So I have a question about the USAMTS grading. Exactly how hard are the graders? Say you make relevant progress on #x and are on the right track, but by some misfortune you undercount or overcount. How many points will you receive? Also, if your solution is mathematically and technically correct but kind of awkward with the wording or length, will you still get full credit?

[quote="The USAMTS website"]Each solution will be given a score of 0, 1, 2, 3, 4, or 5. Solutions will be graded both on mathematical correctness and on writing style.
The grading criteria for each problem will be published on the USAMTS website at the conclusion of the grading.[/quote]

Pretty much they want you to have as clear and complete an argument as possible. My advice would be not to worry about the grading policy and instead worry about making your proof rigorous. If you know that nothing is wrong with your argument, then you'll get a 5 no problems. Just do your best and you'll get exactly what you deserve :)

(And btw I've never heard of someone getting points off because their solution was too long)

I concur with the post above mine, and to be more precise:
[quote="Thunder365"]Also, if your solution is mathematically and technically correct but kind of awkward with the wording or length, will you still get full credit?[/quote]
Probably. To a certain extent, definitely. USAMTS grading is generally lax enough to forgive minor errors in wording, but if it gets to the point where it causes misunderstanding/confusion, you might lose a point or two.
Anyway, USAMTS is supposed to be a good source of practice for solutions--in particular, to work on fixing errors such as these. You should have plenty of time to write up and edit your solutions (unlike in an olympiad setting), so spend the time now to try and remove these minor errors!

Last year I undercounted on R1/#1 by 1, and received a 4 fyi

DEpends. If you are just putting steps that you don't need, you might be docked pts.

If you are just being rigorous, I wouldn't think so although they might comment on that.

[quote="Thunder365"]... but I am talking about a VERY LONG solution (as I said before, possibly 5+ pages).[/quote]
I had to grade some ten-page solutions for Problem 1/1/21. Five pages does not seem long, in contrast.

We graders sometimes joke among each other about taking points off for overly lengthy solutions, because reading them slows down our grading, but we never do. For the most extreme cases I write a comment that we would prefer shorter solutions in the future.

Submissions lose points only because of errors and lack of rigor. Bad writing loses points only when it undermines rigor.

Erin Schram
USAMTS grader

Sorry for reviving, but I have a similar question.

Can points be deducted for a solution that uses a long and tedious approach when there is a slicker approach? You can assume that the long solution is rigorous.

No, as long as your solution is correct. I have in the past brute force-proven (TONS of casework) some of the problems, and I have gotten full credit.

[quote="Thunder365"]Hmmmm now I have a question on the total opposite end of the spectrum. My solution for one of the problems is in depth....VERY in depth...in depth to the point of it being 5+ pages. However, I am sure that my wording is clear and there arent any errors in my proof. Would I still get 5 points because I arrived at the correct answer, or could the graders possibly take off points because of my long solution?
I know cyberspace already answered this, but I am talking about a VERY LONG solution (as I said before, possibly 5+ pages).[/quote]

I would say definitely.  There is merit in brevity, as long as the proof remains rigorous.  However, if a five page solution is necessary to maintain rigor, then by all means, there's nothing wrong with it.

No.. points won't be taken off...
7 of my solutions this year (1st 2 rounds) are 5+ pages
And I have a perfect so far.  :lol:
None of those graders have taken of for length. Obviously.  :P


Let $ F$ be a finite set of real numbers such that every distance between two points in $ F$ (excepting the diameter) achieves at least two times. Prove that if $ a_1,a_2,a_2,a_4 \in F$, then $ \frac {a_1 \minus{} a_2}{a_3 \minus{} a_4} \in Q$.

We induct on the number of elements: with $ 2$ or $ 3$ elements of the set, it's trivial.  If there are more than $ 3$ elements, proceed as follows: without loss of generality the largest element is $ 1$ and the smallest is $ 0$.  If $ x < 1$ is the next-largest element, $ x \neq \frac{1}{2}$ (or else our set is size $ 3$) and so $ x > \frac{1}{2}$, $ 1 \minus{} x \in F$.  Then we can apply the inductive hypothesis to the set $ F \setminus \{0, 1\}$, so all the distances in this set are a rational multiple of $ 2x \minus{} 1$.  It remains to show that $ x$ itself is rational.  Since $ 2x \minus{} 1$ is a distance appearing once in our set, it must appear a second time.  It cannot appear again as the distance between two elements of $ F \setminus \{0, 1\}$, so one of $ 2x \minus{} 1$ and $ 2 \minus{} 2x$ must be an element of $ F$.  Thus $ \frac{(2x \minus{} 1) \minus{} (1 \minus{} x)}{2x \minus{} 1} \equal{} \frac{x \minus{} (2 \minus{} 2x)}{2x \minus{} 1}$  must be rational, so $ x$ itself must be rational, as desired.

[quote="JBL"]  If $ x < 1$ is the next-largest element, $ x \neq \frac {1}{2}$ (or else our set is size $ 3$) and so $ x > \frac {1}{2}$, $ 1 \minus{} x \in F$.[/quote]

I don't understand why $ 1\minus{}x \in F$. Of course, the distance $ 1\minus{}x$ is achieved by two elements in $ F$, $ a$ and $ b$, but not necesarly $ a$ or $ b$ is $ 0$. Maybe i don't get something trivial here...

[quote="JBL"] Then we can apply the inductive hypothesis to the set $ F \setminus \{0, 1\}$.[/quote]

I don't understand this either. Maybe the distance between $ 1$ and an element $ a \in F$ in achieved only twice, so if in the set $ F \setminus {1}$ you will have a distance which will occur once, wlog between $ y$ and $ z$. It is not necesarly that $ y$ or $ z$ is $ 0$ so for $ F \setminus \{0, 1\}$ you cannot aplly the inductive hypothesis. Please explain :) .

For the first objection:

Suppose $ 1 \minus{} x \not \in F$.  The distance $ x$ is obtaned between $ 0$ and $ x$, so it must be obtained again between two other numbers.  We can't have both of these numbers in the interval $ [0, x]$, so one of these numbers must be larger than $ x$ and by the choice of $ x$ it must be between $ 1 \minus{} x$ and $ 1$.

For the second, I have to recreate the process that led to that claim, so watch this spot for new edits ;)

ask a silly question, what is distance between numbers, and what is diameter of numbers?

What do this problem actually mean...

The distance between two numbers $ x, y$ is $ |x \minus{} y|$.  The diameter of a set is the maximal (or supremal, if we're dealing with open sets) distance between two of its elements.

I think my proof can be repaired, but as of right now it seems to be just as much work to deal with the second objection as to solve the original problem.


geometry
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Although both are opinions, I'm sure each one of you has your own opinon on which one is more accurate.

One might argue that they do not conflict, to successfully argue this point, one would have to argue that there is not [i]one, solitary example[/i] of confliction. Since evolution (Charles Darwin) is one, there is clearly [i]a[/i] contradiciton present. I will speak no more of others, but there are certainly many (many is an opinion, so don't argue with whether there are "many" or not).

I personally am an agnostic, and does not believe that God can be either proven or disproven, and therefore I am skeptical of everything I hear. Logically, God doesn't exist (for me), but then again, will logic work when dealing with things such as religion?

Any statistics regarding the possibility are false.

There are (for me) two parts of religion: (1) the facts, and (2) the impact on everyday life.

Although I don't agree with (1), due to certain contradicitons that I have noted (with science as well as self contradictions), but agree wholeheartedly with (2), which I think is actually what makes religion popular. We need a reason to live, and we have a hard time dealing with uncertainty regarding life, death, etc.

One thing that really bothers me is when someone insults the beliefs of others as bieng rubbish, so please don't do that here. I welcome all comments and reasons, but please don't (purposely) actively insult someone else's belief, whatever they might be. I will try to keep to that rule myself.

Open for cross examination...

Science can be tested by experiments, but religion can't be tested(doesn't mean it's false, it's just that we don't know if it's true or not)

[quote]One might argue that they do not conflict, to successfully argue this point, one would have to argue that there is not one, solitary example of confliction. Since evolution (Charles Darwin) is one, there is clearly a contradiciton present. I will speak no more of others, but there are certainly many (many is an opinion, so don't argue with whether there are "many" or not). [/quote]

Yeah - it is your opinion that evolution conflicts with religion. In my opinion, there is no conflict.

We had this discusion in my science class.  Whether evolution exists or not...most of my class said it can't since it contradics the bible, but who says it does???

Evolution is perfectly resonable, it doesn't conflict.

My roommate (who is a math/physics major) likes to say that a proof in physics requires testing the hypotheses in every possible situation. (In particular, this requires testing it on every set of particles. This is why I make fun of physics majors.)

sounds like a good reason to make fun of them to me, ComplexZeta.

[quote="banblink182"]One might argue that they do not conflict, to successfully argue this point, one would have to argue that there is not one, solitary example of confliction.[/quote]

Perhaps a minor quibble, but the burden of proof is on the person making the positive claim that there is a conflict.  You have to show that "science" and "religion" (rather nebulous concepts, stated in such a general way) actually [i]are[/i] in irreconcilable conflict in order to make the point.  (This is sort of an "innocent until proven guilty" idea in argument.  Without it, I could make a claim that, say, pink-and-purple unicorns exist, and then demand that you scour the entire universe for them in order to argue the point that they do not, which is not very reasonable.)  And you haven't exactly explained how evolution "is" a contradiction between "science" and "religion".  The best you can say is that evolution is a hypothesis that is in contradiction with a certain interpretation of Genesis 1 -- not nearly so strong a statement.

Yes, the claim that God exists, or that a certain religion is true, is also a positive claim.  And that is why I think that, since few can clearly see [i]a priori[/i] God's existence and what he demands of us (and I am not one of them), Christianity ought to be presented based on evidence -- philosophical or historical, but evidence.  "The facts" are the most important part of religion.  "Either there is a 'pie in the sky' or there is not," as C.S. Lewis put it regarding heaven.  Either the facts of Christianity, its doctrines, are true or they are not.  In either case, they are the most important part of Christianity, which concerns itself primarily not with the present well-being of individuals or societies but with the individual's (and society's) relation to God.  The same goes (to perhaps a lesser degree in cases) for other religions.  That we should believe these things not because they are true but because, though lies, our belief in them makes us "better" or more comfortable, is ludicrous.

And I suppose that you can fairly well guess what my perspective is.

[quote="Setofallsets"]
Perhaps a minor quibble, but the burden of proof is on the person making the positive claim that there is a conflict.  You have to show that "science" and "religion" (rather nebulous concepts, stated in such a general way) actually [i]are[/i] in irreconcilable conflict in order to make the point.  (This is sort of an "innocent until proven guilty" idea in argument.  Without it, I could make a claim that, say, pink-and-purple unicorns exist, and then demand that you scour the entire universe for them in order to argue the point that they do not, which is not very reasonable.)  And you haven't exactly explained how evolution "is" a contradiction between "science" and "religion".  The best you can say is that evolution is a hypothesis that is in contradiction with a certain interpretation of Genesis 1 -- not nearly so strong a statement.
[/quote]

Correct me if I am wrong...

1. I purposely did not explain conflict, but I assumed most of you would agree. Of course conflict is an opinion [click here to view my argument regarding opinion, even though I did eventually come to the conclusion that it was wrong; nevertheless, an interesting read: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Opinionated+Paradox&t=22851]. My main point was that the bible talks about how men were created by God directly, while science feels that men were descended from apes who evolved from other creatures. I don't read the Bible, so correct me if I am wrong...

2. I don't understand your pink-and-purple unicorn argument, because I gave an example - namely evolution. I didn't just say that my point is right unless you can find an example contrasting it, I'm saying that if one were to say, "there exist [i]no[/i] conflict (which is what one would have to say to attempt to prove that the bible lives in peace with science)", then one would have to disprove my evolution argument (if it can be called that). I hope you get my point.

