what? :mad: i didn't know that. none of our teachers said anything about that. :(
sorry for this probably wrong post. :blush: what can i do now?

[quote="test20"]The Mediterranean Mathematical Competition 2008 is currently running.
In some countries, it is scheduled for mid-April.

It is very very unfortunate that the problems have been posted here.
They should have remained secret till the end of April'2008.
This is damaging the entire competition for 2008.[/quote]That is dumb if it's even true.

It's their fault this happens. In 2008 you can't schedule an international competition across a month. There is nothing to be done.

[quote="Majcena"]I'm sorry, this is not a reply I would expect from you. The whole point of the competition (the way I see it) is that many countries and many students get to participate, not just the best six like at the IMO. In order to do this, it is of course imposible to fix one day for all the countries, even a week is absurd.
I would expect you to encurage posting the problems but only AFTER the competition officially finished (and not saying the month span is stupid and it's their fault). I would expect you to propose what could be done in the future to prevent such a thing rather than saying it's dumb or their fault.

Although in this case apparently the blame is not on the student but the (local) organizer (or the teacher).[/quote]Don't get me wrong I don't support cheating and I don't like it. However I am a realist, and I definitely will not take it upon me or other people to make the job of the organizers for them.

If I were to cheat at this competition, the last thing I would want to do is post it on the boards (so that other people can see my questions and the answers - I would gain nothing). Instead I would PM a couple of people until somebody would show me the questions or agree to discuss them with me. If you have the competition spread over different days and such you definitely cannot ensure that the questions will not be distributed. Or even my friend from another country could help me out with the questions (knowingly or not).

That's why, if you are an organizer, and cannot rally up all the kids at the same place, at least do it in the same day, same hour time frame. All Mediterranean countries are 1 hours apart (maximum) in the timezone, so putting it on a Saturday morning for example would, by all means, not be so difficult.

And, as you said, in this particular case, the organizers fault is that much bigger (especially that kids teacher who didn't tell him other countries are taking it later).

And about my remark above, I was under the impression that this is not true (1 month span). As I said it seems unconceivable that in today's internet  and instant communication age people expect students to keep quiet. That will not happen.

What I propose you do is to establish a weekend (at most) for the competition to take place. Whoever can't make it, well sorry folks, its either this or having 1/2 of the students cheating on the exam.

And also, for future reference, the problems might have been posted in different subforums, not all in one topic here, and as there are over 500 different topics opened every day, it would be almost impossible to make sure the MMO questions aren't among them (unless someone volunteers to keep watch 24/7 - and I doubt there are people that have nothing better to do with their time than to do just that - and even then that person would need to know the questions ahead of time).

Ok, I overreacted, I am sure you ment no harm by your statements. I guess I kind of hated to see this happen when 2 of the problems were proposed by two of my friends.

I partially agree with you but not in whole. It's hard to fix a weekend. For example, all our competitions are fixed in october (the dates have to be approved by the ministry) and there is no way to change the dates afterwards. If this is the case in 3 other countries as well, then maybe choosing a fixed Saturday means that these countries can not participate.
But, I guess it's up to the organizers to figure out the way to go from now on. Maybe it's good that the posting of problems happend on mathlinks since as far as I know, the organizers are now aware of this problem and changed this years problems.

As you said it becomes sort of a local organizing problem once a weekend is chosen to make sure that you can organize it. In the end it shouldn't be THAT hard I mean it's not like everyone in the country has to take the exam right? We're talking about 10-20-30 students per country taking this? (or am I wrong in that assumption)?

PS So what happens to the questions that were posted previously here? if they changed them? :)

It seems that now there is second set of problems that will be used for MMO'2008
in the remaining countries.

For 2009, they clearly will have to find a better way of handling this.

(For instance, APMO asks all participants NOT to circulate the problems
before the problems are listed on the official APMO homepage.  This looks
like a decent solution.)

[quote="kdano"]What I don't see is Why can't the committee choose one single day now, for the MMO 2009. (like we already have when IMO 2009 will be held)
Then all the participating countries could reserve that very day for the competition (And it's a long time till october..), and they would avoid these chaos and mishaps[/quote]That's what I said, but Majcena (if you do not know who she is, she is one of the Team Leaders for the Slovenian IMO Team) said that would be very hard to organize, and possibly some countries would not be able to participate.

Unfortunately, I don't know what is better - once a competition problem gets posted and solved on a forum, to delete (hide) it or to let it stay. The second option makes the competition absurd, but the first one does the same - why should the participants who were lucky enough to access the forum before the topic was deleted (hidden) have an advantage before the ones who have equally good googling skills but tried searching a few days later (maybe because they first tried to solve the problem on their own)?

Valentin (and the other admins as well) - maybe it is time to post an official guideline (for mods and users) about what to do in such cases?

  darij

My guideline was to let the problems stay. Do not create advantages for students who are trying to cheat (this is not the case here, but usually the poster is trying to get the answers so he can cheat).

The competition creators should ensure steps that the competition will not be released online (which basically, outside of having the competition in the same day, is impossible) or at least not on a main site (as a person told me in a PM in the previous years MMO questions were posted ahead of time on a number of different (local) sites, so the whole thing about "telling the students not the post the problems" is kind of futile).

In plus it does not avoid the situation where one has a friend that takes the test first and then gives him the problems. You would think that would not help the person who is giving the problems, but think about this: the person who is giving the problems might have done really bad so he doesn't care if someone else does better than him, and if it can be his friend that gets the gold medal, why no?

I would just like to point out that MMO is not the only competition held in a bigger interval (as I understand, MMO is held in some countries in March and in some countries in May).

[b]Tournament of the Towns[/b], also a well known competition, is also held in the interval of a few weeks, if not more, I think.

On a scale of smaller importance, [b]Math Kangaroo[/b] competition is also (despite its advertising as being held in all the countries at the same time) held in the interval of a few weeks. I believe, for example, Slovenia held it on March 20 while Croatia held it on April 10 due to school holidays.
Also, students in the US, Romania and Croatia might know of [b]American Computer Science League[/b] competition (as the name says, it's a computer science competition, not mathematical). This is also held in the interval of a few weeks, if not a full month.

I'm sure there are many more competitions with that problem. As I see it, it is a rather big problem for the organizers of these competitions and no "please be honest" note will be able to solve it. For example, I try to be honest as possible and I don't ever try to find the problems before the competition (I didn't even want to open this topic before the problems were deleted.  :D ). However, [b]after ACSL competition (Contest #2) was held in my school[/b], I decided to see on Google if problems were available for anyone to see (not that I'd show them to anyone). Of course, it was not hard to find a programming problem (actually, one of the teachers from some school posted it on his web-site). The deadline for the competition was in a week and I'm sure some of the schools didn't hold the competition yet. I e-mailed the organizers informing them and the problem was removed (I guess they e-mailed that teacher).

However, that teacher's web-site is much less popular than Mathlinks and I'm sure fewer dishonest people saw it than here. I'm not pointing fingers, but we all know there is always someone who will try to get the unfair advantage. Why am I rambling on pointlessly, you ask?  :lol:

My point is that Mathlinks can't do anything a day or two after the problem is posted. The only option is to delete it a short time after it's posted (which, Valentin said it, could be hard), limiting the number of people who could have seen it. After that, the competition begins to become pointless. The only option now, I believe, is to change the problems (which I think will happen). However, then we run into some other issues:
1) How can we be sure that these "new" problems will not again get posted here or somewhere else, thus again ruining the competition? As I understand, all further countries will not hold the competition at the same time.
2) What about people who have already written MMO? Will they write again? What if "new" problems are harder/easier than "old" ones (there will always be some nuances)? Then, one group of students will have the advantage at grading, which is also unfair.

Thanks for your time. I'm sure it was boring.  :ninja:

[quote="mornik"]My point is that Mathlinks can't do anything a day or two after the problem is posted. The only option is to delete it a short time after it's posted (which, Valentin said it, could be hard), limiting the number of people who could have seen it.[/quote]Finally somebody understands my arguments for not deleting the thread. Even 4 hours being online on MathLinks means Google gets it in its cache (Google is non stop crawling our site now, because we have such a high traffic), and after that it's pretty pointless.

The organizers could come to me officially, and give me (and a few restricted moderators that are not involved in the competition) the questions before hand and we can stop the spreading when it starts. Other than that, the two possibilities are: don't care about people cheating or hold the competition in at most 1 weekend.

I've heard that the problems for this years MMO have been changed.

I am not involved in MMO so I've also wondered what our local organizer tells our students taking the MMO. There really is no such a thing as letting contestants know that this competition still isn't over and the problems are to be kept secret (there is no "please be honest" at all!!!). I guess there are no general MMO rules on this and it's high time to invent them if they want to keep the competition running. This is the LEAST that should be done.

Otherwise, I just want to add that even if you want to fix a date of MMO 2009, this would not necessarily imply that every country can do it.
There is kind of a general rule when the competitions take place in Slovenia and most of them are in March and April. May is out of the question because of matura (high-school ending exam), which is badly spread over two months. So, what is done in October under the guidance of our ministry is basically just following this basic rules and setting the dates according to the next year's calendar. My guess is that fixing a single day would only work if it is done so that it avoids the general competition season, which are probably in the same months in most countries (March to May maybe?).

Well, concerning the deleted problems, maybe they should be put back (since the MMO now has a second set) so that everyone can see them here and incurage the organizers to think about a good solution to this MMO situation for the future.

For the other questions mornik asked (comparing the two sets of problems, marking...), my answer is: you don't want to know. I didn't know much about rules of MMO until 2 days ago. I no longer think of this as a "fun" competition but.......... I'm not going to say it.

[quote="Majcena"]
I didn't know much about rules of MMO until 2 days ago. I no longer think of this as a "fun" competition but.......... I'm not going to say it.[/quote]

Why not?
I think that now everybody would like to know your opinion...

Strangely the Mediterranean MO does not even have an official website.
(which means: Google did not manage to find one).

Maybe you would like to hear my opinion but I don't have the curage to say it.  :maybe: I am not in any way involved with this competition, it could also be that I misunderstood the person I talked to and things are not as they seem to me.
If they are, I guess you can get similar conclusions if you can ask some (not me) questions about this competition (google is maybe this time not the right "person").  Let's just put it this way, if I knew what I know now, I would not have posted my first post here. Now I regret doing it. Sorry Valentin.

[quote="Majcena"]Let's just put it this way, if I knew what I know now, I would not have posted my first post here. Now I regret doing it. Sorry Valentin.[/quote]It's always like that. No worries, I get bashed all the time for trying to be impartial :D

Now the current theme is ISL 2007, which (obviously) has leaked. As I said to the leaders at the past two IMOs: people if you are using the ISL to select your team for the IMO, then expect some problems to be known in advance. Hell, it's better to just use ISL 1987 rather than this years' shortlist.

Hi, I'm new in this forum.

I'm from Spain and we had our Mediterranean Competition last saturday (May 3rd 2008). The teachers told us about a guy from Turkey who published the problems on mathlinks, so they had to make new ones. Hello, anonymous1173.

I confirm they have been changed, and they were not easy :) But there are some countries who have not done the competition yet, so we must wait before publishing the problems.

anonymous1173 you do not have to worry now, the only problem is that if different countries compete altogether with different problems it could be not fair.

I expect you understand my english. See you!

[quote="Enric"]I'm from Spain and we had our Mediterranean Competition last saturday (May 3rd 2008). The teachers told us about a guy from Turkey who published the problems on mathlinks, so they had to make new ones. Hello, anonymous1173.

I confirm they have been changed, and they were not easy :) But there are some countries who have not done the competition yet, so we must wait before publishing the problems.[/quote]

Mediterranean'2008 is over now.

I would be curious to see the new (second) set of problems.

i'm going be at spain. if you want, you can do you what you want to do to me :oops:
but i must say that i really didn't have any information about its secretness. i think i am very famous now in mediterranean countries. :D

Nobody is blaming you. Or if someone is, that person doesn't know what (s)he is talking about.

hey, it can be understood wrong, i have just joked about punching me. :lol:
valentin thank you for your supporting in this period :)


vector
trigonometry
geometry
3D geometry
sphere
logarithms
calculus

i need someone who took this course recently(last year preferred) to enligthen me with information about the course content, difficulty, and etc.

is that the science physics? i heard it was jokes

I took it.  Good course, but a lot of it will be review if you've done high school physics.  However, you will be exposed to lots of newer material, and the problems will probably be tougher than the crap that gets thrown at you in high school.  Overall a good course, I'd recommend.

The content is primarily mechanics stuff... it starts pretty boring... basic stuff with vectors, force, spring-pulley systems and the like, then energy, momentum, rotational mechanics, torques, moments of inertia, angular momentum, and statics.

The assignments are tougher than what I got in high school, and often you need trig or basic calc to solve the problems... however, they never give you anything overwhelmingly hard math-wise.

A sample problem from one of the assignments is the following:
A particle of mass m rests on top of a frictionless sphere of radius r.  It is given an infinitely tiny push so that it proceeds to slide down the sphere.  At what angle, measured from the top of the sphere, does the particle leave the surface of the sphere and start flying through the air?  Express answer in terms of m, r, and g (acceleration due to gravity).  You have to assume that energy is conserved.

Think that was easy?? (it's not... pm me if you want the answer!)  A variation is to replace the particle sliding down a frictionless surface with a spherical ball of radius r/2, which rolls down the hemisphere without slipping.

hey elyot! in phys 121 we get 2 exams right? 1 for midterm and 1 for final. i've looked at the couple of past exams before the curriculum got changed(phys 121 used to be called Mech, Wave Motion & Heat) and they look pretty ok and not too hard. and is it still multiple choice? and i heard the grades get curved, is it also true?
http://www.scisoc.uwaterloo.ca/exambank/EXAMS/PHYS/121/200009F.pdf
http://www.scisoc.uwaterloo.ca/exambank/EXAMS/PHYS/121/200009M.pdf

I think we had 3 exams... 2 midterms and a final.  The course is called mechanics, wave motion, and heat PART 1.  Phys 122 is part 2 and covers more mechanics, and the wave motion and thermodynamics parts of the course.  Phys 121 is strictly classical mechanics.

