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The German NMO (Deutsche-Mathematik-Olympiade) takes place from 05/02 until 05/05. Here are the links for the problems:

http://www.mathematik-olympiaden.de/Aufgaben/MO43.html
http://www.mo-2004.de

Good luck to Darij.  :)

Thanks :)

Note also the link http://www.mo-2004.de/htmlcontent.htm

This link makes possible to enter the NMO site without having to view the trash... hmm, sorry... Flash video :)

  Darij

2nd prize; actually the problems were from hell and solving them was a hard job to those of us used to elegant problems with elegant solutions  :?  For instance, the sixth problem was: given a cartesian coordinate system, does there exist a circle such that there are exactly 5 integer points (i. e. points whose both coordinates are integers) on the circle? How on the &"!" could I expect that the answer to such a problem is "YES"???

  dg

Well, in fact I think I saw this problem in a NT book :D so it's not that original.

Yeah, it seems that they are lacking original problems  :)

Else, why would they pose a problem like the following one (this one I did solve, of course):

Find all pairs of real numbers (x; y) such that

$x^{4}+y^{4}=17\left( x+y\right) ^{2}$;
$xy=2\left( x+y\right) $.

It's a routine problem, very easy but I find it ugly since one needs to distinguish 3 or 5 cases... I think it's an example of a problem which definitively should not be posed on any olympiad.

[b]PS.[/b] I'll post all problems on the forum if I find the time.

[b]PPS.[/b] The olympiad problems were, as I've said, not the best one could expect, but the PR was excellent... the place around the congress center was full of posters announcing the olympiad, we were photographed and interviewed (although I must say the reporters had not much grip on maths), the first mayor of Essen held a speech, etc....

  Darij

Well, I think creating an original and elegant problem is 10 times more difficult than solving IMO 1996/5  :D

[quote="darij grinberg"]

[b]PS.[/b] I'll post all problems on the forum if I find the time.

[b]PPS.[/b] The olympiad problems were, as I've said, not the best one could expect, but the PR was excellent... the place around the congress center was full of posters announcing the olympiad, we were photographed and interviewed (although I must say the reporters had not much grip on maths), the first mayor of Essen held a speech, etc....

  Darij[/quote] That's the usual stuff here in Romania :) you guys are beginning to get the hang of it. Also it seems that the results came back in time :D

We are waiting for you to post the problems Darij.

PS are you qualified yet for IMO 2004?

I have posted all problems for the 11th, 12th and 13th classes. I have also the problems for the 8th, 9th and 10th class, but they are too hard for me to translate (one problem is accompanied with a 3D figure, some other ones have much text etc.). I'm sorry but I don't think I will manage to translate them.

Here are some technical remarks:

All problems of the German "Mathematik-Olympiaden" are numbered as follows:

Problem OOCCRN is the problem nr. N for the CCth class in the Rth round of the OOth olympiad.

The problems for the 12th and 13th classes are always identical; for the 12th and the 13th class both, we write CC = 13.

The olympiad we are speaking about is the 4th Round of the 43th Olympiad.

This year, the problems were not only identical for the 12th and the 13th class, but also 5 of the 6 problems posed for the 11th class were identic to those of the 12th/13th class.

By the way, on the contest I had to solve the problems for the 11th class. I solved 431141, 431142, 431144 and 431145 completely and got 0 points for 431143 and 431146... second prize!

  Darij

The official results are out now: http://www.mo-2004.de/htmlcontent.htm  :)

And for those of you reading German, the problems are here:

http://www.mathematik-olympiaden.de/Aufgaben/MO43.html

  Darij

[quote="darij grinberg"]Yeah, it seems that they are lacking original problems  :)
[/quote]

Well, there are quite a few known problems now. And more and more come known due to the globalisation of problem-solving. Some decades ago it was not too obvious when the same problems got (sometimes indepently) proposed for different contests.
That is why it gets harder and harder to invent new problems. That is why the ILL is not provided anymore. The ILL problems not included in the ISL are returned to the proposer countries. Hence only the proposer country and IMO Problem Selection Committee know the problems from a certain country and a country can just use the ILL problems (non-ISL) it has proposed for IMO. And most countries do not only use ISL problems in their TSTs because they perfectly emulate IMO style and difficulty but also olympiad people lack ideas for new problems. Maybe it is just because creative problem-creators are missing.
And regarding Germany just look who proposed problems for IMO. And I cannot see Germany on that list. But Darij can contribute some geometry problems.  :)

PS: Slowly Darij gets the most often mentioned name on the forum...  :D

[quote="orl"]Well, there are quite a few known problems now. And more and more come known due to the globalisation of problem-solving. Some decades ago it was not too obvious when the same problems got (sometimes indepently) proposed for different contests.[/quote]

Indeed, now this would cause troubles, not only because of some publishers claiming total copyright for the problems they publish  :D , but also just because olympiad participants are (mostly) smart and look what the people in the other countries have to solve. Hence it is a very bad strategy of an olympiad problem proposer to steal from recent olympiads of other countries (even from Iran, what the Russians do, at least one of them have told me so). But I think it is a much better strategy to use the resources of old olympiads, no matter from which country (well, okay, one shouldn't steal problems from olympiads in one's own country). In fact, we now have such a great amount of published problems that no olympiad participant ever can solve all of them for his preparation. And if somebody really solves them all, he probably would win the olympiad even if the problems were new. But it is really impossible to do all problems of all olympiads in all years. Which problems does a usual IMO participant solve for his training? Well, surely some older IMO problems, then probably some ISL / ILL ones (thanks to Orl, who made them accessible!), then, of course, a number of problems from the olympiads of his own country (this is why I said one shouldn't steal from his own country). Then, of course, some very classic problems, and maybe some from other well-known contests like the MMO, Putnam etc.. And those posted at MathLinks, of course  ;) . But do you really fear that some participants of a regional round of the German NMO have solved problem 2 of the Italian NMO 1986 ? Remember that, as a Russian proverb (or, more precisely, my poor translation of a Russian proverb) says: "The New is the well-forgotten Old." So I think it wouldn't hurt anyone if an olympiad uses some old problems from other olympiads, or from periodicals (e. g. Crux Mathematicorum; few people still have the first issues of this journal). Of course, inventing new problems is better, but a nice and well-forgotten old problem is better than an ugly new one.

[quote="orl"]That is why the ILL is not provided anymore. Thus countries can use these problems in their NMOs.[/quote]

Actually, I know this but I find this a bit strange. It's hard to keep NMO problems secret after the exams, and many students particularly often look for [i]recent[/i] olympiad problems from other countries to train their skills. Hence, after one country has used an ILL problem in its NMO, the problem spreads all over the world and cannot be successfully used by any other country's NMO. This must mean that every ILL problem can be used only by one country, doesn't it?

And you (or one of the people you quoted) also said that the ILL contains lots of ugly and boring problems. Well, I thought it would be a reason against using these problems on any NMO, but well... NMOs can be ugly, yes.

[quote="orl"]And regarding Germany just look who proposed problems for IMO. And I cannot see Germany on that list.[/quote]

Well, Germany indeed has proposed a [i]combinatorics problem of medium difficulty[/i] for the IMO 2004. The problem got into the ISL, but it wasn't selected one of the IMO problems. This is, at least, what I've heard  :)

[quote="orl"]But Darij can contribute some geometry problems.  :) [/quote]

Hmm, who knows?  ;)  But actually, my geometry problems are not really good olympiad problems, at least not really good compared to IMO geometry problems. However, I dare say they are better than this year's German NMO ones, so I'll probably propose some for the German NMO one day (I have heard of some school student proposing problems - he actually proposed a problem for another class; I am not really sure whether this all is true or just a rambling, but I tell you what has been said to me).

[quote="orl"]PS: Slowly Darij gets the most often mentioned name on the forum...  :D[/quote]

Did you search for "Darij" and count the appearances? Hmm... I hope not to become the greatest spammer here...

  Darij

[quote="darij grinberg"]Hence it is a very bad strategy of an olympiad problem proposer to steal from recent olympiads of other countries (even from Iran, what the Russians do, at least one of them have told me so).[/quote]
how about belgium:
3 olympiad problems
1 stolen from canada 85
1 stolen from Holland some years ago
both litterally :P

[quote="darij grinberg"]And you (or one of the people you quoted) also said that the ILL contains lots of ugly and boring problems. Well, I thought it would be a reason against using these problems on any NMO, but well... NMOs can be ugly, yes.[/quote]
Seems that NMO is still quite similar to Flanders MO then :D

Indeed, most competitions are suffering from lack of problems. I couldn't believe how many thanks I get from our olympiad every time I send in a question (and that's only Flanders MO level questions!)
Therefor I think the final result will be that many EU countries start co-operating more and more, sharing questions, making common TST's, etc.

[quote="Peter VDD"][quote="darij grinberg"]Hence it is a very bad strategy of an olympiad problem proposer to steal from recent olympiads of other countries (even from Iran, what the Russians do, at least one of them have told me so).[/quote]
how about belgium:
3 olympiad problems
1 stolen from canada 85
1 stolen from Holland some years ago
both litterally :P [/quote]

I updated my posting regarding ILL problems which are used in NMOs. Well how comes that you know that the other three problems are original ? Maybe you just did not dicover its sources yet.  :D

[quote]
Hence it is a very bad strategy of an olympiad problem proposer to steal from recent olympiads of other countries (even from Iran, what the Russians do, at least one of them have told me so).
[/quote]
   (Foolish but quite natural reaction from me...)
   What do you mean? I am sure than no ARO problem was (consciously) taken
from Iran olympiad - or any other NMO. Perhaps there was a coincidence - then
what problem do you mean?
   The only contest where such borrowing can take place is what in other
countries called TST, but since it is not used for TS (as fiery Fedor Petrov
unpatriotically revealed   :)  :( ), nobody cares...
[quote]
 (I have heard of some school student proposing problems - he actually proposed a problem for another class; I am not really sure whether this all is true or just a rambling, but I tell you what has been said to me).
[/quote]
   This is sometimes done at StPetersburg MO - as far as I remember now,
Shiryaev and Badzyan proposed problems for it when they were still school students.

[quote="Peter VDD"]Seems that NMO is still quite similar to Flanders MO then :D[/quote]

That's what I noted too! And this is why I am trying to solve Flanders MO problems in order to train for the next German NMO...

[quote="guas"](Foolish but quite natural reaction from me...)
What do you mean? I am sure than no ARO problem was (consciously) taken
from Iran olympiad - or any other NMO. Perhaps there was a coincidence - then
what problem do you mean?
The only contest where such borrowing can take place is what in other
countries called TST,[/quote]

Indeed, I was speaking about the TSTs. I don't remember anymore who exactly told me that thing about the Iranian problems, but it seems that they are considered as an inaccessible enough source for "recycling"... of course, MathLinks will disprove this...

  Darij

[quote="orl"]Well how comes that you know that the other three problems are original ? Maybe you just did not dicover its sources yet.  :D[/quote]

The other 3? Orl, [u]2 out of 3[/u] were unoriginal! :P

And the other one yeah, in fact, I know the idea is well-known (flipping over digital numbers) but at least it was not litterally stolen from some well-known olympiad, they gave it another wording etc. :roll: The other problems really had almost the same wording :!:

[quote="darij grinberg"]I have heard of some school student proposing problems - he actually proposed a problem for another class; I am not really sure whether this all is true or just a rambling, but I tell you what has been said to me[/quote]

What's wrong with students proposing problems? :? I also submit questions to our NMO. Of course they can't select a question from a student that will be participating in the event, but if an NMO 12th grader here submits a good question for junior NMO, he has an equal chance of it getting used as anyone else not participating in the junior NMO.

[quote="darij grinberg"][quote="Peter VDD"]Seems that NMO is still quite similar to Flanders MO then :D[/quote]

That's what I noted too! And this is why I am trying to solve Flanders MO problems in order to train for the next German NMO...[/quote]

Well Darij, that'll be a problem then... since I've put up only the nice problems (not the ones with half-a-page wordings :roll: )

If you want to take the tests from belgium, use this link:
http://www.kulak.ac.be/vwo/uk/Olympiad_Problems_Main_Menu_mn.html

The second column (PS files) should have the problem sheets from the past 14 years (1990-2003) on the last page. They still didn't include 2004, 5 months after the contest, it won't surprise me that they'd wait for 2005 to put it up, so the last year is a good one. (2003 was one of the better, 2004 on of the worst, hopefully 2005 will be good again)

PS: a lot of the problems are still on the board, unsolved. ;) (yes, you'll have to enter another forum than geometry and inequalities for them :P)

[quote="Peter VDD"]Well Darij, that'll be a problem then... since I've put up only the nice problems (not the ones with half-a-page wordings :roll: )

If you want to take the tests from belgium, use this link:
http://www.kulak.ac.be/vwo/uk/Olympiad_Problems_Main_Menu_mn.html [/quote]

Ah well, that all looks rather like an IQ test than like a mathematical olympiad. But I'll try some of the more complicated ones, especially the problem with the student training on quadratic equations, cause I like that one  :D

  Darij

Darij, the questions are on the the [b]last page[/b] or [b]last 2 pages[/b] of [b]each document[/b] :D

The others are the multiple choice selection rounds.

