So, I want to start learning some "higher" level math for fun. What math subjects are problem-solving oriented?

In a sense, all of them. There are interesting problems that aren't too big everywhere. Just try something and see if you like it. You can always switch focus if you don't. If you're afraid of the initial time cost getting to know something, combinatorics is perhaps the least technical subject, with the hard problems understandable even with very little background.
In another sense, none of them. They're all just tools, which you can then apply creatively. Problem solving is more fun with a bigger toolbox. For a tool you can use almost anywhere, try linear algebra.

All righty! Do you have any good book reccomendations that have good collections of "nonroutine" problems?

I highly recommend [url=http://www.amazon.com/Proofs-that-Really-Count-Combinatorial/dp/0883853337]Proofs that Really Count[/url] by Benjamin and Quinn.  It discusses a lot of interesting ideas and its exercises are pretty interesting, especially the ones they included because nobody knows a combinatorial proof for them.

[url=http://math.mit.edu/~rstan/ec/ch1.pdf]Chapter 1[/url] of Richard Stanley's Enumerative Combinatorics is also available online and has tons of interesting problems.  He also has another pdf of [url=http://math.mit.edu/~rstan/bij.pdf]~250 bijective proof problems[/url] available, several of which, again, nobody knows a combinatorial proof for.

Of course there's always the books filled with Olympiad and Putnam problems.

I still feel like you aren't asking the right question.  How much time are you willing to spend to learn material before solving interesting problems?  If the answer is "a lot," there are interesting problems in every subject and in many difficult textbooks.

t0r, i've become literally obsessed with math. For instance, today I spent 10 hours working on olympiad problems. So I could definitely see myself spending a solid year or years learning an advanced subject if it meant that I could start solving "hard" problems with that knowledge.

I mean, I like the olympiad/putnam stuff so much now, but I know I can't really become a mathematician whose job is to solve contest problems that rely only on basic knowledge and cleverness. So I'm trying to find more "specialized" fields with equally difficult/clever exercises, so I can eventually take my skill in that area and start solving "real" problems and doing research. I don't know if that makes much sense, but that's as best as I can explain it...:)

EDIT: I guess for fun I could do olympiad problems on the side...there's probably a big enough collection out there to last a long time, right?  :maybe:

You could start by solving the problems [url=http://www.artofproblemsolving.com/Forum/index.php?f=584]here[/url].

Again, pretty much every field has difficult and clever exercises.


Hello,

This is akhil0422's new account (I'll stop talking in third person now).
I promised a whole bunch of AoPS users free tests in a topic but could only get through e-mailing around half of you all.
If I have not given you the tests you wanted please post in this topic as I have been very busy these weeks but still wish to upkeep my promise.

[b]This is NOT another opportunity to REQUEST tests; this only serves as a reminder as to who I STILL HAVE TO SEND TESTS to.[/b]
Do not bother scamming as I will double-check if you have asked in my original topic or not.

Why don't you just go back to that topic and send them from there?  There's no need to make a new thread.

And he can't look through the thread to find emails and requests?

He lost track of who has gotten the tests.

BTW, I got the tests I asked for.

Here is what I posted in the other thread:

[quote="Mathfreak123"]Am I allowed to request for *parts* of tests? For example, the sprint round from the 2002 Nationals. If so, can I have:

1990 State Countdown
1991 State Countdown
1993 State Countdown
1996 State Countdown
1997 State Countdown
2000 State Countdown
2000 National Answers
2001 National Solutions

Note: This is based on the fact that an average test will have 4-5 *parts* (Sprint, Target, Team, Answers, maybe Solutions/Masters), so 2 tests * 4 parts = 8 parts.

My email is mathfreak123@live.com

Thanks a ton![/quote]

You can always check your email sentbox.

My post is on the last page of the locked topic...

And I requested 2008 and 2007 nationals

What is this 'new account' business? I am suspicious your last account died for being a duplicate account and I am suspicious this one will too.

I have not gotten my test. my email is varunkmohan@gmail.com.
DO you supply any new national rounds.


Set systems
combinatorics
Extremal combinatorics
IMO Shortlist

Let $ \alpha < \frac {3 \minus{} \sqrt {5}}{2}$ be a positive real number. Prove that there exist positive integers $ n$ and $ p > \alpha \cdot 2^n$ for which one can select $ 2 \cdot p$ pairwise distinct subsets $ S_1, \ldots, S_p, T_1, \ldots, T_p$ of the set $ \{1,2, \ldots, n\}$ such that $ S_i \cap T_j \neq \emptyset$ for all $ 1 \leq i,j \leq p$

[i]Author: Gerhard Wöginger, Austria[/i]

[quote="mathe760"]Hi,

I really would like to see a elegant combinatorial proof to this beatiful problem :) I just know of an ugly proof which uses some limit computations...
I was thinking about if one could find the biggest such number p in dependence of n?

mathe760[/quote]

Dear [b]mathe760[/b],

You can find an official solution here: artofproblemsolving.com/Forum/viewtopic.php?f=177&t=214623
I know it uses logarithm and limit, but, to me, it is very elegant as a combinatorial proof.

What interests me more is that the approach used in the solution seems to create sets and simply count them.
This is straightforward enough as an enumerative method.
However, it is highly not obvious as to how to come up with such a solution.
I'm eager to know how the proposer came up with it and used it to make the beautiful constant $\frac{3-\sqrt{5}}{2}$ and also how to prove that the constant is sharp.

I believe that there may be some research behind this, which I would love to see.

Sincerely,
Disner.

In the book olympiad combinatorics this is placed in double counting section.The official solution uses only few double counting tricks.Does anyone has a solution using double counting as the main part of the solution?I have some idea but they seem to give us nothing since the result in the problem is quite strong.Here is a sketch of the idea:Assume for country that there doesn't exist such $n,p$ Then for every $p$ distinct subset of the set there are at most $p-1$ subsets that have a common element with all these subsets.Also we can count this in another way for a $k$ element subset there are $2^n-2^{n-k}$ subsets that have a common element with it so we can count this in another way(stars).In fact I used the same method of double counting used in Zarankiewicz’s problem.At the end I got into an inequality If we could show it is wrong for some $p,n$ We are done.Indeed we should use limit and I am not good at all in calculas.Can anyone verify if this works?Anyway if it doesn't I really like to see a solution using double counting(Or even a method that is easier to come up in an exam.

Let $a_0, \dots, a_{k-1}$ be positive integers, and define $s_i = a_0 + \dots + a_{i-1}$ (so $s_0 = 0$); let $s_k = s$. Partition the set $\{1, \dots, s\}$ into segments $I_i = \{s_i + 1, \dots, s_i + a_i\}$ for all $i$, and define two properties on subsets of $\{1, \dots, s\}$ as follows:
[list]
[*] Say a subset has property $P$ if it contains one of the segments $I_0, \dots, I_{k-1}$.
[*] Say a subset has property $Q$ if it contains at least one point from each of the segments $I_0, \dots, I_{k-1}$.
[/list]
Clearly each subset with property $P$ intersects each subset with property $Q$.

By easy counting, we have
\begin{align*}
2^{-s} |Q| & = \prod (1 - 2^{-a_i}),\\
2^{-s} |P| & = 1 - \prod (1 - 2^{-a_i}),\\
2^{-s} |P \cap Q| & = \prod (1 - 2^{-a_i}) - \prod (1 - 2^{-a_i+1}).
\end{align*}
To solve the problem, we let elements of $P$ be the $S_i$s, and elements of $Q \setminus P$ be the $T_i$s. For this to be valid, we need
\[
\prod (1 - 2^{-a_i}) < 1 - \alpha
\quad \text{and} \quad
\prod (1 - 2^{-a_i+1}) > \alpha.
\]
We now specialize to solutions with $a_0 = \dots = a_{k-1} = a$, which requires
\[
(1 - 2^{-a})^k < 1 - \alpha
\quad \text{and} \quad
(1 - 2^{-a + 1})^k > \alpha.
\]
Let $\lambda = \frac{3 - \sqrt{5}}{2}$, so $\alpha < \lambda = (1 - \lambda)^2 < (1 - \alpha)^2$. For each value of $a$, we try the choice $k_a = \left \lfloor -2^a \log (1 - \lambda) \right \rfloor$. Then easy limit computations give
\[
\lim_{k \to \infty} (1 - 2^{-a})^{k_a}
= 1 - \lambda
\quad \text{and} \quad
\lim_{k \to \infty} (1 - 2^{-a+1})^{k_a}
= (1 - \lambda)^2 = \lambda.
\]
It follows that for sufficiently large $a$, both desired inequalities are satisfied.

This is a paper from the 80's by Daykin and Frankl.

[url]https://gilkalai.files.wordpress.com/2022/11/daykinfranklirrationalplusimage.pdf[/url]

In recent progress on Frankl's conjecture this number $\frac{3-\sqrt{5}}{2}$ has also come up.

We prove that $\frac{3-\sqrt{5}}{2}$ is optimal.

Let $[n]=\{1,\cdots,n\}$ and $\mathcal{P}([n])$ to be the set of all subsets (includes $[n]$ and $\emptyset$) of $[n]$. Define an [b]up-set[/b] to be a set $\mathcal{S}$ such that for all $T\in \mathcal{S}$, any set $U$ satisfying $T\subseteq U \subseteq [n]$

[b]Kleitman's lemma:[/b] Let $\mathcal{A}, \mathcal{B}\subseteq \mathcal{P}([n])$ be up-sets, then $$|\mathcal{A} \cap \mathcal{B}| \ge \frac{|\mathcal{A}| |\mathcal{B}|}{2^n}$$

Proof: Induct on $n$. The base cases are clear. To induct, let $\mathcal{A}_1$ be the set of elements containing $n$ in $\mathcal{A}$ and $\mathcal{A}_0$ be the other elements. Define $\mathcal{B}_1$ and $\mathcal{B}_0$ similarly. We can see that $|\mathcal{A}_1|\ge |\mathcal{A}_0|,|\mathcal{B}_1|\ge |\mathcal{B}_0|$. Therefore, by inductive hypothesis on $n-1$ on $\mathcal{A}_1, \mathcal{B}_1$ and $\mathcal{A}_0, \mathcal{B}_0$
\begin{align*}
|\mathcal{A} \cap \mathcal{B}| &= |\mathcal{A}_1 \cap \mathcal{B}_1| + |\mathcal{A}_0 \cap \mathcal{B}_0| \\
&\le \frac{1}{2^{n-1}} \left( |\mathcal{A}_1||\mathcal{B}_1| + |\mathcal{A}_0||\mathcal{B}_0|\right) \\
&\le \frac{1}{2^{n}} \left( |\mathcal{A}_1|+|\mathcal{A}_0|\right) \left( |\mathcal{B}_1|+|\mathcal{B}_0|\right) \\
&= \frac{1}{2^{n}} |\mathcal{A}| |\mathcal{B}|
\end{align*}

Back to the original problem. Let $\mathcal{S} = \{ S_1, \cdots, S_p\}, \mathcal{T} = \{T_1,\cdots,T_p\}$. Let $\mathcal{S'}, \mathcal{T'}$ be the minimal up-sets that cover $\mathcal{S}$ and $\mathcal{T}$. We can see that $$|\mathcal{S}| + |\mathcal{T}| = |\mathcal{S} \cup \mathcal{T}| \le |\mathcal{S'} \cup \mathcal{T'}| $$

$$ = |\mathcal{S'}| + |\mathcal{T'}| - |\mathcal{S'} \cap \mathcal{T'}|$$

I claim for all $A'\in \mathcal{S'}, B'\in \mathcal{T'}$ we have $A'\cap B' \ne \emptyset$. Note there exists $A,B$ such that $A\subseteq A', B\subseteq B'$ and $A \in \mathcal{S}, B\in \mathcal{T}$. Let $X= |\mathcal{S'}| + |\mathcal{T'}|$. This implies that

$$ |\mathcal{S'}| + |\mathcal{T'}| = \sum_{A\subseteq [n]} (1_{A\in \mathcal{S'}} + 1_{[n]\backslash A\in \mathcal{T'}}) \le 2^n$$

By Kleitman's lemma, $$|\mathcal{S'} \cap \mathcal{T'}| \ge |\mathcal{S'}||\mathcal{T'}| 2^{-n} \ge |\mathcal{S}| \left(|\mathcal{S'}| + |\mathcal{T'}| - |\mathcal{S}|\right) 2^{-n} =  |\mathcal{S}| \left(X - |\mathcal{S}|\right) 2^{-n}$$

Since $|\mathcal{S}| \le |\mathcal{S'}|, |\mathcal{T'}|$

It readily follows that $$2|\mathcal{S}| \le X - |\mathcal{S}| \left(X - |\mathcal{S}| \right) 2^{-n} \le 2^n - |\mathcal{S}| \left(2^n - |\mathcal{S}| \right) 2^{-n} $$

Let $\alpha = \frac{|\mathcal{S}|}{2^n}$ then this becomes $$2\alpha \le 1-\alpha+\alpha^2 \iff \alpha^2-3\alpha+1 \ge 0 \iff \alpha < \frac{3-\sqrt{5}}{2}$$

Remark: The condition that $|\mathcal{S}| = |\mathcal{T}|$ is actually weaker than it looks; one can move/add sets "freely" between these two carefully.

