ARML email AoPSwiki Stanford college mathleague.org AMC

So the three biggest problems i've identified with the team selection process as we have implemented it this year are as follows:

Team selection occurs too late in the year. We won't know who is on our A and B teams until less than two weeks before ARML. This prevents us from really developing cohesive teams that know each others' strengths and weaknesses (as i observed while working with the ostensible A team last weekend)..

Team selection is a waste of practice time. We spend over half our time in practices taking team selection tests that, if administered earlier in the year, could free up all sorts of time for us to make better use of our practices..

The process appears to be too constraining. i've had a lot of people tell me they aren't able to make it to enough practices to get a valid score. While i recognize participating in ARML requires some level of commitment, i also understand that a lot of the students we attract are incredibly busy and that offering only four opportunities to accumulate scores might not be sufficient..

So with all that in mind i'd like to hear any suggestions you have for improving the system before i toss in a few of my own..

i would also be interested in hearing your suggestions for improving practices next year. If we can move team selection earlier in the year it will free up considerable time in our practices.

As a coach, my biggest priorities are providing educational opportunities and facilitating long-term team development, but as current competitors you are (rightly) selfish about seeing the team do the best it can in the current year. As such, i suspect several of you will have some good ideas for what we can do to ensure that our team lives up to its potential. Feel free to post your ideas here or, if you're worried about TJ's spies stealing our ideas, email me..  :)

one thing i want to point out
is that the indiv rounds used for the selection had difficulties all over the place
like
last year's arml score could also be counted right
and there was that mathleague test where 7s and 8s were A team and 5s or less were C team
its obvious that the two tests' difficulties are miles apart each other
i wasnt there at the mathleague test
so my last arml score got counted
but
my 5 on the arml was definitely an a team score but because of the mathleague test i got put on the c team the next week
that was dumb

i think it would be better if we just use normal arml indiv rounds instead of searching for other tests of similar format

I agree with junggi in that difficulty was really variable, (consider TRML as opposed to the mathleague thing) which made it unfair because the tests weren't weighted.
But I think a lot of people have the ARML books, and all ARML 2004-2007 are freely available online, so just administering previous ARML contests might not be fair as well.
Sidenote: According to [[Joining an ARML team]] on the AoPSWiki, Minnesota announces their A and B teams the night before ARML.

Maybe we could steal some other team's obscure TSTs or coaching material.
An idea:  Have some tests at the Stanford Math Circle on some weeks in February or March.  This will encourage people to come to the math circle in addition to getting testing done early.

Also, I'm not sure how important the Power Question is in the grand scheme of things, but maybe we should try putting proof problems for maybe the top 30 something students.

Yeah you're right, a lot of regions don't finalize their team selection until just before ARML - particularly in Texas, where i competed and where it is impossible to get the whole team together to practice.  But i think we could really benefit from having our teams chosen earlier, especially given how close together our students are and how easy it is to get these teams together for practice..

i agree that consistency was a problem this year with the material we had; i think that's an even bigger problem than difficulty level.  Basically i have yet to identify a solution that would simultaneously satisfy the needs of consistency, difficulty, and novelty.  mathleague.org pumps out a !@#%load of problems each year which takes care of novelty and consistency, but there has definitely been something lacking in the difficulty department.  of course, since i control mathleague.org i could just tell the testwriters to make the tests harder and maybe that would solve all our problems.  any thoughts?

i'd be interested in knowing what other teams use for TSTs; anyone know about this?  i know when i was on the Texas team they recycled questions, which i see as less than optimal.  That of course is also the problem in using old ARMLs, especially since they are so easily obtainable.  i actually used mathleague.org tests as TST when i coached Missouri and Kansas, but in those states we also didn't have more USAMO qualifiers than we could fit on an A team.  i've heard rumors that Chicago uses a set of tests but i don't know whether they are recycled.  i think some of the New York teams use NYSML results, but we don't have a state championship like that running in California (although it would not be that hard to get one set up)..

Chicago has a special tryout competition that's mandatory for anyone who wants to join the team.  They get everyone together, give them a bunch of individual-round-style tests, and as far as I know, those are the sole factor in team selection.  Minnesota uses exclusively old ARML and NYSML tests in practice, and how you do in practice is the main component in team selection.  They have a formula that they use, but it's very complex (it also takes into account team, power, and relay round performance), enough so that no one ever really knows their score, so they can't compare among each other.  And the coaches don't necessarily hold themselves to the formula.

Minnesota does in fact announce the teams the night before ARML.  However, they're able to get away with that because of a different practice system.  In Minnesota, they invite 30 people and 6-8 alternates to be on the team.  (Alternates get to go if and only if some of the 30 can't make it.)  They only have three practices, each of which runs from early morning to mid-afternoon on a Saturday.  Practices are pretty much mandatory, so you need a good excuse and permission from the coaches to miss practice.  Because of that, by the second practice, we have a good idea of who the Gold team might be, and they practice together for the second and third practices.  The temporary Gold team is chosen by the coaches prior to practice (and sometimes tweaked at lunchtime), so it usually corresponds pretty well to who will eventually be on Gold, minus two or three who aren't at practice.  Thus the teams that are announced on Friday night are only slight variations on teams that have been practicing together for most of the season.

That last point might be something we could try next year.  Instead of assigning teams based on who had the best individual scores at the start of practice, we could have the coaches get together beforehand and rank the students somehow (some combination of past performance and who we think will eventually make it on the A team, perhaps).  Then at the start of practice, see who's there and select an A team from that, possibly tweaking it if results from the day's first event warrant doing so.

I'd also suggest doing a tryout competition like Chicago--I think that's the best way to pick the team if you want a strictly numbers-based method.  However, I think given how willing SFBA people seem to be to ditch the team, I worry that a lot of people would decide they have better things to do on the day of the tryout and give up the entire ARML season because of that.

By the way, it's nearly unheard of in Minnesota for anyone to decline their invitation to ARML without a good reason.  One year, we had a student with graduation Friday night; he went to that, then had his dad drive him to Iowa overnight and meet the team at breakfast the day of the competition.  I chalk this up partly to the fact that an ARML invitation is a huge honor in Minnesota--it marks you as one of the best of the best in the state--and partly to the Minnesota team having a much more cohesive social group built around it than SFBA has.  I'm hopefully going to work on the latter next year, most likely in the form of SUMO-sponsored team socials before or after practices.

Oh, and Tim, if you want harder tests, I can do that.  We could also easily cobble together a TST by picking the hardest problems off of each Mathleague round, and maybe a few good ones that didn't make it onto the SMT tests if we're on top of things enough to have extra problems next year.

The SoCal system: no team selection test. The team selection is based on accumulated knowledge, including both contests like the AMC that the students are doing anyway and the various individual parts of practices. Previous year information is included. There are formulas, but the final decision is subjective.
The process doesn't determine whether a student is invited at all, just which team. SoCal has only rarely turned anyone away who wanted to come. Teams are announced shortly after students confirm they're coming, so there's time for a couple practices with actual teams.

With the high rate of returning students, most of the A team is known far in advance, and usually practices together as a team (with a few B-teamers to cover those missing a practice) during the year.

Washington: I don't know the details, but teams will be announced just before ARML. I don't think there's an organized TST. The practices I've been to have had about one team worth of students each.

North Carolina has no TST and announces the teams the day before the competition. It has not been an issue, as we placed 7, 1, 4 the last three years.

I believe that Chicago ARML has a much more complicated system. There is a mandatory tryout that is about 20 problems. The score on this is a large factor on deciding the team. Each practice, we do like team/relay or power, then an individual. All these scores are used to determine the teams. I'm not sure how the scores are weighted. The practices are mandatory... unless you have a very good excuse. There is also homework due at each practice that is factored it. Also there are other things to get extra points like offering nice solutions to problems or making corrections to problems, discussing the lectures with the teachers, and making problems. I don't really know how all the stuff is factored in. I feel there is a formula, but then the coaches use their discretion to make a final decision. Nobody really knows how it works exactly...It could be much more simple than this.

my state team announces teams on the bus on the way to ARML, we have no selection tests, it's based on several local and national competitions.

[quote="Sly Si"]That last point might be something we could try next year. Instead of assigning teams based on who had the best individual scores at the start of practice, we could have the coaches get together beforehand and rank the students somehow (some combination of past performance and who we think will eventually make it on the A team, perhaps). Then at the start of practice, see who's there and select an A team from that, possibly tweaking it if results from the day's first event warrant doing so.
[/quote]
I like that idea. It saves the time of waiting for results at each practice. In order to make that work, we may need to have a sign-up sheet or roll-call at the beginning of each practice (whichever is less time-consuming).

The team selection is a sorting process:
-Initially, teams can be assigned based upon available data (AMC/AIME/USAMO/last year's ARML, etc.);
-after each practice, some people bubble up and some inevitably drop down;
-this process continues until ready for the event day.

One more point: the previous year's ARML individual score should be weighted more in team selection.

As an anti-example, if we placed the 12th grader in B team, SFBA B team would be the national 1st this year!
(she got 6 in individual round and contributed most to the 35 points in PQ for the F team)

[selfish] I don't like any of these suggestions cause they would all significantly decrease my chances of ever making it to A team. I missed the cutoff for A by very little I think, but I completely died on individual this year and my USAMO index sucks.[/selfish]

Okay, getting over that, I think first of all we need more people to be responsible for coming. Committment to the team should be a larger factor in deciding who gets in. Our current tryout process isn't too bad, it's kind of that we start a bit late and we place too much emphasis on the scoring of different random competitions that aren't scaled for any difficulty. Certain people could have bad days and stuff, right? Er, well, like me.

Maybe if we had one or two longer, fairly difficult tests administered earlier in the year, we could choose teams with them. Then, we should keep the teams together, teach them to work together and stuff before ARML.

I don't know, maybe some teams can do it somehow, but we don't have to choose our teams the day before the competition. Can't hurt to prepare early...

what is all this "commitment to the team" business
define "commitment to the team"
and how it will help our team in securing first place

Actually, I kind of agree with Austin. I missed plenty of academic things this year, including a Quiz Bowl Final, so that I could make each and every practice. I don't think I missed an ARML practice all year, and I'm glad I went to every one. I think that dedication and teamwork, while not the primary factors in team selection, should certainly serve as tipping points in your favor.

Say person A scores, on average, a 6 on the individual round in practice. That's a pretty solid score, and the coaches see that person A also had an 8 AIME score this year, which is decent. It's not quite enough to establish person A as a USAMO participant, though. Person A has shown up to each and every practice.

Now look at person B. Person B has a 7 average on the individual round. Person B also has 11 on the AIME, and scores a 14 on the USAMO. Person B comes to exactly three practices (for the individual round scores to be registered).

Normally, under what I believe is the current system, person A would undoubtably make the B team (and be a pretty solid player, due to the commitment) and person B would most likely make the A team, on the basis of their scores.

I think that these performances are fairly comparable, however, and that person A should be given a chance to make the A team because the attendance record adds tipping points in their favor.

um no
i dont see anything wrong with person b ditching some practices and still making the a team
a 7 is a [i]great[/i] score, and given his performance on the amc series, the arml practices probably wont be too helpful for him ie a waste of time
and uh
person a
why would he make the b team
a 6 on an [i]arml[/i] individual round is definitely an a team score

the thing is
how would this
"commitment to the team"
help in improving our scores

btw whats so great about quiz bowl
from what i know about it, it all seems pointless and a waste of time
no offense but
jw

Commitment to the team is a viable strategy for primarily one reason (and this may be wrong, but I think it's a matter of opinion): the more committed a member is to winning, the harder they will work towards that goal. Hard work and practice usually reveal good results.

ok now lets be honest here
how many of you worked hard for arml specifically between aime/usamo and arml
or did you work hard on math regardless of arml because you either like math or for the amc series
i dont think arml is a serious enough competition to motivate people to work hard
i could be wrong, but at least it is for me

and the arguments i put out that went unrefuted
still stand
better rebut them or no commitment points for you
oh no

okay junggi, you don't have to be an ahole here

I think that the commitment idea definitely contributes something, because oftentimes team members who don't know each other well work very poorly on the team round. I can vividly remember last year's power round, where a whole bunch of the 8th/9th graders who did fairly well individually goof off and do nothing for the last half hour of the test (which probably cost us a dozen points). Someone who's dedicated to the team and wants to do well would most likely put in the entire hour of effort. Also, I don't remember exactly, but I seem to recall that the B team scored very very close to the A team on the power round.

I remember that as well. People started cubing and stuff near the end of the power round, it was pathetic. One should at least should volunteer to check written proofs if they can't make any progress with their problem.
Although I do agree that commitment does play a role in performance at the actual contest, I don't think it should be a main factor in determining the teams. Certainly it should be a factor considered, but just not a vital one, possibly only used when there are multiple people of a comparable level vying for one spot.

hey im only saying this because this is going to hurt my chances of getting on the a team, since going to practices are practically useless in improving my performances on either indivs or teams
and im fairly sure i only got on a team this year because i made mop (scorewise i probably fell in the b team range but i guess that is more of an issue of the actual placement tests but whatever)
well
last year
i dont think thats an issue in commitment
it was bound to happen with
1) a 8/9th grade dominated team
2) an abnormally hard power round
its more an argument for not making a team thats mainly freshmen and middle schoolers, a problem we're not gonna have next year
edit:
oh oh another thing
that "hey last year was a disaster" argument applies [i]only if[/i] the current team actually has a commitment issue
but im sure anyone on the a team this year can tell you, we had no problem working and working for the entire team/power

wait
anyway
lets get this clear first
[i]what exactly is commitment defined as?[/i]
is it measured in terms of number of practices one goes to?
if so, we need a better definition
im sure there are plenty of examples of people who couldnt make it to the practices because of schedule conflicts
one i can think of off the top of my head would be serialkiller
im sure that if they let him take indiv placement tests later on, he would have been one of the more committed team members

I think a team selection test earlier in the year would be a good idea.  The only problem is if people couldn't make it on the test date. You  could have an alternate test date, but if it's the same test a friend could leak answers, and if it's a different test the difficulty may vary (and you can't scale well with a small sample size).