The quadratics is on of the harder indeed, but really there are other ones you should try too.

[quote="Peter VDD"]Darij, the questions are on the the [b]last page[/b] or [b]last 2 pages[/b] of [b]each document[/b] :D[/quote]

Yeah, that's clear to me... but the niveau is generally under the niveau of the German NMO, at least I don't see any problem like the famous [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=966&highlight=numberfully]"Number-fully" (German NMO 2003)[/url].

I'll try the problems which I have a better chance to solve in a nice way...

  Darij

Many of them are really easy... but some pretty hard as well. Quadratics isn't really easy, and the problem on binary words isn't easy at all either. Some others also have decent level.

Besides, easy or not, except for the trivial ones it's a good excercice to do them :)


I looked this up on the resource section, but there wasn't a post, so:

For each positive integer n, let S(n) denote the sum of the digits of n. For how many values of n is n+S(n)+S(S(n)) = 2007?

A)1
B)2
C)3
D)4
E)5


[hide]Well I know at the end of third round the scores must be 120-40-120 (because if A is the amouint of money the winner have originally and B is the amount of money the loser have originally and after a round you want them to equal B-A=2A or B=3A and since B-A=2A=80 A=40 and B=120) so at the end of the round before (the second one) that the could've had either 140-20-140 or 180-100-60. The first set of scores can then be either 140-20-140 or 70-90-210 at the end of first round and the second set can be 30-130-210, 90-150-190, or 50-110-230. The largest score possible is 230.[/hide]

Huh guess the score they started out with didn't matter anyways.

Very close Tare, theres one line of the question you didn't see.

Doh! I mistook three, count 'em three [b]letters[/b]!!! Aaaaaaah! "first" and "final" should be changed to "moley" and "kuru" or something like that :lol:

There is only one possible way this can happen: 90-150-190 then 60-100-180 then 120-40-120 then 80-80-80. Thus the answer is 190.

Ugh.

Syntax Error wrote:Three players play 4 rounds of a game and end with a tie score of 80.  a different player wins each of the final three rounds.  when a player wins, s/he doubles his or her points, and the two losers each must subtract the amount the winner gains from their scores.  what was the highest score at the end of round 1?

[hide]Each person wins one of the last three rounds once.

The scores at the end are 80,80,80. Remove a win from one person to get 40,120,120. Remove a win from the next person: 100,60,180. Remove a win from the third person: 190,150,90. The answer is thus 190.[/hide]


[quote="helpmeplz"]2^-2
------  << in parantheses to the exponent of 1/4  DIVIDED BY
 5^2

2^3
____  <<< in parantheses to the exponent of -2.
5^ -3

I found 2 ^13/2 x 5 ^13/2 but my book sais 2^4 x 5^11/2

Which is right and how do I do it? I'm mostly confused by the fact that I have to flip the 2nd fraction and I donno what to do first to not get mixed up with changing the signs of the exponents and all. Help would be greatly appreciated. :D[/quote]

$ \left(\frac{2^{\minus{}2}}{5^2}\right)^{\frac{1}{4}}/\left(\frac{2^3}{5^{\minus{}3}}\right)^{\minus{}2}$ or what?

The exponent of 2 is $ \frac{\minus{}2}{4}\minus{}3(\minus{}2)\equal{}\frac{11}{2}$
The exponent of 5 is $ \frac{\minus{}2}{4}\plus{}(\minus{}3)(\minus{}2)\equal{}\frac{11}{2}$

[quote="helpmeplz"]I'm mostly confused by the fact that I have to flip the 2nd fraction and I donno what to do first to not get mixed up with changing the signs of the exponents and all. [/quote]

This is understandable.  Solving this problem without confusing yourself is a good exercise in organizing your work carefully.  In this case, what I would do is set the numerator and denominator as separate variables.

Let $ p \equal{} \left( \frac{2^{\minus{}2}}{5^2} \right)^{ \frac{1}{4} }$ and let $ q \equal{} \left( \frac{2^3}{5^{\minus{}3}} \right)^{\minus{}2}$.  What we want to do is find $ \frac{p}{q}$.

The first general rule of simplifying expressions like this is to get rid of negative exponents (the second is to get rid of parentheses).  Since $ 2^{\minus{}2} \equal{} \frac{1}{2^2}$ and $ 5^{\minus{}3} \equal{} \frac{1}{5^3}$ we have

$ p \equal{} \left( \frac{1}{2^2 \cdot 5^2} \right)^{ \frac{1}{4} } \equal{} \frac{1}{2^{ \frac{1}{2} } \cdot 5^{ \frac{1}{2} } }$
$ q \equal{} \left( 2^3 \cdot 5^3 \right)^{\minus{}2} \equal{} \frac{1}{2^6 \cdot 5^6}$

We conclude that

$ \frac{p}{q} \equal{} \frac{2^6 \cdot 5^6}{2^{ \frac{1}{2} } \cdot 5^{ \frac{1}{2} } } \equal{} \boxed{ 2^{ \frac{11}{2} } \cdot 5^{ \frac{11}{2} } }$

Unless you wrote the problem incorrectly, the book's answer is wrong.


Find the radius of the smaller circle which touches the line $ y\equal{}x$ at $ (2,2)$ and also the y-axis.

[hide]There are two possibilities for the placement of the center of the circle; above the $ x$-axis and below it.

Case 1: Above the $ x$-axis. Since the circle touches the line $ y\equal{}x$, the radius from the center to $ (2, 2)$ makes a $ 135$ degree angle with the radius that touches the $ y$-axis. Therefore, we seek $ r$ such that $ r \plus{} r \sin 45 \equal{} 2$. This yields $ r \equal{} 4 \minus{} 2\sqrt{2}$.

Case 2: Below the $ x$-axis. Using similar logic, the two aforementioned radii now have an angle of $ 45$ degrees. We now seek $ r$ such that $ r\sin 45 \plus{} 2 \equal{} r$. This yields $ r \equal{} 4 \plus{} 2\sqrt{2}$.

Therefore, the smaller radius is $ r \equal{} 4 \minus{} 2\sqrt{2}$.[/hide]


function
inequalities
algebra
polynomial
quadratics
calculus
calculus computations

Can you please show how to show functions are positive in given interval.
For example, for the following two functions
\[ g(b) \equal{} \frac {2b^3 \minus{} 5b^2 \minus{} 7b \plus{} 10 \minus{} \sqrt {(4b^4 \minus{} 12b^3 \minus{} 15b^2 \minus{} 12b \plus{} 12)(b \minus{} 1)^2}}{22 \minus{} 4b}
\]

\[ h(b) \equal{} \frac {2b^3 \minus{} 5b^2 \minus{} 7b \plus{} 10 \plus{} \sqrt {(4b^4 \minus{} 12b^3 \minus{} 15b^2 \minus{} 12b \plus{} 12)(b \minus{} 1)^2}}{22 \minus{} 4b}
\]
i want to show that both functions are positive when $ 0\le b \le 1$

Thanks

P.S: I need to solve this for the other question (inequality)

see [url]http://www.mathlinks.ro/viewtopic.php?p=1365733#1365733[/url]

Honestly, think a bit before posting: the denominators do not matter, $ (b\minus{}1)$ can be factored out, squaring the rest results in a third (if not even second) degree polynomial inequality and God knows how many other simplifications are completely straightforward. Bring everything to the form where the real difficulties begin and we'll think of how to take care of them. ;)

Thanks for responding.
Well, i forgot to factor out $ x \minus{} 1$, that is right.
But how would i know that denominators doesn't matter... and also what does it mean "square the result"?
Do you mean square the whole function?
This is what is left after simplification:
\[ g(x) \equal{} \left(x \minus{} 1\right)\left(2x^2 \minus{} 3x \minus{} 10 \minus{} \sqrt {4x^4 \minus{} 12x^3 \minus{} 15x^2 \minus{} 12x \plus{} 12}\right)
\]
and
\[ h(x) \equal{} (x \minus{} 1)\left(2x^2 \minus{} 3x \minus{} 10 \plus{} \sqrt {4x^4 \minus{} 12x^3 \minus{} 15x^2 \minus{} 12x \plus{} 12}\right)
\]
Now how can simplify more?

Thanks.  :)

Anyways, i think i understood myself what you mean by "squaring the result".
So, in the $ g(x)$, since $ x \minus{} 1$ will be negative, we should prove that
\[ 2x^2 \minus{} 3x \minus{} 10 \minus{} \sqrt {4x^4 \minus{} 12x^3 \minus{} 15x^2 \minus{} 12x \plus{} 12}
\]
is also negative.
\[ 2x^2 \minus{} 3x \minus{} 10 \leq \sqrt {4x^4 \minus{} 12x^3 \minus{} 15x^2 \minus{} 12x \plus{} 12}
\]

\[ (2x^2 \minus{} 3x \minus{} 10)^2 \leq 4x^4 \minus{} 12x^3 \minus{} 15x^2 \minus{} 12x \plus{} 12
\]
But we have:
\[ (2x^2 \minus{} 3x \minus{} 10)^2 \minus{} (4x^4 \minus{} 12x^3 \minus{} 15x^2 \minus{} 12x \plus{} 12) \equal{} \minus{} 8(x \plus{} 1)(2x \minus{} 11)
\]
and that is clearly positive for $ 0\le x \le 1$

[b]EDIT:[/b] I don't think i am right. because, i proved the expression is positive. So that means the question is wrong?

the denominators "don't matter" because $ 16 \leq 22 \minus{}4b \leq 22$ on the interval $ [0, 1]$, so it's sufficient to show that the numerators are positive.

from a quick glance, something's wrong
try evaluating $ \sqrt{(4x^4 \minus{} 12x^3 \minus{} 15x^2 \minus{} 12x \plus{} 12)(x\minus{}1)^2}$ at $ x \equal{} 0.99$

I am sorry. i don't think this is correct way of proving either... In Fact, it is the following i am trying to prove...

[b]Problem. [/b]

Given the quadratic equation:
\[ f(x) \equal{} Ax^2 \plus{} Bx \plus{} C
\]
where $ A\equal{}11\minus{}2b$, $ B\equal{}5b^2 \minus{} 2b^3 \plus{} 7b \minus{} 10$, and $ C \equal{} 2b^3 \minus{} 2b^2 \minus{} 2b \plus{} 2$

[color=red]Prove that[/color] $ f(x) \ge 0$ if $ 0 \le x \le \frac{1}{4}$ and $ 0 \le b \le 1$

Thanks for attention!  :)

[quote="shaaam"]I am sorry. i don't think this is correct way of proving either... In Fact, it is the following i am trying to prove...

[b]Problem. [/b]

Given the quadratic equation:
\[ f(x) \equal{} Ax^2 \plus{} Bx \plus{} C
\]
where $ A \equal{} 11 \minus{} 2b$, $ B \equal{} 5b^2 \minus{} 2b^3 \plus{} 7b \minus{} 10$, and $ C \equal{} 2b^3 \minus{} 2b^2 \minus{} 2b \plus{} 2$

[color=red]Prove that[/color] $ f(x) \ge 0$ if $ 0 \le x \le \frac {1}{4}$ and $ 0 \le b \le 1$

Thanks for attention!  :)[/quote]

Sorry For Spamming... But why no one is interested? is it hard? or too easy that everyone ignores? :mad:


inequalities
real analysis
real analysis solved

Let (x_n)n>=2 and (y_n)n>=2 be two sequences of real numbers with the properties:
x_(n+1)+1=(ny_n+1)^(1/n) and 0<y_n<=x_n for any n>=2. Prove that the sequence (nx_n)n>=2 converges iff the sequence (ny_n) converges.

i think u meant (x_n) convergent iff (ny_n) convergent
denote by a_(n+1) = (x_(n+1)+1)^n =>
a_(n+1) = ny_n+1; and b_n = ny_n;
a_(n+1) = b_n+1
"=>"(x_n) convergent results that (a_n) is convergent=>exists limit{n->oo}(a_n)^(1/n) = lim{n->oo}a_(n+1)/a_n = lim{n->oo}(b_(n+1)+1)/(b_n+1)=1(since a_n is convergent)=>(b_n+1)is convergent => (b_n)convergent=>(ny_n) convergent
"<="(ny_n) convergent results lim{n->oo}(nx_n+1)^(1/n) = 1=>
lim{n->oo}x_(n+1)+1 = 1=> lim{n->oo}x_n=0=>(x_n)convergent q.e.d.

hope is good

[quote="Kuba"]i think u meant (x_n) convergent iff (ny_n) convergent
[/quote]

no! I meant (nx_n) converges iff (ny_n) converges

okey :D
then i was wrong :)
i'll try again

"=>"
if (nx_n) is convergent then
lim{n->oo}n(((n-1)y_(n-1)+1)^(1/n)-1) exists and it is finite=>lim{n->oo}(ny_n+1)^(1/n)->1 and from Cauchy D'Alembert critery we get that (ny_n+1)/((n-1)y_(n-1)+1)->1=>(ny_n) is convergent.

