[color=blue]He aquí una lista de enunciados que escogí para quienes aún no se animan a participar del foro principal.
Ruego que los más avanzados se abstengan de enviar sus soluciones, que supongo serán inmediatas.
Recuerden también, en [b][i]Subject[/i][/b], escribir el número del enunciado a resolver.

Éxitos.[/color]

12. [color=blue] Halle todos los enteres positivos $n$ tales que $n+1|n^{2}+1$.[/color]

$m(n+1)= n^{2}+1$

$\frac{m(n+1)}{n}= \frac{n^{2}+1}{n}$

$m(n+1) \equiv{}n^{2}+1 \mod n$

Sabemos que:

$n^{2}+1 \equiv{}1 \mod n$
$n+1 \equiv{}1 \mod n$

Por lo tanto $m=1$, nos queda que:

$n+1\equiv{}n^{2}+1 \mod n$
$n\equiv{}n^{2}\mod n$

$1\equiv{}n \mod n$

Dado que para cualquier numero $a$ se cumple que:

$a \equiv{}a \mod n$

La unica solucion posible es $n=1$

[color=blue]Saludos[/color]

[quote="jcmtnez"]12. [color=blue] Halle todos los enteres positivos $n$ tales que $n+1|n^{2}+1$.[/color]

$m(n+1) \equiv{}n^{2}+1 \mod n$

Sabemos que:

$n^{2}+1 \equiv{}1 \mod n$
$n+1 \equiv{}1 \mod n$

Por lo tanto $m=1$[/quote]

Lo que puedes concluir es que $m\equiv 1 \mod n$, no que $m=1$.

[hide="Mi Solución"] Sabemos que $n+1 \mid n^{2}-1$ luego $n+1 \mid (n^{2}+1)-(n^{2}-1)=2$, de donde $n+1\leq 2$, de esto es fácil ver que $n=1$ es el único valor posible. [/hide]

$Tipe$

[color=blue]Bien aclarado, $Tipe$.
De paso, quisiera recomendar el enunciado $4$. Tiene una solución muy bella y simple.
Si quieren algo un poco más desafiante, el $1$ es el indicado.[/color]

Buenas!
Soy muy nueva en esto de postear en el foro aunque admito que hace pila de tiempo que lo conocia.
Aca mando 2 soluciones de los problemas (que me fue mas facil hacerlos en word que aca)
En fin... espero poder seguir participando activamente de esto.
SAludos a todos
gabi.-

Che RAfa... te pasaste con este topico!

[color=blue]¿Alguno más desea enviar su solución a otro enunciado?
Ya me había encargado de corregir el del último post a la persona correspondiente.[/color]

9. [color=blue]Halle todos los enteros $a \neq{}0$ tales que $a^{a^{n}}=a \forall{}n \in{}\mathbb{N}$[/color]

(1) $a^{a^{n}}=a^{an}=a^{1}$
(2) $an=1 \Longrightarrow{}a= \frac{1}{n}$

Dado que $n \in{}\mathbb{N}\Rightarrow{}n\geq{}1$, por lo tanto la unica solucion es $n=1\Rightarrow{}a=1$ ya que $n>1\Rightarrow{}a\in{}(0,1)$ lo cual es una contradiccion ya que $a\in{}\mathbb{Z}$ y no hay ningun entero en dicho intervalo.

Saludos

[hide]Problemas 6
He encontrado las solucion a este problema pero no hayo la manera de demostrar la formula, alguna idea?[/hide]

[quote="jcmtnez"](1) $a^{a^{n}}=a^{an}=a^{1}$[/quote]
[color=blue]Hay una pequeña definición que te trajo confusiones. No son lo mismo $a^{a^{n}}$ y $(a^{a})^{n}$. Considera el primero como $a^{(a^{n})}$.
¿Cómo resolverías, en ese caso?

[b]P.D.:[/b] Bonita solución del enunciado $13$. Comparando con otras soluciones que hicieron otros, ésta es muy original.[/color]

[quote="jcmtnez"][hide]Problemas 6
He encontrado las solucion a este problema pero no hayo la manera de demostrar la formula, alguna idea?[/hide][/quote]
[color=blue][hide="Ayuda."]Intenta aplicar el método inductivo.[/hide][/color]

[b]A ver, soy nueva acá así que aquí les va la manera en la que resolví la 13., no tengo latex, pero espero que se entienda:[/b]

[b]13. Demostrar que 13 | 2^70 + 3^70[/b]

[hide]2 Ξ 2 (mód 13)
2^2 Ξ 4 (mód 13)
2^3 Ξ 8 (mód 13)
2^4 Ξ 16 (mód 13) y 16 Ξ 3 (mód 13) por lo tanto 2^4 Ξ 3 (mód 13)
2^5 Ξ 6 (mód 13)
2^6 Ξ 12 (mód 13) y 12 Ξ -1 (mód 13) por lo tanto 2^6 Ξ -1 (mód 13)
(2^6)^11 Ξ (-1)^11 (mód 13)
2^66 Ξ -1 (mód 13)
2^70 Ξ -3 (mód 13)

es decir, ambos dejan el mismo residuo al ser divividos entre 13, el cual es 10

3 Ξ 3 (mód 13)
3^2 Ξ 9 (mód 13)
3^3 Ξ 27 (mód 13) y 27 Ξ 14 (mód 13) por lo tanto 3^3 Ξ 14(mód 13) y
14 Ξ 1 (mód 13) por lo tanto 3^3 Ξ 1 (mód 13)
(3^3)^23 Ξ 1^23 (mód 13)
3^69 Ξ 1 (mód 13)
3^70 Ξ 3 (mód 13)

es decir, ambos dejan el mismo residuo al ser divividos entre 13, el cual es 3

	2^70 = 13q1 + r1		donde q1 representa un cociente y r1 el residuo 10
	3^70 = 13q2 + r2		donde q2 representa otro cociente y r2 el resuduo 3

si sumamos miembro a miembro nos queda:

	(2^70 = 13q1 + 10)  +
	(3^70 = 13q2 + 3)
	-----------------------
	2^70 + 3^70 = 13q1 + 13q2 + 13
	2^70 + 3^70 = 13 (q1 + q2 + 1)

	entonces 13 | 13 (q1 + q2 + 1) por lo tanto 13 | 2^70 + 3^70[/hide]

[b]si hay alguna error no duden en decírmelo.

Saludos![/b]

[color=blue]Correcta tu solución. Es útil saber de congruencias para resolver casi instantánteamente estos problemas.

Te recomiendo que, para acortar la solución,utilices apliques también algunas propiedades del sistema de congruencias al final.
Al haber hallado $2^{70}\equiv-3(mod13)$ y $3^{70}\equiv3(mod13)$, puedes directamente asumir que $2^{70}+3^{70}\equiv-3+3\equiv0(mod13)$, con lo cual es inmediato que $13|2^{70}+3^{70}$.

[b]P.D.:[/b] Disculpa mi tardanza al contestar. Tengo tantos deberes adicionales, que tuve que dejar de lado este foro por un muy largo periodo.[/color]

[quote="jcmtnez"][hide]Problemas 6
He encontrado las solucion a este problema pero no hayo la manera de demostrar la formula, alguna idea?[/hide][/quote]
[hide="Nueva ayuda, mejor explicada."][color=blue]Al parecer, tuviste algunos inconvenientes con este enunciado. Halla los primeros valores de $u_{n}$, e intenta inducir una fórmula general para cualquiera de los valores de la serie.Por último, intenta aplicar el razonamiento inductivo, que creo que lo dominas.[/color][/hide]


function
algorithm
number theory
Euclidean algorithm
number theory proposed

Hi eveyybody, I would like to ask whether any one has ideas on this question:

For any prime number p where 4|(p-1), there exist a unique pair of natural numbers x, y such that x^2 + y^2 = p

Now, suppose a, b are some integers where
y*(a^2 - b^2)  + 2*x*a*b = + 1 or -1

Is it possible for a, b be explicitly written as a function of x, y?
And what should that function be in that case?

The first is extremely well-known.  See [url=http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares]here[/url].  There is also a very interesting proof using [url=http://en.wikipedia.org/wiki/Minkowski%27s_theorem]Minkowski's theorem[/url].

I'm not sure I understand the second question.  Are $ x$ and $ y$ given?  If so, the general solution to $ Ay \plus{} Bx \equal{} \pm 1$ is given by the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]extended Euclidean algorithm[/url] but cannot, to my knowledge, be written in closed form as a simple function of $ x$ and $ y$.  We then have the further constraint that $ \sqrt{A^2 \plus{} B^2}$ is a square.

I assume we shall have something like $ (2x)^2\plus{}y^2\equal{}p$ or so...

The main part of the question is that is there any way to express x, y interms of a, b should p is given.

As you know x, y are unique for a certain p, that means a given p would yield a particular pair of x, y. I am just wondering whether the equation yielded after the application of Euclidean Algorithm will be like a function of p, or a particular form related to p.

There is a formula for $ x,y$ depending only on $ p$. Thus you can express them even without given $ a,b$.

You are correct should you know directly after computing the x, y from a given p. But I am just wondering whether the function for a, b could be finally expressed as a form without x and y.

Brutal forces can of course be applied to find out the relationship between a, b and x, y using Euclidean Algorithm once p is given explicitly. However, I want to make one step forward.


Is there a minimum number of solutions we can submit?

We would ask that you submit solutions to a minimum of two problems per round.


Given $ x\equal{}0.\overline{31}_{5}$, find the value of $ x$, expressed as a fraction in lowest terms.


diameter
geometry
point set
combinatorial geometry
IMO
IMO 1965

In a plane a set of $n\geq 3$ points is given. Each pair of points is connected by a segment. Let $d$ be the length of the longest of these segments. We define a diameter of the set to be any connecting segment of length $d$. Prove that the number of diameters of the given set is at most $n$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=distance&t=46728]this[/url].

There's a proof there that if we give our set of points a graph structure by connecting two vertices iff the distance between them is the diameter, then this graph has at most one cycle. Every graph on $n$ vertices with $\ge n+1$ edges, on the other hand, has at least $2$ cycles, so we're done.

Alternatively, in the graph where two points are connected if and only if they have a diameter, it is not hard to see that any two edges must intersect because otherwise a segment with length over $d$ could be formed.  Then we go by induction.  For $n=3$ it is trivial, and we can assume no vertices have degree under $2$, lest we cut them out and reduce it to the $n-1$ case.  If $A$ has $3$ neighbors in this graph $B$, $C$, $D$, with $C$ is the middle (surely $B$, $C$, $D$ lie in some $60$ degree angle), then it is impossible to have an edge with an endpoint at $C$ that meets both $AB$ and $AD$, contradiction.  So we can assume all vertices have degree at most $2$ and then we get the desired result.

Iam really stuck in this problem , Can anyone expalain it simply, please ?


I made a video of me drumming
Im not that great  but check it out

[youtube]WfUQrOnL_LM[/youtube]

remember, im pretty much a beginner so just watch it and comment on it, thanks.  :D

Btw, anyone else here play drums? Can you tell me how to tune some of my drums?

EDIT:
Here's my other video.
[youtube]PxfWLRKHTcU[/youtube]

nice set  :D

Those sticks look sweet too :D

I've played drums for 2 or 3 years. To tune your drums, first tune each drums head knob turning thing so that the drum head is balanced. To do that, just take a drum key and adjust each knob, meanwhile tapping on the drum head in front of it. Do this with every single knob on the drum head, then adjust each knob to make the drum sound how you want it.

Btw, what's your favorite type of drumming? I like rock the best, but I don't really know a lot of types...my teacher is a mainly jazz-style teacher.


arithmetic sequence
number theory proposed
number theory

9.7 Is there an infinite arithmetic sequence $\{a_n\}\subset \mathbb N$ s.t. $a_n+...+a_{n+9}\mid a_n...a_{n+9}$ for all $n$?
([i]V. Senderov[/i])

No, there does not.

Let $a_n=a+nd$. $a_n+...+a_{n+9}=10(a+nd)+45d$, while $a_n...a_{n+9}=(a+nd)...(a+nd+9d)$. Observing that $\gcd (a+nd+md, 2(a+nd)+9d) \leq$
$ \gcd (2a+2nd+2md, 2a+2nd+9d) = \gcd (2a+2nd+9d, (2m-9)d) \leq (2m-9)d$, where $0 \leq m \leq 9$, it is clear that divisibility cannot occur once $n$ exceeds $(10!)d^{10}$.


There's one more possible case- [b]ababa[/b]
There are $9\times 9=81$ numbers to satisfy this condition.

So the total number of 5-digit palindrome is $819+81=\boxed{900}$

[hide="Easier approach"]
The general form of 5-digit palindrome is $abcba$, and let's say $a$, $b$, and $c$ could be the same number.

Then, it's simply $9\times 10\times 10=\boxed{900}$, which is the same answer as we obtained above ;) [/hide]

There was a discussion of this here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52588 .

Basically, there are 9 choices for the first digit. Then for half(rounded up) of the remaining digits, there are 10 choices. In the 5-digit case, this is 900.

[quote]There is a quicker, easier solution. Which digits must we actually consider when counting?[/quote]

I think we only have to think of last three digits....

and using bunch of additions of combinations, we'll be able to get the answer

[quote="MCrawford"]There is a quicker, easier solution.  Which digits must we actually consider when counting?[/quote]
Isn't it just what I said in the previous post? (Look at the hidden section called "Easier approach")

[quote="guile"]I was working on a problem, which came from this forum (but has no answer). Kindly check my work if its correct otherwise post your solution. Thank you in advance.