Solved with [b]megarnie[/b] (at HMMT February!).

Let $a,b$ be (TBD) positive integer parameters and set $n=ab.$ Let $\gamma = \frac{3-\sqrt{5}}{2}$ and the key idea is $1 - 2\gamma + \gamma^2 = \gamma.$

Let the family $T$ be all subsets of $\{1,2, \dots n\}$ that include at least one integer $r \pmod{b}$ for any residue $r.$ Let the family $S$ be all subsets of $\{1,2, \dots n\}$ not in $T$ that include all integers $r \pmod{b}$ for some residue $r$.

First, $|T| = (2^{a} - 1)^{b}.$

To count $|S|$ first subtract from $2^{ab}$ the number of subsets that don't have all the integers $r \pmod{b}$ for any residue $r,$ which is $(2^{a} - 1)^{b}.$ This is $2^{ab} - (2^{a} - 1)^{b}$ and now we have to subtract elements in $T.$ Note the number of subsets of $\{1,2, \dots n\}$ that include at least one integer, but not all integers $r \pmod{b}$ for any residue $r$ is $(2^{a} - 2)^{b}.$ So our final count is $$|S| = 2^{ab} - (2^{a} - 1)^{b} - ((2^{a} - 1)^{b} - (2^{a} - 2)^{b}) = 2^{ab} - 2(2^{a}-1)^{b} + (2^{a} - 2)^{b}.$$

Choose very large $a$ and $b$ so that $\left( \frac{2^{a} - 1}{2^{a}} \right)^{b} \approx \gamma,$ so $\frac{|T|}{2^{ab}}$ is arbitrarily close to $\gamma.$ But now see that $\frac{|S|}{2^{ab}} = 1 - 2\left( \frac{2^{a}-1}{2^{a}} \right)^{b} + \left( \frac{2^{a}-2}{2^{a}} \right)^{b}$ is arbitrarily close to $1 - 2 \gamma + \gamma^2 = \gamma$ as well, noting the approximation $(1-\epsilon )^{2} \approx 1 - 2\epsilon $ for small $\epsilon.$ $\blacksquare$


12 points are placed on a plane such that no three are collinear. Lines are drawn between some of these points such that no two lines intersect. The points are then rearranged (with the lines still connecting them) into a new configuration where none of the points are collinear. What is the maximum number of points of intersection of the lines for the new configuration?

this question seems absurd to me. are we talking lines here or line [i]segments[/i]?


calculus
integration
trigonometry
calculus computations

Evaluate $\int_{0}^{1}(x\tan^{-1}x)^{n}dx$. $n$ runs through all positive integers.

If I denote the given integral by $I_{n,n}$ then I can prove that
$I_{n,n}+\frac{n-1}{n+1}I_{n-2,n}+\frac{n}{n+1}I_{n-1,n-1}= \frac{2}{n+1}(\frac{\pi}{4})^{n}$. I don't know the solution of this difference equation.

I did it with Maple for $n=1, 2, 3, 4$ and the results were
$n=1 \rightarrow \frac{\pi}{4}-\frac{1}{2}$
$n=2 \rightarrow \frac{C+1}{3}-\frac{\pi}{6}-\frac{\pi \ln(2)}{12}+\frac{\pi^{2}}{48}$
$n=3 \rightarrow \frac{\pi}{8}-C-\frac{1}{4}+\frac{\pi^{2}}{32}+\frac{\pi\ln(2)}{4}$
$n=4 \rightarrow \frac{1}{5}+\frac{127\pi^{4}i}{19200}+\frac{\pi^{2}i}{4}-\frac{\pi^{2}}{10}+\frac{3i}{5}\text{polylog}(4,-i)-\frac{\pi}{10}$
$+\frac{\pi^{3}}{80}+\frac{\pi^{4}}{1280}-\pi\ln(1+i)+2C+\frac{\pi^{3}\ln(1+i)}{80}-\frac{3\pi^{2}C}{40}+\frac{3\pi}{10}\text{polylog}(3,-i)$

where $C$ is Catalan's constant. Actually, the last one should be real, so maybe there's a simpler form for it and Maple just won't admit it. But still, this really discourages me from the hope of finding a closed form for this in terms of $n$...

On the other hand, you may note that $(-1)^{n}\left(\frac{2-\pi}{2n+2}\right)$ appears in each of those. I bet it always does. :D

Thanks for your time on this problem. Maple does give some interesting answers.


AMC USA(J)MO USAMO AIME calculus vector inequalities

I got into AIME every year since 8th grade. However i was never able to get into USAMO, but I never really studied for them. This year ( I am a Junior now ) I really want to get into USAMO for at least ONCE in my life. I'm not sure what to study
I bought the AoPS books 3 years go (The authors visited my school) But never really used them.
Even though i took calculus as a sophmore, i can't quite use those skills during math contest.
 So last year for AIME I only got question 1,3,13,15 right, which is weird, since i got all the ones the smartest people in my school got wrong on.
Which chapters should i study from the AoPS books?

1,3,13,15? Are you sure about that? It seems impossible.

Study the whole book if you can; if you can't, I suppose you could ignore th chapters on vectors, inequalities, esoteric geometry, and graph theory.

lol yup ! its possible!
I could have gotten a 6 ,except 2 of the problems i messed up and forgot to count the end and not counting the end
1 and 3 aren't hard
for 13 and 15 i basically wrote out  a lot of stuff anf found a pattern
like for 13 i think i used a binder paper and folded it into 6 columes and wrote out stuff on the entire page and got a patter and got the answer
for 15 i also did something to find the pattern

Yeah...brute-forcing is very possible on many AIME problems...if you're willing to sacrifice actual problem solving time to brute force the brute-forcable problems. 3 hours seems like alot, but in reality it's 12 minutes per problem, alot of brute force could take up to 30 minutes.

Well, you can definitely get into USAMO in one year's work from where you are now (I did last year).  Yes, in the USAMO (and IMO, and most math olympiads for high school students, except for some places like Romania), you shouldn't have to use calculus to solve any problem, and the same goes for qualifying exams like the AIME.  Knowing the [i]material[/i] is definitely different from being able to do the problems.  For material, I would say the AoPS books are excellent.  Probably you can just skim volume 1; for volume 2, everything is helpful for the AIME, except perhaps the limits.  You don't have to do every problem of every chapter, and you probably would become frustrated trying to do so. I guess it is true that vectors don't seem to crop up much except in the occasional 3D geometry, but they are EXTREMELY useful in math in general and I recommend that you learn some stuff about them.  You can actually go in a vaguely linear fashion as long as you don't get hung up on too many individual problems; you have enough time.

For developing the problem-solving ability, I would advise you to work on AIME problems until you get 7s or so, and then start working on easyish olympiad problems.  For this last category, there are plenty of recommendations on this site.  I like Engel's [i]Problem Solving Strategies[/i].  In fact, once you are ready for them, olympiad problems are the best way to get better at easier problems, so that by the time you can comfortably make USAMO, you're probably trying to do well on it, which is interesting, now that I think about it.

And yeah, when you're practicing, you should really try to find elegant solutions.  Brute-forcing does you no good unless you just need another point during an actual competition, and it's pretty unhealthy; if you can't solve it and then look at an elegant solution, and really study it and understand it, and try to understand how you would have gotten it, that will help you.

[quote="Elemennop"][quote="Boy Soprano II"]
And yeah, when you're practicing, you should really try to find elegant solutions.  Brute-forcing does you no good unless you just need another point during an actual competition[/quote]

Really eh

I bruteforce almost everything; most of the time it's quicker for me to brute force a problem than it is for me to do it the elegant way.[/quote]

agreed

thanks! I'm not that great at math problem solve anymore because i didn't really get accutual practice since like 3rd grade( thanks to my dad, i accutally like math), a lot becuase i moved to america and had to learn english and stuff, and my dad got too busy to teach me things also.
i really didn't know how to solve the other problems, that why i was brute forcing the problems that are accutally possible for bruteforce because i basically sat there for half an hour  after i brute force and some people uses like only 2 hours and then they go back to class and get into USAMO

In response to Elemennop and Iversonfan2007, I would like to mention that the title of (at least half of) this site is "The [b]Art[/b] of Problem Solving," not "Getting a Solution at Any Cost."  To be sure, in a competition, the (slightly perverse) goal is to get as good a score as possible.  But in other circumstances, looking for elegant solutions is actually much better for you.  Indeed, I remember one of the instructors saying the following at the end of MOP this year :  "During the competitions, sure, dumb*ss, but at other times, please look for elegant solutions."

And which is more satisfying, after all?

DUMBASS SOLUTION

er

I mean

elegant solution, of course

[quote="rainynightstarz"]any other suggestions on what and how i should study?[/quote]

just do old aimes, like a bunch of old aimes, it helps a lot on amc 12 and aime.

If you got 1,3,13, and 15 right...first of all I'm not sure how that's possible unless you didn't happen to look at 2,4, 5-8 or have terrible accuracy. In the case of the former, I'd suggest you read those questions - they tend to give you points. In the case of the latter, work on that instead of trying to answer more questions.

AIME problems (1-10 at least) tend to have solutions which do not require mathematical knowledge not taught in classrooms. You've taken Cal already; you should be fine. Clearly if you got 13 and 15 right, there's no problem with your ability to find solutions. So work on solving the "easy" ones?

[quote="JSteinhardt"]Of course, not all problems will have the elegant solutions you want them to have. You can sit for 3 hours looking for that beautiful bijection that doesn't exist, or you can just spend the 5 minutes it takes to develop that recursion that is staring you right in the face.

But probably at the AIME level most problems will have a degree of elegance in one form or another that you should try find.[/quote]

I think that Boy Soprano was probably thinking of something like a brute force solution to this year's AIME II #11, where it was possible to find all twenty-some terms and add them.

Yes but if you have to spend lots of time calculating, then that's bad.

And when practicing, going for the elegant solutions makes you better.

And you are less likely to get an elegant solution wrong.

[quote="not_trig"]hmm... not if you're lazy  :ninja:

And elegant solutions... well, when they don't look at your solution, SCREW THAT!!! When they do look at it, work on it. Make it pretty. Math should be aesthetically pleasing, but if you don't have enough time, it's pointless. AMCs and AIME look at [i]if you get the problem right[/i], not how pretty your (possibly wrong?) solution is.[/quote]

As Miles said earlier, look at the top: The [i]Art of Problem Solving[/i]...

