Find the maximum value of $x^{2}y$
For this question, I suggest it is
$({{\frac{(x^{2}+y)}{2}}})^{2}$---[b]a[/b]
However, if there is one constraint given, for example,
$x+y=1$
Then, using the expression created to find the maximum leads towards a wrong answer, why?
Though I know the correct solution for this question, but now I don't have confidence in using AM-GM inequalities.
What's the theorem behind and what does expression[b]a[/b] mean when there's a constraint?

[quote="Command"]Find the maximum value of $x^{2}y$
For this question, I suggest it is
$({{\frac{(x^{2}+y)}{2}}})^{2}$---[b]a[/b][/quote]

It looks like you arrive at this answer using AM-GM:

$\frac{x^{2}+y}{2}\ge \sqrt{ x^{2}y}$

However, when you have an extra constraint, [b]there is no guarantee that you can reach equality.[/b]  Hence $\left( \frac{x^{2}+y}{2}\right)^{2}$ is an [b]upper bound[/b], but is not necessarily the actual [b]maximum.[/b]

Edit:  If you want to know what the theory behind AM-GM in two variables is, it is very simple:

[hide="Proof"] $( \sqrt{x}-\sqrt{y})^{2}\ge 0$
$\Leftrightarrow x+y-2 \sqrt{xy}\ge 0$
$\Leftrightarrow \frac{x+y}{2}\ge \sqrt{xy}$ [/hide]
However, as I have said before, when you have an added constraint you may not necessarily have $x, y$ such that equality is possible.

Edit #2:  I should be clearer.  In this case, the equality case is possible; however, when you use a particular inequality (such as AM-GM) with a given constraint, you have [b]no guarantee[/b] that the equality case of the inequality and the maximal case with the constraint are the same.

Yes it is by AM-GM.
With the expression [b]a[/b],
it is $x^{2}=y$ when the $=$ holds.
This gives an equation,
$x^{2}+x=1$
The positive root$=(-1+\root5)/2$
The true maximum value of the original inequality is $4/27$
where we have $x^{2}=2/3$
while $y=1/3$

With the false x and y, we have the true maximum largest than the false maximum, why?

[quote="Command"]With the false x and y, we have the true maximum largest than the false maximum, why?[/quote]

It is like I said.  When you pick an inequality, that inequality has a particular equality case.  However, when you have a constraint on your problem, it is not necessary that the maximum [b]given the constraint[/b] occurs simultaneously with the equality case.

Think of it graphically:  Imagine that $z = x^{2}y$ is a 3-dimensional surface that we want to find a peak of.

However, our constraint $x+y = 1$ restricts us to a particular "slice" of that surface, and we only want to find the peak of that slice.

When you use AM-GM on $x$ and $y$ you are essentially sketching a curve $z' = \left( \frac{x^{2}+y}{2}\right)^{2}$ that is guaranteed to satisfy $z' \ge z$ - in other words, it is always on top of our original curve.  It will touch our original curve at some certain points.

The key thing to note here is that $x^{2}y$, as is, has no maximal value, since it goes to infinity.

Once we restrict our focus to $x+y = 1$ - our particular slice - we now have some maxima (as long as we restrict our attention to $x,y \ge 0$); however, the intersections of $z'$ with $z$ on that slice are not guaranteed to be maxima in any way.

What about \[1=\frac{x}{2}+\frac{x}{2}+y\geq 3\sqrt[3]{\frac{x^{2}y}{4}}\quad\Leftrightarrow\quad x^{2}y\leq\frac{4}{27}\]


I heard you can buy some books at ARML site.. Which books? Or is it differ from years to year? Also, what kind of books are there?

Yeah i would like to know as well.

It would serve as a nice souvenir as well.

[quote="Sly Si"]At the very least, you can buy the 1995-2003 ARML/NYSML book.  You might also be able to get the one from 1989-1994.  I don't really know what others are available.

You can also buy t-shirts and frisbees.  Those you might want to get early on, though; last year Iowa ran out of frisbees before (or maybe just after) the individual round.[/quote]

Is the store open even on Friday (June 2nd)?

I don't know.  Last year at Iowa, there was a table set up in the front of the individual/relay room selling stuff.  As far as I know, there wasn't anything sold on Friday, but they could have been selling stuff at the math talk or something.

Also, things could be completely different at Penn State.  I've only ever been to the Iowa site.


Hola,

He notado que en general hay muy pero muy poca actividad en los foros en espaniol (perdonen, pero usando linux no tengo acentos) asi que, metamosle candela, hagamos una comunidad hispana que se mueva. De por si hay una buena cantidad de hispanos aqui (bueno, y de gente que no es hispanohablante nativa, pero igual habla!)

Un caluroso saludo,

[quote="Pascual2005"]buena idea, seria del putas que se metiera mas gente latina, no? jaja, lo mejor es que el servidor no reconoce grocerias en español! ajja[/quote]

Pues no se si el moderador entendera grocerias en español, pero yo si! La verdad es bastante desagradable!

Y eso que yo no soy grocero, hablese con Manuel el de puertorrico, ese cangri si es tremendo!

Pascual2005,

Su mensaje original en esta conversacion fue borrado.

ah listo, haga lo que quiera, aqui estamos porque nos gustan las matematicas para que vengan ahorita a molestar por ese tipo de cosas! pues si usted es el moderador, bien lo respeto, pero la autoridad no se impone se gana! ok??? ;)  ;)  ;)  ademas pues creo que debe haber un poquito de libertad, y mi mensaje original no atentaba contra nadie, a ningun miembro estaba insultando!

Pero bueno, si usted se quiere dedicar a remover este tipo de cosas mientras nosotros hacemos problemas es su vida, no? :D , simplmente no me parece necesario...

Pascual,

Aqui estamos en una comunidad abierta al público, el cual, de una u otra forma tiene una imagen que se quiere mantener. Si usted hubiese puesto el mismo mensaje en inglés en un foro cualquier otro foro en el sitio, creo que cualquier otro moderador lo hubiese borrado de inmediato, aquí no tiene por que ser distinto.

Concuerdo con que se debe tener libertad para expresarse, pero una cosa es libertad y otra cosa es libertinaje. Su mensaje fue, en mi opinion, libertino y no liberal. Nuestro idioma es muy rico en forma y articulación, no me parece que limitarnos a un conjunto de palabras que la mayor parte de los hispanohablantes hayan peyorativas.

Tengan en cuenta que aqui, no estamos en un foro formal, pero tampoco estamos en la banqueta de nuestros barrios con nuestros amigos de siempre con los cuales podemos hacer uso de tal vocabulario dado que nadie más lo va a escuchar. Estamos hablando de un foro al que más de dos mil personas visitan todos los dias, y potencialmente un buen número de ellas sera capaz de entender Español (la tercera lengua más hablada en el planeta), y visitar este foro en especifico. Mantengamos una buena imagen, al mismo tiempo que una buena relación.

Recalco, una conversación informal es aceptable, pero deja de serlo en el momento en el que se empiecen a usar palabras obsenas. Al mismo tiempo que no se tolerará ningún insulto o comentario disfémico hacia cualquier persona, sea o no miembro del foro (Que si revisan, he editado unos pocos de esos).

De la misma forma, no me interesa ser despótico. Si alguien considera que alguna acción que tomo es excesiva, esta en libertad de expresarlo, ya sea en el foro (donde talvez no sea lo mas apropiado, pero en fin, pueden hacerlo allí) o en un mensaje privado directamente.

Paz y matematica  :D ,

ok, Jimenez. Pues si se quiere conservar la buena imagen esta bien, pero tambien lo buena persona que sea uno no se ve reflejado ni esta implicito en el uso de su vocabulario! es decir somos jovenes (la mayoria) y no nos interesan mucho esos formalismos, pero si usted (el moderador) le aprecen importantes respeto esa opinion considerando que usted es una persona mayor t depronto ams sabia en estos aspectos de la vida.

Sin resentimientos y solo buscando aclarar lo ocurrido...

Tranquilo Djiminez...
Algo bonito de la cultura latina es la confianza.. estamos entre hermanos.. que es eso de palabras obscenas??? nuestro deber aqui es hacer matematicas y ala misma ves pintar una cara diferente de las personas que le gusta la matematica(no los mamones usuaes, en mi pais esa palabra no es vulgar si lo es para ti, pues perdon) creo que no debe editar los mensajes y mantener esa confianza que tenemos todos los latinos a diferencia de las otras culturas. Pero bien, acuerdese que nosotros tenemos 17 ,18 años.. Usted supongo que ya es mayor.. pero entienda tambien nuestra posicion.. usted con su conocimiento matematico doctoral en donde nos deberia ayudar debe ser en la solucion de problemas.. No es coincidencia que dos jovenes latinos Pascual y Yo interesados en la matematicas nos incomodemos con un pequeño foro en internet donde nos metemos para practicar y aprender mas de matematica pero de forma "latina" . Pero como usted dijo.. Por que la matematica no progresa en los paises del tercer mundo?? empesmos por aqui .. que en nuestro informalismo muchos jovenes latinos se sentiran comodos de aprender mas.. Ok?  :D  Tranquilo, cojalo con calma y vamos a establecer una buena relacion para conocerlo ya que es un honor conocer matematicos hispanohablantes para que nos ayuden anosotros con nuestras curiosidades matematicas. Gracias y que le vaya bien en sus estudios doctorales

Manuel, Pascual,

En algo tienen razon (en estos momentos me encuentro en una terminal linux, no tengo acentos), una de las cosas mas hermosas de la cultura latina es el calor y usual confianza de su gente. Vieran como la extrania uno cuando se encuentra tan lejos.

Primero, AUCHHH!!!, yo no me considero tan mayor  :D  :P , aun cuando, si, en efecto soy un poco mayor que ustedes. A principios de este anio cumpli mis 24 primaveras (literalmente), pero si, tal vez esos escazos seis o siete anios que les llevo de ventaja me hagan ver muchas cosas de forma diferente.

No crean que es que pretendo estar aqui solo molestandolos y cambiando sus mensajes. No, todo lo contrario, me parece que los moderadores (de los cuales empece a formar parte recientemente) deben intervenir solo cuando es estrictamente necesario. Considero que todos deberiamos tener la libertad de expresar lo que queremos expresar, sin embargo el arte esta en la forma de expresarlo. Ciertamente, me parece que la mayor parte de las persoas que activamente usan el foro son efectivamente hispanos de al rededor de 17 anios, sin embargo no son los unicos en accesarla. Recuerden que a diferencia de otras circumstancias, en este caso, sus palabras quedan alli, donde mucha otra gente la vera.

En fin, no quiero iniciar una discusion de nunca acabar, simplemente, sientanse en libertad de expresar lo que quieran, pero por favor, moderen el lexico al hacerlo.


function
group theory
abstract algebra
search
algebra open
algebra

[quote="Dtrung"]Find all function $f : Q$→$Q$ such that :
            $f(f(x)+x+y) = x+f(x)+f(y)$
for all $x,y$ belong to $Q$
[/quote]
Let $S=\{q| \ exist \ x: \ q=x+f(x)\}$, then for $q\in S$:  $f(q+y)=q+f(y)$. Therefore S is set of periods for $g(x)=f(x)-x$, because $g(q+y)=f(q+y)-q-y=f(y)-y=g(y)$.
It mean S is subgroup Q. For all x we have $x=q+t,q\in S, \ t\in Q/S$. Then $f(x)=f(q+t)=g(q+t)+q+t=g(t)+q+t=g(t)+x$.
There are infinetely many solutions. For example.
1. S is not empty. If $S=\{0\}$, then $f(x)=-x$.
2. S=Q, then $f(x)=x+const$.
3. $S=\{bk,k\in Z\}$ $f(x)=-x+b[\frac{x}{b}]+c$, were $c\in bZ$.

