triangle ABC$ with the incircle Let $ w \equal{} C(I,r)$ be the incircle and $ A$ - excircle $ w_a$ of $ \triangle ABC$ .[/color]

[quote] [color=darkred][b][u]Property I.[/u][/b]  Denote$ \left\|\begin{array}{c} M\in (BC)\ ,\ MB \equal{} MC \\
\ D\in BC\ ,\ AD\perp BC\end{array}\right\|$ and $ \left\|\begin{array}{c} X\in AD\cap MI \\
\ Y\in AD\cap MI_a\end{array}\right\|$ . Then $ AX \equal{} r$ and $ AY \equal{} r_a$ .[/color] [/quote]

[quote] [color=darkred][b][u]Property II.[/u][/b]  Denote$ \left\|\begin{array}{c} T\in (BC)\ ,\ IT\perp BC \\
\ T_a\in (BC)\ ,\ I_aT_a\perp BC\end{array}\right\|$ and $ \left\|\begin{array}{c} D\in BC\ ,\ AD\perp BC \\
\ S\in AD\ ,\ AS \equal{} SD\end{array}\right\|$ . Then $ S\in IT_a\cap I_aT$ .[/color] [/quote]

[quote] [color=darkred][b][u]Property IIII.[/u][/b]  The circle $ w$ touches $ [AB]$ , $ [AC]$ at $ E$ , $ F$ . The $ B$ - intenal bisector and $ C$ - internal angle bisector of $ \triangle ABC$

cut $ EF$ in $ M$ , $ N$ . The circle $ w_a$ touches $ AB$ , $ AC$ at $ E_a$ , $ F_a$ . The $ B$ - external angle bisector and $ C$ - external  bisector

of $ \triangle ABC$ cut $ E_aF_a$ in $ M_a$ , $ N_a$ . Then the points $ M$ , $ N$ , $ M_a$ , $ N_a$ belong to the circle with the diameter $ [BC]$ . [/color] [/quote]

[quote] [color=darkred][b][u]Property I.[/u][/b] Denote$ \left\|\begin{array}{c} M\in (BC)\ ,\ MB \equal{} MC \\
\ D\in BC\ ,\ AD\perp BC\end{array}\right\|$ and $ \left\|\begin{array}{c} X\in AD\cap MI \\
\ Y\in AD\cap MI_a\end{array}\right\|$ . Then $ AX \equal{} r$ and $ AY \equal{} r_a$ .[/color] [/quote]

Call M, L the midpoint of CA, AB. Call $ K: AD \cap NL$.

It's know that M,I,K are collinear (the incenter is the nagel point of the medial triangle), then the triangles $ \triangle IDK$ and $ \triangle XAK$ are equal so $ \overline{AX} \equal{} \overline{ID} \equal{} r$

Call $ I_a$ the A-excenter and F the prjection of A on NL. Call $ J: MI_a \cap NL$.

By the Property II it's know that $ I_a$, D, F are collinear and that  $ DA \parallel MJ$ (because the Mittenpunkt is the complement of Gergonne).

Than with an homothety with center F $ \triangle FAD$ goes on $ \triangle FYI_a$ and K goes on J, so $ \overline{JI_a} \equal{} \overline{JY}$ from wich $ \overline{YA} \equal{} r_a$

[quote] [color=darkred][b][u]Property II.[/u][/b] Denote$ \left\|\begin{array}{c} T\in (BC)\ ,\ IT\perp BC \\
\ T_a\in (BC)\ ,\ I_aT_a\perp BC\end{array}\right\|$ and $ \left\|\begin{array}{c} D\in BC\ ,\ AD\perp BC \\
\ S\in AD\ ,\ AS \equal{} SD\end{array}\right\|$ . Then $ S\in IT_a\cap I_aT$ .[/color] [/quote]

w.l.o.g. $ c \ge b$

$ \frac{\overline{DT}}{\overline{TT_a}} \equal{} \frac{\overline{DM} \minus{} \frac{c\minus{}b}{2}}{\overline{DM} \plus{} \frac{c\minus{}b}{2}} \equal{} \frac{a\minus{}2b\cos \gamma \minus{}c \plus{}b}{a\minus{}2b\cos \gamma \plus{}c \minus{}b} \equal{} \frac{a^2 \minus{} (a^2\plus{}b^2\minus{}c^2) \minus{} ac \plus{} ba}{a^2 \minus{} (a^2\plus{}b^2\minus{}c^2) \plus{} ac \minus{} ba} \equal{} \frac{(p\minus{}a)(c\minus{}b)}{p(c\minus{}b)}\equal{}\frac{r}{r_a}$

$ \frac{\overline{SD}\minus{}\overline{IT}}{\overline{SD}}\equal{} 1 \minus{} \frac{2r}{h_a} \equal{} 1\minus{} \frac{a}{p} \equal{} \frac{p\minus{}a}{p} \equal{} \frac{r}{r_a}$

[quote="Virgil Nicula"][color=darkred][b][u]Property III.[/u][/b]  The circle $ w$ touches $ [AB]$ , $ [AC]$ at $ E$ , $ F$ . The $ B$ - intenal bisector and $ C$ - internal angle bisector of $ \triangle ABC$

cut $ EF$ in $ M$ , $ N$ . The circle $ w_a$ touches $ AB$ , $ AC$ at $ E_a$ , $ F_a$ . The $ B$ - external angle bisector and $ C$ - external  bisector

of $ \triangle ABC$ cut $ E_aF_a$ in $ M_a$ , $ N_a$ . Then the points $ M$ , $ N$ , $ M_a$ , $ N_a$ belong to the circle with the diameter $ [BC]$ . [/color] [/quote]