 :)

[quote]Evolution is perfectly resonable, it doesn't conflict.[/quote]

In the beginning God created the heaven and the earth. Genisis 1:1.  Evolution (mostly connected with the Big Band Theory) on the contrary says that a extremely compacted sphere exploded and "created the heavens and the earth".  Even in the first verse of the Bible there is a contradiction with science.

Item 1:  I apologize, banblink182, if it sounded like I was attacking you for not arguing your opinion; my point with my comment about the pink and purple unicorns was to make it clear that in order to argue the opposing opinion, it is not necessary to show that there is necessarily no conflict between the two, but to address and refute any particular example or class of examples that is presented as positive evidence for the conflict.  I'm sorry if I was not clear.

Item 2:  Evolution :!=: Big Bang.

Item 3:  So God couldn't have created the universe by whatever mechanism He chose (such as, as a possibility, the Big Bang), or developed the human body by means of an evolutionary process? :?  Gen. 1:1 states that God created the universe.  It doesn't say what mechanism He used to do so.

Item 4:  I will make the claim that only a literal reading of the early portions of Genesis offers any contradiction to the strictly scientific claims of evolution.  That God is responsible for the creation of the earth and its inhabitants, that He gave to human beings a spirit and a will, and so on, is not in contradiction with scientific theories of by what means this happened.  Only a very literal reading (creation in seven 24-hour days, etc.) of the text offers a contradiction here.  (See next item.)

Item 5:  (Related only to item 4.) It is my opinion that the early portions of Genesis were not [i]meant[/i] to be taken literally, but symbolically.  I won't pretend to be an expert on the subject -- I don't know Hebrew, nor am I a scholar in ancient near east culture -- but it does seem deliberately mythological in character:  "Adam", IIRC, means "man" or "mankind", for instance.  And when you think about it, telling it this way (and being, IMO, fairly clear that it is primarily symbolic -- although I think that we moderns have lost the full meaning of some of the symbols) makes sense if the development of the universe is as complicated as the modern scientific theories assert.  I mean, what if Genesis 1 had read: "In the beginning God created the universe.  First, He caused to come into existence a singularity, which is matter at infinite density.  He then caused this singularity to explode into the matter that is in the universe, creating also space and time.  Gradually, He used the physical laws He created to form many galaxies of billions of stars, and after billions of years, He created the Sun and the Earth which revolves around it.  And at first the Earth was made of molten rock, but gradually it cooled, and it gained an atmosphere and oceans, but still there was no life on it.  And God caused life to be formed from the nonliving elements on the Earth, and this first life was very tiny, so tiny that it is impossible to see.  And over many millions of years He caused this life to develop by gradual mutations into the many forms of life on Earth.  And finally, He caused to develop from this life the human body.  And He then gave to humans a spirit and a will." Etc....   And of course this would have to have been far, far, longer and much more complicated -- because few of those terms would have meant anything to an ancient.  How confusing, not only for the ancients, but for many nonscientists today!  And how useless, when the point is not how exactly God created the universe, but that He did it, and not exactly how He made man, but that He did so, and that He gave him a spirit and a will -- the "breath of life".  The first few chapters of Genesis, as symbolism, were much more helpful to those to whom it was given than a painfully accurate description would have been.  And, as I said, we likely miss many of the deep symbols that were present in the text to its first hearers and readers -- symbols which probably helped reveal truths far better than the type of story we moderns would demand from the text.

Item 6:  If you can't tell already, I believe, tentatively, in Theistic Evolution (ie, evolution as a mechanism used by God).  I'm not saying you have to agree -- but now you can see why I deny conflict between religion and science.  As I've stated before, I don't think they are even trying to answer the same questions.

Setofallsets, my only problem with your reasoning is that God setting the whold BIg Bang into place takes him almost entirely out of the creation.  Why would God need to use a big bang to create the universe if he can simply wish it into place.  God does not want us to understand him he wants us to simply believe.  Why would He care for us to know the science behind creation?  It simply doesn't add up for me.  Plus I understand what you mean about Adam and he 7 days of creation being sybology, but why would Genisis expand so much about Adam's life (Gen 5:3 And Adam lived an hundred and thirty years, and begat [a son] in his own likeness, after his image; and called his name Seth)(Gen 5:5 And all the days that Adam lived were nine hundred and thirty years: and he died.).  I understand what you are saying, but I do not know where you see these verses and can truthfully say, "Adam was a symbol for man, he didn't really exist".  If you can support your arguments about God being involved in The Big Bang, or the like with religious truths or verses from the Bible, then you have room to argue.

[quote="ComplexZeta"]Adam comes from the Hebrew word Adamah, which means ``ground'' or ``Earth.''[/quote]

Confused, I looked it up in Strong's Concordance, which indicates that pretty much all of the words with that root come from a word referring to redness.  Etymology is confusing, especially when I don't know much of anything about the language.  I think I'll drop this and plead lack of expertise.

[quote="JBL"]Hey Setofallsets, long time no see -- how's everything going?[/quote]

I'm doing well; I'm still home on break right now but I head back out to HMC on Sunday to start spring semester.  I'm taking a good bit (well, relatively speaking, as Simon is probably taking a lot more than I) of math this semester, so it should be fun.

I suppose I should have realized that I'd bother some people with my last post.  I'll say up front that while I'll defend my thesis here, I don't think that it's a [i]theologically[/i] important issue.

[quote="somedude"]
my only problem with your reasoning is that God setting the whold BIg Bang into place takes him almost entirely out of the creation. Why would God need to use a big bang to create the universe if he can simply wish it into place.
[/quote]
I disagree that God's using something such as the Big Bang to create the universe takes Him out of the process of creation.  If He is responsible for the creation of the universe, then He is responsible -- no matter how He chose to do it.  As far as why He would do it this way (I agree with you that He did not "need" to), I can only answer that I don't really know.  Perhaps there is some added beauty to the universe being made this way, developing based on a set of natural laws -- or perhaps, done this way, it has a different set of effects on how humans react.  It would certainly be neither more nor less difficult for God to do it this way than any other.  I cannot claim to know the mind of God, though, and so I can offer no other reason for this hypothesis (that God used mechanisms such as the Big Bang and evolution to create the universe and the creatures in it) other than that the evidence seems to indicate that this is how He did it.  If, somehow, evidence turns up that the Big Bang hypothesis is wrong or something of that nature, I would have no theological problem with that, because what is important is that "In the beginning, God created the heavens and the earth", not how exactly He did so.

[quote="somedude"]God does not want us to understand him he wants us to simply believe. Why would He care for us to know the science behind creation?[/quote]
Maybe I'm misunderstanding you, but I don't think I disagree with you here.  But simply because it isn't necessary that we know by what means God created the universe doesn't mean that we can't (through science) try to find out, does it?  In the end, He doesn't care whether we believed in a literal 7-day creation or in the Big Bang and evolution.  But someone who thinks that the evidence for evolution is evidence against God would be hurt if we insisted that if God created the universe, then evolution would be false.  Generally, it's a bad idea to discourage someone's faith by upholding false dichotomies.

[quote="somedude"]Plus I understand what you mean about Adam and he 7 days of creation being sybology, but why would Genisis expand so much about Adam's life(..).  I understand what you are saying, but I do not know where you see these verses and can truthfully say, "Adam was a symbol for man, he didn't really exist".[/quote]
I can't say for sure whether Adam as a particular person actually existed -- he very well may have (and, I agree that he probably did).  But I think that at least during the creation story, "Adam" probably refers to mankind in general, whether or not he [i]also[/i] is a particular person.  (Incidentally, the NIV translates the word as "a man" or "the man" until after the Fall.)  To have something have both symbolic and literal meaning in Scripture is fairly common.  I'm going to withhold judgement on how much of early Genesis is literal and how much is allegorical or symbolic because I have little evidence to support whatever claim.

I knew that Saint Augustine did some writing on the subject, so I did a quick search, and I found that he had this to say about the interpretation of the creation story:

[quote="Saint Augustine"]In matters that are obscure and far beyond our vision, even in such as we may find treated in Holy Scripture, different Interpretations are sometimes possible without prejudice to the faith we have received. In such a case, we should not rush in headlong and so firmly take our stand on one side that, if further progress in the search of truth justly undermines this position, we too fall with it. That would be to battle not for the teaching of Holy Scripture but for our own, wishing its teaching to conform to ours, whereas we ought to wish ours to conform to that of Sacred Scripture. [/quote]
[quote="Saint Augustine"]Usually, even a non-Christian knows something about the earth, the heavens, and the other elements of this world, about the motion and orbit of the stars and even their size and relative positions, about the predictable eclipses of the sun and moon, the cycles of the years and the seasons, about the kinds of animals, shrubs, stones, and so forth, and this knowledge he holds to as being certain from reason and experience. Now, it is a disgraceful and dangerous thing for an infidel to hear a Christian, presumably giving the meaning of Holy Scripture, talking non-sense on these topics; and we should take all means to prevent such an embarrassing situation, in which people show up vast ignorance in a Christian and laugh it to scorn. The shame is not so much that an ignorant individual is derided, but that people outside the household of the faith think our sacred writers held such opinions, and, to the great loss of those for whose salvation we toil, the writers of our Scripture are criticized and rejected as unlearned men. If they find a Christian mistaken in a field which they themselves know well and hear him maintaining his foolish opinions about our books, how are they going to believe those books in matters concerning the resurrection of the dead, the hope of eternal life, and the kingdom of heaven, when they think their pages are full of falsehoods on facts which they themselves have learnt from experience and the light of reason? Reckless and incompetent expounders of holy Scripture bring untold trouble and sorrow on their wiser brethren when they are caught in one of their mischievous false opinions and are taken to task by those who are not bound by the authority of our sacred books. For then, to defend their utterly foolish and obviously untrue statements, they will try to call upon Holy Scripture for proof and even recite from memory many passages which they think support their position, although “they understand neither what they say nor the things about which they make assertion.?[/quote]
[quote="Saint Augustine"]I have thought that each one, in keeping with his powers of understanding, should choose the interpretation that he can grasp. Where he cannot understand Holy Scripture, let him glorify God and fear for himself.  But since the words of Scripture that I have treated are explained in so many senses, critics full of worldly learning should restrain themselves from attacking as ignorant and uncultured these utterances that have been made to nourish all devout souls. Such critics are like wingless creatures that crawl upon the earth and, while soaring no higher than the leap of a frog, mock the birds in their nests above.[/quote]

Hmm, this thread is getting a tad off track.  Sorry about that.

CLASH!

Sorry for not reading the post - the title of the thread caught my attention.

[quote="Setofallsets"][quote="ComplexZeta"]Adam comes from the Hebrew word Adamah, which means ``ground'' or ``Earth.''[/quote]

Confused, I looked it up in Strong's Concordance, which indicates that pretty much all of the words with that root come from a word referring to redness.  Etymology is confusing, especially when I don't know much of anything about the language.  I think I'll drop this and plead lack of expertise.[/quote]

That is possible. The Hebrew word for ``red'' is ``adom.''

[quote="Setofallsets"]I'm taking a good bit (well, relatively speaking, as Simon is probably taking a lot more than I) of math this semester, so it should be fun.[/quote]

Hehe. At the moment I only have three math classes: complex analysis, ring theory, and set theory. I might have analytic number theory at some point later this quarter.

Define 'religion'. Certain interpretations of the Christian religion conflict with evolution. Other interpretations don't--accounting for the theistic evolutionist majority at http://www.christianforums.com (check out the creo/evo debate forum there). I don't know as much about other religions versus science, but my point is that religion conflicts if you think it conflicts, and doesn't conflict if you don't think it conflicts. I mean, there are still a few geocentrists and even flat-earthers around, and they claim heliocentrism and a round earth conflict with their religion, which is fine because that's partially how their individual religion is defined. If you see what I mean.
That wasn't very coherent, but I hope you can get some meaning out of it...
P.S. Maybe we need a separate thread for evolution.