But, ignoring that issue (which we can't ignore, but whatever), we could have a TST for tentative team placement, and then if someone's performance during practice [i]drastically[/i] changes, switch it up.

The AIME I and II successfully ignores that issue, right?

Just to confuse the issue: some data on the correlation of AIME to ARML. This data is from SoCal rather than SFBA, but your data might not be all that different. For the following table, the second column gives the average ARML individual score for people whose AIME score is that given in the first column:
\[ \begin{tabular}{cc}AIME&ARML\\ 11&5.7\\ 10&4.5\\ 9&6.2\\ 8&3.8\\ 7&4.5\\ 6&3.4\\ 5&3.9\\ 4&3.2\\ none&2.2\end{tabular}\]
In case you're wondering what that looks like as a correlation, if we leave off those who didn't take the AIME and consider the others, the correlation coefficient between ARML and AIME was .42. That's a positive correlation, but ...

My point in response to Alumen was, if we make it known ahead of time when the  beastly SFBA TSTs are, there is only a minute chance that anyone will not be able ot make both of the dates, just as with the AIME I and II.  In addition, as we are nice, friendly, and flexible, we can individually work out scheduling issues with those few people, and perhaps arrange to have it on a different date for them.  One issue, however, is that people might not "get" the importance of the TSTs and decide to not come.  How might this be solved?

Well, junggi, I realize you don't think much of the commitment thing. I have no clue what that commitment means, but I meant coming to every practice unless you have a really good reason not to. Also, be active, and don't slack during practice. It's not that hard of a concept. What's the big fuss?

I went to a martial arts promotion test that ran until 12-3 on a Sunday. I did what I could until 1:30, ran out, had my mom drive me like crazy from SJ to Stanford so I could make the damn practice. Ended up bombing the individual practice, which was like the primary reason I didn't make A team. So? I still tried to get there and participate.

If you don't think arml's that important, and you find it boring, just do whatever you want. Skip practice or whatever. I just kinda disagree with your priorities. Of course I didn't work hard for arml specifically. I don't work hard for any math competition. If you don't feel any motivation, though, I don't get why you'd try out for arml.

[quote="Alumen"][quote="archimedes1"]beastly SFBA TSTs[/quote]
What kind of 'beastly'?[/quote]
The challenging kind...  Hopefully like the TRML rounds, in which the average score was around 3ish.  This allows lots of room for potential A-team members to shine without having to get a perfect 8 on the nathleague rounds.

Hmm well it is obvious that Taylor does not want to commit himself to the ARML team. I'd say that whether or not you put Taylor on the A team is a matter of your goals. If you want to have the best possible team, of course you will put him on the team. However, if your primary goal is to encourage interest in mathematics, then you will probably not put him in the A team. For those students who attend practice regularly, it is important for them to feel as if their effort is being rewarded.

what
im not saying that i dont want to commit myself to the arml team
im just saying that going to the practices wont actually help me improve my scores on the arml round
nevertheless, at the actual arml contest, i am going to commit myself to the team and do the best i can on both the individuals and the team/power rounds
and i feel that i did that this year
im just not seeing how not going to practice means not committed to the team

[quote="AIME15"]I was gonna do ARML this year, but I had a piano competition...

What team do you think I woulda made:

AMC 10: PHAIL - 114
AMC 12: PHAIL - 97.5
AIME: yay - 6

for stanford arml team[/quote]
i didnt do much better than that and I got S team (thats one of the fail ones  :D )

dang...

maybe the aops classes im taking this year will help out

MUST NOT GET PHAIL TEAM MUST GET B TEAM (a team yeah right)

A Team was beast last year, hopefully we'll retain our black mopper for once!

Yeah...

doens't evan go to berkely math circle?

WAIT... where are the qualifications? I want to join, except idk if its too late...


function
inequalities
inequalities proposed

Let $ f$ be a convex function on an interval $ I$ and $ a,b,c,d\in I$. Put $ s \equal{} a \plus{} b \plus{} c \plus{} d$ and prove
$ \sum_{6\ terms} f\left(\frac {a \plus{} b}2\right) \plus{} 6 f\left(\frac {s }4\right)\ge 3\sum_{cyc} f\left(\frac {s \minus{} a }3\right)$ :) :)

I don't this one is right.
Let $a >b>c>d$,we can find $b+c>a+d,\frac{s-c}> \frac{s}4$
let $f(x)=0,x\le \frac{s}4$
and $f(x)=k(x- \frac{s}4)$
when k is large enough,you will find the inequality is wrong.
I just can prove:
$\sum_{6\ terms} f\left(\frac{a+b}2\right)+2f\left(\frac{s }4\right)\ge 2\sum_{cyc} f\left(\frac{s-a }3\right)$

Sorry, but I cannot follow you. Take $(a,b,c,d)=(18,12,-6,-24)$. Then $s=0$ and your conditions are fulfilled (I guess you mean $\frac{s-c}3> \frac{s}4$). The inequality reduces to $f(15)+f(6)+f(3)\ge 2[f(8)+f(2)]$ as only the positive arguments contribute to the sums. But as far as I see, this is true. Maybe you have only considered the "leading" terms on each side?
I am quite sure my proof is waterproof ;)

[quote="spanferkel"] The inequality reduces to $f(15)+f(6)+f(3)\ge 2[f(8)+f(2)]$ ?
I am quite sure my proof is waterproof ;)[/quote]
Sorry,the inequality must reduces to $f(15)+f(6)+f(3)\ge 3[f(8)+f(2)]$??
then it is wrong. ;)

[quote="spanferkel"]:blush: Oh seems like you're right. You've found a leak.It's wrong :blush:
But note that I was puzzled by your statement[quote="zhaobin"]
when k is large enough,you will find the inequality is wrong.
[/quote]Replacing $f()$ by $kf()$ doesn't affect the ineq as $k$ factors out[/quote]
Oh,sorry,positive k is enough. :)

[quote="zhaobin"]I just can prove:
$ \sum_{6\ terms} f\left(\frac {a \plus{} b}2\right) \plus{} 2f\left(\frac {s }4\right)\ge 2\sum_{cyc} f\left(\frac {s \minus{} a }3\right)$[/quote]

Sorry, Zhaobin, but can you really prove this?
I set $ f\left(x\right)\equal{}\left|x\right|$, $ a\equal{}1$, $ b\equal{}1$, $ c\equal{}\minus{}1$, $ d\equal{}\minus{}1$ and obtain a wrong inequality...

  darij


What is the smallest positive integer that consists base 10 of each of the ten digits, each used exactly once, and is divisible by each of the digits $2$ through $9$?

No, that's not divisibly by $ 5$... it has to end in 0 for sure, since it's divisible by both 5 and 2.

That is one... proof of minimality/way to come up with that? ;)

1234759680.  :lol:
Solution in chinese:
http://www.mathoe.com/dispbbs.asp?boardID=117&ID=16330&page=1

Can someone translate the solution please?

Suppose this number is $10N$, obviously we only need $7 \times 4\left| N \right.$,where $N$ used $1,2,3,4,5,6,7,8,9$ exactly once.
If the left six digits is $123456$, $N\not  \equiv 0(\bmod 4)$,  contradiction;
If the left five digits is $12345$, $4|N$ force the right two digits must be $68,76,96$, and so $N\not  \equiv 0(\bmod 7)$, contradiction;
If the left four digits is $1234$, $4|N$ force the right two digits must be $56,68,76,96$, suppose the right five digits is $M$, since $123400000\equiv 3(\bmod 7)$, so $M \equiv 4(\bmod 7)$.
a)If $M = \overline {abc68} $, $100\overline {abc}  + 68 \equiv 4(\bmod 7) \Rightarrow \overline {abc}  \equiv 3(\bmod 7)$, so $\overline {abc}  = 759$,$N = 123475968$;
b)If $M = \overline {abc56} $, $100\overline {abc}  + 56 \equiv 4(\bmod 7) \Rightarrow \overline {abc}  \equiv 2(\bmod 7)$, no solution;
c) If $M = \overline {abc76} $, $100\overline {abc}  + 76 \equiv 4(\bmod 7) \Rightarrow \overline {abc}  \equiv 6(\bmod 7)$, so $\overline {abc}  = 895,958$, but $123495876 > 123489576 > 123475968$;
d) If $M = \overline {abc96} $, $100\overline {abc}  + 96 \equiv 4(\bmod 7) \Rightarrow \overline {abc}  \equiv 3(\bmod 7)$, so $\overline {abc}  = 857$, but $123485796 > 123475968$.
So   $10N = 1234759680$ is the best.


[color=darkblue]Let two funtions: $ f: R \to R$, $ g: R \to R$ and $ \begin {cases}f(3x \minus{} 1) \plus{} g(6x \minus{} 1) \equal{} 3x \\
f(x \plus{} 1) \plus{} x^2g(2x \plus{} 3) \equal{} 2x^2 \plus{} x \end {cases}$
Find fomula of $ f$ and $ g$ ?[/color]

[quote="christ"]I think this variable changing may be useful :
u = 3x-1    ,  v= x+1
f(u) + g(u+1) = u+1
f(v)+(v-1)g(2v+1)= 2(v-1)^2 + v-1

u and v are 1 degree expressions so they can be  transformed to each other ,then we can consider only u .

f(u) + g(u+1) = u+1
f(u)+(u-1)g(2u+1)= 2(u-1)^2 + u-1

Solving these equations may lead us to desired formulas .[/quote]

Thank for your post. :maybe:


This has been irritating me for some time and I'm hoping someone can give me a hand.

I work at a manufacturing company where we must add an additional 6% onto our invoices.  If our customers pay in a timely manner, they get to take that 6% off as a "discount".

Sounds simple right?  It's been a long time since I've been in any type of school, but I'm pretty sure if I wanted to add 6% to any given number, I should multiply it by 1.06.  And voila!

My employer seems to think that the correct way to do this is to divide by 0.94 to get the total.  While the difference between the two approaches are virtually negligible at smaller quantities, it becomes substantial when you start billing customers for over $10,000 in merchandise. And the calls come flooding in with Customers thinking we're trying to rip them off.
For Example:
$10,000*1.06= $10,600 vs.$10,000/.94= $10,638
I know my way is right, but I can't seem to find a way to logically explain why it's right. Can someone tell me why dividing by 0.94 will not work so I can pass this along? The best I can come up with is my Jr High word problem solving skills: What (variable) is (equals) 6%(.06) of (*)
$10,000.  But I'm afraid that's not going to convince anyone.....

Ps - Now you know why big business in America is in trouble......

It's not hard to show that you're right.
If $ m$ is the value you want to add 6% to, you're finding $ m\plus{}\frac{6}{100}m\equal{}m\plus{}.06m\equal{}m(1\plus{}.06)\equal{}1.06m$

You can disprove your employer's way like this:
$ \frac{m}{.94}\equal{}\frac{m}{\frac{94}{100}}\equal{}m\cdot\frac{100}{94}\approx 1.064m$

And so you are actually adding 6.4%, which is, as you said, negligible with small amounts but significant with large amounts of money.

You can also consider the example of 100 dollars (because 6% of 100 is pretty obviously 6)

Let $ A$ be the initial amount, and $ p$ be the percent, as a decimal, by which you want to increase $ A$.

Doing it the way your employer says to do it, the final amount would be $ \frac{A}{1\minus{}p}$. Finding the percentage the other way yields $ A\cdot(1\plus{}p)$. This can also be expressed as $ \frac{A}{\frac{1}{1\plus{}p}}$, to make it of the same form as the first result.

Assuming that the methods give the same result, we can say $ 1\minus{}p\equal{}\frac{1}{1\plus{}p}$ for all values of $ p$. However, the equation can be simplified to $ p\equal{}0$, which means $ 0\%$ is the only percentage for which the two methods produce the same final amount. Thus, by contradiction (the original equation had the condition that it worked for all values of $ p$), the two methods do not produce the same result unless $ p\equal{}0$.


Let x be 3a+6b+9c, y be 5a+2b+3c, z be 2a+8b+6c ==>
2(a+b)=(-x+y+z)/2
6(b+c)=(x-y+z)/2
3(c+a)=(x+y-z)/2

So initial inequality is
y/x+z/x+x/y+z/y+x/z+x/y \geq 6 - true!!!

Here is another way to solve it:

Denote:
x=b+c;
y=c+a;
z=a+b; and we get that:
a=1/2(-x+y+z) etc
and the ineq is equivalent to:
6x/(3y+2z)+3y/(2z+6x)+2z/(6x+3y)>=3/2
not denote 6x=X, 3y=Y and 2z=Z and you get a very well known inequality.

cheers! :D :D

This inequality is Nesbitt's inequality :

If a, b, c > 0 then a/(b + c) + b/(c + a) + c/(a + b) >= 3/2.