$ I_a, I_b, I_c$ are the A-, B-, C-excenters of the $ \triangle ABC,$ $ D$ is tangency point of the incircle $ w$ with $ BC.$ Let parallel to $ I_cBI_a$ through $ C$ cut $ BI$ at $ M$ and let parallel to $ I_aCI_b$ through $ B$ cut $ CI$ at $ N.$ $ M, N$ thus defined are on a circle with diameter $ BC$ on account of the right angles $ \angle BNC, \angle BMC.$ Let $ BN, BM$ meet at $ K.$ The $ \triangle KCB \cong I_aBC$ are centrally congruent, having parallel sides and the side $ CB \equal{} BC$ common. $ BCI_bI_c$ is cyclic on account of the right angles $ \angle I_bBI_c, \angle I_bCI_c$ $ \Longrightarrow$ the $ \triangle KCB \sim \triangle I_aI_bI_c$ are oppositely similar with similarity coefficient $ k \equal{} \frac {CB}{I_bI_c} \equal{} \sin \frac {A}{2}$ and common orthocenter $ I.$ Their orthic $ \triangle DNM \sim \triangle ABC$ are similar with the same similarity coefficient $ k$ and so are their concentric incircles $ w', w$ with radii $ r' \equal{} kr, r.$

$ \angle I_bI_cC \equal{} \angle I_bBC \equal{} \angle MBC \equal{} \angle MNC$

which means $ MN \parallel I_bI_c.$ Since $ EF \parallel I_bI_c$ (both $ \perp AI)$, it follows that $ MN \parallel EF.$ Let $ U \in AI$ be the midpoint of $ EF$ and $ U' \in AI$ tangency point of $ w'$ with $ MN.$ Since

$ IU \equal{} IF \sin \frac {A}{2} \equal{} kr \equal{} r' \equal{} IU'$

the points $ U' \equiv U$ are identical and $ E, F \in MN.$

[hide="A-excircle"] The case of A-excircle $ w_a$ tangent to $ BC$ at $ D_a$ is proved similarly: the $ \triangle I_aI_bI_c$ with orthocenter $ I$ and orthic $ \triangle ABC$ is replaced by the $ \triangle II_bI_c$ with orthocenter $ I_a$ and with the same orthic $ \triangle ABC.$ The $ \triangle KCB$ with orthocenter $ I$ and orthic $ \triangle DNM$ is replaced by the $ \triangle K_aCB$ with orthocenter $ I_a,$ where $ K_a \equiv BM_a \cap CN_a,$ and orthic $ \triangle D_aN_aM_a.$ Similarity coefficient of the $ \triangle K_aCB \sim \triangle II_bI_c$ with the common orthocenter $ I_a$ is the same, $ k \equal{} \sin \frac {A}{2}$ and so is the similarity coefficient of their orthic $ \triangle D_aN_aM_a \sim \triangle ABC$ with the common A-excenter $ I_a$ and concentric A-excircles $ w_a', w_a$ with radii $ r_a' \equal{} kr_a, r_a.$[/hide]

Dear Mathlinkers,
revisiting this first result, we can have a nice proof by considering
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=550083
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=358063
Sincerely
Jean-Louis


[size=150]Let $ a,b,c$ be positive numbers,$ \min{(a,b,c)}> t \geq 0$,prove that:
$ {\frac {a \minus{} t}{b \minus{} t}} \plus{} {\frac {b \minus{} t}{c \minus{} t}} \plus{} {\frac {c \minus{} t}{a \minus{} t}}\geq {\frac {a }{b}} \plus{} {\frac {b}{c}} \plus{} {\frac {c}{a}}$

BQ[/size]

Just take $ a \minus{} t \equal{} a' ; b \minus{} t \equal{} b'; c \minus{} t \equal{} c' ;$ so that $ a',b',c'> 0$
Then your inequality equivalents
$ \frac {a'}{b'} \plus{} \frac {b'}{c'} \plus{} \frac {c'}{a'}\geq \frac {a' \plus{} t}{b' \plus{} t} \plus{} \frac {b' \plus{} t}{c' \plus{} t} \plus{} \frac {c' \plus{} t}{a' \plus{} t}$
Which is just your inequality 14 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319100]posted here[/url].
Nice inequalities, congratulations. :)
[edit: Oops, beaten again]

[quote="xzlbq"][quote="xzlbq"][size=150]Let $ a,b,c$ be positive numbers,$ \min{(a,b,c)} > t \geq 0$,prove that:
$ {\frac {a \minus{} t}{b \minus{} t}} \plus{} {\frac {b \minus{} t}{c \minus{} t}} \plus{} {\frac {c \minus{} t}{a \minus{} t}}\geq {\frac {a }{b}} \plus{} {\frac {b}{c}} \plus{} {\frac {c}{a}}$

BQ[/size][/quote]

[size=150]Extended to n -variables,Do the same?[/size][/quote]

certainly.

that's equivalent to your ineq(14).