"<=" i'll think about this one

For the second part use Bernoulli's inequality

nyn+1+1 \leq nxn+1+1 \leq (xn+1+1)n=nyn+1

I guess the rest is obvious :D sandwich theorem :D


Find all values of x:

2csc^2(x) - cot(x) - 5 = 0

One of my answers that I got was (3pi)/4 + pi*K.

ARE THERE ANY OTHERS??

Thanks!!!

[hide="Hint"]\[\frac{2}{\sin^{2}x}-\frac{\cos x}{\sin x}-5 = 0 \Leftrightarrow 5\sin^{2}x+\sin x\cos x-2(\sin^{2}x+\cos^{2}x) = 0 \Leftrightarrow\]
\[3\tan^{2}x+\tan x-2 = 0\] Now trivial.  :wink: [/hide]


$ (x_{n})$ is a squence. For all $ n$ integer , there is a $ M$ so that  $ x_{n\plus{}1}\equal{}\frac{x_{n}^{2}\plus{}10}{7}$ and $ x_{n}<M$.Find all values of $ x_{0}$


find $ x,y\in N$ such $ 5^x\equal{}y^{4}\plus{}4y\plus{}1$

[quote="Leonhard Euler"]find $ x,y\in N$ such $ 5^x \equal{} y^{4} \plus{} 4y \plus{} 1$[/quote]

Modulus $ 3$, RHS can only be $ 0$ or $ 1$ and LHS can only be $ 1$ if $ x$ is even or $ 2$ if $ x$ is odd. So $ x$ is even and LHS is a perfect square.

But, if $ y>2$, $ (y^2)^2 < y^4\plus{}4y\plus{}1<(y^2\plus{}1)^2$ and so RHS can not be a perfest square. So $ y\leq 2$.

For $ y\equal{}1$, RHS is $ 6$ and we have no solution.
For $ y\equal{}2$, RHS is $ 25$ and we have a unique solution $ (x,y)\equal{}(2,2)$


[quote="ch1n353ch3s54a1l"][quote="junggi"]Some of the squares of a $3X3$ board are colored black in such a way that in each row and in each column there is at least one black square. Regarding symmetries of the same coloring as distinct, find the total number of ways in which one can color the board[/quote]

[hide="Maybe..."]Well if there is a black square in every row and column, then there are three black squares that don't share a column or row with each other. These black squares can be chosen in $3!$ ways. Then the other 6 squares have two options each, so the answer is $3!(2^{6})=\boxed{384}$?[/hide][/quote]
[hide="you are..."]over counting.
For example, you count the all-black square a few times[/hide]

[hide="hint"]PIE.
casework works too.[/hide]

do you mean
xxo
ooo is the same as
ooo

oox
oox
ooo ? or do you treat them as different colourings?

someone just solve the problem..unclear

they are different.
every arrangement is counted as different

what's the answer
i get the same as the 2nd person

[hide="ok solution"]Solution 1(this is the one i came up with at first, so yea it's bad):Casework on how the top left 2x2 looks like. Let the squares be numbered as follows:
[img]http://me.yogurttree.com/junggi/untitled.png[/img]
(now isn't this a piece of art?)
Let the top left 2x2 square be called IDONTCARE
There are 4 cases.
Case 1: IDONTCARE is all white. There must be a black square in row with 1,2,3, so 3 must be black. similarly, 6,7,8 must be black also. Then square 9 can be either white or black so $\boxed{2}$ possibilities here.
Case 2: one IDONTCARE square is black. WLOG let it be 1. There must be a black square in row with 4,5,6 so 6 is black. similarly, 8 must be black. 3,7,9 can be either blakc or white so 8 possibilities time 4 because we could have chosen 2,4,5. So $\boxed{32}$ here.
Case 3: 2 adjacent IDONTCARE squares are black. WLOG let them be 1,2.
There must be a black in row with 4,5,6, so 6 is black. also there must be a black in row with 7,8,9, and 3 can be either, so we find 14 cases here times 4 since we could have chosen any 2 adjacent squares. $\boxed{56}$
Case 4: all other possible cases(there's 7 left).We need one black square in last row and last column. If 9 is black, the others can be either, so $2^{4}=16$. If 9 is white, one of 7,8 can be black or both can be black. Same for 3,6, so $3^{3}=9$. Total $9+16=25$ in one case. there are 7, so $25\cdot7=175.$

Adding all up, $175+56+32+2=\boxed{265}$

Solution 2
this is the much more elegant official solution in the contest book thingy(sorry if im bad at explaining).
If we needed a black square in ONLY the columns, then we would have $7^{3}$ possibilities. But this overcounts, so we need to subtract when a whole row is white(say the first row). Then that means the blacks were in the last two rows. The # of possibilities of filling the rectangle with squares 4,5,6,7,8,9 only with at least one black in a column is $3^{3}$. The same could have happened for 2nd and 3rd row, so what we need to subtract from $7^{3}$ is $3\cdot3^{3}$. Now we undercount, because 2 rows could have been completely empty, and we substracted those cases twice for each of the empty rows. There are 3 of those cases so we add back 3.
the final answer is $7^{3}-3\cdot3^{3}+3=\boxed{265}$
[/hide]


f(x) = 0 for x < 0, x^2 for 0 <= x <= 4 and 16 for x > 4.
g(x) = 0 for x < 0, 2x for 0 <= x <= 3 and 6 for x > 3.
Find f(g(x)).

For $ x<0$, $ f(g(x))\equal{}f(0)\equal{}0(0)\equal{}0$.
For $ 0\leq x\leq 3$,$ f(g(x))\equal{}(2x)^{2}\equal{}4x^{2}$.
For $ 3<x\leq 4$, $ f(g(x))\equal{}6^{2}\equal{}36$.
For $ x>4$, $ f(g(x))\equal{}16(6)\equal{}16$ (kind of an abuse of notation, that isn't supposed to be multiplication).

Since 36 isn't in the range of $ f$, perhaps you want to rethink your solution?   :wink:  :roll:

I guess f(6) = 16 means fg(x) = 16 for x > 3

Yup, and JBL is suggesting a crucial thing about compositions: $ \text{range }g\subset\text{domain }f$.


algebra
polynomial
calculus
derivative
algebra solved

Prove that the product of the polynomial $\left(x+1\right) ^{n-1}$ with any other polynomial (which is not the zero polynomial) is a polynomial with, at least, n nonzero coefficients.

[b]PS.[/b] I was asked to post some Kvant problems here. The proposed solution, in Russian, can be found here:

http://kvant.mccme.ru/1992/03/resheniya_zadachnika_kvanta_ma.htm

  Darij

This problem is really easy. All you have to do is to write the fact that all derivatives in -1  from 0 to n-1 must be 0 and then you obtain a linear system whose determinant is of Vandermonde type. If there are at most n-1 non zero coefficients, then this linear system has just the trivial solution and thus the polynomial is 0.


Actually, I was surprised that when I solved this I found out it was a #6.
Anyway...

(not worded exactly)

You and I are playing a game. We start with 2008 bricks. When it is your turn, you can take out anywhere from 1-100 (inclusive) bricks. The person to take out the last brick wins. Assuming that each person plays perfectly and I move first, what move should I make to win?

[hide="Answer and My Solution"]
Observe that the goal is to force the other person to get to 101 bricks left. This is because no matter what he picks, you can win the next turn. So the goal is to get the opponent to 101 bricks left on his turn. To do that, your goal is to get your opponent to 202, because no matter what move he does, you can get him to 101 the next turn. You can keep on applying the same logic, until you get that the goal is to get the opponent to 1919 (2020 is over 2008, so that is impossible). Subtracting 1919 from 2008, you find that on the first turn you should take out

$ \boxed{89\text{ bricks}}$[/hide]
Have fun!

EDIT: Btw, if this isn't the right level, mods feel free to move it. I'm not sure if this is the right place.

[hide]Another way of finding the answer once you know that a total of 101 bricks must be removed total by you and your opponent each turn is to find the number between 1 and 100 that is congruent to $ 2008 \pmod{101}$. Quick division reveals that the answer is 89.

Note that if you started with exactly $ 0 \pmod{101}$ bricks, and both players played perfectly, the second player would always win.[/hide]


geometry
geometric transformation
dilation
circumcircle
angle bisector

In triangle right triangle ABC, let CP and CP' be perpendiculars to hypotenuse AC s.t. that P is the farther from B than P' and CP'=CP=CB. Let Y be the intersection of BP and the angle bisector of A, and let X be the intersection of PP' and the angle bisector of A. Let Z be the intersection of AC and BP. Show that XYZC is cyclic.

[i](Source: Own variation of an old problem)  [/i]

The way I'm drawing I get a quadrilateral XYCZ not XYZC.

Could you point out what's wrong with my diagram? Thanks.
[asy]pointpen = black; pathpen = black + linewidth(0.7); size(200);
real d = 5; /* arbitrary dilation factor for intersections */
pair C = (0,0), A = (1,0), O = (A+C)/2, B = O+.5*expi(pi/6), Pa = (0,0.5), P = (0,1.5), Qa = bisectorpoint(C,A,B), Q = A+1.5*(Qa-A), Y=IP(B--Pa,A--Q), X=IP(P--Pa,A--Q), Z=IP(A+d*(A-C)--C,B+d*(B-Pa)--Pa);
D(MP("A",A)--MP("B",B,N)--MP("C",C,W)--cycle); D(C--D(MP("P'",Pa,W))--D(MP("P",P,W))); D(A--Q,EndArrow(4)); D(MP("X",X,NE)); D(MP("Y",Y)); D(Pa--D(MP("Z",Z)));
D(circumcircle(X,C,Z),linetype("2 2")); D(X--C--Z--Y--cycle,blue+linewidth(1)); D(rightanglemark(A,B,C,1.5)); D(rightanglemark(X,C,A,1.5));[/asy]

Edit: Interestingly enough, I no longer construct $ Y$, since if $ BC \equal{} P'C$, then $ \angle BP'C \equal{} \frac {180 \minus{} \angle P'CB}{2} \equal{} 90 \minus{} \frac {90 \minus{} \angle ACB}{2} \equal{} 90 \minus{} \angle CAX \equal{} \angle AXC$ which implies $ AX \parallel BP'$.  [asy]pointpen = black; pathpen = black + linewidth(0.7); size(200);
real d = 5; /* arbitrary dilation factor for intersections */
pair C = (0,0), A = (1,0), O = (A+C)/2, B = O+.5*expi(pi/6), Pa = (0,(B.x^2+B.y^2)^.5), P = (0,-Pa.y), Qa = bisectorpoint(C,A,B), Q = A+1.5*(Qa-A), Y=IP(B--Pa,A--Q), X=IP(P--Pa,A--Q), Z=IP(A+d*(A-C)--C,B+d*(B-Pa)--Pa);
D(CR(O,1/2),linetype("2 2")); D(MP("A",A)--MP("B",B,N)--MP("C",C,W)--cycle); D(C--D(MP("P'",Pa,W))--D(MP("P",P,W))); D(A--Q,EndArrow(4)); D(MP("X",X,NE)); D(MP("Y",Y)); D(Pa--D(MP("Z",Z)));
D(circumcircle(X,C,Z),linetype("2 2")); D(X--C--Z--cycle,blue+linewidth(1)); D(rightanglemark(A,B,C,1.5)); D(rightanglemark(X,C,A,1.5));[/asy]

Woops I forgot to mention that P'C=PC=BC  :oops:

My bad.