I signed up for Physics 121.  I am tempted to attempt to try to get exempted from it and go straight into Physics 122, but I think that is probably overambitious.  I'll prolly just stay put, considering I already have a heavy courseload 1A.

[quote="WindingFunction"][quote="Elyot"][quote="Sisyphus"]Is the class taught in a more 'ground up' manner than in highschool, or is it more or less the same arithmetic word problem stuff?[/quote]

Probably depends on the prof... it's only a classical mechanics course... there's not that much they can do to make it interesting.  I do remember a few interesting derivations though... like some of the formulas for rocket propulsion required some differential equations and stuff.  But you're not expected to do derivations of that difficulty, at least not until 122![/quote]

Lucky you.  I had to do rocket momentum and the like in my first semester mechanics course.  Ugly stuff.   :huh:[/quote]

Well... we were shown the derivations in class with the differential equations, but all we had to do was be able to apply the formula, which somehow used a natural logarithm of some mass ratio... I can't remember it exactly, but I could probably re-derive it.

The point was, you'd be shown examples in class where you needed integrals, diff eqs, etc to solve the problems, but the questions on the test required minimal calc to solve.  Often we'd be given limits, but they were usually just easy ones to allow us to intuitively check our work (ie, what's the limit as mass 1 goes to infinity, or as gravity goes to zero, or as theta goes to 90 degrees)?

is it easy to get over 90s in physics 121?

hey! if i buy the PHYSICS FOR SCIENTISTS & ENGINEERS WITH MODERN PHYSICS from waterloo bookstore, does it include study guide and the solutions manual? or do i have to purchase them separately?

Not sure... I never bothered getting that book.  If you've done physics before, you probably won't need it, although it's a good book to have, particularly if you're planning on taking future courses in physics, which use the same textbook.

What I did was get a 5th edition used from the used bookstore for 25 bucks.  This is like 80-90% cheaper than buying the new 6th edition, and it won't hurt you in any way.  However, some questions have been renumbered, and some chapters reordered.  The library has copies of all the textbooks you'll ever need anyway, so unless you think you'll be needing the text a lot, you can always just check it at the library if you ever need some supplementary info.

I'll tell you right now that I think buying textbooks for many university courses is a total waste of time/money.  I got through my 1b term without buying ONE SINGLE TEXTBOOK for any of my courses.  Phys 121 is an easy enough course that the text won't help much, unless you lack a good background in physics.

what kind of physics background? i did IB physics HL till gr 11(didn't finish IB) then in grade 12 i came to toronto and did physics 12.

You seem to be really worried about your courses.  Relax, lol, i'm sure you'll do fine!

What Sarah said.

You'll have no problem at all.

[quote="Elyot"]
Well... we were shown the derivations in class with the differential equations, but all we had to do was be able to apply the formula, which somehow used a natural logarithm of some mass ratio... I can't remember it exactly, but I could probably re-derive it.

The point was, you'd be shown examples in class where you needed integrals, diff eqs, etc to solve the problems, but the questions on the test required minimal calc to solve.  Often we'd be given limits, but they were usually just easy ones to allow us to intuitively check our work (ie, what's the limit as mass 1 goes to infinity, or as gravity goes to zero, or as theta goes to 90 degrees)?[/quote]

Ahhh...I needed lots of integrals, diff eqs, etc to solve problems on both my problem sets and my exams.  "Sand is transported through a tube connecting cart A to cart B at a constant rate $b$..."  And detailed analyses of simple harmonic motion...  And yes, I do remember lots of natural logs of ratios, especially in thermodynamics.  And I was stupid, and didn't think to learn any multi-var calc in high school.  Really screwed me over in electicity and masochism second semester...uhhh...that is...magnetism. :P


Yeah, it's just a random game I made in my web design class last year. We were learning flash, and flash action script. It has horrible graphics, but I'm a noob at flash, so it's bad.

Basic Game:
Level 1: Jump over the waffles (got pic from internet)
Get 500 Points to Advance
Level 2: Jump over the syrups. They are faster than waffles and taller, but they are skinnier.
I never got to finish it, so there is no level after Level 2. The red button on the top doesn't work. I tried to hide stuff, but no I couldn't get it. Also it gives some impossible cases, so it can get annoying.
It is very easy still. Rate the game here.

you should see omer's game. you actually shoot and kill stuff. there is sounds when you shoot and stuff. it's like phoenix for ti-84 except it is sideways

[quote="bookaholic"]It's not bad, actually. I never got motivated enough to do anything with Flash besides the most basic move/rotate/change color etc., so it looks pretty impressive to me. Wow, my complete and total ignorance about anything related to computers is really shining through here, isn't it.

EDIT: Darn it, the second level is annoying. I can't finish it. I'm so uncoordinated.[/quote]

hint: duck some times (press down) and jump. sometimes, you can just stay in the corner and duck the whole time

You can duck? Oh. Well. That helps.  :blush:

nope, all info lost over the summer :( the school even changed computers.
[quote=bookaholic]You can duck? Oh. Well. That helps. Blush[/quote]

yup, that makes it way easier. DId you beat it this time?[/quote]

[quote="moogra"]nope, all info lost over the summer :( the school even changed computers. [/quote]
Hm?  I can still get my stuff on satchmo...

[quote="bookaholic"]Yeah, I got it this time. I feel so sad now. I could have done so much cool stuff with Flash when I had a chance, but I goofed off instead and now I don't know the next time I'll get to work with it. Sigh.[/quote]

what exactly did you do in class? Because I slacked off as well by talking to omer and browsing internet. I still got everythign done though... last time I checked so did you.

btw my work wasnt on satchmo, because satchmo=lame, and people like wang steal that stuff

[quote="bookaholic"]Well, I did finish all the projects that we were supposed to to, but I didn't keep playing with Flash in the time left over. I don't know what I was doing. Just general time-wasting, I guess.[/quote]

only my famous page was any good. the rest sucked. I learned from 7th grade (I was lame and put totally random things on "secret" pages that could be interpreted totally nastily i.e. come on, lets do it... guy 2: noooo please don't come closer) I got a C on the project. Then I also made a bad picture edit of willis (and it got removed because I made her a devil )- hey she said we could do anything. And I made a really weird page for my journal that was meant to be a secret page... I got a B+ overall in that semester. I improved a lot in 8th grade in web design because I didn't make dumb pages.

-edit-
I take that back (no random pages) because one page was seriously random. If viewed in internet explorer nothing would be weird, but if viewed in firefox, anything but i.e. you can scroll down and see some pretty funny stuff (no nasty btw). also if you entered the source code, you would see nothing (frames ftw)


[quote="Xevarion"]Is it true: $a, b, c > 0$

$\sum_{sym}\frac{(a+b)b}{(a+c)c}\ge \sum_{sym}\frac{(a+b)c}{(a+c)b}$

?  :maybe:  :lol:[/quote] \[\sum_{sym}\frac{(a+b)b}{(a+c)c}=\sum_{\sigma}\left(\frac{(a+b)b}{(a+c)c}+\frac{(a+c)c}{(a+b)b}\right)\]  \[\sum_{sym}\frac{(a+b)c}{(a+c)b}=\sum_{\sigma}\left(\frac{(a+b)c}{(a+c)b}+\frac{(a+c)b}{(a+b)c}\right)\] By Rearrangement, \[\sum_{\sigma}\left(\frac{(a+b)b}{(a+c)c}+\frac{(a+c)c}{(a+b)b}\right)\ge\sum_{\sigma}\left(\frac{(a+b)c}{(a+c)b}+\frac{(a+c)b}{(a+b)c}\right).\]


invariant
superior algebra
superior algebra unsolved

Let $F$ be a subfield of $\mathbb C$, with $\phi : \mathbb C\to \mathbb C$ an $F$-homomorphism and $\text{trdeg}(\frac{\mathbb C}{F})<\infty$. Prove that $\phi$ is an isomorphism. Prove also that the number of subfields $F$ with $[\mathbb C: F]=2$, is uncountable.

Sorry,Can I ask what is the mean by "homomorphism" ,homorphism or monomorphism?

And what is the mean of $trdeg$? is $[C: F]$?
Thank you for answer

Homomorphisms between fields are always injective. "trdeg" means "transcendence degree". owk

Thank you,owk.

For the first problem:
Let $S$ be the finite trancendental basis of $C$ over $F$,
Then becauae $\phi$ is injective,thus $\phi(S)$ is trancendental independent over $F$,but by the invariant theorem,we have $\phi(S)$ is a trancendental basis of $C$,then $C$ is algebrical over $Im \phi$,but $C$ and $Im \phi$ are also algebrical closed.($Im \phi$ is algebrical close because it is isomorhism:$\phi$:$C \to Im \phi$)
Thus $C=Im \phi$


See if you can do this without any trig.

Given that CD is the diameter with O as the center and A and B are on arc CD, find the measure of arc BD given the measures of the angles as shown in the picture.

The first time I looked at this there wasn't a picture, but suddenly there is one now.  What happened?

I assume you want angle BOD? (50)

Can be quite easily done by remembering OB, OA radii of circle and recognising that BAEO is cyclic quadrilateral (where E is the unnamed point next to O).


inequalities
advanced fields
advanced fields theorems

Let x,y,z,t belongs to R^n where d(x,y)=||x-y||.
Show that(Ptolemy's inequality):
d(x,y)d(z,t)<=d(x,z)d(y,t)+d(x,t)d(y,z)

I have read this [url=http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.1042v1.pdf]document[/url] but I cannot show that the Euclidean space R^n is Ptolemy.
The paper in the second page "2.Preliminaries" says that "To show that the Euclidean space R^n is Ptolemy, consider again four points x,y,z,w. Applying a suitable Mobius transformation we can assume that z is a midpoint of y and w, i.e. |yz|=|zw|=1/2 |yw|. For this configuration...."
But how can we extract from the above paragraph that the Euclidean space R^n is Ptolemy and which is the "suitable Mobius transformation"??

Thanks anyone in advance.

If we use inequalities like Minkowski's or Holder ineq. ?
 (I'm trying to prove it in this way but i don't know if this this is the right way)

- It's easy in $ \mathbb{C}$; we have the identity $ (x \minus{} y)(z \minus{} t) \equal{} (x \minus{} z)(y \minus{} t) \plus{} (x \minus{} t)(z \minus{} y)$.
- Any four points in $ \mathbb{R}^n$ lie in a plane.
- Any plane in $ \mathbb{R}^n$ is isometric to $ \mathbb{C}$.

The equality case is precisely when $ x,z,y,t$ lie on a line or circle in that order.


If $ abc\equal{}1$, prove $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\ge a\plus{}b\plus{}c$.

[quote="bertram"]If $ abc \equal{} 1$, prove $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge a \plus{} b \plus{} c$.[/quote]

Since $ abc\equal{}1$ Let $ a \equal{} \frac{x}{y}, b \equal{} \frac{y}{z}, c \equal{} \frac{z}{x}$

Therefore we obtain

$ \frac{xz}{y^2}\plus{}\frac{yx}{z^2}\plus{}\frac{zy}{x^2} \ge \frac{x^2z\plus{}y^2x\plus{}z^2y}{xyz}$

And

$ x^3z^3\plus{}y^3x^3\plus{}y^3z^3 \ge x^3z^2y\plus{}y^3x^2x\plus{}z^3y^2x$

Which is true by rearrangement

$ a\equal{}2,\; b\equal{}\minus{}\frac{1}{4},\; c\equal{}\minus{}2$ doesn't work.

[quote="JavaMan"]$ a \equal{} 2,\; b \equal{} \minus{} \frac {1}{4},\; c \equal{} \minus{} 2$ doesn't work.[/quote]

I think he also meant $ a,b,c>0$

this is a nice ineq...Let $ a\equal{}xr,b\equal{}yr,c\equal{}zr,x,y,z,r>0$ .We have $ abc\equal{}1 \Leftrightarrow xyzr^3\equal{}1 \Leftrightarrow r\equal{}\frac{1}{\sqrt[3]{xyz}}$ .Instead of $ x,y,z$ we will use \[ x^3,y^3,z^3\], then we will have \[ r\equal{}\frac{1}{xyz} , a\equal{}\frac{x^2}{yz},b\equal{}\frac{y^2}{zx},c\equal{}\frac{z^2}{xy}\] The new form of the ineq looks like : \[ \frac{x^3}{y^3}\plus{}\frac{y^3}{z^3} \plus{} \frac{z^3}{x^3} \geq \frac{x^2}{yz} \plus{} \frac{y^2}{zx} \plus{} \frac{z^2}{xy}\] and with the rearangements ineq we'll have $ \alpha\equal{}(x^2/y^2,y^2/z^2,z^2/x^2)$ and $ \beta\equal{}(x/y,y/z,z/x)$ have the same arangement we have $ S(\alpha,\beta)\geq S(\alpha,\gamma)$


prove:$ n^n\equal{}\sum_{k\equal{}1}^{n} kn^{n\minus{}1\minus{}k}C_{k\plus{}1}^{n\plus{}1},\forall n\in N$


function
topology
advanced fields
advanced fields solved

How can I show that for that for continuous function $ f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ the set of all of its fixed points is closed?