[quote="kuing"][quote="xzlbq"][size=150]Let $ a,b,c$ be positive numbers,$ \min{(a,b,c)} > t \geq 0$,prove that:
$ {\frac {a \minus{} t}{b \minus{} t}} \plus{} {\frac {b \minus{} t}{c \minus{} t}} \plus{} {\frac {c \minus{} t}{a \minus{} t}}\geq {\frac {a }{b}} \plus{} {\frac {b}{c}} \plus{} {\frac {c}{a}}$

BQ[/size][/quote]

both of my proof way in your ineq(14) can be prove this ineq too.

see also: http://www.mathlinks.ro/viewtopic.php?t=319100[/quote]

I see,thanks you!
BQ


in the other topic that just got locked:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143668

that website gave the scores for last year.  i cant seem to find this year's scores.  does anyone have a link.

Well, after the date passed the state discussion window, everyone switched to discussing this year's cutoffs.  You may keep discussing here though, sorry, when I locked the other, i didn't realize that my link was for 05-06 cutoffs.


For $a,b,c$ positive ral numbers such that $abc=1$ prove that

$ \sum(a+b)^2c \geq 4\sum ab $

I remember this problem being given in some MOSP, but I'm not sure. Anyway, it will appear in that variant from MOSP in the book Old and New Inequalities, a book which will appear at IMO (authors Titu Andreescu, Vasile Cartoaje (Vasc), Mircea Lascu (Gil) and Gabriel Dospinescu (harazi)). The problem is not trivial as it sounds, but anyway it is easy:
  Take $ a=\frac{1}{x} $ and so on and reduce it to $ xy+yz+zx+\frac{3}{x+y+z}\geq 4$. But from AM-GM we have ${ xy+yz+zx+\frac{3}{x+y+z}>=4\cdot\sqrt[4]{ \frac (xy+yz+zx)^3}{9(x+y+z)}}\geq 4 $ the last one being very easy.

Sorry but I don't understand..........!!!!!!!!!!!!!!!!!!!!!!!!!!

Can you please detail it better ??

Thank you very much.

He means:

\[ \frac{3}{x + y + z} + xy + yz + zx \geq 4 \sqrt[3]{\frac{(xy + yz + zx)^3}{9(x + y + z)}}\] by the AM-GM inequality so it suffices to prove \[(xy + yz + zx)^3 \geq 9(x + y + z).\]
Now you can use, for example, brute force :D

or you can use $xy+yz+zx\geq$ 3 and $(xy+yz+zx)^2\geq 3xyz(x+y+z)$


calculus
derivative
function
algebra unsolved
algebra

[quote="dayvn"]Assume that  f, g : (0,1)-->R are derivative functions  satisfy
  (i) f, g, f', g' >0   /(0,1)
  (i)f'/g'  is an increasing function
Show that
  h is increasing or decreasing or there exists  0 < :theta: <1 such that h is decreasing in (0, :theta: ) and increasing in ( :theta: ,1)
                  (h= f/g)
 I can solve this problem if  f',g' is continuous, but if f',g' is not continuous , I can't solve it
Could any one help me?
(My English is terrible  :lol: )[/quote]

u know what it is easy
but u have to understand me in azery
or in math lungagge


probability
calculus
geometry
analytic geometry
integration
trigonometry
function

A table of infinite expanse has inscribed on it a set of parallel lines spaced $ 2a$ units apart. A needle of length $ 2l$ (smaller than $ 2a$) is twirled and tossed on the table. What is the probability that when it comes to rest it crosses on a line?

To start out, I think you should place a restriction on where the needle falls; if not, the probability is 0.

[hide="A try"]
We can count the complement, that is, the probability that it does not land on a line, and that it lands in between the lines.
There are 2 things to consider, location of needle and position of needle (angle).
Selecting the center of the needle to be the point that represents its location, we can firstly see that if the needle's center is any farther than L (was that the variable?) units away from a line, it cannot touch the line. If the needle always landed so that it would be positioned at a right angle with the parallel lines, the probability would be L/A. However, the needle can be in any orientation. Let x be the angle the needle is turned from the 0 degree position (parallel to the lines on the table), and let L/d be the distance from the center of the needle to a line (where L/d is between 0 and L). The distance that an end of a needle is from a line is given be Lsinx. We only need to consider {0,90} for x because the needle is rotationally symmetric (in length) about the center.
We have that Lsinx>L/d, and sinx>1/d follows, and then x>arcsin1/d. Since I don't know calculus, I plug that into my calculator (y=(arcsinx)/90), and use it to find the area under the curve, and I get some number 0.36336811 for which I have no clue where it came from. The probability is thus 0.6366189L/A.[/hide]

... let's say the lines have equations $ x \equal{} \minus{}a$ and $ x \equal{} a$. Since you have placed no restriction on where the needle can land, it can land anywhere on the plane; hence, the probability that it lands in a region with an x-coordinate between $ [\minus{}a \minus{} l, a \plus{} l]$ is 0. Hence, the needle will never intersect.

Place a restriction somewhere.

If my argument is incorrect, please point out to me what I did wrong.

No, he's talking about an infinite number of parallel lines equally spaced apart, not just 2.

Crap wait, I think I did something wrong with the calculator trick. Is it L/2A?

I believe that the probability that the needle will cross a line is $ \frac {2l}{a\pi}\int_{0}^{\frac {\pi}{2}}\sin\theta \ d\theta \equal{} \frac {2 l}{a\pi}$.