There are lots of generalisations of this inequality.


probability
algebra
polynomial
modular arithmetic
abstract algebra
Galois Theory
number theory proposed

[quote="vess"]Show that the primes $p$ which divide some number of the form $2^{n}+1$ have natural density $17/24$ in the set of primes.[/quote]
Your condition equavalent, that minimal period $T_{p}: \ p|2^{n}-1\Longrightarrow T_{p}|n$ is even. We call these primes - good primes.
If $p=\pm 3(mod \ 8)$, then $(\frac{2}{p})=-1$, therefore $T_{p}$ is even.
If $p=-1(mod \ 8)$, then $(\frac{2}{p})=1$, therefore $T_{p}$ is odd.
It give, that density of good primes d: $\frac{1}{2}\le d\le \frac{3}{4}$.
But I don't know how calculate densyty good primes for $p\equiv 1(mod \ 8)$.
Probabylity method give $p=1+2^{k}m$, m-odd had $\frac{1}{2^{k}}$ and probabylity, that exist x, suth that $2\equiv x^{2^{k}}(mod \ p)$ is $\frac{1}{2^{k-1}}$. Therefore probability density is
\[\frac{3}{4}-\sum_{k=3}^{\infty}2^{-2k+1}=\frac{3}{4}-\frac{1}{24}=\frac{17}{24}. \]
My calculations: there are $N_{1}=1 439 970$ primes $p\equiv 1(mod \ 8), p<10^{8}$ and for $N_{0}=239953$ of them $T_{p}$ is odd.
Probability $\frac{N_{0}}{N_{1}}=\frac{1}{6}.$
Vess can you prove it?

I apologize for replying only now -- I was away for some time...

Here is a reference where, alongside with the proof of the mentioned result, you will find a proof that the prime divisors of the Lucas numbers have natural density $2/3$ in the set of primes:

J. Lagarias,   Pacific Journal of Mathematics, Vol. 118, No. 2, 1985.

--Vesselin

All one needs to do is find a rigorous justification for Rust's "probabilistic" argument. Such justification follows from Chebotarev's Density Theorem. I made the same computations as Rust, but kept thinking that $\sum_{k\ge 3}2^{-2k+1}$ equals $\frac 1{48}$, and I spent like an hour trying to figure out what had gone wrong :). I'll post the proof, but minus the application of Chebotarev, it's basically outlined above.

We want to show that the density of the set $A$ of primes which divide no numbers of the form $2^{n}+1$ is $\frac 7{24}$. If $2^{k}\|p-1,\ k\ge 1$ (i.e. $2^{k}$ divides $p-1$, but $2^{k+1}$ does not), then $p$ belongs to $A$ iff the polynomial $x^{2^{k}}-2$ has a root modulo $p$. Whenever $2^{k}|p-1$, this latter condition is equivalent to the prime $p$ splitting completely in the field $L_{k}=\mathbb Q(\zeta_{2^{k}},\sqrt[2^{k}]2)$, where $\zeta_{u}$ is a primitive $u$'th root of unity. The density of the set of primes which split completely in $L_{k}$ is, by Chebotarev, $\frac 1{[L_{k}: \mathbb Q]}$. Similarly, the subset of primes $p$ with $2^{k+1}|p-1$ and $x^{2^{k}}-2$ splitting $\pmod p$ has density $\frac 1{[L_{k}(\zeta_{2^{k+1}}): \mathbb Q]}$. It follows that for all $k\ge 1$, the density of the set of primes $p\in A$ such that $2^{k}\|p-1$ is $d_{k}=\frac 1{[L_{k}: \mathbb Q]}-\frac 1{[L_{k}(\zeta_{2^{k+1}}): \mathbb Q]}$. All we need to do now is add up $d_{k}$ for $k\ge 1$ and hope that we get $\frac 7{24}$ :).

The computation of $d_{k}$ isn't difficult. We get $d_{1}=\frac{1}{4},d_{2}=0$, and $d_{k}=2^{-2k+1},\ \forall k\ge 3$. That little hiccup for small values of $k$ happens because $\sqrt 2$ belongs to the cyclotomic field $\mathbb Q(\zeta_{8})$, but then things stabilize because no $2^{k}$'th roots of $2$ belong to any cyclotomic fields for $k\ge 2$ ($x^{4}-2$ already has non-abelian Galois group over $\mathbb Q$). I'll leave the details to the reader :D.

I didn't try to prove the statement about Lucas numbers, but I expect the argument should go along the same lines.

[quote="grobber"]All one needs to do is find a rigorous justification for Rust's "probabilistic" argument. Such justification follows from Chebotarev's Density Theorem. I made the same computations as Rust, but kept thinking that $\sum_{k\ge 3}2^{-2k+1}$ equals $\frac 1{48}$, and I spent like an hour trying to figure out what had gone wrong :). I'll post the proof, but minus the application of Chebotarev, it's basically outlined above.
[/quote]
If $p=1(mod \ 8)$, then $(\frac{2}{p})=1$, therefore for $p=1+m*2^{k}=1+2^{k}(mod \ 2^{k+1}$ probability $T_{p}$ odd is equal to  $2^{1-k}$. For $p=9(mod \ 16)$ densyty equal (without probability) to $2^{-2*3+1}=\frac{1}{32}$. It give
$d=\frac{3}{4}-\sum_{k\ge 3}2^{1-2k}=\frac{3}{4}-\frac{1}{32}(1+\frac{1}{4}+\frac{1}{4^{2}}+...)=\frac{3}{4}-\frac{1}{32(1-1/4)}=\frac{3}{4}-\frac{1}{24}.$

[quote="grobber"]I didn't try to prove the statement about Lucas numbers, but I expect the argument should go along the same lines.[/quote]
I didn't try this one either, but at least your argument readily settles it for the numbers of the form $k^{n}+1,$ where $k \geq 3$ is squarefree.
For example, if you take $k = 3,$ then you get that the desired density is $1-\frac{1}{4}-\frac 1{16}-\ldots = \frac{2}{3}.$


[b]$ax^3 + bx^2 -x-1$ is divisible by $x^2+2x-1$
Then what are a and b?[/b]

[hide]You're looking for a binomial that retains the $-x$ and $-1$ when multiplied by $x^2+2x-1$. It is easy to see that the binomial should be in the form $(ax+1)$ since the $-1$ is kept in both the divisor and the dividend. Now to get $-x$ one needs $-ax+2x=-x$, therefore $a=3$.

$(3x+1)(x^2+2x-1)=3x^3+7x^3-x-1$

$\boxed{a=3, b=7}$[/hide]

[quote="willydavidjr"]Now i see. But if we encounter this kind of problem, how do we approach it?[/quote]

Work out the polynomial division using the variables $a$ and $b$. In this particular case, the final step in the process would be to divide $(b-2a)x^2+(a-1)x-1$ by $x^2+2x-1$. Since the constants match, we have $a-1=2$ and $b-2a=1$, so $a=3$, $b=7$.

[quote="willydavidjr"]Adunakhor how did you get b-2a and a-1? And where is the equation $ax^3+bx^2-x-1$?[/quote]

Polynomial division: I divided $ax^3+bx^2-x-1$ by $x^2+2x-1$.

Another  method whch i can thnk of is finding the roots of x^2+2x-1 as (-1+sqrt2) and (-1-sqrt2)..den finding the values of a and b using two equations..But  its a longer method..


A space station consists of two living modules attached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains $ N$ astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of strength $ g$. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength $ a$. What is the ratio $ a/g$ in terms of $ N$?

let r b radius of station, $ _1$ denote before astronauts move, $ _2$ after move.

$ g \equal{} \omega_1^2 r$, $ a \equal{} \omega_2^2 r$, $ a/g \equal{} (\omega_2/\omega_1)^2$.

Angular momentum is conserved, $ I_1\omega_1 \equal{} I_2\omega_2$. $ I_1 \equal{} 2Nr^2$, $ I_2 \equal{} 2(N\minus{}1)r^2$. Thus, $ (\omega_2/\omega_1)^2 \equal{} (I_1/I_2)^2 \equal{} (N/(N\minus{}1))^2$.

[quote="xsk13"]Could you post your solution?  :)
and to avoid saying this in most of your posts, could you post your solutions to those questions you asked? i'm curious, and i couldn't figure this one out either. thanks![/quote]
It's from a multiple choice problem and the answer is 'none of above'.


real analysis
calculus
calculus computations

Let $ S \subseteq X$ where $ X$ is a metric space. Prove: $ S$ is open $ \Longleftrightarrow$ $ S \cap \partial(S) \equal{} \emptyset$.

For the $ \Leftarrow$ direction: Proceed by contraposition?  Suppose $ S$ is not open. Then $ S$ is closed. And $ \partial(S) \subseteq S$. Thus $ S \cap \partial(S) \neq \emptyset$?

The empty set is both open and closed.
Furthermore, $ (0,1]$ in $ %Error. "field" is a bad command.
{R}$ is not open nor closed.

It's simply not true that a set that is not open is closed or that a set that is not closed is open.  As Rudin puts it, [i]sets are not doors![/i]

To continue the proof by contradiction, "suppose $ S$ is not open.  Then it has a limit point $ p$..."

Here's another way to phrase this. What is $ \partial S,$ anyway? The condition is that $ x\in\partial S$ if and only if every neighborhood of $ x$ contains at least one point in $ S$ and at least one point not in $ S.$

On the other hand, $ S$ is open if and only if for every point of $ S,$ some neighborhood of $ x$ is completely contained in $ S.$ But if some neighborhood of $ x$ is completely contained in $ S,$ then that neighborhood doesn't contain any points not in $ S,$ which by the definition I gave above shows that $ x\not\in\partial S.$

Here's a question: Suppose that $ \partial S\cap S\ne\emptyset$ but $ \partial S$ is not contained in $ S.$ (There's an least one point of $ \partial S$ that is in $ S$ and there's at least one point of $ \partial S$ that is not in $ S.$) What does that tell you about whether $ S$ is open or closed?


geometry
linear algebra
matrix
probability

Ok, this is going to be a RPG-style game, modded by me, gogators! It will also be heavily based on AQW and Aion.

The Story: There is a Road through the world of Asmodia. You will mostly be traveling on the Road, but there will be areas that branch off, and you will always return back to the Road. There will be quests and items along the way. All weapons and items will be derived from the aqwwiki.wikidot.com/
Each attack's power will be determined by random.org, as will the amount of gold a monster drops.
You automatically start out with 10,000 Gold.

There are 5 classes, with 4 skills per class. Each skill requires MP. MP for each character varies, as does health.

Templar-Specialize in attacking and defending, and are not very good with spells and pets. 700HP, 500MP
[hide="Skills:"]1. Strong Attack: Automatically deals 150% damage.-50MP
2. Shield Wall: Decreases the strength of the enemy's attack by 35%.-25MP
3. Rage: Increase attacks by 50% for 4 attacks.-75MP
4. Disable: Does 25% base damage and stuns the enemy.-45MP[/hide]
Berserker-All of their efforts are focused into dealing as much damage as possible. They are very strong, but not very intelligent. 800HP, 400MP
[hide="Skills:"]1. Crosscut: Attack twice.-80MP
2. Fireblade: Each attack does 20 extra damage, ends after either you or your opponent dies.-150MP
3. Call on Tiger: Brings a Tiger(10-20 dmg) to help you in battle.-120MP
4. KO: Has a 5% chance of killing the enemy, does 175% dmg if the enemy does not die.-250MP   [/hide]
Sorcerer-Specialize in casting spells, sorcerers tend to dislike melee attacks. 400HP, 800MP
[hide="Skills:"]1. Lightning Bolt: Does 150% dmg and has a chance to stun the enemy.-75MP
2. Tsunami: (Unblockable) Does 80 dmg and weakens the enemie's attacks for 5 turns.-150MP
3. Nuclear Firestorm: Does two attacks of 50-70 dmg.-200MP
4. Last Resort: Has a 50% chance of missing, but if this attack connects, it will instantly kill the enemy.-800MP[/hide]
Paladin-Warriors of light, Paladins specialize in attacking and healing. 600HP, 600MP
[hide="Skills:"]1. Holy Might: Does 75% dmg and blinds the enemy for 4 turns.-75MP
2. Call on Archers: Does 2 hits of 100% dmg and skips the enemy's turn.-200MP
3. Healing Light: Heals yourself for 25% of your total HP.-100MP
4. Godly Rage: Heals you for 20HP and deals 150% dmg.-150MP [/hide]
Assassin-Silent but deadly, assassins are masters of killing quietly and quickly. 400HP, 700MP
Note:Assassins have less HP, but their attacks are considerably stronger.
[hide="Skills:"]1. Silent Strike: Attack twice, and poison the enemy(Each attack does 15 more damage). -150MP
2. Matrix Moves: Dodge the next 5 attacks used by your opponent.-75MP
3. Throw Ninja Stars: Throw 2 ninja stars, which each deal 75% dmg. The bonus is that each one comes back, making 4 hits of 75% dmg.-200MP
4. Throat Slash: Find the weakest point in your enemy's defenses, dealing 1500% dmg(That is one extremely nerfed attack!).-500 MP
[/hide]
To sign up, post which class, name, and what you wish to buy from the [hide="shop"][url]http://aqwwiki.wikidot.com/yulgar-s-shop[/url][/hide]
In the shop, ignore all damage ranges.
Signups end Monday, around 6 spaces.