This was done before.
If $A$ is any subgroup of $(\mathbb{Q},+)$ with coset representatives $R$ (don't ask me what that is, ask wikipedia or use search),
and $g: R\to A$ any map, then
$f(r+a)=g(r)-r+a$ for all $r\in R,a\in A$
give all the well-defined solutions of the problem.

Examples:
$A=0$, then $f(x)=-x$ for all $x\in\mathbb{Q}$
$A=\mathbb{Q}$, then $f(x)=x+c$ for all $x\in\mathbb{Q}$
$A=\mathbb{Z},R=[0,1),g\equiv 0$, then $f(x)=2[x]-x$ for all $x\in\mathbb{Q}$
You can also take
$A=\{{m\over 2^{n}}: m\in\mathbb{Z},n\in\mathbb{N}_{0}\}$
or
$A=\{{m\over n}: m\in\mathbb{Z},n\in\mathbb{N},n \text{ square-free}\}$
or ...

[quote="Dtrung"]Find all function $f : Q$→$Q$ such that :
            $f(f(x)+x+y) = x+f(x)+f(y)$
for all $x,y$ belong to $Q$

I'm sorry about my poor English!!![/quote]
see http://www.mathlinks.ro/Forum/viewtopic.php?t=127249  :D


How would you find the values of x,y,z when
xyz is a factor of (x+1)(y+1)(z+1)-1?

I know that there are two answers, but how are they obtained is what I want to know.  Could someone please help me?

http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1992
I heard that the question has similar concepts to the question 1 in day1 1992 IMO
Still anyone know how to find the values?


What do you think about the big Swine Flu hype in NC recently? Is it overreacting? Or do you think it is fairly reasoned?

Overreacting, definitely.  Once I got something that MIGHT have been the swine flu.  The doctor ok-ed me to go back to school.  Then when I told everyone I MIGHT have the swine flu students basically quarantined me by staying away and using hand sanitizers.  Seriously, the death rate for normal flu and swine flu is about the same.  Just because people haven't found a cure (and actually the government approved a new vaccine recently) doesn't mean your immune system is too weak to fight a new virus.

Here at NCSSM, people (read administration) are definitely overreacting to swine flu; I see people being sent home with masks on daily.  In almost all of my classes, at least one person is gone from "swine flu," and probably around 10% of the school over all is home.  Note that if you show possible symptoms here, you are sent home and I think you have to stay at home a week before returning.

I'm not sure how many people are just using the swine flu as an excuse to escape from the "stress" caused at NCSSM, but I have heard of people going to the clinic with not entirely legit symptoms so that they can go home and have time to catch up on school work, etc.  However, these people probably do need to go home anyway, but that is primarily due to poor nutrition and pathetic sleeping habits.

Also, the health person here has hand sanitizer [i]everywhere[/i], which I think is a bit much.

Earlier in the year, we were going to have swine flu drills so that classes could figure out what they needed to do in case students got sick, but then enough did that the drills seemed pointless.

In summary, I don't think swine flu itself is causing most of the problems at NCSSM, but is still being over-hyped.

@professordad: I hope you get better! and you really shouldn't have told anyone... lol (sneaky, sneaky :D )
@b-flat: Wow, you really feel strongly about this. Although your little comment about NCSSM stress deos deter ppl (like me) from going to NCSSM....

Nice that we get ppl talking on the NC blog. It's been really slow lately...

I think swine flu has gone from overhyped to ridiculous at my school. We make jokes about it everyday.  Though the teachers act like saying the name is voodoo or something... Overall, good for some laughs.  Don't think any ppl really has it though... I think it's just the media having nothing to talk about.  They have to earn their pay in this bad economy somehow right? Poor us...

Yeah, I [i]did[/i] get better. (Thanks though!  :D ) With medicine, of course, but my dad had the same symptoms as well as symptoms of pneumonia or something, and he got better by himself and Motrin.

But there are quite a lot of cases here, I'll have to admit, THOUGH I still think the flu thing is over-hyped.  I heard that there's 651 students on NC State campus who have flu-like symptoms.  And in my Chinese School there were a few serious cases about flu.[/i]

Yeah like 1/3 of my school is constantly gone cuz of the flu.  I got it a few days ago, but still went. Nobody cared and I infected nobody (which was strange, but lucky i guess...).  Can't afford to miss school, even if I never learn much....

no offense to anyone here, but my friends say that people long ago said that we will have an african-american president when pigs fly.  now, we have the 'swine flu'. :P

Definitely overreaction because its just the flu

Hand sanitizer kills bacteria. Does it kill swine flu virus? xD

ROFLOL at the pig fly joke.

Hm... Supercrash has a point.. From what we learned in Biology Honors, viruses are nonliving and reproduce by controlling your cells.  If that's true, the hand sanitizer can't work because it can't mess with your cells only bacteria.  Anybody taking AP Bio want to prove me wrong?


pigeonhole principle
symmetry
geometry
perimeter
inequalities
geometric transformation
cyclic quadrilateral

Suppose that a triangle can be placed inside a unit square such that the center of the square is not inside the triange. Show that a side of the triangle is less than 1.

You need to precise a little bit by saying that if the center of the square does not belong to the triangle (nor to the inside of the triangle), then one of the sides of the triangle is stricly less than $1$.

I've almost got a proof, but it's not completely rigorous.

First, one can show that there exists a line passing through the center of the square and dividing it into two regions such that the triangle is contained within one of those regions (a rigorous proof is easy to write), the dividing line being parallel to one of the sides of the triangle (the one which is closest to the center, in terms of euclidean distance).

Now I'm pretty sure (but couldn't prove it formally) that it is possible to translate the triangle such that its basis (the one parallel to the dividing line) actually lies on this dividing line, and such that the translate thus obtained still lies in the square.

Afterwards, by symmetry, another isometric triangle can be "glued" to the first one to form a convex quadrangle within the square. But if this quadrangle has each sides with length $\geq 1$, this would lead to a contradiction if the following lemma is true:

[b]Lemma[/b]: Given two convex polygons $A,B$ such that $B$ is inscribed in $A$, the perimeter of $B$ is no more than that of $A$.

Intuitively it seems true to me, but how to prove that ? (surely by some extensive use of triangle inequalities)

Edit: yeah, the lemma seems easy to prove. But what about the translation thing ?

[hide="A hint"]
After you divide the square into two regions by a line through the center such that the triangle is on one side, draw a perpendicular to that line through the center of the square.  The region the triangle is in is divided into two regions.  What can you say about the distance of points in these regions?[/hide]

Forget my last post, this does not work (the glue thing is pure nonsense).

Thank you for the hint, I'm going to think about it on my way to the bus  :D

The two divided regions are either cyclic quads or if the line is a diagonal, a triangle. If it's a cyclic quadrilateral, then the longest distance in the region is less than 1: the sides are obviously less than 1, and so is the diagonal from the centre to a corner. The other diagonal is less than 1 by Pythagoras. If the regions are right angled triangles then the longest distances in each is a side of the square equal to 1.

At least one side of the triangle lies completely in either region (borders inclusive) by PHP. If the line through the centre is not a diagonal, then there therefore exists at least one side with length less than 1. If it is a diagonal, we assume to reach a contradiction that it's possible for a triangle to have no sides less than 1. The only lengths not less than 1 are two adjacent sides of the square. But the third side of the triangle then goes through the centre, reaching a contradiction.

Phelpedo- Take any triangle in a unit square. Draw three lines through the centre of the square, one parallel to each side. If all the sides can cut(touch) the triangle, then the centre must be in(on) the triangle. Can you prove that?

[quote="Phelpedo"]How do you argue rigorously something like the triangle has to be on a certain "side" of the square?[/quote]

You argue rigorously that given any triangle and a point outside the triangle, there is a line through the point which does not intersect the triangle using angles, parallel lines, and a little casework.


There seems to be a small change in the last few days in the link in the email sent for watched topics, with the internal link left out. The result is clicking on the link only goes to the top of the page rather than the post itself.

Thus I got the following link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=679484
instead of
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=679484#679484
so #679484 is missing

Is this deliberate?

Correct, # takes you directly to that post. I don't know why this was changed though...

[quote="Klebian"][quote="7h3.D3m0n.117"]well if you try both links they end up on the same page...however the one with the # after it shows a different post so i think the longer URL is the latest post (to save time browsing through pages of a popular post)[/quote]
Nope, I think the shorter URL gives you the page of that post, the longer URL gets you specifically to that post.[/quote]

thats what i meant lol

That happens to me too and I don't notice anything strange.


What???

What's with all the plus signs???

I don't know what you are trying to say is your solution.

anyway, we want the graph of $ y\equal{}\lfloor \sin x \rfloor$.  Whenever $ \sin x< 0$, $ y\equal{}\minus{}1$; when $ 1 > \sin x \ge 0$, then $ y\equal{}0$, and when $ \sin x \equal{} 1$, $ y\equal{}1$.  Our function is then:

$ y\equal{}\left\{\begin{array}{lr} \minus{}1 & : x \in (\pi \plus{} 2\pi k, 2\pi\plus{}2\pi k)\\
0 &: x \in [2 \pi k, \frac{1}{2} \pi \plus{} 2\pi k) \cup (\frac{1}{2} \pi \plus{} 2\pi k, \pi \plus{} 2\pi k]\\
1 &: x \equal{}\frac{1}{2}\pi \plus{} 2\pi k
\end{array}\right.$

where $ k$ is any integer

That was not my solution. They were the intervals of interest. However you answered my question quite precisely which did tell me that I graphed it correctly. I drew the graph with the '+' signs correctly but when I posted it it became gibberish. But man your good.

How did you do that? Any way, the graph you posted is still inaccurate with my initial graph, you would have to shift this graph horizontally positive pi/2 units, therefore this equation would be y=[sinx-pi/2]. Could you please tell me how you posted the graph?

Use the code button - alternatively, use the syntax: [code][code]Typewriter font stuff here[/code][/code]. There is a quote option (for ex. on the bottom right of this post) which you can use to view previous posts; I copied what you posted, but due to the small horizontal space some things got shifted around.


I'm considering about creating Physics club in my school next year, since my school doesnt have one...
It looks like many high schools already have one.
So I'd really appreciate any kind of recommendations or advice...

One more question: What do u guys do in Physics club? Just solving some Physics pbms.?

Thanks,

Another thing they do is chat and discuss about physiscs e.g. the importance of the relativity theory, the new inventions, energy, etc.... discussing is an important part of  physiscs group..... here I know a group about "nuclear" energy (dont know the exact word) they talk a lot about energy. It's every Saturday. A firnd of mine goes every class!!!!!!!

[quote="quantum leap"]Do you think there is sufficient enthusiasm for physics in your school already or do you plan on generating some?
I personally don't know anyone who would voluntarily do physics problems for fun, but that may be different at your school.  Besides discussing problems and concepts though, I don't know what a physics club would do, although I can't speak from experience.[/quote]
I;m not sure whether there is sufficient enthusiasm....
Yeah, I;m plannig to generate some..

[quote="José"][quote="shadysaysurspammed"]How about starting an online physics club?You know,by chatting...[/quote]

how and where shady? a web page? like MathLinks?????[/quote]

I don't really know..maybe chatting likle msn messenger or yahoo(we can send and receive diagrams n stuff as well.)Only latex problem might be there...we have p.m for that.

I support your physics club. Here, in Azores Islands, Portugal, a physics club is almost an utopy. One or two students interested, no teachers interested.  :lol:

and what about making it here? I mean, start a section here....

ask the admins about that jose,it would be nice if we had discussions on particular topics and compile problems n stuff.

The AoPS forums and AoPSWiki are a good way to discuss, teach, and learn about physics.


algorithm
geometry
3D geometry
tetrahedron
analytic geometry
vector
invariant

It is obvious that we can't have 3 non-planar points. So, in the general case of $n$ non-planar points ( 3 at a time non-planar), how do we find the point such that if it is $P$ and other points are $A$, $B$,.... then $PA +PB +PC+....$ is minimum.