How many 5-digit palindromes are there? (Palindromes are read the same backwards and forward.) ex. 15351

Solution

aaaaa = 9 choices
abbba = 81 choices (a has 9, since 0 is not allowed)
aabaa = 81 choices
acbca = 648 choices ( a has 9, c has 9, and b has 8)

Total is 9 + 81 + 81 + 648 = 819[/quote]

It's much easier if you look at the placements:

$9 \cdot 10 \cdot \cdot 10 \cdot 1 \cdot 1$, the first digit can be any integer $1-9$, can't be $0$, but the next two digits can be, and the final digits are already chosen.


inequalities
logarithms
function
number theory open
number theory

Is the following true or not? Also interested in the proof

Let $ n\#$ denote the product of all primes lesser or equal to $ n$. Than $ n\#\leq e^n$

It is not true, but I don't really have a proof. Notice that this is equivalent to $ \theta(n) \le n$ where $ \theta(n) \equal{} \sum_{p \le n} \ln p$, a famous Chebyshev Function, and the prime number theorem implies $ \theta(n) \sim n$. Googling gave me an article which mentions that $ \theta(x) \minus{} x$ changes sign infinitely often, so it follows that the inequality is false...
[url=http://www.emis.de/journals/JIPAM/images/295_07_JIPAM/295_07.pdf]The article that mentions it.[/url]
I am sure (or hope?) that you can find some $ n$ for which your inequality doesn't hold using a computer... and I might try that later.

[quote="bambaman"]
I am sure (or hope?) that you can find some $ n$ for which your inequality doesn't hold using a computer... and I might try that later.[/quote]
I don't sure, because minimal $ n$, suth that $ \theta (n)>n$ is very big about $ 10^{4000}.$

How to proove $ \theta(n)\sim n$.
if i understand this correctly than this would imply that for every positive $ \epsilon$ we have $ n\#\leq (e\plus{}\epsilon)^n$ for large enough $ n$?


Hi I was just wondering if anyone could help me with factorising this algebraic equation. If possible I would love someone to give me a simple step by step run through so I know how to work out problems like these in the future. Thanks in advance.

[u]Equation[/u] : [b]4x³-x²+12x-3[/b]

$4x^3-x^2+12x-3$
$x^2(4x-1) + 3(4x-1)$
$(x^2+3)(4x-1)$
This is called factoring by grouping
Just practice these enough and you will get used to them...
You have to see that you need to take out the 4x-1 and that you can do that in both *factors*...  :)

it was a typo at the start but it turns out to be the same solution


AMC
USA(J)MO
USAMO
AIME
AMC 10
AoPS Books
AMC 12 A

Okay, so the thing is that I really want to qualify for the USAMO next year. The thing is, I don't know if I could. This year, I got a 120 on the AMC 10 (should've been a 132) and a 4 on the AIME. I am planning on doing about 2 hours of math a day this summer, go over AoPS Volume 1 & 2, and take old AMC's and AIME's. What are my chances of qualifying for the USAMO if I follow this plan? I'm really motivated but if there is a slim chance of qualifying for the USAMO by doing this then I was thinking of maybe using my time to do other things. (like study for the SAT writing and CR)

yeah it is very possible. i went from an 87.5 amc 10 last year to probably qulifying for usamo this year (8 aime but idk about the 2nd index). all i did was do aops 1, some of aops 2, some past amc's, and some past aime problems (never entire tests). i worked whenever i felt like it during the summer, and ~4 hrs every weekend during school (only random aime problems during school days). so i second matt276eagles in saying that it is very possible, but don't force yourself to work.

Its DEFINITELY POSSIBLE! Or atleast, I really want to believe that because I'm sort of in the same situation. But I'm sure you can do it, it'll just be a lot of hard work, and you'll have to cover each topic thoroughly.

it's definitely possible. i had a 97.5 amc 12 last year, with a 5 aime. this year i had a 120 amc 12 and 8 aime, which might get me in. but my aime score could have been much higher - i didnt practice contest math at all between the amc 12 a date and like a week before the aime.
all i did was skim aops vol 1 (most of that stuff is pretty simple) and work through about the first 10 chapters of vol 2. as far as old tests go, i'd say time yourself doing complete tests rather than just do problems. if you do all the problems after each chapter in the aops books, you'll have more than enough practice doing individual problems. the timing will give you an idea on how to pace yourself. however, finish the whole test regardless of time in the end - that way you'll also get practice doing the problems toward the end of the test, which are harder. the timing will essentially give you an idea of what score you would get.
do aops vol 1/2 first, to get the concepts, and then do the tests afterwards. or you could do some concurrently, but i wouldnt advise doing that for aimes, only amcs.

but, as it has been said, make sure you're having fun doing the stuff. because otherwise it really isnt worth it.

Thanks for the feedback. Don't get me wrong, I seriously enjoy math, otherwise, I wouldn't even consider studying hard for the USAMO. So, AoPS Volume 1 & 2 and old contests would be enough? Also, how much studying a day (roughly) should I aim for this summer?

i'd say it would be a good goal to finish AoPS 1 by the end of summer while doing amc 12 tests. If you're going to be in 11th grade or higher, than you need a good amc 12 scroe (I'd shoot for 120+). Take as long per chapter as you need. A lot of the material should be simple, but the problems can take you to a high amc 12 score. then i guess work on aops 2 while doing more amc 12 tests and random aime problems.

I am in like the same position as you. This summer i will begin AoPS volume 1 and do ALL the past AMCs, lol. I am planning to study about 2-4 hours this summer, but i'm not an expert. I cant really say how much you have to study. I guess it all depends on the individual's progress in the practices. Whether it is [b]good[/b]  and productive practices.

practice a lot but make sure you dont get bored. because 2-4 hours a day all summer can get old very fast. its not as much about how much you practice each time as it is about consistency. you want to get into a habit of studying. but ya, aops 1 and 2 and old tests are all i used.

[quote="matt276eagles"]It's [i]possible[/i], and I personally went from a AMC 10 score in the 100-110 range to qualifying for USAMO the next year, but there's no magic formula that guarantees USAMO qualification.  If you want to put a lot of time into contest-type math, I would hope that it's because you enjoy it and not because you just want the achievement of qualifying for USAMO.  Also, if the alternative is SAT-prep, you'll get a whole lot more out of math contests that you will out of preparing for the SAT.[/quote]
when you were a freshman or sophmore you got 110?


SA Se demnostreze ca exista o infinitate de numere prime de forma 4n+1.

[quote="spix"]este de forma $4k+1$,atunci el are un divizor prim de forma $4p+1$ [/quote]
Hm, de ce? $49$ e de forma $4k+1$ si nu are divizori primi de forma $4p+1$ (altfel spus, daca ai un numar par de divizori primi $\equiv 3(mod4)$, atunci produsul lor e un numar $\equiv 1 (mod4)$

Da e gresit ce am facut.
Sper ca e corect acum:
   La fel presupunem ca numai $p_1,p_2,..p_n$ sunt numere prime de forma $4k+1$. Consideram sirul $(q_n)$ in care $q_n$ reprezinta al $n$-lea numar prim,incepand de la $3$.
   Numarul $N=q_1q_2....q_m+4$ pentru $m$ suficient de mare contine toate numerele prime de forma $4k+1$ si in rest numere prime de forma $4k+3$. Daca $m-n$ este par atunci $N$ este congruent cu $1$ modulo $4$,mai mult $N$ nu se divide cu nici un $q_i$ si fiind impar nici cu $2$ deci este prim de forma $4k+1$ si nu apartine $\{p_1,p_2,...,p_m\}$ contradictie.

Sa se demonstreze ca exista o infinitate de numere prime de forma $4n+3$.  :D

[quote="spix"]$N$ nu se divide cu nici un $q_i$ si fiind impar nici cu $2$ deci este prim [/quote]
Din nou eu cârcotesc, dar nu cred că e bine. Asta nu inseamna ca $N$ nu poate sa se divida cu vreun numar prim mai mare ca $q_m$. Tu incerci sa aplici aici demonstratia de la faptul ca exista o infinitate de numere prime, dar nu prea merge:)

Ai dreptate nu e bine. De fapt eu incerc sa aplic principiul extremal(asa am citit pe undeva) si pentru a demonstra ca exista o infinitate de numere prime de forma $4n+3$ iese.

Voi incerca sa demonstrez ca exista o infinitate de numere de forma $8n+5$.
Demonstratia nu e elementara.
Fie $p_n$ al n-lea numar prim, incepand de la $3$.
Fie $a=p_1p_2...p_n$ si $N=a^2+4$.
Cum $a$ este impar, avem $a=2l+1\Rightarrow N=4l^2+4l+5\Rightarrow N\equiv 5 (mod 8)$
Daca $N$ ar avea doar divizori primi de forma $8t\pm 1$, atunci si $N$ ar fi de aceeasi forma, absurd.
Deci $N$ are si divizori primi de forma $8t+3$ si $8t+5$.
Voi arata ca $N$ nu are divizori primi de forma $8t+3$.
Presupunem ca are, deci fie $p|N$ prim, $p=8t+3$.
$p|N = a^2+4\Rightarrow$ -4 este rest patratic modulo $p$.
Deci $\left(\frac{-4}{p}\right) = 1$, unde prin aceasta notatie ma refer la simbolul lui Legendre.
$1=\left(\frac{-4}{p}\right) =\left(\frac{-1}{p}\right)\cdot\left(\frac{2}{p}\right)^2=\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}= (-1)^{4t+1}=-1$, absurd.
Deci $N$ are doar divizori primi de forma $8t+5$, care nu divid pe $4$, deci nici pe $a$, deci orice divizor prim al lui $N$ e mai mare ca $p_n$, de unde rezulta imediat ca avem o inifinitate de numere prime de forma $8t+5$. Acestea sunt si de forma $4t+1$.

Stii cumva vreun link sau o carte unde as putea gasi aplicatii la simbolul lui Legendre?


[quote="Hawk Tiger"]Let $k$ and $m$ be positive integers.Show that $S(m.k)=\sum_{n=1}^{\infty}{\frac{1}{n(mn+k)}}$
is rational if and only if $m$ divides $k$[/quote]

The if part is clear:

Let $k=mp$. Then $\sum_{n=1}^{\infty}{\frac{1}{n(mn+k)}= \frac{1}{m}\sum_{n=1}^{\infty}{\frac{1}{n(n+p)}= \frac{1}{mp}\sum_{n=1}^{\infty}{\left(\frac{1}{n}-\frac{1}{n+p}\right)}}}$
And after telescoping we obtain a finite sum of reciprocals, thus, a rational number.

I don't think there is an elementary solution for the other part of the problem. Using Baker's results on linear forms of  logarithms, the problem is not difficult, but otherwise I think it is impossible.

I found this problem in the Hojoo Lee'  list of elemtery problem ,So I think it's elemetry.  :P

Man, be serious. You can also find Mahler's absolutely non-elementary result on the irrationality measure of $\pi$ in the very same paper!


function algebra domain calculus derivative logarithms search

Does the Riemann zeta function take on any real negative value in the half-plane $ \sigma > 1$?

Hi.  It was wrong for $ 1 < x < \zeta^{ \minus{} 1}(2)$.  Here's what I think is correct for $ x > 2$:
For $ z \equal{} x \plus{} iy$ with $ x > 1$:
\[ \zeta(z) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {1}{n^z} \equal{} \sum_{n \equal{} 1}^{\infty}\left(\frac {\cos(y\ln(n))}{n^x} \minus{} i\frac {\sin(y\ln(n))}{n^x}\right)\]
and so:
\[ \text{Re}\, \zeta(z) \equal{} 1 \plus{} \sum_{n \equal{} 2}^{\infty}\frac {\cos(y\ln(n))}{n^x}\]
but for $ x>2$:
\[ \left|\sum_{n \equal{} 2}^{\infty}\frac {\cos(y\ln(n))}{n^x}\right|\leq \sum_{n \equal{} 2}^{\infty} \frac {1}{n^2} \equal{} \zeta(2) \minus{} 1\]
and $ \zeta(2) \equal{} \frac {\pi^2}{6}\approx 1.66$ so thus:
\[ \sum_{n \equal{} 1}^{\infty}\frac {\cos(y\ln(n))}{n^x} > 0,\quad x > 2\]
So if this is correct, we only need to check the interval $ 1 < x < 2$.

We are not specialists in the field. If you can explain your notation in a line, do it. For example, you could have asked the question this way:

Does the Riemann zeta function $ \zeta(\sigma\plus{}it)$ take on any real negative value in the half-plane $ \sigma>1$?

For $ z \equal{} x \plus{} iy$, I've previously established that $ \text{Re}\,\zeta(z) > 0$ when $ x > \zeta^{ \minus{} 1}(2)\approx 1.728 \equal{} x_0$.  But we can determine additional regions in the range $ 1 < x < x_0$ at which it is positive as well:

Consider again $ \text{Re}\,\zeta(z) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {\cos(y\ln(n))}{n^x}$ and let $ y_k \equal{} \frac {2k\pi}{\ln(2)}$. Then $ \cos(y\ln(2)) \equal{} 1$ and at these points:

$ \sum_{n \equal{} 1}^{\infty}\frac {\cos(y_k\ln(n))}{n^x} \equal{} 1 \plus{} \frac {1}{2^x} \plus{} \sum_{n \equal{} 3}^{\infty}\frac {\cos(y_k)\ln(n))}{n^x}$ and again:
\[ \left|\sum_{n \equal{} 3}^{\infty}\frac {\cos(y_k)\ln(n))}{n^x}\right|\leq \zeta(x) \minus{} (1 \plus{} \frac {1}{2^x})\]
So that if $ 2(1 \plus{} \frac {1}{2^x}) > \zeta(x)$, then at $ \{y_k\}$, we have $ \text{Re}\,\zeta(s) > 0$.  This is at the point $ x_1\approx 1.474$ and by continuity, $ \text{Re}\,\zeta(z) > 0$ in a small neighborhood surrounding each $ x\plus{}iy_k$ for $ x>x_1$ and extending in a horizontal strip to the point $ x_0 \plus{} i y_k$.

I'm wondering how far this line of reasoning can be extended to increase the domain in the strip $ 1 < x < x_0$ at which $ \text{Re}\,\zeta(z) > 0$?