/join

name-Andrew

Don't know what class yet, want to look at paladin and assassin

doesn't buy anything

join.
I play aqw, but my inv. is filled with collectibles and only a few useful items...

Anyway......For some reason I always start as mage. I will be sorcerer (I have rank 10 mage)
My name is also Andrew...

I shall not buy anything yet ,but download all my pets over(can i do that?) :)

I didn't really need your real name, just what you wish to be called ingame.
If you don't purchase anything this is what you look like: [url=http://aqwwiki.wikidot.com/warrior]http://aqwwiki.wikidot.com/warrior[/url]
@Chompy
It would kinda be unfair if you had pets and everyone else didn't.
However, you can get one pet over.(Completely contradicting my last statement)
Can I have you char page?

Your wep: Default Sword-This sword is used by the Knights of Default
Dmg:10-17
Armor: Default Armor-This armor is used by the knights of default. +1 to strength

Will post Paladin and Assasin asap.

I also play AQW....

Can I just use all my weapons and classes from there?

My user name is [b]pitike314[/b] for AQW.

yay!
[url=http://www.aq.com/aw-character.asp?id=BluePerson]blueperson[/url]

hmmm..............one pet.......phoenix or armored frost or armored red or pumpkin or grenwog or chaos thing...
I think ill have armored frost dragon. yay! pet! *sigh* im nonmember...

huh. Ill buy genesis sword and blue scarf and blacksmith apprentice - always nice for a disguise

char page:http://www.aq.com/aw-character.asp?id=qwertythecucumber

No more bringing items from AQW into this game. Not all aspects of AQW relate to this game, and vice versa.
@Chompy:
Armored Frost Dragon does 7-10 dmg.

All beginning monsters will have around 500HP, as the whole team shall mob the monster.

Signups reduced to 4-6 spaces, and I will try to post updates everyday two days or so.
This game is going to move fast, so try to keep up.
 :lol:
pew pew

Edit: If you can, try to keep track of all your gold for now. Level cap is 20, you need 50 exp for lvl 2, each monster drops around 20 exp, you gain more exp from quests.
I will make a chart with all exp and info.

Wait, can we play if we haven't played AQW? If we can, I'll join but I'd like to see the assassin before choosing stuff...

I want to play.

I will be a Sorcerer

Call me spaceguy.

You can play even if you don't have an AQW account.

Signups so far:
1. andrewjjiang97-still choosing class
2. Chompy-sorcerer
3. cyberspace-? I think...
4. spaceguy524-sorcerer

Current stats:
1. andrewjjiang97: 0 exp, x HP, x MP, 10000gold, Items: Default armor, Default sword(10-17)
2. Chompy: 0 exp, 400HP, 800MP, 1500gold, Items: Blacksmith's Apprentice, Blue Scarf, Genesis Sword(20-23), Armored Frost Dragon(5-7)
3. spaceguy524: 0 exp, 400HP, 800MP, 10000gold, Items: Default armor, Default sword(10-17)

I'll be an assassin.  I'll buy a dark scarf and dexterous sword.

I would like to be a Paladin. Um, I have never played, and I dont know the instructions, I cant get on that site for some reason, so please tell me the rules on this forum. I would like to buy a Genesis sword, the Phoenix, and the dark scarf, please.

[quote="gogators"]
Current stats:
1. andrewjjiang97: 0 exp, x HP, x MP, 10000gold, Items: Default armor, Default sword(10-17)
[/quote]

... I need my Piston-Driven Polearm from AQW...

want to be assassin

Cut. No more signups.
I will list all the players, what you look like, and equipment on Monday.

Call me abcak o.o

I used to play AQ.  Let me have my PUMPKIN GOLEM I bought the last time a played AQ since my house got so valauable.

To prove it, I'm AL13250 on AQ.

Edit:  I'm a replacement I guess.

[quote="gogators"]We might have to start on Wednesday.
tennis123 and abcak are replacements(sorry).[/quote]

why?

I have an exam on Wednesday, and I have to study. This is probably the last post I will make.
Sorry.
 :(

[quote="gogators"]Okay.
I'm very very sorry, but we might have to postpone this indefinetly. I will be gone for the next week.
However, I will not let this die. We will start after Christmas.
Sorry.
 :([/quote]

Ugh. :( :( :(

U promised.

So now start the game.

Which you lost.

/bump

The message is too small. Please make the message longer before submitting.

Ok, we are starting.
whoopsie
Players:
1. andrewjjiang97: Assassin 0 exp, 400 HP, 700 MP, 10000gold, Items: Default armor, Default sword(10-17)
2. Chompy: Sorcerer 0 exp, 400HP, 800MP, 1500gold, Items: Blacksmith's Apprentice, Blue Scarf, Genesis Sword(20-23), Armored Frost Dragon(5-7)
3. spaceguy524: Sorcerer 0 exp, 400HP, 800MP, 10000gold, Items: Default armor, Default sword(10-17)
4. cyberspace: Assassin 0 exp, 400 HP, 700MP, 4000gold, Items: Dark Scarf, Dexterous Blade(20-23)
5. Richard_Min: Sorcerer 0 exp, 400HP, 800MP, 10000gold, Items: Default Armor, Default sword(10-17)
6. Butler: Paladin 0 exp, 600HP, 600MP, 4000gold, Items: Genesis sword(20-23), Dark Scarf

All you guys are walking along the road when a mushroom pops up.
"AGH!" screams Chompy.
"AGH!" screams Butler.
cyberspace falls asleep.

The mushroom speaks up. "Prepare to die!"

Teh noob mushroom
500HP
-1000MP
BATTLE
Choose to attack, use a skill, blah blah. If you have a pet, it attacks automatically.

Good luck!

(This battle is required because it shows how attacking, gaining exp, resting, etc. work. All other battles you will have a choice as in what to do)

GAH!!! "Die, O though foul mushroom!" I want to use my Genesis Sword to slash at the mushroom's weak point, right under the "umbrella".

Um.... a few questions. What does the Dark Scarf do? Also, what magic can i use?

Dark scarf multiplies probability to dodge by two.

No spells yet, other than your skills.

THWACK!
Butler hits the mushroom for [color=red]21dmg[/color]!

Poison cloud!
The mushroom counter attacks for [color=red]14dmg[/color]!
The others look on in dumbstruck amazement.

Stats:
1. andrewjjiang97: Assassin 0 exp, 400 HP, 700 MP,
2. Chompy: Sorcerer 0 exp, 400HP, 800MP,
3. spaceguy524: Sorcerer 0 exp, 400HP, 800MP,
4. cyberspace: Assassin 0 exp, 400 HP, 700MP,
5. Richard_Min: Sorcerer 0 exp, 400HP, 800MP,
6. Butler: Paladin 0 exp, 686HP, 600MP,

*Goes check my old AQ (AdventureQuest, right?).*

If I'm in:
I'm a ninja.
I attack with my blade of awe!
HOLY COW IT'S POWER ROLL DIE!
win!
Meh just attack.

But my pumpkin golem can actually do 90-110 damage, so he attacks too.

DIE MUSHROOM DIE!
[color=blue]*Chompy stabs it with sword.[/color]

I hasnt used blade of awe since....last year...
Its been awhile since ive been on AQ (not AQW - that Im pretty active)

[quote="spaceguy524"]Also, I buy Floating Wings of Dark.[/quote]

Ahem, remeber this?

Why am I a drowsy assassin???  :P  If I can wake up, I'll attack with my sword.


complex analysis
complex analysis unsolved

Prove that, for a given $n\geqslant 2$, there exists $C\in\mathbb{C}$ such that:

1) $1^C+2^C+\cdots+n^C=0$;
2) There is no $x\in\mathbb{R}\backslash\{0\}$ such that $\Re(|x|^C)=0$.

The second condition seems always satisfied provided $C$ is not real, which must be the case by the first condition. Are you sure of the wording ?

My friends and I stumped across it while we were solving another problem. All we need is to prove existence of $C$ satisfying the given conditions. If condition 2) follows from the first one, in that better.

Sorry, I meant the second condition is never satisfied if the first one is.

I may be stupid, but I do not see why. Could you please explain it?

$R(|x|^{C})=0$. Writing $C=a+ib,\ (a,b) \in \mathbb{R}^{2}$ gives $R(|x|^{a+ib})=0$, or in other words it is enough to have $b ln(|x|)=\frac{\pi}{2}$. Whenever $b \neq 0$ (which is always the case if 1) is satisfied), it is enough to take $|x|=e^{\frac{\pi}{2b}}$. I may be missing something obvious  :?

Yes, now I see, thank you. I am very weak at complex analysis, but this was really embarrassing.

Let $f(z)=1+2^z+...+n^z$.
The problem is to show that $f$ has zero.
Can Picard's theorem work?
I don't know but it works if $n=2$.


How would you make a 3-4-5 triangle with an inscribed circle and another circle inscribed between the hypotenuse, leg with side 4, and the other circle? Could you also eplain how to make inscribed circles in general? Thanks.

*For those of you in Intro to Geometry, it is the diagram for Week 21, Problem 9.

[asy]import olympiad;
pair A=(0,0), B=(0,4), C=(3,0);
draw(A--B--C--A);
draw(incircle(A,B,C));[/asy]

Can you finish from here?

2 questions:

1) What does import olympiad do?
2) In the incircle part, what do the A,B,C mean? Inscribed in shape ABC?

1) Olympiad is a seperate package enabled on the forum, but you have to download it if you do it on the computer.

2) Do you see the part with pairs? Incircle A,B,C means the incircle of those three points.
You can also do: draw(incircle((1,2),(3,4),(5,6)); or any combination of letters and corridinates.

Hope that helped.  :)

Testing:

[asy]import olympiad;
pairs A=(0,0), B=(0,4), C=(3,0);
//pairs X=(0,2), Y=(1.5,2);
draw(A--B--C--A);
draw(incircle(A,B,C));
//draw(incircle(X,B,Y));[/asy]

Hmm... Something went wrong.

[asy]import olympiad;
pair A=(0,0), B=(0,4), C=(3,0);
pair X=(0,2), Y=(1.5,2);
draw(A--B--C--A);
draw(incircle(A,B,C));
draw(incircle(X,B,Y));[/asy]

Like that?
Something seems wrong.

[quote="dragon96"]I think I got it:

[asy]pair A=(0,0), B=(3,0), C=(0,4);
pair X=(0,4/3, Y=(2,4/3);
draw(A--B--C--A);
draw(incircle(A,B,C));
draw(incircle(X,C,Y));[/asy]

Again, I don't see whats wrong?  :cursing:  For the past diagram had the same code as yours. How come it won't work? Do I need to install something?[/quote]
You missed a parenthesis while defining point X

[asy]pair A=(0,0), B=(3,0), C=(0,4);
pair X=(0,4/3), Y=(2,4/3);
draw(A--B--C--A);
draw(incircle(A,B,C));
draw(incircle(X,C,Y));[/asy]

but I don't think that was the image you wanted.. This is a kinda bruteforce way. A better way would involve using tangents and stuff, which I don't know how to do. :oops:

[quote]
but I don't think that was the image you wanted.. This is a kinda bruteforce way. A better way would involve using tangents and stuff, which I don't know how to do. :oops:
[/quote]
I guess this is what was meant:
[asy]pair A=(0,0), B=(3,0), C=(0,4);
draw(A--B--C--A);
draw(incircle(A,B,C));
pair X=incenter(A,B,C);
real r=inradius(A,B,C), t=abs(X-C), h=(t-r)*r/(t+r);
pair Y=C+(t-r)/(r+t)*(X-C);
draw(circle(Y,h));[/asy]
In principle, it is not hard to add the routines finding the circle(s) tangent to 3 circular or line shapes to the Olympiad package. Do you think they are needed often enough to justify the extension?


algorithm
probability
probability and stats

Here is the problem:

You have a Monte Carlo algorithm that outputs either YES or NO, with probability 2/3 it will be correct.

You choose a large odd k, and have an algorithm B that

Runs the Monte Carlo algorithm k times, and returns the most frequent output (yes or no) from these k runs.

Use the Chernoff bound to show that the output of algorithm B is correct with probability
at least

1 − e−^(k/48).

I really have no idea where to start.

The problem tells you exactly what to do.  First, write down the probability that you're trying to compute.  Then, apply a Chernoff bound.


[hide="hint"]What's the relationship between two consecutive terms?
[/hide]


There is 12 E's among the 100 tiles in Scrabble. What is the probability of selecting 4 E's when selecting 4 tiles?

There are $ \binom{100}{4} \equal{} 3921225$ ways to pick four tiles, since order doesn't matter. Similarly, there are $ \binom{12}{4} \equal{} 495$ ways to pick 4 E's, so the probability is $ \frac{3}{23765}$.


floor function
number theory proposed
number theory

Let's denote $ f(n) $ by the number of divisors of $ n^2+n+1 $.

Prove that we can find infinite numbers satisfying $ f(n)\geq f(n+1) $, where $ n\in N $.

Assume that there exist finite numbers satisfying the condition.

Then there exist appropriately large number $ L $ s.t $ f(n)<f(n+1) $ for all $ n>L $.

from $ n^2 < n^2 +n+1 < (n+1)^2 $, we get $ f(n) $ are all even.

so $ f(n+1) \geq f(n)+2 $ for all $ n>L $.

Therefore, we can find $ m\in N (>L) $ s.t $ f(m)>m $.

