inequalities geometry search perimeter function LaTeX trigonometry

Ok, guys, then probably it is easier if I delete my stupid posts with easy, hard, nice,... and so there will be fewer posts and people will remember. The problem that I see is that there are too many stupid inequalities, which are only copies of old discussed ones. Strangely, the topic names were not a problem earlier  ;) , but suddenly they became. It would be probably easier to distinguish in some way the really useful inequalities from copies. But of course, this is just my opinion...

[quote="harazi"]Strangely, the topic names were not a problem earlier  ;) , but suddenly they became.[/quote]
It was a big problem for ages.

[quote]It would be probably easier to distinguish in some way the really useful inequalities from copies.[/quote]
Hmm... Useful inequalities...  :? Do you mean $x^2\geq 0$?

[quote="harazi"]Ok, if you prefer solving the same inequality 3 times just because some notations differ, then so be it. Personally, I do not have any problem with that, by I just observe that some problems are so insistently repeated that they became myths.[/quote]
I agree. But what can we do? What is your solution for that?

As far as I have seen, there are just a few hundreds very good inequalities, that are extremely useful and that have very nice ideas. So, we (the moderators and anyone who knows well the content of this subforum) could very well try to qualify very good problems( I know, this will be a difficult job, since it involves lot of subjectivity, but we can easily remark which are the problems which really repeat and are extremely discussed and I really think that we can manage to have an objective selection) and move them in a special area, so that it can be easier to search for them. Of course, this includes changing their topic names, but this is not very difficult since we dont have to do this for all 5000 problems o more which were already discussed.  Secondly, I beg the users of the forum to look very well before posting their problems, since it becomes really obsessing to solve 10 times the same problem. Of course, now it is very difficult to find old nice problems, since there are already thousands of problems solved and discussed, but the idea is that between these thousands only a few hundreds are requested each time. I dont know if such a thing can work, but I know that if we dont do something we will have a few hundreds of good problems solved at least 10 times, even if on the site it is written that there are 6000 problems.

I can definitely say that I don't know contents of this subforum.

Yes, Myth, but we need you since you definitely know the level of difficulty of a problem and moreover you know very well which problems are good and useful!

Ok, I've splitted the discussion and made it into an announcement. I agree with Harazi that some order must come into place. Let's discuss all possible solutions and ideas in this thread only ok?

One more thing about posting inequalities.

[b]Please don't write up your problems as follows:[/b]

[quote="A typical geometric inequality"]In a triangle, prove $am_a+bm_b+cm_c\leq \sqrt{bc}m_a+\sqrt{ca}m_b+\sqrt{ab}m_c$.[/quote]

[b]Rather try to write them up as follows:[/b]

[quote="A typical geometric inequality"]Let ABC be a triangle, a, b, c its sidelengths and $m_a$, $m_b$, $m_c$ the corresponding medians. Prove $am_a+bm_b+cm_c\leq \sqrt{bc}m_a+\sqrt{ca}m_b+\sqrt{ab}m_c$.[/quote]

[b]In other words, please explicitely define all notations used instead of hoping that the reader will be able to guess their meaning.[/b] In fact, different people from different countries use very different notations for one and the same thing, and also sometimes denote by one and the same letter different things (for instance, the letter p is sometimes used for the perimeter and sometimes for the semiperimeter of a triangle). Assuming notations to be equal all over the world is nonsense and will only lead to confusion. (Of course, medians are usually denoted by $m_a$, $m_b$, $m_c$, but angle bisectors already cause a Babel tower). But there is also another reason why one should introduce notations. In fact, if one wants to find a special inequality concerning the medians of a triangle on MathLinks, and one uses the "Search" function for that aim, then one gets all the posts where the word "median" is explicitely mentioned - hence, one gets e. g. this post:

[quote="A typical geometric inequality"]Let ABC be a triangle, a, b, c its sidelengths and $m_a$, $m_b$, $m_c$ the corresponding medians. Prove $am_a+bm_b+cm_c\leq \sqrt{bc}m_a+\sqrt{ca}m_b+\sqrt{ab}m_c$.[/quote]

- but one doesn't get this post:

[quote="A typical geometric inequality"]In a triangle, prove $am_a+bm_b+cm_c\leq \sqrt{bc}m_a+\sqrt{ca}m_b+\sqrt{ab}m_c$.[/quote]

since it doesn't contain the word "median" anywhere.