Choose a non-fixed point $ v$.  Then $ |v \minus{} f(v)| \equal{} d > 0$.  Then by continuity of $ f$, there exists $ \delta$ such that ....

true more generally for hausdorff spaces by taking the inverse image of the diagonal under (id, f)

Or, you can define the function $ g(x)\equal{}f(x)\minus{}x$. Now the set of fixed points of $ f$ is exactly the same of the set $ g^{\minus{}1}(0)$. Since $ f$ is continous, $ g$ is continous, $ \{0\}$ is closed in $ \mathbb{R}^n$ and so $ g^{\minus{}1}(0)$ is closed


Given $ a, b, c \ge 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that:
\[ \frac{ab}{\sqrt{b\plus{}c}}\plus{}\frac{bc}{\sqrt{c\plus{}a}}\plus{}\frac{ca}{\sqrt{a\plus{}b}}\plus{}\frac{1}{\sqrt{2}} \ge \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{2\sqrt{2}}\]
When does equality occur ?  :)
PS: [hide]Anh Can? ko dc lam` bai` nay` dau nha :P[/hide]

[quote="nguoivn"]Given $ a, b, c \ge 0$ and $ a + b + c = 3$. Prove that:
\[ \frac {ab}{\sqrt {b + c}} + \frac {bc}{\sqrt {c + a}} + \frac {ca}{\sqrt {a + b}} + \frac {1}{\sqrt {2}} \ge \frac {(a + b)(b + c)(c + a)}{2\sqrt {2}}\]
When does equality occur ?  :)
[/quote]
Let $ x=ab+bc+ca$ and $ y=abc.$ By Holder's Inequality, we have
$ \left( \sum \frac{ab}{\sqrt{b+c}}\right)^2 \left[\sum ab(b+c)\right] \ge (ab+bc+ca)^3=x^3.$
Also,
$ \sum ab(b+c)=(ab^2+bc^2+ca^2+abc)+2abc \le 4+2abc=4+2y.$
From these two inequalities, we deduce that
$ \sum \frac{ab}{\sqrt{b+c}} \ge \sqrt{\frac{x^3}{2y+4}}= \frac{x^2}{\sqrt{2}\sqrt{x(y+2)}} \ge \frac{2x^2}{\sqrt{2}(x+y+2)}.$
It is thus sufficient to prove that
${ \frac{4x^2}{x+y+2}+2 \ge (a+b)(b+c)(c+a)}.$
Since $ (a+b)(b+c)(c+a)=3x-y,$ this is equivalent to
$ \frac{4x^2}{x+y+2}+2 \ge 3x-y,$
or
$ \frac{4x^2}{x+y+2}+(x+y+2) \ge 4x,$
which is obviously true according to the AM-GM Inequality.
Equality holds if and only if $ a=b=c=1,$ or $ a=1,b=2,c=0,$ or any cyclic permutation thereof.

[quote="Dumel"]is there any nice proof for it:
$ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \le 4$
 :?:[/quote]

After homogenization it becomes like the inequality here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306625 (take a look at 7-8 post) :wink:

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c \ge 0$ and $ a + b + c = 3$. Prove that:
\[ \frac {ab}{\sqrt {b + c}} + \frac {bc}{\sqrt {c + a}} + \frac {ca}{\sqrt {a + b}} + \frac {1}{\sqrt {2}} \ge \frac {(a + b)(b + c)(c + a)}{2\sqrt {2}}\]
When does equality occur ?  :)
[/quote]
Let $ x = ab + bc + ca$ and $ y = abc.$ By Holder's Inequality, we have
$ \left( \sum \frac {ab}{\sqrt {b + c}}\right)^2 \left[\sum ab(b + c)\right] \ge (ab + bc + ca)^3 = x^3.$
Also,
$ \sum ab(b + c) = (ab^2 + bc^2 + ca^2 + abc) + 2abc \le 4 + 2abc = 4 + 2y.$
From these two inequalities, we deduce that
$ \sum \frac {ab}{\sqrt {b + c}} \ge \sqrt {\frac {x^3}{2y + 4}} = \frac {x^2}{\sqrt {2}\sqrt {x(y + 2)}} \ge \frac {2x^2}{\sqrt {2}(x + y + 2)}.$
It is thus sufficient to prove that
${ \frac {4x^2}{x + y + 2} + 2 \ge (a + b)(b + c)(c + a)}.$
Since $ (a + b)(b + c)(c + a) = 3x - y,$ this is equivalent to
$ \frac {4x^2}{x + y + 2} + 2 \ge 3x - y,$
or
$ \frac {4x^2}{x + y + 2} + (x + y + 2) \ge 4x,$
which is obviously true according to the AM-GM Inequality.
Equality holds if and only if $ a = b = c = 1,$ or $ a = 1,b = 2,c = 0,$ or any cyclic permutation thereof.[/quote]
Nice proof, can_hang  :lol:


[quote="mihail911"]I'm working through the book "geometry revisited", and I'm finding it a little difficult to work through. I mean, I read the theorems, understand the proofs they give, and do problems, but I still feel that I'm not exactly absorbing all the information.
For example, in reading about the nine-point circle and the euler line, I understood how facts about these two geometric figures were derived, but I still wouldn't feel comfortable using either in a geometry problem.
I guess my question is, how did you people that worked through it or are working through it approach the book?

Thanks![/quote]

haha

thats the same as what i felt....thats why i literally burned that book. i agree that the problems are too simple so just find harder problems to do. but im not that good at geometry so i cant say much.

Thanks!
Anybody who's had success with this book have any input to add?


[quote="fjwxcsl"]Let  $ {a,b,c,m,n,k}$ be positive real numbers. Prove that the following inequality holds:

${ \frac {a^{m + n} + k}{a^{m}b^n + k} + \frac {b^{m + n} + k}{b^{m}c^n + k} + \frac {c^{m + n} + k}{c^{m}a^n + k}}\geq3$[/quote]

${ F(k) = \frac {a^{m + n} + k}{a^{m}b^n + k} + \frac {b^{m + n} + k}{b^{m}c^n + k} + \frac {c^{m + n} + k}{c^{m}a^n + k}} - 3$

m1:=numer(F);
collect(m1,k);

we only prove

$ f_1(a,b,c)k^2 + f_2(a,b,c)k + f_3(a,b,c)\geq 0$

can  prove

$ f_1(a,b,c)=a^{m+n}+c^{m+n}+b^{m+n}-a^mb^n-c^ma^n-b^mc^n\geq 0$

$ f_3(a,b,c)=\sum_{cyc} a^{m+n}b^{m+2n}c^m-3(abc)^{m+n}\geq 0$

I think we only prove

$ f_2(a,b,c)\geq 0$

By AMGM, ${ \frac{a^{m+n}+k}{a^{m}b^{n}+k}+\frac{b^{m+n}+k}{b^{m}c^{n}+k}+\frac{c^{m+n}+k}{c^{m}a^{n}+k}}\geq3\left(\frac{a^{m+n}+k}{a^{m}b^{n}+k}\times\frac{b^{m+n}+k}{b^{m}c^{n}+k}\times\frac{c^{m+n}+k}{c^{m}a^{n}+k}\right)^{\frac{1}{3}}$

So we only have to prove $ (a^{m+n}+k)(b^{m+n}+k)(c^{m+n}+k)\geq(a^{m}b^{n}+k)(b^{m}c^{n}+k)(c^{m}a^{n}+k)$.

From Holder's, $ (a^{m+n}+k)^\frac{m}{m+n}(b^{m+n}+k)^\frac{n}{m+n}\geq a^{m}b^{n}+k$,
$ (b^{m+n}+k)^\frac{m}{m+n}(c^{m+n}+k)^\frac{n}{m+n}\geq b^{m}c^{n}+k$, and
$ (c^{m+n}+k)^\frac{m}{m+n}(a^{m+n}+k)^\frac{n}{m+n}\geq c^{m}a^{n}+k$.

We simply multiply the above three inequalities to get the desired.

This is the proof, dear [b]bakerbakura[/b]!
In fact, this problem represents the generalization for the following inequalities

Inequality MB-29: http://www.mathlinks.ro/viewtopic.php?t=321108
Inequality MB-32: http://www.mathlinks.ro/viewtopic.php?t=321337

I think that [b]Abdek[/b]'s solution to MB-32,  based on Holder's inequality, is the source of inspiration for this nice inequality. Nice remark, dear [b]fjwxcsl[/b]. Congratulations!


for any positive number $ x_{1},\cdots,x_{n}$,can we prove
$ \frac{x_{1}+x_{2}+\cdots+x_{n}}{n}+\frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}}\ge 2\sqrt[n]{x_{1}x_{2}\cdots x_{n}}$

Or is it famous?

[quote="Vasc"]The inequality is not true. See $ a=1, b=c=2$ for $ n=3$.
But, I think the inequality holds
$ (n-1)A_{n}+H_{n}\ge nG_{n}$,
and , moreover,
$ A_{n}^{n-1}H_{n}\ge G_{n}^{n}$.[/quote]
Thank you.
$ A_{n}^{n-1}H_{n}\ge G_{n}^{n}$
because:
$ (x_{1}+\cdots+x_{n})^{n-1}\ge n^{n-2}x_{1}\cdots x_{n}(\frac 1{x_{1}}+\cdots+\frac 1{x_{n}})$
but what is the best number?


integration
quadratics
calculus
trigonometry
geometry
3D geometry
algebra

These are... primes (without 2 and 3 for some reason)?

Or +4 +2 +4 +2 etc?

Under the change of variables $ x\mapsto\frac1x,$ the integral

$ \int_0^1\frac{x^2}{1\plus{}x^2\plus{}x^4}\,dx$  becomes  $ \int_1^{\infty}\frac{x^2}{1\plus{}x^2\plus{}x^4}\,dx$

That means we can write the answer as $ \frac12\int_0^{\infty}\frac{x^2}{1\plus{}x^2\plus{}x^4}\,dx$

or even $ \frac14\int_{\minus{}\infty}^{\infty}\frac{x^2}{1\plus{}x^2\plus{}x^4}\,dx$

These manipulations give us a broader set of tools to use. I just used the residue theorem to get that the answer is $ \boxed{\frac{2\pi}{\sqrt{3}}},$

(One can argue that this use of the residue theorem isn't [i]that[/i] different from factoring the denominator and using partial fractions - the arithmetic is overlapping. And the integral using partial fractions is probably a little easier to compute on $ (\minus{}\infty,\infty)$ than on $ [0,1].$ )

I also just did a Richardson extrapolation on the first 64 partial sums of the series to get an estimate that is within $ 6\times 10^{\minus{}7}$ of $ \frac{2\pi}{\sqrt{3}},$ so I'm pretty confident that this is the right answer.

Hey,
[quote="vijaymenon"]
find the sum of [size=150] [b]1 - 1/5  + 1/7 - 1/11  + 1/13...............[/b][/size] using integration.
[/quote]
[u]Challenge:[/u] Find the sum $ {S}$,
$ {S} \equal{} {{1}\minus{}{\frac{1}{5}}\plus{}{\frac{1}{7}}\minus{}{\frac{1}{11}}\plus{}{\frac{1}{13}}\minus{}\dots}$
without using calculus.

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

Using the concept of Fourier series, we have

$ \frac{\pi}{4} \equal{} \sum_{n\equal{}1}^{\plus{} \infty} \frac{\sin [(2n\minus{}1)x]}{2n\minus{}1}$,

for $ x \in (0, \pi)$. Considering $ x \equal{} \frac{\pi}{3}$ we get

$ 1 \minus{} \frac{1}{5} \plus{} \frac{1}{7} \minus{} \frac{1}{11} \plus{} \frac{1}{13} \minus{} \frac{1}{17} \plus{} ... \equal{} \frac{\pi}{2 \sqrt{3}}$.

[quote="t0rajir0u"]The #1 rule is that $ \frac {1}{n \plus{} 1}$ really means $ \int_{0}^{1} x^n \, dx$.  So this sum is equal to

$ \int_{0}^{1} \sum_{k \equal{} 0}^{\infty} \left( x^{6k} \minus{} x^{6k \plus{} 4} \right) \, dx \equal{} \int_{0}^{1} \frac {1 \minus{} x^4}{1 \minus{} x^6} \equal{} \int_{0}^{1} \frac{1 \plus{} x^2}{1 \plus{} x^2 \plus{} x^4} \, dx$

which has a partial fraction decomposition into two terms with quadratic denominator, so from here it's pretty standard.[/quote]

I'm confused by this post. Can someone explain both the top and bottom?

$\displaystyle\int_0^1 x^n \, dx = \left. \dfrac{x^{n+1}}{n+1} \right|^{x = 1}_{x = 0} = \dfrac{1}{n+1}$
Not sure what more to say than this...

Hence,

$1 - \dfrac{1}{5} + \dfrac{1}{7} - \dfrac{1}{11} + \dfrac{1}{13}
=1 + \dfrac{1}{7} + \dfrac{1}{13} + \dfrac{1}{19} + \ldots
- \dfrac{1}{5} - \dfrac{1}{11} - \dfrac{1}{17} - \ldots$

Then using $\displaystyle\int_0^1 x^n \, dx = \dfrac{1}{n+1}$ we can write
$1 + \dfrac{1}{7} + \dfrac{1}{13} + \dfrac{1}{19} + \ldots = \displaystyle\int_0^1 x^0 \, dx +  \displaystyle\int_0^1 x^6 \, dx +  \displaystyle\int_0^1 x^{12} \, dx + \ldots
= \displaystyle\int_0^1 \sum_{k = 0}^\infty x^{6k} \, dx$
Similarly, $\dfrac{1}{5} + \dfrac{1}{11} + \dfrac{1}{17} = \displaystyle\int_0^1 \sum_{k = 0}^\infty x^{6k + 4} \, dx$

Then, we use the fact that $\sum_{k = 0}^\infty x^{6k}$ is a geometric series, with closed form of $\dfrac{1}{1 - x^6}$. Similarly, $\sum_{k = 0}^\infty x^{6k + 4} = x^4 \sum_{k = 0}^\infty x^{6k} = \dfrac{x^4}{1 - x^6}$.

