Which do you like better? Personally, I like countdown a lot more than written.

countdown, much funner (I know funner is not a word :) )

COUNTDOWN FTW

don't really need to give explanations...

LOL everyone so far put countdown.
Including me.
Yay!

I like the target round best. Not only do you get more time, the problems are easy to do without that much calculation, and calculators are allowed so there's a lot less stress.

I don't really like countdown because I suck at it and it's hard to just get points when your opponent is just buzzing in so quickly without even having to think or perform scratchwork.

I chose sprint, because I don't really like countdown, not only because I suck at it, but because some people could have like bad vision, or be bad at reading or stuff like that, and be at a disadvantage. It is fun though.

I like the sprint because thiry questions in forty minutes seems interesting, and if there are hard questions, it's somewhat challenging. Also, you can still get a few wrong without losing too much opposed to target, where just a couple of questions can basically determine making it or not.

Team because there are pretty hard questions sometimes.

CD is beast.
As long as you can see the problems.

sprint because i suck at cd(i got pwned by a 6th grader.)

i like written better, speed is not my thing.

[quote="ZhangPeijin"][quote="illusiat"]team because alkjash does all the problems :lol: . I just sit around and check.[/quote]
Meh, on my team, Me and Jason just sat there doing 1-5, and Alan and Allison spent their time arguing over the answer to question 10.[/quote]

can you mention that?

I totally agree with ZhangPeijin. Sometimes team round sucks when you're really good and your team is really bad...that means you must do all the problems yourself while the rest of your team just sits around at fights so much. That happened last year when I was a sixth grader in 2007, our team was so bad that an eighth grader and a seventh grader fought a lot and we only got 6 problems right on the team round.

Math Geek, you're so fast at the buzzer, it was really no contest to who was going to win CD.

BTW, I think FTW really helps in CD. Not so much in target or sprint though.

kind of funny, but FTW made me slower at CD

I got faster again when I actually started to do countdown with my team.

(I got 1st place CD at state BTW)

I just buzz after 2 seconds go bye if the problem looks easy, and finish solving in next 4-5 seconds. Then I start to speak. I can get 80% of the problems in 6-7 seconds and 10% in 10 or under.

[quote="gauss1181"][quote="hm29168"]That's a pretty common technique. I think if the MC people make a rule about instantaneous response, it would kill a bunch of people.[/quote]
Why? Are you serious that it's a common technique when people hit the buzzer before the judge even finishes reading the question?[/quote]
Everyone does that.
Anyways, basically, my best advice is to hit the buzzer before you get the answer.

That seems to be too late, even for me and the kids in my chapter. There's this guy from IL named Elliot Chanen who always pressed the buzzer right before the judge was halfway through reading the question. Plus, he got a whole string of problems right and always got 1st countdown.

[quote="Ihatepie"]I hit it before the judge finishes his first sentence  :D[/quote]

I think if the question is easy, I usually buzz in halfway through the judge's first word or so, not to be immodest though.  The questions are just too easy and usually involve multiplying two numbers together to get an answer.  I always pwn at that. :D

It was kind of funny, on one of the problems, I thought it was amazingly easy, so I hit the buzzer, but a second later I realized I had misread it. So when it came time to say the answer I said, "I made a mistake."
 :lol:

That happened again on another question, but that time I got it right.

Countdown round is really high pressure, and for some reason I do really well in math under high pressure (or so I have seen in MC), and its the most fun

Sprint kicks behind.
I always do good in Sprint,terrible in Target, terrible in Team,and OK in countdown

I always do good in sprint, fail target, and become second to alan in CD.
...
That's in our school though.

yeah, your official countdown stats are like...

[quote="numberdance"]I like sprint best.
I get going too fast in countdown - I'm fine against the first two people, then I mess up against the third. Almost always.
I'm never really sure how I did on target after it's over. Two times can feel exactly the same and one time I'll come out with a 16 and another with an 8.
And team round is just...team round. I'd rather make a stupid mistake on an individual problem and get annoyed with myself for it later than make a stupid mistake on a team problem and have three other people know that I made a stupid mistake on that problem.

Sprint doesn't have any of those problems. :lol:[/quote]
bear with me...i mess against person 2 :oops:


Putnam
Support
search
set theory
graph theory

Hi, I was wondering if anybody here is applying, has applied, currently is in Brown/Emory, or just knows this info. I was just wondering, how good are the Math Departments at both these colleges? Also, if anyone has insights on how these colleges did on the Putnam that would be great too...i can't find anything on the web...

There is no comparison as far as faculty research quality goes: it's Brown by a landslide.  As for quality of undergraduate education, who knows... but the students at Brown are probably better (basing my opinion on admissions standards).

I'm a sophomore at Brown. The math department offers many courses, but most notably missing, I think, are set theory and graph theory classes. You can check boca.brown.edu for what courses are offered this and last year. Be sure to also look under the Applied Math and Comp. Sci. depts.; those are separate from the Math dept.

From my experience, the professors are intelligent and engaging. Quite a few are eager to help undergraduates in pursuing research. There are also offices/organizations within the college that support students doing research; search the Brown website for UTRA.

Unfortunately, by skimming recent Putnam results, I couldn't find anyone from Brown placing very well. There are weekly Putnam sessions led by a professor in the fall. He picks interesting problems, from old contests, I guess, on which we work for a week or two and discuss. On good weeks, we have about half a dozen students show up. For the actual contest, I think about 10 people took it, most of them(us) being freshmen.

If you're interested, math.brown.edu has a list of most of the faculty and links to their websites. Many of them put up their course syllabi, worksheets, even textbooks-in-progress, for download. While you're there, also look up the DUG and SUMS for seminars and lectures that were offered or will be offered in the near future.


well i was wondering who is going for the Chem Olympiad this year??

i have asked my teacher to e-mail the coordinator and stuff...

heh, i just guessed that (logically it SHould be United States Chemistry Olympiad) :P


Let $ a,\ b,\ c,\ d$ be postive real numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 10$.

Prove that $ \frac {3}{a^2 \plus{} a \plus{} 1} \plus{} \frac {3b \plus{} 8}{b^2 \plus{} b \plus{} 1} \plus{} \frac {8c \plus{} 15}{c^2 \plus{} c \plus{} 1} \plus{} \frac {15d \plus{} 24}{d^2 \plus{} d \plus{} 1}\geq 10$.

Let $ a_k>0\ (k\equal{}1,\ 2,\ \cdots n)$ with $ \sum_{k\equal{}1}^n a_k\equal{}\frac{n(n\plus{}1)}{2}$

Prove that $ \sum_{k\equal{}1}^n \frac{(k^2\minus{}1)a_k\plus{}k^2\plus{}2k}{a_k^2\plus{}a_k\plus{}1}\geq \frac{n(n\plus{}1)}{2}$

[quote="kunny"]Let $ a_k > 0\ (k \equal{} 1,\ 2,\ \cdots n)$ with $ \sum_{k \equal{} 1}^n a_k \equal{} \frac {n(n \plus{} 1)}{2}$

Prove that $ \sum_{k \equal{} 1}^n \frac {(k^2 \minus{} 1)a_k \plus{} k^2 \plus{} 2k}{a_k^2 \plus{} a_k \plus{} 1}\geq \frac {n(n \plus{} 1)}{2}$[/quote]
$ \sum_{k \equal{} 1}^n \frac {(k^2 \minus{} 1)a_k \plus{} k^2 \plus{} 2k}{a_k^2 \plus{} a_k \plus{} 1}\geq \frac {n(n \plus{} 1)}{2} \equal{} \sum_{k\equal{}1}^n k \iff$
$ \sum_{k \equal{} 1}^n \frac {(a_k\minus{}k)(a_kk\plus{}k\plus{}1)}{a_k^2 \plus{} a_k \plus{} 1} \leq 0 \iff$
If $ a_k > k$ then $ a_kk\plus{}k\plus{}1 < a_k^2\plus{}a_k\plus{}1$ and if $ a_k < k$ then $ a_kk\plus{}k\plus{}1 > a_k^2\plus{}a_k\plus{}1$. Hence $ \frac{(a_k\minus{}k)(a_kk\plus{}k\plus{}1)}{a_k^2\plus{}a_k\plus{}1} \le a_k\minus{}k$ with equality iff $ a_k\equal{}k$. Which concludes the problem since $ \sum_{k\equal{}1}^n a_k\minus{}k \equal{} 0$. Equality iff $ a_i\equal{}i \forall i\equal{}1,2,..,n$.

Good! :)

Here is my solution:

$ a_k\minus{}2k\plus{}\frac{(k^2\minus{}1)a_k\plus{}(k\plus{}1)^2\minus{}1}{a_k^2\plus{}a_k\plus{}1}\equal{}\frac{(a_k\minus{}k)^2(a_k\plus{}1)}{a_k^2\plus{}a_k\plus{}1}\geq 0$

$ \therefore \sum_{k\equal{}1}^n \frac{(k^2\minus{}1)a_k\plus{}(k\plus{}1)^2\minus{}1}{a_k^2\plus{}a_k\plus{}1}\geq \sum_{k\equal{}1}^n (2k\minus{}a_k)\equal{}0$.

[quote="kunny"]Good! :)

Here is my solution:

$ a_k \minus{} 2k \plus{} \frac {(k^2 \minus{} 1)a_k \plus{} (k \plus{} 1)^2 \minus{} 1}{a_k^2 \plus{} a_k \plus{} 1} \equal{} \frac {(a_k \minus{} k)^2(a_k \plus{} 1)}{a_k^2 \plus{} a_k \plus{} 1}\geq 0$

$ \therefore \sum_{k \equal{} 1}^n \frac {(k^2 \minus{} 1)a_k \plus{} (k \plus{} 1)^2 \minus{} 1}{a_k^2 \plus{} a_k \plus{} 1}\geq \sum_{k \equal{} 1}^n (2k \minus{} a_k) \equal{} 0$.[/quote]

This is more efficient than how I would solve this  :maybe:

[quote="kunny"]Sorry, I should have posted my solution after looking at your solution, earldbest.

I would like to see your solution by Cauchy-Schwarz.[/quote]

$ \sum_{k\equal{}1}^{n}\frac{(k^{2}\minus{}1)a_{k}\plus{}k^{2}\plus{}2k}{a_{k}^{2}\plus{}a_{k}\plus{}1}\geq\frac{n(n\plus{}1)}{2}$
$ \Rightarrow \sum_{k\equal{}1}^{n}\frac{[(k\plus{}1)a_{k}\plus{}1]^2}{a_{k}^{2}\plus{}a_{k}\plus{}1}\leq\sum_{k\equal{}1}^{n}(k^{2}\plus{}k\plus{}1)$

for $ k\equal{}1,2,...,n$

$ k \frac{a_k^{2}}{a_k^{2}/k} \plus{} \frac{a_k^{2}}{a_k} \plus{} \frac{1}{1} \ge \frac {[(k\plus{}1)a_{k} \plus{}1]^2}{a_k^{2}\plus{}a_{k}\plus{}1}$

equality holds if $ a_{k}\equal{}k$ for $ k\equal{}1,2,...,n$

it's pretty much the same, that's why I like your solution... :D


[quote="Sunjee"]$ a,b,c$ positive real numbers then
\[ 3\geq 2\left(\frac {2\left(\frac {1}{a} \plus{} \frac {2}{a \plus{} b} \plus{} \frac {3}{a \plus{} b \plus{} c}\right)}{3\left(\frac {1}{a} \plus{} \frac {2}{a \plus{} b}\right)}\right) \plus{} \frac {\frac {1}{a} \plus{} \frac {2}{a \plus{} b} \plus{} \frac {3}{a \plus{} b \plus{} c}}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}\]
[/quote]
Nice inequality, my dear friend. I have a solution but an ugly one. See it here: http://can-hang2007.blogspot.com/2010/01/inequality-54-b-sanchir.html


Ross Mathematics Program
HCSSiM
summer program
Mathcamp

Hi folks,

There is an absolutely outstanding opportunity coming in October. Dr. Dana Randall from Georgia Tech has been invited to give this years Arnold Ross Lecture and even better it's in Augusta! There is also a competition called "Who wants to be a Mathematician?" In this competition you can win up to $ \$$3000. I don't know who is eligible for the competition. It says "local" students. This is one of the top lectures of the year and should be really cool. The date is Thursday, October 29th. I'll post more information if/when I get it, but I'm hoping to go myself.

Past Arnold Ross Lectures and information on this one are located at:
[url]http://www.ams.org/meetings/ross-lect.html[/url]

The flyer for the event is in the attachment.

Tom

Just after I leave for college!  I know of several of the past lecturers.  David Kelly (last year's lecturer) is the founder of [url=http://www.hcssim.org]HCSSiM[/url], an excellent summer program I attended in 2007.  He has a flair for presenting things.  Barry Mazur and Ken Ribet are renowned mathematicians who played major roles in the proof of Fermat's Last Theorem.  And of course Arnold Ross was the founder of the [url=http://www.math.ohio-state.edu/ross/]Ross summer program[/url] for high school students.  I doubt I'll be able to fly down from Bloomington in the middle of the semester, but this sounds really cool.

P.S.  You guys (students) should apply to summer math programs for next summer (e.g., [url=http://www.mathcamp.org/]Canada/USA Mathcamp[/url], where I am right now with Billy and Val and Andrei, and also HCSSiM and Ross).  It is hard to describe how much fun they are, but I've been to four different ones!  Also, rising juniors should apply to [url=http://www.cee.org/programs/rsi]RSI[/url], because it's amazing, you can only go in the summer after your junior year, and it's free if you get in (modulo air fare, spending money, etc.), although that is not the easiest thing to do.