It is true that you should brute force when running out of time majorly. When looking at the AIME, you should NEVER brute force until you have about 30 min left and don't feel like checking your work.

If you "screw that," chances of you doing well on USAMO are negligible.

As Mr. Rusczyk once said, you should put your pencil down and think when you don't see an elegant solution right away. This could save you half an hour of brute forcing, which is boring and error-prone.

[quote="Karth"]until you have about 30 min left and don't feel like checking your work.
[/quote]

If you don't check your work, chances of qualification for the USAMO are also negligible.

In the last 30 minutes you should invariably stop solving problems and instead check the problems you've already solved. If, after checking all those problems, you still have time left, then you can try to solve another problem.

One thing we must realize is that often times what looks like a good solution becomes really messy. When that happens, you neede to stop there and look for another approach.

[quote="bpms"][quote="MysticTerminator"][quote="PenguinIntegral"]Just sit down with AoPS 2 and a drink. Things usually work themselves out from there.[/quote]

"and a drink"? wtf?[/quote]
Maybe penguinintegral focuses better under the influence?[/quote]

i think he meant a soft drink becauase alcohol is BAD right

Recall that the title of his blog is "Vodka Gives Your Liver Superpowers!"

[quote="pianoforte"]but a superpowered liver doesn't help on math[/quote]

REALLY, now?

I beg to differ.  :|

[quote="jhcreinhardt"]this seems to be getting a little off topic..[/quote]

Hey did you go to UGA a couple of days ago?

Completely off topic now. Leave conversations like that to PM or AIM.

This is clearly not very constructive. The topic is trying to make the USAMO. Or perhaps its about actually making it to the USAMO, as in someone needs directions to their test center.

But anyhow, I'll assume its the former, and say that it isn't as difficult as it may seem at first. Having been through the same position, its all about practicing problems. There is absolutely no substitute for practice. Do your AIMEs, those ought to be more than enough.  If you don't make it, oh well, its not the end of the world. Make sure you can qualify for the AIME too. The AIME is pretty predictable, if you're just pretty good at SOMETHING, and decent at everything else, you'll do well. If you're really good at geometry, then you'll have a decent time with it. If you're pretty good at combinatorics, you'll have a decent time with it.

The learning curve is steep, but it gets easier the more you go along.


helo
f an application from$ \mathbb N$ to $ \mathbb N$ such as f is  increasing,proof this:
$ \forall n \in %Error. "mathh" is a bad command.
N: f(n)\ge n$

By induction,

$ f(1)\geq1$
$ f(n\plus{}1)\geq f(n)\plus{}1\geq n\plus{}1$

First, some English:
[i]Hello[/i] is the word commonly used for greetings.
$ f: \bf N \to N$ is a [i]function[/i].
"Proof" is a noun; "Every theorem has a proof."  [i]To prove[/i] is the associated verb.

Now, some math: Jiang's proof is by induction, i.e. you first show it true for $ n \equal{} 1$, then you assume it true for $ n$ and prove it true for $ n \plus{} 1$.  So, $ f(n) \geq n$ is the inductive hypothesis.


http://www.artofproblemsolving.com/Forum/viewtopic.php?p=721505#721505

I didn't understand the solution that was posted.  How did the numbers in form $ a/b$ change to the expression in the solution?

[quote="t0rajir0u"][hide="Discussion"] $ 1000 = 2^{3}\cdot 5^{3}$.  If $ a, b$ are relatively prime divisors of $ 1000$, then only one of them can have factors of $ 2$, and similarly only one of them can have factors of $ 5$.  We then have two cases: either $ a$ is a power of $ 2$ and $ b$ is a power of $ 5$, or else $ b$ is a power of $ 2$ and $ a$ is a power of $ 5$.

Every term $ \frac{a}{b}$ is therefore going to look like $ 2^{i}5^{j}$, depending on which choice of $ a, b$ we make, where $ -3 \le i, j \le 3$.  The expression I gave for $ S$ gives the sum of all possible such terms. [/hide][/quote]

Thanks!  I started out with the same concept, but got a little lost along the way...

[quote]We then have two cases: either a is a power of 2 and b is a power of 5, or else b is a power of 2 and a is a power of 5. [/quote]

Should this case be expanded to those where either a or b can include both powers of 2 and powers of 5, provided that the numerator or denomiator was a power of 0 of 2 or 5?  For example, $ \frac{2^{3}5^{2}}{5^{0}}$ would work, even though it has a power of 5 on both numerator and denominator.

Oh yeah, I worked it out, and when I listed everything out in an organized way, the expression came out.  Nice!

[quote="mathmonster"]For example, $ \frac{2^{3}5^{2}}{5^{0}}$ would work, even though it has a power of 5 on both numerator and denominator.  [/quote]

Right, yes.  I forgot to mention that case  :oops:   Fortunately, it's included in the expression I gave for $ S$  :)


Find $ \lim_{x \to \minus{}\infty} e^{x}(1\minus{}x)$.

[quote="AndrewTom"]Find $ \lim_{x \to \minus{} \infty} e^{x}(1 \minus{} x)$.[/quote]

let $ x\equal{}\minus{}t$

$ \lim_{x\to\minus{}\infty}e^{x}(1\minus{}x) \equal{}\lim_{t\to\infty}\frac{1\plus{}t}{e^t} \equal{}0$

hello, we have $ \lim_{x \to \minus{} \infty}\frac {1 \minus{} x}{e^{ \minus{} x}} \equal{} \lim_{x \to \minus{} \infty}\frac { \minus{} 1}{ \minus{} e^{ \minus{} x}} \equal{} \lim_{x \to \minus{}\infty}e^x \equal{} 0$ by the rules of L'Hospital.
Sonnhard.

Thanks Sonnhard. (Your final $ x \to \infty$ should be $ x \to \minus{}\infty$.)

Hello, thank you for your hint.
Sonnhard.


Each of the 'n' passengers sitting in a bus may get down from it at next stop with probability

P.Moreover, at the next stop either no passenger or exactly one passenger boards the bus. The

probability of no passenger boarding the bus at the next stop being Po . Find the probability that when

the bus continues on its way after the stop, there will again be 'n' passengers in the bus.

The probability that 0 people get on and 0 people get off is _______.  The probability that 1 person gets on and 1 person gets off is _______.  Since only 0 or 1 person will get on (and these two cases are disjoint), __________.  Therefore, the answer is _________.


inequalities
trigonometry
triangle inequality

This is the problem in the section 4 "sdrawkcaB knihT, Write Forwards" of the "How to Write a Solution" [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php]article[/url].  It seems to me the condition that a, b, c are the lengths of the sides of a triangle is [b]not necessary[/b] as long as they are nonnegative.

[quote][b]Problem[/b]: Let a, b, and c denote the lengths of the sides of a triangle. Show that

a2(-a + b + c) + b2(a - b + c) + c2(a + b - c)  :le: 3abc. [/quote]

Adding [tex]a^3,b^3,c^3[/tex] to both sides, the inequality is equivalent to:

[tex]a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\leq a^3+b^3+c^3 + 3abc[/tex]

Is this exactly the Schur's Inequality?

You are correct, that is Schur's Inequality for r=1. The triangle condition is not necessary.

Without having read the article, I would imagine that it is trying to teach a method of thought rather than a base of knowledge.  That is, learning Schur's inequality is nice, but comparatively, knowing how to think about inequality problems is going to be more useful than just knowing Schur's.

Yet another way to do it... (apologies for writing it backwards)

A classic substitution that works for the condition that a, b, c are sides of a triangle is
x = (b + c - a)/2
y = (c + a - b)/2
z = (a + b - c)/2

c = x + y
a = y + z
b = z + x.

This looks especially good for this inequality because you get

2(x(y + z)^2 + y (z + x) ^2 + z (x + y)^2) <= 3(x + y) (y + z) (z + x)

or sigma_sym (x (y + z)^2) >= sigma_sym(3x^2 y + xyz).
The left hand side is also sigma_sym(2x^2 y + 2xyz)

At this point you can just use AM-GM: sigma_sym(x^2 y) >= sigma_sym (xyz).

Since people still like to view it as a triangle inequality, I am posting a pretty complicated proof using [b]trignometry[/b]. See Attached PDF file.

[quote="Alison"]A classic substitution that works for the condition that a, b, c are sides of a triangle is
x = (b + c - a)/2
y = (c + a - b)/2
z = (a + b - c)/2

At this point you can just use AM-GM ...[/quote]

It's nice to see another different proof. Since you are using AM-GM, x, y, z have to be non-negative, that means you need to assume a, b, c are sides of a triangle.


induction
number theory unsolved
number theory

Given m positive integers $x_{1},....,x_{m}$ s.t 1<=$x_{i}$<=n and $ sum_{i=1}^{n}x_{i}$$=k.n!$ with k is positive intege.
Prove that  we can partition $x_{1},...,x_{n} $ into k subsets s.t sum of number in asubset is equal $n!$

We'll show we can get a subset that sums to $n!$, by induction on $n$.

Note that among any $a$ numbers, we can pick a subset of them such that their sum is a multiple of $a$. Indeed, let the numbers be $v_1, v_2, ..., v_a$. Now the $a$ subsets $v_1, v_1+v_2, ..., v_1+...+v_a$ are either all incongruent mod $a$ (so that one will be $\equiv 0$) or two will be congruent mod $a$ (so that their difference will be $\equiv 0$). Now, setting $a=n$, we find that among $n$ numbers there's a subset with its sum a multiple of $n$. Note that this subset has at most $n$ elements.

Among the $x_i$, put away the ones equal to $n$. Among the others, (1) randomly pick $n$, (2) from these choose a subset with its sum a multiple of $n$, and put these away as a group. Then repeat (1) and (2). Since the subsets have at most $n$ elements and each element in them is at most $n-1$, each has sum at most $(n-1)n$.

We can put  the $x_i$ away in subsets leaving at most $n-1$. Now, since each subset we put away is a multiple of $n$, and the total is a multiple of $n$, these last elements add up to a multiple of $n$, so they're another such subset. Now we have groups adding up to some of $n, 2n, ..., (n-1)n$. Just divide by $n$, and it's equivalent to finding a subset adding up to $(n-1)!$ among $x_i$ adding up to $k \times (n-1)!$, each at most $n-1$. Since the base case $n=1$ is trivial, we're done.

Thank you very much.I also have same solution with you.Now , I have posted to http://www.diendantoanhoc.net and hatucdao also have same solution with us.
http://diendantoanhoc.net/forum/index.php?showtopic=7755

It is from Tournament of the towns, a very interesting problem by the way...


factorize $ x^{15}\minus{}y^{15}$
and $ x^{15}\plus{}y^{15}$

someone say there are 8 factors for each of the above expression, but i cant find it

Their factorizations are:

$ (x\minus{}y) \left(x^2\plus{}x y\plus{}y^2\right) \left(x^4\plus{}x^3 y\plus{}x^2 y^2\plus{}x y^3\plus{}y^4\right) \left(x^8\minus{}x^7 y\plus{}x^5 y^3\minus{}x^4 y^4\plus{}x^3 y^5\minus{}x y^7\plus{}y^8\right)$

and

$ (x\plus{}y) \left(x^2\minus{}x y\plus{}y^2\right) \left(x^4\minus{}x^3 y\plus{}x^2 y^2\minus{}x y^3\plus{}y^4\right) \left(x^8\plus{}x^7 y\minus{}x^5 y^3\minus{}x^4 y^4\minus{}x^3 y^5\plus{}x y^7\plus{}y^8\right)$

respectively. Perhaps they meant 8 factors total?