Finally, we can factorise using the difference of cubes and difference of squares formulas:
$(1 - x^4) = (1 - x^2)(1 + x^2)$ and $1 - x^6 = (1 - x^2)(1 + x^2 + x^4)$

Theres a nice trick for this integral:

$ \int^1_0 \frac{1+x^2}{1+x^2+x^4} dx = \int^1_0 \frac{1 + \frac{1}{x^2}}{ x^2 + \frac{1}{x^2} + 1} dx $

$ = \int^1_0 \frac{ d\left(x - \frac{1}{x}\right) }{ \left( x- \frac{1}{x}\right)^2 + 3 } dx = \frac{1}{\sqrt{3}} \left. \arctan \left( \frac{ x- \frac{1}{x} }{\sqrt{3}}\right) \right|^{x = 1}_{x \to 0^+} = \frac{\pi}{2\sqrt{3}} $

This agrees with the answer from the Fourier Series method. Kent, I think you worked out a different integral, you lost the +1 in the numerator!


If a,b,c>0 , then\[ \sqrt{\frac{ab}{(a\plus{}c)(b\plus{}c)}}\plus{}\sqrt{\frac{bc}{(b\plus{}a)(c\plus{}a)}}\plus{}\sqrt{\frac{ca}{(c\plus{}b)(a\plus{}b)}}\ge \frac{3}{2}\]

[quote="Marius Mainea"]If a,b,c>0 , then
\[ \sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}} \plus{} \sqrt {\frac {bc}{(b \plus{} a)(c \plus{} a)}} \plus{} \sqrt {\frac {ca}{(c \plus{} b)(a \plus{} b)}}\ge \frac {3}{2}
\]
[/quote]
Try $ c\rightarrow0^ \plus{} .$ :wink:

Tanks, arqady, is a problem propose not me in SHL-RMO-2005.

The following inequality is true already.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that not two of them not equal to zero. Prove that
\[ \sum_{cyc}\sqrt{\frac{ab}{(a\plus{}c)(b\plus{}c)}}\ge \sqrt{1\plus{}\frac{10abc}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}}\]

[quote="arqady"]The following inequality is true already.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that not two of them not equal to zero. Prove that
\[ \sum_{cyc}\sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}}\ge \sqrt {1 \plus{} \frac {10abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}}
\]
[/quote]

 After square,inequality reduces to\[ \sum {a\sqrt{bc(b\plus{}a)(c\plus{}a)}}\ge 6abc\]wich is true from $ AM\ge GM$

[quote="Marius Mainea"]If a,b,c>0 , then
\[ \sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}} \plus{} \sqrt {\frac {bc}{(b \plus{} a)(c \plus{} a)}} \plus{} \sqrt {\frac {ca}{(c \plus{} b)(a \plus{} b)}}\le \frac {3}{2}
\]
[/quote]
After squaring we need to prove that
$ 8\sum_{cyc}a\sqrt{bc(a\plus{}b)(a\plus{}c)}\leq\sum_{cyc}(5a^2b\plus{}5a^2c\plus{}6abc).$
But $ 8\sum_{cyc}a\sqrt{bc(a\plus{}b)(a\plus{}c)}\leq4\sum_{cyc}a(b(a\plus{}c)\plus{}c(a\plus{}b))\equal{}$
$ \equal{}\sum_{cyc}(4a^2b\plus{}4a^2c\plus{}8abc)\leq\sum_{cyc}(5a^2b\plus{}5a^2c\plus{}6abc).$

[quote="Marius Mainea"]But
[quote="Marius Mainea"]If a,b,c>0 , then
\[ \sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}} \plus{} \sqrt {\frac {bc}{(b \plus{} a)(c \plus{} a)}} \plus{} \sqrt {\frac {ca}{(c \plus{} b)(a \plus{} b)}}\le \frac {3}{2}
\]
[/quote][/quote]

$ \sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}} \plus{} \sqrt {\frac {bc}{(b \plus{} a)(c \plus{} a)}} \plus{} \sqrt {\frac {ca}{(c \plus{} b)(a \plus{} b)}}\le \frac{1}{2}(\frac{a}{a\plus{}c}\plus{}\frac{b}{b\plus{}c})\plus{}\frac{1}{2}(\frac{b}{b\plus{}a}\plus{}\frac{c}{c\plus{}a})\plus{}\frac{1}{2}(\frac{c}{c\plus{}b}\plus{}\frac{a}{a\plus{}b})\equal{}\frac{3}{2}$

$ a\equal{}b\equal{}c$

very easy  :)

[quote="zaya_yc"][quote="Marius Mainea"]But
[quote="Marius Mainea"]If a,b,c>0 , then
\[ \sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}} \plus{} \sqrt {\frac {bc}{(b \plus{} a)(c \plus{} a)}} \plus{} \sqrt {\frac {ca}{(c \plus{} b)(a \plus{} b)}}\le \frac {3}{2}
\]
[/quote][/quote]

$ \sqrt {\frac {ab}{(a \plus{} c)(b \plus{} c)}} \plus{} \sqrt {\frac {bc}{(b \plus{} a)(c \plus{} a)}} \plus{} \sqrt {\frac {ca}{(c \plus{} b)(a \plus{} b)}}\le \frac {1}{2}(\frac {a}{a \plus{} c} \plus{} \frac {b}{b \plus{} c}) \plus{} \frac {1}{2}(\frac {b}{b \plus{} a} \plus{} \frac {c}{c \plus{} a}) \plus{} \frac {1}{2}(\frac {c}{c \plus{} b} \plus{} \frac {a}{a \plus{} b}) \equal{} \frac {3}{2}$

$ a \equal{} b \equal{} c$

very easy  :)[/quote]

 Nice. I think that is original solution.


What are the dimessions of a coin for which the probability of getting a head is 1/3, getting a tail is 1/3 n falling on edge is 1/3?

[quote="Identity"]Well, for such a coin, you must have the areas of the two faces and the surface area of the side all equal.

Area of side of coin = $ 2\pi \times r \times h$, where r is the radius of the coin and h is the thikness.
Area of face = $ \pi r^2$

If you try to equate these, you get:

$ \pi r^2 \equal{} 2\pi \times r \times h$

$ r \equal{} 2h$

And that is the condition that is required, so the thickness must be half the radius.[/quote]

Is that enough??
I think we have to consider the centre of gravity. In that case this is not just mathematics

In which case this should be in the physics forum.

This problem can be solved using vectors; we use a vector to represent the weight of each face and the side of the coin.


[quote="countdownmath"]Let $f \in C^{4}((a,b))$ and let $x_{0}\in (a,b).$  Assuming that $f'(x_{0})=f''(x_{0})=f'''(x_{0})=0, f^{(4)}>0$ show that $f$ has a strict local minimum at $x_{0}$.[/quote]
there was posted problem like this but the question was: prove that there not exist extremum. This proved analogisly by taylor.  :wink:
I mean $f(x)-f(x_{0})=\frac{f^{(4)}(x_{0})(x-x_{0})^{4}}{4!}+o(x-x_{0})^{4}$ wich will be more than 0in some moment :lol: . and after is simpl

[quote="countdownmath"]would this also be true?

$\lim_{x \to x_{0}}\frac{f(x)-f(x_{0})}{(x-x_{0})^{4}}= \frac{f^{(4)}(x_{0})}{24}$

by L'Hopital's rule.. since we are given that $f^{(}4)>0$ and $24>0, (x-x_{0})^{4}>0$, we can conclude that $f(x)-f(x_{0})>0$.

This ensures that $\forall x \in B(x_{0}, \epsilon)$, for small $\epsilon$, $f(x_{0})>f(x)$.

Therefore, $f$ has a stict local min at $x_{0}$.[/quote]
Yes it is true.


8 apples and 9 oranges cost 118 dollars; 9 apples and 8 oranges cost 120 dollars. Find the cost of 4 oranges and 4 apples.

[hide] you can set up the problem by using variables to replace the apples and oranges and you get:
8x+9y=118
9x+8y=120
you add them up and get
17x+17y=238, then you divide by 17 and get
x+y=14, then you multiply by four and get

4apples + 4 oranges =56 dollars
[/hide]

Ok I think I got it.

[hide]

Since 8 apples and 9 oranges cost 118, but 9 apples and 8 oranges cost 120,
we can say that 1 apple = 1 orange + 2
If we turn the first purchase into equation while apple = x, orange = y,
it become $8x+9y=118$. Lets substitute the $x$ with $y+2$ to make the equation
$(16+8y)+9y=118$, then add the terms and subtract 16 on both side

$17y=102$.
$\displaystyle \frac{17y}{17} = \frac{102}{17}$
which becomes $y=6$
$118- 54= 64$, $\displaystyle\frac{64}{8} =8$ Therefore, $x=8, y=6$
$32+24=56$
And the answer is $56$.

[/hide]

But i may be wrong, you never know. :?

Andrew I didn't see I could do it that way. And you posted about 1 min before me!!!

[quote="Andrew619"]you can only do it that way if the number variables in the equation are the same

for example:

3a+b+c+d+e= 28
a+3b+c+d+e= 34
a+b+3c+d+e= 47
a+b+c+3d+e= 25
a+b+c+d+3e=  34

what is the value of a+b+c+d+e?[/quote]

[hide]Sum all the equations and you get $7(a+b+c+d+e)=168$. Then, the answer is $\boxed {24}$.[/hide]

[quote="Knuth's_check"][img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/0c05f4a0607101832ebfd74160f5b42c.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a8aeafd01dee9eca383f633c909f931d.gif[/img]

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d655e5c6a852376cb68c7d12020f85dd.gif[/img], therefore x=8 and answer is 56 ;)[/quote]

Yes, the best way is doing a system of equations (2 equations, 2 unknowns)

Lol, that's expensive.
[hide]
If you add the 2 equations up, you get $17a+17o=238$
So, the answer is 56[/hide]


Prove that the sum of twin primes $ p$ and $ p\plus{}2$ is divisible by $ 12$, provided that $ p>3$.

Since every prime greater than $ 3$ can be written if form $ 6k\plus{}\minus{}1$ we use that.

Since $ p$ is smaller prime then $ p\plus{}2$ it must be $ 6k\minus{}1$ while $ p\plus{}2$ is $ 6k\plus{}1$.

From there we obtain: $ p\plus{}(p\plus{}2)\equal{}6k\minus{}1\plus{}6k\plus{}1\equal{}12k$. Hence, proved.


geometry
ratio
geometric sequence
arithmetic sequence

What is a Geometric Sequence?
Does it have anything to do with Geometry?

A geometric sequence is in the form of: $ a$,$ ar$,$ ar^{2}$,$ ar^{3}$,$ \hdots$.

Let $ n_k$ denote the $ k$th term of a geometric sequence.  Then $ \frac {n_{k \plus{} 1}} {n_k} \equal{} r$, for any $ k$.
In other words, $ r$ is the common ratio of the sequence.
No it doesn't have anything to do with geometry :) :)...

Suggestion: Go look it up in the AOPS wiki for more explicit details.

The sequence is of the form $ a,ar,ar^2,...ar^n$.
[url]http://www.artofproblemsolving.com/Wiki/index.php/Geometric_sequence[/url]

Geometric sequences have NOTHING to do with geometry!

They're just sequences with common ratios (like a, ar, ar^2, ar^3,....)

[quote="mathcrazed"]No it doesn't have anything to do with geometry :) :)...[/quote]

Really now?  [url=http://en.wikipedia.org/wiki/Geometric_progression#Relationship_to_geometry_and_Euclid.27s_work]Are you sure?[/url]  ;)

Another way to think about it:  an arithmetic sequence is a sequence in which every term is equal to the arithmetic mean of the term before it and the term after it.  A geometric sequence is a sequence in which every term is equal to the [url=http://www.cut-the-knot.org/pythagoras/GeometricMean.shtml]geometric mean[/url] of the term before it and the term after it.

oh come on... if you want to go that deep, anything can be related to anything... But usually, you never think of geometry when you see a geometric sequence.

[quote="mathcrazed"]But usually, you never think of geometry when you see a geometric sequence.[/quote]

Did you ever wonder why it was called a "geometric" sequence?  Most mathematical names of this type are not accidents.  (The mathematical names involving mathematicians themselves, on the other hand, are another story...)

In fact, every time I think of geometric sequences I think of the following diagram:

[img]http://static.twoday.net/mahalanobis/images/viewp01.gif[/img]

The relationship between arithmetic/geometric sequences and arithmetic/geometric means is in fact very strong:  a third inclusion is that the [b]harmonic sequence[/b] $ \frac {1}{1}, \frac {1}{2}, \frac {1}{3}, ...$ has the property that each term is equal to the the [url=http://en.wikipedia.org/wiki/Harmonic_mean]harmonic mean[/url] of the term before it and the term after it.  And in case you were wondering whether harmonic sequences and the harmonic mean have anything to do with harmony (that is, music), the answer is a definite [url=http://en.wikipedia.org/wiki/Harmonic_series]yes![/url]

I said usually, but you're an exception.  It's not my fault you're just uber smart... But most people probably wouldn't, unless they have a strong grasp of the concept.  And are you a middle school student, because you're too genius for us T.T.

[quote="t0rajir0u"]Well, what you said is probably true.  Most students are taught geometric sequences in a purely algebraic way, and the connection between geometric sequences and geometry is irrelevant to their algebraic study.  It's unfortunate.

No, I'm not a middle-school student :)  I only post here to clarify misconceptions (especially about $ \infty$).[/quote]

lol. apparently, us middleschoolers, have a lot of misconceptions about that. darn you $ \infty$!
 :D

What is a good source to learn about $ \infty$ in detail?


Gauss
geometry
incenter
conics
hyperbola
geometry theorems

Let a quadrilateral $ABCD$ be circumscribed about a circle with center $O$. Denote by $M$ and $N$ the midpoints of the segments $AC$ and $BD$, respectively. Prove that the points $M$, $O$, $N$ are collinear.

In other words: If $ABCD$ is a circumscribed quadrilateral, then the Gauss line of this quadrilateral (this is the line through the midpoints of its diagonals $AC$ and $BD$) passes through the incenter of the quadrilateral $ABCD$.