[quote="xxxyyyy"]$F(x)$ is the last $4$ digits of $1+2+3+4+\cdots+(x-1)+x$

$G(x)$ is the last $3$ digits of $F(1)+F(2)+F(3)+F(4)+\cdots+F(x-1)+F(x)$

$H(x)$ is the last $2$ digits of $G(1)+G(2)+G(3)+G(4)+\cdots+G(x-1)+G(x)$

Find $H(12345678)$[/quote]
[hide="I think this might be a good idea"]
$\binom{x}{1}=x$+hockeystick[/hide]

what exactly to you mean by "hockeystick"?

I think he means hockey stick Identity. Sorry, but I have no other useful things to post. :(


induction
LaTeX
combinatorics proposed
combinatorics

My site?  I don't remember posting any training documents on my site :?

Sorry! I made a mistake: I took it from a website which had a link leading to it from your site :). My bad :D.

ok, easy problem...
or, as mindflyer says, it's not at all a problem, if i solved it  :D
no good, grobber...
anyway..
it's easy to show that the sum of the elements of T (and, by induction, the elements of T^n) is always 0.
so if x1+x2+x3+x4 doesn't equal 0, for sure T^n won't equal (x1,x2,x3,x4).
then, let's consider x1+x2+x3+x4 = 0.
of course, T equals (x1,x2,x3,x4) if and only if it's (0,0,0,0).
else, let's denote x1+x3 = a, and x2+x4 = b.
we may observe that T = (a,b,a,b), and prove by induction that T^n = (ka,kb,ka,kb), where k = 2^(n-2), for n>1. in fact x1+x2+x3+x4 = 0, so a = -b.
but it's easy to show that x1 = x3 = ka = 2k1 if and only if x1 = x3 =0.
it's the same for x2 and x4...
sorry for notation and for language, but i'm too lazy to learn latex and too italian to use a correct english :D


inequalities
LaTeX
inequalities unsolved

Let a,b,c≥0.Prove that:
$ \sum \frac{b\plus{}c}{2a^2\plus{}bc}$≥$ \frac{6}{a\plus{}b\plus{}c}$

Applying Caushy-Shwartz inequallity on Left side and after calculations... you get  T[300]+2 T[111] >= 3 T[210] I don't think this inequallity is true...

[quote="SUPERMAN2"]Let $ a,b,c≥0.$Prove that:
$ \sum \frac {b + c}{2a^2 + bc}\geq\frac {6}{a + b + c}$[/quote]
After full expanding we obtain:
$ \sum \frac {b + c}{2a^2 + bc}\geq\frac {6}{a + b + c}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^5b+3a^4b^2-4a^3b^3-a^4bc+5a^3b^2c-4a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow(a-b)^2(a-c)^2(b-c)^2+\sum_{sym}(a^5b+2a^4b^2-3a^3b^3+3a^3b^2c-3a^2b^2c^2)\geq0,$ which is obvious.

[quote="arqady"][quote="SUPERMAN2"]Let $ a,b,c≥0.$Prove that:
$ \sum \frac {b + c}{2a^2 + bc}\geq\frac {6}{a + b + c}$[/quote]
After full expanding we obtain:
$ \sum \frac {b + c}{2a^2 + bc}\geq\frac {6}{a + b + c}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^5b + 3a^4b^2 - 4a^3b^3 - a^4bc + 5a^3b^2c - 4a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow(a - b)^2(a - c)^2(b - c)^2 + \sum_{sym}(a^5b + 2a^4b^2 - 3a^3b^3 + 3a^3b^2c - 3a^2b^2c^2)\geq0,$ which is obvious.[/quote]
Thanks.But is there another proof without expanding the ineq?

From CS we have
\[ \sum_{cyc} \frac{b\plus{}c}{2a^2\plus{}bc} \ge \frac{4(a\plus{}b\plus{}c)^2}{\sum_{cyc} 2a^2b\plus{}2a^2c\plus{}b^2c\plus{}bc^2} \equal{} \frac{4(a\plus{}b\plus{}c)^2}{4\sum_{sym} a^2b} \ge \frac{6}{a\plus{}b\plus{}c}\]
\[ \sum_{cyc} a^3 \plus{} 3\sum_{sym} a^2b \plus{} 6abc  \ge 6 \sum_{sym} a^2b\]

which follows from Schur inequality.

Am I right?

No, you are not right, because your inequality is clearly false. Firstly, $ \sum (2a^2\plus{}bc)(b\plus{}c)\equal{}3\sum_{sym}a^2b$ and the last inequality (even with $ 3\sum a^2b$) follows not from Schur.

OMG, surely $ 3\sum_{sym} a^2b$  :oops:
It should be
 \[ 2\sum_{cyc}a^3 \plus{} 6\sum_{sym} a^2b \plus{} 12abc \ge 9 \sum_{sym} a^2b\]
\[ \left( \sum_{cyc} a^3 \minus{}3 \sum_{sym} a^2b \plus{}3abc \right) \plus{} \sum_{cyc} a^3 \plus{} 9abc \ge 0\]

Is it still wrong?

Yes, just take $ c = 0$. Then you get
$ 2a^3 + 2b^3 - 3a^2b - 3ab^2 \ge 0$,
and that's not true.  :wink:

@micf: how do you center the $ \text{\LaTeX}$-formulas?
edit: Thanks  :)

I'll just write down my solution. It's not nice, but I don't expand^^ So maybe Superman2 will like it :p

After some work (not too much) one gets the equivalent inequality
\[ \sum (b - c)^2(b + c)(b + c - a)(2a^2 + bc) \ge 0.\]
Let wlog $ a \ge b \ge c$ and write the inequality as
\[ \sum (b - c)^2S_a \ge 0,\]
where $ S_a = (b + c)(b + c - a)(2a^2 + bc)$. $ S_b$ and $ S_c$ are defined analogously. Clearly $ S_b$ and $ S_c$ are greater or equal to zero. It's left to prove that $ a^2S_b + b^2S_a \ge 0$. The last inequality is equivalent to
\[ (a + c)(a + c - b)(2a^2b^2 + a^3c) \ge (b + c)(a - (b + c))(2a^2b^2 + b^3c).\]
If $ b + c \ge a$, the inequality is clearly true. So assume that $ a > b + c$. The prove follows from
\[ 2a^2b^2 + a^3c \ge 2a^2b^2 + b^3c \Leftrightarrow c(a^3 - b^3) \ge 0\] and
\[ (a + c)(a + c - b) \ge (b + c)(a - (b + c)) \Leftrightarrow (a - b)^2 + ac + 2c^2 + bc \ge 0,\] respectively.

Oh, now I see it... It was so stupid :rotfl:

To center the $ \text{\LaTeX}$ formula use double dollar symbol instead of one


Sorry if this has already been posted.

Compute the exact value: $\tan\left(\frac{3\pi}{11}\right)+4\sin\left(\frac{2\pi}{11}\right)$.

Yeah, posted before : http://www.mathlinks.ro/Forum/viewtopic.php?t=46202


Find the sum of all possible values of $ (x\plus{}y)$ if $ x^2\plus{}y^2\equal{}26$ and $ xy\equal{}5$.

hello
[hide="solution"]
$ x\plus{}y$=+6 or -6
[/hide]
Thank u.

except it asks for the sum of the possible values, which is $ \boxed{0}$


AMC
AIME
number theory
modular arithmetic
\/closed

Do you think that I will be ready for this class? I know pretty much all intro NT I think except for modular arithmetic (which I know is a large part unfortunately).

I never went through the Intro to NT book though. Whatever I know is just the accumulation of what I learned in my math-deprived years of mathcounts.

I really want to take this class, because NT has officially fallen to my worst type of question. I can easily solve a counting AIME #8, but I cannot usually solve a NT AIME #3.

You should also ask yourself if you need the class. I'm still not sure whether to take it or not myself. I already signed up for amc 12 so i dont know...

Yeah, I'm planning on taking the AIME problem class as well. I think this would be a good idea though, given my complete lack of competence when working with NT.

If you don't know modular arithmetic, then you're not ready for Intermediate Number Theory.  However, you can probably pick up what you need of modular arithmetic with careful study of the relevant chapter of the Introduction to Number Theory book for a few weeks.

Does anyone have an idea of when the Intermediate NT book will become available?


Its just a bunch of questions that I am not exactly able to give the proper reasoning or I want to pose to you guys. All little questions but nice concepts involved here.

1) s-orbital has the least energy compared to p and d.
Then why is energy level of sp>sp2>sp3

2) Is Paranitrophenol less acidic than Ethanoic acid? with proper reasoning.

3) Arrange these carbene (derivatives) in decreasing order of stability.

CH2, CF2, CCl2

4) Which is more stable Vinyl Free Radical or Benzyl. In a book, it is given ans is Vinyl.

5) Why is Pyridine a weaker base than aliphatic amines?

i think the answer of the 3rd question is.....
 CCl2<CH2<CF2.........

wrong but i wont answer as shreyas and I discussed this only yesterday  :D

[quote="chemrock"]I think the order might be CH2>CF2>CCl2

It is because CH2 there are no -I groups as such so it is relatively stable.

In case of CF2 there are 2 effects operating the inductive effect and the backbonding. In case of -F it has lone pairs in the similar 2p orbitals as cabon and so the backbonding is much more effective. Though backbonding is there it is only an additional effect so CH2 is more stable.

In CCl2 the backbonding is very low so it is least stable.[/quote]

wrong again  :D

[quote="valeriummaximum"]the stability order is CF2<CCl2<CH2.........[/quote]

Wrong again. Plz dont try all the permutations  :D . And plz give reasonings for your answer  :D

i beleive it's $ \ce{CCl2} > \ce{CH2}>\ce{CF2}$

[quote="pardesi"]i beleive it's $ \ce{CCl2} > \ce{CH2} > \ce{CF2}$[/quote]

I think its wrong again . Now all permutations are exhausted  :rotfl:

now are you going to say that all 3 are equally reactive????

no and stop raving  :D

And btw read the ques properly. Stability is asked  :D

And chemrock had right reason but wrong answer  :P

[quote="valeriummaximum"]i thought that Cf2 being the most reactive will be the least stable followed by CCL2 followed by CH2[/quote]
i really don't like this statement...often told by me friend who infact were told in their tutions....
stability is just a relative comparison...and in some sense though obvious the question is wrong...
it should have been sompare these free radicals from....(this from is important)
for ex. if i compare the stability of dichlorocarbene from dichloro methane  with methylene from benzene then obviously the order may go haywere...
what is important to understand is that stability is a relative copmparison of potential energies of molecules witha certain basic molecule hence sometimes the ctivation energy required for a reaction..thus predicting the rate of a reaction...
valeriummaximum take this example a tertiary carbocation is more stable than a primary as compared to their respective alkane origins hence is formed faster in SN1 while the reverse hold true for SN2 so there is no such rule of the Thumb

CF2>CCL2>CH2 based on chemrock's reasoning.....

P.S added some of my reasonings too

Ok here goes the answer $ \ce{CF2} > \ce{CH2} > \ce{CCl2}$
This is bcos of back bonding as Chemrock said. Ad btw @ chemrock if backbonding is only secondary then why is lewis acid strenth reversed for Boron Halides  :D
Before this ranga even added the 5th wrong permutation :rotfl: .

NOTE: The author of this post is only stating the answer as given by RSM and shall take no credit for wrong information .  :P

for the fifth question i think the answer is because Nitrogen in pyridine shows basic character only due to the presence of lone pair of electrons...
Since these electrons are delocalised and not available for  donating i think pyridine is weaker base than aliphatic amines...

then why don't we use $ \ce{CF3H}$ for Riemer-Tiemann and like reactions....assuming we have a strong base to cleave the $ \ce{C\minus{}F}$ bond....then the produced intermediate should be more stable than due to dichlorocarbene

what do you mean by "" is Benzyl by a mile ""

means there is huge energy difference between the difference of energies between them and their respective origins

[quote="Cyber-Shreyas"]5) Why is Pyridine a weaker base than aliphatic amines?[/quote]

Because pyridine's nitrogen is sp2, while aliphatic amines have a sp3 nitrogen.

[quote="Cyber-Shreyas"]4) Which is more stable Vinyl Free Radical or Benzyl. In a book, it is given ans is Vinyl.[/quote]

Benzyl, of course, since the electron can be delocalized (which cannot happen on a vinyl radical).

[quote="Cyber-Shreyas"]3) Arrange these carbene (derivatives) in decreasing order of stability.

CH2, CF2, CCl2[/quote]

CF2 is the most stable, since fluorines's orbitals are almost the same size as the one's of carbon, and conjugation is most effective.

[quote="Cyber-Shreyas"]2) Is Paranitrophenol less acidic than Ethanoic acid? with proper reasoning.[/quote]

We can say that nitrophenol is more acidic than phenol. To compare p-nitrophenol with acetic acid, consult a table. However, 2,4,6-trinitrophenol (picric acid) is a reasonably strong acid.

[quote="Cyber-Shreyas"]1) s-orbital has the least energy compared to p and d.
Then why is energy level of sp>sp2>sp3[/quote]

Where are the electrons closer to nucleous?