I wish to extend further, the boundary in which $ \text{Re}\,\zeta(z) > 0$ for $ z = x + iy$ in the half-plane $ x > 1$.  Earlier, I showed that for $ x > x_0$, $ \text{Re}\,\zeta(z) > 0$ where $ x_0 = \zeta^{ - 1}(2)\approx 1.73$.  Consider again:
\[ \text{Re}\,\zeta(s) = \text{Re}\,\sum_{n = 1}^{\infty}\frac {1}{n^s} = 1 + \frac {\cos(y\ln(2))}{2^x} + \sum_{n = 3}^{\infty}\frac {\cos(y\ln(n))}{n^x}\]
Since $ \left|\sum_{n = 3}^{\infty}\frac {\cos(y)\ln(n))}{n^x}\right|\leq \zeta(x) - (1 + \frac {1}{2^x})$, then $ \sum_{n = 1}^{\infty}\frac {\cos(y)\ln(n))}{n^x} > 0$ when
\[ f(x,y) = 2 + \frac {1}{2^x}\left[1 + \cos(y\ln(2))\right] - \zeta(x) > 0\]
and I wish to determine when $ f(x,y) = 0$ in some region $ 1 < x < x_0$.  Since $ \zeta(x)$ is a strictly decreasing function on the real axis, the smallest value of x such that $ f(x,y) = 0$ occurs when the expression $ 2 + \frac {1}{2^x}\left[1 + \cos(y\ln(2))\right]$ reaches it's maximum value.  That is, when $ y = \frac {(2n + 1)\pi}{\ln(2)}$ so that $ 2 + \frac {2}{2^x} = \zeta(x)$ or $ x = x_1\approx 1.474$
So that we wish to determine the behavior of the expression:
\[ f(x,y) = 2 + \frac {1}{2^x}\left[1 + \cos(y\ln(2))\right] - \zeta(x) = 0,\quad x_1\leq x\leq x_0\]
Now consider:
\[ \frac {\partial f}{\partial x} = - \frac {\ln(2)}{2^x} - \frac {\ln(2)}{2^x}\cos(y\ln(2)) - \zeta'(x)\]
Since:
\[ \text{Max}\,\left[\frac {\ln(2)}{2^x}\big(1 + \cos(y\ln(2))\big)\right]\leq \ln(2)\]
and:
\begin{align*} \zeta'(x) & = - \sum_{n = 2}^{\infty}\frac {\ln(n)}{n^x} \\
& = - \frac {\ln(2)}{2^x} - \frac {\ln(3)}{3^x} - \frac {\ln(4)}{4^x} - \cdots \\
& < - \ln(2)\sum_{n = 2}^{\infty} \frac {1}{n^x} \\
& < - \ln(2)\left[\zeta(x) - 1\right] \end{align*}
and for $ x_1 \leq x\leq x_0$, $ \zeta(s) > 2$, therefore $ \zeta'(x) < - \ln(2)$ in the range $ x_1 < x < x_0$.  Therefore, for each value of $ x_1 < x < x_0$ at which $ f(x,y) = 0$, $ f(x,y)$ will have a positive derivative in the x-direction and therefore $ f(x,y) > 0$ to the right of the line $ f(x,y) = 0$.
Solving for $ y$:
\[ y_1(x) = \frac {1}{\ln(2)}\arccos\left[(\zeta(x) - 2)2^x - 1\right],\quad 0 < y < \frac {\pi}{\ln(2)}\]
and by periodic extension:
\[ y_2(x) = \frac {2\pi}{\ln(2)} - y_1(x),\quad \frac {\pi}{\ln(2)} < y < \frac {2\pi}{\ln(2)}\]

\[ y_3(x) = \frac {2\pi}{\ln(2)} + y_1(x),\quad \frac {2\pi}{\ln(2)} < y < \frac {3\pi}{\ln(2)}\]
and so forth.
From the above analysis, $ \text{Re}\,\zeta(z) > 0$ for any values to the right of $ \{y_n(x)\}$.  This region is shown as the blue region to the right of a plot of $ \{y_n(x)\}$ (shown in red) below.

I took the liberty to remove the "curled o" discussion, which, in my humble opinion, only cluttered the thread and distracted attention from this really nice question.

The answer is "Yes", by the way. There are real negative values in the half-plane $ \text{Re\,}z > 1$. Moreover, [b]every[/b] complex value except $ 0$ is attained somewhere there infinitely many times.

Thanks for all your efforts, youriens. Thanks for showing up, fedja. I was beginning to suspect that, but hadn't been able to come up with a proof. Thanks.

[quote="fedja"]
The answer is "Yes", by the way. There are real negative values in the half-plane $ \text{Re\,}z > 1$. Moreover, [b]every[/b] complex value except $ 0$ is attained somewhere there infinitely many times.[/quote]

I don't see how you can say "infinitely often" because I don't think the right half-plane $ \text{Re}(z) > 1$ is getting mapped to a deleted neighborhood of the origin under the transformation $ z\to 1/z$.  It remains on the right half-plane $ \text{Re}(z) > 0$ and if you're relying on Picard, I feel the "infinitely often" applies only to a full $ 2\pi$ neighborhood $ 0 < |z| < \epsilon$.

Perhaps though I'm not understanding this correctly.

No, of course I do not rely on Picard. The argument is like the following (the presentation below will have a couple of gaps but they are fillable).

We have $ \log\zeta(s \plus{} it) \equal{} \minus{} \sum_p \log(1 \minus{} p^{ \minus{} s}e^{ \minus{} it\log p})$. Now fix $ s > 1$ and consider this as a function of $ t\in\mathbb R$. Note that the numbers $ \log p$ are independent over rationals. If you add $ 2\pi$ to them, you can get at most one linear dependence involving some primes up to, say, $ P$. Now, the classical simultaneous rational approximation lemma tells you that for every $ t_p\in\mathbb R$, you can choose $ t$ so that $ t\log p\approx 0\mod 2\pi$ for $ p\le P$ and $ t\log p\approx t_p\mod 2\pi$ for any given finite subfamily of $ p > P$ with any precision you wish.

Thus, on this vertical line, you can approximate with arbitrary precision any value of logarithm that you can obtain by choosing numbers $ \xi_p$ that are $ 1$ in absolute value in the expression
\[ \minus{} \sum_{p\le P} \log(1 \minus{} p^{ \minus{} s}) \minus{} \sum_{p > P} \log(1 \minus{} p^{ \minus{} s}\xi_p)\qquad\text{(1)}\]
When $ s$ is close to $ 1$, such sums sweep a huge disk centered at the origin, so you can approximate any number with any precision.

Now, to show that we can not only approximate but also to get an exact value $ a\ne 0$, we need to observe that the derivative of $ \log\zeta(z)$ is uniformly bounded in $ \{\text{Re\,z} > \tfrac{s \plus{} 1}2\}$ and that we can choose the representation of $ \log a$ in (1) so that the derivative at that point, which is
\[ \sum_{p\le P}\frac {p^{ \minus{} s}\log p}{1 \minus{} p^{ \minus{} s}} \plus{} \sum_{p > P}\frac {\xi_p p^{ \minus{} s}\log p}{1 \minus{} \xi_p p^{ \minus{} s}}\,\qquad\text{(2)}\]
is not too small (say, greater than $ 1$ in absolute value). Then we can use Rouche to finish.

The gap is, of course, in showing that for $ s$ close to $ 1$, the expressions (1) and (2) can yield the point $ (\log a,1)$, say. This would immediately follow from the general question I asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308020]here[/url] though this particular case can be handled directly despite the details are somewhat unpleasant.

Thanks for posting that.

In you statement:
[quote="fedja"] . . . $ t$ so that $ t\log p\approx 0\mod 2\pi$ for $ p\le P$ and $ t\log\pi\approx t_p\mod 2\pi$ for any given finite subfamily of $ p > P$ with any precision you wish.
[/quote]
Is the $ t\log\pi$ a typo?  Should it be $ t\log p$?

Is my following interpretation of your analysis correct?

Suppose we start with a known negative value of $ \zeta(z)$:
Assume $ \zeta(s + it) = - 1$ for $ s > 1$.  Therefore $ \log(\zeta(s + it)) = \pi i$.  Now, based on your statements above, there will exist some $ t\in\mathbb{R}$ such that:
\[ \pi i\approx - \left[\sum_{p\leq P_0}\log(1 - p^{ - s}) + \sum_{p > P_0} \log(1 - p^{ - s}e^{ - it})\right]\]
for arbitrary $ P_0$.  For example, if I let $ P_0 = 7$ then there will exist some $ t\in\mathbb{R}$ for fixed $ s > 1$ such that:
\begin{align*} \log(\zeta(s + it)) & = \pi i \\
& = - \sum_{p}\log(1 - p^{ - s}e^{ - it\log p}) \\
& = - \left[\sum_{2,3,5,7}\log(1 - p^{ - s}) + \sum_{p > 7} \log(1 - p^{ - s}e^{ - it})\right] \end{align*}

[quote="youriens"]
Is the $ t\log\pi$ a typo?  Should it be $ t\log p$?
[/quote]
It sure is. I corrected.
[quote]
Is my following interpretation of your analysis correct?
[/quote]
Almost. You get the approximate value for fixed $ s>1$ sufficiently close to $ 1$ (how close depends only on the value) by just varying $ t$ (and you get it with arbitrarily good precision), but to get the exact value, you allow both $ t$ and $ s$ vary a bit around the point where the approximate value is attained. The exact statement we need there is the following. If $ M,r>0$, then there exists $ \delta>0$ such that every analytic function $ f$ in $ D(a,r)$ satisfying $ |f(a)|<\delta$, $ |f'(a)|>\frac 12$, and $ \max_{D(a,r)}|f'|\le M$ has a zero in $ D(a,r)$.

So it goes like that: choose your logarithm $ \ell$ (say, the same $ \pi i$), choose $ s>1$ so that (1) and (2) evaluate to $ (\ell,1)$ for some $ \xi_p$, and, finally, choose $ t$ so that $ \log\zeta(s\plus{}it)\approx \ell, \frac{d}{dz}\log\zeta(s\plus{}it)\approx 1$ with the precision that allows you to apply the above statement with $ r\equal{}\frac{s\minus{}1}2$ and $ M\equal{}\sup_{\{\text{Re\,}z>\frac{s\plus{}1}2\}}\left|\frac{d}{dz}\log\zeta(z)\right|$

I'd like to summarize how I'm interpreting the first part (the approximation) of Fedja's analysis above and propose some questions about it.

For $ z = s + it$ and $ s > 1$ we have from Euler's product formula:
\[ \log(\zeta(s + it)) = - \sum_{p} \log(1 - p^{ - s}e^{it\log(p)})\]
but for sufficiently large $ t$ and for some $ t_{\xi}\in [0,2\pi]$, we can find a sub-set of the primes and write:
\begin{align} t\log(p) & \equiv D(0,\delta)\,\text{Mod}(2\pi), & \quad p\leq P_0 \\
t\log(p) & \equiv D(t_{\xi},\delta)\,\text{Mod}(2\pi), & \quad P_0 < p < P_{m},\quad t_{\xi}\in[0,2\pi] \end{align}
where $ D(0,\delta)$ and $ D(t_{\xi},\delta)$ are small neighborhoods about the points $ 0$ and $ t_{\xi}$ which satisfy the congruences.  And therefore for sufficiently large $ P_m$:
\[ \log(\zeta(s + it))\approx - \left[\sum_{p\geq P_0}\log(1 - p^{ - s}) + \sum_{P_0 < p\leq P_m} \log(1 - p^{ - s}e^{ - it_{\xi}})\right]\]
So that the remaining terms of the last sum are less than $ |\log(1 - \frac {1}{p_{m + 1}})|$ and therefore contribute little compared to the first $ m$ terms and the desired level of accuracy.  Intuitively, one can guess $ t$ is going to be very large with the size being directly proportional to $ P_m$ and inversely proportional to $ |1 - s|$.  For example purposes, suppose $ t_{\xi} = \pi/4$, $ \delta = 0.1$, $ P_0 = 2$, and $ P_m = 10$,which is grossly insufficient for the actual computations but nevertheless just for example purposes, then I would need to find a $ t$ such that the following congruences are satisfied:
\begin{align*} t\log(2) & \equiv\,D(0,0.1)\,\text{Mod}(2\pi) \\
t\log(3) & \equiv\,D(\pi/4,0.1)\,\text{Mod}(2\pi) \\
t\log(5) & \equiv\,D(\pi/4,0.1)\,\text{Mod}(2\pi) \\
& \vdots \\
t\log(29) & \equiv\,D(\pi/4,0.1)\,\text{Mod}(2\pi) \end{align*}
Now, I can make this a little more managable by letting:
\[ t\log(2)\equiv\,0\,\text{Mod}(2\pi)\]
or $ t = \frac {2n\pi}{\log(2)}$ and therefore search for an $ n$ satisfying the follownig set of congruence relationships:
\begin{align*} 2n\pi\frac {\log(3)}{\log(2)} & \equiv\,D(\pi/4,0.1)\text{Mod}(2\pi) \\
2n\pi\frac {\log(5)}{\log(2)} & \equiv\,D(\pi/4,0.1)\text{Mod}(2\pi) \\
& \vdots \\
2n\pi\frac {\log(29)}{\log(2)} & \equiv\,D(\pi/4,0.1)\text{Mod}(2\pi) \end{align*}
Now, at this point I can't prove such an $ n$ even exists so I'll just take it as true for now and note even these ten congruences are going to result in an extremely large value of $ n$.
So I'd like to ask these questions:

(1)  Is there some order of magnitude we can assign to the simultaneous congruences above as a function of $ m$?  That is, how does the variation in the value of $ t$ change as I add additional congruence equations into the set?  I suspect it has some exponential order. That is, if $ t = 10^{10}$ for say $ m = 10$, then perhaps $ t\approx 10^{11}$ when I add another congruence relation.