Here’s a nice justification for it: if $ABCD$ is our quadrilateral and $I$ is the incenter, then $A,C$ lie on a hyperbola (on the same branch, actually) having $B,D$ as foci, and $IA,IC$ are tangents to the hyperbola. This brings the following problem to our attention (which has Newton’s Theorem as a corolary):

Given a conic $\mathcal C$ with center $O$, if $P$ is a point and $PU,PV$ are tangents to the conic, then $P,O$, and the midpoint of $UV$ are collinear (in our case the midpoint of $AC$ was the center of the hyperbola).

This problem is easy to solve: the center of the conic is the pole of the line at infinity, so what we must show is that the polars of $P$ and the midpoint $M$ of $UV$ concur on the line at infinity, i.e. are parallel. The polar of $P$ is $UV$, of course, and the polar of $M$ is the line $\ell$ passing through $P$ s.t. $(PU,PV;PM,\ell)=-1$. It’s clear that it must be $\|UV$.

[url]http://www.cut-the-knot.org/Curriculum/Geometry/NewtonTheorem.shtml[/url]

The proof is nice, but it's pretty difficult to think of if you don't know Léon Anne's Theorem.

I have an elementary solution :
 It's easy to show that : $ S_{OAB} + S_{OCD} = S_{OBC} + S_{OAD} = \frac{1}{2}S $.
 Assume $ AB  \cap CD = E $.
 Take $ G \in AE ; H \in ED $ such that : $EG = AB$; $EH = CD$.
 $ S_{OHG} = \frac{1}{2}S - S_{EGH} $.
 $ S_{NAB} + S_{NCD} = S_{MAB} + S_{MCD} = \frac{1}{2}S $.
 So $ S_{MHG} = S_{KHG} $.
 Hence M ,O , N are collinear .

I didn't read any of this stuff carefully, but if I'm not mistakes, the link posted by mecrazywong contains a proof which could be considered to be precisely this one :).


trigonometry inequalities AMC AIME geometry USA(J)MO USAMO \/closed

I'm wondering which classes I should take to go along with the Independent Study A. I'm not quite sure which two to choose between Intermediate Counting, Intermediate Trigonometry, Olympiad Inequalities, and Olympiad Geometry. I have debate tournaments during Tuesdays, Fridays and Saturdays, which would conflict with perhaps six of the trigonometry classes. At the same time, I feel that trigonometry/complex numbers are my weakest point- something that I hope I can shore up through the classes. For reference, I scored a 137 on the AMC 10 and a 5 on the AIME last year, and would therefore probably be best suited for an intermediate level. My question is would I be best suited for taking the two intermediate classes and  using the class transcripts/forums to piece together what I miss, or would I be better off taking an Olympiad class instead? Thanks for the advice.

My one comment is that I THINK Intermediate Trigonometry would probably help you learn concepts that you could apply to Olympiad Geometry, so I would probably take those two classes in that order.

What my son is doing in Intermediate Independent Study A is taking the two courses LEAST like what he will learn in his curricular math courses this school year.

I'll be interested to see what the course teachers suggest in answer to your question.

I'd suggest using the post-tests as a guide.  Also, use your goals this year to help choose.  The two olympiad courses you mention are much higher level than the two intermediate courses.  If your target is getting to the USAMO, and you're not highly confident already of reaching that goal, you'll be better off with the Intermediates (particularly if you can't currently answer more than ~30-45% of the posttest).

As for missing classes, many of our students follow along with the transcripts rather than attending classes, due to conflicts or being on the other side of the world.  As long as you are disciplined and don't allow yourself to fall behind, you should be fine.

For your reference, I would equate the two intermediate courses in terms of difficulty.  The Inequalities classes are a bit harder, and the problems (on the midterm and final) are much harder.  The Geometry is currently the most challenging class we offer (we had several USAMO honorable mention & winners in the class last spring and nearly all rated the class highly challenging).


A 0.493 kg block of wood hangs from the ceiling by a string, and a 0.0680 kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 5.77 m/s. The wad of putty sticks to the block. How high does the putty-block-system rise above the original position of the block?

Using the law of conservation of momentum to find the velocity of the two mass right after collision

$m_{1}v_{0}=(m_{1}+m_{2})v \Longrightarrow v = \frac{m_{1}v_{0}}{m_{1}+m_{2}}$

The using the law of conseravation of energy to find the answer

$\frac{1}{2}(m_{1}+m_{2})v^{2}= (m_{1}+m_{2})gh \Longrightarrow h = \frac{v^{2}}{2g}= \frac{(m_{1}v_{0})^{2}}{2(m_{1}+m_{2})^{2}g}$

In this problems, $m_{1}= 0.0680 kg, m_{2}= 0.493, v_{0}= 5.77 m/s$


Here's a challenging problem; it took me a long time to solve and I'd be interested in different solutions:

$n$ people participate in Secret Santa one year (they line up in a single file, then randomly select someone else's name out of a box and gives that person small gifts for a week or so; they cannot choose their own names: assume that that never happens, or that if it does, some mechanism is in place to respond to it, e.g. the person puts it back, and if the last person has to pick his or her name, everything starts over). What is the probability that at least one two-cycle (where a person selects the same person who selects him or her) exists?

Sorry if that explanation's too long, but here's an example: let's say Jack, Jane, Joe and John participate in Secret Santa. Here's one possible outcome (Jack picks first, Joe second, John third, and Jane fourth):
Jack picks Jane
Joe picks John
John picks Joe
Jane picks Jack
so there are two two-cycles.

What is the answer for $n=4$? How about $n=5$? Can you generalize?

Someone should give it a shot!  Think derangements and derangements with at least one two-cycle.  For example, for $n=4$, there are 9 (!4; 4 subfactorial) ways for the individuals in Secret Santa to choose each other, and 3 of them involve two-cycles.  The probability is therefore...

Ok, I didn't calculate the numerical thing but this is how I will solve it (I don't have time to work it out).

[hide]
so the total possibility will be the derangement, lets say it's $D_n$.  So instead of counting all the nasty two-cycles, we count the complement, i.e. no two-cycle exist.  This means that, after some thinking, if all the people were nodes in a graph, and edges means that one person is giving a gift to the other, then there will be exactly one cycle connecting the entire graph together.  So we simply have to permutate the string "ABCDE...", because each corresponds to exactly one cycle.  so there are $n!$ permutations.  However, the string $ABCDA$ and $BCDAB$ (the first and the last node are the same to denote cycle) are virtually the same, so there are $(n-1)!$ cycles possible.  So the answer should be $\frac{D_n - (n-1)!}{D_n}$.
[/hide]

am I right?

You're on the right track!  But remember, even if there aren't any two-cycles, there might be three-cycles, or four cycles, etc., so there are more than $(n-1)!$ derangements without a two-cycle.  For example, ABCDEFA and ABCA DEFD are both possible derangements without a two-cycle.

so can the answer be:

[hide]
$\frac{{n \choose 2} * D_{n-2} - 3*1*{n \choose 4}*D_{n-4} + 5*3*1*{n \choose 6} * D_{n-6} - \dots}{D_{n}}$

Because you choose 2 people to have two-cycles, and the rest can permutate anyway
(because it'll be like two separate graph, 2 nodes connected to each other, and the rest of the nodes linked in any fashion), but you over counted and we use PIE to reduce the overcounting.  And we define $D_0 = 1$ and $D_a = 0$ for $a<0$.  The product of odd consecutive integers is to account the number of ways to have two two-cycles in 4 people, three two-cycles in 6 people, etc.
[/hide]

It seems logical...??? (sorry but I don't have paper and pencils with me to work it out, neither do I have the ability to simplify the answer)

Your method seems to generate the number of derangements with at least one two-cycle, but I can't say I really follow your logic.  Could you explain it a little bit more please?

There are nicer formulas for the number of derangements with at least one two-cycle, but I think yours works.  The only question that I have for your approach (other than a fuller explanation) is whether each derangement is equally likely.  You've written the number of derangements with at least one two-cycle divided by the total number of derangements, which would work if each derangement is equally likely.  Are they?

I guess I will try to explain mine better:

So first, there are $n \choose 2$ way to choose two people out of everyone, and then we can say these two people form a two-cycle.  Then, there are $D_{n-2}$ way for the rest of the people to give gifts to each other.  But lets say there are two two-cycles, then this method will count it twice, so we have to subtract the over counting.  For two two-cycles, you can have $n \choose 4$ way to choose 4 people out of everyone, and $3*1$ way to arrange two two-cycles between these four.  The rest of the $n-4$ people can be arranged in $D_{n-4}$ ways.  But then we have subtracted out the counting for three two-cycles, so we add it back.  Doing this alternatively gives the nasty formula I have.

As for your question, since We assume we choose the gifting arrangement randomly, I think each derangement is equally likely.

[quote="kokosai"]OK, I understand now, that makes sense, thanks for the explanation.

It actually turns out that the single file selection method makes it so that each derangement is not equally likely.  Each individual selection is random (although no individual can choose him or herself; assume that if that happens, it all starts over again, or a mechanism is in place to resolve it without changing the probabilities), but the fact that they're choosing in a single file causes some bias.  If we randomly chose a derangement for the gift-giving, that would make each derangement equally likely, but the single file makes it so that it's not.  Surprising, I think...

I think the actual single-file probability is asymptotic to the non-single-file (i.e., each derangement is equally likey) probability, but I don't have an actual proof.  The actual single-file probability for $n=4$ is $\frac{5}{18}$.  Anyone have any ideas on how to prove this rigorously?[/quote]

wow, now I'm confused...can you explain what not all permutations are likely?

OK, let's say that $ABCDA$.  There are then 24 (4!) arrangements (orders) for A, B, C, and D to pick in.  For example, let's say A picks first, B picks second, C picks third and D picks fourth ("->" means "chooses"):
A -> B (Probability: 1/4 that A picked first, 1/3 that A picked B)
B -> C (Probability: 1/3 that B picked second, 1/3 that B picked D)
C -> D (Probability: 1/2 that C picked third, 1 that C picked D)
D -> A (Probability: 1 that D picked fourth, 1 that D picked A)
Note that I'm using conditional probabilities, since it's a single file; for example, there is a 1/3 probability after A picked first that B would then pick second.  Also, no individual can choose him or herself, so C must choose D, because otherwise D would choose him or herself (and everything would restart).  The probability of the arrangement is therefore 1/216.  If you go through all the arrangements, you'll find that the average probability of all the arrangements of $ABCDA$ is $\frac{13}{108}$, and since there are six permutations that are equivalent to $ABCDA$, the probability of a four-cycle is $6*\frac{13}{108}=\frac{13}{18}$.

If you want to do this on your own, note that since you'll dealing with a single four-cycle, you can just assume that A picks first.  You'll then only get six arrangements, but if you did the twenty-four arrangements, each of the six would be repeated four times.


number theory
greatest common divisor
number theory proposed

Let N be an integer $1000 \leq N \leq 9999$, and let W be the numer obtained by writing the digits of N in inverse order.
Finally, let $\displaystyle d_n=gcd(N,W,|N-W|)$ .

For example, if N=1209, then W=9021, and $d_n=93$

Find the maximum of $d_n$

Bye!

[b]Lemma[/b]:  $gcd(n, 0) = n$

Proof:  $gcd(n, n) = gcd(n, n-n) = gcd(n, 0) = n$

Now, $gcd(9999, 9999, 0) = 9999$ is the highest value.  $d_n$ cannot take on a higher value because $d_n | n$ and $n \le 9999$.  QED.

(Is there something wrong with the problem statement, or are you defining $gcd(n, 0) = 0$? )

Yes, I forgot. $gcd(n,0)=0$...
Sorry!

Now, the problem is a bit more challenging...

Bye!


Find all positive integers n for which
$ \cos (\pi \sqrt {n^{2}\plus{}n})\ge 0$

_______________________________________
Azerbaijan Land of the Fire  :P

Put $ \sqrt{n^2\plus{}n}\equal{}2k\plus{}\delta$ where $ k\in\mathbb{N},\delta\in\mathbb{R},0\leqslant\delta<2$. Then by the given condition, $ \delta\in\left[0,{1\over 2}\right]\cup\left[{3\over 2},2\right)$. That also can be written as

$ \left\{{\sqrt{n^2\plus{}n}\over 2}\right\}\in\left[0,{1\over 4}\right]\cup\left[{3\over 4},1\right)\qquad (*)$

It is easily shown that $ n\minus{}{1\over 2}<\sqrt{n^2\plus{}n}<n\plus{}{1\over 2}$ for natural $ n$. Hence

$ {n\over 2}\minus{}{1\over 4}<{\sqrt{n^2\plus{}n}\over 2}<{n\over 2}\plus{}{1\over 4}$

[b]Case 1.[/b] $ n\equal{}2M, M\in\mathbb{N}$. Then we get $ M\minus{}{1\over 4}<{\sqrt{n^2\plus{}n}\over 2}<M\plus{}{1\over 4}$. Obviously, the fractional part of the middle expression satisfies the condition $ (*)$

[b]Case 2.[/b] $ n\equal{}2M\minus{}1, M\in\mathbb{N}$. Then we get $ M\minus{}{3\over 4}<{\sqrt{n^2\plus{}n}\over 2}<M\minus{}{1\over 4}$. Obviously, the fractional part of the middle expression does not satisfy the condition $ (*)$.

Hence we conclude that [b]all even[/b] $ n$ satisfy the initial condition.


Prove that $6^{n}-1$ does not divide $7^{n}-1$.

Assume the contrary that $6^{n}-1|7^{n}-1$ then notice that since $5|6^{n}-1$ so we must have $5|7^{n}-1$

so $2^{n}\equiv 1\mod 5$ gives us $n=4k$

hence $7|6^{4k}-1$ but does not divide $7^{4k}-1$


1. The roots of $ax^{2}+bx+c = 0$ are irrational, but their calculator approximations are $0.8430703308$ and $-0.5930703308$.  If $a$, $b$, and $c$ are integers whose greatest common divisor is 1 and which satisfy $a>0$, $|b|\leq 10$ and $|c|\leq 10$, compute the ordered triple $(a,b,c)$.