To Rangarajan/Sathish/Valeriummaximum
[hide]Before I type what I wanted to type. 3! = 6. Incase you want to know how n! is defined as the continuous product of all natural numbers from 1 to n, where n!=0. (Refer [url=http://en.wikipedia.org/wiki/Factorial]Wikipedia[/url] if you donot  believe me). Back to Topic, My question on the order of stability had only 3 carbenes (or derivatives) and there are only 6 ways to arrange them. You seem have exhausted all of them and conveniently edited your first post to post the correct ans. :(. And how stupid of you to say that the lone pair is delocalized. When a Pi bond and lone pair are present at the same place, the delocalization of lone pairs is suppressed.. If you count the lone pairs too, then the compound will become anti-aromatic. Obviously you were blindly answering the first question with the idea of Inductive effect and second with a blind reason that school questions require. Obviously shows you NV chemistry more than you understand it conceptually. Anybody can remember n reactions but knowing each's significance and answering conceptual questions based on mechanism is very difficult which one may not be able to do if he/she nvs the mechanism. You often tease me as being stupid in chemistry, I have never denied it but whatever I study, I analyze and raise questions to myself. I may not know loads of reactions like Suzuki Coupling or ABC-BBC rearrangement but whatever little I know, I apply. :) Time for introspection, I think. Sorry. :(
[/hide]

That Carbene was a nice trick question.
Ans: CF2>CH2>CCl2. Rakesh, you got the point I wanted but lost it then.

Pyridine: Carculs ans is perfect. sp2, is all that I could think of as well, if anybody has a better ans please post it.

4. Sorry I meant Allyl vs Benzyl. In a book it was given Allyl has greater stability. Wanted to verify that.

2. Ans in book is paranitrophenol. I guessed it was stupid comparing these two, but generally lots of people compare aliphatic and aromatic acids, so whats the general trend.

1. Hmmm...I know you will say sp3 but isn't 2s having less energy than 2p???

In my opinion benzyl radical is more stable than allyl radical, for obvious reasons.

[quote="Cyber-Shreyas"]CF2>CH2>CCl2[/quote]

Wrong. CCl2 is more stable than CH2. Do you know any reaction where free CH2 is involved? You don't. Do you know any reaction where free CCl2 is involved? Yes you do.

To Cyber Shreyas...
[hide]you tell that i nv chemistry isnt rather the other way..................U NV chemistry........
not me.......i may not be good in logic .......but i don Nv any subject..(except English)...........
you Nv chem,maths and in physics,you pose as if you are conceptual at evertything.........
you were so afraid to writer today's FT4...........you ran away you brat and afool of a tukh....[/hide]

@Carcul: Okay, The ans was given the book that CF2>CH2>CCl2. The logical explanation we (me and madness) could make out was because of backbonding in CF2 and lack of it in CCl2.

[quote="valeriummaximum"]
you were so afraid to writer today's FT4...........you ran away you brat and afool of a tukh....[/hide][/quote]

I was afraid to writea?? Thats news to me. I couldn't obviously due to health reasons.... I am not like you da!! :))

Yes, call me NV, call me anything but I think the facts in this thread and in other places speak for themselves.

And lets keep this fight elsewhere, and not spoil this thread.

If u ask me this entire topic is crap because the carbenes will readily undergo dimersiation to give stable products and only their reactivity can be discussed. If you ask me then I will say the proper question should be:

What is order of the ractivity of: [size=150][b][i]CCl2 , CF2 , CH2 [/i][/b][/size] ?

[quote="Carcul"]Reactivity is a measure of stability.[/quote]

Hmmm....Statement I think is slightly ambiguous. We use it as a measure of stability but then again.

I don't think reactivity and stability mean the same. A more stable product can still be highly reactive and we are just talking about relative stability and relative reactivity. For example,
t-butyl alcohol and ethanol undergoes elimination. We all know that t-butyl ion is stable than ethyl carbocation(though not formed), that means you are trying to say that t-butyl ion being more stable should be less reactive in the coming step. So I basically don't agree with REACTIVITY === STABILITY


i) Prove that if $a,b,c\geq 0$, then
\[\frac13(a+b+c)^{2}\geq ab+bc+ca+\frac18\bigl(|(a-b)(b-c)|+|(b-c)(c-a)|+|(c-a)(a-b)|\bigl). \]

ii) Prove that if $a,b,c\geq0$, then
\[\sqrt{3(a^{2}+b^{2}+c^{2})}\leq a+b+c+\Bigl(\frac{\sqrt3-1}2\Bigl)\left(|a-b|+|b-c|+|c-a|\right).\]

Assume $a \geq b \geq c$ for both inequalities. We can replace $1 \over 8$ with $1 \over 6$ in i).

The second one was posted in Unsolved Section long time ago.

The second one is solved now, but the following is still unsovled
\[\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}\leq\frac12(a+b+c+d)+k(|a-b|+|b-c|+|c-d|+|d-a|+|a-c|+|b-d|).\]
Find the least $k$ such that the inequality is true for all $a,b,c,d\geq0$.

Here is my solution for the second one, Thuan.
WLOG, we can assume that $a \ge b \ge c$, then the ineq becomes
\[\sqrt{3}a+b+(2-\sqrt{3})c \ge \sqrt{3(a^{2}+b^{2}+c^{2})}\]
\[\Leftrightarrow (\sqrt{3}a+b+(2-\sqrt{3})c)^{2}\ge 3(a^{2}+b^{2}+c^{2}) \]
\[\Leftrightarrow \sqrt{3}ab+(2-\sqrt{3})bc+(2\sqrt{3}-3)ac-b^{2}-2(\sqrt{3}-1)c^{2}\ge 0 \]
But this ineq is obviously true. Because
\[\sqrt{3}ab+(2-\sqrt{3})bc+(2\sqrt{3}-3)ac-b^{2}-2(\sqrt{3}-1)c^{2}\ge \sqrt{3}b^{2}+(2-\sqrt{3})bc+(2\sqrt{3}-3)bc-b^{2}-2(\sqrt{3}-1)c^{2}=(\sqrt{3}-1)(b-c)(b+2c) \ge 0 \]

We are done. :)

[quote="pvthuan"]The second one is solved now, but the following is still unsovled
\[\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}\leq\frac12(a+b+c+d)+k(|a-b|+|b-c|+|c-d|+|d-a|+|a-c|+|b-d|). \]
Find the least $k$ such that the inequality is true for all $a,b,c,d\geq0$.[/quote]
Naturally, assume $a \geq b\geq c \geq d$, we easily obtain the inequality is true for any $k \geq{1 \over 6}$.


calculus
integration
trigonometry
analytic geometry
search
calculus computations

We have the following double integral:

$ \int_0^5 {\int_0^{\sqrt {25 \minus{} y^2 } } {xe^y \,dx} \,dy} .$

I want to apply a polar transform. Since we all know that $ x\equal{}r\cos\theta$ & $ y\equal{}r\sin\theta,$ what would be the new integration limits?

[quote="Hetidemek"]We have the following double integral:

$ I = \int_0^5 {\int_0^{\sqrt {25 - y^2 } } {xe^y \,dx} \,dy} .$

I want to apply a polar transform. Since we all know that $ x = r\cos\theta$ & $ y = r\sin\theta,$ what would be the new integration limits?[/quote]

[hide]So we have:

$ I = \int\int_D xe^y ~dx~dy = \int_0^5 \int_0^{\sqrt{25-y^2}}xe^y ~dx~dy$

So:

$ D=\{ (x,y) \in \mathbb{R}^2 ~| ~~0<=y<=5, ~~0<=x<=\sqrt{25-y^2} \}$

So what we're integrating over appears to be a quarter of a circle of radius $ 5$ in the positive quadrant.  So:

$ D=\{ (r,\theta) \in \mathbb{R}_0^{+} \text{x} ~\mathbb{R} ~| ~~0<=\theta<=\frac{\pi}{2},  ~~0<=r<=5 \}$

So:

$ I = \int\int_D xe^y ~dx~dy = \int_0^{\frac{\pi}{2}} \int_0^5 r \cos \theta e^{r \sin \theta} ~r~dr~d\theta$
$ I = \int_0^{\frac{\pi}{2}} \int_0^5 r^2 e^{r \sin \theta} \cos \theta ~dr ~d\theta$[/hide]

Okay, thanks.

How'd you work the integration limits with $ x\equal{}r\cos\theta$ and $ y\equal{}r\sin\theta\,?$

I've trying substituting them into the polar coordinates but I don't get something nice.

The thing is: polar coordinates may not to be the nice way to solve this. We can just do it exactly as originally written:

$ \int_0^5 {\int_0^{\sqrt {25 - y^2 } } xe^y \,dx \,dy =}$

$ \int_0^5\left[\left.\frac{x^2}2e^y\right|_{x=0}^{x=\sqrt{25-y^2}}\ \right]\,dy=$

$ \frac12\int_0^5(25-y^2)e^y\,dy$

And then we have an integral that we should know how to do (integration by parts will be involved).

Yes, it's the straightforward way.

It's just that, when I saw [url=http://www.mathlinks.ro/viewtopic.php?search_id=1095939202&t=26197]this[/url] thread, I was wondering how liyi did the polar transform.


Find the function  f: Z--->Z so that

 f (x**2 + y**2) = [ ((f(x) +f(y) +1)/2]

where [x] is the integer part  of x.

From "Arhimede" 2-3 , 2003

I think there was another condition too: $f(xy)=f(x)f(y)$. And I posted it extremely long time ago in the topic Constant, non-constant...
 Here it is: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=constant&t=2354


After completing the Olympiad Problem Solving course , don't we get a certificate? when will this come?

Certificates were mailed out at the end of last week.  They should arrive within the next week.


[quote="aidan"]There are 4 couples and 2 children. How many ways are there to arrange them seated around a round table so that no more than 2 men are seated next to each other?[/quote]

[hide="Ummm..."]I think the number of ways to seat three and four men together are $ \binom{4}{3}\cdot 3!\cdot (8\minus{}1)!$ because choosing three men out of four guarantees three men are sitting together and you treat three of them as one person, seat them in $ 3!$ ways and then there are 8 other "people" and $ (8\minus{}1)!$ ways to seat them around a circle? So then the answer is $ (10\minus{}1)!\minus{}\binom{4}{3}\cdot 3!\cdot (8\minus{}1)!\equal{}\boxed{241920}$?[/hide]

Do the children, if they are male, count as men?

ch1n353ch3S54a1l, your answer is pretty close but differs by about 10000 =P I think you missed out the cases when there are two groups of two men sitting together (while the pairs of men are not together) which isn't allowed by the problem context.

No, the children do not count as men =P


geometry
analytic geometry
algorithm
advanced fields
advanced fields theorems

Hi,
Is there a way to systematically find the area element in an arbitrary coordinate system? For example, how would one get (using some sort of algorithm) the surface element in spherical coordinates or cylindrical coordinates?
Thanks.

Could you please be slightly more explicit? I thought the Jacobian was for the volume element, not the area element.

Two-dimensional volume is called "area".
One-dimensional volume is called "length".
Different names, same concept.


Given positive integers a and b, with a>b, a positive integer n is
called attainable if n can be written in the form xa + yb with x and
y nonnegative integers. If there are exactly 35 non-attainable positive
integers, one of which is 58, find a and b.

 :!:

[hide="Solution"]It is very known that if $ (c,d) \equal{} 1$ and $ c, d > 1$, there are exactly $ \dfrac{(c \minus{} 1)(d \minus{} 1)}{2}$ unattainable numbers related with $ c$ and $ d$, all of them less than $ (c \minus{} 1)(d \minus{} 1)$.

If $ (c,d) > 1$, only multiples of $ (c,d)$ are attainable, so there is an infinity of unattainable numbers related with $ c$ and $ d$.

If $ c$ or $ d$ is $ 1$, there are no unattainable numbers related with $ c$ and $ d$.

Appliying the former results to problem we obtain $ a > b$, $ (a,b) \equal{} 1$ and $ (a \minus{} 1)(b \minus{} 1) \equal{} 70$. So the possibilities for $ a$ and $ b$ are $ 71$ and $ 2$, $ 36$ and $ 3$, $ 15$ and $ 6$, and, $ 11$ and $ 8$. The first couple make $ 58$ attainable, the second and third couples are not relatively prime.

To conclude that the fourth couple is the only solution we need to prove the equation $ 11x \plus{} 8y \equal{} 58$ has no solutions in non negative integers.  Suppose $ x$ and $ y$ do satisfy the equation. It follows that $ 11x\equiv 58\mod 8$, that is $ x\equiv 6\mod 8$. The least non negative integer $ x$ that satisfies that congruence is $ 6$ so $ x\ge 6$ and $ 11x \plus{} 8y > 58$. So there is no solution and $ a \equal{} 11$, $ b \equal{} 8$ is the only solution to the problem.

[/hide]

 :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Consider $ n \geq 2$ groups $ G_1, \ldots , G_n$, each having no non-trivial abelian quotient, and let $ H$ be a subgroup of $ G1 \times \ldots \times G_n$ such that every projection map $ H \to G_i \times G_j$ is surjective. Show that $ H \equal{} G_1 \times \ldots \times G_n$.

Counterexample: $ G_1\equal{}G_2\equal{}G_3\equal{}\mathbb{Z}/p\mathbb{Z}$, $ H\equal{}\{(a,b,c): a\plus{}b\plus{}c\equiv 0\}$.

Sorry, does that satisfy ALL the hypothesis?

Sorry for being dumb  :blush:

It probably doesn't as $ G/\{e\}$ should definitely be considered as a nontrivial quotient.