Can any of those be factored more?  Note that our only restriction is that the coefficients be real.
For example, $ x^4\plus{}y^4$ can be factored as $ (x^2\plus{}y^2)^2 \minus{} 2x^2y^2 \equal{} (x^2 \plus{}\sqrt{2}xy \plus{} y^2)(x^2 \minus{} \sqrt{2}xy \plus{} y^2)$.  Does something similar apply here?

Rewrite the polynomials as $ \left( \frac{x}{y} \right)^{15} \pm 1$ and use [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]Cyclotomic Polynomials[/url], which give the integer coefficient polynomials that factor $ P(x) \equal{} x^n \minus{} 1$ for any $ n$. This gives a quick proof that TZF's (almost called you Zuton) polynomials are irreducible over the rationals.

Likewise, the roots of the original polynomial and its factors are all $ 15$th roots of unity, so to get all $ 8$ factors in real coefficients you'd take each one in pairs except for $ 1$ which is unpaired.

Add 'conjugate'

[quote="MellowMelon"]Likewise, the roots of the original polynomial and its factors are all $ 15$th roots of unity, so to get all $ 8$ factors in real coefficients you'd take each one in [b]conjugate[/b] pairs except for $ 1$ which is unpaired.[/quote]


Hi,

when do you think the new contest starts?

Bye
Daniel Gutekunst

I have decided to go with a bi-annual Spring - Autumn system for the contest.

The 3rd Edition will begin some time at the end of February - beginning of March.

It will be announced with at least 2 weeks in advance.

well the answer to your question is clear now, isn't it? :) it's 22nd of Feburary 2004 :)

Yes, and I am still 19 on that date, so I will try to participate --- if I am not too occupied with the actual Informatics- and Maths-Contests in Germany (or even with school, I hate French, hehe).

Would u please show me some the old mathlink contest problem?

Take a look at http://www.mathlinks.ro/Forum/contest.php

when will 6th edition of Contest begin :?:  :?

Start of october. Hopefully I will find the time to participate this year.

(I usually find time to solve them but not to latex them neatly :P)


[quote="arjunlandes"]All I know is that there has to be an odd number of solutions, because $y=\frac{1}{100}x$ and $y=sinx$ are odd functions and intersect at $x=0$. I'm sure that is very helpful.[/quote]

Here is a hint.

This problem asks you to "visualize" it because without calculator, it is impossible to see which $x$ values will work (other than 0).

[quote="Silverfalcon"]The number of real solutions to the equation

\[ \frac{x}{100} = \sin x  \]

is

$\text{(A)} \ 61 \qquad \text{(B)} \ 62 \qquad \text{(C)} \ 63 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$[/quote]

[hide]
Answer: (E) 65
[/hide]

[quote="kimby_102"][hide="solution"]
It's easy to see that $(-100,-1)$ is the starting point, while $(100,1)$ is the last point that the two graphs $y=\frac{x}{100}$ and $y=\sin x$ meet.
Now we have to count all the points b/w the two points.
On the positive side, excluding $0$, we get 32 points including the ending point. Likewise, we get 32 points on the negative side.
So there are total $32\times 2 + 1=65$points.
[/hide][/quote]

That's not a rigorous solution. How exactly did you "count" the points of intersection in the x interval $(0,100]$?

[quote="kimby_102"][hide="solution"]
... while $(100,1)$ is the last point that the two graphs $y=\frac{x}{100}$ and $y=\sin x$ meet...
[/hide][/quote]
$\sin 100 \neq 1$

Think of it this way. Between x = -100 to x = 100, sin x crosses the line x/100 every time it goes from -1 to 1 and every time it goes from 1 to -1. You can count it—it's 63.

Please explain diagramatically or whatever way it can be understood to show how exactly would intersection occur and how did u count ?

 :oops:

[hide]Consider the graphs of $ y \equal{} \frac {x}{100}$ and $ y \equal{} \sin{x}$ seperately.Since the period of $ \sin$ is $ 2\pi$, the line intersects it every $ \pi$, once in the top, then in the bottom. So, for the $ x>{0}$,  it intersects $ \frac {100}{\pi} \equal{} {31}$ times. Same for the $ x<{0}$ side. That's $ 31 \plus{} 31$, but we count 0 now. $ \boxed{63}$.[/hide].


algebra
system of equations
polynomial
Rational Root Theorem

[quote="marsupilami"] Solve the equation: $ \frac{x^2}{\sqrt{x\plus{}2}}\plus{}1\equal{}2x^2$.[/quote]

We can rewrite this as

$ x^2\plus{}\sqrt{x\plus{}2}\equal{}2x^2\sqrt{x\plus{}2}$

$ \implies x^2\equal{}\sqrt{x\plus{}2}(2x^2\minus{}1)$

$ \implies \sqrt{x\plus{}2}\equal{}\frac{x^2}{2x^2\minus{}1}$

$ \implies x\plus{}2\equal{}\frac{x^4}{4x^4\minus{}4x^2\plus{}1}$

$ \implies 4x^5\plus{}8x^4\minus{}4x^3\minus{}9x^2\plus{}x\plus{}2\equal{}0$

By the Rational Root Theorem, $ x\equal{}\minus{}1$ is a root of the equation. Thus, we may write

$ 4x^5\plus{}8x^4\minus{}4x^3\minus{}9x^2\plus{}x\plus{}2\equal{}(x\plus{}1)(4x^4\plus{}4x^3\minus{}8x^2\minus{}x\plus{}2)\equal{}0$.

However, by the Rational Root Theorem,
$ 4x^4\plus{}4x^3\minus{}8x^2\minus{}x\plus{}2$ does not have any rational roots (it has more real roots, but not rational ones).

Thus, $ x\equal{}\minus{}1$ is the only rational solution.

Thank you but solve it in R  :D
and the system of equations, anyone helps me? :)


i kinda suck at rubix cubes, i just started actually trying 2 figure them out. Does any1 have guide 4 2x2?

I just use a normal method for 3x3's (that is pretty slow in the first place) and get to skip a few steps.  there are specialized methods for 2x2's if you want to get really good :D but I don't know of any

Basically if you learned the Rubiks Cube Booklet way, it's useless for the 2x2. You should learn the Corner Oriented Way for the 3x3 (Or just basically basic CFOP)
Then it's like
Solve the first layer (no f2l needed)
Then switch the 2 corners that ar ein the wrong place
Then solve the cube with OLL


Determine all functions f : R → R such that $ f (x-f (y)) = f (f (y))+x f (y)+f (x)-1$
for all x, y ε R

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19782]IMO 1999, #6.[/url]
(I moved [b]aviateurpilot[/b] and [b]cckek[/b]'s solutions there)


[i]Apologies if this is a bit sloppy, it's a late at night  :oops: [/i]

    I claim that all positive integers $m \leq \frac{n(n + 1)}{2}$ can be written in this form.

    First, we prove that no larger value for $m$ can be obtained,
    The maximum value of $a/b$ for integers $a$ and $b$ occurs when $b = 1$. Thus
    $$
    \frac{1}{x_{1}}+\frac{2}{x_{2}}+\frac{3}{x_{3}}+\cdots+\frac{n}{x_{n}} \leq 1  + 2 + 3 \cdots + n = \frac{n(n + 1)}{2}.
    $$

    Now we prove by induction that all such $m$ can be obtained. For $n = 1$, the maximum is $1$, which can be obtained with $x_1 = 1$, so the base case is true.

    Then, assume that for some $k$ there exists $x_1, x_2, \dots, x_k$ such that all $m \leq \frac{n(n + 1)}{2}$ can be obtained.
    First, note that by adding the term
    $$
    \frac{k + 1}{x_n},
    $$
    we can still obtain any value in $m$, as if $m \neq 1$, we can write for some $x_1, \dots, x_k$,
    \begin{align*}
        \frac{1}{x_{1}}+\frac{2}{x_{2}}+\frac{3}{x_{3}}+\cdots+\frac{k}{x_{k}} &= m - 1 \\
        \implies \frac{1}{x_{1}}+\frac{2}{x_{2}}+\frac{3}{x_{3}}+\cdots+\frac{k}{x_{k}} +\frac{k + 1}{x_{k+1}} &= m - 1 + \frac{k + 1}{x_{k+1}},
    \end{align*}
    so $x_{k + 1} = k + 1$ gives the desired value for $n$. Otherwise, if $m = 1$, we can set $x_1 = x_2 = \cdots x_{k + 1} = \frac{(k+1)(k + 2)}{2}$.

    Now, to obtain all of the values in the range $\frac{k(k + 1)}{2}$ to $\frac{k(k + 1)}{2} + (k + 1)$, we note that if we wish to obtain some value $a$, we can find
    $$
    \frac{1}{x_{1}}+\frac{2}{x_{2}}+\frac{3}{x_{3}}+\cdots+\frac{k}{x_{k}} = a - (k+1)
    $$
    so long as $1 \leq a - (k + 1) \leq \frac{k(k + 1)}{2}$,
    and then setting $x_{k + 1} = 1$ gives
    $$
    \frac{1}{x_{1}}+\frac{2}{x_{2}}+\frac{3}{x_{3}}+\cdots+\frac{k+1}{1} = a - (k + 1) + \frac{k+1}{1} = a,
    $$
    so we can obtain all the values in that range, and so by induction, for any $n$, we can obtain any $m \leq \frac{n(n + 1)}{2}$

I realized that this only works for $x_1 \leq x_2 \leq \cdots \leq x_n$, not $x_1 < \cdots x_n$, my bad.

[quote=moldovan]For each positive integer $ n$ find all positive integers $ m$ for which there exist positive integers $ x_1<x_2<...<x_n$ with:

$ \frac{1}{x_1}\plus{}\frac{2}{x_2}\plus{}...\plus{}\frac{n}{x_n}\equal{}m.$[/quote]

I could show all divisors of $n$ are solutions.But don't know about rest of the cases.

I think the problem is nicer with $x_1 \leq x_2 \leq \cdots$ anyway


probability counting distinguishability search

Does anyone have a solution to this problem:

Bob is making four cookies and has exactly four chocolate chips.  If he distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie?  Express as a common fraction.

I've got an answer, but it doesn't match the one on the anwer key.  Pease post a complete solution.

That's exactly what I got, but it's not what's on the answer key.

I'll post a solution if this is right:
[hide]37/53 ? lol probably wrong cuz it doesn't sound right[/hide]

[quote="mthiyaga"][quote="erdogankerem123"]umm, is it [hide]19/35???[/hide], i guess ill edit on a solution if its right..[/quote]
[hide]Remember, both the cookies and the chips are indistinguishable. So there are only 5 ways to distribute the 4 chips on the 4 cookies as follows:

4, 0, 0, 0
3, 1, 0, 0
2, 2, 0, 0
2, 1, 1, 0
1, 1, 1, 1

So there are only 3 ways to distribute the cookies so that there are no more than 2 chips in any one cookies. So the desired probability is $\frac{3}{5}$.[/hide][/quote]
oh, oops, I didn't realize that  :oops:

o_o .... cookies are indistinguishable...gosh I feel dumb...oh well nothing new...

i know that this is probably cheating.  but if you told us what the answer key says, then we could try to adjust our answers to get that answer.  whenever i get a problem wrong, i try to see how i got it wrong, and then i try to find how they got the answer for the answer key.  of course, having the solutions makes life a whole lot easier

[quote="mthiyaga"][quote="Art of Owna"][quote="ZzZzZzZzZzZz"]actually, none of those answers are right.  I'm tempted to think that the answer key is wrong, since I still can't figure out how to do it.[/quote]

The denominator has to be 35 because its basically counting the number of nonnegative solutions to a+b+c+d=4, which is 7C3=35

there can be a 1 1 1 1-1 way
2 2 0 0-4C2-6 ways
2 1 1 0-4!/2!-12 ways

so 1+6+12=19, so it is 19/35..must be something wrong with the answer key.  What does the answer key say?[/quote]
[hide]This problem is to find the number of ways to distribute indistinguishable objects into indistinguishable places. The number of nonnegative solutions to a+b+c+d = 4 will give you the number of ordered tuples, so obviously you cannot use that here. The correct answer is $\frac{3}{5}$ as I explained earlier, unless I misread something or the problem is asking for something else.[/hide][/quote]
it never says..so assume its distinguishable
plus he said 3/5 was still wrong

[quote="Art of Owna"][/quote][quote="mthiyaga"][quote="Art of Owna"][quote="ZzZzZzZzZzZz"]actually, none of those answers are right.  I'm tempted to think that the answer key is wrong, since I still can't figure out how to do it.[/quote]

The denominator has to be 35 because its basically counting the number of nonnegative solutions to a+b+c+d=4, which is 7C3=35

there can be a 1 1 1 1-1 way
2 2 0 0-4C2-6 ways
2 1 1 0-4!/2!-12 ways

so 1+6+12=19, so it is 19/35..must be something wrong with the answer key.  What does the answer key say?[/quote]
[hide]This problem is to find the number of ways to distribute indistinguishable objects into indistinguishable places. The number of nonnegative solutions to a+b+c+d = 4 will give you the number of ordered tuples, so obviously you cannot use that here. The correct answer is $\frac{3}{5}$ as I explained earlier, unless I misread something or the problem is asking for something else.[/hide][/quote][quote="Art of Owna"]
it never says..so assume its distinguishable
plus he said 3/5 was still wrong[/quote]
[hide]OK,  Assume that both the cookies and the chips are distinguishable.