[hide="Solution."]
The chance of a needle of length $ x$ crossing a line when the lines are spaced $ y$ units apart is $ \frac {x}{y}$.  All that matters is the vertical height, not the horizontal length.  Given an angle $ \alpha$, the vertical height of the needle will then be $ 2l\sin{\alpha}$.  The needle can take any angle between $ 0$ and $ 90$ degrees; the probability of the needle crossing a line as a function of the angle would then be $ P(\alpha) \equal{} \frac {2l\sin{\alpha}}{2a}$.  The average value over the range is:

$ \frac {\frac {\pi}{2}}{\int_{0}^{\frac {\pi}{2}}P(\alpha)d\alpha}$

$ \equal{} \frac {\frac {\pi}{2}}{\int_{0}^{\frac {\pi}{2}}\frac {2l\sin{\alpha}}{2a}}$

$ \equal{} \frac {2l}{a\pi}$.[/hide]


I'm a student and is eager to learn about math..
But, I have not found really an excellent site for mathematic for little bit advanced but below level of AME.. Since there are smart instructors and other people who are good at math, please help me if they know any..

why isn't this site excellent? AOPS has everything from books, free jams, courses, free circles?

This website is a bit advanced, but if you are sure you are advanced try Mathlinks Everyone (which is similar but is Olympiad level).

Of course, it is... I mean except this site.. Besides, MathJams are far away from me now and I don't take virtual classes...


algebra
polynomial
trigonometry
algebra proposed

Determine all polynomials $ P(x)$ with rational coefficients such that for all $ |x| \le 1$, $ P(x) \equal{} P(\frac { \minus{} x \plus{} \sqrt {3 \minus{} 3x^2}}{2})$.

$ P(x) \equal{} a(4x^3 \minus{} 3x)\plus{}b, a,b \in \mathbb{Q}$

Can you explain the method of your solution please?

We have $ P(\cos \theta) \equal{} P(\cos \left(\theta \plus{} \frac{2 \pi}{3} \right)$ where $ \theta \in [0, \pi]$

This tells us that in general any polynomial of the form $ aP(x) \plus{} b$ where $ P(\cos \theta)$ is the polynomial corresponding to $ \cos 3k \theta$ expanded in terms of $ \cos \theta$ will satisfy the given equation.


Have you found an inequality true also for a general positive $ k$ coefficient?

$ \frac{a^{k}+1}{b^{k}+c^{k}}+\frac{b^{k}+1}{c^{k}+a^{k}}+\frac{c^{k}+1}{a^{k}+b^{k}}$

Thank you very much.

However, very nice inequalities. :)


http://www.kalva.demon.co.uk/aime/aime92.html

I am having trouble understand the last step a sub n = ½n^2 + an + b,  but since a19 = a92 = 0, so the polynomial must be (n-19)(n-92)/2.

Since the coefficient of the n^2 term is 1/2 of a sub n, why is it also in the second polynomial?

n=19 and n=92 are roots of the polynomial, so when plugged those two values in, you need to get 0.

[quote="Calculus"]Right, I understand that part, sorry that I didn't make my question clear.

Why is the 1/2 carried over? Since its 1/2 n^2 in the first equation why is it like that in the second equation too?[/quote]
Why wouldn't it be like that?  When you expand the factored polynomial you have to have the coefficient 1/2 before the $n^2$, so you need the 1/2 in the factored form.


(1/(x^2 - a)) + (1/(x^3 - b)) = 1

Your goal is to write x in terms of a and b.

Please explain how as well.

hello, by simplifying $ \frac{1}{x^2\minus{}a}\plus{}\frac{1}{x^3\minus{}b}\minus{}1\equal{}0$ we get
$ \frac{a\plus{}b\plus{}ab\minus{}x^2\minus{}bx^2\minus{}x^3\minus{}ax^3\plus{}x^5}{(a\minus{}x^2)(x^3\minus{}b)}\equal{}0$ and so
you have to solve the following polynomial equation in $ x$
$ x^5\minus{}ax^3\minus{}x^3\minus{}bx^2\minus{}x^2\plus{}ab\plus{}a\plus{}b\equal{}0$ which is not possible with square roots in the general case.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, by simplifying $ \frac {1}{x^2 \minus{} a} \plus{} \frac {1}{x^3 \minus{} b} \minus{} 1 \equal{} 0$ we get
$ \frac {a \plus{} b \plus{} ab \minus{} x^2 \minus{} bx^2 \minus{} x^3 \minus{} ax^3 \plus{} x^5}{(a \minus{} x^2)(x^3 \minus{} b)} \equal{} 0$ and so
you have to solve the following polynomial equation in $ x$
$ x^5 \minus{} ax^3 \minus{} x^3 \minus{} bx^2 \minus{} x^2 \plus{} ab \plus{} a \plus{} b \equal{} 0$ which is not possible with square roots in the general case.
Sonnhard.[/quote]General quintics are not solvable, but ones in this form are if the problem is correct.

So it is impossible to write x in terms of a and b?

Quintics of that form can't generally be solved, let a=0, b=1 you get $ x^5 \minus{} x^3 \minus{} 2x^2 \plus{} 1 \equal{} 0$, not factorable by [url]http://mcis.jsu.edu/faculty/desmith/polyfactor-cgi.html[/url]. So it is impossible to write x in terms of a and b.