$ ABCDEFG$ is a regular heptagon inscribed in a unit circle centered at $ O$.  $ L$ is the line tangent to the circum-circle of $ ABCDEFG$ at $ A$, and $ P$ is a point on $ L$ such that $ \Delta AOP$ is isosceles. Let $ p$ denote value of $ AP\cdot BP\cdot CP\cdot DP\cdot EP\cdot FP\cdot GP$. Determine the value of $ p^2$.

[hide]Your value $ p^2$ is a symmetric polynomial in the $ 7$ roots of the polynomial $ z^7\minus{}1$.  So write it in terms of the coefficients of that polynomial.[/hide]

Please help a liitle further... I did not know how to proceed..


alright so i really don't know how to go about solving this question(source AMC 10):

How many three-digit numbers have at least one 2 and at least one 3?

Answer: 52

seems simple enough but i can't figure out where to start...

thanx

We will use a technique known as [i]complementary counting[/i].  That is we will find all numbers that [b]don't [/b]have any two's or three's.  (Our clue to try this is the words "at least").  Consider the first digit of the number.  It can be 1,4,5,6,7,8,9 for a total of 7 digits.  The second digit has 8 possibilities (can be a 0) and the third digit has 8 possibilities.  The total number of numbers that [b]don't[/b] satisfy the conditions is 7*8*8=448.  The number of total 3-digit numbers is 900, so our answer is :

total number of 3-digit numbers - number of 3-digit numbers that don't work = number of 3-digit numbers that do work

900-448=[b]452[/b]

This technique is very useful in counting and can also be applied to probability.

We see that there are a total of 900 3 digit numbers. Lets use complementary counting.
There are 8*9^2 numbers that don't contain a 2 or a 3.
There are 7*8^2 numbers that contain neither 2 nor 3.
Now the union of all these elements is
2*8*9^2(sets)-7*8^2(intersection)=842.
Now we subtract this from the total which is 900-842=56.
Or you could just brite force it.
if the number starts with 2 then there are 10*2 ways of having it.
If a number starts with 3 then there are 10*2 ways of having it.
If it ends in 2 and 3 then there is 9*2 ways of having it.
This adds up to 58.
We see that we double count:233,322,223,232,332,323.
So that is 58-6=52

First of all 452 is not the answer.
Secondly, You miss a lot of steps in your counting procedure.

varunrocks I don't understand how you found the union when using complementary counting. I understand that its sets - intersection. But why is it $ 2(8)(9^{2}) \minus{} 7(8^{2})$ , where does the 2 in front of the sets come from?

Personally i think this problem is easier to brute force.

if you are organized this is not that hard.

well we have to do case word

Case 1:1_ _
we would have 2 choices

Case 2:2_3

well since we already have a 2 in the hundreds digit we can have 203 213 223... till 293

case 3:23_

that would be 10-1=9 because we cannot coun 233.all the way from 230,231...239

case 4:3_ _

well this is going to be the same thing as the previous scenario so it is 10 +9=19

case 5:well this will be the same as Case 1 4_ _,5_ _,6_ _,7_ _,8_ _,9_ _ that would be to for eeach 32 and 23 so it would be 2x6=12

you will see that if you add 2 +10+9+19+12=52

as you can see with case work you can easily solve this problem

@billy314
Well-written solution, but I think you made a mistake:
[quote="billy314"]We will use a technique known as [i]complementary counting[/i].  That is we will find all numbers that [b]don't [/b]have any two's or three's.[/quote]
The problem asks how many three-digit numbers have at least one 2 and at least one 3, so the numbers we are looking for are the three-digit numbers do not have a 2 [b]or[/b] do not have a 3 (or both).

The number of three-digit numbers do not have a 2 is $ 8\cdot9\cdot9\equal{}648$, and the number of three-digit numbers do not have a 3 is also $ 648$. The number of three-digit numbers that do not have a 2 or a 3 is $ 7\cdot8\cdot8\equal{}448$. Thus, by PIE, we have
$ \text{\# in union}\equal{}\text{\# without a 2}\plus{}\text{\# without a 3}\minus{}\text{\# without a 2 or a 3}$.
$ \text{\# in union}\equal{}2\cdot648\minus{}448\equal{}848$.

Now we subtract to get the answer: $ 900\minus{}848\equal{}\boxed{52}$.
(This is the same as varunrocks' first solution.)

I [b]wouldn't[/b] recommend complementary counting for this one, since you have to think about more cases.

Case 1 : $ x23$

There are $ 9$ choices for $ x$ and $ 2$ ways to arrange the $ 2$ and $ 3$, therefore we have $ 18$ numbers.

Case 2: $ 2x3$

We have $ 10$ choices for $ x$ and also $ 2$ ways to arrange the other 2, so we have $ 20$ numbers.

Case 3: $ 23x$

$ 10$ choices fore $ x$, 2 for the other $ 2$, therefore $ 20$.

There are $ 6$ numbers that repeat so we get $ 18\plus{}20\plus{}20 \minus{} 6 \equal{} \boxed{52}$.


MATHCOUNTS HCSSiM AMC USA(J)MO USAMO email

Please post you score and state after the nats.

[hide="my EMBARRASSING RESULLTTSSS"] :(  :(  :(  :(  :(  :(  :(  :(  :(  :(  :(  :(  :( Individual Place :(  :(  :(  :(  :(  :(  :(  :(  :(  :(   :(  one  :(  :(  :(  :(  :(  :(  :(  :( hundred :(  :(  :(  :(  :(  :( eighty :(  :(  :(  :(  :(  :( four.  :(  :(  :(  :(  :(  :(  :(  :(  :(  :( Team :(  :(  :(  :(  :(  :(  :(  :(  :( forty :(  :(  :(  :(  :(  :( eighth :(  :(  :(  :( *cry really loud* *cry really loud* *cry really loud* *cry really loud* i know they SUCK I got 1 target dont know sprint or score. 4 team dont know score. State=delaware[/hide]

Well, Delaware, at least your pins were really nice! :) You should be happy to be where you are at. Plus, listen to this:

Sprint:18*1=18 points
Target:3*2=6 points
Individual score:18+6=24 points
Individual rank: 58th

Okay, I know, you're thinking "what's so bad about 58th?" Try being one place away from getting on the top 25% list. And what's worse: I tied with Amy Tai, who got 57th. (At least, according to the Oklahoma coach.) :(

Top 5 scores:
1.Sergei Bernstein-MA-38
2.NATHAN BENJAMIN-IN-36
3.Mark Zhang-TX-35
4.Patricia Li-CA-35
5.DAVID BENJAMIN-IN-35

GO MARK!!!!
He won the tiebreakers becuase he got a 25/5

Team: Nebraska

Individual: 26 points : 4 target, 18 sprint

Rank: 49th

Rank in team: 1st :lol:

Team rank: 31st -sigh-

The competition was real close(as I imagine it always is), I know for a fact that 39th place had one point higher than me.

30 points, MA, 4th place team, 14th place individual

Hmm..., maybe if I tried the nationals, I would get 1 sprint, 0 targets?
rank: #last for me?

[u][b]ALABAMA[/b][/u]
sprint:22
target:6
total:28
place: 42nd

team:37th

Hey, mathwizard, you got the same place as I did last year. :lol:

Team: Texas

Individual: 28 points : 4*2 target, 20 sprint

Rank: 30th

Rank in team: Tied for 3rd

Team rank: 1st :first:  :first:  :first:  :first:  :first:  :first:  :first:  :first:  :first:  :?  :?  :?  :?  :?  :?  :?  :?  :?  :?  :?  :?  :D

Overall i am relatively happy but...there was a slim possibility of making CD if i got one more target

[quote="yif man12"]Sheesh, what's up with the low scores? I remember getting 69th place with 33 one year...[/quote]
33 would have gotten you in top 10 (like 7th or 8th)

I made 20/3=26 (a stupid mistake and a misunderstanding kept my target down from 5, which would have been top 20). I got 45th. Team got 10th (NC). Hooray.

Awww, that's unfair.  If only my year's test were that hard... :(

I did that on at least two. Like I got brain-dead on the target about the point in the square and the distances to the vertices, and I guessed 19 at the last second because it looked right but the answer was 18. Or vice versa.

well if people knew my state score they would be like.."oh yeah i definitely shouldve beat her..this is so messed up"..but like....its not all proportional....like someone who does well at states shouldnt necessarily be the best at nationals...nationals is harder math, so maybe some people make mistakes at state and at nationals...but when other people cant do some naitonal problems that they can do that raises their score..u kno?

but yeah of course people who make 46 at states get in top 12..

Hey Frost13, you proctored me. I was the one that was in the 7th grade and i sat next to the girl.

My embarassing rank: 166th

Score: 14 ( 10/2 )
team round : 5 team rank : 44th

at least i wasn't the worst on the team.

wat did everybody get in state? just curious...

-rorri :?

Indiana had 46(pardha),45(me),41(nathan),41(yixin-man :D).

=(

63rd, 24.

I made 12 points of careless errors.

in state??!? or nat'ls? my ny team was:
andrew: 44,
varun: 37 or 38 i think
me: 36
woo-hyung: 35...
5. kevin, my teammate: 34...

-rorri :?

what state are you.
I got 40, 5th in IN, not complaining

Only? That's... insane.

Heh, I got 3 AIM usernames :? . "Only" would be appropriate there.

I was captain. Captain Daesun got one of his problems wrong, but we still got a 9. I corrected three problems, though, which kinda makes up for the one I got wrong.

[quote="jb05"][quote="solafidefarms"]Yes, that was me collecting emails.

If you know anybody's email address who was a national competitor, PM me with it please. I only got 132 emails.[/quote]

Only 132 emails??? At what point do you stop using the word only? And what's the point anyway? I might be interested, but it's not like I'm going to send out an email to all participants at 2005 national mathcounts. Anyway if you want mine I suppose I'll give it to you. But I won't tell you anybody else's on my team.[/quote]

You're the only one on your team I didn't get. :P

Why did I do it? Public service. Tell all of them to come be AoPSers. And in case someone wanted to keep in touch with someone else and didn't get their email address. It's something I do at every national I'm at. Great way to meet people too. You're the 3rd person to ask me why!

Billy

Treething, we tied at 24, but I got a better bit of the tie, 58th. One off of the list! And 57th also got 24, according to the Oklahoma coach (that is Amy Tai). Bummer.


Good luck to anyone taking the USACO US Open tomorrow!

I'll be taking Green, which is now even harder than before since the addition of the blue division :).

Yes. C /  C++ / Pascal / Java.

They give you a problem (like finding shortest paths or something), and you write a program that takes input data and outputs the result.

Of course, the problems are a lot harder than just shortest path, but you get the idea ;).

[quote="Nukular"]Yay results are in.

Training camp for me ;).[/quote]

Congratulations!

Does one need to know all those languages, or would it just be helpful for certain problems.  Could a person get by just knowing C++ or Java?

I only know C++.

C++ differs from Java in many respects, but an OOP is an OOP is an OOP, when you get into the gory details of everything. You can do the same things with both.

However, Java gets a 5x runtime bonus, because its Java, its interpreted, and its slower.


geometry
rectangle
modular arithmetic
induction

Sorry if I repost, but I haven't got any answer.
You've got some pieces, whose shape is shown in the attached picture. Prove that the rectangles you can build with it have at least one side which is multiple of 4.

2nd question: $2^kn-7$, for any k positive integer, you can find infinite n so that the result it's a perfect square...
For example, with k=1, n=8,16,... with k=2, n=4,8...
Thank you!

For the first: look at the columns of the big rectangle $\pmod 4$. Obviously they must all be congruent, since they're equal. If they're congruent to $0 \pmod 4$, we're done. Otherwise, note that each piece covers either $3$ squares of column $i$, $2$ squares of $i+1$ and $1$ square of $i+2$ or vice-versa ($1, 2, 3$), depending on how it's rotated. We say that a piece is on column $i$ if it covers a square of column $i$ but none of column $i-1$. We also say a piece is $1, 2, 3$ or $3, 2, 1$ intuitively, according to the above.

Now, working $\pmod 4$, we only need to consider the residue $\pmod 4$ of the number of pieces of each type. Note also that a $1, 2, 3$ piece and a $3, 2, 1$ piece on the same column add up to $0, 0, 0 \pmod 4$, so we can ignore these pairs. So on each column we have at most one type of piece, and we have at most $3$ pieces of this type. Also, $2$ pieces of type $1, 2, 3$ add up to $2, 0, 2$, just as $2$ pieces of type $3, 2, 1$ do. And $3$ pieces of type $1, 2, 3$ add up to $3, 2, 1$, while $3$ pieces of type $3, 2, 1$ add up to $1, 2, 3$.

So the pieces on column $i$ contribute to the overall sum as one of the following: $0$ for columns $i, i+1, i+2$, or: $1$ for column $i$, $2$ for $i+1$, $3$ for $i+2$, or: $3$ for column $i$, $2$ for $i+1$, $1$ for $i+2$, or: $2$ for $i$, $0$ for $i+1$, $2$ for $i+2$. We say column $i$ is of type $A, B, C$ or $D$ in each case respectively.

Now the only column affecting the residue of column $1$ is column $1$ itself, so it can't be of type $A$ (or else the side would be a multiple of $4$). If it is of type $B$, then all columns must be $\equiv 1$. Then since column $1$ contributes $+2$ to column $2$, column $2$ must be of type $C$. Then column $3$ gets $+3$ from column $1$ and $+2$ from column $2$, so it must be of type $A$, to be $\equiv 1$. And column $4$ only gets $+1$ from column $3$, so it must be of type $A$ too. Then column $5$ is unaffected, and it must thus be of type $B$; by induction we get blocks of length $4$, so the total length of the rectangle must be a multiple of $4$. We proceed similarly if column 1 is of type $C, D$.