[quote="Anaxerzia"] P'C=PC=BC
[/quote] [color=darkblue]Is it possible? Then $ P' \equal{} P$.[/color]

Yeah it does. And even so, angle Y is reflex if azjps's diagram is correct, and cyc quads must have opp. angles equaling 180...

ARGH. one more mistake: Y is the intersection of BP, not BP'. And Z is the intersection of AC and BP.

Could you please write the whole thing a bit more clearly again?

Yay - so restatement of problem:

Let $ \triangle ABC$ be a right triangle, with $ \angle B \equal{} 90^{\circ}$. Let $ \overline{CP}, \overline{CP'}$ be perpendicular to hypotenuse $ AC$ such that $ CP \equal{} CP' \equal{} BC$, and $ P'$ is closer to $ B$ than $ P$. Let $ X$ be the intersection of $ \overline{PP'}$ and the angle bisector of $ \angle A$, and let $ Y$ be the intersection of $ BP$ and the angle bisector of $ \angle A$. Let $ Z$ be the intersection of $ \overline{AC}, \overline{BP}$. Show that $ XYZC$ is cyclic.

[hide="Solution"]
$ CB \equal{} CP \equal{} CP'$ implies $ C$ is the circumcenter of $ \triangle PBP' \Longrightarrow BP \perp BP'$, and from $ AX \parallel BP'$ (angle chase: $ \angle BP'C \equal{} \frac {180 \minus{} \angle P'CB}{2} \equal{} 90 \minus{} \frac {90 \minus{} \angle ACB}{2} \equal{} 90 \minus{} \angle CAX \equal{} \angle AXC$), we have $ AX \perp BP$, so $ \angle XCZ \plus{} \angle XYZ \equal{} 90^{\circ} \plus{} 90^{\circ} \equal{} 180^{\circ} \Longrightarrow XYZC$ is cyclic.
[asy]pointpen = black; pathpen = black + linewidth(0.7); size(250);
real d = 5; /* arbitrary dilation factor for intersections */
pair C = (0,0), A = (1,0), O = (A+C)/2, B = O+.5*expi(pi/4), Pa = (0,(B.x^2+B.y^2)^.5), P = (0,-Pa.y), Qa = bisectorpoint(C,A,B), Q = A+1.5*(Qa-A), Y=IP(B--P,A--Q), X=IP(P--Pa,A--Q), Z=IP(A+d*(A-C)--C,B+d*(B-P)--P);
D(MP("A",A)--MP("B",B,N)--MP("C",C,W)--cycle); D(C--D(MP("P'",Pa,W))--D(MP("P",P,W))); D(A--Q,EndArrow(4)); D(MP("X",X,NE)); D(MP("Y",Y));
D(circumcircle(X,C,D(MP("Z",Z))),linetype("2 2")); D(X--Y--Z--C--cycle,blue+linewidth(1)); D(rightanglemark(A,B,C,1.5)); D(rightanglemark(X,C,A,1.5)); D(Pa--B--P);[/asy][/hide]

[hide="Angle-chasing"]All angle measures are in degrees. $ \triangle BCP$ is isosceles, so setting $ BCA \equal{} \theta$, $ CBP \equal{} CBZ \equal{} \frac {180 \minus{} (90 \plus{} \theta)}{2} \equal{} 45 \minus{} \frac {\theta}{2}$. Since $ CBA \equal{} CBZ \plus{} ABZ \equal{} 90$, $ ABY \equal{} ABZ \equal{} 45 \plus{} \frac {\theta}{2}$. $ A \equal{} 90 \minus{} \theta$, so $ BAY \equal{} 45 \minus{} \frac {\theta}{2}$, and $ XYZ \equal{} BYA \equal{} 90$. Now we have $ XYZ \plus{} XCZ \equal{} 180$, so $ XYZC$ is cyclic, which was what we wanted.[/hide]

Follow up: What is the circumradius of XYZC? (in terms of the sides)


When trying to add myself as a contributor to my blog, I get:

[code]
SQL requests not achieved

DEBUG MODE

SQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 2

SQL Request: INSERT INTO phpbb_weblog_contributors (weblog_id, user_id) VALUES (728, )

Line : 297
File : class_blog_options.php
[/code]

Is it possible to instead make that into "You cannot add yourself as a contributor to your own blog" or something? Very trivial, though.


If $ \triangle ABC$ has points $ D, E$ and $ F$ on sides $ AB$, $ BC$, and $ AC$,respectively, and the cevians $ AE$, $ CD$, and $ BF$ are concurrent at point $ P$, then how do you find the ratio, using mass points, of $ \frac {AD}{DB}$ if you know the lengths of $ CP, BP, DP, FP$ and $ FC$?

I know how to use mass points to find the ratio of the segments the cevians are split up in given the ratios of the segments the sides are split up in, but I am getting confused on how to use the ratio of the cevians segments to find the ratio of the segments the sides of the triangle are split up in.

I would appreciate any help.
Thanks.

Honestly, I think that this question could go to Geometry subforum (mass point geometry and barycentric coordinates are considered a common geometrical technique, so they're a bit closer to mathematicians than to physicists in this case).

Pardesi, could you please move it?


abstract algebra
group theory
superior algebra
superior algebra unsolved

we have one definiction:
A left A-module N containg M as a submodule is called a related extension of M if every non-zero submodule of N has a non-zero intersection with M. A related extension N of M is called a maximal related extension if it is impossible to find another related extension N' of M and an isomorphism $ \psi$ of into N' with $ \psi m\equal{}m, m\in$M such that  $ \psi$N $ \neq$N'

If N isn't a maximal related extension what is mean? Does it mean that exist A-module P - related extension of M  properly containg N? If it is true (exist A-module P - related extension of M  properly containg N) I know that N isn't maximal related extension. But if I know that N isn't  maximal related extension is this implication true?

I think the answer to your question is yes, P exists.

The english term is "essential extension" rather than "related extension".

A maximal essential extension of M is also called an injective envelope of M, E(M).  An essential extension of M that is not maximal is some proper submodule N, that is M ≤ N < E(M).  The original paper is very nice:

Eckmann, B.; Schopf, A. "Über injektive Moduln."
Arch. Math. (Basel) 4, (1953). 75-78.
http://www.ams.org/mathscinet-getitem?mr=55978
http://dx.doi.org/10.1007/BF01899665

Lam's textbook Lectures on Modules and Rings describes how to explicitly calculate injective envelopes of modules over a variety of interesting rings.  This might help you calculate examples.

The easiest example takes R=Z, M cyclic of order 2, then every cyclic group of order 2^n is an essential extension, so none of them are maximal.  How do you get a maximal one? Just union.  The result is called the Prüfer quasi-cyclic 2-group, Z[1/2]/Z ≅ Q/Z_(2).  You can do a similar calculation for finitely generated modules over a Dedekind domain.


geometry
3D geometry
tetrahedron
trigonometry
AMC
AIME

Faces $ABC$ and $BCD$ of tetrahedron $ABCD$ meet at an angle of $30^\circ$. The area of face $ABC$ is $120$, the area of face $BCD$ is $80$, and $BC=10$. Find the volume of the tetrahedron.

[hide]Because the volume of the tetrahedron is $\frac{1}{3}(\text{base})(\text{altitude})$, we simply need to find the altitude. Notice that the height of triangle $BCD$ to side $BC$ is $2\cdot\frac{80}{10}=16$. Now, we apply our knowledge of trigonometry to find the altitude of the tetrahedron to face $ABC$. $h_{ABC}=16\sin30^\circ=8$. Therefore, the volume of the tetrahedron is $\frac13(120)(8)=\boxed{320}$[/hide]

ugh, the AIME recycles problems?!  this is virtually identical to 1984 #9, save a few change in side lengths...

i just hope they do that again this year and i will be set  :lol:

[quote=droid347]Could someone provide a diagram?  I don't see how the $16\sin 30^{\circ}$ applies...[/quote]

Same, can someone explain how the height is that, possibly by providing a diagram. Thanks

let K the foot of the D-altitude in triangle BCD and H the foot of D-altitude in tetrahedron

justify why the plane (DHK) is orthogonal to line (BC)
then angle $\widehat{DKH}$ is by definition the angle of the two planes (BCD) and (ABC)


analytic geometry
geometry
parallelogram

[quote="bachikarn"]So I remember an old trick to find the centroid of an arbitrary polygon on a coordinate plane to be averaging the x and y coordinates of each vertex. I know this definitely works for a triangle, but someone has told me this trick might not actually be valid. So I'm wondering if you guys have heard of this trick and if so, how would you prove it?[/quote]

Well, it doesn't work for everything.

I don't know how to prove it but I remember seeing an example which doesn't satisfy the conditions

i think it has something to do with moving  it around so that its base is the x axis and it has a vertex on the y axis, then using a specific sum of averages or something...
i know something like that works

Take a look at this example.

First, consider the square with vertices $(0,0),(2,0),(2,2),(0,2).$ clearly, the centroid of this is also the average of its vertices, namely $(1,1).$ So far, so good.

But now, consider the pentagon with vertices $(0,0),(1,0),(2,0),(2,2),(0,2).$ The average of the vertices is now the point $(1,0.8).$ However the region is just the same square as before and still has centroid $(1,1).$

It won't do any good to complain that that's "not really a pentagon." I could tweak that fifth point by some very tiny increment, and that tweak would move the mean of the vertices and the true centroid by only a tiny amount each - certainly not enough to make them the same.

Consider it this way: you can "triangulate" an arbitrary polygon: turn it into a union of nonoverlapping triangles, and you can do that with the original vertices. You can calculate the centroids of each of those triangles by averaging the vertices of each triangle, but when it comes to computing the centroid of the whole polygon, what do you do? It's not the "flat" average of the centroids of the triangles, it is a weighted average, weighted by area. To do that, you must also calculate the areas of the polygons.

It would be possible to construct a formula for centroid based on the vertices, since there are formulas for areas in terms of vertices. But it can't be as simple as a straight average.

[quote="bachikarn"]So I remember an old trick to find the centroid of an arbitrary polygon on a coordinate plane to be averaging the x and y coordinates of each vertex. I know this definitely works for a triangle, but someone has told me this trick might not actually be valid.[/quote]
It depends on what you mean by centroid.  If you mean the centroid of just the vertices, then your method works fine.  If, more usually, you mean the centroid of all the points in or on the polygon, then your method won't work.

For a quadrilateral, Wittenbauer's theorem will find the centroid of all the points inside:
[url]http://mathworld.wolfram.com/WittenbauersTheorem.html[/url].
It involves trisecting each side of the quadrilateral, and forming a parallelogram out of the trisection points.


Gauss
geometry
real analysis
real analysis unsolved

let $f:R^n->R$ be a $C^1$ function. Suppose that $E=\{(x,y) s.t. f(x,y)=0\}$ is non-empty bounded.
Prove that $D:E->S$,$D(x)=\frac{grad(f(x))}{|grad(f(x))|}$ is surjective where $S$ is the unit sphere.

$E=\{x\in R^n ; f(x)=0\}$

We should add $grad(f(x))\neq 0$ on $E$

Is there any condition on $n$  such as $n \geq 2$ ?

$D$ look like Gauss map well known in differential geometry or Riemaniann geometry


Jared has nine coins in his pocket.  They all look alike, but one coin is counterfeit and weighs less than the others.  What is the least number of weighings on a balance scale needed to guarantee that the counterfeit coin is found?

Im not sure on this one, but i think its two.

[hide="Solution"] You only need two; The first weighing will have three on each side and three not being weighed. If the scale tips, take the three that are lightest, while if the scale balances, take the three not being weighed. Take these three and weigh one on each side.Repeat the same steps. If only one is counterfeit, it will take only two weighings. [/hide]

[hide]Take 3 on each side to start with. If tips, weigh 2 of the coins. If tips again, the light one is it. If no tip, the one off the scale is it. If the 3/3 does not tip, do the same as if it tips, just use the coins not weighed the first time.
[/hide]
Now if we were to have 2 counterfeits...
.
Billy

[hide="my somewhat logical solution"]
okay. this is what i said.
take one form the pile and separate the leftover pile in half and weigh them. if they weigh the some, the one you took out of the pile is the light one. so in that case, you need only one weighing. if they don't wigh the some, the one you hold is abviously not the counterfet so take the lighter pile, cut it in half and reapeat. if you do it until the end, you do it three times[/hide]

[hide="solutiom"]i just did this problem at school today!

the answer is 2[/hide]

[quote="zmli"]Now, what if the problem is:

Jared has nine coins in his pocket. They all look alike, but one coin is counterfeit and weighs [b]different [/b]than the others. What is the least number of weighings on a balance scale needed to guarantee that the counterfeit coin is found?[/quote]

[hide="Solution to Problem by zmli"]I think the answer to this question is three. You put three on each side and set aside three, and see if they balance. If they do, go to section A, while if they don't go to section B. Either way, you will have three weighings for the minimum.