Irreducibility is not a sufficient criterion to guarantee unsolvability by radicals.  The nature of Galois extensions is [url=http://en.wikipedia.org/wiki/Quintic_equation#Examples_of_solvable_quintics]more subtle[/url] than that.

Besides, "impossible" is a misstatement of Abel's theorem.  It [i]is[/i] possible to solve quintic polynomials - if one is willing to use certain [url=http://en.wikipedia.org/wiki/Quintic_equation#Beyond_radicals]tools[/url] other than root extraction.

In any case, it is likely that there is no "elementary" solution to this problem.

So is anyone here able to write x in terms of a and b?

Have you been reading the above posts? It's not whether "anyone" can do it - you [b]can't[/b] do it with ordinary operators! See [url=http://en.wikipedia.org/wiki/Quintic_equation]Quintic equation[/url] to see why you can't solve some quintic equations. It means that you really cannot express $ x$ in terms of the ordinary operators - it's just not expressible in those terms.

Of course, if you introduce $ \text{BR}(x)$ (the [url=http://en.wikipedia.org/wiki/Bring_radical]Bring radical[/url]), then one can express $ x$ in terms of $ a$ and $ b$.

Sorry, I was just reading t0rajir0u's post and hoping.


trigonometry
geometry
trig identities
Law of Cosines
Law of Sines

Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find \[ \frac{\cot \gamma}{\cot \alpha+\cot \beta}. \]

[hide="hint:"]extended law of sines[/hide]

Here's what I got, anyhow...

[hide]Answer: $\boxed{994}$.
Solution: \[ \cot{\alpha}+\cot{\beta}=\frac{\cos{\alpha}}{\sin{\alpha}}+\frac{\cos{\beta}}{\sin{\beta}}=\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}}= \] \[ \frac{\sin{\alpha+\beta}}{\sin{\alpha}\sin{\beta}}=\frac{\sin{\gamma}}{\sin{\alpha}\sin{\beta}} \]
By the law of cosines, \[ a^2+b^2-2ab\cos{\gamma}=c^2=1989c^2-2ab\cos{\gamma} \implies ab\cos{\gamma}=994c^2 \]
Now \[ \frac{\cot{\gamma}}{\cot{\alpha}+\cot{\beta}}=\frac{\cot{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\gamma}}=\frac{\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin^2{\gamma}}=\frac{ab}{c^2}\cos{\gamma}=\frac{ab}{c^2}\frac{994c^2}{ab} \] \[ =994 \][/hide]

$ cotA$=$ \frac{cosA}{sinA}$=$ \frac{(b^{2}\plus{}c^{2}\minus{}a^{2})/2bc}{sinA}$

=$ \frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{4S}$

this will make it much more easier.....

Draw altitude CH length h. $ Cot A \equal{} AH/h$ and $ cot B \equal{} BH/h$, so $ cot A \plus{} cot B \equal{} c/h$.
Also, by Law of Cosines, $ 994c^2 \equal{} ab cos C$. But the area $ K \equal{} 1/2 ab sin C$, so dividing, $ 497c^2/K \equal{} 497c^2/(ch/2) \equal{} 994(c/h) \equal{} cot C$.
So the answer is $ 994(c/h)/(c/h) \equal{} 994$.

Let $a_{}^{}$, $b_{}^{}$, $c_{}^{}$ be the three sides of a triangle, and let $\alpha_{}^{}$, $\beta_{}^{}$, $\gamma_{}^{}$, be the angles opposite them. If $a^2+b^2=1989^{}_{}c^2$, find $\frac{\cot \gamma}{\cot \alpha+\cot \beta}$

[hide="Solution"]

Use the hint in the book to turn all the $\cot$ into $\cos$ and $\sin$.  Then, use Law of cosines to give us $c^2=a^2+b^2-2ab\cos(\gamma)$ or therefore $\cos(\gamma)=\frac{994c^2}{ab}$.  Next, we are going to put all the sin's in term of $\sin(a)$.  We get $\sin(\gamma)=\frac{c\sin(a)}{a}$.  Therefore, we get $\cot(\gamma)=\frac{994c}{b\ sin a}$.

Next, use Law of Cosines to give us $b^2=a^2+c^2-2ac\cos(\beta)$.  Therefore, $\cos(\beta)=\frac{a^2-994c^2}{ac}$.  Also, $\sin(\beta)=\frac{b\sin(a)}{a}$.  Hence, $\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}$.

Lastly, $\cos(\alpha)=\frac{b^2-994c^2}{bc}$. Therefore, we get $\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}$.

Now, $\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sin a}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}$.  After using $a^2+b^2=1989c^2$, we get $\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}$.

[/hide]

This is so troll

Rearranging we have

$a^2=1989c^2-b^2$.  let us have a right triangle so
$a^2=c^2+b^2$

Easy to solve we get $a=c\sqrt{995}, b=c\sqrt{994}$ and from here it is a matter of plugging in to get

$\boxed{994}$ as the answer

[quote=sicilianfan]@above You cannot really say "[b]without[/b] loss of generality" when in fact, you lose every single bit of generality by assuming $a^2=c^2+b^2$.[/quote]

ok, ok true dat

[quote=AruKasera][quote="K81o7"]$\frac{\sin{\alpha}\sin{\beta}}{\sin^2{\gamma}}=\frac{ab}{c^2}$[/quote]

How do you get that?[/quote]

Law of sines. $\frac{a}{\sin{\alpha}} = \frac{b}{\sin{\beta}} = 2R$. WLOG let $R = \frac{1}{2}$ (just scale down triangle), which makes $a = \sin{\alpha}$ and $b = \sin{\beta}$.