However, $ f(m) \leq 2 \tau(\lfloor \sqrt{m^2 +m +1} \rfloor) $. ($ \tau(a) $ is the number of divisors of $ m^2+m+1 $ less than $ a $)

cuz $ m^2+m+1 $ is odd, $ \tau (\lfloor \sqrt{m^2 +m+1}\rfloor ) \leq \frac{m}{2} $.

so we get $ f(m)\leq m $. Contradiction for $ f(m)>m $

$ Q.E.D. $


linear algebra
matrix
LaTeX
linear algebra unsolved

For any positive integer $ n$ compute $ A^n$, where
$ A \equal{} \left( \begin{array}{ccc} 1 & 1 & 3 \\ 1 & 3 & 1 \\ 3 & 1 & 1 \end{array} \right) \in \mathcal{M}_3(\mathbb{R})$.

$ A\equal{}J\plus{}2B,$ where $ B\equal{}\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}.$ We have that $ B^2\equal{}I$ and $ JB\equal{}BJ\equal{}JI\equal{}IJ\equal{}J.$ We also have $ J^2\equal{}3J.$

$ A^2\equal{}3J\plus{}2J\plus{}2J\plus{}4I\equal{}7J\plus{}4I$

$ A^3\equal{}(J\plus{}2B)(7J\plus{}4I)\equal{}21J\plus{}4J\plus{}14J\plus{}8B\equal{}39J\plus{}8B$

$ A^4\equal{}(J\plus{}2B)(39J\plus{}8B)\equal{}117J\plus{}8J\plus{}78J\plus{}16I\equal{}203J\plus{}16I$

The pattern is a multiple of $ J$ plus a multiple of $ I$ (for even powers) or $ B$ (for odd powers). What remains is determining the coefficients.

Notice the row sums. If $ x\equal{}[1\ 1\ 1]^T,$ then $ Ax\equal{}5x,$ so $ A^nx\equal{}5^nx.$ Since we also have the coefficient of $ B$ or $ I,$ we can see how to finish:

For $ n$ odd, $ A^n\equal{}\left(\frac{5^n\minus{}2^n}{3}\right)J\plus{}2^nB.$

For $ n$ even, $ A^n\equal{}\left(\frac{5^n\minus{}2^n}{3}\right)J\plus{}2^nI.$

[quote="coquitao"][quote="Kalle"]It has three different eigenvalues. Wouldn't be much simpler to just diagonalize and perform the power?[/quote]

[color=red]Is a Romanian 11th-grader supposed to know that [i]much[/i] Linear Algebra?[/color]  :blush:[/quote]

yep

Moubinoul: the matrix and array commands in $ \text{\LaTeX}$ do work here:

$ P=\begin{bmatrix}\frac12&\frac16&\frac13\\ 0&-\frac13&\frac13\\ -\frac12&\frac16&\frac13\end{bmatrix},\ D=\begin{bmatrix}-2&0&0\\ 0&2&0\\ 0&0&5\end{bmatrix},\ P^{-1}=\begin{bmatrix}1&0&-1\\ 1&2&1\\ 1&1&-2\end{bmatrix}.$

But I'm not going to apologize for my solution above - I still like it. (I did use the eigenvalue $ 5$ and its corresponding eigenvector without explicitly saying that was what I was doing.)


Prove that  ( :Sigma: x_i 2 )n :le: ( :Sigma: x_in+1 )*( :Sigma: x_i )n+1 and find when the equality holds

Sorry but your inequality is wrong. It actually can't be true since the left hand side has power 2n in $x_i$, while the right hand side has power 2n + 2. I think you made a typo: One of the two "n + 1"'s on the right hand side should be an n - 1. Or you forgot a factor. Moreover, are $x_1$, $x_2$, ..., $x_n$ meant to be real numbers or positive real numbers?

  Darij

[quote="yurik"]Sorry, I meant ( :Sigma: x_i 2 ) n  :le: ( :Sigma: x_i n+1)( :Sigma:  x_i) n-1 , all x_i are positive, n natural[/quote]

By Holder, we have

$\displaystyle \sum x_i^2 = \sum (x_i^{n+1})^{1/n}\cdot x_i^{(n-1)/n} \leq \left(\sum x_i^{n+1}\right)^{1/n} \cdot \left(\sum x_i\right)^{(n-1)/n}$.

Taking n-th power on both sides and we are done.


I was reading some old topics, and apparently many people suggested not doing math for the week before a big math contest. Well, the AMC is not "big" for me, but I'd like to get a good score. But at the same time, I'd like to study hard for the AIME, which is certainly "big" for me, and I imagine taking a week off would not help me much in my quest to get a good score on the AIME. In the week before the AMC, if I was not taking a week off, I'd be doing AIMEs, not AMCs. So what is your strategy the week before a math contest? Any advice for me? Thanks.

[quote="nsato"]I would offer these two pieces of advice:

First, for any test, write previous versions of the test in the same conditions (or as similar as possible), meaning same time of day, same duration, etc.  So on the day of the test itself when you walk in, it will seem completely familiar.  Read the rules and instructions, so there are no surprises.

Second, just before the test, do whatever helps you relax the most.  It might be doing math, not doing math, video games, etc., but don't cram the night before.  Get a good night's sleep, and have a good breakfast.  Try not to deviate too much from your regular schedule.

The idea behind these is to be as comfortable as possible when you write the test.[/quote]

By write, you mean take a practice AMC, and not make up questions, right?

but I'm always TOO excited the day before and can't get asleep............lol

[quote="h_s_potter2002"]By write, you mean take a practice AMC, and not make up questions, right?[/quote]

Yes, take a practice AMC, of course :roll:

Anyway, for me it would be doing math to relax :D

[quote="h_s_potter2002"]By write, you mean take a practice AMC, and not make up questions, right?[/quote]
I mean contests from past years and mock versions.

EDITED because people don't seem to understand what it means to be "stop."

[quote="Silverfalcon"]Here is one warning for the week before AMC or AIME:

DO NOT TRY TO LEARN SOMETHING NEW!
[/quote]

Hmm... I disagree.  For most people here, we're constantly learning all the time, so why stop just because a contest is coming up?  Well, maybe it's just me but that just makes everything more unnatural and adds to the stress.  So yeah, I'd agree with what nsato said: "Try not to deviate too much from your regular schedule."  For most of us here, that's doing math and constantly learning... (plus, who knows? maybe the technique you picked up a few days before the contest helped you solve a problem, not that it'd make a significant difference in general)

well the most important thing i have learned from my past personal experiences is that,
try to be calm and just be yourself. you will easily score 120+  :)

[quote="beta"][quote="Silverfalcon"]Here is one warning for the week before AMC or AIME:

DO NOT TRY TO LEARN SOMETHING NEW!

Most times, people think learning few formula can help them greatly but when it comes to AIME/AMC, the formulas don't really matter. [/quote]

Well memorizing formulas is not exactly "learning". I do agree with ThAzN1, if you are doing math almost everyday, why stop doing math just because of the competition?[/quote]

Okay I take it back. Please stop commenting on this though. I didn't mean that you should STOP doing math at all but just try to review more. Sorry for being unclear.

[quote="probability1.01"]NO Silverfalcon! You can not take it back! It's too late now! Now I'm going to randomly disagree for no reason!

Oh yeah hmm you should also think about test-taking strategy or something silly like that.[/quote]

yea...do all the problems and get them right? that usually works out I think


algebra
polynomial
induction
algebra unsolved

Suppose $ P,Q$ are monic complex polynomials such that $ P(P(x))\equal{}Q(Q(x))$.

Prove that $ P\equal{}Q$

Assume Q(x)=x^n+ a(n-1)x^(n-1)+...+a(0), and P(x)=x^n+b(n-1)x^(n-1)+...+b(0).
Q(Q(x))=Q^n+a(n-1)Q^(n-1)+..., P(P(x))=P^n+b(n-1)P^(n-1)+...
We compare the coefficient of x^(n*n-1) between P(P(x)) and Q(Q(x)) will get a(n-1)=b(n-1),
then compare the coefficient of x^(n*n-2) between P(P(x)) and Q(Q(x)) will get a(n-2)=b(n-2),
...,
compare the coefficient of x^(n*n-n) between P(P(x)) and Q(Q(x)) will get a(0)=b(0).

[quote="xxp2000"]
We compare the coefficient of x^(n*n-1) between P(P(x)) and Q(Q(x)) will get a(n-1)=b(n-1),
then compare the coefficient of x^(n*n-2) between P(P(x)) and Q(Q(x)) will get a(n-2)=b(n-2),
...,
compare the coefficient of x^(n*n-n) between P(P(x)) and Q(Q(x)) will get a(0)=b(0).[/quote]

Can you show in more detail?

$ P(P(x))\equal{}\sum_{i\equal{}1}^n b_iP(x)^i$. We already have $ a_n\equal{}b_n\equal{}1$. Let us assume $ a_{n\minus{}i}\equal{}b_{n\minus{}i}$ holds for $ i<\equal{}k$. We need to show it also holds for $ i\equal{}k\plus{}1<n$.

If we look at the coefficient of $ x^{n^2\minus{}k\minus{}1}$ in $ P(P(x))$, only $ P(x)^n$ contributes because $ P(x)^{n\minus{}1}$ has degree $ n^2\minus{}n<n^2\minus{}k\minus{}1$. Moreover, only $ (\sum_{i\equal{}n\minus{}k\minus{}1}^n b_ix^i)^n$ contributes.

This coefficient comes from two parts:
1) $ b_{n\minus{}k\minus{}1}*n$, and
2) coefficent of $ x^{n^2\minus{}k\minus{}1}$ in $ (\sum_{i\equal{}n\minus{}k}^n b_ix^i)^n$.
By induction assumption, the second part is equal between $ P$ and $ Q$. So we have $ b_{n\minus{}k\minus{}1}*n\equal{}a_{n\minus{}k\minus{}1}*n$, i.e., $ b_{n\minus{}k\minus{}1}\equal{}a_{n\minus{}k\minus{}1}$.

The above argument works up to $ i\equal{}n\minus{}1$. For $ i\equal{}n$, $ b_{n\minus{}1}P(x)^{n\minus{}1}$ contributes $ b_{n\minus{}1}$ to the coefficient of $ x^{n^2\minus{}n}$. Then we get $ a_0\equal{}b_0$ by using similar argument.

Hopefully this is detailed enough.


function
calculus
derivative
symmetry
ratio
real analysis
real analysis unsolved

Let $f: \mathbb{R}\rightarrow\mathbb{R}$ be a differentiable function with continuous derivative, that satisfies $f\big(x+f'(x)\big)=f(x)$. Let's call this property $(P)$.

a) Show that if $f$ is a function with property $(P)$, then there exists a real $x$ such that $f'(x)=0$.

b) Give an example of a non-constant function $f$ with property $(P)$.

c) Show that if $f$ has property $(P)$ and the equation $f'(x)=0$ has at least two solutions, then $f$ is a constant function.

a) Consider $f$ at $0$. $f(0+f'(0))=f(0) \implies f(f'(0))=f(0)$, by hypothesis $f'(0)$ is not zero, so we apply Rolle's theorem.
b) You should note that the function only has first derivative 0 at one point, and thus requires some type of symmetry, since between any two points such that $f(a)=f(b) \implies f'(c)=0$ After some thought, one example is $-(x^{2})$, and I can't think of any others...

[quote="gka"](b) You should note that the function only has first derivative 0 at one point, and thus requires some type of symmetry, since between any two points such that $f(a)=f(b) \implies f'(c)=0$ After some thought, one example is $-(x^{2})$, and I can't think of any others...[/quote]
If I'm not mistaken, $-x^{2}+bx+c$ also works.

(c) Since $f^\prime$ is continuous, there are two cases:
- we can find a bounded interval $[a,b]$ such that WLOG $f^\prime (x) > 0, \, \forall x \in (a,b)$ and $f^\prime (a) = f^\prime (b) = 0$. Now you can find a number $u$ very close, but bigger than it, such that $f^\prime (u) \in (a,b)$. But $f$ is strictly increasing on $[a,b]$, contradiction. When $f^\prime (x) < 0$, do the same thing near $b$;
- $f^\prime$ is $0$ on an interval and has constant sign on a neighbourhood of $\pm \infty$. The contradiction can be found pretty easily;
- $f^\prime \equiv 0$ :)

I'm not really sure about the above solution, but I hope it's correct.

[quote="perfect_radio"]
- we can find a bounded interval $[a,b]$ such that WLOG $f^\prime (x) > 0, \, \forall x \in (a,b)$ and $f^\prime (a) = f^\prime (b) = 0$. Now you can find a number $u$ very close, but bigger than it, such that $f^\prime (u) \in (a,b)$. But $f$ is strictly increasing on $[a,b]$, contradiction. When $f^\prime (x) < 0$, do the same thing near $b$;
- $f^\prime$ is $0$ on an interval and has constant sign on a neighbourhood of $\pm \infty$. The contradiction can be found pretty easily;
- $f^\prime \equiv 0$ :)
I'm not really sure about the above solution, but I hope it's correct.[/quote]

If I understand correctly, for your first case, you're claiming that you can find a $u$ closed to $a$, such that $(u+f'(u))\in (a,b)$, so $f(u)=f(u+f'(u))$, but $f$ is strictly increasing, which is the contradiction.  Is there a rigorous way to justify that claim, though? With regards to the second claim, $f'=0$ on an interval and has constant sign on a neighborhood of $\pm \infty$, maybe I'm being obtuse, but what exactly is the contradiction?