ALso, for the simple case of the fermat problem in say a triangle , suppose we don't know that the fermat point is the one which subtends an angle of $120^{\circ}$ at all sides. How do we find it before actually prooving it is the required point?? i.e. what is the algorithm??

Let the tetrahedron be ABCD with corner coordinates XA=(xA,yA,zA) etc. and the unknown Fermat point be X0=(x0,y0,z0).
Writing (X0-XA) as UA and the length of AO as LA etc. (where LA=|UA|, vector length of UA) then the condition for the Fermat point is:
          UA/LA + UB/LB + UC/LC + UD/LD = 0
With the U's being vectors this is 3 equations in the 3 unknowns: x0, y0 & z0.
I know of no way of getting an explicit solution to this but it is simple to set up a Newton-Rapphson solution which gives 15 s.f. in 4-5 iterations depending on starting point.

Thanks for replying!!! Though I am aware of the analytic solution (algorithmic approach) you have given for a general $\mathbb{R}^{n}$ space, I was looking for a 'purely geometric ' solution to the 2D and 3D case. No offence plz. I really appreciate your posting this wonderful solution! I hope you understand what I mean by 'purely geometric'? Basically, it shouldn't inlvolve algoritms and equations, Do you think such a solution is possible? Are you aware of one!?
Thanks again!!!

I’m new to Math Forum and am not sure I have got the hang of posting replies.
No, I don’t know of any solutions for this.
What I have found - and it is presumably well known stuff - is that for tetrahedron ABCD with Fermat point O:
Angle AOB = COD
Plane AOB bisects angle COD
The Fermat point is unaltered when point A moves along AO (on the same side of O as A).  (This means that a simpler case - some sides equal - can be analysed and then generalised.)
All the above are independent of the labelling order of the vertices.
The solid angles subtended by the faces at O are all equal (to pi steradians).
There is a paper by Rubinstein et al. “Minimum networks for four points in space”, Geometriae Dedicata, 93, 57-70, 2002, which is looking at Steiner trees rather than Fermat Points - but this is closely related.  They say that for the general tetrahedron there is no solution that does not involve radicals.

Do you know where I can find a proof for : [quote="John Brew"]What I have found - and it is presumably well known stuff - is that for tetrahedron ABCD with Fermat point O:
Angle AOB = COD
Plane AOB bisects angle COD

The solid angles subtended by the faces at O are all equal (to pi steradians).
[/quote]

SO, basically, the fermat point of a tetrahedron is the point where all the faces subtend equal angles??? Or did I misunderstand you! So basically, the solution si sort of a 'logical extension' of the result for 3 points in 2D to 4 in 3D ... actually, my conjectures seem to indicate that this might be true for the general case also!
Regarding the above proof, are you aware of a 'purely geometric' one. Or is it possible only through computation????
Thanks for your help!

Hi - yes faces subtend equal angles - just like the triangle case but everything is up a dimension.

I have no geometric proofs for these!
I am an engineer looking at cable nets which are essentialy chains of tetrahdra and Fermat points.  (The vertex of one tetrahedorn is the Fermat point of the next!)
Results about the point being a Fermat Point come from minimum energy principles.
Results about angles and planes come from analysis of force equilibrium of this solution.
Result about equal angles comes from numerical analysis of actual cases.  All the ones I have looked at have angle pi.

If any of these results are not known then I wish you good luck with the proofs!

Thanks for clarifying the result!!! I'll see if I can make the proof!!!  :)  :lol:

Right now, I am finding the results so that I can prove them later. If possibly, you have the results for the general case or some other case (say 5 nonplanar points etc) please post them!! I'll tell you the proof if I find one!
Thanks! Keep posting!
Can you cite some online source where I can get 'the minimum energy principle' proof you told about. Maybe I can get some geometric idea from that! Anyways, thanks!

I’m afraid I don’t have any online sources on the minimum energy principle.  However it is a very standard approach in engineering and mechanics which I have simply applied to my cable nets.  There should be sources that cover it.  Taking the smallest possible element cables AO, BO, CO, DO running from fixed points and under constant tension have minimum energy when their total lengths are minimised.  (Think of the linkage all being connected ‘round the back’ into a single cable and a weight hung on its free end.  The minimum energy is when the weight is lowest, i.e. when the total lengths AO+... etc. are shortest.)
After that we know the cable tensions are equal and that O is in equilibrium so that forces, resolved in any direction, sum to zero.  Hence the results about equal angles and planes bisecting angles.
The result about moving any point without affecting the location of the Fermat Point comes directly from the first vector solution I posted.
This suggested that the invariant here is solid angle.  I checked some results numerically and found that in all cases the solid angles are equal.  I suppose you would call this experimental maths!
I don’t have much further practical interest in this so I suppose I will let it quietly drop (although I would be glad to know of simpler ways of calculating the Fermat Point).
Another interest of mine is tension membrane structures which have the same mathematical form as soap films.  The cable nets are a type of discrete analogue of these.  One condition for these membranes is that the mean curvature is everywhere zero.  I was interested to find that for the cable nets the corresponding measure - the mean kink in the cable - is also zero.  (This is the result that angle AOB = COD in the tetrahedron.)  Also of practical interest for the membranes is the Gaussian curvature (which affects stiffness and ease of manufacture) and I am wondering what is the cable net (tetrahedron & Fermat point) equivalent.  Possibly the product of the cosines of the semi-angle of AOB and COD?
Maybe there is a mathematical connection here for you between the Fermat point and Plateau’s problem.
As you see this is a far cry from proper maths and I feel a fraud on this site.
I don’t know of any of the generalisations you suggest but there is a wide practical interest in this problem in Geography (facility location & transport network planning) but they are mainly concerned with the plane.  Maybe the space scientists and computer designers (minimising cable connection lengths) have the same problem in 3-space and have done work on it.
I haven’t got much more I can add on this.
Again, good luck.

I have just realised that the proof of the equality of the solid angles could be quite straight forward.
IF the pairs of opposite angles I mentioned earlier ARE equal:
AOB=COD= p, say;  AOC=BOD=q;  AOD=BOC=r
then each solid angle (O_ABC, O_ABD, O_ACD and O_BCD) is a three-way corner comprising the same three different angles: p, q and r.
Then (I assume that there is a theorem that states this explicitly) if this condition applies the enclosed solid angles are equal.

John, do you think you can find the 'Fermat point' of a Hexagon and an Octagon using your computation system. i'd be obliged if you can do this. It will allow me to make some conjectures that I mighjt try to prove. Thanks!!!

The solution for an n-gon is the same in principle as for a tetrahedron.
For all points P (whether 2D, or any higher dimension) the Fermat Point is given by:
Sum for all P of UP/LP is the zero vector.
UP is the vector (x-xP, y-yP etc.) for point P (with as many dimensions as needed for the space the figure is in) and LP is the scalar distance from Fermat Point to the point P.
However this simply leads to a numerical method for finding the numerical answer.  Is that any help to you?
Using the corresponding equality for the tetrahedron in my first posting leads directly to ‘opposite angles are equal’.  Split the equation so that A & B are on one side and C&D on the other, square each side (i.e. scalar product with itself), yields 2+2.cosAOB=2+2.cosCOD.  Of course the equation can be split two other ways so that all three pairs of these opposite angles are equal.  Hence equal solid angles.
Coming back to geometrical methods have you looked at ellipsoids constructed on opposite sides of the tetrahedron? e.g. AB and CD with A & B and C & D as foci?  With appropriate minor axes these just touch at the Fermat Point I think.  Not clear what the condition on the minor axes should be though.

First quick look at hexagons suggests the following (but I’ve not checked this in detail).
For the hexagon ABCDEF with Fermat Point O there are 10 such equations as:
cosAOB+cosAOC+cosBOC=cosDOE+cosDOF+cosEOF
One for each 3 and 3 partition of the corners.
There are 15 such equations as:
cosAOB=1+cosCOD+cosCOE+cosCOF+cosDOE+cosDOF+cosEOF
One for each 2 and 4 partition of the corners.
There may also be useful equations in the 1 and 5 partitions.  (Although having said that I’m not sure how useful any of the above are.)
The above results hold whether the figure is a hexagon, an octahedron or a simplex in 5-space.
The same sorts of equations - but many more of them - apply to octagons.

I dopn't have the computational methods for solving the above! Can you please do this and give me the result! really appreciate

How did you arrive at the fact that the Fermat point is the one st.t the sum of unit vectors from it to vertices is zero???

CAn you plz explain what you meant by constructing ellipsoids

Hi.
I don’t think there is much point in solving these equations with all the cos terms in.  However they might, or might not, be useful for deriving other relationships.
For example in the case of the tetrahedron there are so few angles that it is clear - angles subtended at O by opposite sides are equal.  Hence it can be shown that the solid angles subtended by all the faces are equal.  This is a nice result and doesn’t depend on the numerical or algebraic solution of the equation.
For the hexagon it is less clear.  Many of these angles are not independent (e.g. AOC=AOB+BOC).  It may be possible to select certain equations and expand some of the cos terms and obtain some nice relationships.  For the octahedron it may be useful to take pairs of opposite faces for the 3x3 partitions of the vertices (in which no angles overlap).
It’s like a  lot of maths - you have to play about with messy equations and hope that it simplifies or that you can see a pattern.
The ellipsoid is an ellipse rotated about its major axis so that it is circular in section.
If you construct an ellipse so that its foci are at A and B then the ellipsoid is the shell of revolution that you get by rotating the ellipse about the axis AB.  The length of the ellipse is the major axis, 2m, say and its width (diameter here) is 2n.  The distance between the foci, the length of AB, is  2d where d^2=m^2-n^2.
The interesting property of the ellipse in this context is that a line drawn from one foci to the perimeter and back to the other foci has a constant length for all points on the perimeter and is equal to 2m, the major axis.  The same relationship holds for the ellipsoid but now there is an extra dimension.
So, if you draw an ellipse with foci A & B through the Fermat Point O the length of the line AOB is 2m.  A line AXB to any point X inside the ellipsoid will have a shorter length.
Now suppose you have drawn another ellipsoid with foci C & D (the opposite side of the tetrahedron) through the Fermat Point.  The same relationship holds for the line COD: its length is constant for all points on the ellipsoid and it would be shorter for a line CXD if X were inside that ellipsoid.
Now consider the two ellipsoids, both passing through O.  If they overlap then there will be point X that is inside both ellipsoids.  The line AXB would be shorter than AOB, and CXD would be shorter than COD.  But by definition of the Fermat Point this cannot be: the Fermat Point IS the point at which the sums of the lengths is minimised.  Therefore there can be no such point X inside both ellipsoids.  Therefore the ellipsoids cannot overlap.
However we know they both pass through O so they must just touch there (have a common tangent plane).
We don’t know everything about these ellipsoids unless we know the position of the Fermat point.  (But either of the major or minor axes defines the other since we know the inter-focal lengths AB and CD.)  However it is an interesting relationship and would bear further investigation.  The joint normal to the two ellipsoids must intersect AB and CD and it might be interesting to find out where.
The sum of the Fermat distances, AO+BO+CO+DO, is the sum of the major axes of these ellipsoids.  So another way of looking at the Fermat problem on a tetrahedron is to find the minor axis lengths of ellipsoids with AB and CD as the pairs of foci which minimise the sum of the two major axis lengths.  Of course this may be no easier than the initial problem - but it is another way into it.

Anyways, I obvoisly don't expect you or me to solve these equations!!!! I'm just asking you , if possible, to find a possible approximate solution to the location of th Fermat point of a hexagon and octagon. Maybe you are aware of somebody else having done this. ANyways, I'll be obliged if you can provide this result so that I can make some conjectures, etc.!!!!