So, I would ask everyone to define all involved notations verbally in the problems.

Thanks!

[b]PS.[/b] About the most important inequalities which are often used and posted 10 times - why not just move them into Inequalities Theorems and Formulas?

  Darij

Yes, I agree. Moreover, following idea came in my head some hours ago: we can define for each (or for almost each) problem list of its keywords. It is the most widespread method, for example, for scientific papers.

Your opinion?

A lot of the inequalities can be reflected in the title of the thread (that is, to just use the ineq, or part of it, as the title). Such way of naming topics is a lot more informative than "nice and hard".

I mentioned this quite a number of times now, if the search engine would allow us to display only the first post of each thread (with LaTeX enabled), then finding a particular inequality would be much easier.

[quote="billzhao"]I mentioned this quite a number of times now, if the search engine would allow us to display only the first post of each thread (with LaTeX enabled), then finding a particular inequality would be much easier.[/quote]
Of course, if you have unlimited traffic :D

[quote="billzhao"]I mentioned this quite a number of times now, if the search engine would allow us to display only the first post of each thread (with LaTeX enabled), then finding a particular inequality would be much easier.[/quote]It will be able soon. I will do that, when all the other stuff is bug-free (which is not, some latex stuff is still broken in different places).

I like Myth's idea, however I think it would be hard to apply. It would be indeed easier to move some of the most well-known inequalities, those that come up over and over again to the Theorems and Forumlaes section ..

[quote="Megus"]I think Myth idea with keywords is really good. We could add in brackets this keyword for example "[trig] in triangle" - or something like that. With for example 10 of such keywords finding inequality would be MUCH easier. What do you think of it ??[/quote]

My proposal: Extending this idea to other subjects than inequalities.  :)

As Peter DVD suggested once but slightly changed. Experienced users might be allowed to rank the level of the problems. As these are just a few selected people we will know appropriate difficulty level. We know the skills of these people.  :)

[quote="orl"]As Peter DVD suggested once but slightly changed. Experienced users might be allowed to rank the level of the problems. As these are just a few selected people we will know appropriate difficulty level. We know the skills of these people.  :)[/quote]Ranking was planned post-wize, but then I had to rush into creating the new board, thus that has moved (once again) further in the future.

[quote="orl"]My proposal: Extending this idea to other subjects than inequalities.  :)[/quote]
Actually, I didn't mean inequalities at all.

So wat about proper naming of topics ? - I see nothing has changed so far, but maybe we could start to change it somehow... still topics named "easy,nice,hard" appears in a great number.   :?

[quote="Myth"]Unfortunatelly, users don't hear me... :( Though I say them from time to time to change titles.[/quote]

I know. Most people are not aware that it is difficult to find old threads easily. I suggest that we just rename improper thread subjects. We also could add keywords as harazi does recently. I also suggested once that you guys make suggestions how to rename the ISL/ILL project titles as I named the source and title the same for convenience matters.  :)

Here is my solution to this problem:

First we need to catagorize all the old inequalities, so a user can easily see if the inequality he/she wants to post already exists or not....  So here are the important catagories:

1.  Inequality Class - First he/she defines the class of inequalites this inequality belongs to: is it algebraic, trigonometric, geometric, number-theoritic/combinatoric (ex. 1*3*5*...*(2n-1)<n^n)

2.  Inequality Sign- What type of sign does the inequality have >, >=, >> ?

3.  Number of Variables- How many variables does it have:
1( inequalities like x^6+2 >= x^4 +2x), 2, 3, 4, 5 or n-variables?

4.  Type of variables? Do the variables belong to the set of integral, non-negative, real or complex numbers? etc.

5.  Type of Inequality- This is the most important catagory, after the user replies to catagory 1 and 3, they can then select the type of inequality it belongs to:  For example if an inequlity is algebriac and has 3 variables, the user can select from the list below to narrow the search:

a. Symmetric Inequality- ex. a + b + c > abc.
b. Inequality with Roots- ex.
$\displaystyle  x^2+y^2+z^2+\sqrt\r\n{3xyz(x+y+z)\r\n}\geq2(xy+yz+zx)$
c. Inequality with Fractions- ex.
$\displaystyle \displaystyle \frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2} \ge \frac{a+b+c}{3}$
d. Power Inequality- Inequalites where the variables are powers. Ex.
$\displaystyle  \\ac^b+\frac{b}{c^a}\geq a+b$

etc.