For the second: you want to show that for any $k$, we have infinitely many pairs $n, a$ such that $2^kn=a^2+7$, in other words, we want to show that for any $k$ and infinitely many $a$ we have $2^k \mid a^2+7$ (then $n$ just comes from division). Obviously if this happens for some $a$, it also happens for $a+2^kr, r=1, 2, ...$ so we need just one $a$.

Indeed, the result is true for $k=1, 2, 3$, taking $a=1$. Note that $a$ is always odd. Now suppose $a, k$ are such that $2^k \mid a^2+7$ but $2^{k+1} \nmid a^2+7$. Then $a^2+7 \equiv 2^k \pmod{2^{k+1}}$. Finally,
\[(a+2^{k-1})^2+7=a^2+2^ka+2^{2k-2}+7 \\
(a+2^{k-1})^2+7 \equiv 2^k(a+1) \pmod{2^{k+1}}\]
Since $a$ is odd, $a+1$ is even and the last expression is congruent to $0$. So we have a good $a$ for $k+1$. The result is true by induction.


Thanks(for IMO 1996 inequality).It's very easy  :rotfl: . Who help me prove the inequality from Costa rica 2006?

Zhas, you don't need to post same problems 5 times. Stop it!

Japanese Communities Moderator

kunny


[quote="nayel"]does there exist three perfect squares which are three consecutive terms of an arithmetic progression?[/quote]

Besides for what ZeTaX said, the theorem is that you cannot have 4. Hence 4 is a minimial number.

[quote="ZetaX"]$1,25,49$[/quote]

are there any others except $k^{2},25k^{2},49k^{2}$?

[quote="sylow_theory"]Besides for what ZeTaX said, the theorem is that you cannot have 4. Hence 4 is a minimial number.[/quote]

could i get a proof? thanks.

Write $(1+\sqrt 2)^{2k+1}= b+a \sqrt 2$, then you have the progression $1^{2}, a^{2}, b^{2}$.
You get lots of others if you don't want the first term to be $1$, too.

i see, you turned this into a pell equation. but what happens for higher powers instead of squares?

i appreciate your concearn. :omighty:

This get's a lot harder, e.g. there is no nontrivial (not of type $-a^{3},0,a^{3}$ or constant) arithmethic sequence of cubes.
The general equation is $x^{n}+y^{n}= 2 z^{n}$ which remembers me to just one thing...


geometry
parallelogram
analytic geometry
geometric transformation
reflection
vector

Parallelogram ABCD with A(2,5), B(4,9), C(6,5), and D(4,1) is reflected across the $ x$-axis to A'B'C'D' and then A'B'C'D' is reflected across the line $ y\equal{}x\plus{}1$ to A''B''C''D''. This is done such that D' is the image of D, and D'' is the image of D'. What is the ordered pair of D'' in the coordinate plane?

After the first reflection, $ D' \equal{} (4, \minus{} 1)$. Then, since we know how to reflect about the line $ y \equal{} x$, we can translate one unit down so that our temporary point is $ (4, \minus{} 2)$, and we can now reflect about the line $ y \equal{} x$ to get $ ( \minus{} 2,4)$. Finally, we have to translate one unit back up, so our answer is $ \boxed{( \minus{} 2,5)}$.

In general, to reflect $ (x,y)$ across $ Ax \plus{} By \plus{} C \equal{} 0$ where $ A$ is positive, if the point is below the line, the reflection is $ (x',y') \equal{} \left(x \minus{} \frac {2A|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2},y \minus{} \frac {2B|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2}\right)$, and if the point is above the line, the reflection is $ (x',y') \equal{} \left(x \plus{} \frac {2A|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2},y \plus{} \frac {2B|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2}\right)$. Of course, this can be somewhat unwieldy, but I'm just posting it for fun. :P A proof can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1367473#1367473]here[/url]. It's very tedious without vectors, though.


Let $ ABCD$ be a quadrilateral and $ (I)$ be the incircle of $ ABCD$.Let $ P$ be apoint inside the quadrilateral $ ABCD$ and $ R,S$ be intersection of $ DP,AP$ with $ BC$ respectively.Prove that the incentre of incircle of triangles $ ABS$,$ DCR$,$ PAD$,$ PRS$ lie on same circle.

Take a look at: http://www.mathlinks.ro/viewtopic.php?t=247411 . This problem was created by Tran Quang Hung. I am in a hurry so if I am not mistaken then it's a small generalization of [url=http://www.mathlinks.ro/viewtopic.php?t=218636]this problem[/url]. He has submitted it for a Maths magazine in Vietnam, so I will present his nice and elementary solution later.


An arbitrary circle can intersect the graph $ y \equal{} \sin x$ in

$ \textbf{(A)} \text{ at most 2 points} \qquad \textbf{(B)} \text{ at most 4 points} \qquad$
$ \textbf{(C)} \text{ at most 6 points} \qquad \textbf{(D)} \text{ at most 8 points} \qquad$
$ \textbf{(E)} \text{ more than 16 points}$

When the radius of the circle tends to $ \infty$, arc of the circle tends to be a straight line. So the answer is $ E$.


If $ f(\overline{z}) \equal{} \overline{f(z)}$, throughout the complex plane, prove that $ f(z)$ is real everywhere on the real axis.

If $ z$ is real, then $ \overline{f(z)} \equal{} f(\overline{z}) \equal{} f(z)$, so $ f(z)$ is real.


Points $ B \equal{} B_1, B_2,.......,B_{n \plus{} 1} \equal{} C$ ARE chosen on side $ BC$ of a triangle $ ABC$ in that order.Let $ r_j$ be the inradius of trinagle $ AB_jB_{j \plus{} 1}$ for $ j \equal{} 1,2,3,.....n$ and $ r$ is the inradius of trinagle $ ABC$. Show that there exists a constant $ \lambda$ independent of $ n$ such that
$ (\lambda \minus{} r_1)(\lambda \minus{} r_2)....(\lambda\minus{}r_n)  \equal{} \lambda^{n \minus{} 1}(\lambda \minus{} r)$.

This constant is equal to $ \frac{_h}{^2}$, where $ h$ is length of the A-altitude of the $ \triangle ABC.$ See:
[url]http://www.mathlinks.ro/viewtopic.php?t=123243[/url],
[url]http://www.mathlinks.ro/viewtopic.php?t=270926[/url].


Two of the sides of a triangle have lengths of 20 and 24.  If the triangle is inscribed in a circle with a diameter of 30, what is the altitude to the third side?
I'm stuck with this problem.  Are there specific equations I can use?

You could use the fact that $ [ABC] \equal{} \frac {abc}{4R}$ where a, b, and c are sides of a triangle, R is the circumradius, and [ABC] is the area of the triangle.

[hide="solution"]

Let $ c$ be the third side of the triangle and let $ h_c$ be the altitude to c.

$ \frac {(20)(24)c}{4(15)} \equal{} \frac{h_c(c)}{2}$

Thus, $ h_c \equal{} 16$[/hide]

Haha...yeah...forgot to divide by 2 :oops: changed solution now...

lol i do that all the time too

Esp when I calculate the area of triangles based on their coordinates :lol:


Do you mean that we should construct a square inscribed in the quadrilateral using the ruler and the compass?

And if so, what does "inscribed" mean? I think we can't always find a square having a vertex on each of the four sides of $ABCD$, but I might be wrong.

yes I meaned that. but all vertex must be on each side of ABCD

[color=red][Moderator edit: This problem is clearly a special case of the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?p=129848 .][/color]


floor function
inequalities
algebra unsolved
algebra

Giving this a separate thread:

Let $\lambda > 1$ be any real number. Show that there exists an $n \in \mathbb{N}$ such that the number $\lfloor \lambda^n \rfloor$ is composite.

This problem can be reduced to showing that, for any real $\lambda > 1$, there exists a natural number $n$ such that $2 \leq \lambda^n < \sqrt{\lfloor \lambda^n \rfloor^2 + 1}$.

If this second inequality is true, then \[2\lfloor\lambda^n\rfloor\{\lambda^n\} + \{\lambda^n\}^2 < 1\] so \[\lfloor \lambda^{2n} \rfloor = \lfloor\lfloor\lambda^n\rfloor^2 + 2\lfloor\lambda^n\rfloor\{\lambda^n\} + \{\lambda^n\}^2\rfloor = \lfloor\lambda^n\rfloor^2,\] which is composite.

I don't have any idea whether this is easier or harder to prove than the original problem. I'll try it in the morning.

The inequality $2 \leq \lambda^n <\sqrt{\lfloor\lambda^n\rfloor^2 + 1}$ is equivalent to $4\leq\lfloor\lambda^{2n}\rfloor = \lfloor\lambda^n\rfloor^2$. I'm not going to write out the entire proof of this because it's not particularly difficult to figure out.

I wrote a computer program to test my theory that for every $\lambda > 1$ there exists $n \in \mathbb{N}$ such that $4\leq\lfloor\lambda^{2n}\rfloor = \lfloor\lambda^n\rfloor^2$. For most $\lambda$ values I tested, the smallest $n$ value that worked was a single digit, but for $\lambda = 1.4$ and $\lambda = 1.7$ I tested $n$ values from 1 to about 20 and none worked. I don't know what this means, but if anyone wants to explore it more it would probably be interesting.

[quote="James"]This problem can be reduced to showing that, for any real $\lambda > 1$, there exists a natural number $n$ such that $2 \leq \lambda^n < \sqrt{\lfloor \lambda^n \rfloor^2 + 1}$.[/quote]
Clearly, it is untrue. Put for example, $\lambda=2+\sqrt{3}$.
Sorry. It was nice idea.

[quote="vess"]
Let $\lambda > 1$ be any real number. Show that there exists an $n \in \mathbb{N}$ such that the number $\lfloor \lambda^n \rfloor$ is composite.[/quote]

Anyone willing to try and prove this nice result? :)


calculus
integration
limit
geometry
calculus computations

For the following problems, show that the following definition holds. \[ \int^b_a f(x)\text{ d}x \equal{} \lim_{n\to \infty} \sum^{n}_{i\equal{}0} f(x_i)\Delta x_i\] [hide="#1"]Equal subintervals
area below $ y\equal{}x^2$,   $ 0\le x\le2$[/hide]
[hide="#2"]Unequal subintervals (hint: $ \Delta x_i \equal{} x_{i\plus{}1}\minus{}x_i$)
area below $ y\equal{}\sqrt{x}$,   $ 0\le x\le2$[/hide]

Choose a partition $ P_{n}\equal{}\{t_{0},\ldots,t_{n}\}$ with $ t_{i}\equal{}2i/n$.
Then
$ \sum_{i\equal{}0}^{n}f(t_{i})\Delta t_{i}\equal{}\sum_{i\equal{}0}^{n}t_{i}^{2}\cdot\frac{2}{n}\equal{}\frac{2}{n}\sum_{i\equal{}0}^{n}t_{i}^{2}$
$ \equal{}\frac{2}{n}\sum_{i\equal{}0}^{n}\frac{4i^{2}}{n^{2}}\equal{}\frac{8}{n^{3}}\sum_{i\equal{}0}^{n}i^{2}\equal{}\frac{8}{n^{3}}\cdot\frac{n(n\plus{}1)(2n\plus{}1)}{6}$
$ \equal{}\frac{8}{6}\cdot\frac{(n^{2}\plus{}n)(2n\plus{}1)}{n^{3}}
\equal{}\frac{4}{3}\cdot\frac{2n^{3}\plus{}3n^{2}\plus{}n}{n^{3}}.$
As $ n\rightarrow\infty$, the second fraction becomes more and more like 2, so the sum tends to $ \frac{4\cdot 2}{3}\equal{}\frac{8}{3}$, which we know to be correct.


quadratics
Gauss
number theory
relatively prime
number theory unsolved

Let $a,b,c$ be positive integers that are pairwise relatively prime, and that satisfy $a^{2}-ab+b^{2}=c^{2}$

Show that every prime factor of $c$ is of the form $6k+1$

-----------------------------------------------------------------------
I found another problem that looks related to that problem..

there exist a triangle which has sides whose length is $a,b,c$ , and the angle $C$ is $120$. when this is satisfied, we call $(a,b,c)$ 'semi-pythagorian numbers'. Prove that, if $(a,b,c)$ is semi pythagorian number, then $c$ has a prime factor which is bigger than $5$.

Let $p|c$, then $p|(2a-b)^{2}+3b^{2}$. Therefore Lejander symbol $(\frac{-3}{p})=1\Longrightarrow p=1(mod \ 6)$.

since $gcd(a,b)=1$. $a, b$ are not both even.
therefore $c$ is odd.

$4(a^{2}-ab+b^{2})=4c^{2}$
$(2a-b)^{2}+3b^{2}=4c^{2}$

Let $p$ is prime which dividing $c$.
then, $(2a-b)^{2}\equiv-3b^{2}(mod p)$
Using Legendre symbol
$(1 = (\frac{(2a-b)^{2}}{p}) = (\frac{-3b^{2}}{p}) = (\frac{-3}{p})= (\frac{-1}{p}) (\frac{3}{p})$
Using Gauss ~~~ (I can't remember it's name...  :oops: )
$1=(-1)^{(p-1)/2}\times (\frac{p}{3}) \times (-1)^{(3-1)(p-1)/2}=(\frac{p}{3})$

$\therefore p \equiv 1 (mod 6)$

Well...
semi pythagorian numbers is root of $a^{2}+ab+b^{2}= c^{2}$
not $a^{2}-ab+b^{2}= c^{2}$
but the proof is almost same.