A. If they balance, take the third pile and put one on each side.  If they balance, then you know that the third one is counterfeit, but if they don't, replace one of the coins with the leftover one. If the balance stays the same, then the unchanged side is counterfeit, but if the result is a balance, the one you took out is the counterfiet.

B. If they don't balance, replace one of the piles with the leftover pile on the balance. If the result doesn't change, then the side that you left alone is the counterfeit pile, but if the scale balances, the pile that you took out is the counterfeit side. Also record whether the counterfeit side went up or down, which would tell you whether the coin is lighter or heavier. Then proceed to put one on each side of the balance. If the scale balances, then the leftover coin is the counterfeit, but if the scale tips to one side, then you have to go back to your recording to determine if the coin is lighter or heavier. If the coin was lighter, go with the side that tipped up, while if the coin was heavier, go with the side that tipped down.

<<Answer: 3 Weighings Minimum to Guarantee Finding the Counterfeit>>[/hide]

[hide] Yeah, I think it is 3.

Different weighings.

4 coins vs. 4 coins.

2 coins vs. 2 coins.

1 coin vs. 1 coin. [/hide]

Oops I just realized that the answer wasn't three after I consulted my father. The answer is two. I goofed  :D


function
trigonometry
algebra unsolved
algebra

Two functions f,g have definement sets symetric through 0 satisfying:
$f(x+y)=f(x).g(y)+f(y).g(x)$
and $g(x+y)=g(x).g(y)-f(x).f(y)\ (when\ f(-x)=-f(x) \forall \ x \in R)$
Prove that:
a) $g(x)=g(-x)$ and $g(0)=1$
b) $f^{2}(x)+g^{2}(x)=1$, $f(3x)=3f(x)-4f^{3}(x)$, $g(3x)=4g^{3}(x)-3g(x)$

Let $F(x)=g(x)+if(x)$, then F(x) satisfyed to
(1) $F(x+y)=F(x)F(y)$ (it is rquavalent to equation).
One is solutions $F(x)=0$ or $f(x)=0=g(x)$.
For another solutions $F(0)=1$ and $G(x)=-i*ln(F(x))$ satisfyed
(2) $G(x+y)=G(x)+g(y)$.
If $G(x)=(a+ib)x$, then $f(x)=e^{-bx}\sin{ax}, g(x)=e^{-bx}\cos{ax}$. But from $f(-x)=f(x)$ we have b=0.
All solutions (2) are linear Q functions $G(ax+by)=aG(x)+bG(y), \ a,b\in Q$. From $f(-x)=-f(x)$ we have $G(x)\in R$. It give $f^{2}(x)+g^{(}x)=1$ and $g(x)=g(-x)$.


Let $ a,b$ be positive integers with $ a$ odd. Define the sequence $ \{ u_n \}$ as follows: $ u_0\equal{}b$, and for $ n \in \mathbb{N}$, $ u_{n\plus{}1}\equal{}\frac{1}{2} u_n$, if $ u_n$ is even, and $ u_{n\plus{}1}\equal{}u_n\plus{}a$ otherwise.

$ (a)$ Show that $ u_n \le a$ for some $ n \in \mathbb{N}$.
$ (b)$ Show that the sequence $ \{ u_n \}$ is periodic from some point onwards.

[quote="moldovan"]Let $ a,b$ be positive integers with $ a$ odd. Define the sequence $ \{ u_n \}$ as follows: $ u_0 \equal{} b$, and for $ n \in \mathbb{N}$, $ u_{n \plus{} 1} \equal{} \frac {1}{2} u_n$, if $ u_n$ is even, and $ u_{n \plus{} 1} \equal{} u_n \plus{} a$ otherwise.

$ (a)$ Show that $ u_n \le a$ for some $ n \in \mathbb{N}$.
$ (b)$ Show that the sequence $ \{ u_n \}$ is periodic from some point onwards.[/quote]

[hide="My solution"]
Let $ v_n$ the subsequence of $ u_n$ defined as : $ v_0\equal{}b$ and, for $ n\in\mathbb N$, $ v_{n\plus{}1}\equal{}\frac 12v_n$ if $ v_n$ is even, and $ v_{n\plus{}1}\equal{}\frac{v_n\plus{}a}{2}$

It's clear that if $ v_n>a$, then either $ v_{n\plus{}1}\leq a$, either $ v_{n\plus{}1}\minus{}a < v_n\minus{}a$ hence the first statement $ (a)$

Using point $ (a)$, we know that $ \{u_n\}$ has infinitely many elements in $ [1,a]$, so at least two equal, hence the second statement $ (b)$

[/hide]

$ a)$ we take $ v_1=u_1$ and $ v_{n+1}={v_n/2\ if\ v_n\ is\ even  or =(v_{n}+a)/2^{h}\ if\ 2^h\parallel{}v_n+a}$
we have if $ v_{n}>a$ we have $ v_{n+1}<v_{n}$ so $ \exists k: \ v_{k}\le a$ and $ v_k=u_h$ for some $ h$

$ b)$ the result a) give $ S_n=\{i> n: \ u_i\le a\}\neq \emptyset$
we take $ f(0)=min(S_0)\ and\ f(n+1)=min(S_{f(n)})$, and obviously $ \exists n>m: \ u_{f(n)}=u_{f(m)}$
because $ \{u_{f(i)}\}\subset [1,a]$
so $ \forall k: \ u_{f(n)+k}=u_{f(m)+k}$
so $ \forall q\ge f(m): \ u_{k}=u_{k+T},where\ T=f(n)-f(m)$


integration
real analysis
real analysis unsolved

i've got this question:

does the equation  y' = 1 + y^3 has a solution defined in (-inf,inf)?
explain.

I guess i should answer it without solving the equation,
i've no idea how.

any help is welcome

thanks! :)

[quote="The_Seer"]i've got this question:

does the equation  y' = 1+y^3 has a solution defined in (-inf,inf)?
explain.

I guess i should answer it without solving the equation,
i've no idea how.

any help is welcome

thanks! :)[/quote]

Consider $x=x_{0}+\int_{y_{0}}^{y}\frac{dy}{y^{3}+1}$.
If $x\to+\infty$ then $\int_{y_{0}}^{+\infty}\frac{dy}{y^{3}+1}=+\infty$...


limit
trigonometry
calculus
calculus computations

Find the limit

[img]http://img18.imageshack.us/img18/1064/sanstitreni.png[/img]

 :)

hello, do you mean
$ \lim_{n\to \plus{}\infty }\sin(2\pi \sqrt{n^2\plus{}1})$?
Sonnhard.

Well, that didn't dd anything of value.

OK, now how would we solve it? First, $ \sqrt{n^2\plus{}1}$ is about $ n$ when $ n$ is large. So, the argument goes to $ \infty$- but $ \sin$ stays bounded. If that were just $ \sin(2\pi n)$, every term of the sequence would be zero. By the periodicity, we even have that $ \sin(2\pi n\plus{}t)\equal{}\sin t$.
Now we need to know more about that $ \sqrt{n^2\plus{}1}$.  What does $ \sqrt{n^2\plus{}1}\minus{}n$ look like as $ n\to\infty$?


Two players in turn replace asterisks in the matrix $\left(\begin{smallmatrix}* &* & \ldots & * \\ * &* & \ldots & * \\ \ldots & \ldots & \ldots& \ldots \\ * & * &\ldots & * \\ \end{smallmatrix}\right)$ of size $10\times10$ by positive integers $1,\ldots,100$ (at each turn one may take any number which has not been used earlier and replace by it any asterisk). If they form a non-singular matrix then the first player wins, else the second player wins. Has anybody of players a winning strategy? If somebody has, then who? (V. Brayman)

See all the problems from [color=orange]Kyiv Mechmat Competition, 2007[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=135749]here[/url]

The second player has the strategy
We can simply pair-up the cells like this, the $(2i,j)$ cell with $(2i+1,j)$ cell.
Now when the first player puts the number $n$ in some cell, we can put the number $101-n$ in it's twin cell (The cell paired with it)
After the game finishes, the sum of $2i$ and $2i+1$ rows is always $(101,101,\ldots,101)$
So if $r_{i}$ is the $i$th row, then $r_{1}+r_{2}-r_{3}-r_{4}$ is always zero, so the matrix is always singular


geometry
inequalities
perimeter
trigonometry
geometry unsolved

The circumference  of a convex quadrangle is 2p

find the max area of this quadrangle

According to the isoperimetric inequality for polygons: the quadrilateral that maximum area is a square. Since perimeter of square $ 2p$, area is $ \frac {p^2}{4}$

second solution: Let $ ABCD$ convex quadrilateral and $ |AB|\equal{}a, |BC|\equal{}b, |CD|\equal{}c, |DA|\equal{}d.$ We know that: if Area$ (ABCD) \equal{} S, \angle ABC \equal{} \alpha, \angle ADC \equal{} \beta$ then $ 16S^2 \plus{} (a^2 \plus{} b^2 \minus{} c^2 \minus{} d^2)^2 \equal{} 4a^2b^2 \plus{} 4c^2d^2 \minus{} 8abcd. \cos(\alpha \plus{} \beta)$.

According to the this area formula, when $ \alpha \plus{} \beta \equal{} \pi$, area of $ ABCD$ will be maximum. That is, $ ABCD$ must be a cyclic quadrilateral.

Now, we yields Brahmagupta formula: $ S^2 \equal{}(p \minus{} a)(p \minus{} b)(p \minus{} c)(p \minus{} d)$. From $ A\minus{}G\minus{}M$ inequality

$ \sqrt[4]{(p \minus{} a)(p \minus{} b)(p \minus{} c)(p \minus{} d)} \leq \frac{(p \minus{} a)\plus{}(p \minus{} b)\plus{}(p \minus{} c)\plus{}(p \minus{} d)}{4}\equal{}\frac{4p \minus{} (a\plus{}b\plus{}c\plus{}d)}{4}\equal{}\frac{p}{2}$

So $ \sqrt{S}\leq\frac{p}{2}$ and finally $ S\leq \frac{p^2}{4}$.

Equality satisfy when $ p \minus{} a \equal{} p \minus{} b \equal{} p \minus{} c \equal{} p \minus{} d$. That is $ a \equal{} b \equal{} c \equal{} d$ and $ ABCD$ is square.


probability
geometry
3D geometry
tetrahedron
real analysis
real analysis unsolved

If you choose randomly 6 positive numbers (uniformly distributed) what is the probability of being the lenghts of a tetrahedron?

Is that uniform from $0$ to $A$ for some fixed $A$ (which might as well be 1)?

The answer would depend on the shape of the allowed region; that's why I ask. There's no such thing as uniform on all positive numbers.

Then consider all $6$ positive real numbers uniformly distributed in some interval $[0,M]$ - or assume $[0,1]$ for simplicity.
But I think it doesn't matter since you may fix the greatest as $M$ and take the other $5$ uniformly distributed variables in $[0,M]$.
Fix $M=1$, and look in the $5$-th dimensional space inside the hypercube $[0,1]^5$ for the volume of the allowed region which gives our probability and it is generated by the $5$-tuples $(x_1,x_2,x_3,x_4,x_5)$  that satisfy all the conditions like (intersect them):
"any $x_1$, $x_2 + x_3 \geq 1$ and $x_4 + x_5 \geq 1$"

I think the right answer is $\frac{5}{8}$.

If you choose randomly $12$ positive numbers (uniformly distributed in $[0,1]$) what is the probability of being the lenghts of some geometrical shape?


trigonometry
function
real analysis
real analysis unsolved

Can someone help me with this?

An infinitely long plate of width L has its two parallel sides maintained at temperature zero and its other side at constant temperature T. What is the rate of change?

I have that Laplace's equation holds and u(x,0) = T, u(0,y) = 0, u(L,y) = 0 with the lim as y goes to infinity of u(x,y) finite.

I know how to do the separable equations stuff, but can someone help me with the eigenfunctions (i.e. what are X(x) and Y(y) for u(x,y) = XY). Thanks.