6.  Constraints/Conditions - To help narrow down the search the user can select what types of conditions the problem has, for example for an algabraic inequality with 3 var.  we can have a + b +c =1, or a + b +c <=1, or abc=1, or abc<=1, or a+b<c, or another inequality, etc.

7.  Origin of the Inequality- Is the inequality from the IMO? AMO, BMO, CMO, ... etc.
or is it from journals? ... etc.

In all I think these 7 catagories can narrow down the inequality correctly.  But this requires a lot of time to go through all the old inequalities.  Further, I recommend we set rules on variables... for example for all 3-var. algabraic inequalities we have to use only the variables a, b, c.  So that the search engine can easily match equations.  I hope this well help.

[quote="manlio"]It is a long job  :?[/quote]

I agree with you. People don't post sources properly nor do they care about putting them into right kind of inequality section. I think your suggestions are nice but they contradict reality and I guess you want to do all the sorting and renaming,

As nothing was done yet why not to do something that was mentioned [unfortunately I don't remember who mentioned it] in some other topic. I mean: create additional subsection in every Olympiad Section subforum (essentialy in Inequalities  :) ) where all the famous problems would be stored (like Iranian 1996 ineq, etc). It'd rather won't help in bringing the order (proper naming of the problems :)), however it'd be easier to find particular links.

Well, I have still one week of holidays and after that I won't be able to help. But if there is anything to be done in this issue this week, I'm ready. You know, I think it is a little bit ridiculous to see each time the same problem and to post each time the same solutions. If there are different solutions, ok, but most of the time they are all the same. So, why shouldn't we put in a special place all popular or difficult or nice or call them whatever problems? It would be much easier.

If only I knew... but I believe it is not my work, but rather the work of a superiour worker here... ;)

[quote="jensen"]Harazi how did you post your message as stikies?can you tell me?Thanks a lot. :lol:[/quote]Harazi is a moderator on this forum and he can make posts sticky if he wants, but in general I stickfied his posts :D

It seems like this site can do searches with latex input.

Maybe we (i.e. the forum program) can do a search for the formulas when some user tries to post a new topic. If we find matches, we don't let the user post, or if that is too harsh  :D , at least notify him/her that this problem might have been posted before and provide a link to the search results.

I checked some latex formula and it seems like the current search using latex formulas is not that good: there is too much noise. Maybe we can improve the search by having a latex formula specific search function, not sure how hard would that be to implement, but I think we could get a reasonable latex formula matcher implemented. If the matcher is smart enough, maybe it could even cope with renamed variables etc.

This matcher could then also be used to save the space used by the images too. I noticed that, even though we have the exact same looking formulas in different posts, the image files are different.

Just a thought.

These issues have been addressed many times. The problem is that it is hard to create such a program that will "recognize" $\LaTeX$ as users do. I have a working feature right now on the site, however it doesn't cover all the possible cases and outcomes. That is why (for now) it is harder to search in latex posts as well.

I think there is a challenge for all math forums: how to index/categorize a post?

We can certainly have rough categorization for the posts like sub forum etc, but this will not solve the duplication problems. My thinking is to have a database column called tags where people can put multiple words for the nature of the math topic. that way a old post may be easily found.

Here is a suggestion. It may take a while to implement though, but it's possible with time and effort.

For each of the Olympiad Unsolved and Proposed subforums, have mandatory categorization for the type of problem one is posing. But not difficulty level, as this could easily change if someone finds a new easy method, not to mention difficulty has nothing to do with identifying repeated problems.
For example, for Inequalities, it can be Algebraic or Geometric.
For either, how many variables are there? Is there a popular constraint such as the product and/or sum of certain variables equal certain constant/s?
For geometric inequalities, are there medians involved? Heights? Sides of a triangle? Quadrilateral? Cyclic polygons? How many circles? Isosceles properties?
Of course, the tags would only refer to the problem, not the solution.
I think it is not too difficult to develop [b]intelligent algorithms [/b]to take input (mostly [b]cleverly-designed multiple choice[/b], and [b]even the Latex[/b]) and match with previously tagged problems, and then ask if the proposed problem is the same as ones that seem to be similar.
Every problem proposer should have to go through the (not THAT tedious) process of standardized tags, and checking off that his/her proposed problem does not match with ones that the search algorithm shows are most similar.
I'm thinking that for any post (not just the first post of a topic) that proposes a problem (or proposes AND solves it), a set of input is required, followed by more fields, depending on what was input previously, and ending with checking it is not a post that is the same as a previous post.