What about prrof for the throem you stated - sum of unit vector from FP(Fermat point - suggest you use this shortform too!!!) to the vertices is zero?!?!?!?!?!?

The proof of the condition for the Fermat Centre uses analytic geometry.
We need to minimise the sum of distances OP from unknown Fermat Point O to various vertices which I denote by P for a general term.  The result holds for any number of P and any dimensionality.  I shall assume 3D here.
Let O be (x,y,z) and points P be (xP,yP,zP) then by Pythagoras:
distance OP is LP=sqrt[(x-xP)^2+(y-yP)^2+(z-zP)^2].
Partially differentiate w.r.t. x:
dLP/dx=(x-xP)/LP
The total distance to be minimised is L=sum of all such LP so that
dL/dx=sum for each P of (x-xP)/LP.
The condition for a minimum is that dL/dx (and for other dimension coordinates dL/dy etc) is zero.
So for the Fermat point for each dimension such as x the sum of (x-xP)/LP for all P is zero.
This is written more concisely in vector terms where UP is the vector (x-xP,y-yP,z-zP) for as many dimensions as there are and the condition for the Fermat Point is that sum of UP/LP for all P is the zero vector (0,0,0).
This is 3 equations for 3 unknowns (if 3D) and can be solved in principle.

A straightforward numerical solution to this is an iterative method which is very easy to program on a computer or programmable calculator.
Step 1.   Make an initial guess at (x,y,z) at O.  (e.g. the average of all the P vertices).
Step 2.   Calculate all the distances LP=sqrt[sqrt[(x-xP)^2+(y-yP)^2+(z-zP)^2].
Step 3.   Calculate the next guess at O: x=[Sum of (xP/LP)]/[Sum of (1/LP)] and similarly for y and z.
Step 4.   Repeat from Step 2 until the changes in the values of (x,y,z) become negligible.
This is not the fastest solution method but it is the simplest.  If you have access to some form of programming the computation time will be negligible.

As I have said I’m afraid I don’t know of any geometric solutions to the problem.
Curiously there is a well-known mechanical solution for 2D.  If you draw the position of the points on a board and make holes at these points, tie the appropriate number of pieces of string together at one end (one for each vertex), pass each free end through one of the holes and tie an equal weight on to each end, then the with the board held horizontal (for convenience) the knot comes to rest at the Fermat Point of the vertices.  Of course this assumes frictionless contact of string and board.  This would work for 3D too if you position the holes for the vertices P in space.
It would be difficult in 4D!

Do you know the FP for a hexagon and octagon?

No.  I don’t know any geometric solutions for any figure other than the triangle.  (Napoleon’s construction, I think it’s called, using equilateral triangles constructed on the sides of the triangle in question.)
I only know how to find the numerical solution of the Fermat Point for any other figure (last posting gives the condition and one algorithm to solve it).
The relationships I have suggested in previous postings may give you the conditions necessary for the hexagon (or octagon) but I have not tried to do this myself.  It looks messy.

I've heard of Weitzfield's Algortihm. Is your the samew???

If you solve the equations above numerically, is it not possible to find the 'geometric' lovcation of FP of hexagon etc??? I am asking for a NUMERICAL SOLUTION NOT THE GEOMETRIC one - that's for me to find out!!!! :lol:

I’ve not heard of Weitzfield’s algorithm.  The one I described is a general method for finding the zero of a function.  It is usually the first one described in any text book on the numerical solution of equations and is often called Fixed-point iteration.
Basically, to solve x=g(x), iterate so that x[n+1]=g(x[n]) where x[n]is the nth approximation to x.
In the case of Fermat Point x, y and z are treated like this simultaneously as I described.
Yes, it will find the numerical coordinates of the Fermat Point of any hexagon (octagon, or any other figure).  This is the solution you want.

I do not have the available computational programs etc. to carry out this algorithm you have given for a hexagon and octagon. Since you work in a university/computer 'place', is it possible for you to carry out these numerical calculation for the hexagon and octagon and give me the result, please!!!!!  :)  :(  :lol:  :)  :(  If you can recommend some downloadable program or some which you can send me on a p2p server, I'd be grateful.... Thanks!!!!!!! :) I might be able to conjecture somethings on the basis of your response!

Regarding the statement you made regarding the general Fermat Problem, is it equivalent to the following. Can you plz explain where the result I have given below comes from?????
[quote]
Tedenat in 1810 found that for $n$ points the necessary condtion for the minimum solution point is that the sum of cosines of angles between any arbitrary line in the plane and the set of lines connectong the $n$ given points with the minimum point should be zero. Later,1837, Stiener proved that the necessary and sufficient condition for the minimum solution are that the sum of the cosines and sines of the above mentioned angles must be zero.
[/quote]

Also, if one can prove that it is impossible to construct a point( in our context a Fermat point for $n \geq 5$ , $n$ = no. of points) with the help of a ruler and compass only, is [b]it equivalent [/b]to saying that we CANNOT characterise the point, i.e. we CANNOT find the properties of a point, e.g. what angles the sides subtend at that point etc... Hopw you understand me!!! I can clarify if required! :)  :lol:

There are several things here.
a)	I could find the Fermat Point for specific hexagons or octagons but these are not general solutions.  They would be a specific set of (x,y) pairs for a specific set of 6 (x,y) pairs for each hexagon and 8 (x,y) pairs for each octagon.
b)	I don’t know what a p2p server is.  Would that be by direct email?  However I think I probably can’t send you a program unless you have access to Excel or some similar spreadsheet or to other computational software.  Let me know what you have access to.  (I mostly use Mathcad - and work from my own PC.)
c)	Tedenat & Steiner’s statements are equivalent to the one I gave you (in 2 dimensions), but I don’t know where they got it from I’m afraid.
In 2D I stated that (i) sum over all P of (x-xP)/LP=0, (ii) sum of (y-yP)/LP=0.
This is equivalent to (i) sum of all cos(aP)=0, (ii) sum of all sin(aP)=0, where aP is the angle OP makes with the horizontal.  (From definition of sin & cos.)
If a constant is added to each angle aP so that bP=aP+c  then the sine and cosine sums are still zero when applied to bP rather than aP.
Using sin and cos expansions for the sum of two angles gives for the left hand sides of each:
	(i)  cos(c).[sum of cos(aP)]-sin(c).[sum of sin(aP)],
	(ii) cos(c).[sum of sin(aP)]+sin(c).[sum of cos(aP)].
Each of the terms in square brackets is zero (from above) so each of these left hand sides is zero for all values of c.  Hence the angles of each OP can be measured relative to the same arbitrary line in the plane (at angle c to horizontal here).
d)	The question regarding ‘ruler and compass only’ constructions is not something I know much about.  I don’t think it limits ‘characterisation’ of the point by other means.  Basically the use of ‘ruler and compasses only’ limits you to a small set of geometrical transformations.  It depends on who is setting the rules on what is allowable.

(i) I too believed that any computer program would only give specific answers!!! Is there anyway I can find the FP of a hexagon and/or Octagon!!??? I haven't been able to think of one off hand, are you aware of the answer??? Can you possibly plase contact someone who might know!!! As I said before, this result might help me to conjecture and poroove somethings!!! Thanks!!!
(ii) P2P servers are often used in universtites etc. by students to transfer some programs / papers/journals that might otherwise be NOT available frelly for doenload, but they can be transferred securely among people who say know each other well. It's sort of like e-mail but more secure, reliable and able to take large bandwidth. A hot topic in Computer research these days are p2p servers and Algebaic coding! I have excel but not Mathcad or any other Computer Algebra System(CAS) Can you send me one ( a good CAS)???
(iii) Ok, so you took the arbitary line 'tedanet' mentioned to be the horizontal and he proved it for an arbitaray line!!! It's quite obvious that since your result and Steiner's results are the same, they won't help me much in my specific problem!
(iv)Actaully, in Galois theory in Mathematics, all possible constructions using ruler and compass (simple ones from which all the complicated ones can be built up!) are defines (obvoiusly) as algebraic transformations. Then, we associate a particular polynomial with a certain point/line /angle to be constructed and find whether that polynomial is reducible over $\mathbb{Q}$ ( a bit technical!). If it is, we can construct the point and vice-versa. e.g. it can be proved that we canNOT trisect an angle of $60^{\circ}$C alone with a compass and ruler!!! I read a result and proof in C.L. Bajaj's paper that we CANNOT contrsuct the FP for a polygon with $n \geq 5$ . I was worried whether that might contradict my 'possible' result and that is why I wanted to clarify it! I think its clear now that there is no clear connection between construction and characterization - of course , without proof ---- it will be exceedingly difficult to find this!!! :lol:

Hi,
I work for a university - but from home.  I could send you an Excel spreadsheet which would generate the Fermat Point solutions.  What version of Excel do you have, and does it have ‘Solver’ in its drop-down list of tools?  I will need an email address to send it to.

I have Excel 2000 ( Version 9.....) I have the 'Solver' Add-in in Tools. my e-mail is rushilgoel@hotmail.com . I'd be obliged if you can include some instructions for runninjg the program you will be sending. Thanks!!!! :)

If you can find the FP of hexagon/octagon etc. for the general case (i.e. the geometric characterization of the FP) , please inform!!!! :lol:

I seem to have some problems with opening and doenloading Excel files I recieve through Hotmail (I've had this problem before that is why I beleive your file may NOT be corrupted and the problem might be with my setup!)

Do you mind sending the revised versions of the FP programs for hexagon and ocatgon to rushilgoel@yahoo.com !!!!  :lol:  :)  :lol:  :)  :lol: Thanka  a lot!!!! really appreciate if you can include any instructions (even for the general case if possible!!!_

 :D  :D  :D

Please see the following document! Do you think the result proved here contradicts the solution I am trying to reach??? What other views do oyu have on it in context of our discussions???

What is 'the following document'?
Did the Excel program work?
I shall be away for a few days.
Good luck.

The document refers to the PDF I have posted!!!

I could not download the Excel program through hotmail!!!

Mind sending it to rushilgoel@yahoo.com PLEASE!!!!

Wow, John!!! Excellent program!!!
You just need to enter the points and it automatically generates the Fermat point as well as the lengths of the vectors AS Well as a approx figure. Great!!

Could you please explain how you made it and its actual working!!! Thanks!!
Just another thing. Is it possible to change the program so that it also measures the angles subtended by the sides at the Fermat point. I'd be greatly obliged if you can briefly state as to how to make this program for the general case!!! Thanks a lot!!!!!!! :lol:  :lol:  :lol:  :lol:  :lol:  :lol:
P.S> Do I need to enter an initial guess or just the points???

Thanks a lot!!!  :)  :)  :)

I’ve had a look at Bajaj’s paper and most of this is beyond me, I’m afraid, although it certainly looks very interesting.  I would take issue with his statement in the 1st paragraph of Section 5.  I think there IS a regular point within the tetrahedron corresponding to lines subtending equal angles at the Fermat point.  With the tetrahedron the angles are solid and subtended by the 4 faces rather than the simple angles subtended by the 3 sides of the triangle -  however the analogy is there.  The reasons for this I have given above and I have checked it  numerically for a number of cases.  (The engineers approach!)


Find all $ f: R\to R$ such that $ f(x\plus{}cos(2009y))\equal{}f(x)\plus{}2009cos(f(y))$
It is one of the most difficult problem in our test,and I couldn't solve,so I think I will fail  :(  :(
Best wishes to our site and all members

This is an interesting problem. Here is my solution.
We let $ c\equal{}2009$.
1) $ f(x\plus{}rq)\minus{}f(x)\equal{}r(f(x\plus{}q)\minus{}f(x))$ for any $ q$ and rational $ r$
Let $ r\equal{}\frac ab$, where a and b are coprimes. We can always find big integer $ n$ such that $ \cos(cy)\equal{}\frac q{bn}$ has some solution, say $ y_0$. Let $ d\equal{}c\cos(f(y_0))$.
We have $ f(x\plus{}rq)\minus{}f(x)\equal{}an*d$ and $ f(x\plus{}q)\minus{}f(x)\equal{}bn*d$. Obviously we have what we want.