If we set $ u \equal{} X(x)Y(y),$ then we have $ \frac {X''}{X} \plus{} \frac {Y''}{Y} \equal{} 0,$ so $ X'' \equal{} \lambda X$ and $ Y'' \equal{} \minus{} \lambda Y.$

The eigenvalue problem $ X'' \equal{} \lambda X; X(0) \equal{} X(L) \equal{} 0$ is pretty well-known. You would solve exactly this in a number of other PDE contexts. The eigenvalues are $ \lambda \equal{} \minus{} \frac {k^2\pi^2}{L^2}$ and the eigenfunctions are $ X \equal{} \sin\left(\frac {k\pi x}{L}\right).$

From that we get that for that $ \lambda,$ $ Y(y) \equal{} Ae^{k\pi y/L} \plus{} Be^{ \minus{} k\pi y/L}.$ The restriction that the function be bounded as $ y\to\infty$ forces $ A \equal{} 0.$ That gives us the following general form of the solution:

$ u(x,y) \equal{} \sum_{k \equal{} 1}^{\infty}b_ke^{ \minus{} k\pi y/L}\sin\left(\frac {k\pi x}L\right).$

If we now set $ y \equal{} 0$ we get the "initial" condition $ u(x,0) \equal{} T.$ This (the "square wave") has a pretty well known Fourier expansion - in fact, I have it memorized. The final answer:

$ u(x,y) \equal{} \frac {4T}{\pi}\sum_{j \equal{} 0}^{\infty}\frac1{2j \plus{} 1}e^{ \minus{} (2j \plus{} 1)\pi y/L}\sin\left(\frac {(2j \plus{} 1)\pi x}L\right).$

At least that's what you get if you assume that you want the steady-state temperature. If you want to know how things change with time if it's not the steady-state being asked for (and the question about rate of change makes me wonder about that), then we'll have to solve the heat equation instead of Laplace's equation, and that will take some more work.


so....How is everybody, AMSPers???  Is school BUSY??
I still miss AMSP!!
Anybody plannig to go to any math/science contests??  We should have a reunion!

School just started today, and it wasn't as bad as I feared.

AMSP reunion might be a good idea if there's a contest we all participate in. Maybe several AMSP iTest teams? Idk. The only science olympiads I know I'm doing is physics.

well im sorry for playing starcraft too much EXCEPT I ACTUALLY DONT

We are all caught up in a horrible place called school.

I will not be at any math contests because I simply fail. The utterest failure there is: that which is spelled correctly and not as "phail" or any other sorts of variants.

...wow...?=^..^=(its a cat)...i dont get this emoticon...? :arrow: :rotfl: guess who it is?

I am so busy at college... (but not working that much lol)...

[quote="dannyhamtx"]dangit i don't do itest anyways

ARML may be a good idea except there are 4 sites.

There won't be enough places in MOP for all of us to go

You can't meet people in AMC, AIME or USAMO

many of us finished MC

and some of us *cough* ccy and pythag01 *cough* are annoying.

and some of us play starcraft *cough* junggi *cough* too much

some of us are now very inactive on AoPS *cough* (can't think of anyone) *cough*

*sigh* there is only little hope left......[/quote]

how was i not included in the annoying category?

[quote="dannyhamtx"]they have been censored due to the nudity

some of us are too busy killing little kid's rating on FTW *cough* mustafa *cough*[/quote]

HEY

STOP THAT

schooll =beeeeeeeeeeeeeeeeeeeetttchhh.
first two days went by,
so much effin work
yet i still waste my time away on fb.
already looking forward to summer 09. damn.

[quote="pythag011"]How many people are doin HMMT?

..........
..........
..........
..........
..........[/quote]
I am planning to go to HMMT.  Are you going??

There're going to be alot of "mini-reunions" as far as I can tell.  Ppl from Bay area, people from NYC, people from TX, ppl from Romania.

Eric, Meena and I are taking a NYC math class together.

There's really no practical way to all meet except all making MOP.  Even then, I'm not sure that's even possible.   :stink:  I wish there was...

Yes. I already met quite a few AMSPers through ARML.

Btw David, I'm going to take Ali's classes too. It's called AMSP too  :huh:

[quote="zephyredx"][quote="cheeesec4k3"]Ohh!
Yeah, we should. But most people probably don't know who I am...  :|[/quote]

We wouldn't mind an introduction.

And I plan on attending WOOT on Mondays.[/quote]

Ahaha well I never give out my real name on the internet due to my paranoia. =p

But yeahh I'm in the yearbook... and I think I have a pretty distinct username. x] Oh, and if you are who I think you are, I think you're one of my friends on facebook. Hmm... I'll PM you. o.o

I'll probably be doing WOOT on Mondays. Except for this one cause I have to stay after school until 7 or so to lay out the newspaper. :(

[quote="1=2"]P must be monic. continue from there[/quote]

This fact is trivial. It is not used at all (at least in the solution I have.)

hint. Guess the solutions. HInt: Infinitely many solutions.

ahahhaa. nice vid 1=2
ARGHHHH. like nobody goes to the iowa arml :'( i wish arml was one big nats. competition... thatd be amazinggg =]]

[quote="pythag011"]lol. That's interesting. PROBLEM TIME!!!!

If Q(n) = $ n^2 \plus{} 1$, find all polynomials P such that P(q(n)) = Q(P(n)) for all n.[/quote]

[hide=" :rotfl: "]
[quote="pythag011"]I have solved it; define the sequence $ P_n$ as $ P_1 \equal{} x$ and $ P_{n \plus{} 1} \equal{} P_n(x^2\plus{}1)$ those are the soluitons. So there are infinitelty many solutions.[/quote]

I actually edited the quote a bit because I didn't copy the LaTeX. It might be wrong, but you get the idea.  :P Good use of resources![/hide]

dont wry leoxnlin... i have no clue what he's talkin' bout either...

[quote="leoxnlin"]I am too stupid to understand what might be so funny about pythag's problem.  :([/quote]

It's funny because pythag posted it in his blog, and I just copy-pasted the solution that he found to his own problem. Hence "using resources"  :P

prove it.

I have a proof that your solution is correct, but I will not poste it until you give up.


geometry
perimeter
number theory
least common multiple

three people a,b,c have speeds og 4m/hr,7m/hrand 14m/hr resp and are runnin on a circular track of perimeter 12m.when will they meet?i couldnt get any method other than hit and trial. :mad:

do they start at the same place?

and where do you want to calculate that they meet? at their start position or at the first place that is possible?  :?

they start fromm the same position..so find where will they next meet.

Well, since the second one runs only half as fast as the third one, the only position they can meet is the starting point
Just visualise it, and you'll see: after each lap of the second one, the third one will have exactly two laps, and they won't interfer anywhere in between...

This may be off but I'm thinking perhaps it'll be the least common multiple of 4.5, 7, and 14?

4.5*7*2= 63

They'll meet in 63 miles.

We've already established that B and C will intersect every 2 laps.

lcm (4.5, 14)= 63

Its a little too early in the morning for me to be visualizing anything...

It's not 4.5, it's 4. As Psycho said, the only place the second and third runners can meet is at the starting point, so we need to find when the first runner will be at the starting point at the same time as the second runner. Since it's a 12-mile track, the first runner takes $3$ hours to run the whole thing, and the second runner takes $\frac{12}{7}$ hours to run the whole thing, so we find $LCM(3,\frac{12}{7})=12$, so the first time they will all meet is in 12 hours, which, upon checking, shows that the first runner will be at the starting line.

If it's 4, 5, and 14, it's a little more difficult. It's a similar strategy, but with more work. $4$ runs the 12-mile track in $3$ hours, $5$ runs the 12-mile track in $\frac{12}{5}$ hours, and $14$ runs the 12-mile track in $\frac{6}{7}$ hours. $LCM(3,\frac{12}{5},\frac{6}{7})=12$ hours, still, but it's a little more difficult to calculate.

HOW CAN THAT BE HIT AND TRIAL! it is the only actual way without l.c.m it is not possible.

[quote="shadysaysurspammed"]i may be dumb... :D but whats the reason for taking the l.c.m?thats sort of hit and trial.isnt there any other proper mathematical method? :([/quote]

Why would you say it's guess-and-test?

[quote="shadysaysurspammed"]i'm not sayin it actually guess n test.i just thought there might be another way to solve it,yoyknow,using equations n syuff...
welll,if thats the way its done,i guess i'll have to take the method.
THANX FOR UR REPLIES GUYS... ;)[/quote]

Of course there is a way with equations:

[hide]
When they meet, obviously they will have gone for the same amount of time, but will have gone different distances.  Using $d=rt$ we get:

\begin{eqnarray*}
d_a &=& 4t\\
d_b &=& 7t\\
d_c &=& 14t
\end{eqnarray*}

In order for two of them to be at the same spot one of these two must have gone some multiple of 12 miles farther than the other.  We can thus say

\begin{eqnarray*}
d_c &=& d_b +12k_1\\
d_b &=& d_a+12k_2
\end{eqnarray*}

for some integers $k_1,k_2$. Plugging in for the d's and simplifying we get

\begin{eqnarray*}
7t &=& 12k_1\\
3t &=& 12k_2
\end{eqnarray*}

Thus $k_1=\frac 73 k_2$.  Remember that both of these must be integers and we want them to be as small as possible to find the first time that the people meet.  Thus we take $k_2=3$ which makes $k_1=7$.  Thus $t=\boxed{12.}$
[/hide]


geometry
cyclic quadrilateral
power of a point
radical axis
geometry unsolved

Let $ ABC$ be a triangle, $ D$, $ E$ and $ F$ the points of tangency of the incircle with sides $ BC$, $ CA$, $ AB$ respectively. Let $ P$ be the second point of intersection of $ CF$ and the incircle. If $ ABPE$ is a cyclic quadrilateral prove that $ DP$ is parellel to $ AB$

Invert about $ F$.  Let $ A,B,C,D,E,P$ be transformed into $ A^\prime,B^\prime,C^\prime,D^\prime,E^\prime,P^\prime$.  $ AC$ is transformed into circle $ \Gamma_1\equal{}A^\prime C^\prime F$, and $ BC$ is transformed into circle $ \Gamma_2\equal{}B^\prime C^\prime F$.  $ D^\prime$ and $ E^\prime$ are on $ \Gamma_2,\Gamma_1$ respectively so that $ D^\prime E^\prime$ is a common tangent.  Since this tangent is the transformation of the incircle, it is parallel to $ A^\prime B^\prime$.  Then $ P^\prime$ is the intersection of the radical axis $ FC^\prime$ and the common tangent $ D^\prime E^\prime$.  Thus, $ P^\prime D^\prime\equal{}P^\prime E^\prime$.  Since $ ABPE$ is cyclic, so is $ A^\prime B^\prime P^\prime E^\prime$.  Since $ P^\prime E^\prime\parallel{}A^\prime B^\prime$, $ A^\prime B^\prime P^\prime E^\prime$ is an iscoceles trapazoid.  Since $ D^\prime E^\prime$, the common tangent is parallel to $ A^\prime B^\prime$, it follows that the projection of $ E^\prime$ onto $ A^\prime F$ bisects the segment $ A^\prime F$.  Thus, $ D^\prime P^\prime\equal{}P^\prime E^\prime\equal{}A^\prime B^\prime\minus{}A^\prime F\equal{}B^\prime F$.  Since the projection of $ D^\prime$ onto $ B^\prime F$ bisects the segment, it follows that the projection of $ F$ onto $ P^\prime D^\prime$ bisects the segment.  Thus, the circle $ D^\prime P^\prime F$ is tangent to $ A^\prime B^\prime$ and thus $ DP$ is parallel to $ AB$, as desired.

$DE\bigcap AB={M}$
Then MP is the tangent of cyclic (PEFD).
$PM\bigcap CA={K}, PB\bigcap DE={H}$
ABPE is cyclic then $\angle PBM=\angle KEP=\angle PDE=\angle KPE$
Hence PDBA is cyclic.
Then $\angle BDE=\angle BPM=\angle DEA$
KPHE is cyclic. Hence $\angle PHD=\angle PKE \Rightarrow \angle HPD=\angle HDP$
Hence We have $  PD\left |  \right |AB$ :)


MATHCOUNTS
number theory
modular arithmetic

These are problems I need to find easier solutions to.  Please give me a formula or something to find any problem of the same type.

1.  What is the units digit of 6^2004?

2.What is the greatest prime factor of 10!?

Thanks for the help!

1) Well, 6^x will always end in 6, so you don't have to worry about that.