(2) What is the order as a function of $ \delta$?  Again, I suspect it's exponential.

(3) What is the order as a function of $ |1 - s|$?  In this case, the closer $ s$ is to one, the more terms are needed in the logarithm sum for a desired accuracy, and in turn, the more congruences needed, leading to a larger value of t.  Again, I suspect this is exponential.

(4) This method of locating negative values of $ \text{Re}\zeta(z)$ would require the locations to be very far from the real line, I would guess on average somewhere beyond $ t = 10^{100}$.  However, I do not believe it precludes the existence of negative values closer that are just not accessible by this method.  Is this correct?

The order of magnitude for $ t$ in this solution is astronomical. For the existence result, we need the following classical lemma. If $ \alpha_j$ are independent over rationals, then the line $ t\mapsto L(t)\equal{}(\alpha_1t,\alpha_2t,\dots,\alpha_n t)$ is dense in the cube $ [\minus{}\pi,\pi]^n$ modulo $ 2\pi$. Indeed, just note that for every trigonometric polynomial $ P$ of $ n$ variables, we have $ \lim_{T\to\infty}\frac 1T\int_0^T P(L(t))\,dt\equal{}(2\pi)^{\minus{}n}\int_{[\minus{}\pi,\pi]^n}P(x)\,dx$ and we can approximate the characteristic function of any small ball with such polynomials. (So, we do not really need to care about possible dependences between logarithms of primes and $ 2\pi$ here, that part was there just because I was overly cautious).

Of course, the least $ z$ satisfying $ \text{Re\,}z>1$, $ \zeta(z)\equal{}a$ should be much less than the one guaranteed by this approach. But that is a completely different question, to which I do not have a ready answer at this moment.

Ok.  Thanks for taking the time to explain this solution Fedja :).    I thought it was beautiful and I think others too were interested.  The principle then is one of existence and is not amendable to computational scrutiny.

Has anyone ever definitively located a region in the half-plane $ s > 1$ where the real part of $ \zeta(s \plus{} it)$ dips below the s-t plane?  I edited this from my initial comment since I realized that the real part is actually a function of two variables, a surface, and with the exception of the case where it just touches the s-t plane, in general, $ \text{Re}\,\zeta(s \plus{} it) \equal{} 0$ over some contour where the surface intersects the plane.  So apparently, such contours can be defined all about the region.  Would be interesting to characterize them or at least the very first one encountered.  What does it look like and where is it?

I'd like to improve my estimate of the region in the half-plane $ \sigma > 1$ where $ \text{Re}\, \zeta(s)$ takes on negative values.  The following analysis is based on Titchmarch, "The Theory of the Riemann Zeta Function" (Ch. 11) where he details the analysis of this problem along the lines described by Fedja above.  It is one of the most interesting chapters of basic complex analysis I have ever studied and I recommend it to everyone interested in the subject.

In this chapter, Titchmarsh derives an expression for the range of $ \log \zeta(s)$ in the half-plane $ \sigma > 1$.  These ranges are in the form of circles with centers at $ 1/2\log \zeta(2 \sigma)$ which of course, as $ \sigma\to 1$, the radius grows to infinity.  However, away from one, the range is finite and at some point, below the threshold at which $ \log\zeta(s)$ can take a value at which the real part of zeta would be negative.  The formula giving the set $ U\equal{}\log\zeta(\sigma\plus{}it)$ is:
\[ U\equal{}\left\{w: w\in \text{D}\left[1/2\log \zeta(2\sigma), \frac {1}{2}\log\frac {\zeta^2(\sigma)}{\zeta(2\sigma)}\right]\right\}\]
Note this is a monotonic function of $ \sigma$ in the indicated range.  For example, when $ \sigma \equal{} 2$, then $ R\approx .5$.  This means the values of $ \log\zeta(\sigma \plus{} it)$ for $ \sigma\geq 2$ are all (approximately) contained in a circular region of radius $ .5$ and center at approximately $ (0.04,0)$.  I say approximately because Titchmarch makes some simplifying albeit still accurate assumptions. And the larger $ \sigma$ becomes, the smaller this circular region becomes.  Note what this is saying:  For all values of $ \sigma > 2$, the absolute value of $ \log\zeta(s)$ is smaller than $ 0.5 < \pi/2$.  Consider $ \zeta(s) \equal{} \minus{} a \plus{} i\gamma$ somewhere in the half-plane $ \sigma > 1$, then $ \log( \minus{} a \plus{} i\gamma) \equal{} \ln| \minus{} a \plus{} i\gamma| \plus{} i(\theta \plus{} 2k\pi)$ with $ |\theta \plus{} 2k\pi| > \pi/2$.  This value then is never contained in the circle of radius $ 0.5$ and in order for zeta to attain a negative real part, $ \text{Im} \log\zeta(s)\geq \pi/2$. This implies:
\[ R \equal{} \frac {1}{2}\log\frac {\zeta^2(\sigma)}{\zeta(2\sigma)}\geq \pi/2\]
or $ \sigma < 1.2$ and once we cross this boundary, the Titchmarsh circle extends beyond the $ \pi/2$ line and so $ \text{Re}\,\zeta(s) < 0$ is attained infinitely often.  With this in mind, I'd like to update my graphics of the region in the half-plane at which $ \text{Re}\zeta(s) > 0$:
\[ \text{Re}\zeta(\sigma \plus{} it) > 0,\quad \sigma > 1.2\]
and make the following statement:  In the range $ 1 < \sigma$, The real part of zeta takes on negative values only in the red strip (for all t) of the plot below.
(edit:  corrected calculations and plot)

[quote="youriens"]Would be interesting to characterize them or at least the very first one encountered.  What does it look like and where is it?[/quote]
Hi guys.  I've been working on this.  Would anyone here be willing to confirm for me, or suggest to me where I could turn to to relibably verify the values in the following table using some method other than Mathematica.   I used the Mathematica command:
\[ \texttt{N[Zeta[1 + it],\{Infty,20\}]}\]
to compute the value of $ \zeta$ at the indicated points but I have some doubts.  In the table, $ k = \frac {\pi}{\log(3) - \log(2)}$ and I've written the values of $ t$ in arbitrary-precision so that $ \zeta(1+it)$ can be calculated using multi-precision arithmetic.  For example, the first one would be:
\[ \zeta\left[1 + \left(\frac {4378640 \pi}{\log(3) - \log(2)} - \frac {2}{5}\right)i\right]\]
Thanks :)
\begin{tabular}{|c|c|l|l|} \hline ID & Type & t & \zeta(1 + it) \\
\hline 1 & C_3 & 4\,378\,640k\phantom{111k} - 2/5 & - 0.0094 - 1.22i \\
2 & C_4 & 415\,782\,314k\phantom{1k} - 2/5 & - 0.024 - 1.23i \\
3 & C_6 & 20\,430\,730\,768k - 1/10 & - 0.0154 - 1.076i \\
4 & C_6 & 25\,705\,015\,862k - 1/5 & - 0.00163 - 1.198i \\
5 & C_6 & 47\,668\,373\,108k - 1/10 & - 0.0268 - 0.961i \\
6 & C_6 & 53\,761\,507\,682k & - 0.014 - 1.202i \\
7 & C_6 & 62\,484\,882\,686k - 1/5 & - 0.0133 - 1.134i \\
8 & C_6 & 65\,421\,460\,042k - 1/10 & - 0.0027 - 0.914i \\
9 & C_6 & 97\,190\,286\,104k - 1/10 & - 0.0176 - 0.99i \\
10 & C_6 & 99\,154\,858\,182k - 3/10 & - 0.0264 - 1.246i \\
\hline \end{tabular}


[color=darkred]The incircle of the triangle $ABC$ touches the its sides in the points $A'\in BC,\,B'\in CA,\,C'\in AB.$ Define the points $K\in B'C'$ so that $AA'\perp A'K$
Prove that $BC$ bisect $AK$[/color]

I found a solution to this very nice problem .
Let $AA'$ meets the circle at $L$ and $B'C'$ meets $BC$ at $P$ and let $I$ the incenter and let $H$ the midpoint of $B'C'$ .
It is well-known that PL is tangent to the circle .(by pole and polars)
We have $AH\cdot AI=AC'^{2}=AL\cdot AA'$ so $LHIA'$ is cyclic and so
 $\angle{IA'L}=\angle{AHL}$ so $90-\angle{IA'L}=90-\angle{AHL}$ so
$\angle{LHP}=\angle{LA'P}$ so $PLHA'$ is cyclic so $\angle{PLA'}=\angle{PHA'}$(1)
But $AHA'K$ is also cyclic and so $\angle{KHA'}=\angle{KAA'}$ (2)
(1)=(2) so $\angle{PLA'}=\angle{KAL}$ so $\angle{KAL}=\angle{LA'P}$
so if $BC$ meets AK at N then $AN=NA'=NK$ and so N is the midpoint of  AK

Hope you liked it . :)

[size=150][color=red]Nice solution, [b]silouan[/b]![/color][/size]

[color=indigo]And here is my proof.

[b]Proof.[/b] We have: $B'C'$ is the polar of $A$ w.r.t $(I)$
So the circle with diameter $AK$ is orthogonal circle of the circle $(I)$
On the otherhand, $BC$ is a tangent of $(I)$, which implies that $BC$ passed through midpoint of segment $AK$ (center of circle $(AA'K)$)[/color]


Edges of triangle is given by equations $ a_ix \plus{} b_iy \plus{} c_i \equal{} 0\ (i \equal{} 1,2,3)$.
Prove that the area $ P$ of triangle is given with
\[ P \equal{} \frac {{\Delta}^2}{2|{\Delta}_1{\Delta}_2{\Delta}_3|}, \text{where}\   \Delta \equal{} \vmatrix a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \endvmatrix ,\]
and $ \Delta_i$ are algebric complements of elements  $ c_i$.

can't anybody solve this trivial problem


1) Mark Twain and Harry Trumann were born here

2) The population center of the U.S. is in Missouri

3) Brad Pitt is from Missouri

4) Bob Costas lives in Missouri

5) Missouri was the northernmost slave state

6) Missouri is the only state to have a street named Barack Obama Blvd

7) Missouri has the spammiest forum

8) Missouri has the most combined multiple accounts, FTW mutes, and pisses off the admins the most

9) If you are reading this then there is a 90% chance that you dont live in Missouri because that's how popular this forum is

10) You are tempted to post something bad about Missouri right now but the awesomeness of Missouri prevents you from doing so.

Yes, I couldn't have said it better tiny.  Exactly why Missouri pwns so much.

Haha, I count as one of those 10%.

Wait, tiny...

I still count right....?

I'm Missourian enough, right?? :read:

Yeah... didn't Missouri annex Indiana? :roll:


Does anyone know when the results/statistics for the 2005 12A will be posted?

Thanks
JB

[i]Moderator note: re-titled to clarify[/i]

Not for a while. I think they post the A and B test results together.

I suppose that makes sense.  They probably don't want the B kids having an advantage going in by knowing what they need to get ...

JB

I hope that we can start sending the emailed results about 2 weeks from now.  (That would mean about 3 weeks from the contests).  The paper-mailed results will start going shortly after that.  Global statistics and analyses will follow the paper mailing.  This depends on everything functioning well, even perfectly.

The timing of sending the results has nothing to do with the B contests, and everything to do with the process of sorting the mail and scanning the answer forms.  Right now, we have about 2,500 packages which are being opened as I write this.  Each package has to be checked in and logged that it was received with all proper documentation for the school.  Then the scanning of approximately 150,000 pieces of paper has to take place.  After that, editing has to be done on the electronic records, particularly if you did not fill in the bubbles for your name correctly.  (We have a surprising number of school high scores filled out by, e.g J*H* S**TH because we cannot correctly scan the bubbles indicated by * ) The more editing there is, the longer the scoring process takes, so filling out your answer form carefully and completely actually speeds you getting your results.   After that, things go reasonably fast.  But please imagine the care that must be taken with 150,000 pieces of paper from 2,500 schools world-wide so that we do not lose yours, and accidentally not qualify you for the award you deserve.

Steve Dunbar
AMC Director

Oh, I suppose I was mistaken then. Nice to know though.

Mr. Dunbar --
Thank you for your reply.  I understand the amount of work and care that goes into grading the AMCs, and was not trying to hurry you, or to sound impatient, if that was the impression you received.

Thank you very much
JB

My testing group's scores just arrived by email in the last few hours. They must be emailing today and the next few days until all the scores go out to contact persons for whom they have email addresses.

It's official now [brag mode on:] A+MATH qualifies for AIME for the second year in a row. [brag mode off]

I see the original poster's concern in this thread was more along the lines of the official statistics Web pages that appear on the AMC site. Thus must be coming along soon, but I haven't checked yet today. By the end of February or sooner, it may be.

Our schools scores have come as well...
Tokenadult, what did your son score?


I was wondering what happened to the link to purchase Art of Problem Solving Shirts. Can we still buy them, or are they discontinued? Thanks

Sorry for reviving a topic, but I really would like an answer. Maybe I'll just pm an administrator.

Sold out? Huh? Can't you just manufacture more?

Oh, okay. Will you get more anytime soon? Thanks


Is this like Simon's Favorite Factoring Trick or something stupid like that?

[hide="Hint that may actually mislead you"]Add/Subtract (up to you to decide which is right because I sure don't know) the first equation from the second and factor[/hide]

[quote="anirudh"][hide]$\frac{19}{4}$[/hide]

[size=75][color=darkred]Don't forget hide tags.
-nebula42[/color][/size][/quote]
Don't post an answer that has already been posted without any work whatsoever... It isn't good...

Extension. If a+b=x and $a^{2}+b^{2}=y$ solve for $a^{3}+b^{3}$, $a^{4}+b^{4}$, and $a^{5}+b^{5}$ in terms of x and y.