2. Given the four-digit base 10 number $ABCD$ with $A \neq 0$, let the palindromic distance $L$ be defined by $L = |A-D|+|B-C|$.  Compute the number of positive four-digit numbers where $L=1$.

[hide="1"] The roots add up to $0.25$.  This suggests that $a = 4, b =-1$.  The product of the roots is somewhere around $-0.5$, which suggests that $c =-2$.  Hence our triple is $\boxed{ (4,-1,-2) }$. [/hide]
[hide="2"] [b]Case:[/b]  $A, D$ differ by $1$, and $B = C$.  There are $10$ possibilities for $B$, while for $A = 1, ... 8$ there are $2$ possibilities for $D$ and for $A = 9$ there is only one.  Then there are $10 \cdot 17 = 170$ such numbers in this case.

[b]Case:[/b]  $B, C$ differ by $1$, and $A = D$.  There are $9$ possibilities for $A$, while for $B = 1, ... 8$ there are $2$ possibilities for $C$ and for $B = 0, 9$ there is only one.  Then there are $9 \cdot 18 = 162$ such numbers in this case.

In total there are $\boxed{ 332 }$ such numbers. [/hide]


geometry
geometric transformation
induction
vector
linear algebra
matrix
graph theory

each student in a class has 0,1,2,3,4,5 or 6 candies.at each step the teacher chooses some students and gives one candy to these students and to each of the friends of any of these students.When a student has 7 candies he/she eats all of them. there is a third student for every pair of students such that the third one is the friend of exactly one of them.Prove that the teacher can make each student have any number of candies independent from the number of candies a student has at the beginning

you mean that the teacher is able to make the students have $(a_1,a_2...a_n)$ candies for every such a n-tuplet.
I realise u can't say this. I don't understand. And if you give a candy to 2 friends, they will both receive 2 candies each?

i mean what you've understood.why can't i say this and why is giving two candies to two friends is a problem.The exam test was  Turkish and i translated it.There may be some problems about the translation but i think there is not.

it says one candy to each of those students;i think this means only one candy to each student from the set of those students

I think I got it know, but still I want to make sure that what is requested is to show that the teacher can reach every n-tuplet, 'cause you haven't said anything about this.

I think I can solve, if I understand correctly. Let me reformulate what I understood in graph theory language:
We are given a graph $G$ where every vertex has a label in $Z_7$. For any set $S$ of vertices denote $F(S)$ bet the set $S$ in union with all vertices which have at least one neighbour in $S$. We are allowed to increase by $1$ the label of every vertex in $F(S)$ for any $S$. It's known that for every two vertices there is a third vertex acquainted to exactly one of them. Prove we can reach every labelling.

Proof: Let's prove for any pairwise dictinct vertices $u_1,\cdots u_m, v_1,\cdots, v_n$ we can increment by $1$ the labels exactly on vertices from $F(u_1,\cdots,u_m) / F(v_1,\cdots,v_n)$.
Proof is by induction on $m+n$.
For $m=0$ or $n=0$ we have nothing to prove.
For $m=1,n=1$ we change $F(v_1)$ $6$ times, then $F(u_1,v_1)$ to get the result.
Now induction step:
If $n>1$ then we change $F(u_1,\cdots,u_m)/F(v_1,\cdots,v_{n-1})$ and $F(u_1,\cdots,u_m)/F(v_2,\cdots,v_{n})$ (one time) and then change $F(u_1,\cdots,u_m)/F(v_2,\cdots,v_{n-1})$ $6$ times.
If $n=1$ then $m>1$ and hence we change
$F(u_1,\cdots,u_{m-1})/(F(v_1,\cdots,v_n)$ and $F(u_2,\cdots,u_{m})/(F(v_1,\cdots,v_n)$ (one time) and then change
$F(u_2,\cdots,u_{m-1})/(F(v_1,\cdots,v_n)$ $6$ times.
The induction is finished.
Now for any vertex $V$ with neighbours $u_1,\cdots,u_m$ and non-neighbours $v_1,\cdots,v_n$ we note that
$F(u_1,\cdots,u_m)/F(v_1,v_2,\cdots,v_n)=\{V\}$ (indeed, if a second vertex $U$ belongs to it, then there is no third vertex connected to exactly one of $U$ and $V$).
Thus we can change every vertex independently on others and hence every labelling is possible.

I dont know if this works, but what about this: every process can be identified with an n-tuple $(a_1,a_2...,a_n)$ such that $a_i$ is the number of times student $i$ is choosen in the initial set. Obviously the important is to look this n-tuple modulo 7. Now there are $7^n$ n-tuples, and there are $7^n$ posible configurations of numbers of candies, so we can try to prove that different n-tuples give different numbers of candies, or something like that, it would be very nice.

I have some ideas I want to share and i know someone very good in linear algebra can finish:

for every kid $i$, define a vector $v_i=(t_1,t_2...,t_n)$ where $t_j=1$ iff $i$ knows $j$ (including i=j) and $t_j=0$ otherwise.

Now, let $A=(a_1,a_2...,a_n)$ be defined as in my previous post and let $f_i$ be the number of candies kid $i$ has. Obviously, everything is modulo 7. We must have:

$f_i=A.v_i$ for $i=1,2,3...,n$ so we get that $A$ is unique, given the $f_i$'s, if the vectors $v_1,v_2...,v_n$ are lineary independent. This is what i have until now. I think here we must use the other condition, that for every 2 guys there is a third one aquainted with exactly one of them.

Actually, Pascual, I think you are wrong because we can also take entire sets of students and the sets are quite many, $2^n - 1$. And they are some kind of vectors that represent the students that will receive candies. Now, among all these $2^n - 1$ sets/vectors there exists $n$ lineary independent, otherwise the system, as you wrote, will have the $det(A) = 0$.
I hope that from the fact that among any $n$ of them, one of the sets can be represented as the sum of others in $\mathbb Z_7$ should follow a contradiction, but I think it's hard.

I dont know, there are some conditions you are ignoring, for example consider the matrix with the vectors as rows. it is simetric, and also it has an extra condition: for every two guys there is a third one aquainted with exactly one of them!! I think this makes its determinant be nonzero!

I think you are wrong Pascual. That's because you are trying to prove a stronger assertion: That one can reach any labelling by taking ONE student and giving candies to him and each of his friends. However, the problem text says you can take MORE students and give candies to them and each boy who is friends to each of them.
To your statement a counter-example can be easily found:
$11100000001$
$11010000001$
$10101000001$
$01010100001$
$00101010001$
$00010101001$
$00001010101$
$00000101011$
$00000010101$
$00000001011$
$11111111111$
The last row is the sum of other rows multiplied by $5$ in $Z_7$ and you see the condition is fulfilled.[/b]


Let $ \left(a_{n} \right)_{n \equal{} 1}^{\infty }$ be a sequence on real numbers such that $ a{}_{n \plus{} 1} \equal{} a_{n} a_{n \plus{} 2}$ for every $ n\ge 1$. The number of elements in the set $ \left\{a_{n} : n\ge 1\right\}$ cannot be

$\textbf{(A)}\ 2   \qquad\textbf{(B)}\ 3   \qquad\textbf{(C)}\ 4   \qquad\textbf{(D)}\ 5   \qquad\textbf{(E)}\ \text{None}$

By calculating first few values of $ a_n$, we find that it's periodic with period $ 6$, and that the value set is $ \left\{a_1,a_2,{1\over a_1},{1\over a_2},{a_1\over a_2},{a_2\over a_1}\right\}$. Hence there can't be $ 5$ different values, so the answer is $ D$.

no it is not. from $ a_1\equal{}a$ and $ a_2\equal{}a$, we get $ \{a,\minus{}a,\minus{}1,1/a,\minus{}1/a,\minus{}1\}$. So $ 5$ can be got.

I think that it is E-none.

For 2: put $ a_1\equal{}1; a_2\equal{}\minus{}1$
For 3: put $ a_1\equal{}a_2$ ,not equal to 1
For 4: put $ a_1\equal{}a_2^2$ or  $ a_1\equal{}\frac{1}{a_2}$
For 5: previous post.


Λοιπόν μιας και μπηκαμε στην τελευταία εβδομάδα του Προκριματικού Διαγωνισμού παραθέτω τα θέματα του 2003.

[color=red]1[/color] Να αποδείξετε ότι η εξίσωση $x^2+y^2+z^2=x+y+z+1$
δεν έχει λύση στους ρητούς αριθμούς .

[color=darkblue]2[/color] Δίνεται τρίγωνο ΑΒC με  $BC=CA+\frac{AB}{2}$. Θεωρούμε σημείο $P$ πάνω στην ΑΒ ώστε  $\frac{BP}{PA}=\frac{1}{3}$.
Nα δείξετε οτι $\widehat{ CAP}=2\widehat{CPA}$

 [color=yellow]3[/color] Έστω $a,b,c\geq 0$ με $a+b+c=3$ .Να αποδείξετε ότι
$\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\geq\frac{3}{2}$

[color=green]4[/color]Θεωρούμε ένα σύστημα μεταλλικών σφαιρών απείρου πλήθους με κέντρα σε όλα τα σημεία . Λέμε ότι το σύστημα είναι ευσταθές, αν η θερμοκρασία κάθε σφαίρας ισούται με το μέσο όρο των θερμοκρασιών των  πιο γειτονικών της σφαιρών. Υποθέτουμε ότι η θερμοκρασία κάθε σφαίρας είναι μεταξύ των 0 και 1 βαθμών. Αν το σύστημα των σφαιρών είναι ευσταθές, να αποδείξετε ότι όλες οι σφαίρες έχουν την ίδια θερμοκρασία.

 :lol: Χαίρετε[/i]

Σόρρυ που βάζω δεύτερο συνεχόμενο ποστ αλλά θέλω να σας ανακοινώσω ότι έχω( μου τα έδωσε πριν από λίγο ένας πολύ καλός φίλος ) όλα τα θέματα ΜΕΣΟΓΕΙΑΚΩΝ ΔΙΑΓΩΝΙΣΜΩΝ από το 1998 μέχρι και το 2005 .
Πείτε μου από ποια χρονιά θέλετε να βάλλω τα θέματα .

Σιλουανέ,

   σκανάρισέ τα και βάλτα στο φόρουμ ,αν γίνεται.Θα κερδίσεις κόπο να τα γράφεις.

  Μπάμπης

Δεν μου παίρνει και πολύ χρόνο . Αλλά έτσι και αλλιώς τα έχω σε pdf

ΣιΙΛΟΥΑΝΕ ΠΕΣ ΜΟΥ ΣΕ ΠΑΡΑΚΑΛΩ ΑΜΕΣΑ ΠΟΤΕ ΕΙΝΑΙ ΟΙ ΠΡΟΚΡΙΜΑΤΙΚΟΙ?
ΣΕ ΕΥΧΑΡΙΣΤΩ ΚΑΙ ΧΑΙΡΕΤΩ

ΛΥΣΕΙΣ ΑΣΚΗΣΕΩΝ ΟΣΕΣ ΕΧΩ ΛΥΣΕΙ  :lol:
1 ΑΚΡΙΒΩΣ Η ΙΔΙΑ ΜΕΘΟΔΟΛΟΓΙΑ ΜΕ ΑΡΧΙΜΗΔΗ 2 ΦΕΤΙΝΟ
3 ΕΙΝΑΙ ΓΝΩΣΤΟ ΟΤΙ
Β^2+1=<(Β+1)^2 ΚΙ ΚΥΚΛΙΚΑ
ΑΡΑ ΑΡΚΕΙ ΝΔΟ
Α/(Β+1)^2+ Β/(Γ+1)^2+ Γ/(Α+1)^2>=3/2  (1)
ΑΠΟ B-C-S ΛΑΜΒΑΝΟΥΜΕ  ΟΤΙ ΓΙΑ ΝΑ ΙΣΧΥΕΙ Η (1) ΠΡΕΠΕΙ ΚΑΙ ΑΡΚΕΙ
6>=ΑΒ+ΒΓ+ΓΑ
ΑΝ ΤΩΡΑ ΑΒ+ΒΓ+ΓΑ>6  2*(ΑΒ+ΒΓ+ΓΑ)>12 (2)
(3) Α^2+Β^2+Γ^2>=0 (3)
(2)+(3)
(Α+Β+Γ)^2>=12
9>=12 ΑΤΟΠΟ ΑΡΑ ΙΣΧΥΕΙ ΤΟ ΖΗΤΟΥΜΕΝΟ

[quote="ambros"]ΛΥΣΕΙΣ ΑΣΚΗΣΕΩΝ ΟΣΕΣ ΕΧΩ ΛΥΣΕΙ  :lol:
1 ΑΚΡΙΒΩΣ Η ΙΔΙΑ ΜΕΘΟΔΟΛΟΓΙΑ ΜΕ ΑΡΧΙΜΗΔΗ 2 ΦΕΤΙΝΟ
3 ΕΙΝΑΙ ΓΝΩΣΤΟ ΟΤΙ
Β^2+1=<(Β+1)^2 ΚΙ ΚΥΚΛΙΚΑ
ΑΡΑ ΑΡΚΕΙ ΝΔΟ
Α/(Β+1)^2+ Β/(Γ+1)^2+ Γ/(Α+1)^2>=3/2   Η ΑΚΟΜΑ ΠΙΟ ΙΣΧΥΡΑ >=3(1)
ΑΠΟ B-C-S ΛΑΜΒΑΝΟΥΜΕ  ΟΤΙ ΓΙΑ ΝΑ ΙΣΧΥΕΙ Η (1) ΠΡΕΠΕΙ ΚΑΙ ΑΡΚΕΙ
6>=ΑΒ+ΒΓ+ΓΑ
ΑΝ ΤΩΡΑ ΑΒ+ΒΓ+ΓΑ>6  2*(ΑΒ+ΒΓ+ΓΑ)>12 (2)
(3) Α^2+Β^2+Γ^2>=0 (3)
(2)+(3)
(Α+Β+Γ)^2>=12
9>=12 ΑΤΟΠΟ ΑΡΑ ΙΣΧΥΕΙ ΤΟ ΖΗΤΟΥΜΕΝΟ[/quote]

Η λύση σου στην ανισότητα είναι λάθος. Καταλήγεις σε άτοπο γιατί χρησιμοποιείς πολύ ανίσχυρες ανισότητες .Αν λυνόταν ετσι δεν θα έμπαινε σε Μεσογειακό αλλά στον Θαλή ή τον Ευκλείδη .όσο για την πρώτη αν δε βάλλεις πλήρης λύση δεν μπορείς να λες ότι τη λύνω σαν την 2η του φετινού Αρχιμήδη .Αλλιώς όλες τις ασκήσεις θα τις λύναμε έτσι.
Θα λέγαμε:'' Αυτή μοιάζει με την τάδε άσκηση από τον τάδε διαγωνισμό άρα λύθηκε :lol: ''
Δεν κάνω σε καμιά περίπτωση τον έξυπνο .Αν κάτι δεν το έγραψα καλά συγχωρήστε με.