$\cot\alpha+\cot\beta$=$\frac{sin\alpha cos\beta+cos\alpha sin\beta}{sin\alpha sin\beta}=\frac{sin{\alpha+\beta}}{sin\alpha sin\beta}=\frac{sin{\gamma}}{sin\alpha sin\beta}$

[hide=Solution]Rewrite as $\frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}}$, or $\frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\sin(\alpha + \beta)}{\sin \alpha \sin \beta}}=\frac{\cos \gamma \sin \alpha \sin \beta}{\sin(\alpha + \beta)\sin \gamma}$, or $\frac{\cos \gamma \sin \alpha \sin \beta}{\sin^2 \gamma}$. From LoC on our original triangle, we have $1989c^2-2ab \cos \gamma = c^2$, or $ab \cos \gamma = 994c^2$. But $\frac{\sin \alpha \sin \beta}{\sin^2 \gamma}=\frac{ab}{c^2} = \frac{994}{\cos \gamma}$, so the answer is $\cos \gamma \cdot \frac{994}{\cos \gamma}=\boxed{994}$.[/hide]

Note that $\sin A = t\cdot a$ for some constant $t$, and similarly for the other vertices. Also, $\cos A = r\cdot \frac{(b^2 + c^2 - a^2)}{2bc}$ for some constant $r$, and similarly for the other vertices. Letting $X$ be the expression we want, we have
\[ X = \frac{\frac{1}{tc}\cdot r\cdot \frac{a^2 + b^2 - c^2}{2ab}}{\frac{1}{ta}\cdot r\cdot \frac{b^2 + c^2 - a^2}{2bc} + \frac{1}{tb}\cdot r\cdot \frac{a^2 + c^2 - b^2}{2ac}} = \frac{a^2 +b^2 - c^2}{(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2)} = \frac{1989c^2 - c^2}{2c^2} = \boxed{994}.\]


I thought just like the Asymptote or the $ \text{\LaTeX}$ forum, there should be a geogebra forum. Some people have already asked me/other people questions.

Private message an administrator because asking other people in forum won't create the forum. But I would also like a Geogebra forum.

Yeah I don't think that will happen. It's so simple that there aren't that many questions to ask. I'll probably move all the discussions down in the Asymptote forum (and maybe call that Asymptote and Geogebra forum.)


How many whole numbers from 10 to 99 have units digit greater than the tens digit?

There is $ 12,13,14,15,16,17,18,19,23,24,25,26$.... an easier way to do this is to notice that while each tens digit goes up by one, the number of these numbers decrease by one. So it is $ 8\plus{}7\plus{}6\plus{}5\plus{}4\plus{}3\plus{}2\plus{}1\equal{} \boxed{36}$


quadratics
floor function
modular arithmetic
Quadratic Residues
pen

Let $p$ be an odd prime number. Show that the smallest positive quadratic nonresidue of $p$ is smaller than $\sqrt{p}+1$.

If $a,b$ are quadratic residues, then so is $ab$. Thus it suffices to show that the set $S = \{ s \in \mathbb Z | 0 < s < \sqrt p +1 \}$ generates all residues $\neq 0$.

Assume this to be false and let $x$ be the smallest positive integer whose residue class is not represented by $S$. Then $x \geq \sqrt p +1$ and we consider $a := 1+\left\lfloor \frac{p-1}{x} \right\rfloor$ and observe:
a) $0<a<1+ \frac{p-1}{x} \leq 1+ \frac{p-1}{\sqrt p +1} = \sqrt p$, thus $a \in S$.
b) $p-1 =  \frac{p-1}{x} \cdot x < ax \leq \left( \frac{p-1}{x} +1 \right) \cdot x = p+x-1$, thus (clearly $p \nmid a,x$) $ax-p \in S$ (remember that $x$ was minimal).

But now $x=1 \cdot x \equiv a^{p-1} \cdot x = a^{p-2} \cdot (ax) \mod p$, so $x$ is a product of elements of $S$ seen $\mod p$, contradiction. And we are done

[quote="ZetaX"]
b) $p-1 =  \frac{p-1}{x} \cdot x < ax \leq \left( \frac{p-1}{x} +1 \right) \cdot x = p+x-1$, thus (clearly $p \nmid a,x$) $ax-p \in S$ (remember that $x$ was minimal).[/quote]

I have only one question. You say that $ax-p \in S,$ but is it always true? Actually, I think that it is enough to say that $ax-p$ is a quadratic residue since it is strictly less than $x.$

[quote="ZetaX"]$S$ should be replaced with the residues generated by $S$ there.[/quote]
It is okay now.

Let $a$ be the least quadratic nonresidue $\pmod{p}$, and let $b=\left\lfloor\frac{p}{a}\right\rfloor+1$. Since $0<ab-p<a$, $ab-p$ must be a quadratic residue. Thus, \[1=\left(\frac{ab-p}{p}\right)=\left(\frac{a}{p}\right)\left(\frac{a}{p}\right)=-\left(\frac{b}{p}\right),\] so $\left(\frac{b}{p}\right)=-1$, so $b$ is a quadratic nonresidue $\pmod{p}$, which means $a\le b<\frac{p}{a}+1$, and the result trivially follows.