My solution begins the same as your's(for c)), perfect radio, but I finished it in another way. We can find a point $c$ in $(a,b)$ such that $c+f^{'}(c)$ be in $(a,b)$, too, and therefore, having $f$ strictly increasing on $(a,b)$ we have that $f(c+f^{'}(c))>f(c)$ which contradicts the given relation.It is the same rationament I used for a) where supposing $f'>0$ we have again $f$ strictly increasing on $R$ and therefore for all real x we have $f(x+f^{'}(x))>f(x)$ contradiction.
 It is easy to prove that the desired $c$ exists using a continuous function $g: R\to R g(x)=x+f^{'}(x)$ and the assumptations about $[a,b]$.

gka, looks like you posted in the same time I was writting and I haven't seen your question before I submited my post.the existence of $c$ can be justifed by using be function $g$ given above, for which we have $g(a)=a$and $g(b)=b$ . If such$c$ wouldn't exist then $g(x)\geq b$ or $g(x)\leq a$ for all x in $(a,b)$ .Therefore $g([a,b])$ will be the reunion between $g(a,b]$ and g(a) in the first case, or between $g[a,b)$ and $g(b)$ in the second case. but in both cases $g([a,b])$ won't be an interval, contradiction.

And one more thing I forgot to mention above in order to finish the demonstration. Because there is no interval of the form $[a,b]$ with $f^{'}(a)=f^{'}(b)=0$ and $f^{'}>0$ on $(a,b)$ we obtained that there exists a dense set of points in $R$ such that $f^{'}=0$ on this set, and $f^{'}$ being continuous we have $f^{'}=0$ on $R$, as desired.

a somewhat harder challange : at a course a few days ago, a teacher told us that the more "general" statement can be solved :
find all functions $f$ with the property $(P)$.

Okay, let's suppose that there's exactly one solution to $f^\prime (x) = 0$. Note that $f$ is a solution iff $x \mapsto f(x+A)$ is a solution, so suppose that $f^\prime (0) = 0$. It's easy to see that $x f^\prime (x) < 0, \, \forall x \neq 0$. Now note that $h(x) =-\left( x+f^\prime (x) \right)$ is an increasing bijection from $J = \left[ 0, \infty \right)$ to itself.
So we have that $f(x) = f(-h(x)), \, \forall x \geq 0$. Substituting $y = h(x)$, $f \left( h^{-1}(y) \right) = f(-y)$, so
\[f(x) = f \left( h^{-1}(-x) \right), \, \forall x \leq 0 . \]
Since the $LHS$ is differentiable, the $RHS$ must be too. If we play a little with the ratios, we should get that $h^{-1}$ is differentiable. Hence,
\[f^\prime (x) =-f^\prime \left( h^{-1}(-x) \right) \cdot \left( h^{-1}\right)^\prime (-x), \, \forall x \leq 0 . \]
Plugging into $f \left( x+f^\prime (x) \right) = f(x)$, we get that
\[f \left( x-f^\prime \left( h^{-1}(-x) \right) \cdot \left( h^{-1}\right)^\prime (-x) \right) = f(x), \, \forall x \leq 0 . \]
By the way $h$ was constructed we get that
\[x-f^\prime \left( h^{-1}(-x) \right) \cdot \left( h^{-1}\right)^\prime (-x) = h^{-1}(-x), \, \forall x \leq 0 . \]
Substituting $z = h^{-1}(-x)$,
\[-h(z)-f^\prime (z) \cdot \left( h^{-1}\right)^\prime (h(z)) = z, \, \forall z \geq 0 , \]
i.e.
\[f^\prime (z) =-\frac{z+h(z)}{\left( h^{-1}\right)^\prime (h(z))}, \, \forall z > 0 . \]
But
\[f^\prime (x) =-\left( x+h(x) \right), \, \forall x \geq 0 , \]
so
\[\left( h^{-1}\right)^\prime (h(x)) = 1, \, \forall x > 0 . \]
That implies $h^{-1}(x) = x+a$, so $h(x) = x-a$. Since $h$ is a bijection from $J$ to $J$, $a=0$. Hence, $f^\prime (x) =-2x, \, \forall x > 0$, so $f(x) =-x^{2}+c, \, \forall x > 0$. Using the formulas above, we extend this to the whole real line.
The only functions that remain standing in this case are those of the form $-x^{2}+bx+c$.

PS: I really hope I didn't mess it up.


Here is the problem. Please help me

The base angles of an isosceles trapezoid are 60 degrees and one of the diagonals bisects the base angle. If the median is 15 inches long, find the area of the trapezoid.

PLEASE HELP ME
thank you

[hide="Answer"] $75\sqrt{3}$

Actually, let me check it first.
[/hide]

[hide]First of all, the area of a trapezoid is the median times the altitude.

We know the median is $15$, so we just need to find the altitude.

I'm too lazy to draw this in Geometer's Sketchpad, so just try to visualize. You have an isosceles trapezoid. Draw the diagonal, so that the $60^{\circ}$ base angle is split into two $30^{\circ}$ angles. The other base angle is also $60^{\circ}$, so you have a right triangle.

The top angles of the trapezoid are $120^{\circ}$, so the other triangle is isosceles (two $30^{\circ}$ angles). Let the length of the top base be $2a$. The length of the side is also $2a$.

Draw altitudes from the two top angles. You form two 30-60-90 triangles, so the other base has length $a+2a+a=4a$.

$\frac{2a+4a}{2}=15$, since the median is the average of the two bases.

Thus $a=5$.

Again, from the 30-60-90 triangle, you get the altitude is $5\sqrt{3}$.

The area is thus $15\cdot 5\sqrt{3}=75\sqrt{3}$. [/hide]

Thank you very much
I finaly understand this problem


induction
combinatorics unsolved
combinatorics

For a set $ S$, let $ |S|$ denote the number of elements in $ S$. Let $ A$ be
a set of positive integers with $ |A| = 2001$. Prove that there exists a
set $ B$ such that

(i) $ B \subseteq A$;
(ii) $ \dsp |B| \ge 668$;
(iii) for any $ u, v \in B$ (not necessarily distinct), $ u + v \not\in B$.

Uh... can't we just have $ B \equal{} \{1,3,5,7,...,2001\}$ which has way more elements than requested and clearly satisfies the condition (by parity analysis)? I think you either misstated or forgot a condition.

Edit: crap nevermind... I didn't realize that A could be any set of positive integers.

I think this is a quite well known problem which I heard somebody says appeared in the mathlink contest before.
[Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=5966 .]

[quote="gauravpatil"]Does anyone have something concrete over the problem?[/quote]

By the way, what is something concrete :roll:

[quote="greentreeroad"]I think this is a quite well known problem which I heard somebody says appeared in the mathlink contest before.

By the way, what is something concrete :roll:[/quote]
any good ideas as how to approach the problem ?any hints?

if more than 668 elements have same largest power of 2 (for example, 668 odd)
but it seems nothing to do with this problem, don't know if it can bring you any thought...

even if numbers of the form $ 3k \plus{} 1$ or $ 3k \plus{} 2 are \ge 667$then we are done
but i see noting useful in this :(

A question: when you see a problem says "a set of positive integers", does that infer that the elements are all distinct. Or could there exists same element?

I remember hearing that elements in a set should be distinct.....

Unless its specified as a "multiset", it is assumed that the set $ \{1,2,3,4\}$ is equivalent to the set $ \{1,1,1,1,1,2,2,3,3,3,4,4,4,4,4\}$ or such. A good example of this is the set of consecutive differences of the sequence $ 1,2,3,4...$ is equivalent to that of $ 1,2,3$.

So can the set in this problem consists of 2001 same numbers (I know it will be trivial if so, but I still like to know) :?:

No. In saying that $ |A|\equal{}2001$ we imply that A contains 2001 distinct elements.

thanks that does change my understanding of the question

[quote="gauravpatil"]try replacing 2001 by 3k and 668 by k+1 and use induction it really works :roll:  :D[/quote]

You have a solution, great! Show it!

[quote="greentreeroad"][

You have a solution, great! Show it![/quote]
sorry i made a grave mistake and hence i dont have a solution

does that mean noone has the solution to this one ? :o  :|  :(  litterally noone!!!!!!!!!!!!!!

see this link greentreeroad
http://www.mathlinks.ro/viewtopic.php?p=1097446#1097446
i dont know how to fix it so i am giving the whole link :D

Hi everybody,

This is my first posting so please bare with me. A generalization of the problem you are talking about has been proved by Alon and Kleitman in their 1990 paper "Sum-free subsets". (see theorem 2). I am including the Math Review link below.

[url=http://www.ams.org/mathscinet/pdf/1117002.pdf?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&s4=&s5=sum-free%20subsets&s6=&s7=&s8=All&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=7]a loooooooooooooooooooooooooooong URL that has been edited by a moderator in order not to stretch the page[/url]

could you make us available that paper
the link needs subscription
many thanks

the best I can do is attach the review itself.
Unfortunately, I do not have the paper.

Thank you for all of your thinking

Unfortunately, I think these papers are not elementary enough to fit to a problem for high school students. I think the only way we can know that is to ask staffs or students around year 2001 to see if they know the official solution.


integration
calculus
trigonometry
complex numbers
geometric series
complex analysis

Evaluate $\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{1+nk}$ as a function of $k$.

My idea is ${\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{1+nk}=\sum_{n=-\infty}^{\infty} (-1)^n \int^1_0 x^{kn} dx=\int^1_0 \sum_{n=-\infty}^{\infty}(-x^k)^n} dx$

Especially for $k=2,\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}=\frac{\pi}{4}$. [b]Libnitz series[/b].By the way do we have another condition for $k$?

That could be useful, but there's a flaw: $\sum_{n=-\infty}^{\infty}x^n$ diverges for all $x$. I'm being vague about $k$: the series diverges when $k=\frac1n$, but that's trivial. It may be easier to assume $k>1$.

For $k=2$, you get $\dots-\frac17+\frac15-\frac13+1+1-\frac13+\frac15-\frac17+ \dots=\frac{\pi}{2}$: the terms for negative $n$ match the positive ones. In general, you can't evaluate the sums for $n>0$ and for $n<0$ independently, although they both converge.

i guess we could use the poisson identity $\sum_{n\in\mathbb{Z}} f(n)=\sum_{k\in\mathbb{Z}} g(k)$, where $g(k)=\int_{-\infty}^{\infty} f(z)e^{-2\pi ikz}dz$.

if we take $f(n)=\frac{(-1)^n}{1+xn}=\frac{e^{\pi in}}{1+xn}$, we get $g(k)=\int_{-\infty}^{\infty} \frac{e^{\pi iz(1-2k)}}{1+xz}dz$. now, we get the identity $\phi (x)=\sum_{n\in\mathbb{Z}} \frac{(-1)^n}{1+xn}=\sum_{k\in\mathbb{Z}} \int_{-\infty}^{\infty} \frac{e^{\pi iz(1-2k)}}{1+xz}dz$. now, assume $x=a+ib$ with $b>0$. then the integrand has one pole at $z=-\frac{1}{x}$ with imaginary part $>0$. if $k<1$, we integrate over the real line and lower complex-halfplane(the value goes to zero there), and there is no pole, thus the integral is 0. if $k>0$, then we integrate over the real line and the upper complex-halfplane(mathematically positive directed). the pole hat $z=-\frac{1}{x}$ is $\frac{1}{x}e^{\frac{\pi i(2k-1)}{x}}$; the value of the integral is thus $\frac{2\pi i}{x}e^{\frac{\pi i(2k-1)}{x}}$. summing this for $k>0$ gives $\frac{2\pi i}{x}e^{\frac{\pi i}{x}}\left(e^{\frac{2\pi i}{x}}-1\right)^{-1}$. by analytic continuation, this is the value for all complex $x$.

Peter

btw, this simplifies to $\frac{\frac{\pi}{x}}{\sin(\frac{\pi}{x})}$. the series converges iff $\frac{1}{x}\not\in\mathbb{Z}$.

BTW, it is the same as $\int_0^{+\infty}\frac{dt}{1+t^x}$ for $x\geq 1$, $x\in \mathbb{R}$.
Is there some relation?

Probably Complex Integration will do well. We offer to make Kent Merryfield's entrance.

kunny

You don't need me to make an entrance. I would have said something about the Poisson summation formula, but Peter Scholze took care of that. Yes, Myth, it was inspired by that problem, but jmerry can explain it himself without needing my help.

Nice solution, Peter. I didn't have that method, although I do have one that uses analytic continuation to get from the real line to everywhere else.

Now, on the problem itself:
First, the sum does come from that integral:

Assuming $\Re(k)>1$
$\int_0^{\infty}\frac1{1+x^k}dx= \int_0^1(1-x^k+x^{2k}-x^{3k}+\dots)dx + \int_1^{\infty}(x^{-k}-x^{-2k}+x^{-3k}-x^{-4k}+\dots)dx$., using the geometric series formula to expand $\frac1{1+x^k}$ in two different ways.
The first integral is $\sum_{n=0}^{\infty}\frac{(-1)^n}{1+nk}$ and the second is $\sum_{n=1}^{\infty}\frac{(-1)^n}{1-nk}$, so the whole integral
$\int_0^{\infty}\frac1{1+x^k}dx= \sum_{n=-\infty}^{\infty}\frac{(-1)^n}{1+nk}$

My initial method was to evaluate this integral in another way (assuming that $k$ is real and $k>1$): take the line integral around the wedge with spokes on the real line and at an angle of $\frac{2\pi}{k}$ above, completed by a distant circle back. Since $k>1$, the integral over the distant circle goes to zero and we have $(1-e^{2\pi i/k})\int_0^{\infty}\frac1{1+x^k}dx=2\pi i\cdot$(residue at the pole).