Let $ ABCDE$ be convex pentagon such that $ \angle BAC \equal{} \angle EAD$ and $ \angle BCA \equal{} \angle EDA$ and let the lines $ CB$ and $ DE$ intersect at point $ F$. Prove that the midpoints of $ CD$ , $ BE$ and $ AF$ are collinear.

Let $ AB\cap ED\equal{}\{K\}, AE\cap BC\equal{}\{J\}, I, H, L, G$ be the midpoints of $ KJ, BE, CD, AF$.
By Gauss's theorem we get $ I, H, L$ are collinear.
So we prove $ I, H, G$ are collinear. (*)
Since $ \Delta JAC\sim \Delta KAD$ and $ \Delta BAC\sim \Delta EAD$ we have $ \frac{JB}{JC}\equal{}\frac{KE}{KD}$
So $ (*)$ is true.
We are done!


A finite set of points X and an equilateral triangle T are given on a plane. Suppose that every subset X' of X with no more than 9 elements can be covered by two images of T under translation. Prove that the whole set X can be covered by two images of T under translation. :D

Let the length of the side of the equilateral triangle be $ k$.

Lemma 1. One of the sides of a triangle formed by selecting any three points ( possibly a degenerate triangle) is not more than $ k$.
Proof. If all the three points do not lie in a equilateral triangle, by pigeon hole principle, two of the points lie in either the equilateral triangle or in its translation.

By similar reasoning,
Lemma 2. Among any given 9 points, we can find a subset of at least of 5 points so that any two points in the sub set are not separated by more than $ k$.


rotation
integration
calculus
calculus computations

let R be the region by the curves y=x^2 and x=y^2. suppose the region is rotated about the line y=-1

A: compute the volume of the generated solid using shells method.

B: compute the volume of the generated solid using washers method.

thanks for helping...


[quote="Adriana N."]Solve the following equations:
1)$ 2^x \minus{} (\frac {1}{2})^{2x} \minus{} (\frac {1}{2})^x \plus{} 1 \equal{} 0$
2)$ 8*4^{\frac {1}{x}} \plus{} 8*4^{ \minus{} \frac {1}{x}} \minus{} 54*2^{\frac {1}{x}} \minus{} 54*2^{ \minus{} \frac {1}{x}} \equal{} \minus{} 101$
3)$ 3^{2x} \minus{} \frac {18}{3^{2x}} \minus{} (3^x \plus{} \frac {6}{3^x}) \equal{} 2$
4)$ 27*2^{ \minus{} 3x} \plus{} 9*2^x \minus{} 2^{3x} \minus{} 27*2^{ \minus{} x} \equal{} 8$
5)$ 3 \plus{} 5^{\frac {x}{3}} \equal{} 2^x$
6)$ 2^{x \plus{} 1} \plus{} 20 \equal{} 6^{x \minus{} 1}$;
7)$ 2^x \minus{} 3^{\frac {x}{2}} \equal{} 1$
Please try to solve these equations because I want to say how I can solve them.   :P[/quote]

These equation seem not to be olympiad ones :(

1)$ 2^x \minus{} (\frac {1}{2})^{2x} \minus{} (\frac {1}{2})^x \plus{} 1 \equal{} 0$
Let $ z\equal{}2^x$. The equation becomes $ z \minus{} \frac {1}{z^2} \minus{} \frac {1}{z} \plus{} 1 \equal{} 0$
So $ z^3 \plus{} z^2 \minus{} z \minus{} 1 \equal{} 0$
So $ (z\minus{}1)(z\plus{}1)^2\equal{}0$
So the only positive root $ z\equal{}1$ and so the only root of the original equation : $ x\equal{}0$

2)$ 8*4^{\frac {1}{x}} \plus{} 8*4^{ \minus{} \frac {1}{x}} \minus{} 54*2^{\frac {1}{x}} \minus{} 54*2^{ \minus{} \frac {1}{x}} \equal{} \minus{} 101$
Let $ z\equal{}2^{\minus{}\frac 1x}$. The equation becomes $ 8\frac{1}{z^2} \plus{} 8z^2 \minus{} 54\frac 1z \minus{} 54z \equal{} \minus{} 101$
Let $ u\equal{}z\plus{}\frac 1z$. The equation becomes $ 8(u^2\minus{}2)\minus{}54u\equal{}\minus{}101$
So $ 8u^2\minus{}54u\plus{}85\equal{}0$
So $ u\in\{\frac 52,\frac{17}{4}\}$
So $ z\in\{\frac 14,\frac 12,2,4\}$
So $ x\in\{\minus{}1,\minus{}\frac 12,\frac 12,1\}$

3)$ 3^{2x} \minus{} \frac {18}{3^{2x}} \minus{} (3^x \plus{} \frac {6}{3^x}) \equal{} 2$
Let $ z\equal{}3^x$. The equation becomes $ z^2 \minus{} \frac {18}{z^2} \minus{} (z \plus{} \frac {6}{z}) \equal{} 2$
So $ z^4 \minus{} z^3 \minus{} 2z^2 \minus{}  6z \minus{} 18 \equal{} 0$
So $ (z\minus{}3)(z^3\plus{}2z^2\plus{}4z\plus{}6)\equal{}0$
$ z^3\plus{}2z^2\plus{}4z\plus{}6$ can only have negative roots, so none available for us ($ z\equal{}3^x$).
So the only interesting root is $ z\equal{}3$ and so $ x\equal{}1$

4)$ 27*2^{ \minus{} 3x} \plus{} 9*2^x \minus{} 2^{3x} \minus{} 27*2^{ \minus{} x} \equal{} 8$
Let $ z\equal{}2^x$. The equation becomes $ \frac{27}{z^3} \plus{} 9z \minus{} z^3 \minus{} \frac{27}{z} \equal{} 8$
So $ (\frac 3z\minus{}z)^3\equal{}8$
So $ \frac 3z\minus{}z\equal{}2$
So $ z^2\plus{}2z\minus{}3\equal{}0$
So $ z\in\{\minus{}3,1\}$
And the only interesting value (must be positive) is $ z\equal{}1$, and so $ x\equal{}0$

5)$ 3 \plus{} 5^{\frac {x}{3}} \equal{} 2^x$
Let $ f(x)\equal{}5^{\frac {x}{3}} \minus{} 2^x \plus{} 3$
Obviously $ f(3)\equal{}0$
$ f'(x)\equal{}5^{\frac {x}{3}}\frac{\ln(3)}{5}\minus{}2^x\ln(2)$
Obviously $ f'(x)$ has a unique root $ x_0$ and is $ >0$ over $ (\minus{}\infty,x_0)$ and negative over $ (x_0,\plus{}\infty)$
And since $ \lim_{x\to \minus{}\infty}f(x)\equal{}3>0$ :
$ f(x)>0$ $ \forall x\in(\minus{}\infty,x_0]$
$ f(x)$ is strictly decreasing for $ x>x_0$ and so may have at most one root.
So $ x\equal{}3$ is the only solution.

6)$ 2^{x \plus{} 1} \plus{} 20 \equal{} 6^{x \minus{} 1}$
Let $ f(x)\equal{}2^{x\plus{}1}\plus{}20\minus{}6^{x\minus{}1}$
Obviously $ f(3)\equal{}0$
Same as above :
$ f'(x)$ has a unique root $ x_0$ and is $ >0$ over $ (\minus{}\infty,x_0)$ and negative over $ (x_0,\plus{}\infty)$
And since $ \lim_{x\to \minus{}\infty}f(x)\equal{}20>0$ :
$ f(x)>0$ $ \forall x\in(\minus{}\infty,x_0]$
$ f(x)$ is strictly decreasing for $ x>x_0$ and so may have at most one root.
So $ x\equal{}3$ is the only solution.

7)$ 2^x \minus{} 3^{\frac {x}{2}} \equal{} 1$
Let $ f(x)\equal{}2^x \minus{} 3^{\frac {x}{2}} \minus{} 1$
Obviously $ f(2)\equal{}0$
Same as above :
$ f'(x)$ has a unique root $ x_0$ and is $ <0$ over $ (\minus{}\infty,x_0)$ and positive over $ (x_0,\plus{}\infty)$
And since $ \lim_{x\to \minus{}\infty}f(x)\equal{}\minus{}1<0$ :
$ f(x)<0$ $ \forall x\in(\minus{}\infty,x_0]$
$ f(x)$ is strictly increasing for $ x>x_0$ and so may have at most one root.
So $ x\equal{}2$ is the only solution.

Thank a lot!These were not olympiads ones,you are right!I wanted help because I still lern.Thank!   :P  :P

$ f(x)\equal{}2^x \mbox{ increasing function and: } \\
g(x)\equal{}\left( \frac{1}{2}\right)^{2x}\plus{}\left( \frac{1}{2}\right)^x\minus{}1 \mbox{ decreasing function!}\Rightarrow \\
\Rightarrow \mbox{equation } f(x)\equal{}g(x) \mbox{ has a unique solution: } x\equal{}0.$


function
number theory unsolved
number theory

Cases, cases, cases, ....

I know 5 solutions. And I haven't analyzed all cases yet...
Maybe there are infinitely many solutions?

[quote="darkmaster"]Find all functions $ f:Z\longrightarrow Z$ satisfying $ f(m+n)+f(m)f(n)=f(mn+1) \forall m,n \in Z$[/quote]

PNT seems to be the soure of adapted problems...

Maybe! but he give me those problems for self-solving, not for posting here.
The only problem he asks me to post here is the problem "Prime divisors". ( He doesn't know Internet, so he can't post by himself)

And by the way, please tell me what problem this one adapted from?Maybe I can find a good solution from it.


Using only the digits 1, 2, 3, 4 and 5, two players A, B compose a 2005-digit number N by selecting one digit at a time as follows: A selects the first digit, B the second, A the third and so on, in that order. The last to play wins if and only if N is divisible by 9. Who will win if both players play as well as possible?

I found one solution to this, but i'm fairly sure there are others...

Don't just post the answer, post a proof...

Since it is $A$ to play last, it is $B$ who has a winning strategy :
Note that $1+5=2+4=3+3=6$. Thus at each turn $B$ plays the integer which add up to $6$ with the last one written by $A$.
After his last turn, the sum of the $2004$ first digits will be $1002 \times 6 = 0\mod [9].$
Therefore, whatever $A$ chooses as last digit the sum of the $2005$ digits will never be divisible by $9$, so the final number won't be divisible by $9$ either.

Pierre.

Wish i got that in the exam! Allthough my method was right, i just got a bit mixed up with who was who... where'd you come stephen? Oh yeah, me the new Stephen Casey!


hello, your inequality is equivalent with
$ \left(1\plus{}\frac{1}{n}\right)^n<n$ and this is true for $ n\geq3$ because
$ \left(1\plus{}\frac{1}{n}\right)^n\le e<3\le n$
Sonnhard.


Let m>=2n be two positive integers. Find a closed form for the following expression:

$E(m,n)=\sum_{k=0}^{n}(-1)^k\frac{(m-k)!}{n!(m-k-n)!}\frac{n!}{k!(n-k)!}$

----------------------------------------------------------------------------
I know that one polynomial should be $(x-1)^k$ but I cannot find the second one.

Probably because there exists a very very easy solution using repeated differences. Just start to make the repeated differences of the polynomial $ f(x)=\frac{ x(x-1)(x-2)...(x-n+1)}{n!}$.

thanks Harazi !  Out of curiosity, does a solution using polynomials exist [surely it is, stupid question]? If so, can somebody post it [I've promised myself recently to do as many this kind of problems using polynomials as I can  :cool: ]

I'm sorry to disappoint you, but I think that a combinatorial approach is more efficient here, since it does not use any computation :

First, just rewrite the desired sum as $\dsp E(m,n) = \sum_{k=0}^n (-1)^k{ {n} \choose {k}} { {m-k} \choose {m-k-n}} $.

Now, let's consider a group of $n$ boys and $m-n$ girls. You want to form a group of $m-n$ persons, and among them you have to select some leaders but the leaders have to be boys.
You may first select some leaders, say $k$. This leads to $ { {n} \choose  {k} }$ such choices. Then, you complete the gorup with $m-n-k$ people among the $m-k$ remaining ones, which which can be done in ${ {m-k} \choose {m-k-n} }$ way.s

It follows that $E(m,n)$ is the number of such groups with an even number of leaders minus the number of such groups with an odd number of leaders.

But, there is another way to compute this sum :
First, we select a group of $m-n$ persons among the $m$ people. Let $g$ be the number of boys in this group. You have to select the leaders in this group among these $g$ boys. If you want to have $k$ leaders, this can be achieved in ${ {g} \choose {k}}$ ways. For this particular group, the number of groups with an even number of leaders minus the number of groups with an odd numbers of leaders is $\dsp \sum_{k=0}^g (-1)^k { {g} \choose {k}} = (1-1)^g = 0^g$, from the binomial theorem.
Therefore, if $g>0$ there as many groups with an even numbers of leaders than there are groups with an odd number of leaders, for this particular group of $m-n$ chosen persons. Summing over all the groups of $m-n$ persons we easily deduce that $E(m,n)$ is just the same sum but over the group with $g=0$ bys. But, since there are $m-n$ persons in a group and there are $m-n$ girls in the whole population, this can be done only if you choose the $m-n$ girls (one unique way to form the group). In that case, there is only one way to select the leaders (no one is a leader), which correspond to an even number of leaders (namely 0). Therefore, this has to be considered with a $+$ in the sum.