[quote="me@home"][quote="Peter"]Let p be an odd prime number. Show that the smallest positive quadratic nonresidue of p is smaller than \sqrt{p}+1.[/quote]
Let $S=\{m\in \mathbb{F}_p|m^2<p\}$.
The statement is equivalent to showing that $S$ contains a non-square.
Suppose not, for contradiction; then all of $S$ are squares. The crucial observation here is that $T=\left\{\prod_i s_i | s_i \in S\right\}$ generates all of $\mathbb{F}_p$.
But this yields a contradiction, since we supposed that all of $S$ was square, hence all of $T$ is square, contradiction.[/quote]

Sorry to revive, can  you show how $T$ generates all of $\mathbb{F}_b$? I don't quite see

Instant by [url=http://www.artofproblemsolving.com/community/c6h1146084p5404925]Thue's Lemma[/url]


RSI's application materials are now availabe on their website: http://www.cee.org.  For those of you not familiar with RSI, it is a summer camp devoted to developing a research project.  RSI only accepts applications from students just finishing their junior year, and is free to all participants.

Do I have any chance at all if I have never done any research before?

Bah! they require PSAT scores, I didn't take it, and the next date to take the SAT is 12 days before the deadline is due so theres no way I would get the scores in time.  Oh well, not like I would get in anyway.

the SAT Reasoning Test isn't required is it?

[quote="K81o7"][quote="paladin8"]@xqpx: Yes. AFAIK, most people there hadn't done research prior to RSI.

@nateharman1234: The PSAT isn't the most important part of the application, so I would consider applying anyway.

@krustyteklown: No, the SAT is not required (though a solid number of people had taken it already - just make sure you don't put a bad score on your application).[/quote]

IS a good PSAT score really helpful?  I got a score above the cutoff they recommended, but not by much.  Does it really affect the decision much?[/quote]

Well, as I am not a part of the admissions committee, I really can't say for sure. IMO, it is more of a standardized indicator of aptitude (though clearly flawed in many respects) so doing well would show that you are capable. You would then have to indicate somehow that this capability can translate into research/thinking skills in some field (e.g. doing well on the USAMO, already having done research, etc.).

How critical is it to pick a "right" field/subfield? I like physics, bio, and chem. It doesn't really matter which field/subfield that I end up with (if I could get in) as long as within these three. But I heard that if one picks a very "popular" field, he or she might have less chance to get in.

Does anyone know, in the past, what were the more popular or applied-for fields, and what were the least applied-for fields?

Regarding first-choice and second choice research fields, can they be the same field with two different subfields? For example, bio/neurobio and bio/molecular&cell.

Are there any FAQs about RSI app online?

Richard, is there any chance that we could have a Jam on RSI?
Thanks!

[quote="xqpx"]How critical is it to pick a "right" field/subfield? I like physics, bio, and chem. It doesn't really matter which field/subfield that I end up with (if I could get in) as long as within these three. But I heard that if one picks a very "popular" field, he or she might have less chance to get in. [/quote]

They want people who know what they're doing in their field. They want you to have extensive experience/knowledge already; if you are confident in your abilities in multiple fields, it's possible to get a project involving all of them. As far as I know, picking a popular field does not influence your chances too much; this summer, we had easily 30+ bio projects. But... who knows?

[quote="xqpx"]Does anyone know, in the past, what were the more popular or applied-for fields, and what were the least applied-for fields? [/quote]

As I mentioned, many people did bio projects. There were considerably less of pretty much everything else; math was probably the next most popular, followed by physics (just off the top of my head). Chemistry, computer science, engineering, and astronomy were at the lower end.

[quote="xqpx"]Regarding first-choice and second choice research fields, can they be the same field with two different subfields? For example, bio/neurobio and bio/molecular&cell.[/quote]

That's a good question. I don't have an answer, but I selected two different fields completely.

[quote="xqpx"]Are there any FAQs about RSI app online? [/quote]

Not that I know of. Your best bet is probably talking to alums.

I'm with Jeffrey on all of these.  If you missed the PSAT date, definitely take the SAT as soon as you can, and even if you get the scores in after the application deadline, CEE will probably understand, although I'd contact them ASAP to let them know about your situation.

From what I hear, PSAT scores don't matter that much provided you're above some reasonably high number (the one they toss around is 220, but a bit below that is probably fine too).  Lower scores are going to cause them to ask more questions.  But if you get a really high score--great, so did hundreds of the roughly 1200 applicants.  Now you need to show that you have taken initiative; doing research or succeeding in math/science contests are some ways to do that.  I think they look for leadership too, so other (not necessarily math and science-related) extracurriculars will probably help here.

As for picking the right field/subfield--I have no idea.  On one hand, everyone and his grandmother applies for bio (I did).  On the other hand, there may be more bio mentors than, say, quantum physics mentors out there.  Given that mentors aren't exactly dime a dozen, I'm willing to bet that the field breakdown is determined by the availability of mentors rather than the availability of students, in which case it's probably a good idea to apply for bio (since if you apply for physics, there are fewer spots, so you're more likely to get edged out by some...olympiad winner or something).

Since this is a math forum, maybe we need a disclaimer about math projects: they're hard, and I suspect the competition for math spots is pretty tough.  And random?--about half the math people were MOPpers, and the other half had never taken USAMO, which is kind of mysterious.  (One of them was because Sherry was accepted but turned them down, so they took someone off the waitlist; not sure about the rest.)  Anyway, the point is, unless you're really good at math, it might be a good idea to apply for something else.