Some specific ideas for the logic behind matching two Latex expressions:
1. Consider $n\sqrt[n]{x}$ the same as $x^{\frac{1}{n}}$.
2. Recognizing two equivalent 3-variables expressions, where one uses $a,b,c$ and the other uses $x,y,z$.
3. Consider $\frac{a}{b}$ the same as $a/b$.
4. Consider $[$ brackets the same as $($.
5. Standardize for this community that $[x]$ does NOT mean a floor or ceiling function, but the nearest integer.
6. Consider $a>b$ the same as $b<a$.
7. Consider $(ab)^2$ the same as $a^2b^2$
And the list goes on and on...

To ensure that the algorithm catches as many fish as possible, non-Latex problems on the Olympiad forums should be deleted or edited.
And at one point or another, people would have to go back and complete the titanic task of categorizing all past posts.

It would take a lot of collaboration and testing through time to perfect the program's parameters and such. But together, we can make it happen.
If the administrators are interested, and if I can help in any way, like developing some of the logic, just ask.

How about a sticky page with the most common inequalities and stuff related. 1 for alg., 1 for geometry, and 1 for NT/misc.


modular arithmetic
induction
number theory proposed
number theory

Given a sequence $(a_n)$ of real numbers such that the set $\{a_n\}$ is finite.
If for every $k>1$ subsequence $(a_{kn})$ is periodic, is it true that the sequence $(a_n)$ must be periodic?

Hmm I think I misunderstand something since this is too easy.
I think you should more clearly define the word "[b]periodic[/b]"

if a_n goes 1,3,2,5,1

then a_2n goes 1,2,1 which is periodic (?)
a_3n and a_4n only got one element so are periodic by definition (?)

but a_n as listed there is not periodic (?)

Though, I think this will be around the word periodic we're twisting.

No, the sequence is infinite, but only take a finite number of different values.

Pierre.

I don't think $(a_n)$ needs to be periodic. With a bit of effort it can be shown that we may take any value we want for $a_p$ with $p$ prime, and then we could employ Dirichlet's theorem: for any $T$ we make sure that there are two primes $p_1,p_2=nT+1$ s.t. $a_{p_1},a_{p_2}$ have different values.

I think it's enough to work with only two numbers :$0$ and $1$.

Can anyone please post the full solution for this problem! I think that 1 and 0 are enough, too.

I was looking through some older problems and I remembered this one, so I decided to give it another shot :).

We want to prove that there is a non-periodic sequence $\mathcal S_1=(a_n)_{n\ge 1}$ with elements $0$ or $1$ s.t. for all $k\ge 2$ the subsequence $\mathcal S_k=(a_{kn})_{n\ge 1}$ is periodic.

First let's prove that if $(p_m)_m$ is the sequence of primes, then we can choose $a_{p_m}$ as we wish so that our sequence satisfies the given condition (i.e. $\mathcal S_k$ is periodic is periodic for all $k\ge 2$). Then, as I said above, if we choose for each $T$ a pair of primes $q_T^1\equiv q_T^2\pmod T$ and set $a_{q_T^1}\ne a_{q_T^2}$; then $T$ cannot be a period of the sequence $\mathcal S_1$, and we're done.

First of all, notice that if $\mathcal S_m$ is periodic, then so is $\mathcal S_{mn}$ (for any $n$), so in the construction of our sequence it suffices to ensure the fact that $\mathcal S_{p_n}$ is periodic for all $n$. Now assume we've chosen arbitrary values for $p_1,\ldots,p_{n-1}$ s.t. $\mathcal S_{p_i}$ are periodic for $i=\overline{1,n-1}$, and we want to do the same for $p_n$. The sequences $\mathcal S_{p_np_i},\ i=\overline{1,n-1}$ have some periods $T_i$ respectively. We set $T_n=\prod_{i=1}^{n-1}p_iT_i$, choose all the elements of $\mathcal S_{p_n}$ with index $<p_n+T_n$ arbitrarily (I mean the index that they have as elements of the initial sequence $\mathcal S_1$), and complete the sequence $\mathcal S_{p_n}$ s.t. $T_n$ is its period. This way, we complete the induction step.