If a*b=22 and a+b=12,
what is absolute value of b-a?
A)2 sqrt 66
B)2 sqrt 14
C)5 sqrt 66
D)2 sqrt 10
E)5 sqrt 14

[hide="hint"]
Find $ a^2 \plus{} b^2$, and use that to find $ (b\minus{}a)^2$, take square root and you have your answer.
[/hide]

$ ab\equal{}22$ so $ 2ab\equal{}44$
$ a\plus{}b\equal{}12$
$ (a\plus{}b)^2\equal{}144$
$ a^2\plus{}2ab\plus{}b^2\minus{}2ab\equal{}144\minus{}44$
$ a^2\plus{}b^2\equal{}100$
$ a^2\minus{}2ab\plus{}b^2\equal{}100\minus{}44$
$ (a\minus{}b)^2\equal{}56$
$ |a\minus{}b|\equal{}|b\minus{}a|\equal{}\sqrt{56}\equal{}\sqrt{2^3*7}\equal{}2\sqrt{14}$
So the ansvwer is B


MATHCOUNTS
algebra
difference of squares
special factorizations

Umm i bet this is wrong but

[hide]y^3 = 2 + x^3[/hide] ?

Nuh, uh: [hide]First of all that's not y in terms of x (It's supposed to be y= something so in your case it would be y=(2+ x^3)^(1/3)

Also the answer's actually really simple...really simple

Just use difference of squares and you'll be fine.

Difference of Squares: (x+y)(x-y)=x:^2:-y:^2:[/hide]

You know it's kind of odd posting questions from my math class onto the MATHCOUNTS forum...it's either AoPS is a real advanced place (which it is) or my math class is a total waste of my time (which partly is  )

Don't feel discouraged por favor

Oh poo no wonder it's so hard...I forgot the 1/ part to the right side of the equation, apologies all around. Now it should be easy :)

I'd still like to here how both of you got that answer if it isn't too much of a trouble since the problem was much more interesting before ((y-x)/(y-x)=x-y)

In that case... [hide]y=x-1

oh solution sorry...

edit: (y-x)/(y^2-x^2) = (x-y)/(x^2-y^2)... i multiplied both sides by (x^2-y^2) to get x-y=1 so y=x-1[/hide]

Tare here is my solution:  [hide]With the difference of squares theorem, I found that (y-x)/(y :^2: -x :^2: )  1/(x+y)  (Have you noticed you wrote that theorem as (x+y)(x-y)= x :^2: +y :^2: .  That can be misleading to some.)  Anyway, I multiplied the remaining of the equation and the 2nd equation by x+y to get 1=1/(x-y).  Next, I multiplied both sides by x-y to get x-y=1.  Finally I subtracted x on both sides and then multiplied both sides by -1

Ergo, y, in terms of x,=x-1[/hide]

[quote="MysticTerminator"]I don't think all of that was necessary; try to find an easier solution.

[quote="frodo"](Have you noticed you wrote that theorem as (x+y)(x-y)= x :^2: +y :^2: .  That can be misleading to some.)  [/quote]

I edited it a while ago.[/quote]

was mine the easier solution?

i believe so. good example not to always routinely bust out factoring tools, although the other solution was still short.

but factoring is so fun!!! and expanding is .. ugh.

Yup yup. Can anyone post a solution to the original problem? ((y-x)/(y-x)=x-y)

hmmm (x-y)/(x-y)=y-x therefore y-x=1 and y=1+x. What exactly did chinaboi do, i am confused.

Yea Tare i did that before u gave the 1/ on right side ... thats how i got my answer.

[quote="Tare"]Mmm, I'm not so sure myself...If you multiplied both sides by (x^2-y^2) shouldn't you get x-y=x-y?[/quote]

What Chinaboi said was that the left hand side (y-x)/(y^2-x^2) = (x-y)/(x^2-y^2). (negative of top and bottom). So you get (x-y)/(x^2-y^2) = 1/(x^2-y^2) and then you multiply both sides by that..


Find all solutions to the equation:
$5(x+\frac1x)=12(y+\frac1y)=13(z+\frac1z)$
$xy+yz+zx=1$

From the first equalitz all the variables must have the same sign. The negative solutions are just -1-multiples of the positive ones, so suppose x,y,z>0.
Letting $x=tg(\alpha/2)$,$y=tg(\beta/2)$,$z=tg(\gamma/2)$, the second condition is equivalent to $\alpha+\beta+\gamma=\pi$ so the three angles must be inner angles in a triangle. We have
$x+\frac{1}{x}=\frac{sin(\alpha/2)}{cos(\alpha/2)}+\frac{sin(\alpha/2)}{cos(\alpha/2)}= \frac{sin^{2}(\alpha/2)+cos^{2}(\alpha/2)}{sin(\alpha/2)cos(\alpha/2)}=\frac2{sin(\alpha)}$.
Similarly $y+\frac{1}{y}=\frac2{sin(\beta)}$,
$z+\frac{1}{z}=\frac2{sin(\gamma)}$.
From the first eq we have
$\frac5{sin(\alpha)}=\frac{12}{sin(\beta)}=\frac{13}{sin(\gamma)}$,
so the triangle is a Pythagorean one, $\gamma=\frac{\pi}2$, so z=1. So $x+\frac{1}{x}=\frac{26}{5}$, which has solutions x=5 and x=1/5. For x=5 we have 6y+5=1 from the second eq, y=-2/3, contradiction. For x=1/5 we have 1.2y+0.2=1, y=2/3.
So the only sols are [1,1/5,2/3] and [-1,-1/5,-2/3].

It's marvellous! But how do you think of the transformation that $x=\tan(\frac{\alpha}2)$...?

[quote="Kondr"]
Letting $x=tg(\alpha/2)$,$y=tg(\beta/2)$,$z=tg(\gamma/2)$, the second condition is equivalent to $\alpha+\beta+\gamma=\pi$ [/quote]

How did this equivalence come from? :(

There are several well konown substitutions. I don't remember them all, but
$xy+yz+xz=1$ leads to x=tan(a/2)etc. and
$x^{2}+y^{2}+z^{2}+2xyz=1$ leads to x=cos(a) for x,y,z>0
with angles a,b,c in a triangle.
It can be proven by some formulas for sin(u+v) and cos(u+v)... I don't think it's hard to prove that for each triangle these equalities hold. And since tan can have arbitrary value, if we have such x,y,z we can find a suitable triangle for the first condition. If we have positive x,y,z fullfilling the second condition, it"s obvious that each of the terms on LHS is smaller than 1, so -1<x,y,z<1 and x,y,z can be cos's of angles in some acute triangle.

possibly because:

$\tan{\alpha}+\tan{\beta}+\tan{\theta}=\tan{\alpha}\tan{\beta}\tan{\theta}$

when $\alpha+\beta+\theta=k\pi$

[quote="t0rajir0u"][quote="Kondr"]$x^{2}+y^{2}+z^{2}+2xyz=1$ leads to x=cos(a) for x,y,z>0[/quote]

I've always found this identity remarkable.  Could someone show or link me to a proof?[/quote]

To go from $x^{2}+y^{2}+z^{2}+2xyz = 1$ to $x = \cos{a}$, $y = \cos{b}$, $z = \cos{c}$ with $a+b+c = \pi$:

First $0 \le x, y, z \le 1$ so let $x = \cos{a}$, $y = \cos{b}$. It remains to show $z = \cos{(\pi-a-b)}$. We solve the equation as a quadratic in $z$.

$z^{2}+2xyz+(x^{2}+y^{2}-1) = 0$

$z = \frac{-2xy+\sqrt{4x^{2}y^{2}-4(x^{2}+y^{2}-1)}}{2}=-xy+\sqrt{(1-x^{2})(1-y^{2})}$ (we omit the case $z < 0$)

$z = \sin{a}\sin{b}-\cos{a}\cos{b}=-\cos{(a+b)}= \cos{(\pi-a-b)}$

as desired. The other direction follows from the substitution $\cos{c}=-\cos{(a+b)}$ and some algebra.


inequalities
calculus
derivative
trigonometry
inequalities unsolved

(Without using partial differentiation) Find the largest value of
cosx+cosy+3cosz
where 0<=x,y,z<=pi/2 and x+y+z=pi

It can be done by vector.The general form is:
[tex]ab \cos x+bc \cos y+ca \cos z \leq \frac{a^2+b^2+c^2}{2}[/tex]
And luckily,in this problem the equality occurs when [tex]x,y,z \in [0,\frac{\pi}{2}][/tex] :)

Anh Cuong, Maybe you ought to use "$" instead of "[tex].


Can someone tell me what the time complexity of the funtion "pos", from Pascal, which search for a pattern on a string, is?

It's N x M. We checked this in IOI last year :).

[quote="yangzhe1990"]So, FP make us write anything by ourselves :([/quote]

Unfortunately, it's true... What string-matching algorithm do you recommend one to learn? KMP, Rabin-Karp, Boyer-Moore?


With the three digits of the integer 142, there are six two-digit positive integers that can be formed. When the sum of those two-digit integers is divided by the sum of the three digits of the original three-digit integer, the result is 22. When the same thing is done with the [i]k[/i] digits of a [i]k[/i]-digit integer, and each of the [i]k[/i] digits is different and non-zero, the result of dividing the sum of all of the possible two-digit integers by the sum of the [i]k[/i] digits of the original [i]k[/i]-digit integer is 66. What is the value of [i]k[/i]?

To pinksky:  Please do not post problems in more than one forum.

To all others:  The 4 problems that pinksky posted here can also be found in the High School Basics forum.


geometry
rectangle
trigonometry
analytic geometry
calculus
integration

What is the area of the region bounded by the axes and the circle with center $(4,4)$ and radius $7$?

[hide="Solution"][img]http://img443.imageshack.us/img443/4345/tempid3.png[/img]

The area of the square is $4^{2}=16$.

The two triangles can be combined into a rectangle. The shorter side of the rectangle is $4$, and the longer side is $\sqrt{33}$. The area is $4\sqrt{33}$.

The angle of the sector is $360-\left(90+2\cos^{-1}\frac47\right)\approx159.7^{\circ}$. The area is $7^{2}\cdot\frac{159.7}{360}\cdot\pi\approx68.29$.

$A\approx\boxed{107.27}$[/hide]

[quote="CircleSquared"]Aren't there 4 regions bounded by the axes and that circle?
I'm just getting that off i_like_pie's diagram.[/quote]
The actual problem starts out with only the two long lines and the circle.

I added the radii and the other two lines to split the region into 4 smaller ones.

Umm, that's not what I meant.
I meant isn't there a region bounded by the axes and circle in each quadrant of the coordinate plane?

EDIT: I meant quadrant, not sector.  :|

Oh... In that case, I'm not sure what jli meant, but if he clarifies, I can provide a solution for the other regions as well.

oops...i guess the ones in the I, II and III quadrant

srry about that.

fot the 3rd quadrant, just find the area under the x-axis and then you can multiply that by two and add the area of the first quadrant and find the overlap.

[quote="7h3.D3m0n.117"][quote="jli"]oops...i guess the ones in the I, II and III quadrant[/quote]

Why can't you solve for the area in the fourth quadrant? Is it because there is no algebra way to find the area? (of course integration works?).[/quote]
The area in the fourth quadrant is the same as in the second.

[hide="I"]As shown before, $A\approx\boxed{107.27}$.[/hide]

[hide="II, IV"]The area in quadrants II and III is a sector minus a triangle.

The angle of the sector is $2\cos^{-1}\frac47\approx110.3^{\circ}$.

The area of the sector is $7^{2}\cdot\frac{110.3}{360}\cdot\pi\approx47.16$.

The area of the triangle is $4\sqrt{33}\approx22.98$.

The area in quadrants II and III is $\approx24.19$.