The pole is at $e^{\pi i/k}$, and it has residue $-\frac{e^{\pi i/k}}{k}$, so $\int_0^{\infty}\frac1{1+x^k}dx=\frac{-2\pi i e^{\pi i/k}}{k(1-e^{2\pi i/k})}=\frac{\pi/k}{\sin(\pi/k)}$

Once we have this, the general version follows by analytic continuation.

Alternately, we can work directly with the sum and never invoke integrals at all.
Let $g(z)=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{z+n}$. If $f(z)=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{1+nz}$, $g(z)=\frac{f(\frac1z)}{z}$.
We can easily see that $g$ is analytic; the alternate form $g(z)=\sum_{n=-\infty}^{\infty}\frac1{(z+2n)(z+2n+1)}$ converges absolutely. Obviously $g$ has simple poles at every integer, with residue $(-1)^n$. Also, $g(z+1)=-g(z)$, and $g(z)$ tends to 0 as the imaginary part of $z$ tends to $\pm \infty$ (use the alternate form).
A function that shares these properties is $\frac{\pi}{\sin(\pi z)}$; we want to show that it is the same. Consider the function $g(z)-
\frac{\pi}{\sin(\pi z)}$ on the vertical strip $|\Im(z)|\le \frac12$. Matching the Laurent series at the pole, the remaining terms of the difference all have nonnegative exponents, so this singularity is removable; remove it. The function must then be bounded on the whole strip, since it becomes small far enough away. By periodicity, this makes the difference bounded on the whole plane. A bounded entire function is constant, and this one goes to zero somewhere, so it must be zero.

Integral $\int_0^{\infty}\frac{dx}{1+x^k}$ is calculated somewhere on forum by me  ;) That is why I know it by sight.


A collection of sets S1 , S2,.........Sm from a space of [d] elements is called a (L, a)design if

1.    For all i,  |S i | =  L

2.     For all i, j,  |S i  :cap:  S j | < a

Prove that if m < [((d choose a)/((L choose a) 2))] then there exists a (L, a) design.

Clearer stating of question

http://web.uvic.ca/~gsrivast/problem.jpg


algebra
polynomial
logarithms
limit
function
Gauss
number theory

Chebyshef's theorem gives that $a f(x) \leq \pi(x) \leq b f(x)$ with $f(x)= \frac{c}{\log (x)}$ and some constants $a,b$.
Writting $a \sim b$ if $\lim_{x \to \infty} \frac{a(x)}{b(x)}$ we have $P(x) \sim ax^p$ and $Q(x) \sim bx^q$ for some $a,b,p,q$.
If the above equality would hold, $\pi(x) \sim c x^r$ for $c=\frac ab$, $r=p-q$.

But it's clear that $\lim_{x \to \infty} \frac{\pi(x)}x =0$ by Chebyshef's theorem, thus we need $r\leq 0$, which again would say that $\pi(x)$ is bounded, surely wrong.

It's not difficult to show that any real $r$ is impossible using the other half of Chebyshef's theorem.

We don't really need Chebyshev's bounds. By contradiction, admit there're polynomials $ P, Q \in \mathbb{C}[x]$ such that $ \pi(x) = \frac{P(x)}{Q(x)}$, for any $ x \in \mathbb{N}= \{1, 2, \ldots\}$. Then let $ m = \deg(P)$ and $ n = \deg(Q)$. As $ \lim_{x \to+\infty}\pi(x) =+\infty$, obviously $ m \ge n+1$. Now consider the function $ f: \mathbb{R}^{+}\mapsto \mathbb{C}: x \to P(x)-\pi(x) Q(x)$ and observe that $ f(x) = 0$, whenever $ x \in \mathbb{N}$. Then recall there exists $ k \in \mathbb{N}$ such that each integer in the $ (m+1)$-set $ \{k, k+1, \ldots, k+m\}$ is composite. This means the function $ g: \mathbb{C}\mapsto \mathbb{C}: x \to P(x)-\pi(k) \cdot Q(x)$ has at least $ m+1$ distinct zeros. But actually $ g$ is a complex polynomial of degree $ m \ge 1$ and hence it has at most $ m$ distinct zeros in $ \mathbb{C}$, by Gauss' fundamental theorem of algebra (absurd). So the claim.

[quote="Peter"]Prove that there do not exist polynomials $ P$ and $ Q$ such that
\[ \pi(x)\equal{}\frac{P(x)}{Q(x)}\]
for all $ x\in\mathbb{N}$.[/quote]

We first prove the following claim.

 Theorem. Let $ l_{0}\in\mathbb{N}$. Suppose that $ \zeta :\mathbb{N}\rightarrow\mathbb{R}$ be a nonconstant function having the property that, for all $ l\geq l_{0}$, there exists a sequence of distinct positive integers $ a_{1}$, $ \cdots$, $ a_{l}$ with
\[ \zeta\left( a_{1}\right)\equal{}\cdots\equal{}\zeta\left( a_{l}\right).\]
Then, there is no rational function $ f$ satisfying $ \zeta(n) \equal{} f(n)$ for all $ n\in\mathbb{N}$.

 Proof.  Assume to the contrary that there exist a rational function $ f$ such that $ \zeta(n) \equal{} f(n)$ for all $ n\in\mathbb{N}$. Since $ f$ is rational, one may write $ f(x) \equal{}\frac{P(x)}{Q(x)}$ for some polynomials $ P(x)$, $ Q(x)\in\mathbb{R}[x]$. Take a sufficiently large positive integer $ l$ so that $ l > max(deg\; P,\; deg\; Q)$. By the assumption, one may write
\[ \zeta\left( a_{1}\right)\equal{}\cdots\equal{}\zeta\left( a_{l}\right)\equal{}\lambda\]
for some positive integers $ a_{1}$, $ \cdots$, $ a_{l}$ and $ \lambda\in\mathbb{R}$. Introduce the polynomial $ \Omega(x)\in\mathbb{R}[x]$ defined by
\[ \Omega(x)\equal{}P(x)\minus{}\lambda Q(x).\]
Since $ \lambda \equal{}\zeta( a_{k}) \equal{} f\left( a_{k}\right) \equal{}\frac{ P( a_{k}) }{ Q( a_{k}) }$ , we find that $ \Omega\left( a_{k}\right) \equal{} 0$ for all $ k\leq l$, which implies that the polynomial $ \Omega(x)$ has $ l$ distinct roots. Since $ deg\;\left( P(x)\minus{}\lambda Q(x)\right)\leq max(deg\; P,\; deg\; Q) < l$ or $ deg\;\Omega(x) < l$, the fundamental theorem of algebra implies that $ \Omega$ is identically zero so that $ P(x)\equiv\lambda Q(x)$. Since $ f(x) \equal{}\frac{P(x)}{Q(x)}$, this implies that $ f(x)\equiv\lambda$. It follows that $ \zeta(n) \equal{} f(n) \equal{}\lambda$ for all $ n\in\mathbb{N}$ so that $ \zeta :\mathbb{N}\rightarrow\mathbb{R}$ is a constant function. This is a contradiction.

 Now, we prove the main result. We need to show that the function $ \pi$ satisfies the conditions in the theorem.

 Proposition. Let $ \pi :\mathbb{N}\rightarrow\mathbb{Z}_{\geq 0}$ be the function defined by
\[ \pi(n)\equal{}\left\vert\;\{ p\leq n\;\vert\; p\; is\; prime.\}\;\right\vert\]
for all $ n\in\mathbb{N}$. Then, $ \pi$ is nonconstant. However, for any $ l\in\mathbb{N}$, there exists a sequence of consecutive positive integers $ c\plus{}1$, $ \cdots$, $ c\plus{}l$ such that
\[ \pi\left( c\plus{}1\right)\equal{}\cdots\equal{}\pi\left( c\plus{}l\right).\]
Proof.  Since there are infinitely many primes, it's immediate that $ \pi$ is not a constant function. The second assertion also follows from the fact  that there exist are arbitrary length sequences of consecutive composite numbers.

 Lemma. Let $ l\in\mathbb{N}$. There are infinitely many $ c\in\mathbb{N}$ such that $ c\plus{}1$, $ \cdots$, $ c\plus{}l$ are all composite numbers.

 First Proof. Since there are infinitely many primes, we can pick distinct primes $ p_{1}$, $ \cdots$, $ p_{l}$. Consider the system of linear congruences
\[ \begin{cases}x\equiv\minus{}1\;\left( mod\; p_{1}\right),\\ x\equiv\minus{}2\;\left( mod\; p_{2}\right),\\ \;\;\;\;\;\;\;\;\;\;\;\cdots\\ x\equiv\minus{}l\;\left( mod\; p_{l}\right).\\ \end{cases}\]
Since $ gcd\left( p_{i}, p_{j}\right) \equal{} 1$ for all $ i\neq j$, Chinese Remainder Theorem implies that it has a unique solution modulo $ p_{1}\cdots p_{l}$. Hence, one may take a sufficiently large integer $ c > p_{1}\cdots p_{l}$ so that
\[ p_{k}\;\vert\; c\plus{}k\]
for all $ k\leq l$. Since $ c\plus{}k > p_{k}$ and and $ c\plus{}k$ is divisible by $ p_{k}$, we find that $ c\plus{}k$ is always composite.

  Second Proof. The  $ l$ consecutive integers
\[ (l\plus{}1)!\plus{}2,\cdots, (l\plus{}1)!\plus{}l\plus{}1\]
are all composite because $ (l\plus{}1)!\plus{}k$ is divisible by $ k$ for all $ k\in\{2,\cdots, l\plus{}1\}$.

[quote="Peter"]Prove that there do not exist polynomials $ P$ and $ Q$ such that
\[ \pi(x)\equal{}\frac{P(x)}{Q(x)}\]
for all $ x\in\mathbb{N}$.[/quote]

 We prove the following claim.

 [b]Proposition 1.[/b] Suppose that $ \eta :\mathbb{R}\rightarrow\mathbb{R}$ is a function having the property that
\[ \lim_{x\rightarrow\infty}\eta(x)\equal{}\infty\;\;\text{and}\;\;\lim_{x\rightarrow\infty}\frac{\eta(x)}{x}\equal{}0.\]
Then, $ \eta$ is not a rational function.

 [b]Proof.[/b] Assume to the contrary that $ \eta(x) \equal{}\frac{P(x)}{Q(x)}$, where $ P(x)$, $ Q(x)\in\mathbb{R}[x]$. On the one hand,
\[ \lim_{x\rightarrow\infty}\frac{P(x)}{Q(x)}\equal{}\infty\]
implies that $ \deg P\geq 1\plus{}\deg Q$. On the other hand,
\[ \lim_{x\rightarrow\infty}\frac{P(x)}{xQ(x)}\equal{}0\]
implies that $ \deg P\leq\deg Q$. This is a contradiction.

 Now, we establish that the function $ \pi$ satisfies the conditions.

 [b]Proposition 2.[/b] Let $ \pi :\mathbb{R}_{\geq 0}\rightarrow\mathbb{Z}_{\geq 0}$ be the function defined by
\[ \pi(x)\equal{}\left\vert\;\{ p\leq x\;\vert\; p\; is\; prime.\}\;\right\vert\]
for all $ x\in\mathbb{R}_{\geq 0}$. Then
\[ \lim_{x\rightarrow\infty}\pi(x)\equal{}\infty\;\;\text{and}\;\;\lim_{x\rightarrow\infty}\frac{\pi(x)}{x}\equal{}0.\]

[b] Reference. [/b]  A Classical Introduction to Modern Number Theory by Kenneth Ireland, Michael Rosen

 [b] Lemma 1.[/b] When $ x\geq 1$, we have
\[ \theta(x)\equiv\sum_{\substack{ p: prime,\\ p\leq x}}\ln p\leq\left( 4\ln 2\right) x.\]
[b]Proof[/b]   Since $ \binom{2n}{n}\equal{}\frac{(n\plus{}1)\cdots (2n) }{n!}$ is divisible by all primes $ p$ with $ p\in (n, 2n)$, we get
\[ \prod_{\substack{ p: prime,\\ n<p<2n}}p\leq\binom{2n}{n}\leq\sum_{k\equal{}0}^{2n}\binom{2n}{k}\leq 2^{2n}.\]
or
\[ \theta(2n)\minus{}\theta(n)\leq 2n\ln 2.\]
Since $ x\geq 1$, we can find a positive integer $ k$ so that $ x\in\left(2^{k\minus{}1}, 2^{k}\right]$. It follows that
\[ \theta(x)\leq\theta( 2^{k})\equal{}\sum_{i\equal{}1}^{k}\left(\theta( 2^{i})\minus{}\theta( 2^{i\minus{}1})\right)\leq\sum_{i\equal{}1}^{k}2^{i}\ln 2\equal{}\left( 2^{k\plus{}1}\minus{}2\right)\ln 2 < 2^{k\plus{}1}\ln 2 <\left(4\ln 2\right)x.\]
[b]Lemma 2.[/b]
     When $ x\geq 2$, we have
\[ \pi(x)\leq\left( 2\plus{}8\ln 2\right)\frac{ x }{\ln x}.\]
[b]Proof[/b] Let $ x\geq 2$. It follows that
\[ \theta(x)\geq\sum_{\substack{ p: prime,\\ \sqrt{x}<p\leq x}}\ln p\geq\sum_{\substack{ p: prime,\\ \sqrt{x}<p\leq x}}\ln\sqrt{x}\equal{}\left(\pi(x)\minus{}\pi(\sqrt{x})\right)\ln\sqrt{x}\geq\pi(x)\ln\sqrt{x}\minus{}\sqrt{x}\ln (\sqrt{x})\]
so that
\[ \pi(x)\leq\frac{2\theta(x)}{\ln x}\plus{}\sqrt{x}\leq\left( 8\ln 2\right)\frac{x}{\ln x}\plus{}\frac{2x}{\ln x}\equal{}\left( 2\plus{}8\ln 2\right)\frac{ x }{\ln x}.\]
We now prove Proposition 2. Since there are infinitely many prime numbers, we immediately find that $ \lim_{x\rightarrow\infty}\pi(x) \equal{}\infty$. Since $ \frac{1}{\ln x}\rightarrow 0$ as $ x\rightarrow\infty$, \textsc{Lemma 3.4} implies that
\[ \lim_{x\rightarrow\infty}\frac{\pi(x)}{x}\equal{}0.\]
[b] Note [/b] We proved that $ \frac{\pi(x)}{\frac{ x }{\ln x}}\leq 2\plus{}8\ln 2$.  In fact, it is known that  \begin{thm} \textsc{Prime Number Theorem} We have
\[ \pi(x)\sim\frac{ x }{\ln x}\]
or
\[ 1\equal{}\lim_{x\rightarrow\infty}\frac{\pi(x)}{\frac{ x }{\ln x}}.\]

 In 1896, Hadamard and Valle'e Poussin independently proved  [b]Prime Number Theorem[/b] using complex analysis. Several alternative proofs were known, including the  elementary proofs of Paul Erd"os and Atle Selberg.