Using complementory counting, there are 4 ways to place all 4 chips in 1 cookie (i.e: 4, 0, 0, 0 can be permuted in 4 ways). Similarly 3, 1, 0, 0 can be permuted in 12 ways. So this gives us a total of $4 \times 12+4 = 52$ ways. So the number of ways to distribute 4 chips into 4 cookies so than there are no more than 2 chips per cookie is $4^{4}-52 = 204$. Hence our desired probability is $\frac{204}{256}= \frac{51}{64}$.[/hide]

[quote="nhansen"]If you were to actually perform this experiment (with four actual chocolate chips and four cookies), the chips and cookies would certainly be distinguishable.  When I think of the problem this way, I get the answer 51/64.[/quote]

oh my bad..the chips are also distinguishable.

Yeah, 51/64 is right.  Thanks.

By the way, this problem came from the 2004-2005 handbook, WU 18 #6.

[hide]Edit: I read the question wrong, here's new solution.

if you have 4000 then theres only 4 possibilities
if you have 3100 then there are 4*3(Make sure not to divide by
two: AAAB is different then BBBA)*4(ways to organize)

and you get 52/256  256-52=204 204/256= 51/64

[/hide]


geometry
geometric transformation
rotation
function

Hi Silverfalcon,

I would like to know what it means for the electrons to be "quantized."

Anyway, I think electron configuration is how the electrons are arranged in an atom, like into orbitals, etc. It's also a study of how electrons move and work (like when they gain enough energy to move up another principal energy level).

I'm just learning this too. What grade are you in (because i want to know when they actually learn real "chemistry")?

I think the site it's quite clear but it forget to say something about the others quantum numbers.
Quantum numbers are $n,l,m,m_{s}$
$n$ determine the energetic level that an electron can have
$l$ which assume all values from $0$ to $n - 1$ determine the form of the orbital.
(for example if $l = 0$ the orbital is spherical)
electrons with $l = 0$ are called elctrons $s$
                     $l = 1 \rightarrow p$
                     $l = 2 \rightarrow d$
                     $l = 3 \rightarrow f$
$m$ that can assume all integers values from $-l$ to $l$ determine the number of the orbital's orientations in the space.
$m_{s}$ determine the verse of rotation of the electron around the axes
and assume only the values $\pm \frac{1}{2}$.

Tell me if it's more clear.
If not later I can prepare something  :)

Thanks I'll try to look on my chemistry book (I have simple one.. :D ) and see if I still have a question.

I'll be sophomore this year and well, I do remember chemistry from physical science in 8th grade. I remember electrons and like you, quantum and those terms seem to confuse me...

[quote="Silverfalcon"]Hi,

I'm trying to learn chemistry before the school starts and everything is okay till I came across electron configuration. What is an electron configuration? How do I use it? I especially don't understand it as it goes to higher levels.

I tried googling and got this site

http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson36.htm

But I still don't understand.  :( Thank you![/quote]

Electron Configuration is chemist's shorthand for the "quantum address;" the set of quantum numbers, that uniquely describes an atom.

The site you gave has a nifty chart in upper-middle left of a triangle listing the principal quantum number and its possible orbitals. The red diagonal arrows indicate the order in which the orbitals fill. If you write down a specific electron configuration you identify a specific atomic (or ionic) species, e.g. $1s^22s^22p^63s^1$ is the electron configuration of $Na$. Each orbital can take a maximum of two electrons. There are 1 s-orbitals, 3 p-orbitals, 5 d-orbitals, 7 f-orbitals, etc. per principal quantum number. The orbitals have different shapes according to whatever spherical harmonic function describes them, and correspond to the orbital (second) quantum number $l$, i.e., s corresponds to $l = 0$, p to $l = 1$, d to $l = 2$, f to $l = 3$, etc.

[quote="Silverfalcon"]Thanks but I'm still stuck.  :(

My book has calcium(Ca)'s electron configuration is:

\[1s^2 2s^2p^6 3s^2p^6 4s^2\]

In another word, it has 2 electrons in $1s$ subshell, 8 electrons in the $2sp$ subshell, 8 electrons in the $3sp$, and 2 electrons in the $4s$ subshell.

But for oxygen, why is it represented as

\[1s^2 2s^2 p^4\]

In another word, it has 2 electrons in $1s$ subshell, 2 electrons in $2s$ subshell, and 4 electrons in the $p$ subshell.

How come? The first one's third one has $3sp$ subshell but the second one's third one has $p$ subshell.  :?

This is confusing..  :(  :(  :([/quote]

My chemistry book makes you put the number of the subshell before each letter, as in

\[1s^2 2s^2 2p^6 3s^23p^6 4s^2\]

It makes it easier to understand.

Oxygen is \[1s^2 2s^2 2p^4\]. The exponent for 2p is 4 since oxygen has only 8 electrons. It would have a higher exponent if it had more; for example, \[O^{3-}\] I think would be \[1s^2 2s^2 2p^6 3s^1\]. Though I'm not sure about that example.

And thanks for reminding me about the fact that 3d orbitals require more energy than 4s ones; I had forgotten that.
 :blush:

Oh, thanks a lot!

I should remember this from physical science class but I can't (It's been two years  :blush: ), how many electrons are in second, third, and fourth layer?

I know the first one is only 2. But I can't remember anything further. Any help?

just count them
we had $1s^2$ for the first one so thats 2 for the first one.
then wehavE $2s^22p^6$ so thats $8$
then you have $3s^23p^63d^10$ so thats $18$.

you do the rest.

Thuogh remember that they don't just fill up the first layer, then the second layer, then the third layer, etc., because of the energy difference between, say, $4s$ and $3d$.

Also, just to make things clear, solafidefarms's example was $\text{O}^{3-}$ (the electrons carry a negative charge).  $\text{O}^{3+}$ would probably be $1s^22s^22p^1$.

oooops.... :roll:

There is 1 s-orbital per subshell.
3 p-orbitals starting at the second
5 d-orbitals starting at the third, but they take up more energy than a s-orbital of the next larger subshell
Some number of larger ones; you should learn what their letters are and what their numbers are in AP Chemistry.

When writing configurations, here are two rules;

1. Electrons enter the lowest level first. That means 4s after 3d and 2p after 2s and etc.

2. An orbital describes at most 2 electrons (pretty obvious, that's why p can't be more than 6 as its exponent, it has 3 orbitals).

Rows correspond to periods on the periodic table:

1s
2s2p
3s3p
4s3d4p
5s4d5p
6s4f5d6p
7s5f6d7p

^Exactly!  Just count them from the periodic table and its groups.

[quote="solafidefarms"]236!, 3d takes more energy than 4s.[/quote]

Oops. going off task again  :oops:


geometry
geometric transformation
reflection
rotation
geometry unsolved

how can i show that any isometry in $R^2$ is product of no more than 3 reflections?

First, there are two kinds of isometries: those which preserve orientation (let's call them isometries of the first kind) and those which reverse orientation (isometries of the second kind)
Now, here are the steps of the proof:

1. Show that an isometry of the first kind is either a translation, or a rotation.
To do this, assume $T$ is such an isometry, take an arbitrary point $A$ and let $A'=T(A)$ and $A''=T(A')$. We obviously have $AA'=A'A''$. If points $A,A',A''$ are collinear, show that $T$ is a translation. If not, show that it is a rotation.

2. Show that an isometry of the second kind is the product of a reflection and a translation.

3. Show that a translation is the product of two reflections (with parallel axes) and that a rotation is the product of two reflections (whose axes intersect at the center of rotation)


http://i2.photobucket.com/albums/y44/x-disturbed-x/CalorimetryEnthalpyChangeQuestion.jpg

Would be grateful uf some one told me what each part of the question is asking me to do. I do the calculations but i get the feeling i'm not doing it right. I miss out one important piece of information, the 1.00g in part c because I think it's asking me to use the energy transfered = mcptemp change forumla..HELP!!

a)
    _          +
OH    +   H            -------------->   H20

B)

1*50/1000 = 0.05 moles

C)
H=mc(dT)
 = 1*(50+50)g  *   4.18 J/g K  *     6.5 K

D)

Standard enthalpy change
= H/number of moles of WATER formed


let $ f,g : X\to Y$ be homotopic maps, prove that $ f_{*}\equal{}g_{*} : \tilde {H}_{n}(X)\to \tilde {H}_{n}(Y)$
 thanks

Is this given that standard homology is homotopy invariant? Because $ \widehat{H}_n(X) \equal{} H_n(X,x)$ where $ x \in X$ in that case; if not, usually you proceed by subdividing $ \Delta^n \times I$ into simplices in some canonical way, and then pushing forward this construction to show that homotopies of maps induce chain homotopies of chain complexes. I know an alternate proof using the Eilenberg-Zilber theorem via the method of acyclic models, which is a bit high-tech for this application but ends up being the cleanest expression of the intuitive concept.

The idea is to give a map $ C(X) \otimes C(I) \to C(X \times I) \to C(Y)$, where $ C$ is the reduced chain complex functor. The map $ C(X \times I) \to C(Y)$ comes from the homotopy; the chain map $ C(X) \otimes C(I) \to C(X \times I)$ comes either by pushing through the dissection $ \Delta^n \times I$ into $ \Delta^{n \plus{} 1}$'s or through Eilenberg-Zilber. And it's easy to check that a map $ C(X) \otimes C(I) \to C(Y)$ is the same as a chain homotopy $ C(X) \to C(Y)$.

I can give more detail upon request, but that's the basic idea. Then again, homotopy invariance is proven in almost every algtop text, so I don't see the need to completely reconstruct it here, unless you know it for standard homology and would like to prove it for reduced.

You should also replace the singular chain complex $ C(I)$ by the simplical chain complex with respect to the canonical simplical structure of $ [0,1]$. Calling this chain complex $ {\mathbb I}$, a homotopy of chain complexes is indeed simply a map $ {\mathcal C}\otimes{\mathbb I}\to{\mathbb D}$.- very nice  :roll:


continued fraction
number theory unsolved
number theory

The HJ continued fraction of a number $ a/b$ is $ [a_0, \dots, a_n] \equal{}a_0 \minus{}
\displaystyle\frac{1}{a_1 \minus{}
\displaystyle\frac{1}{\ddots \minus{}
\displaystyle\frac{1}{a_n}}}$

The partial convergents are $ a_i/b_i \equal{} [a_0, \dots a_i]$ in lowest form.