2)For this one, 10! has a greatest prime factor of 7. You don't even need to prime factorize this one because you know 11 isn't a factor of 10!(10*9*8..*1), and 7 is the greatest prime less than 11.

These are likely candidates for future countdown problems.

That's where I got them, practice packet for MathCounts, old Countdown Round.  Thanks, that helps a lot.

In fact, it might be useful to memorize the units digits of integer powers.  Notice:

2^1 has a units digit of 2
2^2 ........ 4
2^3 ........ 8
2^4 ........ 6
2^5 ........ 2
2^6 ........ 4
and so on.  Thus, the units digits of powers of 2 cycle 2-4-8-6.

Similarly, units digits of powers of 3 cycle 3-9-7-1
Units digits of powers of 4 cycle 4-6

Try and find the patterns for all digits up through 9.  These will be useful.  Guaranteed.

(I guess I could explain how this is justified by modular arithmetic base 10, but we'll leave it at that for now, eh?)


Hello,

What do you think about this one,

Proof or find a counter example: Given two integers $p,q$ such that
\[\frac{p}{q}<\sqrt7\]
then
\[\frac{p}{q}+\frac{1}{pq}<\sqrt7\]

Best regards,

we have: $7q^2-p^2=m>0$ so we have: $7q^2p^2-p^4=mp^2$ and we want to prove $mp^2>2p^2$ or $m>2$.

Sow e have to prove $7q^2-p^2$ is never $1$ or $2$. locking modulo 8 we have it is congruent to $-q^2-p^2$ wich is congruent to $3, 6, 7, 0$ modulo 8, so in fact 1 and 2 are not ateinable.


floor function
number theory unsolved
number theory

[quote="Aravind Srinivas L"]Determine the number of 5-digit positive integers $ ABCDE$ such that sum of 3 digit numbers $ ABC$ and 2 digit number $ DE$ is divisible by 11.[/quote]
8181 its such numbers as 11|ABCDE

Sorry....I thought ofposting a tougher one...never knew this one was so easy...by mistake pposted itin the unsolved....Will post a stronger one in the propsed soon.

[quote="Mathias_DK"][quote="Aravind Srinivas L"]Determine the number of 5-digit positive integers $ ABCDE$ such that sum of 3 digit numbers $ ABC$ and 2 digit number $ DE$ is divisible by 11.[/quote]
Let $ x \equal{} ABC$ and $ y \equal{} DE$. and $ z \equal{} ABCDE \equal{} x \plus{} 100y$.
Then $ x \plus{} y \equiv x \plus{} 100y \equal{} z \bmod 11$. So we only have $ 11 \mid z$. The number of natural numbers not exceeding $ 99999$ which is divisible by $ 11$ is exactly $ \left \lfloor \frac {99999}{11} \right \rfloor \equal{} 9090$. The number of natural numbers not exceeding $ 9999$ which is divisible by $ 11$ is exactly $ \left \lfloor \frac {9999}{11} \right \rfloor \equal{} 909$. So the number we are looking for is $ 9090 \minus{} 909 \equal{} 8181$.[/quote]

Let $ x \equal{} ABC$ and $ y \equal{} DE$. and $ z \equal{} ABCDE \equal{} x \plus{} 100y$.

should be

Let $ y \equal{} ABC$ and $ x \equal{} DE$. and $ z \equal{} ABCDE \equal{} x \plus{} 100y$.

But the final result is right.


It's a long-debated issue, but I have a healthy opinion on it: yes, we [i]should[/i] keep exploring space.

I think it's just not really about any survival tactics or trying to improve technologies here on Earth - it's simply about the human desire for exploration.

Sure, Earth has a lot of problems, like starvation of the poor - but why should that stop us from exploring space? We already spend enough money as it is trying to fix our problems, and it's not just a problem of money, it's a problem of political will.

(e.g. There are some religious fundamentalists that are firmly opposed to space travel because they believe it contravenes God's will for people to be subservient and faithful to him, like the tower of Babel did.)

Of course, I'm talking actual exploration here, not the stuff that happens at the ISS, which is just sedentary research. I'm talking sending probes and eventually people "out there", perhaps to explore new planets or new stars.

What do you think?

Space exploration is one of those places where we need to take the "long view." Some may say that it does nothing to improve life on earth, but they are too shortsighted. The potential benefits are just too great to stop right here, to give up. It''s too late at night for me to go into what I think are the underlying philosophical issues.

[quote="12markkram34"]What are some of the potential benefits? How would they balance the costs?[/quote]

Well, let's forget potential benefits for a moment and see what benefits it's [i]already[/i] given us:

Better health care
Some materials we use everyday - where do you think chip bag wrappers and microwaves came from?
A better understanding of Earth itself
Some new theories in physics

Further potential benefits include a permanent moon colony. Yes. For sure. And at a fraction of the cost it takes to have a war in Iraq!

Just make sure those "short-sighted conservatives" don't get in the way...

I don't think any of the arguments above are valid - a colony on the moon is not yet a good idea, and the other goals are either exaggerated or could have been achieved more cheaply without the added baggage of space exploration. I think the idea has its own merits, though. If we're talking as percentage of world GDP, space exploration is a pittance. Even as a percentage of US GDP. Eventually, larger-scale stuff will be viable, but we have to start now. The inspirational value is more difficult to quantify, but it exists.

I thought microwaves were invented when a person realized that the bar of chocolate in his pocket melted?

[url]http://www.cnn.com/2007/LIVING/worklife/10/04/nasa.everyday/index.html[/url]
Here's a site with a list of inventions that came out as a result of space.

[quote="zapi2007"]I thought microwaves were invented when a person realized that the bar of chocolate in his pocket melted?[/quote]

Sorry, wrong invention.

That's a good reason, but are there activities on Earth that have similarly stimulated technological advancement?

[quote="ZetaX"](there in fact is no overpopulation problem in any western state).[/quote]

That's... correct, actually. But we should still try to rectify the problem in developing countries.

And if we can't terraform the Moon, what about Mars? Might that happen within the next 60 years?

[quote="ZetaX"][quote="Ihatepie"]I personally would love to see a permanent moon base within my lifetime. Colonizing the moon would not only provide a new frontier, but would get rid of our overpopulation crisis, and give a new area for scientists to use their knowledge (basically how to make the moon get an atmosphere, good soil, water, etc.) Plus, almost anyone can do a backflip there.  :lol:[/quote]
I think you are getting some things wrong. Terraforming the moon is out of reach (not only for this century), if not impossible, and colonizing it means having some bases with 5 till some thousend people on it, thus not solving any overpopulation problem (there in fact is no overpopulation problem in any western state).[/quote] No I don't think I am. I never said any of those things. I only said I wanted [i]a[/i] [b]permanent moon base[/b] within my lifetime. I know that terraforming and completely colonizing it are out of reach. (at least within the century)

Also,I do think it could help solve the overpopulation problem. In a few hundred years, when we actually have the technology to completely colonize the moon, we could have some displaced people go there, solving our overpopulation crisis. (Note that there isn't much of a problem now with it, but there will be in the future.) But we can never get even close to these things if we don't start working on it. I might not see results in my life time, but over time, humanity as a whole will benefit.

I honestly do not care about problems that can't be solved, or won't even arise, within my lifetime.
I'm not shortsighted, just not immortal.

The problem with overpopulation is the distribution of a finite amount of [i]resources[/i]; lack of physical space on which to live is irrelevant at the world scale.  The idea that we will be able to live sustainably on the moon is silly; the idea that it makes more sense to do that then, say, to increase population density in Utah or the Australian outback is really strange.

In my opinion,we should continue space exploration.[b]To solve a problem, one cannot divert all the energy  to its solution[/b].Space exploration may bring out facts that may help mankind in the long run.

[quote="JBL"]The idea that we will be able to live sustainably on the moon is silly.[/quote]

And people thought that machines flying was silly, 100 years ago.

And if not the Moon, how about Mars? We need to start [i]somewhere[/i].

How about Earth?   :roll:

Edit: Also, your rejoinder is just wrong: kites and hot air balloons have been in use for more than 2000 years, according to [url=http://en.wikipedia.org/wiki/History_of_flight]Wikipedia[/url], and the first human glider flight was apparently in the 9th century CE.  The first flight by a steerable aircraft was in the 1850s.  A moon colony that could support enough human life so as to "help solve the overpopulation problem" is to current space flight technology as a 747 is to a helium balloon.

[quote="JBL"]How about Earth?   :roll:[/quote]

But we've already finished here. Sure it won't be sustainable in the long run, but what I mean is that "we need to start extraterrestrial exploration somewhere".

[quote]Edit: Also, your rejoinder is just wrong: kites and hot air balloons have been in use for more than 2000 years, according to [url=http://en.wikipedia.org/wiki/History_of_flight]Wikipedia[/url], and the first human glider flight was apparently in the 9th century CE.  The first flight by a steerable aircraft was in the 1850s.  A moon colony that could support enough human life so as to "help solve the overpopulation problem" is to current space flight technology as a 747 is to a helium balloon.[/quote]

I meant heavier-than-air flight, but alright.

And I didn't say that the moon colony had to solve the overpopulation problem immediately when it was built, or directly help in any way. But if we were really able to build a self-sufficient colony, we'd learn more about how to recycle resources here on Earth.

It's not like the problem here on earth is a lack of some technology. What fails here are politics and the people itself. Any argument of type "space colonialisation will solve our current ressource/recycling/clima/population/whatever problems" is just wrong, as such a thing can be archieved much easier and much cheaper with current technology, but just no one does (and the colonies will give them no reason to start doing so!).
The only good argument for colonies in the short run is curiousity. In the very long run (some thousend years maybe), ressources may come into play, but not for current issues.


Prove that $ (ab\plus{}bc\plus{}ca) (\frac{1}{(a\plus{}b)^2} \plus{} \frac{1}{(b\plus{}c)^2} \plus{} \frac{1}{(c\plus{}a)^2}) \ge \frac{9}{4}$

[quote="hsbhatt"]Prove that $ (ab \plus{} bc \plus{} ca) (\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}) \ge \frac {9}{4}$[/quote]
$ (ab \plus{} bc \plus{} ca) \left(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right) \ge \frac {9}{4}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(4a^5b \minus{} a^4b^2 \minus{} 3a^3b^3 \plus{} a^4bc \minus{} 2a^3b^2c \plus{} a^2b^2c^2)\geq0,$ which true by Muirhead and Schur.

It's Iran 1996
See here : http://www.mathlinks.ro/viewtopic.php?search_id=1425734671&t=3547


Compute the minimum $ n$ such that there exist $ n$ distinct odd integers $ a_1$ through $ a_n$, none equal to $ 1$, such that $ \displaystyle\sum_{i \equal{} 1}^n{\displaystyle\frac {1}{a_i}} \equal{} 1$.

$ n\equal{}9$ works. 3, 5, 7, 9, 11, 15, 35, 45, 231 work.

I didn't figure this out myself, I got it from a book. I'm not sure how we could actually deduce this by ourselves.

[quote="Zhero"]$ n \equal{} 9$ works. 3, 5, 7, 9, 11, 15, 35, 45, 231 work.

I didn't figure this out myself, I got it from a book. I'm not sure how we could actually deduce this by ourselves.[/quote]

Yeah, you're right.. I think it'll be very difficult to find.. :maybe:

I made a proof for $ n < 9$, but it's .. little 'dirty'.. maybe there's another simple proof.

[hide]First, because $ \frac {1}{3} \plus{} \cdots \plus{} \frac {1}{13} \equal{} 0.955134 < 1$ , $ n > 6$.

If $ n \equal{} 7$, because $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{13} \plus{} \frac {1}{15} \plus{} \frac {1}{17} \equal{} 0.989715 < 1$, $ (a_1 , a_2 , a_3 , a_4 , a_5 )$ must be $ (3,5,7,9,11)$ (WLOG $ a_1 < a_2 < \cdots < a_n$)

Therefore, let $ a_6 \equal{} x,  a_7 \equal{} y$, $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} 1 \minus{} (\frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{11})$. It cannot be done, definitely.

If $ n \equal{} 8$, because $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{9} \plus{} \frac {1}{11} \plus{} \frac {1}{13} \plus{} \frac {1}{15} \plus{} \frac {1}{17} \plus{} \frac {1}{19} \equal{} 0.990398 < 1$,  $ (a_1 , a_2 , a_3 )$ must be $ (3,5,7)$ .

Therefore, there must be a multiple of 35(=5*7) in $ (a_4 , a_5 , a_6 ,a_7 , a_8 )$.
Similarly, there must be a multiple of 15(=5*3) and 21(=3*7) too.