Of course, the most important thing is, as Jeffrey said, experience.  And passion--pick what you like, or you might be bored.

[quote="krustyteklown"]ackkk, PSAT score wasn't so hot; however...

when they say 2 recommendations, is that a cap, or a minimum?[/quote]

I suspect that it's an exact number... I think it's best to get your two strongest recommendations; that way even if you have another "good" one there won't be the possibility that it detracts from your application by not being "great."

That's a real shame because I'm much more into quantum physics than biology.

They say from math and/or science teachers.  Is that required?  I'm assuming that's best?

I think the general idea here is to present yourself as best you can.  If you really like quantum physics more than biology, then you probably want to try to do research in quantum physics rather than biology.  There is no magic way to "game the system" by submitting X number of recommendations or choosing X field / subfield.  And for many things that you can't change, like your PSAT score, your previous research experience, or what have you, well, you can't change them.

K81o7: I'd generally recommend against non-math/sci teachers, but then again I don't know your situation.  I don't think anything is required per se, so do what you think will present you best.

krustyteklown: I think its a minimum.  I definitely know people who submitted 4-5ish.  But also many with just two.  I had three myself.

About the math projects: They are hard, but if you stick to what you feel you know and are comfortable with, you should be ok.  Plus the mentors are largely very good.  So don't be discouraged from applying to math if you really want to :)

Good luck with the apps!

[quote="calc rulz"]Although I'm not applying until next year, do they have a way to give AMC/etc scores?[/quote]

You can just put it somewhere on the application.

[quote="JSteinhardt"]So do they let you come late to RSI? I.e. is it possible to do MOP and then come to RSI a week late? Also is the decision deadline before or after USAMO-type stuff occurs?[/quote]

Last summer, I believe they did not let you come late. There were several students who qualified for MOP but chose to attend RSI instead. The decision deadline last year was March 31, a couple of weeks before the USAMO, so you should have plenty of time to decide.

[quote="K81o7"]The deadline is nearing, and so I'm getting pretty worried about something:

What kind of score verification is needed?  For APs, etc. I can just make a copy of the score report, but what about AMC scores, etc?

Another question: I asked one teacher for a recommendation, and the next day, a close family member passed away.  What to do in this situation?  Would RSI listen if I had to send an application a little late?  I can probably ask another teacher, but if not?[/quote]

No score verification is necessary, it's like college applications (as far as AP/AMC scores go). They will take your word for it; obviously, if you lie and they find out, it's not going to be pretty...

Hmm, as for your other situation, I'd try contacting them directly about it. I'm sure that if you give them sufficient reason to accept a late recommendation, they'll be willing to deal with it. That said, they need to finish the application process by the end of March (something like that), so it can't be too long.

I hope none of the information I provided is incorrect; I would check with them to make sure.

I actually think that you need to send a copy of your ACT/SAT scores...not from College Board, but just a photocopy of the report sent to your house.  Other than that, I agree with paladin8: call them if you have any questions!  Good luck!  :)

I assumed that the AP and SAT II scores would be on my transcript, so I didn't send anything else. I'm applying for math, but I'm going to be blown away by all the MOPpers.

hmmm, some schools do that, but you should send them copies of your scores just in case.

Well thanks for the input...

The teacher came back and said he was fine.  Nevertheless, I've asked another teacher just in case, so I guess 3 recommendations won't hurt...

I guess I'll just send print outs of score reports, but for contest scores, I have no real proof (unless they want pictures of trophies, medals, etc...

[quote="white_horse_king88"]They said keep the answers to a 2 paragraph maximum. But don't you need basically a paragraph to explain each of your questions/problems for application question number two? Because I certainly can't sum up why I think three different questions are important in a short little paragraph. Not to mention rather that it's six questions in two different subfields. I'm a little confused.[/quote]

I interpreted the 3 page total to be the more relevant limit.  Obviously the response to question 2 (1st and 2nd field and subfield) should be longer than two paragraphs.

[quote="xqpx"] Regarding first-choice and second choice research fields, can they be the same field with two different subfields? For example, bio/neurobio and bio/molecular&cell. [/quote]

I e-mailed them about this and they said they prefer two (field, subfield) pairs with different fields.

also, on page 2 of the application, where it asks you to state
Research Interest Primary Field
Research Interest Sub Field

it should juts be something like
Math for the first one
and Combinatorics for the second one;

as opposed to listing out both your choices for field/subfield

is that right?

Yes, that's how I interpreted it. Although I did not chose math for either of my fields.  :roll:

Darn, I chose math for both my fields. I should have chosen biology for one; I always wanted to fiddle with genes.

I know I'm double posting...but considering it's been a few weeks, I assume this is okay.

So decisions came out! Who made it from AoPS?

Hi.  :D

I'm surprised considering I only wrote about one field and one subfield (number theory).  Perhaps I was able to turn that into a good thing?   :P

You guys alerady got replies?  I'm still waiting on my rejection letter...unless by some freak accident I was accepted...

Edit: I got the letter, rejection.

Edit 2: Btw, I applied for NT and group theory.

t0rajir0u, I did the exact same thing and made it last year.  Number Theory FTW.

I applied nt too, but graph theory was my first choice. All I really said about nt was "i might want to work with primes."