The area of the whole circle minus the area and minus the area in quadrant I is the area in quadrant IV.

$A=7^{2}\cdot\pi-(107.27+24.19)\approx\boxed{22.48}$[/hide]

[hide="III"]The area of the region in quadrant II is known, and the area of the region in quadrants II and III combined is known.

The area of the region in quadrant III is simply one subtracted from the other.

$A=24.19-22.48\approx\boxed{1.7}$[/hide]

Note: Although the way I typed my solutions suggests that I used intermediate rounding, I actually never rounded except for the final numbers, so they are accurate.


limit
logarithms
function
real analysis
real analysis unsolved

I wonder if we can prove:
$(n-k)^{k+1} \ge (n-1)^2,k=1,2,\cdots,n-2$
when $n$ is large enough... :)

Yes. $n\ge 3$ is large enough.

Consider $f(x)=(n-x)^{x+1}$. We have
$f'(x)=f(x)\cdot(\ln(n-x)-\frac{x+1}{n-x})=\frac{f(x)}{n-x}\cdot((n-x)\ln(n-x)-x-1)$
Since $t\ln t$ is an increasing function for $t>1$, $(n-x)\ln(n-x)-x-1$ is decreasing and $f'$ has at most one zero between $0$ and $n-1$. It has one zero by the intermediate value theorem, since it is clearly positive at $x=0$ and negative at $x=n-1$. If $f'(a)=0$, we have $f$ increasing for $x<a$ and decreasing for $x>a$, so the minimum of $f$ on the closed interval $[1,n-2]$ occurs at an endpoint. Testing the endpoints, $2^{n-1}\ge (n-1)^2$ whenever $n\ge3$, and $(n-1)^2$ is the minimum.

This is easy.

nice,jmerry.
and the limit is sloved in a nice way,too :)


Ok so I kinda skimmed over the chapter in AOPS volume I that dealt with mods and I am utterly and completely lost. How do I solve things in different bases, and how is it helpful when solving complex algebra? Like I'm seeing stuff such as "13^13=1mod(6)" how does this work? Thanks!

13^13=1(mod 6) means the following. When divided by 6, 13^13 has a remainder of 1. This is true because 13 has a remainder of 1 when divided by 6, and 1^13=1. In general, a=b(mod c) means that when divided by c, a has the same remainder as b. You can also add, subtract, and multiply congruences.

[b]@ OP:[/b] Hunt around for stuff about "modular arithmetic". Wikipedia has a very understandable introductory article. Also, make sure you are typing "modular arithmetic" or "modulo" and not "modulus" -- modulus is an unrelated concept.

[b]@ mathemonster:[/b] I've seen accidental double posts, but this, never...


Find all functions $f: \mathbb R \to \mathbb R$ such that
\[ f(f(x)\plus{}2y)\equal{}x\plus{}f(2y)\]
for all $x, y \in \mathbb R$.

[quote="TRAN THAI HUNG"]Find f(x) R to R
and $ f(f(x) \plus{} 2y) \equal{} x \plus{} f(2y)$ :)[/quote]

Let $ P(x,y)$ be the assertion $ f(f(x) \plus{} 2y) \equal{} x \plus{} f(2y)$
Let $ a \equal{} f(0)$

$ P(0,\frac x2)$ $ \implies$ $ f(x \plus{} a) \equal{} f(x)$
$ P(x \plus{} a,y)$ $ \implies$ $ f(f(x \plus{} a) \plus{} 2y) \equal{} x \plus{} a \plus{} f(2y)$ and so $ f(f(x) \plus{} 2y) \equal{} x \plus{} f(2y) \plus{} a$ and so $ a \equal{} 0$

Then $ P(x,0)$ $ \implies$ $ f(f(x)) \equal{} x$
And $ P(f(x),\frac y2)$ $ \implies$ $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$

And so all solutions are involutary solutions of Cauchy's equation.
And it's easy to check that all involutary solutions of Cauchy's equation are indeed solutions.

Hence the answer : The set of solutions is the set of all  involutary solutions of Cauchy's equation (infinitely many).
If we add the continuity constraint, we get only $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{} x$

Is it valid???

Let $ x_1,x_2 \in \mathbb{R}$ so that
$ f(x_1) = f(x_2) \Longleftrightarrow f(x_1) + 2y = f(x_2) + 2y \Longleftrightarrow f(f(x_1) + 2y) = f(f(x_2) + 2y) \Longleftrightarrow x_1 + f(2y) = x_2 + f(2y) \Longleftrightarrow x_1 = x_2$.

So f is one to one function.

So if we put $ x = y = 0$ in the equation we take:

$ f(f(0)) = f(0) \Longleftrightarrow f(0) = 0$.

Now setting $ y - - > 0$ we take:

$ f(f(x)) = x$.

Now setting $ x - - > f(x)$ and $ w - - > 2y$ we take:

$ f(x + w) = f(x) + f(w)$......

[quote="pco"][quote="Dimitris X"]Is it valid???[/quote]

Sure. You just miss the conclusion[/quote]

I did a preview before posting and i saw your solution with the conclusion....There is no need to write the same..... :lol:  :)
I've post it because it is a different(and easier i think :) ) way.....

Thank you for your reply....

[quote="Dimitris X"]
I did a preview before posting and i saw your solution with the conclusion....There is no need to write the same.....
I've post it because it is a different(and easier i think ) way.....[/quote]

Ahhh OK.
 :)

[quote="Dumel"][quote="pco"]And it's easy to check that all involutary solutions of Cauchy's equation are indeed solutions. [/quote]are you sure that each solution of Cauchy's equation satisfies $ f(f(x)) \equal{} x$  :?:[/quote]

I'm sure of the contrary. That's the reason for which I wrote "all [u][b]involutary [/b][/u]solutions of Cauchy's equation are indeed solutions"  :)

it was a very stupid question  :blush: I wanted to change it but you were faster  :wink:
It took me a few minutes to find what 'involutary' means
my english...


Find the remainder when $ 17^{17^{17}}$ is divided by 7.  Use modulo.

Go to High School Basic.

$ 17^{17}\equal{}(\minus{}1)^{17}\equal{}5(mod 6)$
So $ 17^{17^{17}}\equal{}17^5\equal{}3^5\equal{}3 \times 9^2\equal{}3 \times 2^2\equal{}5(mod 7)$ by Fermat's little theorem.


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181077

When I press on the link posted by Pardesi, windows continuously pop up.

Please post which operating system you are using ( Windows XP, Mac OS 10.4.3 etc.)
AoPS or Mathlinks website
Which skin, default is AoPS but can be changed to mathlinks ( if AoPS)
Which browser (FireFox 2.0.0.11 ...)
That helps people decide what is wrong.
:)


Does anyone know any website that has problems and solutions to Poland Olympiad?

http://www.om.edu.pl

It is a official website, but it is in polish [though enter section "linki" and there is a link somewhere to english translations of some problems and solutions i guess]

some are also on kalva.demon.co.uk

 :)


This problem appears as a generalization in Gazeta Matematica for a problem proposed in Romania NMO 2006,belonging to Dan Schwarz.Here is the generalization:
Prove that for all real numbers x,y,then we have:
$(x^3+(x+y)^3)(y^3+(x+y)^3)\leq 4(x+y)^6$.Equality occurs when $x^2+3xy+y^2=0$ But how can one prove this nice inequality?


topology
function
real analysis
real analysis unsolved

Does there exist a closed set F contained in R1\Q such that m(F)>0?

The affirmative answer to the first question follows from the regularity of the measure: the measure of a set is the supremum of the measures of the closed sets contained in it. We apply this to the set of irrationals, and we're done.

For the second question, if we prove that each set $X$ with $0<a<m(X)$ contains a closed subset of measure $a$, then we're done, since the empty interior can be achieved by using the above for $X=E\setminus\mathbb Q$ instead of $E$. The subset we extract will automatically have empty interior.

We now have to prove the last part, namely that from each $X\subset\mathbb R$ with $0<a<m(X)$ we can extract a closed subset with measure $a$. We start with a closed set $F$ of measure $>a$ (this is possible, again, due to the regularity of the measure). The collection of closed real sets included in $F$ and having measure $\ge a$ is non-void partially ordered by inclusion. By [url=http://www.absoluteastronomy.com/encyclopedia/H/Ha/Hausdorff_maximality_theorem.htm]Hausdorff's maximality theorem[/url], we can extract a maximal totally ordered subcollection, and it's clear now that the intersection of all the sets in this subcollection has measure $a$.

We may omit Hausdorff theorem here. Just define $F(x)=m(F\cap [-\infty,x])$. Note that it is a continuous function (even Lipschits), hence it takes intermediate values.


Let $a_{1},a_{2},...,a_{n}\in\mathbb{R}$ such that $a_{1}\cdot cosx+a_{2}\cdot cos3x+...+a_{n}\cdot cos(2n-1)x\geq 0$ ,$x\in\mathbb{R}$. Prove that:

     $a_{1}\cdot cosx+a_{2}\cdot cos3x+...+a_{n}\cdot cos(2n-1)x= 0$ ,$x\in\mathbb{R}$.  ;)

[irrevelant counterexample]. I'm sorry, I didn't got that it must hold for all [tex]x[/tex] in [tex]\mathbb{R}[/tex] ;)

I suppose the both conditions are satisfied for each x in R.
If f(x) = Σk=1,nak*cos(2*k - 1)x.
Then f is continuous and periodical with period 2*∏ and satisfies f(x) ≥ 0, for each x ε [ 0, 2*∏ ].
We need to show that f(x) = 0, for each x ε [0, 2*∏ ].
Let's suppose there is a point x0 in (0, 2*∏) such that f(x0) > 0.
Because f is continuous we find a > 0 with: f(x) > 0 , for each x ε [ x0 - a, x0 + a].
Then: 0 = ∫0;2*п f(x) dx ≥ ∫x0-a;x0+a f(x) dx > 0 contradiction.
So f(x) = 0 for each x in (0, 2*∏) so in [0, 2*∏ ] (f is continuous).


calculus
derivative
function
inequalities
logarithms
inequalities proposed

Let's just think, this is not a solution

the minimum value of  $\frac{1}{x}$ is when $x\to \infty$

but, since the product of the variables need to be $1$

so, we can make one variable $x\to 0$
if we make this, so, the anothers variables can tend to infinity

if we take the partial derivatives, we will find that  $x\to \infty$ fo all variables

but this is impossible, since one variable need to $x\to 0$

so we will get $1+0+0+0+0...=1$

this is not a proof

it's justa analisis

OK. $a_{1}=0,a_{2}=a_{2}=...=a_{9}=+\infty$ is also one case of equality (different from the obvious case $a_{1}=a_{2}=...=a_{9}=1$.)

$a_{1}=1,a_{2}=a_{2}=...=a_{9}=+\infty$

tis does not make sense

the true is

$a_{1}=0,a_{2}=a_{2}=...=a_{9}=+\infty$

Assuming that this product is
$\infty . 0 = 1$

but need to use limit to make this strange math hehe

It's unexpected! I don't know that this function is convex. Is your solution valid for
\[\sum \frac{1}{(a+1)^{2}}\ge 1\]
if $abcd=1$?

[quote="GacKiem"] Is your solution valid for
\[\sum \frac{1}{(a+1)^{2}}\ge 1 \]
if $abcd=1$?[/quote]
Easy to check. But it's a very known inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98967

Ok. My solution for both problem of 4 numbers and problem of 9 numbers bases on Cauchy. If convex function is valid, these problems seem to be simpler than I though.