[quote="bpms"][quote="Ignite168"]Is this like Simon's Favorite Factoring Trick or something stupid like that?[/quote]
Not Simon's favorite factoring trick. No need to insult it.[/quote]
What is this factoring trick?

this has been asked many times around the forum.
simon's favourite factoring trick is
$xy-nx-my+mn=(x-m)(y-n)$

To clarify on Simon's factoring trick
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108384
Look at the solutions...


Here is something interesting for people just learning calculus.

If we construct an equilateral triangle $ABC$, and continue to construct equilateral triangles from the midpoints of the sides of the larger triangle (as shown), we have $AB + BC = 2 AC$,  $AD + DI + IE + EC = 2AC$, and $AF + FB + BH + HI + IJ + JK + KL + LC = 2AC$.  If we continue this process infinately many times, then does the segment $AC$ approach twice its length?

Sorry for the poor wording and if this has already been posted.

You might as well ask it this way: if $f_n$ and $f$ are functions on $[a,b]$ that have arclengths and if $f_n$ tends to $f$ uniformly, does the arclength of $f_n$ tend to the arclength of $f$ ?


Given $x,y,z,a,b,c,p,q,r,d>0$ such that $ax^{n}+by^{n}+cz^{n}=d$ prove that
$\frac{p}{x^{m}}+\frac{q}{y^{m}}+\frac{r}{z^{m}}\ge (\frac{1}{d})^{\frac{m}{n}}(\sqrt[m+n]{a^{m}p^{n}}+\sqrt[m+n]{b^{m}q^{n}}+\sqrt[m+n]{c^{m}r^{n}})^{1+\frac{m}{n}}$
hope you can do it  :D

Is it really trivial ?, try to post all solution  :D

We need prove $(ax^{n}+by^{n}+cz^{n})^{m}(\frac{p}{x^{m}}+\frac{q}{y^{m}}+\frac{r}{z^{m}})^{n}\geq (\sqrt[m+n]{a^{m}p^{n}}+\sqrt[m+n]{b^{m}q^{n}}+\sqrt[m+n]{c^{m}r^{n}})^{m+n}$. Trivial by Holder, again! :D


linear algebra
matrix
calculus
integration
combinatorics proposed
combinatorics

All graphs are directed graphs, possibly with loops or multiple edges.  The adjacency matrix of such a graph has nonnegative integral entries.  Also, read the last paragraph first :)

Given a graph $ G \equal{} (V, E)$ with adjacency matrix $ A$, let $ G^n$ be a graph with vertex set $ V$ and adjacency matrix $ A^n$, i.e., the edge $ (v, w)$ appears in $ G^n$ exactly as many times as there are paths of length $ n$ from $ v$ to $ w$ in $ G$.

Given a graph $ G \equal{} (V, E)$, define $ \widehat{G}$ to be the graph whose vertex set is $ E \times \{0, 1\}$ with edges going from $ (e, 0)$ to $ (e, 1)$ for each $ e \in E$.  (This graph is a perfect matching that we get by "pulling apart" the graph $ G$.)

Given $ G$, we take countably many copies of $ \widehat{G^n}$ for each $ n$, where we assign to each edge of $ G^n$ the weight $ t^n$.  Now choose an $ n$ and choose an edge in $ G^n$ and consider the (countably many) copies of this edge in our construction so far.  Let these copies be $ e_1, e_2, \ldots$, with $ e_i \equal{} (a_i \to b_i)$.  Add to the graph also a set of sources $ v_1, \ldots$ and sinks $ w_1, \ldots$, regular vertices $ t_1, \ldots$  and the following edges (all of weight 1): the edges $ (v_i \to a_i)$, $ (b_i \to t_i)$, $ (t_i \to w_i)$, $ (v_i \to t_{i \plus{} 1})$ and $ (t_{i \plus{} 1} \to w_i)$.  (The purpose of this last bit is to ensure that the only noncrossing path systems connect $ v_i$ with $ w_i$ for each $ i$, and that the set of the $ e_i$ that we actually traverse is of the form $ \{e_1, \ldots, e_k\}$ for some $ k$.  This should allow us to choose $ k$ copies of $ e_i$ for any $ k$, without allowing us any other choices.  If we don't somehow tie them together, then all the coefficients will work out to be infinite, I think.)

I claim that this is the graph we want.

All of the details here are provisional.  I think instead of $ \widehat{G^n}$ I want a graph with an edge for every [i]cycle[/i] in $ G$ of length $ n$, right?  Also, I'm not sure that any of the other technical details of my construction work out right.  But I'm completely certain that this is the right sort of approach.   :)

Wait, here's the non-stupid way to do this.  We're given $ G$.  For each (directed) cycle $ c$ in $ G$ of length $ n$, give our graph vertices $ v_c, w_c, a_{c, 0}, a_{c, 1}, a_{c, 2}, \ldots, b_{c, 0}, b_{c, 1}, \ldots$ and edges $ v_c \to a_{c, i} \to b_{c, i} \to w$, where edge $ a_{c, i} \to b_{c, i}$ has weight $ t^{ni}$ and all others have weight 1.

I think this is actually the graph I want.

Please give your posts better titles in the future. This one could have been called "Directed graphs and their adjacency matrix" or something.

Just kidding!!!! :rotfl:  :D  :rotfl: (hope it's no offence)


The [b]geometric mean[/b] for any numbers $ a$ and $ b$ is simply $ \sqrt {a \cdot b}$.

The geometric mean is (x*y)^1/n (where n is the number of terms, in this case n is 2 because there are only two numbers.) so its just (27*12)^(1/2)=18.


Elemennop wrote:Hint

[hide]

Substitute  and find a  so the constant term dissapears. Then just use the characteristic polynomial and the basic techniques to find the general  which leads to the general .

[/hide]

Yea I'm too lazy to actually do the problem

Actually, that hint doesn't work in this particular case - try it and you'll see why.

SolutionRearrange the equation by subtracting  from both sides:

which can be written as

Now substitute  where  to get

This is a geometric sequence, hence

Now we have

By summing up the expressions for , where clearly some telescoping takes place, and using the formula for the sum of geometric sequence, we get

With  we get

[quote="justdudxit"]I keep seeing the term "characteristic polynomial" everywhere, but I have no idea what it is (gap in education).  Can someone enlighten me?[/quote]
It is a way to find the general terms of a sequence. I think the stuffs of Characteristic Equation can be found in almost every contest book ( at least they appear in every book that I have ).

For example, a sequence is defined as: $pa_{n+1}+qa_{n}+ra_{n-1}=0$, with $p,q,r$ being constant values. Let $x_{1}$ and $x_{2}$ be the roots of the equation $px^{2}+qx+r=0$, then if $x_{1}\neq x_{2}$, we shall have $a_{n}=Ax_{1}^{n}+Bx_{2}^{n}$. If $x_{1}=x_{2}$, then $a_{n}=(A+Bn)x_{1}^{n-1}$.

[quote="shobber"]Given $\{ a_{n}\}$ such that $a_{1}= a_{2}=1$ and $a_{n+2}= \frac{1}{2}a_{n+1}+\frac{1}{2}a_{n}+1$. Find $a_{n}$.[/quote]

Lets try and eliminate the constant a different way: define $b_{n}=a_{n}+kn$ (we think of this becuase adding a constant $n$ cancels out, but it is more than a constant each time, namely 2,1,0 added to each term), then use the recursive relationship: \[b_{n+2}+k(n+2)=\frac{1}{2}\cdot(b_{n+1}+b_{n})+\frac{k}{2}\cdot(2n+1)+1\] Now equate the constants from both sides so they cancel out: \[k(n+2)=\frac{k}{2}\cdot(2n+1)+1\implies k\cdot\left(2-\frac{1}{2}\right)=1\implies k=\frac{2}{3}\] at which point we say that $b_{n}=a_{n}+\frac{2}{3}\cdot n$ so $b_{1}=\frac{5}{3}$ and $b_{2}=\frac{7}{3}$

Now we can easily use characteristic equations to find $b_{n}$ explicitly: \[2b_{n+2}-b_{n+1}-b_{n}=0\equiv 2x^{2}-x-1=0\implies x=1,-\frac{1}{2}\] Then we can say that \[b_{n}=c_{1}\cdot 1^{n}+c_{2}\cdot \left(-\frac{1}{2}\right)^{n}\] Now we can use the values of $b_{1}$ and $b_{2}$ to solve for $c_{1}$ and   $c_{2}$ (which are constants) using a system of linear equations. Solving gives: \[c_{1}=\frac{19}{9}\qquad c_{2}=\frac{8}{9}\] It follows that: \[a_{n}=\frac{19}{9}+\frac{8}{9}\cdot\left(-\frac{1}{2}\right)^{n}-\frac{2}{3}\cdot n\]

[b]Altheman[/b], I'm sorry but I have to point out that you made an initial mistake which led you to the wrong solution.

First, if you define $b_{n}=a_{n}+kn$, then $a_{n}=b_{n}-kn$ and your rearranged equation should be

$b_{n+2}-k(n+2)=\frac{1}{2}\cdot(b_{n+1}+b_{n})-\frac{k}{2}\cdot(2n+1)+1$

instead of [quote="Altheman"] \[b_{n+2}+k(n+2)=\frac{1}{2}\cdot(b_{n+1}+b_{n})+\frac{k}{2}\cdot(2n+1)+1\] [/quote]

Alternatively, if you want to stick with your equation, the definition of $b_{n}$ should be $a_{n}=b_{n}+kn$ - because THAT is your actual substitution - hence with your results we would have $b_{n}=a_{n}-kn=a_{n}-{2n\over 3}$ and that would give $b_{1}={1\over 3}, b_{2}=-{1\over 3}$, which means that your system in $c_{1}, c_{2}$ is also wrong.

Hope you're not angry :)

PS. Since your answer is quite different from mine, I was curious and reviewed your work step by step.

[quote="Farenhajt"][b]Altheman[/b], I'm sorry but I have to point out that you made an initial mistake which led you to the wrong solution.

First, if you define $b_{n}=a_{n}+kn$, then $a_{n}=b_{n}-kn$ and your rearranged equation should be

$b_{n+2}-k(n+2)=\frac{1}{2}\cdot(b_{n+1}+b_{n})-\frac{k}{2}\cdot(2n+1)+1$

instead of [quote="Altheman"] \[b_{n+2}+k(n+2)=\frac{1}{2}\cdot(b_{n+1}+b_{n})+\frac{k}{2}\cdot(2n+1)+1\] [/quote]

Alternatively, if you want to stick with your equation, the definition of $b_{n}$ should be $a_{n}=b_{n}+kn$ - because [i]that[/i] is your actual substitution - hence with your results we would have $b_{n}=a_{n}-kn=a_{n}-{2n\over 3}$ and that would give $b_{1}={1\over 3}, b_{2}=-{1\over 3}$, which means that your system in $c_{1}, c_{2}$ is also wrong.

Hope you're not angry :)

PS. Since your answer is quite different from mine, I was curious and reviewed your work step by step.[/quote]

ok i was looking for what mistake i made when i saw your answer, thanks for pointing that out  :)


1. Is there a fourth of July in England?

2. How many birthdays does the average man have?

3. Some months have 31 days, how many have 28?

4. How many outs are there in an inning?

5. Is it legal for a man in California to marry his widow's sister?

6. Divide 30 by 1/2 and add 10. What is the answer?

7. If there are 3 apples and you take away 2, how many do you have?

8. A doctor gives you three pills telling you to take one every half hour. How many minutes would the pills last?

9. A farmer has 17 sheep, and all but 9 die. How many are left?

10. How many animals of each sex did Moses take on the ark?

11. A clerk in the butcher shop is 5'10" tall. What does he weigh?

12. How many two cent stamps are there in a dozen?

13. You are participating in a race. You overtake the second person. What position are you in?

14. If you overtake the last person, then you are...?

15. Very tricky math! Note: This must be done in your head only. Do NOT use paper and pencil or a calculator. Try it: Take 1000 and add 40 to it. Now add another 1000. Now add 30. Add another 1000. Now add 20. Now add another 1000. Now add 10. What is the total?

16. Mary's father has five daughters: Nana, Nene, Nini, Nono. What is the name of the fifth daughter?

What was your score??? Are you an idiot?!!Post your answers to me and i'll tell u what ur score was.

[hide]1.Of course
2.1
3.1
4.6
5. Uhh... Yes because a widow has no husband.
6.70
7.1
8.30mins.
9.9
10.Moses didn't take any animals on the ark, noah did.
11.You can't tell.
12.12
13.2nd
14. You can't overtake the last person because there's no one after last.
15.4100
16.Mary.[/hide]

[hide]1. yes Duhhh
2. 1 duhhh
3. 1 duhhh
4. 3 duhhh
5. No for her to be a widow it means her husband is dead (Dark Knight got this wrong)
6. 70 Duhhh
7.  1 duhhh
8. 90 minutes Duhhhh
9. 9 Duhhhhhhhhhh
10. 1
11. Meat
12. 12 duhh
13. 2nd duhhhh
14. Not posible
15.  4100
16. Mary
[/hide]

A+:

[hide]
7) If you take away 2 apples then you now have two.[/hide]

[quote="djimenez"]The ones that I think everyone have wrong so far,

3. Some months have 31 days, others 30, but ALL the months in the (Gregorian) calendar have at least 28 days.
7. You have 2 not one, because YOU took 2 and left one.
8. Cmon, this one was easy, how did you get it wrong. 60 minutes of course!!, suppose he gave them to you at noon, you took one at 12:00 sharp, the next at 12:30 and the last at 1:00 sharp.
9. Cmon, seventeen, nine dead, eight alive, but seventeen in total.