It follows that $E(m,n) = 1$ for all $m,n$ and we are done.

Pierre.

Nice solution Pierre - I like your fable  :P but anyway I won't give up - I'll find those polynomials  :maybe:

I have another combinatorial solution:

For $i = 1, 2, \ldots, n$, let $S_i$ denote the set of all $n$-element subsets of $\{1, 2, \ldots, m\}$ that does not contain the element $i$. That is
\[
S_i = \left\{ A \subset \{1, 2, \ldots, m\} : |A| = n, i \notin A \right\}
\]

Since $S_1 \cup S_2 \cup \cdots \cup S_n$ contains all $n$-element subset of $\{1, 2, \ldots, m\}$ except for the subset $\{1, 2, \ldots, n\}$, we have
\[
|S_1 \cup S_2 \cup \cdots \cup S_n| = \binom{m}{n} - 1
\]

Also, the intersection $S_{a_1} \cap S_{a_2} \cap \cdots \cap S_{a_i}$, where $1 \leq a_1 < a_2 < \cdots < a_i \leq n$, contains all $n$-element subsets of $\{1, 2, \ldots, m\} \backslash \{a_1, a_2, \ldots a_i\}$. So
\[
S_{a_1} \cap S_{a_2} \cap \cdots \cap S_{a_i} = \binom{m-i}{n}
\]

By the Inclusion-Exclusion Principle,
\[
|S_1 \cup S_2 \cup \cdots \cup S_n| = \sum_{1 \leq i \leq n} |S_i| - \sum_{1 \leq i < j \leq n} |S_i \cap S_j| + \sum_{1\leq i < j < k \leq n} |S_i \cap S_j \cap S_k| - \cdots -(-1)^n |S_1 \cap S_2 \cap \cdots \cap S_n|
\]
which is equivalent to
\[
\binom{m}{n} - 1 = \binom{m-1}{n} \binom{n}{1} - \binom{m-2}{n} \binom{n}{2} + \binom{m-3}{n} \binom{n}{3} - \cdots - (-1)^n\binom{m-n}{n} \binom{n}{n}
\]
and rearranging gives
\[
\sum_{k=0}^n (-1)^k \binom{m-k}{n} \binom{n}{k} = 1
\]
Hence $E(m,n) = 1$ for all $m, n$ satisfying the given conditions.

And let me just extend on Harazi's comment regarding common differences (might as well, since I already wrote it up for the contest.)

Let $f$ be the polynomial
\[
f(x) = \binom{m-x}{n} = \frac{(m-x)(m-1-x) \cdots (m-n+1-x)}{n!}
\]
Then
\[
E(m,n) = \sum_{k=0}^n (-1)^k \binom{n}{k} f(k) = \triangle^n f(0)
\]
Since $\deg(f)=n$, by the theory of finite difference, $\triangle^n f(x)$ is independent of $x$. So
\[
E(m,n) = \triangle^n f(0) = \triangle^n f(m-n)  = \sum_{k=0}^n (-1)^k \binom{n}{k} f(m-n+k)
\]
Since $f(m-n+1) = f(m-n+2) = \cdots = f(m) = 0$, we have
\[
E(m,n) =  \sum_{k=0}^n (-1)^k \binom{n}{k} = f(m-n) = 1
\]
So the desired answer is 1.

Personally, I like the finite differences solution better.


limit
trigonometry
algebra
function
domain
calculus
calculus computations

calcul this limit with proof :

$ \lim_{x\to0} \sin{{\frac{\pi}{x}}}$

thanx.

It doesn't exist.  Are you familiar with the definition of a limit?

yes I know that no exist,but is there a proof?
thanx

$ \lim_{x \to x_o} f(x) \equal{} b$ if and only if for any sequence $ x_n$ of numbers belonging to the domain of f(x), with $ x_n \neq x_o$ for any natural n, such that $ x_n$ converges to $ x_o$ then $ f(x_n)$ converges to $ b$ (we assume that $ x_o$ also belongs to the domain of f(x)).


linear algebra
matrix
superior algebra
superior algebra unsolved

Find all muliplicative groups $(G,\cdot)\subseteq\mathcal M_{3}(N)$ with the property: $A^{3}=I_{3}, \ \forall{A\in G}$

Let $A \in G$. Then we have that both $A$ and $A^{-1}= A^{2}$ are in $\mathcal M_{3}\left( \mathbb N_{0}\right)$. I proved [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=teachers&t=102435]here[/url] that all such matrices must be permutation matrices (i.e. with exactly one $0$ on every line and every column).

Since $A^{3}= I$, we must have $\left( \det A \right)^{3}= 1$, so $\det A = 1$. It is easily seen that all permutation matrices of determinant $1$ satisfy $A^{3}= I_{3}$. They are $I_{3}, \ U = \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \ V = \begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$.

The possible groups are: $G_{1}= \left\{ I_{3}\right\}, \ G_{2}= \left\{ I_{3}, U, V \right\}$.

This one was really easy... My solution used eigenvalues, I then proceeded to analyse the cases trA=0 and trA=3....The groups are those in perfect radio's solution...solved it in 10 mins and got 7 for it. :wink:

[quote="perfect_radio"]Let $A \in G$.
Since $A^{3}= I$, we must have $\left( \det A \right)^{3}= 1$, so $\det A = 1$. It is easily seen that all permutation matrices of determinant $1$ satisfy $A^{3}= I_{3}$. They are $I_{3}, \ U = \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \ V = \begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$.

The possible groups are: $G_{1}= \left\{ I_{3}\right\}, \ G_{2}= \left\{ I_{3}, U, V \right\}$.[/quote]

For this I took 1 point :) also saying that G1 and G2 are groups(with proof) and they are the only ones without proof :(


geometry
rectangle
geometric transformation
rotation

$ P$ is an internal point of a rectangle $ ABCD$ such that $ \left\|\begin{array}{c}AP\equal{}3\\CP\equal{}5\\DP\equal{}4\end{array}\right\|$ . Find $ PB$ .

I've tried various rotations, but none of them help...

[hide="Solution"]By [url=http://en.wikipedia.org/wiki/British_flag_theorem]British Flag Theorem[/url], we have $ 3^2 \plus{} 5^2 \equal{} 4^2 \plus{} BP^2\implies BP \equal{} \sqrt {18}$.

Answer: $ 3\sqrt {2}$[/hide]

I see, interesting and simple theorem, thank you very much!


conics
parabola
quadratics
function
special factorizations

Hello everyone,

   I am going through the quadratic function in College Algebra Demystified, but I am in high school. I am on the section where it shows me that I can find the vertex of a parabola by completing the square of y = ax^2 + bx + c.

I understand how you get h and k:

h = - b/2a

k = c - b^2/4a

However, I don't understand why one can also find k by substituting x = h into y = ax^2 + bx +c.

Could someone please explain?

Thanks in advance.

Well the vertex is the point (h,k). And for any point (x,y) on the graph of a function f, we know that f(x)=y. Thus f(h)=k.

Hope that helps.


How does mathscounts figure out what is the team score?

They take the sum of the individual scores of the people the team and divide by 4, and add twice the team round score. If a team has less than 4 students, they still divide the sum by four.

I saw the score of my team today and the average per student, and it appeared that they did not double the team score , they just left it how it was.

meh, I've always heard it as the average of the team scores plus twice the team score.

they add all of the people on the team's scores, divide it by 4, and then add twice the team.

[quote="gauss1181"]Why do they still divide the sum by 4 even if a team has fewer than 4 students? That doesn't seem fair for smaller teams, though... :maybe:[/quote]

It's because a team is supposed to be four students--no more, no less.  If they divided by the number of students on the team, then some teams would have an incentive to enter a one-person team consisting of a single superstar student.


What is the coefficient of $ x^2y^6$ in the expansion of $ \left(\frac{3}{5}x\minus{}\frac{y}{2}\right)^8$?  Express your answer as a common fraction.

Using the binomial theorem, we have $ \binom82\left(\frac{3}{5}\right)^2\left(\minus{}\frac{1}{2}\right)^6x^2y^2\equal{}28\left(\frac{9}{25}\right)\left(\frac{1}{64}\right)x^2y^2\equal{}\boxed{\frac{63}{400}}x^2y^6$

hello, we have $ \binom{8}{6}\left(\frac{3}{5}\right)^2\cdot \frac{1}{2^6}\equal{}\frac{63}{400}$.
Sonnhard.

Oof I got $-\frac{63}{400}$ because for some reason my brain thought that $(-\frac{1}{2})^6$ was negative. :wallbash:

It is. $-\frac{(1^6)}{2}=-\frac{1}{2}$.

@below I'm following order of operations. It's an important lesson to learn.

olympus i think you know what he meant don't do this


(i)Prove that:$\forall m\in \mathbb{N^*},\exists f_m(x)\in \mathbb{R}[X]$ such that:
\[\sum_{i=1}^ni^{2m+1}=f_m(n\cdot (n+1)),\forall n\in \mathbb{N^*}\]
(ii)Prove that:\[x^2|f_{1001}(x)\]

This is a problem on Crux Mathematicorum, some issue in 1997, which are free to access!

PS: We use the same problem to chose our National Oympiad Team, 2003 :D . There is another terrible solution submit by huakhacnam , using Newton fomula.


geometry
geometric transformation
national olympiad

\[\text{Day 1}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=142728
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=808480#808480
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=142583
\[\text{Day 2}\]
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=142749
5. http://www.mathlinks.ro/Forum/viewtopic.php?t=142750
6. http://www.mathlinks.ro/Forum/viewtopic.php?t=142753

How many days are there? Are there two days?

Still day 1!How do you think?Post or no? :lol: Thanks very much!

I will try, because he translation problem 1 of Day 1 is not good, I think so.


inequalities
function
inequalities proposed

Let $a,b,c$ be nonnegative real numbers such that $ab+bc+ca=1$, prove that
\[\frac{1}{\sqrt{a^{2}+bc}}+\frac{1}{\sqrt{b^{2}+ca}}+\frac{1}{\sqrt{c^{2}+ab}}\ge 2\sqrt{2}\]
:)

@: A nice solution exists!

I would have a very nice solution if only $-8x^{5}y^{3}-24x^{4}y^{4}-8x^{3}y^{5}+16x^{3}y^{3}-8x^{2}y^{2}+x^{2}+2xy+y^{2}$ was always positive for positive $x,y$... is it?

If I am not wrong the inequality can be improved to :
$E_{1}=\sum{\frac{1}{\sqrt{a^{2}+bc}}}\geq 3 > 2\sqrt{2}$.

Denote $a+b+c=s$, thus the inequality is equivalent with:
$\sum{\frac{1}{\sqrt{a(2a-s)+1}}}\geq 3$

Now consider the function $f(x)=\frac{x}{\sqrt{x(2x-s)+1}}$, which is convex, thus by applying Jensen, we have that:

$\sum{(bc \cdot \frac{a}{\sqrt{a(2a-s)+1}})}\geq \frac{3abc}{\sqrt{3abc(6abc-\sum{a})+1}}$.

$\iff E_{1}\geq \frac{3}{\sqrt{3abc(6abc-\sum{a})+1}}$.

Thus it remains us to prove that $\frac{3}{\sqrt{3abc(6abc-\sum{a})+1}}\geq 3$.
By squaring it is equivalent with:
$3\sum{a^{2}bc}\geq 18 a^{2}b^{2}c^{2}\iff 3\sum{a}\geq 18abc$, but $3\sum{a}=3\sum{a}\sum{ab}\geq 27 abc \geq 18 abc$.

The equality occurs for $a=0, b=c=1$ and the other cyclic permutations.


The topic name can be misleading, as the story itself won't be anywhere near serious. What will is the attempt. This is different from the erstwhile never ending story as you can't just come in add something to it.

First I need five main writers to sign up. [b]Do not sign up unless you are regular - or you will soon end up getting the boot.[/b] The main authors may then recruit any assistant writers under them. We will then proceed planfully, like a real writer does, by choosing the setting, the characters, the plot and so on. The real story may not be underway for months.

If you like to be creative this is the place for you. Even if you start as assistant, if you show enough flair you will get promoted.

So start registering everyone! If successful this may be the greatest project ever on AoPS!

Oh well, I'll sign up as an assistant writer.
(I at least hope someone will choose me.)

I'll be either a real or assistant)preferably assistant

pick me for assistant.
im not very creative though.

You could directly sign up as main writers everyone.
But still, what we need is a workforce, so keep registering!
Meanwhile I will think up some rules and guiding principles.

(I hope this does not have the same fate as the Hitchhiker club!)

Still too few. I am quite surprised that none of perfectnumber, ignite, B-tone party, madatmath, rem, acidpop, Chocoboom and the other quite active members are yet to sign up! Please hurry up everybody! Only two main writers till now ...

i'm also in as main and preferably, i'm pitting my wits against acidpop.

OOoh I'd like to sign up! As a main writer if I can.

I would like to sign up as an assistant writer

yup i agree to team up with mbabbitt2003. let us begin!!
 :rotfl:

ok..... so what do you vote? yes on the idea of voting, or no?

oh. Well, If no one likes the idea ecept me, then we will drop the voting idea...........

hey! i'll get back in two days time!!
ba bye till then!!
 :(   :)

fine. go on. i will sit back and watch the fun! as for voting you have my thumbs up.