[quote="Peter"]Prove that there do not exist polynomials $ P$ and $ Q$ such that
\[ \pi(x) \equal{} \frac {P(x)}{Q(x)}
\]
for all $ x\in\mathbb{N}$.[/quote]
suppose that $ k\equal{}Max\{degP,degQ\}$.
we have $ \pi((k\plus{}2)!\plus{}2)\equal{}\pi((k\plus{}2)!\plus{}3)\equal{}...\equal{}\pi((k\plus{}2)!\plus{}k\plus{}2)\equal{}t$  therefore at $ k\plus{}1$ point we have $ P(x)\minus{}tQ(x)\equal{}0$ and
 $ deg(P\minus{}tQ)\leq Max\{degP,degQ\}\equal{}k$ therefore $ P\minus{}tQ$ is zero polynomial and for all $ x\in\mathbb{N}$ we have $ \pi(x)\equal{}\frac{P(x)}{Q(x)}\equal{}t$  but $ \pi(2)\equal{}1,\pi(3)\equal{}2$!contradiction!

[quote="ali666"][quote="Peter"]Prove that there do not exist polynomials $ P$ and $ Q$ such that
\[ \pi(x) \equal{} \frac {P(x)}{Q(x)}
\]
for all $ x\in\mathbb{N}$.[/quote]
suppose that $ k \equal{} Max\{degP,degQ\}$.
we have $ \pi((k \plus{} 2)! \plus{} 2) \equal{} \pi((k \plus{} 2)! \plus{} 3) \equal{} ... \equal{} \pi((k \plus{} 2)! \plus{} k \plus{} 2) \equal{} t$  therefore at $ k \plus{} 1$ point we have $ P(x) \minus{} tQ(x) \equal{} 0$ and
 $ deg(P \minus{} tQ)\leq Max\{degP,degQ\} \equal{} k$ therefore $ P \minus{} tQ$ is zero polynomial and for all $ x\in\mathbb{N}$ we have $ \pi(x) \equal{} \frac {P(x)}{Q(x)} \equal{} t$  but $ \pi(2) \equal{} 1,\pi(3) \equal{} 2$!contradiction![/quote]Nice, this is the way to go, but it is the same as s.tringali

For all positive integers $ n$, $ \pi(4n) \equal{} \pi(4n \minus{} 1)$. Hence it would follow that $ P(4n)Q(4n \minus{} 1) \equal{} P(4n \minus{} 1)Q(4n)$ for all positive integers $ n$. Since both sides are polynomials, it follows that $ P(x)Q(x \minus{} 1) \equal{} P(x \minus{} 1)Q(x)$ for all real numbers $ x$. In particular $ P(3)Q(2) \equal{} P(2)Q(3)$, so $ \pi(2) \equal{} P(2)/Q(2) \equal{} P(3)/Q(3) \equal{} \pi(3)$.


Given that 11 is the mean of the set $ \{19, 9, 11, 5, 14, x\}$, what is the value of $ x$?

There are 6 numbers in that set, including $ x$, and since the mean is 11, we know the sum is $ 6 \times 11 \equal{} 66$. The sum of the 5 integers we know is $ 19 \plus{} 9 \plus{} 11 \plus{} 5 \plus{} 14 \equal{} 58$, so $ x$ must be $ 66 \minus{} 58 \equal{} \boxed{8}$.

The mean of this data set is defined as the sum divided by total n number of terms, where n here is 6.

$ \dfrac{19\plus{}9\plus{}11\plus{}5\plus{}14\plus{}x}{6}\equal{} 11\Longrightarrow 58\plus{}x\equal{} 66\Longrightarrow x\equal{}\boxed{8}$

[hide="this is my answer"]the mean is adding all the #'s and divide by the amt of #'s you added

doing so you get the equation $ \frac{19+9+11+5+14+x}{6}=$ simplifying and using simple algebra we get the answer of

8 or $\boxed{8}$[/hide]

[b][i][u]please rate my post thanks ![/u][/i][/b]

please be fair though!!

 :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:


[u]time[/u]
Saturday, April 23, 5:10 pm ~ 6:30 pm, 5 session total

[u]Participants[/u]
Peak
Long
Hanpin

Tao
Maria
Stan

Kevin
Alan
Sho

[u]Timer[/u]
Geehoon

[u]Problems From[/u]
google..-.-

[u]Team Score[/u]
Team1: 0 0 0 0 4
Team2: 0 2 4 4 4
Team3: 2 4 4 4 4

[u]Next Practice Session[/u]
TBA

yay! that was so much fun, and went pretty well... we should do this more often!

eep, double posting

it is fun though....definitely should do more...
only problem,when do we get the list for ARML,it could be very frustrating if we did a lot of parctice and yet not on same team....

it was really messy in the beginning
but after it went smoothly i guess
sorry about my mitake as a timer here and there
but hopefully we'll have a better timer next time

the first two sets of problems are from polya; it is supposed to be harder
but i think we should try practice hard problems also.

I'll post up problems and solutions as soon as possible.
Good job guys!!

everyone should be able to participate in online relays! even if the teams turn out to be very different for ARML, extra practice doesnt' hurt anyone


algebra
polynomial
number theory proposed
number theory

Here is a modular version of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26551&highlight=]this[/url] thingy:

Let $p$ be a prime and $a_1,\ldots,a_k \in \mathbb{F}_p^*$ be pairwise different non-zero elements of the finite field $\mathbb{F}_p$ of $p$ elements. If $f_1,\ldots,f_k \in \mathbb{F}_p[x]$ are polynomials of degrees at most $p-1$ such that
\[
f_1(n)a_1^n + f_2(n)a_2^n + \cdots + f_k(n)a_k^n = 0 \quad \forall \, n \in \mathbb{Z},
\]
then $f_1 \equiv \cdots \equiv f_k \equiv 0$.

It's enough to show it for the case when the $f_i$'s are constant:

Looking first at $n=0,p-1,2(p-1),\ldots,(p-1)^2$, we find that $\sum f_i\equiv 0$. Then we take $n=1,1+(p-1),1+2(p-1),\ldots,1+(p-1)^2$ to find $\sum a_if_i\equiv 0$. We repeat this to get $\sum a_i^n f_i\equiv 0,\ \forall n$. Now, we can apply the problem for the case when the polynomials are constant to the terms containing $x^{p-1}$ in our polynomials. We find that all of these are null. Then we apply it to the terms containing $x^{p-2}$, and so on.

For the case when the polynomials are constants, we can use the Vandermonde determinant constructed with the powers of the $a_i$'s from $0$ up to $k-1$.


function
geometry
circumcircle
geometry unsolved

1)Let  A,B,C given points.Find the locus of set of P points that the distance of $PA^2+PB^2+PC^2=r^2$ is constant.

2)Let  A,B,C given points.Find the locus of set of P points that the distance of $PA+PB+PC=r$ is constant.Has it any name?.Can that curve has a "sharp" point?

$1)$ [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=54736[/url]

Suppose that $A,B,C$ are distinct

Let $G$ be the centroid of $\triangle ABC \Rightarrow  \overline{GA} + \overline{GB} + \overline{GC} = \overline{0}$

Let $f : {\Re}^2\to \Re,$ with $f(P)=PA^2+PB^2+PC^2$

It's obvious that $f(P)>0 , \forall P$

$\overline{PA} = \overline{PG} + \overline{GA}$
$\overline{PB} = \overline{PG} + \overline{GB}$
$\overline{PC} = \overline{PG} + \overline{GC}$

$PA^2 = PG^2+GA^2+2 \cdot \overline{PG} \cdot \overline{GA}$
$PB^2 = PG^2+GB^2+2 \cdot \overline{PG} \cdot \overline{GB}$
$PC^2 = PG^2+GC^2+2 \cdot \overline{PG} \cdot \overline{GC}$

$f(P) = 3PG^2 + (GA^2+GB^2+GC^2) +2 \cdot \overline{PG} \cdot (\overline{GA} + \overline{GB} + \overline{GC})$
$f(P) = 3PG^2 + f(G)$

Since $f(G)$ is constant, $f(P)$ is a function depended only on the distance $PG$ from $P$ to the barycenter of the triangle

So the minimum value of $f(P)$ is when $P \equiv G$

$f(P)=r^2 \Leftrightarrow$
$3PG^2 + f(G) = r^2 \Leftrightarrow$

$PG^2 = \frac{r^2-f(G)}{3}$

If $r^2 \ge f(G)$ then we have a solution for $PG$

The locus for $f(P)=r^2$ is a circle centered at $G$ with radius $\sqrt{\frac{r^2-f(G)}{3}}$.

$*$ If $r^2 = f(G)$ then the locus is the point $G$ itself

$2)$ Let $C(r)$ the set of the points $P$, for which $PA+PB+PC=r$.

I don't know this curve, but I'm sure that

$C(3R)=O$, where $(O,R)$ is the circumcircle of $ABC$ ($O$ is the only point with this property).

Also, we know that $C(r)=\emptyset,\ \ \forall r<3R$.

This is not true.
If the three angles of ABC are <120°, the minimum value of PA + PB + PC is
k = root((a^2+b^2+c^2)/2 + 2.root(3).area(ABC)); thus
if r<k, C(r) is empty
if r = k, C(r) = {first Fermat point}
if r>k, C(r) has infinitely many points; this is the case when r = 3R

Oh you are right, jpe  :(

For a moment, I thought that the point $O$ is the one with the minimum sum of distances.

Now I see that, if $A$ is extremely obtuse angle, then $O$ is too far away from the points $A,B,C$, and we can find easily a $P$ close to $\triangle ABC$ such that $PA+PB+PC<3R$

Sorry.

By the way, how did you obtain

$k = \sqrt{\frac{a^2+b^2+c^2}{2} + 2 \cdot \sqrt{3} \cdot (ABC)}$ ?

[u]A generalization of (1):[/u]

Let $n\in N^*,\ P_k,\ k\in \overline {1,n}$ be any fixed points and let $p_k\in R^*,\ k\in \overline {1,n}$.
Ascertain the geometrical locus of the point $L$ for which $\sum\limits_{k=1}^{n} p_k\cdot LP^2_k=l^2$ (constant).

[b]I will come back soon.[/b]

If F is the Fermat point minimizing PA + PB + PC,we have
<BFC = <CFA = <AFB = 120°
Thus
a^2 = FB^2 + FC^2 + FB.FC and two similar relations.
Adding, we get
(1) : a^2+b^2+c^2 = 2(FA+FB+FC)^2 - 3(FB.FC+FC.FA+FA.FB)
As area(BFC) = root(3)/4.FB.FC, we get
3(FB.FC+FC.FA+FA.FB) = 4.root(3).area(ABC)
and, with (1),  the value of k = FA+FB+FC

I suppose $p_1+p_2\ne 0$. Then, for the point $P_{12}\in P_1P_2$ which respects the relation $\frac{\overline {P_{12}P_1}}{\overline {P_{12}P_2}}=-\frac{p_2}{p_1}$ there is the relation $p_1LP^2_1+p_2LP^2_{2}=(p_1+p_2)LP^2_{12}+\frac{p_1p_2}{p_1+p_2}P_1P^2_2$. Therefore, we will find the geometrical locus of the point $L$ for which $(p_1+p_2)LP^2_{12}+\sum\limits_{k=3}^n LP^2_k=l^2_{12}$, where $l^2_{12}=l^2-\frac{p_1p_2}{p_1+p_2}P_1P^2_2$ (now we have $(n-1)$ points: $P_{12}$ and $P_k$, $k\in \overline {3,n}$) a.s.o.