Best,[/quote]

CRAP!!!! I read so many of the questions wrong. :( and it thought i wasn't an idiot.  :P

extension

17.
What is the answer:
a:a
b:d
c:b
d:a

18.
What is 2*24+1?
a. peanut butter
b. 30a
c. 49
d. infinity

[hide]Yes.
1
1
6
no, the man is dead
70
2
90
9
1? dunno, i'm not religous
meat
12
second
second to last
4100
Mary
Edit: sorry, i thought it said number of hours, not minutes
[/hide]

[hide]1. Yes
2. one
3. twelve
4. three... right? >>
5. No.
6. 70
7. two! right?
8. 90?
9. nine.
10. A lot... right? one of each sex for every animal. I don't know how many animals he took...
11. A lot? :?
12. 12?
13. second.
14. ummm... not in the race?
15. 4100
16. Mary[/hide]

Umm... yeah. I think I got a lot wrong xD

edit: :O I finally understand number 10. Haha... I wasn't thinking straight.

[hide]
1. Yes
2. 1
3. Every month has 28 days
4. 6
5. Impossible cause the husband is dead
6. 70
7. 1
8. 90?
9. 9
10. Moses didnt take any animals on the ark. Noah did
11. Meat
12. 12
13. 2nd
14. Second to last
15. eh, 4100, is there a trick or sumthin?
16. Mary
[/hide]

Good Job!!!
Some answers-

[hide]
6.   70, 30/1/2= 60+10=70
8. 60 min. you take on right when he gives it to you, and a half hour later, and another one a half hour later.
11. Meat, a butcher weighs meat!!!
14. You cant overtake the last person.....so that dosnt make sense.[/hide]

Ok here's the scale-

0 wrong- good job
1-3 wrong- average
4 or more- IDIOT

Ok heres your scores-
[hide]DarkKnight- IDIOT

A+Math-.....IDIOT duhhhhhhhh

jli-IDIOT

eryaman- IDIOT

MathWhiz2004-IDIOT

Sheepwarrior- Average. NICE! :D

Sran- Average. Well done :lol:

Paladin8-  Good Job

So far the top scorer is Paladin8, who got all of them right. [/hide]

Hmmm... I thought this place was for smart people........but it seems most of you are IDIOTS

Here is what I got. I'm curious to see if I got any right. This was pretty creative.

[hide]
1. Yes.
2. 1
3. 12
4. 0? (Sounds like sports, and I am very ignorant. So I'll just pretend it wants to know about the letters)
5. No specific law exists that prohibits it; I don't think the state of California has yet extended it marital policies to the afterlife.
6. 70 (This is math, hence I will miss it; it must require creativity)
7. 2
8. 60 (Assuming he takes one immediately, and we wish the time interval he has at least one pill intact outside his body)
9. 9 (Assuming left=left alive, else 17)
10. 1? (Not Christian, don't know Bible)
11. Meat
12. 12
13. 2nd
14. Impossible to answer. If we assume that 1) the track is a closed loop, and 2) that you are the first person to overtake the last person, then you are the 1st person.
15. 4100, by grouping?
16. Nunu?[/hide]

Wait a moment, I just thought about number 16 some more. I would like to change my answer. If this is allowed, my answer to number 16 is below:
[hide]
16. Mary :blush: .[/hide]

[hide]
1.  There's a 4th of July everywhere.
2.  1.
3.  They all do (12)
4.  6.
5.  Yes...
6.  70
7.  2.
8.  60
9.  9...
10. Noah did, not Moses!
11.  Meat.
12.  12
13.  Second
14.  You can't.
15.  4100
16.  Mary
[/hide]


[quote="tang zy"]This question occur to me when I solve a hard ineq and this can be part of it if it is right:
$ (\sum a^{2})^{3}\geq 9\sum a^{3}b^{3}$
Anyone can help me? :?:[/quote]

If both sums are cyclic then it's wrong:
Put (1;1;0) and you will get :
$ 8 \ge 9$ which isn't true.


linear algebra
matrix
calculus
derivative
function
abstract algebra
linear algebra unsolved

Let $ S\in\mathcal{M}_n(\mathbb{R})$ be a given skew symmetric matrix. Does there exist an orthogonal matrix $ Q\in\mathcal{M}_n(\mathbb{R})$ s.t. $ Q \minus{} Q^T \equal{} S$ ?

Certainly not for all possibilities. Orthogonal matrices are a compact set and skew-symmetric matrices aren't.

All skew-symmetric matrices sufficiently close to zero? Yes. The derivative of $ f(Q)\equal{}Q\minus{}Q^T$ at the identity is invertible (with $ f$ interpreted as a map from the orthogonal group to its Lie algebra the skew-symmetric matrices), and we have the inverse function theorem.

a example for n=2) for $ \alpha \in R$
 Take $ S\equal{}\left[\begin {array} {cc} 0 & 2sin(\alpha) \\ \minus{}2sin(\alpha) & 0 \end {array} \right]$ is skew-symmetric matrix
when the exist Q matrix is orthogonal matrix as : $ Q\equal{}\left[\begin {array} {cc} cos(\alpha) & sin(\alpha) \\ \minus{}sin(\alpha) & cos(\alpha) \end {array} \right]$ Such that $ Q\minus{}Q^{T}\equal{}S$

Hi jmerry,
you had a very pretty argument (using function $ f$) but you can do better.
If I give you $ S$ you can say if, yes or no, such a matrix $ Q$ exists
Moreover if yes, then you can calculate a solution $ Q$.
Hint: use the spectral radius of $ S$.

Then you should have asked that question in your initial post. Be more specific.

Hint:
Calculate: $ (Q\minus{}\dfrac{1}{2}S)^2\equal{}?$


geometry
circumcircle
geometric transformation
3D geometry
sphere
symmetry
incenter

1. In the square $ABCD$ is given  a  point $E$, and  out of the square a point $F$, such that $\triangle{ABE}=\triangle{BCF}$.Find the angles of the triangle $ABE$, if  $EF$ equal to the side of the square and $\angle{BFD}=90^{0}$. $\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500755#500755]Discussion[/url]

2.In the triangle $ABC$,$O$-is the circumcentre ,$A',B',C'$-are such  a point that $CA'=CA$ and $AA'\bot{BC}$,$CB'=CB$ and $BB'\bot{AC}$,$BC'=BC$ and $CC'\bot{AB}$,
$A_{1}\in{OA'}\cap{BC}$,$B_{1}\in{OB'}\cap{AC}$,$C_{1}\in{OC'}\cap{AB}$.
Prove that the lines $AA_{1},BB_{1},CC_{1}$ intersecting in the same point. $\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500765#500765]Discussion[/url]

3. In the triangle $ABC$, $l_{a}=\frac{h_{a}+m_{a}}{2}$ and $\angle{A}>90^{0}$,then $AB=AC$.$\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500768#500768]Discussion[/url]

4. In the plane are given two lines $a$ and $b$ ,and two points $A$,$B$.$X$ moves in the line $a$, and point $Y$ in the line $b$, such that $AX\|{BY}$.Find the locus of the intersection points between $AY$ and $XB$. $\clubsuit$
 [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500770#p500770]Discussion[/url]

5. In the plane is given a circle and a point $P$ which doesn’t lie on that circle. Find the locus of the orthocenters of the triangles $PAB$, where $AB$  is any diameter of the circle. $\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500773#500773]Discussion[/url]

6. The triangle $ABC$  circuscribed in the circle and in the $ABC$ iscribed a circle, which tangents the sides $AB,BC$ and $AC$ at $A_{1},B_{1}$ and $C_{1}$ respectively .$P\in{B_{1}C_{1}}\cap{BC}$,and $M$ is the midelpoint of  $PA_{1}$.
Prove that the touch lines segments from $M$ to the circumscribed and inscribed circles are equal.$\clubsuit$
 [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500776#p500776]Discussion[/url]

7. Through the orthocenter $H$ of the triangle $ABC$, is driven two perpendicular lines, one of them meets side $BC$ at $X$, second one meets the side $AC$ at $Y$. The lines $AZ,BZ$ are parallels to the lines $HX,HY$ respectively. Prove that the points $X,Y,Z$ are collinear.$\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500779#p500779]Discussion[/url]

8.In the plane are given 4 points  $A,B,C,D$. The points $A_{1},B_{1},C_{1},D_{1}$ are orthocenters of the triangles $BCD,CDA,DAB,ABC$.$A_{2},B_{2},C_{2},D_{2}$ are orthocenters of the triangles $B_{1}C_{1}D_{1},C_{1}D_{1}A_{1},D_{1}A_{1}B_{1},A_{1}B_{1}C_{1}$ and ....Prove that the circumcircles  of all  triangles formed by middelpoints of  this triangles are passing through the same point.$\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500781#500781]Discussion[/url]

9. Let $ABC$ is a triangle and $A'\in{[BC]}$,$B'\in{[AC]}$ and $C'\in{[AB]}$.
Prove that
\[S_{ABC}S_{A'B'C'}^{2}\geq 4S_{AB'C'}S_{A'BC'}S_{A'B'C}\]
and equality holds only when the lines $AA',BB'$ and $CC'$ intersecting in the same point. $\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500786#500786]Discussion[/url]

10.Through the point $P$ inside of the given circle are driven two perpendicular lines which meets the circle at $A$ and $B$. Find the locus of the projections of $P$ onto $AB$. $\clubsuit$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=500789#500789]Discussion[/url]

[quote="shobber"][quote="Tiks"]I have solved and sent them solutions of 13 problems(from 20)[/quote]
So are there still 10 problems which are not posted yet?  :)[/quote]
Yes $shobber$ ther are,but my English is not so good to translating them,anyway I will try to do that next time and I will add them here.

[quote="Megus"]Tiks - so maybe you can write these problems in Russian or give some link ? - surely someone will be able to handle translating - myself I could read some Russian (especially geometry as I read some chapters in original of Prasolov's book)  :)[/quote]
Dear $Megus$ here I have attached the file with all problems(in Russian),I hope you will be able to understend problems.Prosolov's book is a very nice book to study geometry ;) ,thanks Prosolov!:lol: .

[quote="Megus"]Thanks  :)  I try now with one - hope to do more tommorow -please tell me if it's correct  :)

23. $ABCD$ - convex quadrilateral, $G$ - point which is an intersection of two lines connecting centroids of triangles obtained by dividing $ABCD$ by its diagonals (ie line connecting centroids of $ADC$ and $ABC$ and line connecting centroids of $ADB$ and $BCD$) [I'm not quite sure of this - is it correct ?  :? ]

a) $ABCD$ is cyclic with center of its circumcircle in $O$. We define $H$ analogically as $G$ with orthocenters instead of centroids. Prove that points $O,G,H$ are collinear and $HG: GO=2: 1$

b) In $ABCD$ can be inscribed circle with center $I$. Nagel point $N$ of circumscribed quadrilateral is a point which is an intersection of two lines connecting points on opposite sides of quadrilateral, symmetrical to tangency points [my guess...] of inscribed circle with respect to midpoints.    Prove that $N,G,I$ lies on one line and $NG: GI=2: 1$[/quote]
Megus your translation is correct!
And also you have translated the best broblem from all ;),I solved this problem too,and I liked it much others.

[quote="Megus"]24. a) Through a fixed point $P$ inside given circle there are drawn two perpendicular radiuses intersecting the circle in points $A,B$. Find locus of perpendicular projections of $P$ onto line $AB$.
     b) Analogical as above but it is given a sphere instead of a circle

After all I think my Russian sux :blush:  I have difficulties with translating other problems (mostly some single words) - any help from someone more fluent in Russian ?  :)[/quote]
Megus,mybe you havn't seen but I have translated this problem :lol: ,look at the laste problem from my first post,it even have been solved :lol:.

[quote="Megus"]Oh, pardon - I haven't noticed  :) But, [b]Tiks[/b], by the way, do you know where can I get problems from previous years?[/quote]
Of cours I know,but this is the second olimpiad,and here is the link for the first one ;)
[url=http://olympiads.mccme.ru/sharygin/2005/final.htm]olympiads.mccme.ru/sharygin/2005/final.htm[/url]
I hope your Russian is good enought to anderstand this site.

Here are the rest of the problems:

1. Two lines intersect forming a $46^o$ angle. They are axes of symmetry of a planar figure $F$. How many other axes of symmetry can this figure have?

2. The points $A$ and $B$ are moving along two congruent circles with identical velocity. Prove that the perpendicular bisectors of a line segment $AB$ pass through a fixed point.

3. We have three segments of three different roads depicted on a map. These roads are connecting three villages that are located beyond the sides of a map. Furthermore there is one special point on a map, namely the point that is equidistant from the three villages. Can we find this point with the help of a compass and a straightedge when we can perform constructions within the map only?

4. a) We are given the squares $ABCD$ and $DEFG$. Moreover the point $E$ lies on the line segment $CD$ and the points $F$ and $G$ lie beyond the boundary of $ABCD$. Find an angle that is formed by the straight lines $AE$ and $BF$.
b) We are given two regular pentagons $OKLMN$ and $OPRST$. Moreover the point $P$ lies on the line segment $ON$ and the points $R$, $S$ and $T$ lie beyond the boundary of $OKLMN$. Find an angle that is formed by the straight lines $KP$ and $MS$.

5. a) Tile the square $10\times 10$ with the help of a rectangular strip $1\times 118$.
b) Tile the square $10\times 10$ with the help of a rectangular strip $1\times (100+9\sqrt{3})$.
You can bend the strips, but you can not cut them.

6. a) We are given the line segment $AB$ with a point $C$ on it. This segment is a chord of a circle having its radius equal to $R$. Construct a circle tangent to both the circle of radius $R$ and the line segment $AB$ at the point $C$.
b) We are given the line segment $AB$ with a point $C$ on it. This segment touches the circle of radius $r$ at $C$. Construct a circle passing through $A$ and $B$ and tangent to the circle of radius $r$.

8. The line segment $AB$ divides a square into two parts in both of which one can inscribe a circle. Radii of these circles are $r_1$ and $r_2$ respectively with $r_1>r_2$. Find the length of $AB$.

9. $L(\alpha)$ is a line connecting two points on a unit circle that correspond to the angles $\alpha$ and $\pi-2\alpha$. Prove that if $\alpha+\beta+\gamma=\pi$ then the lines $L(\alpha),\ L(\beta),\ L(\gamma)$ pass through a common point. (I really don't get this problem. Can someone explain how the lines $L(\alpha)$ are defined?)