2) $ f(x)$ is continuous at $ 0$.
We notice that $ |f(x\plus{}t)\minus{}f(x)|\leq c$ whenever $ |t|\leq1$. To prove the continuity, we need to show $ f(x_n)$ converges to $ f(0)$ for any given series $ x_n$ which converges to 0. For any $ x_n$, we can find integer $ y_n$ such that $ \frac12\leq |z_n\equal{}x_n2^{y_n}|<1$. Apparently $ y_n$ goes to infinity. Now we have
$ |f(x_n)\minus{}f(0)|\equal{}|f(2^{\minus{}y_n}z_n)\minus{}f(0)|\equal{}2^{\minus{}y_n}|f(z_n)\minus{}f(0)|\leq c2^{\minus{}y_n}$. Obviously the RHS converges to 0.

3) $ f(x)\equal{}ax\plus{}b$ for some constants a and b.
Let $ a\equal{}f(1)\minus{}f(0)\equal{}c\cos(f(0))$, $ b\equal{}f(0)$. We know $ f(x)\equal{}ax\plus{}b$ for all rational x, by 1). To prove it works for any $ x$. We fix $ x$ and construct series $ y_n$ such that $ y_n$ converges to 0 and $ r_n\equal{}\frac x{\cos(cy_n)}$ are all rationals. Now we have
$ f(x)\minus{}f(0)\equal{}f(r_n\cos(cy_n))\minus{}f(0)\equal{}r_n(f(\cos(cy_n))\minus{}f(0))\equal{}r_nc\cos(f(y_n))$. Obviously the RHS converges to $ xc\cos(f(0))$ by 2). We proved $ f(x)\equal{}ax\plus{}b$ for all $ x$.

4)
Now we have $ a\cos(cy)\equal{}c\cos(ay\plus{}b)$ for any $ y$. Let $ y\equal{}0$, we have $ a\equal{}c\cos(b)$.
Case 1) $ \cos(b)\equal{}0$. So $ a\equal{}0$, $ f(x)\equal{}b$, where $ \frac b{\pi/2}$ is any odd number.
Case 2) $ \cos(b)\neq0$. So $ \cos(b)\cos(cy)\equal{}\cos(ay\plus{}b)$. Since $ a\neq0$, RHS can achieve $ 1$. $ \cos(b)$ has to be either $ 1$ or $ \minus{}1$, i.e., $ b$ is a multiple of $ \pi$. So we have either $ f(x)\equal{}2009x\plus{}b$, where $ \frac b{\pi}$ is even; or $ f(x)\equal{}\minus{}2009x\plus{}b$, where $ \frac b{\pi}$ is odd.

[quote="xxp2000"]This is an interesting problem. Here is my solution.
We let $ c \equal{} 2009$.
1) $ f(x \plus{} rq) \minus{} f(x) \equal{} r(f(x \plus{} q) \minus{} f(x))$ for any $ q$ and rational $ r$
Let $ r \equal{} \frac ab$, where a and b are coprimes. We can always find big integer $ n$ such that $ \cos(cy) \equal{} \frac q{bn}$ has some solution, say $ y_0$. Let $ d \equal{} c\cos(f(y_0))$.
We have $ f(x \plus{} rq) \minus{} f(x) \equal{} an*d$ and $ f(x \plus{} q) \minus{} f(x) \equal{} bn*d$. Obviously we have what we want.

2) $ f(x)$ is continuous at $ 0$.
We notice that $ |f(x \plus{} t) \minus{} f(x)|\leq c$ whenever $ |t|\leq1$. To prove the continuity, we need to show $ f(x_n)$ converges to $ f(0)$ for any given series $ x_n$ which converges to 0. For any $ x_n$, we can find integer $ y_n$ such that $ \frac12\leq |z_n \equal{} x_n2^{y_n}| < 1$. Apparently $ y_n$ goes to infinity. Now we have
$ |f(x_n) \minus{} f(0)| \equal{} |f(2^{ \minus{} y_n}z_n) \minus{} f(0)| \equal{} 2^{ \minus{} y_n}|f(z_n) \minus{} f(0)|\leq c2^{ \minus{} y_n}$. Obviously the RHS converges to 0.

3) $ f(x) \equal{} ax \plus{} b$ for some constants a and b.
Let $ a \equal{} f(1) \minus{} f(0) \equal{} c\cos(f(0))$, $ b \equal{} f(0)$. We know $ f(x) \equal{} ax \plus{} b$ for all rational x, by 1). To prove it works for any $ x$. We fix $ x$ and construct series $ y_n$ such that $ y_n$ converges to 0 and $ r_n \equal{} \frac x{\cos(cy_n)}$ are all rationals. Now we have
$ f(x) \minus{} f(0) \equal{} f(r_n\cos(cy_n)) \minus{} f(0) \equal{} r_n(f(\cos(cy_n)) \minus{} f(0)) \equal{} r_nc\cos(f(y_n))$. Obviously the RHS converges to $ xc\cos(f(0))$ by 2). We proved $ f(x) \equal{} ax \plus{} b$ for all $ x$.

4)
Now we have $ a\cos(cy) \equal{} c\cos(ay \plus{} b)$ for any $ y$. Let $ y \equal{} 0$, we have $ a \equal{} c\cos(b)$.
Case 1) $ \cos(b) \equal{} 0$. So $ a \equal{} 0$, $ f(x) \equal{} b$, where $ \frac b{\pi/2}$ is any odd number.
Case 2) $ \cos(b)\neq0$. So $ \cos(b)\cos(cy) \equal{} \cos(ay \plus{} b)$. Since $ a\neq0$, RHS can achieve $ 1$. $ \cos(b)$ has to be either $ 1$ or $ \minus{} 1$, i.e., $ b$ is a multiple of $ \pi$. So we have either $ f(x) \equal{} 2009x \plus{} b$, where $ \frac b{\pi}$ is even; or $ f(x) \equal{} \minus{} 2009x \plus{} b$, where $ \frac b{\pi}$ is odd.[/quote]

Too long!!

To prove $ f(x) \equal{} ax \plus{} b$,we just need to prove$ g(x) \equal{} f(x) \minus{} f(0)$ is a additive function on R and $ f(x)$is continuous at $ 0$

Let$ x \equal{} 0$,we have: $ f(cos2009y) \equal{} 2009cosf(y) \plus{} f(0) (1)$
and with $ g(x) \equal{} f(x) \minus{} f(0)$ we have $ g(x \plus{} cos2009y) \equal{} g(x) \plus{} g(cos2009y) \forall x,y \in R$
And $ g(x \plus{} y) \equal{} g(x) \plus{} g(y) \forall x \in R,y \in [ \minus{} 1,1]$
Since then,very easily,we have $ g(x \plus{} y) \equal{} g(x) \plus{} g(y) \forall x,y \in R$
From $ (1)$,we have $ g(x)$ is a bounded function on $ [ \minus{} 1,1]$
So,$ g(x)$ is continuous at $ 0$

Hence,$ g(x) \equal{} ax ,\forall x \in R$
And $ f(x) \equal{} ax \plus{} b ,\forall x \in R$

Thank you, ducpbc. I do not like my old proof either. I would always be more than happy to see an efficient proof.


calculus
integration
function
derivative
calculus computations

Suppose we have $ g(x)\equal{}\int_0^1\frac{e^{\minus{}x^2(1\plus{}t^2)}}{1\plus{}t^2}\,dt.$

What are the conditions so that we can differentiate $ g$?

Hi Null. Is this any good to you?: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

They're going to parallel conditions for switching the order of integration, because this can be rewritten in that form. Generalize this to $ F(x)\equal{}\int_I f(t,x)\,dt$; we ask if $ F'(x)\equal{}\int_I \frac{\partial f}{\partial x}(t,x)\,dt$. Assume that the defining integral of $ F$ converges, so we're talking about a function that actually exists.

One sufficient condition: the derivative inside is continuous in both $ t$ and $ x$, and $ I$ is compact. That's equivalent to the version of Fubini's theorem you see in a first multivariable class, or to the integral of a uniformly convergent limit being the limit of the integral.

Stronger sufficient conditions: either $ \frac{\partial f}{\partial x}(t,x)$ is positive everywhere, or $ \int_I \frac{\partial f}{\partial x}(t,x)\,dt$ converges absolutely for all $ t$.

To justify any of these, write $ f(t,y)\equal{}f(t,x_0)\plus{}\int_{x_0}^y \frac{\partial f}{\partial x}(t,x)\,dx$. This gives
$ F(y)\equal{}\int_I f(t,x_0)\,dt\plus{}\int_I\int_{x_0}^y \frac{\partial f}{\partial x}(t,x)\,dx\,dt$
$ F(y)\equal{}F(x_0)\plus{}\int_{x_0}^y\int_I \frac{\partial f}{\partial x}(t,x)\,dt\,dx$, (Justify interchange appropriately)
Then, by the FTC, we differentiate the integral in $ x$:
$ F'(x)\equal{}\int_I \frac{\partial f}{\partial x}(t,x)\,dt$


logarithms
inequalities
inequalities proposed

We write ${\log_{ab}^{2}{c}} =\left (\frac{\log {c}}{\log {a} + \log {b}} \right )^{2}$

Now from Cauchy and Nesbitt inequalities

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

 :)

[quote="Ramanujan"][quote="socrates"][quote="darkdevil"]$a,b,c\geq 0$, $a+b+c=1$ prove that
$\sum \frac{\log_{ab}^{2}{c}}{a+b}\geq \frac 98$[/quote]

CS+nesbitt!

 :lol:[/quote]

ooops i was writing the solution and didn saw this  :roll:  :blush:   :)[/quote]

no problem....... :lol:


algebra
polynomial
Vieta
algebra proposed

If alpha,beta and gama are the roots of the eqn.

(x-a)(x-b)(x-c)+1=0

then show that (alpha-x)(beta-x)(gama-x)+1=0
[/code][/url]

Regardless of what $ \alpha, \beta, \gamma$ are, $ (\alpha \minus{} x)(\beta \minus{} x)(\gamma \minus{} x) \plus{} 1$ is a cubic polynomial in $ x$, so it is surely not identically $ 0$.  Did you mean something else? Is $ x$ not meant to be a variable in the last equation?  If so, what is it?