The case $n=4$ is a Chinese inequality. I would like to dedicate the following solution to my Chinese frieinds, Ji Chen, Zhao Bin:

Let $a,b,c,d$ be positve real numbers such that $abcd=1$, prove that
\[\frac1{(1+a)^{2}}+\frac1{(1+b)^{2}}+\frac1{(1+c)^{2}}+\frac1{(1+d)^{2}}\geq1. \]
Put
\[a=\frac{yzt}{x^{3}},\; b=\frac{xzt}{y^{3}},\;\; c=\frac{xyt}{z^{3}},\;\; d=\frac{xyz}{t^{3}}. \]
Let $L$ be the left hand side of the given inequality, then the desired inequality reads
\[L=\frac{x^{6}}{(x^{3}+yzt)^{2}}+\frac{y^{6}}{(y^{3}+xzt)^{2}}+\frac{z^{6}}{(z^{3}+yxt)^{2}}+\frac{t^{6}}{(t^{3}+yzx)^{2}}\geq1. \]
By Cauchy Schwarz inequality,
\[L\geq\frac{(x^{3}+y^{3}+z^{3}+t^{3})^{2}}{x^{6}+y^{6}+z^{6}+t^{6}+(xyz)^{2}+(yzt)^{2}+(ztx)^{2}+(txy)^{2}+2xyzt(x^{2}+y^{2}+z^{2}+t^{2})}. \]
It is now sufficient to prove that
\[\frac{(xy)^{3}+(yz)^{3}+(zx)^{3}+(xt)^{3}+(ty)^{3}+(zt)^{3}}2\geq \sum_{\rm cyclic}(xyz)^{2}, \]
and
\[\frac23[(xy)^{3}+(yz)^{3}+(zx)^{3}+(xt)^{3}+(ty)^{3}+(zt)^{3}]\geq xyzt(x^{2}+y^{2}+z^{2}+t^{2}), \]
both of which can be solved using AM-GM inequalities say
\[(xy)^{3}+(yz)^{3}+(zx)^{3}\geq 3(xyz)^{2}. \]
and
\[x^{3}(y^{3}+z^{3}+t^{3})\geq3x^{3}(yzt). \]

This is the end of the proof.

Dear Thuan! This is also my solution basing on Cauchy as I've said above and this solution helps me create the problem of 9 numbers. Anyway, it's easy.

With the same substitution, now we have a new solution for Vietnam Team Selection Test 2005.
\[ \left(\frac a{a \plus{} b}\right)^3 \plus{} \left(\frac b{b \plus{} c}\right)^3 \plus{} \left(\frac c{c \plus{} a}\right)^3\geq\frac38.
\]
Let $ b/a \equal{} yz/x^2$, $ c/b \equal{} zx/y^2$, $ a/c \equal{} xy/z^2$. The desired inequality reads
\[ \frac {x^6}{(x^2 \plus{} yz)^3} \plus{} \frac {y^6}{(y^2 \plus{} zx)^3} \plus{} \frac {z^6}{(z^2 \plus{} yz)^3}\geq\frac38.
\]
By Cauchy, ............. we need to prove
\[ 5(x^6 \plus{} y^6 \plus{} z^6) \plus{} 13(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3)\geq 9xyz(x^3 \plus{} y^3 \plus{} z^3) \plus{} 27(xyz)^2.
\]
This must be simple.

Now, can you prove?

[b]Problem.[/b] Let $ a,b,c,d,e,f$ be postive real numbers such that $ abcdef \equal{} 1$. Prove that
\[ \frac1{(1 \plus{} a)^3} \plus{} \frac1{(1 \plus{} b)^3} \plus{} \frac1{(1 \plus{} c)^3} \plus{} \frac1{(1 \plus{} d)^3} \plus{} \frac1{(1 \plus{} e)^3} \plus{} \frac1{(1 \plus{} f)^3}\geq\frac34.
\]

[quote="arqady"]Let $ f(x) \equal{} \frac {1}{(2e^{x} \plus{} 1)^{2}}.$ [/quote]Then $ f''(x) \equal{} \frac {4e^{x}(4e^{x} \minus{} 1)}{(2e^{x} \plus{} 1)^{4}} < 0$ when $ x<\minus{}2\ln{2}\equal{}\minus{}1.3862\cdots$ :wink:
[quote="hitek"][b]Problem[/b] Let $ a,b,c,d,e,f$ be postive real numbers such that $ abcdef \equal{} 1$. Prove that

$ \frac1{(1 \plus{} a)^3} \plus{} \frac1{(1 \plus{} b)^3} \plus{} \frac1{(1 \plus{} c)^3} \plus{} \frac1{(1 \plus{} d)^3} \plus{} \frac1{(1 \plus{} e)^3} \plus{} \frac1{(1 \plus{} f)^3}\geq\frac34.$[/quote]See http://www.mathlinks.ro/viewtopic.php?t=32332

Cauchy-Schwarz is applied wrongly. Look at your algebra more closely.

[quote="mehdi cherif"]yes thanks. it's just a wrong of innatention  :D[/quote]
It's edited  :)


若一個數只由0 6 8 9四個數碼組成稱為何理數
前16個
何理數有小到大依序為
6 8 9 60 66 68 69 80 86 88 89 90 96 98 99 600
則第2008個正何理數是多少?

[color=darkblue]Translate[/color]
If an integer comes from four digits 0,6,8,9 , we call it Holi Numbers.
The first 16 Holi Numbers list in the ascending order as the following:

6 8 9 60 66 68 69 80 86 88 89 90 96 98 99 600

How about the 2008th Holi Number ?


[quote="Phelpedo"]Let $p$ be a prime. Compute the number of subsets $T$ of $\{1,2,...p\}$ such that $p$ divides the sum of the elements of $T$.

[hide="eh?"]
The generating function we want is
\[F(x) = (1+x)(1+x^{2})...(1+x^{p}) = \sum a_{k}x^{k}\]
Now we let $\omega = e^{\frac{2\pi}{7}i}$ and use the root of unity filter:
\[\sum_{i=0}^{p-1}F(\omega^{i}) = p \sum_{p|k}a_{k}\]
So all we have to do is find $F(\omega) = F(\omega^{i})$. So
\[F(\omega) = (1+\omega)(1+\omega^{2})...(1+\omega^{p}) \]
Define $G(x) = (x-\omega)...(x-\omega^{p}) = x^{p}-1$. Plugging in $-1$ for x,
\[(-1-\omega)...(-1-\omega^{p}) =-(1+\omega)(1+\omega^{2})...(1+\omega^{p}) =-2 \]
so $F(\omega)=2$. Additionally, $F(1) = (1+1)(1+1^{2})...(1+1^{p})=2^{p}$. So
\[\sum_{i=0}^{p-1}F(\omega^{i}) = 2^{p}+2p = p\sum_{p|k}a_{k}\]
Only problem is, $2^{p}+2p$ isn't divisible by $p$...what?[/hide][/quote]

It looks like it should be
[hide="this"]
\[\sum_{i=0}^{p-1}F(\omega^{i})=2^{p}+2(p-1)\]
[/hide]
at the end


calculus
integration
calculus computations

A company purchases a new machine for which the rate of deprecation is:

$ \frac{dV}{dt} \equal{} 10000(t\minus{}6)$ for $ 0 \leq t \leq 5$.

where V is the value of the machine after t years. Evaluate the definite integral that yields the total loss of value of the machine over the first 3 years.

[quote="math92"]rate of deprecation[/quote]  How quickly it [url=http://dictionary.reference.com/browse/Deprecation]expresses earnest disapproval[/url]?  :P


vector
function
Support
superior algebra
superior algebra solved

Find an orthogonal basis of $\mathbb R^{2^n}$ s.t. the components of the vectors are $\pm 1$.

This is fairly well-known. Start with the "Rademacher vectors": Let $r_0$ be the vector of all 1's. Let $r_1$ have $2^{n-1}$ 1's (halfway across) followed by the same number of -1's. In general, let $r_k$ have alternating blocks of $2^{n-k}$ 1's and blocks of the same number of zeros. To illustrate this with $n=3$:
$r_0 = (1,1,1,1,1,1,1,1)$
$r_1 = (1,1,1,1,-1,-1,-1,-1)$
$r_2 = (1,1,-1,-1,1,1,-1,-1)$
$r_3 = (1,-1,1,-1,1,-1,1,-1)$
This set of vectors is not merely orthogonal; it is probabilistically independent. (You can think of it as recording the results of each of n coin flips.) The probabilistic independence allows us to do what we're going to do next.

Take the set of all subsets of {$r_1,r_2,\dots ,r_n$}. For each such subset, take the component-wise product of all the vectors in that subset. Use the convention that if the subset is empty, the result is $r_0$. This gives us $2^n$ orthogonal vectors composed of 1's and -1's. Think of the set of vectors as a list corresponding to the entire sample space for the experiment of flipping the coin n times. It might be worth your while to write out the 8 vectors you get in the $n=3$ case just to convince yourself that it works.

[quote="liyi"]wavelet basis.[/quote]

Related, yes, but not exactly. The basis functions for a wavelet basis would have limited support, and would hence have 1's, -1's, and 0's, instead of all 1's and -1's.

That's cool. I did the following: for $\displaystyle R^2$ take $(1,1),(1,-1)$. Then, to go from one step to the other, replace all $1$s by $1,1$ and $-1$s with $1,-1$ at first, getting half of the vectors needed on the next step, then replace the $1$s with $1,-1$ and the $-1$s with $-1,1$ to get the other half.


Find the sum of the 60 positive odd whole numbers beginning with 5.

[quote="math154"]Let $ S_n$ be the sum of the first $ n$ odd whole numbers. It is well known that $ S_n\equal{}n^2$, so our desired sum is $ S_{62}\minus{}S_2\equal{}62^2\minus{}2^2\equal{}(62\minus{}2)(62\plus{}2)\equal{}60\cdot64\equal{}\boxed{3840}$.[/quote]

How did you get $S_{62}\minus{}S_2$?

He added the first 62 odd numbers, which includes the 60 odd numbers after 5 as well as 3 and 1, then subtracted the 3 and 1 out.


Veo k ace rato k no se ponen desigualdades x aki, asi k pongo esta:

Si a, b, c son reales positivos k cumplen a+b+c >= 1/a+1/b+1/c
demostrar que a+b+c >= 3/abc

Si necesitan algun tip, favor d pedirlo!!!

 :)

Supongamos que dicho resultado no es cierto, es decir existen números reales positivos a,b,c tales que:
i) $abc(a+b+c)<3$
ii) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le a+b+c$

Elevando la segunda desigualdad al cuadrado, y multiplicandola por la primera obtnemos:

$abc(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}<3(a+b+c)$
$(ab+bc+ca)^{2}<3abc(a+b+c)$
$(ab)^{2}+(bc)^{2}+(ca)^{2}<abc(a+b+c)$
$(ab-bc)^{2}+(bc-ca)^{2}+(ca-ab)^{2}<0$
Esto último es una contradicción.


What is the sum of the reciprocal of every number in row n of pascal's triangle (the top row is row 0) in terms of n?

AFAIK, there is no closed form expression for this sum. But there are some interesting identities like \[\sum_{k=0}^{n}{n\choose k}^{-1}= \frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^{k}}{k}.\] For further details see [url=http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/sum_reciprocal_binomials][1][/url] and [url=http://www.arxiv.org/abs/math.CO/0104026][2][/url].


probability function articles Support algebra domain

Over the years I've gone back and forth a few times on whether or not I accept some form of the [url=http://en.wikipedia.org/wiki/Doomsday_argument]Doomsday argument[/url], and I'm curious what the mathematically-oriented community here makes of it.  At least as a formal calculation, the probabilistic formulation of the statement and all the underlying assumptions seem to be reasonable.  At the same time, it seems to me the conclusion implies at least some a priori knowledge of things that, intuitively, I feel we should not and do not have a priori knowledge of.  Yet, the formulations I've seen are all vague enough to resist my attempts to find a specific assumption or logical step that I feel to be clearly unintuitive or incorrect.  I've read some of the common philosophical and probabilistic objections, but I haven't found any to be thoroughly convincing.

So what do you think, does it make any sense?  If not, exactly which assumption do you reject, and why?