ΦΙΛΕ ΜΟΥ ΣΙΛΟΥΑΝΕ ΜΕ ΤΗΝ ΙΔΙΑ ΛΟΓΙΚΗ Η ΛΙΓΟ ΧΑΖΗ ΛΥΣΗ ΣΟΥ ΟΠΩΣ ΤΗΝ ΑΠΟΚΑΛΕΣΕΣ ΔΕΝ ΕΙΝΑΙ ΛΥΣΗ
ΤΗ ΑΝΙΣΟΤΗΤΑ ΤΗΝ ΑΠΕΔΕΙΞΑ ΜΕ ΜΙΑ ΑΡΚΕΤΑ ΙΣΧΥΡΗ ΣΧΕΣΗ ΚΑΙ ΑΝ ΤΝ ΔΕΙΣ ΛΟΓΙΚΑ ΚΑΛΥΠΤΕΙ Η ΟΛΗ ΔΙΑΔΙΚΑΣΙΑ ΟΛΟΥΣ ΤΟΥΣ ΑΡΙΘΜΟΣ ΠΟΥ ΙΚΑΝΟΠΟΙΟΥΝ ΤΗΝ ΣΥΝΘΗΚΗ ΔΕΝ ΜΠΟΡΕΙ ΝΑ ΛΕΣ ΜΙΑ ΛΥΣΗ ΛΑΘΟΣ ΕΠΕΙΔΗ ΕΙΝΑΙ ΠΟΛΥ ΧΑΖΗ Η ΔΕΝ ΜΟΙΑΖΕΙ ΠΕΡΙΠΛΟΚΗ  ;)  ΑΛΛΑ ΜΟΝΟ ΑΝ ΕΙΝΑΙ ΛΟΓΙΚΗ ΟΚ? :D
ΧΑΙΡΕΤΩ

Σόρρυ Αμβρόσιε δεν είπα ότι η λύση συ είναι χαζή αλλά ότι είναι λάθος .
Παρακαλώ και τους υπόλοιπους έλληνες mathlinkers να επιβεβαιώσουν την ορθότητα ή όχι της λύσης. Δείξε μου για ποιους αριθμούς χρησιμοποίησες την B-C-S για να δω αν έχω πράγματι δίκιο.Νομίζω ότι εκεί κάνεις το λάθος. Πάντως ωραία η χρήση του άτοπου στο τέλος. Ευχαριστώ

ΣΙΛΟΥΑΝΕ ΜΑ ΟΥΤΕ ΕΓΩ ΕΙΠΑ ΟΤΙ ΕΙΠΕΣ ΤΗΝ ΛΥΣΗ ΜΟΥ ΑΛΛΑ ΤΗΝ ΔΙΚΗ ΣΟΥ
B-C-S
(Α/(Β+1)^2+ Β/(Γ+1)^2+ Γ/(Α+1)^2)(A+B+Γ)>=(Α/(Β+1)+ Β/(Γ+1)+ Γ/(Α+1))^2(ΝΔΟ)>=9/2
Η 3/2 ΑΠΟ ΤΗΝ ΣΤΙΓΜΗ ΠΟΥ ΕΧΟΥΜΕ Α+Β+Γ=3 ΒΕΒΑΙΑ ΠΙΟ ΙΣΧΥΡΑ ΑΡΚΕΙ ΝΑ ΔΕΙΞΩ Α/(Β+1)+ Β/(Γ+1)+ Γ/(Α+1)>=3/2 Η ΠΙΟ ΙΣΧΥΡΑ ΜΕ ANDRESCU ΚΑΙ ΜΕ ΚΑΤΑΛΛΗΛΕΣ ΜΕΤΑΤΡΟΠΕΣ ΤΗΝ ΑΝΙΣΟΤΗΤΑ ΠΟΥ ΠΡΕΠΕΙ ΚΑΙ ΑΡΚΕΙ 6>=ΑΒ+ΒΓ+ΓΑ ΤΩΡΑ ΓΙΑ  ΑΥΤΗΝ ΤΗΝ ΑΝΙΣΟΤΗΤΑ ΤΗΝ ΑΦΗΝΩ ΣΕ ΣΕΝΑ ΝΑ ΤΗΝ ΒΡΕΙΣ ΚΑΘΩΣ ΔΕΝ ΤΗΝ ΘΥΜΑΜΑΙ ΟΚ? ΣΟΡΡΥ
ΑΛΛΑ ΕΙΝΑΙ ΣΙΓΟΥΡΑ ΣΩΣΤΗ  :D
ΧΑΙΡΕΤΩ

Είναι

   $\frac{a}{b^2+1} = a-\frac{ab^2}{b^2+1} \ge a- \frac{ab^2}{2b} = a- \frac{ab}{2}$

Προσθέτουμε τις τρεις αυτές σχέσεις.Όμως:

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

  κλπ. Η ισότητα ισχύει μόνο αν $a=b=c = 1$

  [color=red]ΔΕΣ και << Στεργίου - Σκομπρής, Αλγεβρικές Ανισότητες , άσκηση 3.154>>[/color]

[u] Μπάμπης[/u]

Αμβρόσιε έγραψες
[quote="ambros"]
ΑΡΑ ΑΡΚΕΙ ΝΔΟ
Α/(Β+1)^2+ Β/(Γ+1)^2+ Γ/(Α+1)^2>=3/2  (1)
[/quote]

όμως για $a=b=c=1$ η (1) δεν ισχύει άρα την ελλάτωσες πολύ και δεν αποδεικνύεται με αυτόν τον τρόπο άρα σου απέδειξα ότι κάνεις λάθος.

[quote="ambros"]ΟΚ ΣΙΛΟΥΑΝΕ ΜΟΝΟ  ΒΑΛΕ ΣΤΗΝ ΑΝΙΣΟΤΗΤΑ >=3 ΚΑΙ ΘΑ ΒΓΕΙ ΑΜΑ ΤΟ ΔΕΙΣ ΚΑΛΑ ΟΚ?
ΧΑΙΡΕΤΩ[/quote]

μα γιατί να βάλλω >=3 αφού πρέπει να είναι >=3/2.
Η λύση σου είναι λάθος σόρρυ ψάξε να βρεις κάτι καλύτερο

ΓΙΑ ΝΑ ΙΣΧΥΡΟΠΟΙΗΣΩ ΤΗΝ ΑΝΙΣΟΤΗΤΑ ΑΦΟΥ >=3>=3/2
ΔΕΣ ΚΑΙ ΒΓΑΙΝΕΙ ΕΔΩ ΔΕΝ ΠΡΕΠΕΙ ΝΑ ΕΠΙΜΕΝΕΙΣ ΠΡΙΝ ΤΟ ΔΕΙΣ
ΟΚ?
ΧΑΙΡΕΤΑΙ

δηλαδή μου λες ότι αρκεί να δείξεις ότι
ΑΡΑ ΑΡΚΕΙ ΝΔΟ
Α/(Β+1)^2+ Β/(Γ+1)^2+ Γ/(Α+1)^2>=3  (1). Σωστά ???

όμως και πάλι τότε για  για $a=b=c=1$ η (1) δεν ισχύει άρα την ελλάτωσες πολύ περισσότερο . :lol:
Κατάλαβες ?Δοκίμασε κάτι άλλο σου είπα

ΔΕΝ ΜΠΟΡΕΙ ΡΕ ΣΙΛΟΥΑΝΕ ΚΑΠΟΙΟΣ ΑΡΙΘΜΟΣ ΘΑ ΠΡΕΠΕΙ ΝΑ ΙΣΧΥΕΙ ΩΣΤΕ ΝΑ ΑΡΚΕΙ ΝΔΟ ΟΤΙ Η ΠΑΡΑΣΤΑΣΗ Η ΓΝΩΣΤΗ >=Χ ΑΥΤΟΥ ΜΗΠΩΣ ΜΠΟΡΕΙΣ ΝΑ ΤΟ ΒΡΕΙΣ ? ΔΙΟΤΙ ΑΣΧΟΛΟΥΜΑΙ ΜΕ ΑΛΛΟ ΠΡΟΒΛΗΜΑ ΑΥΤΗΝ ΤΗ ΣΤΙΓΜΗ/ ΘΑ ΜΕ ΥΠΟΧΡΕΩΣΕΙΣ ΕΥΧΑΡΙΣΤΩ ΚΑΙ ΧΑΙΡΕΤΩ

[quote="silouan"][color=red]1[/color] Να αποδείξετε ότι η εξίσωση $x^2+y^2+z^2=x+y+z+1$
δεν έχει λύση στους ρητούς αριθμούς .[/quote] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=788347#p788347]εδώ[/url] κι [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1732937#p1732937]εδώ[/url]


So, the John O'Bryan Competition was today... here are the final five questions of the individual test.

1)
Find the sum of the solutions to $|x^2-2|=|2x-1|$

2)
Triangle $ABC$ satisfies $\angle BAC = \angle ABC + 30 ^\circ$ and $D$ lies on $BC$ such that $AC = CD$.  Find $\angle DAB$

3)
A list of consecutive integers starting with 1 is written on the blackboard.  One of the numbers is erased.  If the average of the remaining numbers is $35 \frac{6}{17}$, what number was erased?

4)
Some of you have read the delightful story Holes about a boy named Stanley who must dig holes.  Stanley digs at a constant rate and knows that he can dig to a depth equal to his height in exactly 3 hours.  $a$ hours after starting a new hole he finds that his head is below ground level a distance three times the distance his head was above the ground $b$ hours after starting.  If $a + b = 7 \frac 1 4$, find $a - b$.

5)
A group of $N$ people agree to play a game $N$ times.  The loser (no ties) of each game must give to each of the non-losers the amount of money that each non-loser currently has - that is, the non-losers will double their holdings.  Initially the players each have a different number of dollars (possibly fractions of dollars), and at the end of $N$ games they each wind up with $64$ dollars.  Sally lost only the third and fourth games and Joe lost only the sixth game.  If Sally had to pay Joe $73$ dollars at the end of game 4, then how many dollars did she start with?

Problem 1:
[hide]You can set them both positive or you can set one negative.  Both positive results in x^2-2x-1 which has a sum of 2.
One negative results in x^2+2x-3 which has a sum of -2 so the total sum is 0.[/hide]

Problem 3:  [hide]The fraction in the denominator is the biggest hint.  Since 17 is a fairly large prime, we know we're dividing by 17 or a multiple of 17 so the number of numbers is 17k+1.  Also, the fraction in improper form is (35*17k+6k)/17k.  This is 601k/17k, we don't want to cancel out the k because the number that existed when one number was erased is 601k.  Trying n(n+1)/2 for 17k+1, we remove 18,35,52.  At 69, however, we have it.  69*35=2415 and 601*4=2404 so the number removed was 11.[/hide]

[hide="Number 2"]
Note that $\angle ABC+\angle BAC=\angle ADC+\angle DAC(=180^\circ -\angle ACB)$.

$\angle ADC+\angle DAC=2\angle ADC=x+(x+30^\circ)=2x+30^\circ$

$\therefore \angle ADC=x+15^\circ$

Hence $\angle DAB=\angle ADC-\angle ABD= (x+15^\circ)-x=\boxed{15^\circ}$

[/hide]

Problem 4:
[hide]We know it takes Stanley 3 hours to dig a hole to his head.
Let d be how deep his head is in the dirt.  Let h be how tall he is.
b=3+3d/h
a=3+9d/h.
Adding that tells us that 12d/h+6=7+1/4 or 12d/h=5/4 so d/h=5/48 so a-b is 6d/h or 6(5)/48 or 5/8.[/hide]

Good solutions... I want to see one for 5, that is the one I did not get on the test.  All my work gave me major discrepancies in all my answers... weird.

Dengmi, your answer for 4 is wrong.  You're making it harder than it needs to be by using the h variable.

Oops, I didn't read it correctly.
[hide]a=3+3x where x is how much time it took him to dig a hole to the distance specified.
b=3-x since his head was above ground.
Since this equals 5/4=2x, x=5/8 so (3--1)=4x which is a-b=5/2[/hide]

Closer, but still wrong... You don't need a third variable.  Try writing an equation given what you know about a and b, using only a and b.

Alright, sorry I don't know how to hide, but the answer to #4 is this, I think

Since after 3 hours his head is even with the ground, the height above ground can be written as $3-b$ and since a is 3 times greater than b you can setup up this system of equations:
$a+b=7\frac{1}{4}$
$3(3-b)=a$

after solving you have $b=\frac{7}{8}$ and $a=\frac{51}{8}$

therefore $a-b=\frac{11}{2}$

thanks for the good bit of help on this one deej21, so props to you

[hide="Is number 5 this...?"] 4665/32[/hide]
I just did some stuff while I was half asleep so I hope I did it right.

Stiffler, I think you wrote down my help wrong because that is wrong.  Your first equation is close to what you need there, but not quite it.

FMako, that is not the correct answer.