Ok. two yeses.
one more yes and then we can start voting on topics, if Madatmath doesn't join.

[size=200][color=green]Yes.[/color][/size]
i'm Ok with the voting idea.
but post your idea quickly.
even if it has weak areas, we can always refine it.

 one more thing.
 when we do actually start writing, there'll
 always be posts on improving certain ideas,
 and on changing certain parts. so all in all it's goin'
 to be a mess. to pm the ideas is a headache, as there are too many
 people involved. ideally start the "Spot of serious writers forum'"
 i think i'll do that myself on behalf of bubka. if you have any ideas, additions ,
 objections etc. after we begin writing, or right now, if you have
 an anecdote, a spot of humor you think should be moulded into the
 story , then post there, and it will be stored. Note ; the main story gets
 recorded ONLY in this thread. once it starts, we will post here ONLY after
 the paragraph has been finalised. till then , correspondence in the other forum .
 i'm not taking voting on this because this is probably the best thing to do.
 besides, if we find it's tedious, we can always forget the other thread, and
 MithsApprentice will be glad to have a one more topics to delete.    :lol:

there are enough yeses. I will change the voting to the Serious story writing room.

Everyone, listen to me(the writers here don't have to.) We need more writers!!! We are very short on them. Anyone can sign up(accept authors of books :rotfl: ) and you will need to be present while we are writing this book. You need to post here if you want to become a writer. You can be an assistant or a main writer. Once you do become a writer, please read the "Serious story writer's room" topic, because that is where we work. Also tell us that you are a writer once you post in the "Serious story writer's room" topic, so we know that you are an official writer. Please, sign up now!

Can you explain with detail what do a writer have to do?? Thanks!

*sigh* You vote on a topic, and then you write stories about that topic, and then we vote on those. Since NO ONE has voted for a topic yet, all you have to do is just sign up and vote! The first vote wins!(now that we are impatient)

i would like to join, though in the summer, i probably wont have friequent internet access from july to mid august

that's all right. Do you want to be an assistant writer, or a main writer? I do need an assistance writer.

I am back in control friends. And yes star99, you can easily join as an assistant or as a main writer if you want to.


Hi guys, it's my first time come to this forum...
I am looking for somebody to help me to solve this question:

let A be an m x n matrix, where m < n. Then， (the transpose of A) X A is an n X n matrix. Show that det((transpose of A) X A)= 0        (m is row and n is column)

please help me ,cheers...

Are you sure these are all the conditions ?

Take $A = [1,1]$

and than you equation is not true!

[quote="perfect_radio"]In your case we have $\det \left( A^T A \right) = \left| \begin{matrix} 1 & 1 \\ 1 & 1\\ \end{matrix} \right| = 0$.[/quote]

I do not understand! Please explain!

[quote="perfect_radio"]In your case we have $\det \left( A^T A \right) = \left| \begin{matrix} 1 & 1 \\ 1 & 1\\ \end{matrix} \right| = 0$.[/quote]

sorry, i don't understand ,can you explain it please?

$rank(A)\leq m , rank(A^TA)\leq rank(A)\leq m <n$

Is this claim true:
$Rank(A*B) \leq Rank(A),Rank(B)$


I need help connecting some java applets.  In one, I have a game.  In another I have a menu where you can read the rules and pick your colors and then click start when you're ready to play.

First off: how can you call another applet?  like it doesn't really have a constructor does it?  I have an init() method and paint() and all the actionListeners and such.
I just need the actual call, I have the button working fine.

Secondly: know how I would pass it two Color variables?

Thanks, I can't find anyone who uses applets :(

Google for it. First hit:

http://forum.java.sun.com/thread.jspa?threadID=588488&messageID=3045964

EDIT: Ah, I forgot, Happy Coding  :D

I'm just running it in TextPad for a project, not embedded in HTML so none of those solutions worked.  Thanks anyway though.

But new plan of attack : JOptionPane.  When my game applet runs I put a pop-up with the rules in it in the init() method so that works fine.  Now I'm just thinking could you use JOptionPane to select a color from a list of three?  I'll look it up, but my one book didn't have anything so Ijust wondering if people know any specs off the top of their head

EDIT:  Yay, I did the JOptionPane and my game is workin beautifully.  only thing is when you win, it doesn't stop.  you can just keep playing.. the other person can win too!  :o  its no big deal, I already turned it in and you can just close it when you win, but does anyone know how to stop an applet prematurely?


I was wondering if there's a formula for the following counting problem:

There are x students who need to be seated in a row (for whatever purposes...). Unfortunately there are 2 boys each who cannot possibly sit together or else they will fight and will need to go to the hospital. In how many ways can this happen? What if there is an adult sitting in the center (x/2 students on each side) and will obviously not fight with anyone?

And what if there are 4 boys? What about y boys in the group? What if there are x students and y boys, and no two boys can be seated together?

It recurs a lot in MATHCOUNTS and other math competitions, and I was wanting a formula. I found one on a website but it was ugly and didn't use LaTeX.

the really easy formula is ways that the students can be seated next to each other - ways that 2 boys are seated together, but i dont know how to simplify that

[quote="dwx314"]o_O how'd you get that?

couldn't this be another one of those 'partition probs'?
I don't know[/quote]

Well...  :blush: that's what superman said, and I didn't know if it was right or not.

Its gets exponentially harder as y gets higher

[quote="mathnerd314"]I saw this on a math competition and ended up with second place because of it.  :(

I was thinking that you could find the number of ways that the boys [b]are[/b] seated together and subtract that from x!.[/quote]
You can, and thats what I did

[quote="nat mc"]It would be
$x!-2(x-2)!=(x^2-x-2)(x-2)!=(x+1)(x-2)(x-2)!$

With y it would be more compilcated because it wouldn't work if there were 2 together, 3 together, 4 together up to y together.  I'm thinking you would have to use summation[/quote]
Whoops, I screwd up.
It should be

$x!-2(x-1)!=(x-2)(x-1)!$

because the two boys can be in $x-1$ spots

no, before I said how many ways, if the two boys only sat on the first and second seats, I think my new way is correct

Thanks a lot!!! You just made me copy the wrong formula into Word!!!  :mad:

Just kidding.  :D  :D  :D  :D

But seriously, folks. I'm confused :what?: - I've never been good at counting and probability. Is it (x-2)(x-1)! or (x+1)(x-2)(x-2)! ? And please, please, please give a good explanation of your position.

Many thanks to all!!!!

Just find how many ways $x$ students can be arranged, $x!$, and substract the number of ways the 2 people can sit together. Treat them as 1 person, AB. Then there are $(x-1)!$ ways to arrange them. But there's also BA. Then there are $(x-1)!$ to arrange them that way. $x!-2(x-1)!$ is the answer.

When I got home from the math competition, that was what I thought too.

Thanks chess64!!!!!

[quote="nat mc"]its definitely
$(x-2)(x-1)!$ (same as chess64)[/quote]

Yes, that's a better way to write it.

It looks prettier, and it would be easier to multiply


Lý thuyết trường và lý thuyết Galois :
 Cau 1 :  Đạt $Z_{13}$ là tập hợp các lớp đòng dư mod 13, hãy định nghĩa pháp cộng, phép nhân trên $Z_{13}$để nó là 1 trường
Câu 2 : hãy xây dựng 1 trường có 9 phần tử
 Câu 3 : Tìm 1 tập mở rộng vô hạn của truờng số hữu tỉ Q
Câu 4 ; Tìm bậc trên Q của trường nghiệm của đa thức $f(x)=(x^{2}-5)(x^{2}-7)$Câu 5 : cho đa thức $f(x) =x^{3}-7$ xác định cấp của nhóm Galois của đa thức$f(x)$ trên Q

Bay gio anh ay di hoc o Nga hay sao ma thay co bai bang ca tieng Nga? :lol:


inequalities
calculus
function
derivative
logarithms
algebra unsolved
algebra

The sequence $\{a_n\}$ is defined by: $a_1=\frac{21}{16}$, and for $n\ge2$,\[ 2a_n-3a_{n-1}=\frac{3}{2^{n+1}}. \]Let $m$ be an integer with $m\ge2$. Prove that: for $n\le m$, we have\[ \left(a_n+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left(m-\left(\frac{2}{3}\right)^{{\frac{n(m-1)}{m}}}\right)<\frac{m^2-1}{m-n+1}. \]

I found a horrible, ugly solution using basic calculus, and it requires manually checking 2<=m<=5.

$2a_n-3a_{n-1}=\frac{3}{2^{n+1}}\Longleftrightarrow a_n+\frac{3}{2^{n+3}}=\frac{3}{2}\left(a_{n-1}+\frac{3}{2^{n+2}}\right)$,so we have $a_n+\frac{3}{2^{n+3}}=\left(\frac{3}{2}\right)^{n-1}\left(a_1+\frac{3}{2^4}\right)=\left(\frac{3}{2}\right)^n$.
Therefore we can rewrite the given inequality as follows.

\[\left(\frac{3}{2}\right)^{\frac{n}{m}}\left[m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right]<\frac{m^2-1}{m-n+1}\]

$1-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}=1-\left[\left(\frac{2}{3}\right)^{\frac{m-1}{m}}\right]^n<n\left[1-\left(\frac{2}{3}\right)^{\frac{m-1}{m}}\right]$

$\Longleftrightarrow m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}<m+n-1-n\left(\frac{2}{3}\right)^{\frac{m-1}{m}}$

[quote="kunny"]
Therefore we can rewrite the given inequality as follows.

\[\left(\frac{3}{2}\right)^{\frac{n}{m}}\left[m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right]<\frac{m^2-1}{m-n+1}\][/quote]

And from here, note that when $n=0$, we have equality. So think of it as a function in n, for $m \geq 6$, it suffices to show that this is monotonic increasing. This is a straight forward derivative check. One can then manually check when $2\leq m \leq 5$

[quote="Erken"]Using [b]Bernoulli[/b] inequality:
${ (\frac {n}{m - n + 1} + 1})^{\frac {m}{2n}}\geq(\frac {m}{2m - 2n + 2} + 1)$,but as we know $ m\geq n\geq 1$.[/quote]

It is wrong,because Bernoulli inequality requires ${\frac {m}{2n}}\ge 1$.:)


calculus
geometry
rectangle
analytic geometry
graphing lines
slope
calculus computations

A ball thrown downward at 10 feet per second from the top of a 100 feet building hits the ground in less than 3 seconds.  How can you prove it?

You do not need calculus for this problem.  Use the elementary school formula $ d\equal{}v_it\plus{}\frac{at^2}{2}$.

Though that formula IS derived through calculus (integrating $ a$ twice with respect to $ t$)

Chickendude, you don't need calculus to derive that expression (otherwise, projectile motion wouldn't be learned in high school physics, right?). For a constant acceleration motion, consider a graphic representation v = v(t). Then the distance traveled is given by what?

The "non-calculus" derivation depends on the relation between antiderivatives and areas, which is definitely a calculus theorem. Doing mechanics without basic calculus is seriously unsatisfying.
(What "high-school physics"? I went straight to the calculus-based AP class.)

This is a standard exercise at the introduction to antiderivatives: the point is understanding how the initial conditions relate to the arbitrary constants in the antiderivative formula.

I don't agree with that. Do you need a calculus based reasoning to consider the area of a rectangle?

You don't need calculus for rectilinear motion I believe. I remember deriving various high school physics formulas using average slopes and areas under straight lines, though this might all be wrong when curvature is introduced.

Of course you can calculate areas. They just don't mean anything without the FTC, and this exercise is given before the FTC is stated.

[quote="Jmerry"]They just don't mean anything without the FTC[/quote]

Forgive my ignorance, but why do we need the FTC to give areas a meaning?

Going from velocity to position is an antiderivative. Finding the area under the graph of velocity appears at first to be a different problem; the FTC relates the two.


Does anybody have either the solutions or the answer key to the 2002 Chapter Sprint Round?  I have done the problems but the link to the solutions can now "not be found"  :wallbash_red: .  Could anybody provide a link or give me a score?
[hide="my answers"]
1. 180
2. 61
3. 364
4. 4
5. 2
6. 2
7. 1.55
8. 21
9. 15
10. 5
11. 21

12. -3
13. 199
14. 72
15. 7/50
16. 16
17. 80
18. 9/17
19. 132
20. 9
21. 5
22. 2,664
23. 3/5
24. 24
35. 79
26. 7
27. 75
28. 12
29. 18
30. 15 [/hide]

How did I do(I hope I got at least 26 :))

You can trade for them in the MathCounts Resources HERE! Forum. Is it okay to post the answer in a forum?

[quote="yaofan"]gj. you got a 28[/quote]

Which ones did I get wrong(I bet one of them was 27)?
I don't have anything of value to trade, so that's why I posted here.

Oddly enough, that's the same score I got on the State Sprint too(but with state I used all my time and here I did it in 33 minutes)

Also, I think it's OK to post an answer key here.  But I'm not sure if your allowed to post the detailed solutions(probably OK, but best to be safe on this one).

you got [hide="#14"]120[/hide] and [hide="#27"]72[/hide] wrong.

Darn, 14 was an easy one.  I accidentally visualized it as if the 4 smaller pieces were the bigger pieces.  The ironic thing was, my first instinct was 120, but then I changed it. :(  But it was still a decent score anyway. :D.

Also, I made a typo on my last post.  I got a 28 on the 2003 State Sprint Round, not 2002(I got a 21 on the 2002 one, not one of my finer moments).

And thanks yaofan!