Well, it merely says that if the remaining interval is much larger (or smaller) than the elapsed one, then you observe an event of small probability. Now, next time you are at the airport, look at the person standing in front of you in the security line and ask what is the probability (assuming that everybody has equal chance to travel on this day) to see just this person at that particular time. If it is still large enough for you, look at the person in front of that person and ask what is the probability of seeing them both (assuming independence), and so on. By the tenth or so person, you'll reach the event whose probability is so small, that if you allow the same probability to be observed in the Doomsday argument, the remaining interval will far exceed what we expect to be the lifespan of the Universe.

So, the only thing it tells is that if each human being estimates the remaining time as about equal to the elapsed time, then you get quite a lot of people right about their estimate and there is no much better strategy in the collective game of guessing that time. More precisely, if we play this game in the sense of minimax, i.e., you are allowed to design the lower bound $ f(t)$ and the upper bound $ g(t)$ depending only on $ t$ and differing at most $ k$ times, say, and tell me these functions, and I choose the $ T>0$ and draw a random sample $ t$ from $ [0,T]$, your task being to make the probability that $ T\in [t\plus{}f(t),t\plus{}g(t)]$ as large as possible for the case when I choose the worst for you $ T$, it can be shown that your optimal strategy is to choose $ f(t)\equal{}ct$ and $ g(t)\equal{}kct$ where $ c$ is the root of $ (1\plus{}kc)^2\equal{}k(1\plus{}c)^2$ (whence we always have $ c<1<kc$, so $ t$ is always a good guess for the remaining time). Thus, if every human being ever born is required to estimate the remaining time in this way and the God awards the humanity the bonus proportional to the portion of right guesses in the afterlife, you know how to make your guess and this is what the Doomsday argument is really about.

The only little problem is that this collective guessing game is, probably, not what you are interested in.

It is logical, however, it's very hard to tell if we're in the 5%, or if we're in the 95%. In addition, if we're in the 95%, it's very hard to tell if we're in the part that gets us extinct in the next 4000 years, or in the later 5120 years.

What might be more useful of that would be a graph of chance verses the greatest amount of years for extinction based on that chance.

fedja, I agree with most of what you wrote, but it's not really what I was getting at.  I have little doubt that the argument is correct in the academic sense -- if all the assumptions hold, then the conclusion is valid.  That is merely a matter of calculation, and not particularly interesting.  Rather what I'm not sure about is that all the assumptions are even remotely reasonable, or that there is a relevant probability space on which we can make calculations to begin with.

For example, in your airport security case, there is absolutely no way we can calculate any probabilities unless we make some assumptions -- e.g., any person (from some set of people we need to define) is equally likely to be standing in the line in front of me.  There are many other assumptions, some minor and some not, that we would need to make.  The restrictions imposed by the set of assumptions and definitions that we choose to make in turn lets us define the probability space that we use to make calculations.

Now, back to Doomsday.  First, the observer assumes that his birth is uniformly distributed among the birth of all humans.  How reasonable is this assumption?  Maybe not too bad at first glance.  So let's say we accept it and get some implied probability distribution on the end of the human race.  But now let's see what happens if we enrich the class of observers and say that an observer may be either a human or a dolphin (who we stipulate, too, can observe whether or not humans exist).  If the periods of existence of dolphins and humans do not overlap perfectly, the implied probability distribution will change.  For example, if most dolphins exist during a period in which very few humans have existed so far, then the dolphin observers will estimate the expected time to human extinction as being much sooner than will the average human observer.

So the calculation depends on who the observer is; this isn't surprising.  But it reveals a major question about what at first seemed a perfectly reasonable assumption -- we assumed that the observer's birth is uniformly distributed among the birth of all [i]humans[/i].  Why should we decide our existence is undefined unless we are human?  This assumption seems to have an almost theological slant, and conceals a bias that will clearly affect the result of the calculation we're ultimately interested in.  If we have no clearly motivated way of assigning the set of observers, does the calculation even have any meaning?

This is not the only, and perhaps not even the most important, problem that I have with the argument.  But there are of course counterarguments, some rather convincing as well, and I admit the whole realm of discussion is much closer to philosophy than it is to mathematics.

Well, the only assumption that is really made is that there will be a finite number of humans (which makes sense). After that the task is to guess the length of the chain beyond you knowing only how many people are in front of you. The conclusion is merely that the guess "about as many as there are in front" will be close to the truth for a large portion of the people in the chain and if you try to guess in any other way, that portion may go down if the chain length is chosen in the appropriate way. That's all; the probability language is there just to confuse the matters because it raises a lot of senseless questions like "am I a random person, and, if yes, according to what distribution?", etc. There is absolutely no "philosophical" content in the entire story, it is just a simple strategy game. If you change "humans" to anything else ("Joe's", "living beings", etc.) and require everybody in the chain make a guess (well, it is rather interesting how you can make a frog to make a guess, but it is a different story), you'll get the same conclusion: the strategy "about as many as there are in front" is optimal in the sense that it gives the right answer for the largest portion of the participants (but it may be a completely different portion because you changed the set of the game participants). Everybody can invent dozens of such guessing games, but for some strange reasons this particular choice of the chain and the participant set attracted so much attention that even made its way to Wikipedia. The whole trick is restating the correct statement "If there are 60 people in the chain, then only $ 5$% of them have numbers below $ 4$" as a requirement that the third person in the chain should estimate the chance of the chain being longer than 60 people by below 5% (whatever that means). You see, it doesn't really hold any water. The reasons we are talking like that in statistics are mainly historical and everybody understands what is really meant.

[quote="fedja"]If you change "humans" to anything else ("Joe's", "living beings", etc.) and require everybody in the chain to make a guess (well, it is rather interesting how you can make a frog to make a guess, but it is a different story), you'll get the same conclusion: the strategy "about as many as there are in front" is optimal in the sense that it gives the right answer for the largest portion of the participants (but it may be a completely different portion because you changed the set of the game participants).[/quote]That's exactly right, the strategy will be exactly the same, but the answer will be completely different.  But we're interested in the [i]answer[/i], not the strategy. The whole point of asking the question was to get an answer, not to engage in a guessing strategy exercise.  And this is the the only claim I'm making: it's not clear that the Doomsday argument carries any valuable information value, insofar as the calculation is highly sensitive to assumptions that are made arbitrarily (e.g. that having been born human is the only possible state of observation).

I don't think we're disagreeing about anything, just talking about different things.  You're looking at it and asking whether or not the doomsday argument gives an optimal guessing strategy within a framework of certain assumptions.  And I think you're right that it does.  I'm looking at the doomsday argument and asking whether or not it has predictive value not merely within its assumptions, but within [i]objective reality[/i].  And I'm not sure that it does.

Well, as I said, it gives you the optimal strategy in some particular guessing game with restricted information and certain objective [b]but nothing else[/b] for that simple reason that it has never been really intended to be anything else. So, it has no "objective value" beyond that if something in the human life depends upon correctly guessing that remaining period [b]and their behavioral choices do not change it [/b] (which I find highly unrealistic), the best thing you can do [b]in absence of other information[/b] is to behave like it equals the elapsed time.

A question for JoeBlow: as stated in the Wikipedia article, the conclusion of this argument is something to the effect of, "There is a 95% probability that humanity will be extinct in 10,000 years."  (In fact, I think it should be more like, "There is a 95% probability that humanity will suffer a population crash of sufficient proportions to render this calculation no longer valuable in the next 10,000 years," since the computation is quite sensitive to the expected future stable population before extinction.)  Does this seem like an unreasonable conclusion to you?  To me, it sounds completely consistent with other things I know about the world, e.g., plausible estimates for the biological carrying capacity of the planet, current trends in species extinction, the general rarity of charismatic megafauna, ongoing climactic changes (and, in the longer run, the likely return to an ice age in the next dozen or two millenia), etc.

It sounds a bit unreasonable to me :). I believe that the civilization will either end much sooner or last much longer (do not really have a clear idea which is more likely). We consume the resourses and destroy the planet at much higher rate than would allow us to last for the next 10,000 years without changing our current lifestyles. On the other hand, we gradually come up with inventions that may make us independent of the currently used resources and much more resistant to the geological and climatic events (I assume here in particular that the current blood and flesh human form is of no particular value to the civilization; that will have to be abandoned eventually if we want to survive the next ice age or something like that). The question is what will come first, the population/climate/war/whatever crisis or something like what currently goes under names like "singularity". For all I can tell, if you give the civilization a thousand years of uninterrapted development, it will reach the point where it could easily exist on biologically bare planets with harsh climatic conditions (I do not say that it would justify turning the Earth into one of those either before or after that, of course). So, I'm quite optimistic in the sense that beyond this 1,000 years, we may have eternity. :) On the other hand I'm rather pessimistic in the sense that I suspect we may not have this 1,000 years :(. In either case, I believe that it is either a beginning, or an end, not the middle. Of course, I may be completely wrong here; it is really just another guessing game with insufficient or hard to interpret information.

Anyway, I have another question for JoeBlow: suppose that tomorrow somebody (time traveler from the future, or whoever else who really knows) will tell you that the remaining time is 5,329 years, 4 months, 2 weeks, 1 day, 3 hours, 5 minites and 6 seconds. What will you do with that information? (I mean is it of any more importance to you, really, than knowing the exact (or approximate) number of the needles on a pine tree that grows outside the office building in Seattle I'm now sitting in?). If your next question to that person will be "What can I do to increase/decrease this number?", it can be answered without knowing the number itself and, in my humble opinion, that is all that really matters...

the doomsday argument seems surprising familiar from personal experience. certainly, when I join some facebook group, it's usually pretty reasonable for me to expect that it's long-term membership will be of the same order of magnitude or maybe one higher than the number of people when I first join. so very roughly, I don't have any problems believing doomsday in this respect. we run into problems when trying to define what exactly humans are, since on these long timescales, there's obvious potential for change. I don't think it's necessarily theological to confine the results to humans -- the argument would seem to apply equally well to "black-haired mathematicians", "humans", "mammals", or "multicellular organisms" -- it merely happens that the case of particular interest is the second one here, in some sense that has to be defined a bit more clearly. The other issue is whether one says "hmm it is roughly equally likely that I am within the first half versus the second half of humans that have ever lived" or "hmm it is roughly equally likely that I am within the first half versus the second half of human civilization", i.e. with respect to what independent variable (population, time, or so forth) one formulates this mediocrity principle. I guess this is really what the choice of the prior probability distribution is all about, and I suppose that's based on some sociological data somewhere ?

Believe it or not, I still think this world will end before I'm old enough to receive Social Security (I'll be 65 in 50 more years).

Most of the prophecies in Revelation are either already fulfilled or going to be fulfilled in the next few years. That means Jesus's coming back, and ending this world.

One more thing regarding this (on topic): These chances are only if it's completely random when each human is born, and the thing is, those humans born already are already born. In other words, you have no idea what fraction of the way it's already been.

However, that's also why they have to use probability. However, probability is something so hard to determine, especially if you don't know what determines which chance comes out. Something with 5% A and 95% B has the same amount of possibilities as something with 50% A and 50% B. In other words, you can't tell if it's going to be A or B, even if it's more percentage for B.
So while the 95% might limit us to 9120 years later, extending it to 98% can extend the year of sure-we're-extinct A LOT further, and extending it to 100% will extend the year of sure-we're-extinct to infinity, since 1 over an infinitesimal number is infinity.
And having 98% chance of a later year, or 92% chance of an earlier year, is still using probability and thus you can't know if it's in the true or the false.

And in support of my I-won't-live-to-get-social-security, we have a chance (like 10% or something) that it ends in the next 50 years.

yeah, we should probably keep religious prophecies out of this. I'm not convinced that the argument presented here is based on rigorous mathematics, but it certainly makes an attempt to at least stay within the confines of logic and hence does not lie within the domain of religion.

also: thanks for explaining probability to us ?