I missed something in my question.
I am retyping it.
If alpha,beta and gamma are the roots of the eqn.
then prove that a,b and c are the roots of

(alpha-x)(beta-x)(gamma-x)+1 = 0

Please use latex. It's difficult to understand this way.
What I interpret your problem:

[quote]If $ \alpha, \beta, \gamma$ are the roots of the equation $ (x\minus{}a)(x\minus{}b)(x\minus{}c)\plus{}1\equal{}0$, then prove that:
$ a,b,c$ are the roots of $ (\alpha\minus{}x)(\beta \minus{}x)(\gamma \minus{}x)\plus{}1\equal{}0$[/quote]

In that case

[hide="Hint"]Use Vieta's formulla[/hide]
[hide="Solution"]The first eqn. can be rewritten as $ x^3\minus{}(a\plus{}b\plus{}c)x^2\plus{}x(ab\plus{}bc\plus{}ca)\minus{}(abc\minus{}1)\equal{}0$
By vieta $ \alpha\plus{}\beta\plus{}\gamma\equal{}a\plus{}b\plus{}c,\alpha \beta\plus{}\beta\gamma\plus{}\gamma\alpha\equal{}ab\plus{}bc\plus{}ca,\alpha\beta\gamma\equal{}abc\minus{}1\implies abc\equal{}\alpha\beta\gamma \plus{}1$
Again by vieta: $ a,b,c$ are the roots of $ x^3\minus{}(a\plus{}b\plus{}c)x^2\plus{}x(ab\plus{}bc\plus{}ca)\minus{}abc\equal{}0$
(Alternately it can be said $ a,b,c$ are roots of $ (x\minus{}a)(x\minus{}b)(x\minus{}c)\equal{}0$ and then expanding, it's actually the same thing.)
But $ a\plus{}b\plus{}c\equal{}\alpha\plus{}\beta\plus{}\gamma,ab\plus{}bc\plus{}ca\equal{}\alpha \beta\plus{}\beta\gamma\plus{}\gamma\alpha,abc\equal{}\alpha\beta\gamma \plus{}1$
Replacing in the last eqn., we have that $ a,b,c$ are roots of:
$ x^3\minus{}(\alpha\plus{}\beta\plus{}\gamma)x^2\plus{}x(\alpha \beta\plus{}\beta\gamma\plus{}\gamma\alpha)\minus{}(\alpha\beta\gamma \plus{}1)\equal{}0\iff \left[x^3\minus{}(\alpha\plus{}\beta\plus{}\gamma)x^2\plus{}x(\alpha \beta\plus{}\beta\gamma\plus{}\gamma\alpha)\minus{}\alpha\beta\gamma\right] \minus{}1\equal{}0$
$ \implies (x\minus{}\alpha)(x\minus{}\beta)(x\minus{}\gamma)\minus{}1\equal{}0\iff (\alpha\minus{}x)(\beta\minus{}x)(\gamma\minus{}x)\plus{}1\equal{}0$
QED[/hide]
If I have interpreted it correctly, it's not in right forum either.


function
trigonometry
algebra unsolved
algebra

[color=blue]Find all function $f: \mathbb{R}\rightarrow\mathbb{R}$ satisfying
$f(2x)=f(x)\cos x$[/color]

It have only one solution for f(0)=a if and only if f(x) continuosly in x=0. This solution is
$f(x)=\frac{a\sin x}{x}.$

Let $f(x)=a(y)\frac{\sin x}{x}, y=lnx/ln2$. We have a(y+1)=a(y). Therefore $f(x)=a(lnx/ln 2)\frac{\sin x}{x}.$


What does something like $a^{-1} \equiv k(\mod p)$ mean?

it is the so called Multiplicative Inverse in the ring of modulo p.
u can find k: $a^{-1} \equiv k \pmod p$ iff $\gcd(p,a)=1$
it can be found by the extended euclid algorithm

[quote="inteluser"]I'm not sure it means much of anything, but if I had to guess, $ka\equiv 1 \ (\text{mod }p)$ when $a$ is non-zero mod $p$?[/quote]

That's correct. :) Some people even write $\frac{1}{a} \equiv k\ \textrm{mod}\ p...$


[quote="GameBot"]The length and width of a given rectangle are 3 cm and 4 cm. respectively. A square is formed by using a diagonal of the rectangle as a side. Then, a circle is circumscribed about the square. How many square cm are in the area of the circle? Express your answer as a common fraction in terms of $ \pi$.[/quote]

     By the Pythagorean Theorem, the diagonal of the rectangle has length five, so that's the length of a side of the square. The length of the diagonal of the square, which is also the circle's diameter, is $ 5\sqrt{2}$. Thus the area of the circle is $ \frac{25\pi}{2}$.


analytic geometry
graphing lines
slope
quadratics
trigonometry
algebra

1. Let $x_1$ and $x_2$ be such that $x_1\not= x_2$ and $3x_i^{2}-hx_i=b$. $(i=1,2)$. Then $x_1+x_2$ equals...

2. The equatoins of L1 and L2 are y=mx and y=nx, respectfully. Suppose L1 makes twice as large an angle with the horizontal (measured counterclockwise from the positive x-axis) as does L2, and the L1 has 4 times the slope of L2. If L1 is not horizontal, then m*n is...

3. How many positive integers are factors of $70^{5}$?

[hide="1"]This is really asking for the sum of the two roots of that equation - because $x_1$ and $x_2$ are not equal.  This is always $-b/a$ for a quadratic equation $ax^2+bx+c$.  So the sum of $x_1$ and $x_2$ is $h/3$.[/hide]

[quote="xxreddevilzxx"][hide="3"]
$70^5 = 2^5 \cdot 5^5 \cdot 7^5$

$(5+1)(5+1)(5+1) = 216$
[/hide][/quote]

One question, when you got that, why did you do $(5+1)^3$?

[quote="JesusFreak197"]The number of distinct integer factors of a number is found by prime factorizing in that way. Say a number has three prime factors, $a$, $b$, and $c$, so the number prime factorizes to $n=a^xb^yc^z$. The number of distinct integer factors of that number is $(x+1)(y+1)(z+1)$, because that is all possible combinations of each prime factor, from $a^0$ to $a^x$, etc.[/quote]

Thank you Jesusfreak :)

[hide="2"]You can begin by labeling the angle between the x-axis and $L_2$ as z.
Then the angle between the x-axis and $L_1$ is 2z.
If you draw a vertical line from (1,0) to $L_2$, you can solve for the angle z with trigonometry by using the slopes.
$4\tan{z}=\tan{2z}$
$4\frac{\sin{z}}{\cos{z}}=\frac{\sin{2z}}{\cos{2z}}$
Using double angle formulas:
$4\frac{\sin{z}}{\cos{z}}=\frac{2\sin{z}\cos{z}}{\sin^2{z}-\cos^2{z}}$
$2\sin^2{z}=3\cos^2{z}$
Taking the square root of both sides:
$\sqrt{2}\sin{z}=\sqrt{3}\cos{z}$
$\tan{z}=\frac{\sqrt{6}}{2}$
Looking back at the graph, you can see that the slope of $L_2$ must be $\frac{\sqrt{6}}{2}$ and $L_1$'s slope is four times that or $2\sqrt{6}$
mn then equals $6$[/hide]


on a n*n table we place natural numbers so that any two neighboring numbers difer at least 1 prove that there is a number that appears on a table at least n time

From any cage to other length less than n. Therefore if difference less than n. If all numbers integer, then $n^2$ numbers have $\le n$ distinct.

[quote="Rust"]From any cage to other length less than n. Therefore if difference less than. If all numbers integer, then $n^2$ numbers have $\le n$ distinct.[/quote]
  are you sure that it is right

I don't understand this problem, nor Rusts' solution. Clearly, it is possible to fill $n^2$ cells with naturals so that any two neighbouring cells will differ by at least $1$ - and no number will appear even twice. Just use the numbers from $1$ to $n^2$ and you're there.

What mean "difer at least 1"?
Let x(i,j) is number in cell (i,j).
1.If it is mean $|x(i_1,j_1)-x(i_2,j_2)|\ge 1$, when $|i_1-i_2|\le 1, \ |j_1-j_2|\le 1$, then it is not true. For example $x(i,j)=i+(j-1)*n$.
2. If it mean $|x(i_1,j_1)-x(i_2,j_2|\le 1$, when $|i_1-i_2|\le 1, \ |j_1-j_2|\le 1$, then it is true for integer x(i,j), and I prove it.

yes .this mean
[quote="Rust"]Let x(i,j) is number in cell (i,j).
2. If it mean $|x(i_1,j_1)-x(i_2,j_2|\le 1$, when $|i_1-i_2|\le 1, \ |j_1-j_2|\le 1$, then it is true for integer x(i,j), and I prove it.[/quote]
         but it can have more than n distinct integer numbers ( I can prove it)
   can you post your complete solution,please


probability
conditional probability
combinatorics proposed
combinatorics

A box contains n red balls and n black balls.
We remove two balls at a time repeatedly and :

a) Discard them if they are of the same color ;

or

b) Discard the black one and return to the box the red  one if they are different colors.

What's the probability that this process will terminate with one red ball in the box?

[hide]Well, the easy case: if $n$ is even, the probability is 0 as the number of red balls can only change by 0 or 2 on any given draw.

Actually, for that matter, the odd case is just as easy: we obviously can't be stuck with more than 1 red ball (as we would eventually draw two reds if they existed), and since red balls always change parity by 2, we end up with one red ball with probability 1 when $n$ is odd.  So the desired probability is just $n \mod 2$.[/hide]

$n$ is odd or $n$ is even ?
Think once again.
Only one solution to the problem  is asked.
One doesn't know wether n is odd or even and every assumption on that condition must be equally valid for both cases becouse nothing else was mentioned.
$Conditional$ $probability$ !

Your response doesn't make any sense to me.  If $n$ is even, the probability is 0.  If $n$ is odd, the probability is 1.  That is the answer.  If you prefer a closed-form expression, I could write this as $\frac12 - \frac12(-1)^{n}$, but the meaning is identical (so long as it's acceptable to assume $n$ is an integer, of course).

This is sort of a trick question.
 The point is in info,or better say lack of it,as regards knowledge of $n$ (odd or not).
The only knowledge given in the text about number of red and black balls is equal number of these two.
Pondering over it,one may assume ,in the best case, equal probability of n being odd or even ($p1=p2=0.5$).


4^32 x 1/625 ^15 x 25^60
How many zeroes in answer?

x=times using a labtop...

oh oopsies :blush:  625 is 5^4 not 5^3 :wallbash_red:

The answer is 60, I'm pretty sure.  Are you sure you posted the problem right?

are you? but i'm gonna have to side with him on this one the answer has to be 60

[quote="Ihatepie"][quote="humblemastermind"][hide="solution"] OK, you can express this as $2^{64}\cdot 5^{120}\cdot \frac{1}{5^{60}}= 2^{64}\cdot 5^{60}$. Thus, we have 60 pairs of 5 and 2, so there will be $\fbox{60}$ 0s. [/hide][/quote]

this is wrong[/quote]
No, it's not. His reasoning is correct, and so is his answer.

[quote="i_like_pie"]$\frac{1}{625^{15}}=\frac{1}{\left(5^{4}\right)^{15}}=\frac{1}{5^{4\cdot15}}=\frac{1}{5^{60}}$[/quote]

sorry messed up its entire thing with 15 as exponent

(1/625)^15

and anyways wouldn't it be 61 zeroes because it is 2^64*5^60=2^60*2^4*5^60=2^60*16*5^60=10^60*16 which has 61 zeroes?

Well, if the whole fraction
$\left(\frac{1}{625}\right)^{1}5$
is exponentiated, we know by a rule of exponents that this is equal to
$\frac{(1^{15})}{(625^{15})}$, which is just
$\frac{1}{625^{15}}$

[quote="Ihatepie"]wouldn't it be 61 zeroes because it is $10^{60}\cdot16$ which has 61 zeroes?[/quote]
$10^{60}$ has 60 zeroes. Multiplying by 16 adds another [b]digit[/b], not another zero.


analytic geometry
graphing lines
slope
calculus
calculus computations

consider the curve y^3+(3x^2)y-9y=-5
Find the rate of change with respect to x of the slope of the curve at the point (1,1).

Let $ F(x,y)\equal{}y^3\plus{}3x^2y\minus{}9y\plus{}5$, so the equation is $ F\equal{}0$ The slope is $ \frac{dy}{dx}\equal{}\minus{}\frac{F_x}{F_y}$ (implicit differentiation). You are asked to differentiate this expression again, wrt $ x$. Just use implicit diff. again.