If there isn't a multiple of 105, $ a_6 \equal{} > 15, a_7 > \equal{} 21, a_8 > \equal{} 35$

Therefore, because  $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{11} \plus{} \frac {1}{13} \plus{} \frac {1}{15} \plus{} \frac {1}{21} \plus{} \frac {1}{35} \equal{} 0.98688 < 1$, $ a_4$ must be 9. Therefore there must be at least one other multiple of 9. (which is bigger than 27)

It makes contradiction, ( $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{27} \plus{} \frac {1}{15} \plus{} \frac {1}{21} \plus{} \frac {1}{35} \equal{} 0.967196 < 1$ ), so there must be a multiple of 105.

WLOG let $ a_8 \equal{} > 105$. because $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{13} \plus{} \frac {1}{15} \plus{} \frac {1}{17} \plus{} \frac {1}{105} \equal{} 0.999239 < 1$, $ (a_1 , a_2 , a_3 , a_4 , a_5 )$ must be $ (3,5,7,9,11)$ (WLOG $ a_1 < a_2 < \cdots < a_7$)

Therefore, there must be another multiple of 11 in $ (a_6 , a_7 , a_8 )$.

Because $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{11} \plus{} \frac {1}{13} \plus{} \frac {1}{33} \plus{} \frac {1}{105} \equal{} 0.994961 < 1$, $ a_6$ and $ a_7$ cannot be another multiple of 11. Therefore $ a_8$ must be a multiple of 11.

Now, we get $ (a_1 , a_2 , a_3 , a_4 , a_5 ) \equal{} (3,5,7,9,11)$ and $ a_8 \equal{} > 1155 \equal{} 11*105$.

Next, if there aren't 13 or 17 in $ (a_6 , a_7)$, it makes contradiction ($ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{11} \plus{} \frac {1}{15} \plus{} \frac {1}{19} \plus{} \frac {1}{1155} \equal{} 0.998375 < 1$

Therefore 13 or 17 must be included in $ (a_6 , a_7)$. First, if 13 is included, $ a_6 \equal{} 13$. and similarly, we can show that $ 13|a_8$. Because $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{11} \plus{} \frac {1}{13} \plus{} \frac {1}{23} \plus{} \frac {1}{13*1155} \equal{} 0.998679 < 1$ , $ a_7 < \equal{} 21$. In each case, $ a_8$ cannot be an integer (maybe... :?:  I don't have enough time to check all)

Finally, if 17 is included, $ a_7 \equal{} 17$ (if $ a_6 \equal{} 17, \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{9} \plus{} \frac {1}{11} \plus{} \frac {1}{17} \plus{} \frac {1}{19} \plus{} \frac {1}{1155} \equal{} 0.990532 < 1$ -> contradiction!)

therefore, $ a_6$ can be 13 or 15. and in each case, $ a_8$ cannot be an integer (maybe again)

∴ n>8.

(What a dirty sol :!: )[/hide]

or, for the $ n \equal{} 8$ case, we could employ a simple parity argument. :P

In fact, we will show that 1 cannot be written as the sum of the reciprocals of $ 2n$ distinct odd integers $ a_1, a_2, \ldots, a_{2n}$. Suppose it can. Then

$ a_1a_2 \ldots a_{2n} \equal{} \sum_{i \equal{} 1}^{2n} \prod_{j \neq i} a_j$. Each of the addends is a product of odd numbers and is thus odd. The sum itself of $ 2n$ odd numbers must therefore be even. But the left-hand side, the product of odd numbers, must be odd.


Every time I try installing TeXnicCenter, it says "Error: Access is denied." Can anyone please help? Thanks! :) I tried it for almost 45 minutes and still no success!!! :(

Are you using Vista? Then if so ,right-click on the icon and choose Run as Administrator.

No, that happens when I'm installing it...I can never install it!

That is what I said. You [b]install[/b] TexnicCenter by using Run as Administrator.


I am poor and starving now, I wish to find some money to feed myself and do something good for freedom of the world. :P

http://www.ned.org/grants/06programs/grants-asia06.html

[quote]
The Endowment is especially interested in proposals that originate with indigenous democratic groups. It is also interested in nonpartisan programs seeking to strengthen democratic values among all sectors of the democratic political spectrum. Note that the Endowment does not make grants to individuals.
[/quote]
So, it seems like the first step would be to become a member of "indigenous democratic group" (my God, I had to look up the first word in the dictionary!) :( and a good initial move in this direction will be to stop spamming in the Round Table :P.


Just curious .. has anyone else not yet received their round 3 checkmark?

Thanks for the help,
JB

[quote="Xantos C. Guin"]I have recieved my round 3 check mark a week ago.

BTW: When will Round 4 problems be up?[/quote]

So did I.  I believe they said the next round will be posted in early February.

I received mine (by snail mail) almost immediately.

My snail mail ones are in also, maybe they were the first ones to be recorded. How did you send yours in?

I got mine about two days after submitting, I submitted via email.
By the way don't the admins need to change the usamts home page since it right now says that you can still do round 3?

[quote="Xantos C. Guin"]I have recieved my round 3 check mark a week ago.

BTW: When will Round 4 problems be up?[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127449]See Rusczyk's post[/url]


topology
inequalities
function
limit
calculus
integration
IMO Shortlist

Pascual, please use  \leq and \geq for $\leq$ and $\geq$ rather than <= and >=! Thanks.

What does this source mean ? If it means IMO ShortList than you will better for you reading the following two threads first before unnecessarily reposting problems:

- [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=119224]IMO ShortList 2001[/url]
- [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=153484]What are IMO ShortLists ?[/url]

[quote="Pascual2005"]Let $A$ be a set of real numbers. There exist integers $b_1,b_2,b_3...b_n$ and $c_1,c_2,c_3...c_n$ such that the $n$ sets: $b_iA+c_i$ are included in A and also are disjoint.

Prove that: $1/b_1+1/b_2...+1/b_n<=1$[/quote]

This is a problem about the fractal (Hausdorff) dimension of $A$.  Say $d = dim(A)$, $0 \leq d \leq 1$.
Assume there is a d-dimensional volume measure on $A$ for which the transformation $t \rightarrow bt + c$
multiplies volume by $b^d$.  Then
$1/b_1^d + 1/b_2^d + ... + 1/b_n^d  \leq 1$  because of the disjointness.
For the actual proof you don't need to discuss any measure, only to look at coverings by small
intervals and prove the inequality for all $d + \epsilon$.  But the $d$-dimensional volumetric idea
is the basic intuition.

To fleeting_guest:
Honestly, I didn't understand anything out of that solution  :?  :(
About the solution, I think something like iterating the function $f_i(x)=b_i\ldots x + c_i$ will give us some bound, but I forgot it completly..... :(

well i also tried considering thah functions, but i dont really now hoe to solve it, and i didnt understand neither the solution in the post above, i was waiting for a nice olympiad styloe solution because it is a problem from ISL, so there must be a solution like that

My solution goes as follows. Consider the power series $f(x) = \sum_{a \in A} x^a$; because $A$ is and infinite set and $f(x)$ is majorized by $\frac{1}{1-x}$, $f(z)$ has radius of convergence $1$. By the assumption of the problem, $f(x)$ majorizes the series
\[
\sum_{1 \leq i \leq n} \sum_{a \in A} x^{b_ia+c_i} =  \sum_{1 \leq i \leq n}x^{c_i}f(x^{b_i}).
\]
Hence, for every $t \in (0,1)$ we have the inequality
\[(*)
f(t) \geq  \sum_{1 \leq i \leq n}t^{c_i}f(t^{b_i}),
\]
i.e.
\[
\sum_{1 \leq i \leq n} \frac{t^{c_i}f(t^{b_i})}{f(t)} \leq 1.
\]
On the other hand, (by L'Hospital's rule or otherwise), one easily sees that, for every $b\geq 1$,
\[(**)
\lim_{t \to 1-0} \frac{f(t^b)}{f(t)} \geq \frac{1}{b}.
\]
Putting $b=b_1,b_2,\ldots,b_n$ in (*), we obtain the desired inequality $\frac{1}{b_1} + \cdots + \frac{1}{b_n} \leq 1$. Q.E.D.

It's not too difficult, is it?  :lol:

[b]Remark 1. [/b] It is clear from the solution that the following generalization of the problem holds. If $A \subset \mathbb{N}$, the sets $b_iA+c_i \subset A$ are pairwise disjoint, and every element of $A$ belongs to at most $k$ of the sets $b_iA+c_i$, then
\[
\frac{1}{b_1} + \cdots + \frac{1}{b_n} \leq k.
\]

[b]Remark 2. [/b] If $A$ has a positive upper density, then the inequality in the limit (**) becomes equality. Thus, it is clear from the solution that we have the following related result. Let $A \subset \mathbb{N}$ be a set with a positive upper density, and let the sets $b_iA+c_i$ form a disjoint partion of $A$. Then,
\[
\frac{1}{b_1} + \cdots + \frac{1}{b_n} = 1
\]
(with equality. )

--Vesselin

Vess, your solution assumes A is a set of integers.   Above the problem is for A real.
Which one does the problem intend?

I think you can prove a similar statement for real A if you consider a sort of universal
space for this problem, i.e. the pairs of integers  $(b,c)$ so that $bx+c$ is a
composition of linear functions from the given list.
Because of the disjointness any such expression as a composition must be unique.
Then form a zeta function such as $\Sigma 1/|b|^s$ over all such pairs, for any s where it converges (or has some pole --- the nonvanishing is the important thing) one gets
an inequality.   I will try to work out details when I have some time.  Note that this is not unlike
the zeta-function computation of fractal dimension for subsets of R...

Oh, sorry, I didn't see that Pascual asks for subsets of the reals; the IMO Shortlist problem (as it appears, for instance, on the [url=http://www.kalva.demon.co.uk/short/sh02.html]kalva[/url] site), concerns sets of positive integers.

In any case, the problem, as formulated by Pascual ($A \subset \mathbb{R}$ and $b_i,c_i \in \mathbb{Z}$) is easily reduced to the integral case $A \subset \mathbb{Z}$ by considering a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$. But what if the $b_i$ are not necessarily rational? And what if they're not necessarily finitely many?

[b]Conjecture. [/b] Let $A, B \subseteq \mathbb{R}$ be sets of reals, and let $\lambda : B \to \mathbb{R}$ be an arbitrary mapping. Assume that the sets $Ab+\lambda (b), b \in B$, are subsets of $A$ and are pairwise disjoint. Then,
\[
\sum_{b \in B} \frac{1}{|b|} \leq 1.
\]
In particular, $B$ is countable.

fleeting_guest, I'm looking forward to seeing your proof for the real case. Can it be extended to prove this conjecture?

--Vesselin


analytic geometry
geometry
pigeonhole principle
modular arithmetic
number theory
least common multiple
combinatorics unsolved

Let $ S$ be a set of all points of a plane whose coordinates are integers. Find the smallest positive integer $ k$ for which there exists a 60-element subset of set $ S$ with the following condition satisfied for any two elements  $ A,B$ of the subset there exists a point $ C$ contained in  $ S$ such that the area of triangle $ ABC$ is equal to k .

I think the best possible is $k=210$.

Lemma: For $A,B\in S$ such that the absolute coordinates of $\vec{AB}$ are $(p,q)\in\mathbb N_0\times \mathbb N_0$, the smallest area $\triangle ABC$ with $C\in S$ is $\frac 12\gcd(p,q)$ if $pq\ne 0$ and $\frac p2$ if $q=0$ (or $\frac q2$ if $p=0$). Furthermore, all other triangle areas are multiples of this.
Proof: This follows easily from $\triangle OAB=\frac 12 |a_1b_2-a_2b_1|$ where $A=(a_1,a_2)$ and $B=(b_1,b_2)$.

Now, among 60 points of $S$, by the pigeonhole principle, we can find for each $n\le\sqrt{60}$, i.e. $n=2,...,7$, two points $A,B$ such that $a_i\equiv b_i\pmod n$ for $i=1,2$. So $a_1-b_1\equiv a_2-b_2\equiv 0\pmod n$. By the lemma, the triangles containing these pairs have areas that are multiples of $\frac 22,...,\frac 72$. As $k$ is requested to be an integer, we get $k\ge lcm (3,2,5,7)=210$.

For the 60-element subset we can take an $8\times 8$ grid without the four corners. Actually, we can keep one of the corners, so have a 61-element subset. :D

nice problem,very beautiful solution
spanferkel