JBL: I don't think the conclusion is unreasonable.  But whether or not the conclusion is correct has no bearing on the validity of the argument, nor the validity of the assumptions underlying it.  Consider the following argument: (all pink elephants are not grey)$ \Rightarrow$(the odds are 19:1 that the world will end within ten millennia).  It's evidently true that all pink elephants are not grey, but the conclusion (though it may very well be true) does not follow.  But I suppose what you were asking is what reasons I have to doubt the validity of the argument.  The reason I doubt the argument (or, more precisely, doubt either the argument, or its assumptions, or both) is not that the conclusion seems to me to be clearly false, but rather that the conclusion seems to me to hold more information than was at first glance implied by the stated assumptions alone.

fedja: I would travel the world.  But if you're implying that thinking about the question is not interesting because even an exact answer to it wouldn't be practically useful, I strongly disagree.  Much of philosophy and mathematics has revolved around, and been driven by, seeking answers to questions that have no clear utility.  What would you do with it if, for example, you were told that there are infinitely many twin primes?  What would change if you were told that the Riemann Hypothesis is true?  How would it change life on earth if you discovered with absolute certainty that the many-worlds interpretation of QM (or insert here any arguably unfalsifiable theory) is or isn't 'true'?  These questions may not be useful (and perhaps in the latter example an absolute 'answer' will never be within reach), but I think many on these fora will disagree that they are not worth thinking about.

Let me narrow down my question to something more specific: The Doomsday argument seems to depend critically on the following assumption (in fact I would dare to say it's [i]equivalent[/i] to it):
[quote]"I could have been born the $ k$th human being with probability exactly $ \frac {1}{N}$ for each $ 1\leq k\leq N$, where $ N$ is the total number of human beings that will ever exist."[/quote]Under what conditions is this assumption reasonable (given that we agree on some precise definition of 'human being')?

Like I said, "those humans born already are already born."


In order to be better prepared for the next MIMC, we are going to go ahead and start writing it.  It will not be released for a while yet, but here is the appeal for problems.  If you wish to submit problems, and do not plan on taking the MIMC II test, you may submit them via PM to [b]MIMC Director[/b].  Please note that if you submit problems, you cannot take the MIMC II test.  All problem authors with 3 or more problems included in the test will be recognized on the problem document.  Thanks!

Current MIMC Staff:
MIMC Director
Iversonfan2005
mysmartmouth
joml88
arkmastermind
elemmnop
altheman

Just a suggestion, to limit the problems on the test- why don't you have a couple beta-testers? That way you can identify problems beforehand from the viewpoint from different people.

[quote="mysmartmouth"]The MIMC Staff will be the beta testers.  However, an idea we were given is:

Have a small group of extremely smart people take the contest a month early, turn in answers a month early, and get feedback from them.  They would be just like the AoPS community taking the test, with the same testing conditions, just during a different time window.  If you place in the Top 16 on MIMC I, and want to be a beta tester for MIMC II, PM MIMC Director immediately.  Thanks![/quote]
Ehm...that's basically what I meant.
It really would make no sense to have the people who make the test to beta-test it. I have made many tests, and it didn't work.

Eh, well I'm sure it would go better this time if the staff were given a little more time to look over the test before it was released...

More people will participate this time because it's summer.

[quote="Xantos C. Guin"]Would $\beta$ testers still be eligible to win awards?

Also, will the MIMCI solutions PDF be compiled, or is that dead now?[/quote]

Beta testers would be eligible for everything, and no distinction will be made between beta testers and regular testers on the test results.

MIMC Solutions will be compiled as soon as we get all the solutions together, so keep posting the solutions to the problems listed.

[quote="not_trig"]I think beta testers should be in their own category if scored :D[/quote]

What would be the reasoning behind that?  Beta testers would not have any advantages, in fact, they would be taking a potentially flawed test, putting them at a disadvantage.  But all beta testers would be forewarned about that, and it would be their decision to participate as a beta tester.

Since the beta testers were high scorers on the original  :D


logarithms
function
geometry
inequalities
3D geometry
rectangle
probability

Here are the rules

1. If you want to pist a question you must answer a quistion.
2.To answer a question you must post a question
3.No impossible ones
4.No offensive language

I'll start with a easy one
$ (3^2)*2$

18

Hmmm...

4=|x^2|

What are the possibilities for x?

Pretty easy, although there are some that people forget about.

$ x\equal{}2, \minus{}2$

$ \log_2{16,384}$

Let's see if anybody can get this.

I'll do Richard_Min's problem because it's easier.

$ \boxed{x \equal{} \sqrt [3]{81}, \minus{} \sqrt [3]{81}}$

I want to do an easy one:

$ 6! \equal{}$

$ 720$.

And to answer spaceguy's question, $ \log_{2}{16,384}\equal{}13.9999998$

I'll do a hard one.
Find the critical number of the function $ g(t) \equal{} | 6t \minus{} 9 |$.

9/2 is the critical number, here's a hard one,

What is $ 100!$?

Oh this is hard...
2.
I think. Not quite sure.

Humm...
Let's be easy..
$ \text{Please state the prime factorization of}$  $ \sqrt92034$

um
i'm just gonna do my own problem..
2..
:)
Ok
um
2+2

4

If $ a! b!\equal{}c!$ for positive integers $ a$, $ b$, and $ c$, find the smallest sum $ a\plus{}b\plus{}c$. (This should be trivial.)

3

If $ a!b!<c!$ for positive integers $ a<b<c$, find the smallest value of $ a\plus{}b\plus{}c$. Also trivial.

6

If $ a_1!a_2!a_3! \ldots a_{2009}! < n!$ for non-negative integers $ a_1 < a_2 < a_3 < ... < a_{2009} < n$, find the smallest value of $ a_1 a_2 a_3 \ldots a_2009$ (i.e. the product of all $ a_k (1 \leq k \leq 2009)$).

 :roll:

0

The sum of 2 numbers is 34. What is the greatest value of their product?

If by "greatest" you mean greatest possible, 289.

Consider right triangle ABC, with <C a right angle. The sides have length AC=5, BC=12, and AB=13. Draw altitude CD onto AB. Find the area of triangle ACD.

[hide="solution"]We have $ \dfrac{b}{a}\equal{}\dfrac{10a\plus{}b}{10}\Longrightarrow 10b\equal{}10a^2\plus{}ab\Longrightarrow b\equal{}\dfrac{10a^2}{10\minus{}a}$.

Then, we get $ a\equal{}2,b\equal{}5$, for $ \dfrac{a}{b}\equal{}\boxed{\dfrac{2}{5}}$.[/hide]

[hide="New Problem"]Show that $ \sin^2{2x}\equal{}(2\cos{x}\plus{}2\cos^2{x})(2\cos{x}\minus{}2\cos^2{x})$.[/hide]

:mad:  :oops: = rotfl since a double negative is a positive.
How many zeroes are at the end of 10000000!

less than 15!,15!!, and 15!!!.
it's actually 2499999

show that 1+1=2

1+1=2
I just showed you
Now
what is blah blah blah blah blah blah +pizza pizza pizza * pizza pizz +blah blah 0 * alskd /45?

45?

Convert 2989 to hexadecimal and decode the message.

2989 = $ BAD_{16}$
which decodes as "Nosy" (i.e. No C)
(fail)

You have an impenetrable sphere, a compass, a ruler, and a rigid flat fixed piece of paper. The compass can be used to construct a circle on either of the sphere or the paper, or to transfer distances. The ruler can only be used to draw a straight line on the paper. Under these conditions, prove that, for any two reals $ a$ and $ b$, at most two of them are integers. Show that this upper bound can be obtained. You may assume that the sphere is of uniform density, frictionless, uncharged, an insulator, literally impenetrable, rotatable, massless, and red.

x=2

what is the smallest three digit number in pascal's triangle?

 :D  :P  :ninja:

9.86076*10^71

What is the answer to AIME 2007 II, number 13?

$ 5 \cdot 2^7 \equal{} 640$

What's the prime factorization of the cube root of 24,389?

15 :rotfl:

what is the inverse of 28 mod 29


geometry proposed
geometry
U sviti matematyky

Circles $ \omega_1$ and $ \omega_2$ intersect at points $ A$ and $ B.$ Diameter $ BP$ of $ \omega_2$ intersects the circle $ \omega_1$ at point $ C$ and diameter $ BK$ of the circle $ \omega_1$ intersects the circle $ \omega_2$ at point $ D.$ The straight line $ CD$ intersects the circle $ \omega_1$ at point $ S\ne C$ and the circle $ \omega_2$ at point $ T\ne D.$ Prove that $ BS\equal{}BT.$

we have the quadrilateral $ PDCK$ is cyclic so $ \angle CKD\equal{}\angle CPD$. Thus $ \angle BST\equal{}\angle BTS\Rightarrow BS\equal{}BT$
It's easy problem :D  :D


Could anyone tell me about a good book on NUMBER THEORY.

Well books come in a lot of different levels of sophistication. So it depends on what grade you're in, and what level of style and rigor you're looking for. It also depends if you're looking for textbooks or problem solving books.

If you're looking for "textbooks", for starters I'd recommend the book [u]The Higher Arithmatic[/u] by H. Davenport. It's a bit of an older book, so some of the information on computers and record primes and unsolved problems is outdated, but it's a classic and the content and explainations are very good. Another good book for starters is [u]Number Theory[/u] by G. Andrews. Both of these books are probably appropriate for high school students of an upperclassman level.

For slightly more advanced books, I recommend [u]An Introduction to the Theory of Numbers[/u] by Niven, Zuckerman, and Montgomery (this is used in a lot of college introductory number theory courses) and [u]A Course in Number Theory[/u] by H.E. Rose (this book ranges from the basics to the very deep. I like it because you can learn from it, but it will still keep you busy for a while.)

My daughter enjoyed the flavor of "An Adverturer's Guide to Number Theory" by Richard Freiburg.  More imaginative, readable, and playful than most texts of at this level - still very good and honest mathematics.  Lots fo good problems.

Years ago, I used "Elemtary Number Theory" by Armendariz and Stephen McAdam - a good college-level text.

The afforementioned one by Andrews is great.  It's easy for self-study.  And of course, someone has to say it, if you want a real challenge, go Hardy/Wright.

A further recommendation, in addition to the excellent ones above, is "A Pathway into Number Theory" by R.P. Burn, which is especially suitable for self-study.

This is an unusual book which provides an excellent introduction to number theory. It is more of a course in structured thinking than a normal textbook. Very little prior knowledge is assumed, and it contains no expositions or formal proofs - instead each chapter consists of a sequence of questions and problems, building one on top of another. These exercises lead the reader step by step to discovering key results in number theory, from the infinitude of primes through to quadratic reciprocity.

Each chapter starts with numerical examples to establish motivation and background, and ends with brief solutions to all exercises and a summary of key concepts - but a reader who skips straight to the answers and the summary will gain very little from this book.

I was studying at the University Math and Computer Science. But now I know that what is really interesting for me and what I like the most, is Mathematic...I know mathematic, I mean advance mathematic, but I realize that I have to learn more about number theory. At this moment I'm not studying, Thus, I think I'll change to Mathematic......but I don't know about competition, but I like it, I used to compete for Computers Program competition in my country and they taught me Number Theory, but not at all...so I want to go deep thru Number Theory right now...So guys, what book do you recomend me....?

[quote="Magnara"]If you really, honestly want depth, Hardy/Wright is the place to go.[/quote]

I agree. Hardy/Wright is simply amazing.

[quote="Magnara"]If you really, honestly want depth, Hardy/Wright is the place to go.[/quote]

I would actually (somewhat) disagree. It's not that Hardy and Wright isn't a great book - it's just that there are books with more modern treatments of number theory nowadays. I would actually recommend either [u]A Course in Number Theory[/u] by H.E. Rose, or [u]A Classical Introduction to Modern Number Theory[/u] by Ireland and Rosen as better first choices than Hardy and Wright.