It was down for a couple of minutes, just a couple of updates.


function
algebra
functional equation
algebra unsolved

Find all function for every x,y satisfy that

                 f(f(x)^2+f(y))=x.f(x)+y

[quote="mathmaniack"]Find all function for every x,y satisfy that

                 $ f(f(x)^2\plus{}f(y))\equal{}xf(x)\plus{}y$[/quote]

Find all $ f: \mathbb{R} \to \mathbb{R}$ such that $ f(f^2(x)\plus{}f(y))\equal{}xf(x)\plus{}y\ \forall x,y \in \mathbb{R}$

Let $ P(x,y)$ the assertion $ f(f^2(x)\plus{}f(y))\equal{}xf(x)\plus{}y\ \forall x,y \in \mathbb{R}$

$ P(0,y) \iff f(f^2(0)\plus{}f(y))\equal{}y\ \forall y \in \mathbb{R}$

So $ f(x)$ is surjective. Therefore $ \exists u \in \mathbb{R}$ such that $ f(u)\equal{}0$

Moreover, if $ f(a)\equal{}f(b)$ then $ f(f^2(0)\plus{}f(a))\equal{}f(f^2(0)\plus{}f(b))$ and then $ a\equal{}b$.

So $ f(x)$ is a bijection of $ \mathbb{R}$

$ P(u,u) \iff f(0)\equal{}u$

$ P(0,u) \iff f(f^2(0))\equal{}u \iff f(u^2)\equal{}u$ and then $ u^2\equal{}0 \iff u\equal{}0$

So $ f(0)\equal{}0$ and $ f(f(x))\equal{}x\ \forall x \in \mathbb{R}$

$ P(f(x),y) \iff f(f^2(f(x))\plus{}f(y))\equal{}f(x^2\plus{}f(y))\equal{}xf(f(x)) \plus{} y \equal{}x^2\plus{}y$

$ P(f(x),0) \iff f(x^2)\equal{}x^2 \qquad (*)$

So $ f(x^2\plus{}f(y))\equal{}f(x^2)\plus{}f(f(y))\ x,y \in \mathbb{R}\qquad (**)$ (Cauchy functional equation)

The line $ (**)$ is Cauchy functional equation

Since $ f(f(x))\equal{}x$ then $ f(x)$ is monotonous and then $ f(x)\equal{}cx$ is the only solution

Using the $ (*)$ we have that $ c^2x^2\equal{}x^2 \iff c\equal{} \pm 1$

So we have

$ f(x)\equal{}x\ \forall x\in \mathbb{R}$

$ f(x)\equal{}\minus{}x\ \forall x\in \mathbb{R}$

and these are indeed solution of the equation


[hide]

$\displaystyle A_k=\frac{k(k-1)}2\cos\frac{k(k-1)\pi}2 = (\sum_{i}^{k-1}i)(\cos\frac{k(k-1)\pi}2)$
the sign in front of the calculated term is positive if $4|k$ or $4|k+1$

looking at the first four terms, we get
$-\sum_{i}^{18}i+\sum_{i}^{19}i+\sum_{i}^{20}i-\sum_{i}^{21}i$
which is equal to $19-21=-2$
this kind of telescoping goes on for 19 more reiterations, giving us
$|-2(20)|=\boxed{40}$ as the answer..(right?)

[/hide]

[hide="Solution"]After calculating the first 4 terms, from $A_{19}$ to $A_{22}$ it becomes apparent that each set of four numbers will yield -2. (The cosines switch the positive/negative one after the other and the $\frac{k(k-1)}{2}$ increases by k from each odd term to even term. Making the net $k-(k+2)=-2$.) Since there are $\frac{98-18}{4}=20$ sets of 4, then $|20 \times -2|=|-40|=40$.[/hide]

[hide="Answer"]We have triangular numbers, negative if odd and positive if even. Notice that the sum of every $4$ terms, beginning with the $3^{rd}$, is $-2$. Therefore, since we have $20$ sets of $4$ terms, our answer is $|-2(20)|=\boxed{40}$.[/hide]

p9: Note that we can always reduce cosine $\pmod{2\pi}$. Also, $k(k-1)$ is always even, so the cosine can only reduce to $\cos 0, \cos \pi$. This gives a result of $\frac{-18 \cdot 19 + 19 \cdot 20 + 20 \cdot 21 - 21 \cdot 22 - 22 \cdot 23 + \ldots - 97 \cdot 98}{2}$, equal to $\frac{19 \cdot 2 - 21 \cdot 2 + 23 \cdot 2 \ldots - 97 \cdot 2}{2} = -\frac{80}{2} = -40 \implies \boxed{40}$.

[quote=OlympusHero]Also, $\frac{k(k-1)}{2}$ is always even[/quote]
note that $k(k-1)$ is always even, not $\frac{k(k-1)}{2}$.

Now, notice that you say that the result is $-18 \cdot 19 + 19 \cdot 20 + 20 \cdot 21 - 21 \cdot 22 - 22 \cdot 23 + \ldots - 97 \cdot 98$.
This is almost correct. It is this divided by 2.

Finally, I'm not sure how you got $19 \cdot 2 - 21 \cdot 2 + 23 \cdot 2 \ldots - 97 \cdot 2 = -160$. First notice that there are 40 terms here. If you pair them up, we see that we have $-4-4-4-4\cdots-4$ $\frac{40}{2}$ times, which is -80. Since we forgot to divide by 2 in the first place, we have the answer of $|\frac{-80}{2}|=\boxed{040}$


There are $ n$ markers , each with one side white and the other side black, aligned in a row so that their white sides are all up in the beginning. In each step, if possible, we choose a marker with the white side up(but not one of the outermost markers), remove it, and reverse (turn the other face up) the neighbouring two markers (to the left and the right) . Prove that one can acheive the state with only two markers remaining if and only if $ n\minus{}1$ is not divisible by $ 3$.

Please help!

ISL? Can you give me the link please? you mean imo shortlist?

http://www.mathlinks.ro/Forum/viewtopic.php?t=90046

  darij


calculus
integration
logarithms
trigonometry
calculus computations

\[ \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 \plus{} \tan x} \right)dx}
\]

You should post it in Calculus Computations and Tutorials. :)

hey isnt that just

((sec(x))^2)/(1 + tanx) ?

from pi/4 to 0 of course

I think you did the derivative.  :D
you should get  u=1+tan
ln u/(sec(x)^2), and this is hard to simplify.

But then note that sec^2=1+tan^2=(1+tan)^2-2tan=u^2-2u+2, and thus the integral is
$ \displaystyle \int_1^2 \frac {\ln{u}}{u^2 \minus{} 2u \plus{} 2} \ du$. Integrate that and you're done.

EDIT: Agh! Forgot the du!

Someone please check me.
[b]Can a mod move this to Calc Tutorials?[/b]

Actually, does anybody know how to evaluate MY integral?  :blush:  I ran it through with Maple, and got this:

$ \displaystyle \int_1^2 \frac {\ln{u}}{u^2 \minus{} 2u \plus{} 2} \ du\equal{}\frac{\pi\ln(2)}{4}\minus{}K$, where K is Catalan's constant.

$ I \equal{} \int_0^{\pi/4}\ln(1 \plus{} \tan x)\,dx \equal{} \int_0^{\pi/4}\ln\left[\frac{\sqrt{2}\sin\left(x \plus{} \frac{\pi}{4}\right)}{\cos x}\right]\,dx$

$ \equal{} \int_0^{\pi/4}\ln(\sqrt{2})\,dx \plus{} \int_0^{\pi/4}\ln\left[\sin\left(x \plus{} \frac{\pi}{4}\right)\right]\,dx \minus{} \int_0^{\pi/4}\ln\cos x \,dx$

$ \equal{} \frac{\pi \ln 2}{8} \plus{} \int_0^{\pi/4}\ln \cos t\,dt \minus{} \int_0^{\pi/4}\ln\cos x \,dx \equal{} \frac{\pi \ln 2}{8}$,

where the substitution $ x \equal{} \frac{\pi}{4} \minus{} t$ was made.

This forum should be locked.

Japanese Communities Moderator

kunny


http://www.mathlinks.ro/Forum/viewtopic.php?t=83640 , $ n\equal{}3$


A set S is defined as follows:

1. 2 is in S
2. for every number x in S, 3x and x+7 are also in S

What is the lowest number above 2004 that is not in S

[hide]
Consider the numbers that are in the set because of the $3x$ rule and are less than 2004.
They are
2,6,18,54,162,486,1458
Thus you must be able to get to one of these numbers by subtracting 7 multiple times.
For 2005, 2005-486 is divisible by 7. Thus you can get to 486 by subtracting 7 multiple times.
For 2006, 2006-18 is divisible by 7.
For 2007, 2007-54 is divisible by 7.
For 2008, 2008-6 is divisible by 7
For 2009, you can't subtract any of the above numbers to get a multiple of 7.
Thus it is not in S[/hide]

[hide]
2005-Subtract 7 and divide by 3...
2006-Subtract 7, then again, divide by 3...
2007-Divide by 3...
2008-Subtract 7, divide by 3...
2009-It isn't divisible by 3, and subtracting 7 won't get to a multiple of 3 or to 2.
So, $\boxed{2009}$ is the answer.[/hide]

Pascal- according to the rules of the set, you can add 7 any number of times, so that solution doesn't work (subtracting 7 from 2009 gives 2002; subtracting 7 again gives 1995, which is a multiple of 3).

The answer key says 2009, but there is no solution

[hide]If I remember correctly, the key to that problem is that you can never get a multiple of 7, so you are just looking for the smallest multiple of 7 larger than 2002, and since 2002 is divisible by 7, 2009 is the smallest that you can't make that fits the rules.

I looked at the problem by noting all members of S had to be of the form 3^x + 7y.  Since 7y is always divisible by 7, when you add 3^x(which is never divisible by 7) you always get a number not divisible by 7.

Hope that makes sense
[/hide]

Actually frost they should be in the form $2\cdot3^x+7y$ (doesnt make a difference though). But good explanation :thumbup:

Good catch Biffanddoc.  Say hi at Nats, I'm the NH coach.


Let $ a,b$ and $ c$ in $ \mathbb{R}$ such that:
(for all $ x$ in $ \mathbb{R}$)   $ |ax^2 \plus{} bx \plus{} c|\leq 1$
1)Prove that $ |c|\leq 1$
2)Prove that  $ |a \plus{} c|\leq 1$
3)Prove that $ a^2 \plus{} b^2 \plus{} c^2\leq 5$.

[color=red]Edited by: N.T.TUAN[/color]

How can the condition be satisfied? $ ax^{2}\plus{}bx\plus{}c$ is a quadratic and so it can take values arbitrarily far away from 0 depending on the sign of $ a$.

I guess that the statement should be : $ \forall x \in [\minus{}1,1],\  |ax^2\plus{}bx\plus{}c|\leq 1$ ;  $ x\equal{}0 \Longrightarrow |c|\leq 1$
$ x\equal{}1\Longrightarrow \minus{}1 \leq a\plus{}b\plus{}c \leq 1 \ \ ,\ \ x\equal{}\minus{}1\Longrightarrow \minus{}1 \leq a\minus{}b\plus{}c \leq 1$ . You add and then $ \minus{}1 \leq a\plus{}c \leq 1$
$ (\minus{}1\leq a\plus{}c \leq 1)\& (\minus{}1\leq c\leq 1)\Longrightarrow \minus{}2\leq a \leq 2\Longrightarrow (i)\  \minus{}4ac \leq 8$
$ \minus{}1 \leq a\plus{}b\plus{}c \leq 1 \Longrightarrow  (ii)\ a^2\plus{}b^2\plus{}c^2 \plus{}2ab\plus{}2bc\plus{}2ac \leq 1$ ; $ \minus{}1 \leq a\minus{}b\plus{}c \leq 1 \Longrightarrow  (iii)\ a^2\plus{}b^2\plus{}c^2 \minus{}2ab\minus{}2bc\plus{}2ac \leq 1$
You add   $ (i),(ii),(iii)$ and finaly you obtain $ a^2\plus{}b^2\plus{}c^2\leq 5$
 :cool:


analytic geometry
graphing lines
slope
probability and stats

For families who live in apartments the correlation between the family's income and the amount of rent they pay is r=0.6.  Explain why on average, the families spend 60% of their income on rent.

I mean seriously... how is it possible?

Most likely, it isn't. The correlation could be 1, but the slope might be zero. Correlation is not equal to slope.