Prove that if $ 0 < a/b < 1$, then $ a_i$ is weakly increasing and $ b_i$ is strictly increasing.

that was the most useless response i've seen in a while


Hi..., there is a problem:

Let's consider (n x n) chessboard. At begenning there are some coloured fields, now let's imagine that we are dyeing the fields of this chessboard as follows: We can dye the field if and only if it has at least 2 coloured neighbours (by edge, not by corner). How many fields at least must be coloured at begenning so that we can dye all fields of the chessboard?

My observation:
It's easy to see that n coloured fields at begenning is enough (for example all diagonal fields). But how to prove that n - 1 inital fields is too little?

Thanks

Consider all colored cells. They form some dyed figures.
Let p be the sum of their perimeters.
It's easy to observe that p is nonincreasing when we are dying cells
as in the problem.
When all cells are dyed p = 4n.
At the beginning p<= 4k, where k is the number if cells.
So 4k >= 4n, k >= n


vector
function
algebra
polynomial
quadratics
group theory
abstract algebra

Let K be a field included in R, which is different from R. Prove that there is no real number a s.t. R={p+q*a | p,q are in K}.

Consider R as a vector space over K. The condition in the problem turns to proving that {1,a} is not a basis of R for any a in R.
Your problem is thus equivalent to proving that there is no subfield of R such that dim{over K}(R)=2 or [R:K]=2.

I'll be back 4 this.

I had another approach for this, but yours might turn out to be cooler  :D

Ok suppose we have a basis {1,a} of R over K.

a 2  is in R, so there exist u,v in K such that a 2 =(2u)*a+v.
Then we have (a-u) 2 =u 2 +v=v' which is an element of K.
It is easy to see that {1,a-u} is also a basis of R for any u in K.

So we can chose a basis {1,a} for R such that a 2 is in K.
If {1,a} is a basis, then so is {1,-a}. From this we can assume a>0.

Now  \sqrt a is in R, so there exist x,y in K such that  \sqrt a=x*a+y, so
a=x2 *a 2 +2xy*a+y 2 , so
a(1-2*xy)=x 2 *a 2 +y 2 .
The right side is an element of K since x,y and a 2 are in K.
In the vector space R: (x 2 *a 2 +y 2)*1+(2*xy-1)*a=0, so
2xy=1 (1) and
x 2a 2+y 2 =0. (2)
Because a is in the basis, a <>0 and since a,x,y are reals, (2) must mean x=y=0. But then (1) is not possible. We've got our contradiction.

I don't think you have a conceptually different solution. Mine might just be easier to understand (for someone with vector space knowledge of course  :D). If your solution is different, please post it.

[b]I have 2 more questions I don't have a proof for yet.
1. Does there exist a subfield K of R such that [R:K] (the dimmension of R as a vector space over K)=n for an integer n \geq 2?
2. Doest there exist a subfield K of R such that R has a countable (of course infinite) basis over K? (it would not be a good idea to try Q. A basis of R over Q si called a Hammel basis and is not countable)[/b]

Thx for ths solution. here's the outline of what I did: a^2 is in R, so a^2=p*a+q, so a^2-p*a-q=0. This is a second degree equation. It's obvious that it must have 2 real solutions, because if it had only one, then a would be in K, so a*p+q would be in K=/=R for all p and q in K. Let the other root of the eqn be b.

We assume that every real has form p*a+q. Then it can be written like this uniquely. We define the function f(p*a+q)=p*b+q. It's not that hard to prove that f is an automorphism of R, different from the identical function, but it's a well-known fact that the only automorphism of the field (R,+,*) is the identical function. We have a contradiction.

This is actually not my solution. My teacher showed me the problem, and then the solution (he gave me something like 5 minutes to try to solve it!
:D ).

I can't prove my first problem even for n=3.

I have managed to prove that if there exists K a subfield of R such that dim{R}(K)=3, then there exists x such that {1,X,X 2 } is a basis for R over K.

Suppose {1,x,x 2 } is not a basis for R/K for all x in R.
Let {1,a,b} be a basis for R over K.
Since {1,a,a 2 } is not a basis, its elements are not linearly independent. Then there exist u,v,w in K, not all 0 such that u+va+wa 2=0. w<>0 or else a is in K and then 1 and a are linearly dependent (a contradiction). Then we get some u',v' in K, just as in my previous post, such as (a+u') 2=v'. It is clear that {1,a+u',b} is a basis for all u' in K.
Analogously we get a basis for R/K {1,a,b} where a 2, and b 2  are in K.

ab is in R, so there are u,v,w in K such that ab=u+va+wb. Or
a 2 b=ua+va 2 +wab=ua+va 2+w(u+va+wb), or
(va 2+wu)+(u+wv)a+(w 2 -a 2)b=0, so w 2=a 2 , so a= \pm w, so a is in K (contradiction).
I don't know how this helps, but it should.
---------------------------------------------------------------------------------
Please tell me if there is a simpler proof for what I did. Somethig like: by ...'s theorem we get ...my statement.

[quote="amfulger"]2. Doest there exist a subfield K of R such that R has a countable (of course infinite) basis over K? (it would not be a good idea to try Q. A basis of R over Q si called a Hammel basis and is not countable)[/quote]
Anyway, $|K|=|\mathbb{R}|$ (because of http://www.mathlinks.ro/viewtopic.php?p=129743, thanks to sam-n).
Is it possible?

This is a neat problem.  I showed that there is no proper subfield $K$ of $\mathbb{R}$ such that [$\mathbb{R}$ : $K$] is finite.  Here's what I have.

Suppose (on the contrary) $K$ is proper subfield of $\mathbb{R}$ with $n = [\mathbb{R} : K]$  finite.  We may assume $K$ is maximally proper, that is, there is no field $L$ lying strictly between $K$ and $\mathbb{R}$ (Otherwise find a maximal proper subfield lying between $K$ and $\mathbb{R}$.)

First I show that if $r$ is in $K$ with $r > 0$ then $r^{1/n}$ is also in $K$.  Then I show that $r^{1/m}$ is in $K$ for every $m \geq 1$.  So that $K$ is "very close" to being $\mathbb{C}$, its algebraic closure.  $\mathbb{R}$ is also "very close" to $\mathbb{C}$ so that $K = \mathbb{R}$.  Of course I make this rigorous below.

Take any $r$ in $K$ with $r > 0$.  Let $f(x) = x^{n^2} - r$ and $g(x)$ be the minimal polynomial of $r^{1/n^2}$ over the field $K$.  Since $f(x)$ is in $K[x]$ and $f(r^{1/n^2}) = 0$, $g(x)$ divides $f(x) = (x - u)(x - uw)(x - uw^2)...(x - uw^{n^2 - 1})$ where $w$ is primitive $n^2$-th root of unity and $u = r^{1/n^2}$.

Since there is no field between $K$ and $\mathbb{R}$, and deg $g$ = [$K(r^{1/n^2})$ : $K$], deg $g$ = $1$ or $n$.  If deg g = $1$ then since $g \mid f$, $uw^i$ is in $K$ for some $i$ (by looking at the constant coefficient of $g(x)$, which is in $K$).  Since $w^i$ is nonreal number $w^i = 1$ or $-1$ so $u$ is in $K$.  So $u^n = r^{1/n}$ is in $K$.

If deg g = $n$ then from $g \mid f$ and the factorization of $f$ in $\mathbb{C}[x]$, we get that the constant coefficient of $g(x)$ is $u^n w^k$ for some integer $k$.  Since $g(x)$ is in $K[x]$, $u^n w^k$ is in $K$ and so $r^{1/n} = u^n$ is in $K$.

Next take $m \neq n$, $m \geq 1$.  Consider $x^m - r$ which is in $K[x]$.  It factors in $\mathbb{C}[x]$ as $(x - v)(x - vq)(x - vq^2)...(x - vq^{m-1})$ where $v = r^{1/m}$ and $q$ is primitive $m$-th root of unity.  Like before, the minimial polynomial of $r^{1/m}$ over $K$ divides $x^m - r$ so (by looking at the constant coefficient of our minimal polynomial) $r^{n/m} = v^n$ is in $K$.  From the previous paragraph we have $(r^{n/m})^{1/n} = r^{1/m}$ is in $K$.

Now note that if $p(x)$ is quadratic polynomial in $K[x]$ then $p(x)$ has a root in $K(i)$ ($i^2 = -1$) (just use a quadratic equation and the above result shows that $\surd{r}$ is in $K(i)$ for $r$ in $K$).  Likewise every cubic polynomial $q(x)$ in $K[x]$ has a root in $K$ (look at the expression for roots in terms of radicals given by Cardano's method, for example).

We have all the ingredients we need to show that $K(i)$ is algebraically closed now.  The next is immitation of proof of algebraic closure of $\mathbb{C}$, the complex numbers.

Let $f(x)$ be in $K(i)[x]$ and suppose $c$ is a root of $f$ in some extension field of $K(i)$.  Let $F$ be a galois closure of $K(i)(c)$ over $K$ and let $G$ = Gal($F$ / $K$).  $|G| > 1$ of course.  Take $H$ a sylow 2-subgroup of $G$ and let $L$ be the fixed field of $H$.  So $[L : K]$ is odd.  In fact $L = K(l)$ for some $l$ in $L$ (characteristic 0, or just that there are only finitely many intermediate fields).  But the minimal polynomial of $l$ over $K$ has a root in $K$, so $L = K$.  This shows that $H = G$ so that $G$ is 2-group.  Let $A$ = Gal($F$/$K(i)$), which is a subgroup of $G$ so that $A$ is also a 2-group.  If $|A| \neq 1$ then let $M$ be any maximal proper subgroup of $A$, which has index $2$ in $A$.  Let $E$ be the fixed field of $M$.  So that $[E : K(i)] = 2$.  But every polynomial of degree 2 has a root in $K(i)$, contradiction.  So $|A| = 1$, hence $F = K(i)$.  So $K(i)$ is algebraically closed.

Now the concluding paragraph.  Take y in R\K.  So R = K(y) (since no intermediate fields between K and R).  Let f(x) be the minimal polynomial of y over K.  Since K(i) is algebraically closed f(x) splits in K(i)[x] so that y has the form a + bi where a,b are in K.  Since y is in R, b = 0 so y = a is in K.  This is contradiction.  qed

[quote="Myth"]It is quite hard to read it here. Can you prepare latex document (dvi, pdf)?[/quote]

I hope that is more legible.

There's a problem in the middie of the next to last paragraph of my proof, where I say that the minimal polynomial of $l$ over $K$ has a root in $K$ (where [K(l) : K] is odd).  This is not justified.  Here's a patch.

Again suppose $K$ is contained in $\mathbb{R}$ with $1 < [\mathbb{R} : K] < \infty$.  So $[\mathbb{C} : K] \geq 4$.  Since $\mathbb{C}$ is algebraically closed, $\mathbb{C}$ is galois over $K$ ($\mathbb{C}$ is normal and separable (characteristic = 0) over $K$).  Let $G =$ Gal($\mathbb{C} / K$).  If $G$ is a $2$-group then $H$ = Gal($\mathbb{C} / K(i)$) is again a $2$-group.  If $|H| \neq 1$ then let $M$ be any maximal proper subgroup of $H$, which has order $|H|/2$.  We take $L$, the fixed field of $M$ so that $[L : K(i)] = 2$.  I showed in the first few paragraphs before my false proof that $K(i)$ is algebraically closed that $r^{1/4}$ is in $K$ for every $r \in K$ with $r > 0$.  This shows furthermore that $\sqrt{r}$ is in $K(i)$ for every $r \in K(i)$, by explicitly calculating $\sqrt{a + bi}$ in terms of roots of $a,b$ where $a,b \in K$.  So we obtain a contradiction if $[L : K(i)] = 2$.  So $|H| = 1$ and $K(i)= \mathbb{C}$, hence $K = \mathbb{R}$, contradiction.