Ireland and Rosen in particular is a GREAT book. It gives good historical information and it gets into some deep and fairly new results. Some parts of it are a bit advanced (it's billed as a Graduate Text), but advanced high school or undergraduate students who take the time can follow it with few prerequisites.

Anyone who's seriously interested in number theory should at some point read Hardy and Wright, but it's not necessarily the best place to start.

Edit: I was just looking at my Ireland and Rosen book and maybe it is a bit too advanced to begin with. I would probably just recommend Rose for a comprehensive introduction to the basics of number theory, then follow it later by Ireland and Rosen.

I find [i]Fundamentals of Number Theory[/i] by William LeVeque quite good. It also has the benefit of being a Dover book and hence being about a third the price of most of these other books. Hardy/Wright is terrific for theory, but alas there are no problems, and so it will be difficult for the reader to get a complete understanding. (One doesn't learn mathematics by watching other people doing mathematics.)


Does anyone know how to program commands that will change based on the font size?  I know there are different measurements, such as "em" and "ex,"  but there is still some funky stuff that happens.

I'm not sure what you mean but section 6.2 of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]A (Not So) Short Introduction to LaTeX2e[/url] may help.


I see, like, the \mathbb and the \mathcal things on the Instructions, but when I try doing that, it doesn't work. I even tried downloading the amsfonts package. Might it be that i'm not utilizing the package right? When I do the mathcal, i just get some random symbols. Like this $ \mathcal{hello. my name is mathboy. see?}$.

You need the amsmath and amssymb packages, which it's worth using in all your documents.
Here is a minimal example document:
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
$\mathbb{R},\mathcal{R},\mathfrak{R}$
\end{document}[/code]


linear algebra
matrix
calculus
derivative

For 1):
Let $ J$ be the upper triangular nilpotent Jordan matrix of dimension $ n$ ($ J^n\equal{}0,J^{n\minus{}1}\not\equal{}0$). $ A\equal{}I\plus{}2J\plus{}\cdots\plus{}nJ^{n\minus{}1}$.  We do this formal calculus:let $ x$ s.t. $ x^n\equal{}0$; $ f(x)\equal{}x\plus{}x^2\plus{}\cdots\plus{}x^n\equal{}x(1\minus{}x^n)/(1\minus{}x)\equal{}x/(1\minus{}x)$; then $ f'(x)\equal{}1\plus{}2x\plus{}\cdots\plus{}nx^{n\minus{}1}\equal{}1/(1\minus{}x)^2$. Thus $ A\equal{}(I\minus{}J)^{\minus{}2}$ and $ A^k\equal{}(I\minus{}J)^{\minus{}2k}$. We conclude using binomial formula.

Remark: the result is correct but I must calculate the derivative of $ f(x)$ before doing $ x^n\equal{}0$.

[quote="loup blanc"]For 1):
$ A \equal{} (I \minus{} J)^{ \minus{} 2}$ and $ A^k \equal{} (I \minus{} J)^{ \minus{} 2k}$. We conclude using binomial formula.[/quote]
Thanks! Yes I do similar ,too


Prove that if $ a_i $ are reals and positive then:
$ \prod\limits_{i=1}^n (a_i^3+1)\geq \prod\limits_{i=1}^n (a_i^2a_{i+1}) $ where $ a_{n+1}=a_1.$

Dear caugust,did you make any mistake?If i am right,our inequality is:
[tex] (a_1^3+1)...(a_n^3+1) \geq a_1^3...a_n^3 [/tex] which is right clearly

I think he wants to prove the inequality $ (1+a^3_1)\cdot...\cdot(1+a^3_n)\geq (1+a^2_1\cdot a_2)(1+a^2_2\cdot a_3)\cdot...\cdot(1+a^2_n\cdot a_1) $. And this comes from some applicatins of Cauchy-Schwarz inequality:
  $ (1+a^2_1\cdot a_2)^2\cdot...\cdot(1+a^2_n\cdot a_1)^2\leq (1+a^3_1)(1+a^2_2\cdot a_1)\cdot...\cdot (1+a^3_n)(1+a^2_1\cdot a_n) $. Similarly, we have $ (1+a^2_2\cdot a_1)^2\cdot...\cdot (1+a^2_1\cdot a_n)^2\leq (1+a^3_2)(1+a^2_1\cdot a_2)\cdot...\cdot (1+a^3_1)(1+a^2_n\cdot a_1) $

Harazi, I'm sorry but I don't exactly understand how you show this:

[quote="harazi"]$ (1+a^2_1\cdot a_2)^2\cdot...\cdot(1+a^2_n\cdot a_1)^2\leq (1+a^3_1)(1+a^2_2\cdot a_1)\cdot...\cdot (1+a^3_n)(1+a^2_1\cdot a_n) $.[/quote]

Another remark. Just a wild conjecture, most probably it's wrong, but maybe...

Let $a_{1},$ $a_{2},$ $...,$ $a_{n},$ $b_{1},$ $b_{2},$ $...,$ $b_{n}$ be positive numbers such that $a_{1}+a_{2}+...+a_{n}\geq b_{1}+b_{2}+...+b_{n}$ and $a_{1}a_{2}...a_{n}=b_{1}b_{2}...b_{n}$ (or, maybe, even $a_{1}a_{2}...a_{n}\geq b_{1}b_{2}...b_{n}$ would be enough). Then,

$\left( 1+a_{1}\right) \left( 1+a_{2}\right) ...\left( 1+a_{n}\right) \geq
\left( 1+b_{1}\right) \left( 1+b_{2}\right) ...\left( 1+b_{n}\right) $.

In case this is true, this would be a generalization of caugust's inequality above, and also of the "Huygens inequality" stating that

$\left( 1+a_{1}\right) \left( 1+a_{2}\right) ...\left( 1+a_{n}\right) \geq
\left( 1+g\right) ^{n}$,

where $g=\sqrt[n]{a_{1}a_{2}...a_{n}}$ is the geometric mean of the numbers $a_{1},$ $a_{2},$ $...,$ $a_{n}$.

Proof or counterexample?

  Darij

[quote="darij grinberg"]Let $a_{1},$ $a_{2},$ $...,$ $a_{n},$ $b_{1},$ $b_{2},$ $...,$ $b_{n}$ be positive numbers such that $a_{1}+a_{2}+...+a_{n}\geq b_{1}+b_{2}+...+b_{n}$ and $a_{1}a_{2}...a_{n}=b_{1}b_{2}...b_{n}$ (or, maybe, even $a_{1}a_{2}...a_{n}\geq b_{1}b_{2}...b_{n}$ would be enough). Then,

$\left( 1+a_{1}\right) \left( 1+a_{2}\right) ...\left( 1+a_{n}\right) \geq
\left( 1+b_{1}\right) \left( 1+b_{2}\right) ...\left( 1+b_{n}\right) $.[/quote]

I'm sorry, this is wrong: Take, for instance, $a_1 = 8$, $a_2 = 2$, $b_1 = 4$, $b_2 = 4$. Hence, most probably we can throw the above "generalization" into the trash unless somebody finds an interesting condition under which it is valid.

  Darij

Dear Darij, the inequality I've written is exactly Cauchy-Schwarz applyied n-times and then take the product.

Oh yes, harazi, I was really blind... thanks...

  Darij

Dear Harazi, sorry, but I don't understand your solution?
I have understood the first step but I cannot undestand how you can get the desidered result.
Can you, please, complete it?

Thank you very much.

Hello everybody.I have some strange ideas about this one.Can this one be right:
$ \prod\limits_{i=1}^n (a_i^3+1)\geq \prod\limits_{i=1}^n (a_i^2+a_{i+1}) $  :?

Dear Manlio, I think we have discussed this method when solving the following problem:
  If $ x+y+z=2 $ then $ x^2y+y^2z+z^2x\leq x^3+y^3+z^3\leq 1+\frac{x^4+y^4+z^4}{2} $. Do you remember?  NOw let us denote $ A=(1+a^2_1\cdot a_2)\cdot...\cdot (1+a^2_n\cdot a_1), B=(1+a^2_2\cdot a_1)\cdot...\cdot (1+a^2_1\cdot a_n), C=(1+a^3_1)\cdot...\cdot (1+a^3_n) $. We must prove that $ A\leq C$. We have shown that $ A^2\leq BC$ and $ B^2\leq AC$. If $ B\leq C$ we are done with the first. Otherwise, the second saves us.
  And yes, Anh Cuong, I think the inequality you posted is right and can be deduced from Cauchy, using this method.

Sorry nickolas, my net doesn't work. The year was 2003 or 2004


quadratics
number theory proposed
number theory

show that there is  infinitely many x such that

$x/5 , (2^x)-1/(2x+1)$  are natural numbers.

Are you sure that's right?  You're asking for 2x+1 to divide, and so be, a power of 2.  The only odd number which is a power of 2 is 1 and so x would have to be zero.  The condition that x be a multiple of 5 is redundant.

hmm, it seems that you want to find many infinitely $x$ in which $5\mid x$ and
\[2x+1\mid 2^x-1\]
uhh?  :roll:

In that case just chose as following $p=2x+1|2^{5^k}-1$ is prime ($k$ is .... any positive integer) , and then $p\mid 2^{(p-1)/2}-1$.

Indeed, $p\mid 2^{5^k}-1$ then $\gcd(p-1,5^k)>1$ so $5^s\mid p-1$ ($s>0$) and then $p-1$ is divisible by $2\cdot 5^s$, ie... $5\mid x$. Morever , from $p\mid (2^{p-1}-1,2^{5^k}-1)=2^{5^s}-1$ we have
\[p\mid 2^{5^s}-1\mid 2^{(p-1)/2}-1\]

Take $k$ any you want and then we're done  :D

If $2x + 1 = p$ is prime then $p \mid 2^x-1$ iff 2 is a quadratic residue of $p$ iff $p \equiv 1,7 \bmod 8$ by quadratic reciprocity.  If $5 \mid x$ then $p \equiv 1 \bmod 5$.  By Dirichlet's Theorem in primes in AP there are infinitely many such $p$.  Composite values of $x$ with the same property include 170 and 280.


What is the $ 87^{\mathrm{th}}$ odd positive integer?

[hide="Solution"]It's like I said before. Use the formula 2x-1, where x is the _ odd postive integer.Thus, 2(87)-1=173[/hide]

[hide="Solution"]I started a list of the odd positive integers 1, 3, 5, 7, 9,,,
then I numbered the first few in sequence        1, 2, 3, 4, 5,,,,87
I noticed that each of the odd integers was equal to its sequence number plus the previous sequence number.
So the 87th number would be 87 plus 86 (equals 173).[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide your solutions[/url].[/color]

[hide]As you can see there is a pattern. It goes the first odd number is 1. The second, 3. The third, 5. The fourth, 7. Subtract the term by 1 and then add the result to the term. This gives you the answer. Therefore, 87 - 1 = 86 and 87 + 86 = 173.[/hide]

[hide=Sorry for bump but this is a great solution]First, we multiply $87$ by $2$. This gives us $174$. Since this is even, the answer must be either $175$ or $173$. Since you are allowed guess two times on Alcumus, first guess $175$. If it is wrong, you know that the answer is $\boxed{173}$.[/hide]

[quote=ARay10][quote=prestonh]No, it's Friday.[/quote]

Did u have to bump up a 9 year old thread[/quote]

[quote=franzliszt](S)he's been doing quite often now....[/quote]

:P

[hide=Solution]
If you add $1$ to the list you will get $2, 4, \ldots$. Notice that the number of terms doubled is the number on the list. (For example the first term is $2$ and $1\times2=2$.) So you just have to subtract one from the number of terms doubled to get the odd positive integer you want. So $87\times2 - 1= 174-1=173$.
[/hide]