By the way, to hide things, you just place them inside hide tags.  I disabled code in this post so it'll show up right for you.  So saying

[hide]
test
[/hide]

will encase the word "test" inside the hidden part.  If you want the hide-toggler thing to say something specific, you do this

[hide=click here to see the answer]
answer is here
[/hide]

Okay, last night after posting i think i realized my problem and it haunted me all night.  I believe it should be this system:
$a+b=7\frac{1}{4}$
$3(3-b)=a-3$
anyone agree?

Ok, using 3-b is a better idea than mine.  [hide]This is when his head is above ground.
However, a is when Stanley's head is 3(3-b) length below ground so a=3+3(3-b).
Adding these yields b+3+9-3b=7+1/4.  THis is -5+1/4=-2b.  This is -4-3/4=-2b or 2+3/8=b.  a=5/8*3+3 or 4+7/8.  We subtract a and b to get 2 and a half.[/hide]

Does the answer to problem 5 depend on N?  I'm getting that it does, but if it doesn't could you post your solution?


[b]Problem.[/b] [i]Suppose that $f: \mathbb{N}\rightarrow\mathbb{N}$ satisfies $f(1)=1$ and for all $n$,
\[\\\text{(a)}\ 3f(n)f(2n+1)=f(2n)(1+3f(n)),\\\text{(b)}\ f(2n)<6f(n).\]
Find all solutions to the equation $f(k)+f(m)=293$.[/i]


geometry
3D geometry
sphere
geometric transformation
rotation
vector
linear algebra

hi guys,

how can i show that if A is an isometry of sphere then A is a rotation???

Thanks,

Rosa

It's a product of two reflections. Now just visualise that and it should be obvious that it's a rotation around the common line of the two planes, with angle equal to twice the anle between the two planes.

I mean how to show that if A is orientation preserving then A is rotation, not the converse.

is this your question : why is every isometry of three dimensional euclidean space preserving orientation a rotation?
well, consider this :
all eigenvalues have modulus 1
at least one eigenvalue is real
suppose one is not an eigenvalue, then there are two cases :
they are all real, impossible as our determinant must be 1 and $(-1)^3=-1$
if one is real and the others are imaginary the determinant would again be $-1$
so there is an eigenvector with eigenvalue 1, take this vector as first basis vector, and now extend it with two orthogonal vectors to form a basis
in this twodimensional complementary space you have an isometry with determinant one : a rotation thus


Given: Infinite supply of apple juice, Infinite supply of water, a 5 L bucket, a 3 L bucket.
Make: A 50% mix of water & apple juice

It IS possible

[hide="Hint"][hide="A really big hint"][hide="A really really big hint"]
The buckets themselves have no volume.[/hide][/hide][/hide]

Oh, you MEAN an infinite SUPPLY of water and an infinite
SUPPLY of apple juice.

This word can't be overlooked.

Make sure you don't leave out necessary words and that words
or abbeviations you've included aren't misleading; I looked at your tree of hints.

Ummm, just put the same number of apple juice  and water?

[hide="Maybe"]
It is easy to transfer 1L water into the 5L bucket.
Then turn the 3L bucket upside down and push it to the bottom of the 5L one.
Finally, fill the 5L bucket with juice
 :D
[/hide]

You're close, however I forgot to put this in the original problem: we need 4L of half-half, not 2.

OK, based on modelers strategy...

[hide="My Guess"]
Fill 5L with apple juice.
Pour into 3L. You now have 2L of apple juice in the 5L jug.
Empty the 3L jug and put the 2L in it.
Place the empty 5L jug over the 3L jug like a cap, and shove the 3L jug to the top, (because we are assuming that the bottom of the large jug and the top of the small jug have the exact same shape...).
While firmly holding the 3L jug to the top, flip the contraption over. You now have a 2L space above the 3L jug, and 2L of apple juice contained beneath the 3L jug.
Fill the space above the jug with water. You now have 2L of water and 2L of apple juice.
Now, drill a hole in the 3L jug and push down so that a liter of water goes in and fills the space. You can now remove the 3L bucket without displacing the water, so do so.
Ta da?
[/hide]

That's it, except to subtract the 3L from the 5L, I just would carefully press the 3L bucket (with 2L AJ) into the 5L bucket.

aaaaa ---
aa--- aaa
aa--- ---
----- aa-
----- aaw
aaw-- www
aawaa www

Pour contents of both containers to the floor and there is your 50% concentrated stuff. Be more specific pls.


conics
ellipse
geometry
angle bisector
geometry unsolved

when given an ellipse and given a random point on the ellipse,
a)construct the two foci
b)construct the tangent to the ellipse from the point

a) you can use this fact: let a line r that meet the ellipse on A and B and let a line t parallel to r that meet the ellipse on C and D. Call M the midpoint of AB and N of CD. Then MN pass trought the center of the ellipse. When you have the center you can easy costruct major and the minor axis, from wich it's easy to costruct the foci $ F_1$ and $ F_2$ using $ c^2 \equal{} a^2 \minus{} b^2$ and pytagora.
b) when you have the foci the tangent from P is the external angle bisector of $ \angle F_1PF_2$.
if you don't have the foci, let 4 point A, B, C, D and call $ E: AD \cap PC$, $ F: BC \cap PD$ and $ P' : EF \cap AB$. Then PP' is the tangent from P.


trigonometry
analytic geometry
graphing lines
slope
function

Explain the graphs:

$y=\sin^{-1}(\sin x)$

$y=\cos^{-1}(\cos x)$

$y=\tan^{-1}(\tan x)$

[quote="Altheman"]wtf you totally changed the problem...that was a good one too...[/quote]
OMG you caught me. But I realized the other problem had been posted before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mosp+sets+disjoint&t=43383

:dry:

[hide="The first one"]
As you may have noticed, $y = \sin^{-1}(\sin x)$ has a tricky feature.

Take an angle $x$, and suppose $-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}$. We take its sine; call that $s$.

The arcsine of $s$ is the angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ whose sine is $s$.

But that's just the angle we started with. So for $x$ in this interval, $y = x$.

Now suppose $\frac{\pi}{2}\leq x\leq\frac{3\pi}{2}$.

Then the arcsine of $s$ will be the angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ that has the SAME sine as this angle $x$. The angle will be $\pi-x$.

For example, if $x=\frac{3\pi}{4}$, then $\pi-\frac{3\pi}{4}= \frac{\pi}{4}$ has the same sine, and is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, as required by the range of the arcsine function.

So for $x$ in this interval, $y = \pi-x$.

In every interval, we will have a straight line whose slope is either 1 or -1, making a zigzag.

In general, the composition of any function and its inverse is the line $y=x$, since the inverse undoes what the original function does, and takes you back to $x$. The zigzags are caused by the fact that we have to choose just one of many angles when we find the inverse trig function.[/hide]


algorithm
inequalities
three variable inequality

$\sum\frac{a^{2}}{b}=\sum\frac{a^{4}}{a^{2}b}$

Cauchy gives:

$\sum\frac{a^{4}}{a^{2}b}\geq\frac{(\sum a^{2})^{2}}{\sum a^{2}b}$

So it suffices to show that

$\frac{(\sum a^{2})^{2}}{\sum a^{2}b}\geq\ 3(\sum a^{2})\leftrightarrow\sum a^{2}\geq\ 3\sum a^{2}b\leftrightarrow\sum a(a-b)^{2}\geq 0$

Done. :P

10maths_tp0609, how do you come up with such transformations? Is there some kind of algorithm or what?

Behemont, can you explain the last step to me? I understand that we should prove
$a^{2}+b^{2}+c^{2}\ge 3a^{2}b+3b^{2}c+3c^{2}a$, but I don't understand how is this equivalent to $a(a-b)^{2}+b(b-c)^{2}+c(c-a)^{2}\ge 0$.
Also, where did you use $a+b+c=1$?

Thank you. I would never think of something like that.

http://www.mathlinks.ro/viewtopic.php?p=515092#p515092
this problem was last year in romanian JBTST  :)

It is true the strogner one - $ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\ge \frac{\left(a\plus{}b\plus{}c\right)\left(a^2\plus{}b^2\plus{}c^2\right)}{ab\plus{}ac\plus{}bc}$. After clearing the denominators and putting all the stuff on the left, the rest is equivalent to $ \frac{\left(bc^2\minus{}a^2c\right)^2\plus{}\left(ab^2\minus{}bc^2\right)^2\plus{}\left(a^2c\minus{}ab^2\right)^2}{2}\ge 0$.

only from Chebyshev inequality we have
$ (\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x})(y+z+x)\geq3(x^{2}+y^{2}+z^{2}) $

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that:
\[ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge \frac{93(a^2+b^2+c^2)-19}{12} \]

Assume $b$ between $a$ and $c$
We have $\sum \frac{a^2}{b}=\sum (\frac{a^2}{b}-2a+b)+(a+b+c)^2=\sum \frac{(a-b)^2}{b}+(a+b+c)^2\geq \frac{(a-b+b-c+a-c)^2}{a+b+c}+(a+b+c)^2=4(a-c)^2+(a+b+c)^2\geq 3(a^2+b^2+c^2)\Leftrightarrow (a-c)^2+(b-a)(c-b)\geq 0$(True with assume)

[quote=mornik]Let $a,b,c>0$ such that $a+b+c=1$. Prove: \[\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2}) \][/quote]

By computer,we have
$$LHS-RHS=\sum_{cyc}{\frac{(c+a)(a-b)^2}{b}}\ge{0}$$

Clearing denominators and homogenizing, it suffices to prove that \[(a^3c+b^3a+c^3b)(a+b+c)\geq 3a^3bc+3ab^3c+3abc^3\iff a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^3bc+ab^3c+abc^3)\iff \sum c(a^2-bc)^2\geq 0,\] which is true, so we're done.

[quote=AlgebraFC]$a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^3bc+ab^3c+abc^3)\iff \sum c(a^2-bc)^2\geq 0$ which is true, so we're done.[/quote]

Why?
$\sum c(a^2-bc)^2\geq 0\iff a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^2b^2c+a^2bc^2+ab^2c^2)$

Clearing the denominator and homogenizing, we get

$a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^3bc+ab^3c+abc^3)$

By AM-GM inequality,

$\frac{1}{2}(\sum_{cyc}a^4c)+(\sum_{cyc}a^3c^2)\geq \frac{3}{2}(\sum_{cyc}a^3bc)$

So it is enough to show that

$\sum_{cyc}a^4c\geq \sum_{cyc}a^3bc\iff \sum_{cyc}a^3c(a-b)\geq 0$

WLOG, we suppose that $a\geq b\geq c$ or $a\geq c\geq b$.

In either case, it is easy to show that this inequality holds :D


geometry
rectangle
3D geometry
sphere
calculus
trigonometry
calculus computations

[1] What is the maximum possible area for an isosceles triangle with two sides equal to $a$?

[2] Find the dimensions of the rectangle of largest are that can be inscribed in an equilateral triangle of side $L$ if one side of the rectangle lies in the base of the triangle

[3]A right circular cylinder is inscribed in a sphere of radius r.Find the largest possible surface area of such a cylinder

Thanks

#1 is trivial without calculus: use the area formula $K=\frac{ab\sin C}{2}$.
#2 is also very easy; by (weighted) AM-GM, the rectangle should have half the height and half the base of the triangle.
#3: You might actually want to use calculus on this one. Set up the problem (a cross-section perpendicular to the flat bases of the cylinder gives you a rectangle inscribed in a circle) and do it.


Let a sequence $\{u_n\}$ be defined by $u_1 = 5$ and the relation $u_{n+1} - u_n = 3 + 4(n-1)$, $n=1, 2, 3, \cdots$.  If $u_n$ is expressed as a polynomial in $n$, the algebraic sum of its coefficients is:
The number of points equidistant from a circle and two parallel tangents to the circle is:
\[ \text{(A)}\ 3\ \ \ \ \text{(B)}\ 4\ \ \ \ \text{(C)}\ 5\ \ \ \ \text{(D)}\ 6\ \ \ \ \text{(E)}\ 11 \]

would it be two for the second one? It has to be equidistint from each of the two lines, aka the line running midway between them, and the distance from this line to the other two lines has to be the same as the dist from the circle to this new line, giving us two points only?

I thought of that but then said, its the circle, not the circles circumference thats in the question... the circle would be the whole enclosed space aswell...

What about the first problem?  I have some ideas, and I might know the answer but I'm not sure.

Nice.... I got the last one, here goes:

[hide]Consider the polynomial function that determines u sub n as P(n).  Then in order to find out the sum of the coefficients you must find P(1) thus the sum of the coefficients is equal to u sub 1, which is equal to 5.  That or you could write out a messy system of equations. [/hide]


Hi,
I have a really huge problem.

I need to construct a Turing machine which calculates sum [b]S[/b]:

[url=http://www.imagesforme.com/viewer.php?id=2330untitled.GIF][img]http://www.imagesforme.com/images/2330untitled.GIF[/img][/url]

Turing machine never halts, and it writes on output tape the result of every iteration.

So the output tape should look like this:

1 3 9 33...etc

It isn't important what sign(s) is between the numbers...

Thank you for your time :)

1. [url=http://www.answers.com/topic/turing-machine]Turing machines[/url], by definition, cannot be constructed? Especially if you want it to "never halt"?

2. I think the output should look like this:
[i]1 1 3 9 33 [/i]...
but of course it's up to you.

3. Can a computer be used? A printer(to print on tapes)?

Thynks paganinio, yes, I was wrong about the output. It should look like:

1 2 4 10...

But I already constructed it, so...thanks anyway :)

[quote="paganinio"]1. [url=http://www.answers.com/topic/turing-machine]Turing machines[/url], by definition, cannot be constructed? Especially if you want it to "never halt"?[/quote]

In what crazy sense are Turing Machines designed to be something you cannot construct? Even if it was just specifying the mechanics behind the concept, isn't specification alone a sense of construction?

And that a Turing Machine doesn't halt is no harder to create than one that does.