Consider positive random variables I, R, S, where I is monthly income, R is monthly rent, S is the remainder. We have I = S+R.

$ \text{cov}(I,R) \equal{} \text{cov}(R \plus{} S,S) \equal{} \text{var}(R) \plus{} \text{cov}(R,S)$

Equivalently: $ \rho _{(I,R)} \sigma _I \sigma _R \equal{} \sigma _R ^2 \plus{} \sigma _R \sigma _S \rho _{(R,S)}$

$ \implies \boxed{\frac {3}{5} \sigma _I \equal{} \sigma _R \plus{} \sigma _S \rho _{(R,S)}}$.

That's the best you can do with the given information. I have no idea what they want you do after that -- there's no way you can get to the desired result, namely $ \frac {3}{5} E[I] \equal{} E[R]$, without imposing additional assumptions.

For example, if you assume:
:: $ E \propto \sigma$
:: $ \rho _{(R,S)} \equal{} 0$
Then you get the desired result. Things to think about:
1. What conditions are implied by those two assumptions
2. Are these assumptions valid?


calculus
integration
trigonometry
LaTeX
analytic geometry
geometry
rectangle

Find the volume bounded by cylinder $ x^2 \plus{} y^2 \equal{} 9$ and the upper half of the cylinder $ y^2 \plus{} z^2 \equal{} 9$.

Where I'm at:
I've arrived at the following integral for an answer but don't see how to solve it.
$ \int_{0}^{2\pi}{\int_0^{3}}\sqrt {9 \minus{} r^2sin^2{\theta}}dr d\theta$
Any ideas? I've been working with trig substitutions from the above integral. All help is greatly appreciated.

Well, for one thing, you're missing the Jacobian of the change of variables to polar coordinates. This should be
\[ \int_0^{2\pi}\int_0^3\sqrt {9 - r^2\sin^2\theta}\,r\,dr\,d\theta\]
($ \text{\LaTeX}$ tip: check the code of what I've written, both for the proper use of \sin and for the value of the spacer code \,)

Edit:

Not to mention that polar coordinates is the hard way anyway. Stick to rectangular coordinates, but do it in the smart direction:

$ \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\sqrt{9-y^2}\,dx\,dy=\int_{-3}^32(9-y^2)\,dy=72.$

Plausibility check: this fits inside a $ 6\times 6\times 3$ box. The box has volume $ 108.$ The volume just computed is $ \frac23$ of the volume of the box, which seems plausible. For an estimate on the other side, this intersection of cylinders is bigger than a hemisphere. The hemisphere has volume $ 18\pi,$ which is between $ 56$ and $ 57.$ Our answer of $ 72$ is still plausible.

When we cut the solid by a plane that is parallel to $ xy$ plane hight of $ z$, the cross section is a square and it's side length is $ 2\sqrt{9\minus{}z^2}$. So, finding the volume by disk method :

$ \int_{0}^{3}\left(2\sqrt{9\minus{}z^2}\right)^2dz\equal{}72$.

J.Y.Choi:

Wrong axis; the section you get slicing the way you said does not give squares.

To do what you want, you have to slice parallel to the $ xz$ plane, and you get rectangles instead of squares - in fact, you would get essentially the same thing as my rectangular coordinates integral above.

Sorry, I had a mistake. I wanted to say slicing parallel to $ zx$ plane. If we consider the volume bounded by two cylinders $ x^2\plus{}y^2\equal{}9$ and $ y^2\plus{}z^2\equal{}9$, each cross sections are 'square'. The volume we want to evaluate is half of this solid, given by the disk method I mentioned above:

$ \frac{1}{2}\int_{\minus{}3}^{3}\left(2\sqrt{9\minus{}y^2}\right)^2dy\equal{}72$.


How many distinct arrangments are there in SYNTAXERROR?

For those of you that do not know, distinct is another word for different, but it is commonly used in many MATHCOUNTS tests.

Oooh, brownie points if someone can tell me why MATHCOUNTS is in all caps?

The are 11 letters and the R appears 3 times.  Therefore, there are 11!/3! or 6,652,800 arrangements.


I don't quite understand the proof for the following question in my book .

[i]Let $ D$ be a denumerable set,and$ f: D\rightarrow S$ a surjective mapping .Then $ S$ is denumerable or finite.[/i]

[i]Proof.[/i] For each $ y\in S$ ,there exist an element $ x_{y}\in D$ such that $ f(x_{y})\equal{}y$ since $ f$ is surjective (This part is ok with me) . The association $ y|\rightarrow x_{y}$ is an injective mapping of $ S$ into $ D$ .(Why?)

Let $ g(y)\equal{}x_{y}$ .The image of $ g$ is a subset of $ D$ and is denumerable or finite.Since $ g$ is a bijection between $ S$ and its image ,it follows that $ S$ is denumerable or finite .(This part will be clear if i can understand the "Why?" part above.)

Why can we say that mapping of $ S$ into $ D$ is injective ?

Q2 ) What is the different between [i]mapping into[/i] and [i]mapping onto[/i] ?

Thanks  :)

[quote="shyong"]The association $ y|\rightarrow x_{y}$ is an injective mapping of $ S$ into $ D$ .(Why?)[/quote]
Suppose that there are $ u\neq v$ such that $ x_{u}\equal{} x_{v}: \equal{} x$. Then $ f(x) \equal{} u\neq v \equal{} f(x)$ :huh:

Q2. [i]mapping into[/i] - specifies the [url=http://en.wikipedia.org/wiki/Codomain]codomain[/url] of a function;
[i]mapping onto[/i] - specifies the [url=http://en.wikipedia.org/wiki/Codomain]codomain[/url] of a function and also tells us that the map is surjective.


if $ f(x)\equal{}2x^3\plus{}7x^2\plus{}2x$ then find $ f^{\minus{}1}(\minus{}7)$

$ f(x)\equal{}2x^3\plus{}7x^2\plus{}2x\equal{}\minus{}7 \Leftrightarrow (2x\plus{}7)(x^2\plus{}1)\equal{}0$

So $ f^{\minus{}1}(\minus{}7)\equal{}\minus{}\frac{7}{2}$ if you only want the real one.

why does it equal zero in the second equation?
what happened to the -7?

plug in  y as x (definition of inverse, y=x).

x = 2y^3 + 7y^2 + 2y

Plug in x = -7

-7 = 2y^3 + 7y^2 + 2y

0 = 2y^3 + 7y^2 + 2y + 7

0 = 2y ( y^2 + 1 ) + 7 ( y^2 + 1)

0 = ( 2y + 7 ) ( y^2 + 1 )

y = {-7/2 , i}

Since we're looking for only real solutions, i doesn't count so y = -7/2.


anyone doing Harvard-Mit math tournament?  Registration is soon (feb. 3 for early registration).  I'm going to see if my school will register.

Or maybe we could organize an AOPS team or something... that'd be neat.

i was kind of hoping that you weren't... i can't think of many other people here that are from massachusetts.  What rounds are you going to do?

If anyone else lives around MA, i'd be interested to see if we could get enough for a team.

I'll be there (Albany Area team).  An AOPS team would be a cool idea, but it's a bit late now.  Maybe next year?

yeah it is pretty late.  I just found out the registration date today... i thought it would be a lot later.  Anyways, was just a thought.

Well there's me (not sure if it counts :lol:), Fiertycoon, you, Joel (not sure if he'll qualify), TheSouthpaw (whom I forgot his name; the Japanese guy that looks like Brad Pitt who haven't come here in a long, long time :lol:), namstap (though he/she only posted once), and paladin8 (though he could be anywhere in New England). How many do we need anyway?

joel's in college.  does not count.  Maybe we could sneak him in though.  fierytycoon is going with his school, so does not count.  I thought paladin8 was from seattle... could be wrong though.  We'd be really rushed to get people together this year; we should shoot for next year.

8 kids is a full team.

[quote="Alison"]I'll be there (Albany Area team).  [/quote]

Uh oh...we're dead. :D

cats..., I am going to do Algebra and probably Geometry.

Oooh, not so good looking...still I'll do anything I can since I'm too lazy to registrate/research this thing...I don't even know what this competition is! :lol:

Ballerina6's in New England (7 posts, last post in July 9th), xyz123 (2 posts, September 26th) and timhlu (active; recently posted January 19th)'s in Massachusetts, and that's about it. (excluding all those who never posted) Which will add up to a grand total of...8! Whoo hoo! Course this is under the assumption that all those whom I mentioned joins...

http://web.mit.edu/hmmt/

ask your teacher if they'll register your school.  That's what i plan on doing, even though it's probable that we wont actually register.

OMG!!! Only three more...*looks at calender*ONE MORE DAY! AAAAAARGGGHHHHHAAAAAHHHH! :lol:

Grrr should've known earlier...ah well, I'll get to go to MIT anyway, I hope :wink:

baltimore will likely be sending 2 teams, though .. if anyone is from MD area and not from Howard County, or TJ, or M Blair... you can contact me :)

What is the date of this tournament? I may come. (My debate team is going to a Harvard tournament about the same time, I could get a ride with them :))

I am from Mass. Cats, I'd be interested in doing an AoPS team if you want to.

Otherwise, can I go as an individual if my school doesn't want to do it? I have asked around at my school already, and it doesn't seem like there is any interest.

You can go alone but you will be placed on a *hybrid* team along with other students without a formal team.

This Harvard-MIT tournament seems rather difficult compared to what I've done in the past.  The problems are quite fun though...  Southpaw would be on a high school team (Lexington High) if he made it.

I'm going... AAST freshman team... it'll be fun! :)

I'll be going. Got in the Top 10 last year, and I know I made some stupid mistakes.

Last year's individual tests were pretty easy, but the team round was really hard.

what rounds are you guys planning on doing?

I did general round last year, and will probably do it again. It isn't worth as much as if you take two seperate tests, but it can still net you a high score.

Big news! (at least for me :)) Now that I've joined my high school math team I can go to HMMT!!! Yay! I'll get to see all of you participating there for the first time, wow how exciting! :)

I'm not sure what "team" I'll be on though...


If  A $ \equal{} \{1,2,3, \ldots 9\}$. If x is the sum of the composite numbers in A and y is the sum of the prime numbers in A, what is the value of x - y?

Composite numbers are $ 4\plus{}6\plus{}8\plus{}9\equal{}27$ and primes are $ 2\plus{}3\plus{}5\plus{}7\equal{}17$, subtracting gives $ \boxed{10}$.

Composite numbers in 1-9 are 4,6,8,9 and prime numbers are 2,3,5,7.

Remember that 1 is neither.\[ (4\plus{}6\plus{}8\plus{}9)\minus{}(2\plus{}3\plus{}5\plus{}7)\equal{}\boxed{10}\]


Let a,b,c be positive real numbers such that $ abc\equal{}1$.Prove that
\[ \sum_{cycl}\frac{a^3(b\plus{}c)}{\sqrt{b^5}\plus{}2\sqrt{c^5}} \geq 2\]

[quote="apollo"]Let a,b,c be positive real numbers such that $ abc \equal{} 1$.Prove that
\[ \sum_{cycl}\frac {a^3(b \plus{} c)}{\sqrt {b^5} \plus{} 2\sqrt {c^5}} \geq 2
\]
[/quote]
By AM-GM and Cauchy-Schwarz inequalities, we have:
$ LHS \ge \sum \frac{2\sqrt{a^6bc}}{\sqrt {b^5} \plus{} 2\sqrt {c^5}}\equal{}\sum \frac{2\sqrt{a^5}}{\sqrt {b^5} \plus{} 2\sqrt {c^5}}$
$ \ge 2\frac{(\sqrt{a^5}\plus{}\sqrt{b^5}\plus{}\sqrt{c^5})^2}{3(\sum \sqrt{a^5}\sqrt{b^5})} \ge 2$