Therefore $G$ is not a $2$-group.  This means there is some odd prime $p$ that divides $|G|$ because $|G| \geq 4$.  Let $\sigma \in G$ be an element of order $p$.  Let $F$ be the fixed field of $<\sigma>$.  So $[\mathbb{C} : F] = p$, an odd prime, and Gal($\mathbb{C} / F$) $= <\sigma>$.  I will show that this is impossible.

Let $\omega$ be a primitive $p$-th root of unity in $\mathbb{C}$.  Let $\alpha = l + \omega \sigma (l) + \omega^2 \sigma^2 (1) + ... + \omega^{p-1} \sigma^{p-1} (l)$ where $l \notin F$.  $\alpha$ has the property that $\sigma (\alpha) = \omega^{-1} \alpha$.  Since $\mathbb{C} / F$ is galois, $\alpha \notin F$.  Thus $\mathbb{C} = F(\alpha)$.  Furthermore, $\sigma^{p}(\alpha) = \alpha$ so $\alpha^p \in F$.  Now we consider any $\beta \in \mathbb{C}$ such that $\beta^p = \alpha$.  Since $\beta$ is a root of $x^{p^2} - \alpha^p \in F[x]$ it follows that $\sigma(\beta) = \zeta \beta$ where $\zeta^{p^2} = 1$.  Now $\zeta^p \neq 1$, for otherwise $\sigma(\beta^p) = \beta^p$ hence $\alpha = \beta^p \in F$, contradiction.  So $\zeta$ is primitive $p^2$-th root of unity.  Now I study where $\sigma$ sends $\zeta$ to.  Note that $\zeta^p$ is primitive $p$-th root of unity.  Therefore the minimal polynomial of $\zeta^p$ over $F$ has degree less than or equal to $p-1$ since it divides $1 + x + x^2 + ... + x^{p-1}$.  So $\zeta^p \in F$ because $[\mathbb{C} : F] = p$, an odd prime.  So $\sigma(\zeta)^p = \zeta^p$.  So $\sigma(\zeta) = \zeta^{1 + pk}$ for some integer $k$.  Now $\sigma^p$ is the identity map so $\beta = \sigma^p (\beta) = \zeta^{1 + 1 + pk + (1 + pk)^2 + ... + (1 + pk)^{p-1}}$ from $\sigma(\beta) = \zeta \beta.$

Therefore $p^2$ must divide $1 + 1 + pk + (1 + pk)^2 + ... + (1 + pk)^{p-1}$.  By binomial theorem $p^2$ divides $1 + (1 + pk) + (1 + 2pk) + (1 + 3pk) + ... + (1 + (p-1)pk) = p + p^2 (p-1)k/2$.  So $p$ divides $1 + p(p-1)k/2$.  But $p$ is an odd prime, so that $p(p-1)k/2$ is divisible by $p$, contradiction.

[i]Thanks to zeno[/i]

What rik basically outlined was a proof of Artin-Schreier, google folks.[/i]


I am part of my school's math team, and I'm wondering how to help improve other members' math skills.  I feel that the main problem is that they are not used to the type of thinking that is necessary for these contests.  For example, one question that was given to us was to compute an infinitely nested radical, and only one other person besides me was able to solve this (because I showed him a problem like this last year).

So the question is, how can I help my friends get better at these types of problems?

Definitely no easy answer to this question.

[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]This article[/url] can help you attain the perspective you need to do problems where simple tricks and brute force fail. That's not going to be any immediate boost in ability, but it may increase the effectiveness of practice.

In your post, you say "they are not used to this type of thinking". To get used to something, you typically expose yourself to it a lot. Doing lots of problems that require the type of thinking you're talking about may do the job.

This thread might help
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179530 .
 :)

[quote="7h3.D3m0n.117"][quote="lingomaniac88"]So the question is, how can I help my friends get better at these types of problems?[/quote]

Positive peer pressure might help. This doesn't include making fun of other mathletes because they don't know how to solve a problem  :wink: .[/quote]

I don't think anyone would do that.  I know I wouldn't, and the advisor of Mathletes at my school said that even if a person is terrible at math, he/she won't get kicked out of the club.  He/she just won't be on the "A team."

In terms of resources for problems, I think that some obvious places to look are at the AoPS books, previous problems from our local contests, and this forum itself.  I am actually considering showing this forum to the chairperson of the math department at my school.  Do you know any other sources that can be utilized?

I think the first main step is to get people involved.  I was looking at how the teams did on the local contest (same one as you... go Nassau at ARML and NYSML!).  Most teams have a certain lack of participation.  In terms of your school, out of the 28 who have competed, 16 have missed at least one contest.  Once people get committed, only then can you truly start to have them get better at math.  I have the same problem at my school.  13 out of the 23 kids who have competed have missed at least one contest.  And that does not help at all.  If you are able to get people excited about math, there will be immediate improvement among your team's scores.  And if you find out how to do that, please let me know.

By the way, if you need problems from even further back than the website says, I think my school has some resources.  So if 9 contests from the past is not enough, I might be able to help.  Just PM me...


modular arithmetic
number theory unsolved
number theory

Let $ n$ a integer greater than 1.Prove that if $ \frac{b^{n}\minus{}1}{b\minus{}1}$ is a power of prime for some positive integer $ b$, then $ n$ is prime.

[quote="msecco"]Let $ n$ a integer greater than 1.Prove that if $ \frac {b^{n} \minus{} 1}{b \minus{} 1}$ is a power of prime for some positive integer $ b$, then $ n$ is prime.[/quote]

If $ n$ is not a power of a prime, then we can write $ n \equal{} pq$ such that $ p,q > 1$ and $ \gcd(p,q) \equal{} 1$.

Then $ b^p \minus{} 1, b^q \minus{} 1|b^n \minus{} 1$ gives $ \frac {b^p \minus{} 1}{b \minus{} 1},\frac {b^q \minus{} 1}{b \minus{} 1}$ are power of a prime $ r$.
But it's well known that $ \gcd(b^p \minus{} 1,b^q \minus{} 1) \equal{} b^{\gcd(p,q)} \minus{} 1 \equal{} b \minus{} 1$.
So $ \frac {b^p \minus{} 1}{b \minus{} 1}$ and $ \frac {b^q \minus{} 1}{b \minus{} 1}$ , both being a power of the same prime $ p$, are coprime to eaxh other, contradiction.

So $ n \equal{} p^k$ for a prime $ p$ and a positive integer $ k$.

Now let $ \frac {b^{n} \minus{} 1}{b \minus{} 1} \equal{} r^y$.
If $ k \ge 2$,  $ \frac {b^{p} \minus{} 1}{b \minus{} 1},\frac {b^{p^2} \minus{} 1}{b \minus{} 1}|\frac {b^{p^{k}} \minus{} 1}{b \minus{} 1}$

gives $ r|\gcd(b^p \minus{} 1,\frac {b^{p^2} \minus{} 1}{b^p \minus{} 1}) \equal{} \gcd(b^p \minus{} 1,p)|p$,
(because $ \frac {b^{p^2} \minus{} 1}{b^p \minus{} 1} \equal{} b^{p(p \minus{} 1)} \plus{} ... \plus{} b^p \plus{} 1 \equal{} p \pmod{b^p \minus{} 1})$ so $ r\equal{}p$, contradiction. So $ k \equal{} 1 \Rightarrow n \equal{} p$ is prime.

If we take a look at [url=http://mathworld.wolfram.com/ZsigmondyTheorem.html]ZsigmondyTheorem[/url],the problem becomes trivial :)[/url]


Paula can have pennies, nickels, dimes and quarters in her purse. The average value of the coins in her purse is 20 cents. If she adds one more quarter to her purse, the average value of all of the coins will be 21 cents. How many nickels does she have in her purse?

[hide="Solution"]Let $ s$ be the value of coins in her purse, and $ n$ be the number of coins present. $ s \equal{} 20n$, and $ s\plus{}25 \equal{} 21(n\plus{}1)$. This means that $ n \equal{} 4$ and $ s \equal{} 80$. The only way to have $ 80$ cents through $ 4$ coins is with $ 3$ quarters and $ 1$ nickel. Therefore, she has $ \boxed{1}$ nickel in her purse.

[/hide]


Write balanced overall ionic equations for the following reactions:
1.aqueous solution of manganese(II)sulfate undergoing hydrolysis
2.solid cobalt(II)hydroxide+hydroiodic acid(I wonder whether CoI_2 is precipitate or not)
3.NH4Cl+CoSo4(no reaction?)
4.tin(IV) nitrate+potassium sulfite
5. Rubidium phosphate+ titanium(IV) nitrate

I am told that a lot of AP Students botch this particular aspect in the AP Final.

But here are some things you should consider:

1.) Solubility rules: You must have these memorized in order to identify the precipitate and the net ionic equation.

2.) Each one is a double replacement reaction. Therefore, given the reactants, one can easily find the products.

So for example, let's take the first one:

\[ \text{MnSO}_4 \plus{} \text{H}_2 \text{O} \to \text{H}_2 \text{SO}_4 \plus{} \text{MnO} \text{(s)}
\]

Now, examine the products. Sulfates are generally soluble, but there are exceptions. However, sulfuric acid is not one of them. Oxides are generally insoluble with the exception of group IA metals. However, manganese is not a group IA, and therefore, manganese oxide precipitates out of the reaction.

To write a net ionic equation, we seperate the components into their respective ions:

\[ \text{Mn}^{2\plus{}} \plus{} \text{SO}_4 ^{2\minus{}} \plus{} 2\text{H}^{\plus{}} \plus{} \text{O}^{2\minus{}} \to \text{H}_2 \plus{} \text{SO}_4 ^{2\minus{}} \plus{} \text{MnO} \text{(s)}
\]

Crossing out the spectator ions, we are left with:

\[ \boxed{\text{Mn}^{2\plus{}}\plus{} \text{O}^{2\minus{}} \to \text{MnO}}
\]

[hide="Latex tips"]Same as other places: The hard part is:

[code]$MnSO_4$[/code]will give $ MgSO_4$
But it's better as [code]$\text{MgSO}_4$[/code]It gives $ \text{MnSO}_4$
And [code]$\boxed{\text{MgSO}_4}$[/code]will give $ \boxed{\text{MgSO}_4}$
And for ions [code]$\text{H}^{+}$[/code]works as $ \text{H}^{ \plus{} }$
These are required for other uses also, while writing math formulas etc.

Tip: To learn something from another person's post, click on the quote button on bottom-right of the post, you need not really post it, see preview and stuff...[/hide]

Also, check out [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176537]this[/url].


calculus
integration
trigonometry
inequalities
limit
absolute value
calculus computations

Determine if the following integrals converge or diverge.

i) $ \int_{1}^{\infty}\tan{\frac{1}{x}}dx$

ii)  $ \int_{0}^{1}\frac {1 \minus{} \cos{x}}{\sin^3{2x}}dx$

iii) $ \int_{0}^{\infty}x^{ \minus{} \frac {1}{5}}\sin{\frac {1}{x}}dx$

1) Limit comparable to $ \frac1x$ as $ x\to\infty.$ The integral diverges.

2) Limit comparable to $ \frac1x$ as $ x\to0.$ The integral diverges.