10. For what $n$ a regular polygon with $n$ sides can be dissected by nonintersecting diagonals into $n-2$ isosceles (some of which may be equilateral) triangles.

16. Regular triangles are erected outwards on the sides of the triangle $ABC$. It is known that their "distant" vertices are the vertices of an equilateral triangle. Is it true that $\triangle ABC$ is regular.

17. Two parallel chords $A_1B_1$ and $A_2B_2$ are drawn in two circles that intersect at points $A$ and $B$. The lines $AA_1$ and $BB_2$ intersect at a point $X$, while the lines $AA_2$ and $BB_1$ intersect at a point $Y$. Prove that $XY\parallel A_1B_1$.

19. Through the midpoints of the sides of a triangle $T$ the lines perpendicular to the corresponding angle bisectors of a triangle pass. These lines bound a triangle $T_1$. Prove that the circumcenter of $T_1$ is the midpoint of the line segment that joins the incenter and the orthocenter of $T$.

22. We are given a circle, points $A,\ B$ that lie on this circle and a point $P$. $X$ - any point on the circle while $Y$ is the intersection point of the lines $AX$ and $BP$. Find the locus of circumcenters of the triangles $PXY$ as $X$ varies.

25. In a tetraedron $ABCD$ two-sided (is this the right term?) angles at edges $BC,\ CD$ and $DA$ are equal $\alpha$ while the angles at the other three edges - $\beta$. Find the ratio $AB/CD$.

26. We are given the four conics with a common vertex but various radii of the bases. Each one of them is tangent to two others. Prove that the four points of tangency of the bases of the conics lie on a circle. (I'm not sure how precisely I've translated this problem, there may be discrepancies).

[EDITED due to discussion below.]

[quote="nsato"][quote="Xixas"]19. Through the midpoints of the sides of a triangle $T$ the lines perpendicular to the corresponding angle bisectors of a triangle pass. These lines bound a triangle $T_1$. Prove that the circumcenter of $T_1$ is the midpoint of the line segment that joins the circumcenter and the orthocenter of $T$.[/quote]

Can you check this?  It doesn't seem to be true.  On the other hand, it does seem to be true that the two triangles have the same nine-point centre.[/quote]
This translation is a little incorreec,here is the correct version.

[color=blue]Through the midpoints of the sides of a triangle $T$ the lines perpendicular to the corresponding angle bisectors of a triangle pass. These lines bound a triangle $T_1$. Prove that the circumcenter of $T_1$ is the midpoint of the line segment that joins the center of the inscribed circle and the orthocenter of $T$.[/color]

[quote="Tiks"][quote="indybar"]Hmm... a geometry olympiad? Strange competition!  :)

How did you do on the test?[/quote]
Why is it  strange?I would say this is a very-very nice olimpiad :) ,I like geometry very much :blush: .
This is still correspondence round of this olimpiad,and I have solved and sent them solutions of 13 problems(from 20),I hope that will be enought and they will invite we to the confrontation round(usually 9 problem is enought :oops: ),confrontation round will be this summer in Moscow ;) .[/quote]
Hi guys! ;)
The results of correspondence round is already known and I was invited to the Dubna(Russia) for the  confrontation round,it will take place from 30 of July till 1 of August.
So guys after I will be back from Slovenia(IMO 2006),I will travel to Russia for this olimpiad ;) ,July will be wery busy monthe :(   :D .

So guys the final round is over too.
I took first place in it :first:  :wink: .
I think you would like to see the problems and here they are. :wink: \[Final Round\]  \[31.07.06\] [b]1.[/b]In the plane is given a circle with radius $R$. Two other circles with sum of radiuses equal to $R$, touch the circle internally. Prove that the line passing through the touch-points include one of the common points of those circles.
http://www.mathlinks.ro/Forum/viewtopic.php?p=606885#606885

[b]2.[/b]In the plane is given a circle, a point $A$ on it and a point $M$ in it. A chord $BC$ is passing through $M$. Prove that the circles passing through middle-points of all triangles $ABC$ is touch a fixed circle.
http://www.mathlinks.ro/Forum/viewtopic.php?p=606889#606889

[b]3.[/b]The triangles $ABC$ and $A_{1}B_{1}C_{1}$ are similar and have opposite orientation. On the segment $AA_{1}$ is chosen a point $A'$, such that $\frac{AA'}{A_{1}A'}=\frac{BC}{B_{1}C_{1}}$.The points $B'$ and $C'$ are constructing analogously. Prove that the point $A'$,$B'$ and $C'$ are collinear.

http://www.mathlinks.ro/Forum/viewtopic.php?p=606890#606890

[b]4.[/b]In the not convex hexagon every angle are equal either $90^{0}$ or $270^{0}$.Is that true that for some length of sides the hexagon can be cat into the two not equal hexagons similar to him?
      http://www.mathlinks.ro/Forum/viewtopic.php?p=607151#607151

[b]5.[/b]The line passing through the circumcentr and the orthocenter of the not equilateral triangle $ABC$, divides it’s perimeter and area in the same relation. Find that relation.
http://www.mathlinks.ro/Forum/viewtopic.php?p=607153#607153

[b]6.[/b]In the plane is given a convex quadrilateral $ABCD$. The point $A',B',C',D'$ are orthocenters  of  the triangles $BCD,CDA,DAB,ABC$. Prove that in the quadrilaterals $A'B'C'D'$ and $ABCD$ the corresponding diagonals in the points of intersection are divides in the same relation.
http://www.mathlinks.ro/Forum/viewtopic.php?p=607158#607158
[u][i]Start  at 9:00:Finish at 14:00   [/i][/u]

So, Tiks, can you post the problem from the first olympiad as well?

Dear Jumbler here is a link for the final round of the first olympiad
[url]http://olympiads.mccme.ru/sharygin/2005/variant.htm[/url]
But it is in Russian.Now I have no time to translate them,maybe next time :| .


calculus
integration
superior algebra
superior algebra unsolved

Let R and S be two commutative rings and R is contained in S, in addition S is integral extension over R.
Prove if x belongs to S and has an inverse element in S, then it has an inverse element in R.

I don't think I understand

let R be Z
and S be Z[i], the Gaussian integers, the latter is an integral extension of the first

now i is in S, has an inverse element in S, but none in R?

I'm guessing this should read "if x belongs to [i]R[/i] and has an inverse element in S, then it has an inverse element in R".
If $x \in R$, $xy = 1$ and $y \in S$ then $y$ satisfies a monic $y^n + a_{n-1}y^{n-1} + \cdots + a_1y + a_0 = 0$ over $R$.  So $y = x^{n-1}y^n = - x^{n-1}(a_{n-1}y^{n-1} + \cdots + a_1y + a_0) = - a_{n-1} - a_{n-2}x - \cdots - a_0x^{n-1}$ and this is in $R$.


What year would you say had the hardest AMC 12?  The easiest?  I'm an officer for my school's math club next year (school year is starting soon  :( )so this info would greatly help us!

A quick search on the AMC website shows that it's probably 2001, the AIME qualifying score was only 84.  :o

Ordered from most difficult to least difficult, based on average score:

2001
2000
2002 A
1999
1998
2004 A
2002 B
2003 B
2003 A
2005 A
2005 B
2004 B
2006 B
2006 A
:D

I think what happened in 2001 was that the overall problem difficulty didn't cover a huge range. If I recall correctly, the last 5 problems weren't as difficult as usual, but the easier problems were harder than they normally are.

Maybe I'm just imagining things, though...

[quote="mathgeniuse^ln(x)"]AMC 2001 was the hardest based on average.

One person told me, it was a joke.  An 84.

But I think 2002 B was easiest as well as 2006 A.[/quote]

Well, there can only be one easiest year :D

well, the 2006 AMC 12 A had the easiest first 18 problems (which were quite frankly around SAT II level difficulty, making it easy to actually qualify for AIME), but I think about three of the last problems, including #19, were more challenging than the last few problems on previous AMCs.

So being an AIME qualifier would've been relatively easier than in other years, but getting a score of over a 135, i think was more difficult.

I personally think 2003 B was just as easy as 2006 A, for getting a high score that is, not AIME qualification.

And also, I agree with 2001 being the hardest.


I think this shd be it  :D .
$ \ce{PhCOPh} < \ce{PhCOMe} < \ce{MeCOMe} < \ce{MeCOH} < \ce{PhCOCH2COPh}$
Tell me if its right otw i wont give the explanation  :D .
[hide]
Bcos i always write big explns and then the author of the prob says no u r wrong and i get irritated  :rotfl:
[/hide]

Is it devastatingly wrong or is it only a small mistake?  :maybe:   :huh:

Well isnt the order of Deuterium exchange the order os acidity or reactivity or whatever?  :huh:

let others attempt
well aready given a hint by telling that urs is devastatingle wrong

yes that was obviuos enuf but still lacks correctness :P

See no one will try hard to answer this when there is a lot of vetti talk in the indian forum  :D

that's not vetti...justing freeing the mind of all the tension an d preparing tos tudy

i don't get out f my room most of the day...so ...

[quote="madness"].... u've become fat???  :D  :D   (Just continued ur sentence)  :rotfl:[/quote]
it should have been i not u :P  :P  :P

Pardesi - Have you ever seen Rohit? Then you wouldnt have said that at all... :D

no i live far far far away(no not that far :P ) in orissa....anyway was expecting that too happen :rotfl:

so how r u at school u talk a lot(i get this impression)  :)

[quote="shruthibaskaran"]Lol.... Seriously, one wouldnt get the impression about how the people here are really like by reading the posts at all... For one, Rohit barely talks to people in school, but look here and you suddenly find him quintessentially talking... :D[/quote]

You ask Shreyas and others abt this. Shreyas might even kill u for saying that  :D   :D   :P

Shre has told me about that a couple of times, but hell, I can talk about what I used to think cant I?

@Pardesi - I completely agree

And as for Me? I am the perennial talkative kinda person who gets into loads of trouble for talking... :) N the kind of person who likes having fun, a lot.... And not someone who has their nose buried in a book whenever you look at them.... (like ranga and to a certain extent even rohit ;)

Maybe you should ask these guys about me..... Lol. I am sure they will have great opinions.. Esp Shre... Poor guy has put up with me for more than 10 years..

[quote="pardesi"]ava apodha pathi pesural...avalaku enna theriyum namma rohit anna eppo epidi mari peyetanu[/quote]

Translate that plz  :D  Ur tamil is becoming unrecognisable pardesi  :D

i said she was talking  baout the old rohit she knew...u have changed now :P

Frankly i say no  :D  I dont think i have changed (Except for coming on AOPS and haing more friends than before and studying more than before  :D )

Yep... I like this forum :D Friends and studyin together, thats a rare combo most of the time...

ye me too...have been longing for people here ... :)


It is right but i also think the problem is  wrong since it doesnt seem possible to factor it any better... :maybe:

[quote="not_trig"]Um...
Is the second one right?  :huh: \[2q^{2}+14qw-22w^{2}= 2(q^{2}+7qw-11w^{2})\] :maybe: The problem seems wrong...[/quote]

thats what it says..i like checked it a couple times...mayb a trick question or something..thanks


Given any triangle with sides a, b, and c.

Let x  be the altitude of the triangle perpendicular to side a
Let y  be the altitude of the triangle perpendicular to side b
Let z  be the altitude of the triangle perpendicular to side c

How do you express the radius of circumcircle R in terms of the altitudes x, y, and z?

It is easy...
Using Heron and $ ah_{a} \equal{} 2S \equal{} bh_{b} \equal{} 2S \equal{} ch_{c}$ you may express sides of the triangle and it's surface in terms of heights. Then for R you may use: 4R=abc/S. Hope this helps.


geometry
circumcircle
geometric transformation
geometry proposed

Given a triangle $ ABC$. Let $ Fe$ be its Feuerbach point. Let $ P$ be the tangency point of $ A$-mixtilinear incircle of triangle $ ABC$ with its circumcircle. Show that if the three points $ A, Fe, P$ are collinear then $ AP\perp BC$.

(1) According to Monge-D'Alembert's circle theorem (read the first part of my post from [url=http://www.mathlinks.ro/viewtopic.php?t=214649]here[/url]) applied to the incircle, nine-point circle and circumcircle of $ ABC$, the points $ F_{e}$, $ H$ and $ X_{56}$ are collinear ($ X_{56}$ is the exsimilicenter of the incircle $ (I)$ and circumcircle $ (O)$).

(2) Since $ A$ is the exsimilicenter of $ (I)$ and the mixtilinear $ A$-incircle $ \mathcal{K}_{A}$ and $ P$ the exsimilicenter of $ (O)$ and $ \mathcal{K}_{A}$, by the same Monge-D'Alembert theorem, it follows that $ A$, $ X_{56}$ and $ P$ are collinear.

In this case, if $ A$, $ F_{e}$, $ P$ are collinear, by (2), $ A$, $ F_{e}$, $ X_{56}$ are collinear, and thus from (1) $ H$ also lies on this line. Thus, $ AP \perp BC$. Conversely, if $ AP \perp BC$, then $ A$, $ H$, $ X_{56}$ are collinear, and thus $ F_{e}$ also lies on this line. Hence, $ A$, $ F_{e}$, $ P$ are collinear. This proves our problem.


calculus
derivative
calculus computations

Given f(x)=-2a(x^3)+3(a^2)(x^2). Find the maximum value on the interval
0 (less than or equal to) x (less than or equal to) 2.
Assume a is greater than 0

(sorry: it's x^3 and a is greater than 0)

Take the derivative and set equal to 0 to get $ f'(x)\equal{}\minus{}4ax\plus{}6a^{2}x\equal{}x(6a^{2}\minus{}4a)\equal{}0$, which equals 0 when $ x\equal{}0$ (or $ a\equal{}0$ or $ a\equal{}2/3$ but that doesn't tell you anything about $ x$ and it would yield a degenerate curve anyway).  So plugging in $ x\equal{}0$ yields $ f(x)\equal{}0$.  We also need to check the other endpoint, which is $ f(2)\equal{}\minus{}4a\plus{}12a^{2}$.  Is this positive or negative?  We don't know since you never specified any conditions on $ a$.


function
algebra
domain
limit
calculus
calculus computations

I am taking my first analysis course and I have a question regarding the definition of continuity.
The the definition of continuity of a real-valued function f at a point c can be stated

1)using the epsilon-delta formulation
2)using sequences

Can someone gives me the definition using sequences? Why are they equivalent?