3) Limit comparable to $ x^{\minus{}6/5}$ as $ x\to\infty.$ As $ x\to0,$ wildly oscillatory but has absolute value bounded by $ x^{\minus{}1/5}.$ The integral converges absolutely.

im sorry but i dont understand how you can compare the given integrals with the shown limits.
are you using some trig inequality?  :maybe:

I'm using the following fundamental trigonometric limit: $ \lim_{\theta\to0}\frac{\sin\theta}{\theta}\equal{}1.$

Since $ \cos\theta\to1,$ we also have $ \lim_{\theta\to0}\frac{\tan\theta}{\theta}\equal{}1.$

And then we can derive (by various means - I remember it because of power series) this:

$ \lim_{\theta\to0}\frac{1\minus{}\cos\theta}{\theta^2}\equal{}\frac12.$

----

So for i): Since $ \frac1x\in(0,1]$ is always in the first quadrant, the integral is only improper in one place, as $ x\to\infty.$ From the second limit that I wrote above,

$ \lim_{x\to\infty}\frac{\tan\frac1x}{\frac1x}\equal{}1.$ Hence my claim that this integral is limit comparable to $ \int_?^{\infty}\frac1x\,dx,$ which diverges.

For ii), we do have to check that $ 2x$ does not approach the next zero of the sin, which it doesn't, since $ 2<\pi.$ This integral is only improper in one place, as $ x\to0.$

By the third inequality I wrote above, $ 1\minus{}\cos x$ looks like a constant times $ x^2$  as $ x\to 0.$ But since $ \sin 2x$ looks like $ 2x$ as $ x\to 0,$ then $ \sin^32x$ looks like a constant times $ x^3.$ That makes the whole fraction look like a constant times $ \frac{x^2}{x^3}\equal{}\frac1x.$

Another way to phrase what I'm saying is this:

$ \lim_{x\to0}\frac{\frac{1\minus{}\cos x}{\sin^32x}}{\frac1x}\equal{}A,$ where $ A$ is neither $ 0$ nor $ \infty.$

(In fact, $ A\equal{}\frac1{16},$ but we can proceed without nailing down that detail.)

Hence the integral in question is limit comparable to $ \int_0^?\frac1x\,dx,$ which diverges.

For iii): This time the integral is improper in two places and each must be dealt with separately. By the first limit above, $ \sin\frac1x$ looks like $ \frac1x$ as $ x\to\infty.$ That means that $ x^{\minus{}1/5}\sin\frac1x$ looks like $ x^{\minus{}6/5},$ which I could rephrase as

$ \lim_{x\to\infty}\frac{x^{\minus{}1/5}\sin\frac1x}{x^{\minus{}6/5}}\equal{}1.$

That makes that part of the integral limit comparable to $ \int_?^{\infty}x^{\minus{}6/5}\,dx,$ which converges.

As $ x\to0,$ the only useful thing we can say about $ \sin\frac1x$ is that $ \left|\sin\frac1x\right|\le 1.$ Hence $ \left|x^{\minus{}1/5}\sin\frac1x\right|\le x^{1/5}.$

Since $ \int_0^?x^{\minus{}1/5}\,dx$ converges, this means that our integral converges absolutely by comparison.

----

Whatever you do, don't try to find exact antiderivatives for these functions. That would waste your time and distract you from looking at how to answer the question asked.


solve the following equation:
$ \displaystyle \frac{1}{x^1}\plus{}\frac{1}{x^3}\plus{}\frac{1}{x^5}\plus{}\frac{1}{x^7}\plus{}\frac{1}{x^9}\plus{}\frac{1}{x^{11}}\plus{}\frac{1}{x^{13}}\equal{}6$

hello, this equation has only one real solution:
$ x \approx 1.02286244...$.
Sonnhard.

This equation is the same as $ 0\equal{}6x^{15}\minus{}x^{14}\minus{}6x^{13}\plus{}1, |x|\not\equal{}1$.
Don't know how to solve that though.

I agree that this equation can only have one real solution (for $ x<0$, it clearly has none; for $ x>0$ the term on the LHS is strictly decreasing). Also, it has to have a solution, since the LHS assumes all values between zero and infinity. However, I'm not so sure about the value of $ x$ you got, since AM-GM shows that $ x>\sqrt[7]{\frac{7}{6}}\approx 1.02$...

Mathematica didn't find a nice form for the answer for either of your problems; it just gave a polynomial and said it was the root of it. The answer in the first reply is fine except for an extra $ 0$. The root of the second equation is about $ 1.26985$.

Okay, but... who would possibly get that without a computer algebra system?

Well, obviously you have to guess it and plug it in... after all, the problem really suggests that this is a solution...

Seriously, problems which can't be proven without a CAS aren't... that great.


inequalities
function
analytic geometry
geometry
rectangle
trapezoid
college contests

i'm having trouble with the left inequality.
Prove
$\frac{n\pi}{4}-\frac{1}{\sqrt{8n}} \leq \frac12+\sum_{k=1}^{n-1} \sqrt{1-\frac{k^2}{n^2}}\leq \frac{n\pi}{4}.$
for every integer n>=2.

I have an idea of how to prove the right-hand -side inequality, it seems to me this is a geometric inequality. Here is the proof. The inequality to prove reads:

$\frac{1}{2n}+\frac{1}{n}\sum_{k=1}^{n-1}\sqrt{1-\frac{k^{2}}{n^{2}}}\leq \frac{\pi}{4}$.

Let consider the function $f: [0,1]\rightarrow R$, $f(x)=\sqrt{1-x^{2}}$, and consider the division, I will  denote the points on the real axis by $C_{k}$ and the coresponding points on the unit circle by $A_{k}$.

$C_{0}(0)$, $C_{1}(\frac{1}{n})$, $C_{2}(\frac{2}{n})$, $\cdots$, $C_{n-1}(\frac{n-1}{n})$, $C_{n}=A_{n}(1)$.

On each of the intervals $[k/n,(k+1)/n]$ consider the intermediary points as the right -end-points. The perpendiculars through the points $C_{k}$ meet the unit circle at the points $A_{k}$, please observe that according to this notation $A_{0}$ is the point located on the y axis whose coordinates are $(0,1)$ and $A_{n}$ is located on the x axis, i.e, it is the point $(1,0)$.
On the other hand, the perpendiculars through $A_{k}'s$  parallel to x axis meet the perpendiculars through $C_{k}'s$ paralel to y axis at points $B_{k}'s$, $k=1..(n-1)$, here we consider that $B_{1}$ is located on the y axis, $B_{2}$ is located on the segment $A_{1}C_{1}$, the last point of this construction being $B_{n-1}$ which is on $A_{n-2}C_{n-2}$.

We have that,

The Rieman sum above plus $\frac{1}{2n}$ is nothing else but the sum of the areas of the n-1 rectangles plus the sum of the areas of the n triangles $A_{0}B_{1}A_{1}$, $A_{1}B_{2}A_{2}$, $\cdots$, $A_{n-2}, B_{n-1}A_{n-1}$, $A_{n-1}C_{n-1}A_{n}$  which is definitely less than a quarter of the area of the unit disk, hence the inequality to prove follows.
for solving the left hand side inequality I have no clue, I think it is also a geometric construction maybe something involving trapezoids...

didilica, there is $\frac{1}{2}$ not $\frac{1}{2n}$, and the right inequality follows immediately from trapezium formula.


Could someone please explain to me the usage of brackets in chemical equations with a couple of examples if possible? I read that they are used within compounds, but I have seen compounds written without any brackets.

Any help would be greatly appreciated, cheers;)

Brackets, like parentheses, are used for grouping.  Usually, people like to use parentheses (for example, $Mg(OH)_{2}$ denotes that there are two copies of the group OH).  Brackets are used much less commonly; generally I have seen them only used for complex ions or other places where you would have to use too many parentheses (for example, $[Co(NH_{3})_{6}]^{3+}$).  In general, if you have to have a double layer of parentheses the outer layer can be written using brackets instead.

Oh, and for another usage of brackets, $[X]$ is commonly used to denote the molar concentration of some species $X$.

I've seen brackets used in the two ways that greg has outlined above.

The square brackets for molarity, for indicating concentration [H+] like in an acid.

Or for compounds with more than one polyatomic ion. If the coefficient is one for the polyatomic ion, there is no need for brackets but if there it's not, then the brackets help distinguish the group.


[quote="LuCy4EvA"]1. Sleepy, Doc, Grumpy, and Sneezy play with a special deck of 32 cards. Sneezy deals them out unequally, then says, "if we want to all have the same number of cards, Sleeepy should divide half his cards between Doc and Grumpy, then Doc should do the same with Grumpy and Sleepy, then Grumpy should divide half of his cards between Sleepy and Doc." How many cards did Sneezy deal himself.

2. If it takes 16 Munchkins 10 hours to fill 8 tubs of lollipops, how long will it take 12 Munchkins to fill 9 tubs?

3. What is the positive difference in the area of a square with side length square root of 10 and the area of an isosceles right triangle with leg 10?

4. The product of two consecutive numbers is 3 more than three times the sum of the numbers. What is the positive difference of the numbers?

5. rolling two standard dice, what is the probability that you will get two numbers whose sum is a prime number?[/quote]

[hide="Hints"]1.Work backwards, starting from everyone having the same amount of cards. Or you could notice an something about the different things that have to be done to equalize the number of cards...

2.Look at how long it takes one munchkin to fill one tub.

3.Just simple multiplying.

4.Lots of extraneous info. LOTS.

5.Just computation.[/hide]

most of these questions are "trick" question... think I just gave the answers away by saying that... look at the question REALLY carefully before you do them.... and especially look at number 1....

[hide]
I agree, LuCy4Eva!
   But I think #4 is the funniest one.

I see no way around #5 except a brute-force listing.

Prime sums:
$1$ way to get a sum of 2.
$2$ ways to get a sum of 3.
$4$ ways to get a sum of 5.
$6$ ways to get a sum of 7.
$2$ ways to get a sum of 11.

There are 15 prime sums out of the possible $6^2 = 36$ outcomes.

$P(\text{prime sum}) = \frac{15}{36} = \frac{5}{12}$[/hide]

Heh, I almost did the algebra for number 4 :P .


algebra
polynomial
vector
combinatorial geometry
absolute value
linear algebra
linear algebra unsolved

Let $ A$ be a stochastic matrix.
Show that $ 1$ is a root of multiplicity $ 1$ of the minimal polynomial of $ A$.

This is equivalent to saying that the Jordan blocks corresponding to the eigenvalue $ 1$ all have size $ 1$. Suppose this is not the case. Then we can find non-zero vectors $ x,y$ such that $ Ax\equal{}x$ and $ Ay\equal{}y\plus{}x$. It follows that $ A^{n}y\equal{}y\plus{}nx$ for all positive integers $ n$. This is impossible: the components of the vectors $ A^{n}y$ are contained in the convex hull of the components of $ y$ for all $ n$, whereas the greatest absolute value of a component of $ y\plus{}nx$ increases indefinitely with $ n$ (because $ x\ne 0$).


geometry
geometric transformation
rotation

[hide="Hint"]The angular momentum of the asteroid is $mvr_{\epsilon}$ where $r_{\epsilon}$ is the radius of the earth.  The angular momentum of the earth is \[I\omega_{\epsilon}=\frac{2}{5}m_{\epsilon}{r_{\epsilon}}^{2}\omega_{\epsilon}.\] Now use conservation laws.[/hide]

[quote="nigma"][quote]An asteroid of mass 1.0 * 10^5kg, traveling at a speed of 30km/s relative to the Earth, hits the Earth at the equator tangentially, and in the direction of Earth's rotation. Use angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision.[/quote]

I can't figure out how to express the angular speed of the Earth after the collision without referencing the radius of the asteroid (for figuring out the moment of inertia of the asteroid), which isn't provided. Would someone be willing to point me in the right direction?[/quote]Treat the asteroid as a point mass.

$I_\text{asteroid}= mr^{2}\omega$, where $r$ is the distance between the asteroid and the center of rotation (the radius of the Earth at the equator).