Sequential definition:

$ f$ is continuous at $ a$ iff for all sequences $ x_n$ in the domain of $ f$ such that $ \lim x_n\equal{}a,$ then $ \lim f(x_n)\equal{}f(a).$

-----

Suppose $ f$ is continuous at $ a$ by the $ \epsilon$-$ \delta$ definition. Suppose $ x_n$ is a sequence (in the domain of $ f$) that tends to $ a.$

Given $ \epsilon>0,$ there exists $ \delta>0$ such that if $ |x\minus{}a|<\delta,$ then $ |f(x)\minus{}f(a)|<\epsilon.$ But by the definition of the convergence of a sequence, $ \exists\,N$ such that for $ n>N,\ |x_n\minus{}a|<\delta.$ Then it follows that for $ n>N,\ |f(x_n)\minus{}f(a)|<\epsilon.$ Again quoting the definition of convergence of a sequence, we have that $ \lim f(x_n)\equal{}f(a).$

-----

We'll prove the converse by actually proving the inverse. Suppose $ f$ is [b]not[/b] continuous at $ a$ by the $ \epsilon$-$ \delta$ definition. Then $ \exists\,\epsilon>0$ such that $ \forall\,\delta>0,\ \exists\,x$ such that $ |x\minus{}a|<\delta$ and $ |f(x)\minus{}f(a)|\ge\epsilon.$

For that $ \epsilon,$ let's take a specific sequence of such $ \delta$ - say, $ \delta\equal{}\frac1n.$ Then we have $ x_n$ with $ |x_n\minus{}a|<\frac1n$ such that $ |f(x_n)\minus{}f(a)|\ge\epsilon.$ Thus we have a sequence $ x_n$ with $ \lim x_n\equal{}a$ but $ \lim f(x_n)\ne f(a).$ That means that $ f$ is not continuous at $ a$ by the sequential definition.

-----

You could replace all expressions like $ |x\minus{}a|<\delta$ in the above argument by their distance equivalents, [i]e.g.[/i] $ d(x,a)<\delta.$ That is, the equivalence of definitions holds in an arbitrary metric space.


What is the largest three-digit multiple of 9 whose digits' sum
is 18?

Going through the largest multiples of $ 9$, we get $ 999$, $ 990$, $ 981$...
$ 990$ is the first one where the digits add up to $ 18$.


geometry
circumcircle
geometric transformation
reflection
parallelogram
cyclic quadrilateral
geometry proposed

Given a triangle $ ABC$ with $ \angle C \equal{}120^{\circ}$ and circumcircle $ w$. $ M_{a}, M_{b}$ are midpoints of $ CB, CA$ respectively, $ AM_{a}, BM_{b}$ meet $ w$ again at $ A', B'$ respectively, $ A'', B''$ are reflections of $ A', B'$ on $ M_{a}, M_{b}$ respectively, $ H$ is a foot of perpendicular from $ C$ to $ AB$. Prove that points $ C, A'', B'', H$ lie on a circle.

hello it's not to hard
see that CB"AB' and CA"BA' are parallelogram
and it's very easy to prove that AMbB"H and BMaA"H are cocyclic
now easily we can prove that B"CA"+B"HA"=180°
sorry am tired i can't write the full solution

[quote="saif"]sorry am tired i can't write the full solution[/quote]

Could you write up a full solution later?  Some of the things you say are easy, I cannot see.  I would greatly appreciate an elaboration.

[b]Lemma.[/b] In any triangle $ \triangle ABC,$ the orthocenter $ H,$ its othogonal projection on the median $ AM$ and the vertices $ B,C$ are concyclic.

[b]Proof.[/b] Let $ D,E$ be the feet of the altitudes from $ B,C$ and  $ Q \equiv DE \cap BC.$ Line $ AQ$ is polar of $ H$ WRT the circle $ (M)$ with diameter $ BC$ $\Longrightarrow$ $ QH$ is polar of $ A$ WRT $ (M)$ $ \Longrightarrow$ $ AM \perp QH.$ Let $ K \equiv AM \cap QH.$ Note that $ H$ is common orthocenter of $ \triangle AQM$ and $ \triangle ABC,$ thus $ KH \cdot HQ = DH \cdot HB = CH \cdot HE.$ Hence, inversion with center $ H$ and power $ KH \cdot HQ$ takes $ K,B,C$ into the collinear points $ Q,D,E$ $\Longrightarrow$ $ B,C,H,K$ are concyclic. Thus if  $ AM$ cuts the circumcircle of $\triangle ABC$ again at $ L,$ then $ KCLB$ is a parallelogram, due to $ \angle BKC = \angle BLC = 180^{\circ} - \angle A.$
______________________________________________

Using the lemma, we get that $ A''$ and $ B''$ are the orthogonal projections of the orthocenter onto the medians $ AM_a$ and $ BM_b,$ since the quadrilaterals $ CA'BA''$ and $ CB'AB''$ are parallelograms. Let $ H_c$ denote the foot of the altitude from $ C$ and $ H$ the orhocenter of $ \triangle ABC.$ By angle chase, we have:

$ \angle B''H_cA'' = \angle HAM_a + \angle HBM_b = 180^{\circ} - 2\angle AHB + \angle CBM_b + \angle CAM_a$

$ \angle B''H_cA'' = 180^{\circ} - 120^{\circ} + \angle CBM_b + \angle CAM_a$

$ \angle B''H_cA'' = 60^{\circ} + (120^{\circ} - \angle B''CA'') = 180^{\circ} - \angle B''CA'' $ $\Longrightarrow$ $ CA''H_cB''$ is a cyclic quadrilateral.

Your solution is much more easier then mine. I proved that these points lie on a circle with diameter $ CL$ where $ L$ is a foot of bisector of $ \angle ACB$.
Now I see that the problem is not hard.

I did not notice that the foot $ L$ of the bisector of $ \angle ACB$ also lies on this circle. Using the same notations, we can prove it easily by angle chase

$ \angle A''CL \equal{} \angle BCL \minus{} \angle BCA'' \equal{} 60^{\circ} \minus{} \angle CBA' \equal{} 60^{\circ} \minus{} \angle CAA'  \ \ (1)$

From the cyclic quadrilateral $ HA''H_cA$ we have

$ \angle CH_cA'' \equal{} \angle HAM_a \equal{} \angle CAA' \plus{} \angle HAC \equal{} \angle CAA' \plus{} 30^{\circ} \ \ (2)$

From $ (1)$ and  $(2)$ we get $\angle A''CL \plus{} \angle CH_cA'' \equal{} 90^{\circ}$ and the conclusion follows.

dear cosinator i think that you're meaning $ AM_{b}B"H$ and $ BM_{a}A"H$ are cocyclic
it's easy because
$ \angle M_{b}HA\equal{}\angle M_{b}AH\equal{}\angle CB'B\equal{}\angle M_{b}B"A$
and by a simple angle chase we can get the result. :D

[quote="saif"] $ \angle M_{b}HA \equal{} \angle M_{b}AH \equal{} \angle CB'B \equal{} \angle M_{b}B"A$[/quote]
Dear Saif, could you justify this step?. I'm not seeing where <C=120 is used.

the condition $ \angle C \equal{} 120^{o}$ we will use it later in our angle chase to prove that  $ \angle B"CA"\plus{} \angle B"HA"\equal{}180^{o}$ but

$ \angle M_{b}HA\equal{}\angle M_{b}AH\equal{}\angle CB'B\equal{}\angle M_{b}B''A$ to prove that $ AM_{b}B"H$ is cyclic I think it's clear
now ill prove that $ \angle B"CA"\plus{} \angle B"HA"\equal{}180^{o}$
we have $ \angle M_{b}HM_{a}\equal{}120^{o}$
so to prove the result $ \angle B"CA"\plus{} \angle B"HA"\equal{}180^{o}$ we need to prove
$ \angle M_{b}CB'' \plus{} \angle M_{b}HB'' \plus{}\angle M_{a}CA'' \plus{}\angle M_{a}HB'' \equal{} 60^{o}$
it's easy because $ \angle M_{b}CB''\equal{} \angle CBB''\equal{} \angle B''AM_{b}$
and $ \angle M_{b}HB'' \equal{} \angle M_{b}AB'' \equal{} \angle B'CA \equal{}\angle B'BA$ so
$ \angle M_{b}CB'' \plus{} \angle M_{b}HB'' \equal{} \angle B$
and   $ \angle M_{a}CA'' \plus{}\angle M_{a}HB''\equal{} \angle A$ ....

[quote][color=darkred][b][u]Lemma[/u].[/b]  In a triangle $ ABC$ the orthocenter $ H$ , its othogonal projection on the median $ AM$ and the vertices $ B,C$ are concyclic.[/color][/quote]

[b][u]A nice and short proof[/u].[/b]Denote $ E\in BH\cap AC$ and $ K\in AM$ , $ HK\perp AM$ . Prove easily that $ E$ and $ K$ belong to the circle $ w$ with diameter $ [AH]$  and $ ME$ is  tangent to $ w$ . Thus, $ ME^2\equal{}MK\cdot MA$ , i.e. $ MC^2\equal{}MK\cdot MA$ $ \implies$ $ MCK\sim MAC$ $ \implies$ $ \widehat {MKC}\equiv\widehat {BCA}$ , i.e. $ m(\widehat {HKC})\plus{}m(\widehat {HAC})\equal{}$ $ \left(90^{\circ}\plus{}C\right)\plus{}$ $ \left(90^{\circ}\minus{}C\right)\equal{}$ $ 180^{\circ}$ , i.e. $ BHKC$ is cyclic.

[b]virgil[/b], you killed the problem outright, your solution is very simple and direct.


If c=ax^2+by^2, a,b,c>0 and x,y are real numbers ax+by<=sqrt(c(a+b))

[i]unknown source[/i]

straight Cauchy

(ax^2 + by^2)(a + b) \geq (ax + by)^2

implies  \sqrt c(a + b) \geq (ax + by)


calculus
integration
calculus computations

Evaluate $ \int_0^1 (x \plus{} 3)\sqrt {xe^x}\ dx$.

hello
I m getting $ 2\sqrt{e}$.Is it the answer?


If 1000 > p > 2000/3 then it appears twice in 2000! and twice in $1000!^2$.  So p < 2000/3.

The largest such p is p = 661.  Then it appears 3 times in 2000! but only twice in $1000!^2$.  Hence answer.

[/img][quote="RISH2909"]i did not get it[/quote]

[quote=Rushil]Determine the largest $3$ digit prime number that is a factor of ${2000 \choose 1000}$.[/quote]

[b]A useful lemma:[/b]
If $p$ is a prime such that $\dfrac {2n}3<p\leq n$, then $p$ doesn't divide $N=\binom{2n}n$.

For proof, note that, $\nu_p(N)=\nu_p((2n)!)-2\nu_p(n!)=2-2\cdot1=0$. This lemma is used as part of a proof Erdos's proof of Bertrand's postulate. As for the problem, this lemma can do the trick to find such prime.

A very similar question asked in AIME 1983. Just instead of 2000 & 1000 are 200 & 100.

[quote=aayush-srivastava]661 is the answer...
see that square of any 2 digit prime number is greater than 2000. highest power of p that divides 2000 is $[\frac{2000}{p}]$...
similarly for 1000 is $[\frac{1000}{p}]$..
the highest power of p that divides ${2000 \choose 1000}$ is $[\frac{2000}{p}]$ - [b]2[/b] $[\frac{1000}{p}]$...
If p>666 then $[\frac{2000}{p}]$ - [b]2[/b] $[\frac{1000}{p}]$ = 2-2(1)=0..
i.e. p doesn't divide  ${2000 \choose 1000}$..
the required prime number is the largest one less than 666..
661 is the anwer and clearly $[\frac{2000}{p}]$ - [b]2[/b] $[\frac{1000}{p}]$ becomes $3-2$=$1$ as required...
[b]Answer = 661[/b]

(here $[\frac{a}{b}]$ represents the greatest integer function of $\frac{a}{b}$)[/quote]

Square of any 3 digit prime number , not 2 digit.

By definition,
$$\binom{2000}{1000} = \frac{2000!}{1000!\cdot1000!} = \frac{2000\cdot1999\cdots\cdot1001}{1000!}$$
So, we want a prime number whose multiples come twice in numerator and once in denominator (for largest result). This means we have the following inequalities:
$$1000 < 2x, 3x < 2001$$
$$0 < x < 1000$$
We don't need the second inequality. From, the first inequality, we can get these inequalities.
$$500 < x < 1001$$
$$333 < x < 667$$
This means that:
$$500 < x < 667$$
So, we want the largest prime in this range. Testing, we see that our answer is $\boxed{661}$.

how to do the cdots latex ;-; it looks horrible :/

[hide=Solution]We rewrite $\binom{2000}{1000}$ as $$\frac{2000!}{1000! \cdot 1000!}=\frac{2000 \cdot 1999 \cdot 1998 \ldots \cdot 1001}{1000!}.$$ The prime number obviously has a factor in the denominator, so it must have two factors in the numerator. To make this happen, the number must be greater than $\frac{1001}{2}$ rounded up, or $501$. It also must be less than $667$, since that number only has $1$ factor. Testing, $665, 663$ are not prime but $\boxed{661}$ is, making that the answer.[/hide]