I hope it's Ok.


I want to register for the AMC 8 and 10/12 for the 2008-09 school year. It's my first time. How do I do that if my school doesn't offer it?

Is it possible that some colleges offer test sessions for AMCs so that students can take them there? If so, do they have to miss school?

Also, what's the cost for one student taking the AMC? And do students need to register again for the AIME/USAMO if they qualify? Or do initial AMC registrations include AIME/USAMO admittance if a student qualfies?

[quote="worthawholebean"]All these questions are answered on the [url=http://www.unl.edu/amc]AMC website[/url].[/quote]
I didn't really understand it. :(

Enumerate what you din not understand and we will try to make it clearer.

But first, re-read Frequently Asked Question 4 at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


The Mock MOEMS Contest Round 1/Meet 1 has been posted

Please do not use calculators. This is based on the honor system.

[hide="Scores"] \begin{tabular}{|l|c|c|c|c|c|c|c|c\parallel c|l|}\hline & Meet 1 & Meet 2 & Meet 3 & Meet 4 & Meet 5 & Total\\ \hline 1. SplashD& 4 & & & & &\\ \hline 2. Arvind\_{s}n& & & & & &\\ \hline 3. kheg\_{k}& & & & & &\\ \hline 4. i\_{l}ike\_{p}ie& & & & & &\\ \hline 5. mimar& & & & & &\\ \hline 6. ragnarok23& & & & & &\\ \hline 7. meenamathgirl& & & & & &\\ \hline 8. goldendomer& 3 & & & & &\\ \hline 9. now a ranger& 5 & & & & &\\ \hline \end{tabular} [/hide]

[quote="anirudh"]yes...I just need [u][b]1[/b][/u] volunteer to grade the papers[/quote]
I'll do it. :D :D

ok. so we have some participants...
lotrgreengrapes7926 will coach...
i guess that means that yoou pm him the answers, and then he grades them... i'll pm the answers to him

They are from various sources...
I'll try to make them original

How many rounds of 5 questions will there be? :maybe:

You could make up your own problems. I wrote the 50 problems for MOEMS Competition in about 2-3 days. Of course, that was over the summer. :D

I don't know. My plan is 5, if it succeeds, and have some awards :rotfl:

OK. I have posted meet 1. PM the answers to lotrgreengrapes7926; he has the answers. Everyone will then do Meet 2.

[size=200][b]It's not too late, until roudn 2 has been posted[/b][/size]

ok. Answers are due by the end of this month

[quote="anirudh"]The Mock MOEMS Contest Round 1/Meet 1 has been posted[/quote]
They're not posted anymore. :D

[quote="anirudh"]I dont know what happened to it[/quote]
Could you post it again? :lol:

[u]People who are going to do this:[/u]

SplashD
Arvind_sn
kheg_k
i_like_pie
mimar
ragnarok23?
meenamathgirl
goldendomer

If you have a question mark next to you, please verify by posting here. There are going to be 5 rounds of 5 questions each

The question sets I received and scores have been posted on scores on the first post. Anyway, I will be PM'ing anybody who hasn't submitted their Meet 1 Paper. If you do not submit it, you will receive 0 points for it. You can still try out for Meets 2-5.

Meet 1 Deadline-September 25. Exceptions Negotiable.

OK, so when are we going to do the next round. It's been awhile.

I am verrrrrrrrrrry interested hehe  :lol:

You don't have to do meet 1 to do meet 2.  :)

kk =) and it is 5 PM right now. Is it the same for you all?

oooh I am from the state which Vestavia is located in... lol... which  people probably know. I guess you must live in a location east of alabama such as atlanta or ny haha. After all, central time states are small.

Round 2 has now been posted. Please show methods. If methods are not shown, no points will be given. PM answers to anirudh.

[b][color=blue][size=200]DUE NOVEMBER 7, 2006![/size][/color][/b]

[quote="now a ranger"]haha i already pmed it in.[/quote]
Do meet 1. If you're still can.

you have to do meet 1 still (unless you're willing to get a 0)


AMC USA(J)MO USAMO AIME trigonometry geometry MATHCOUNTS

I'm in 10th grade, and I haven't really done much since middle school. I got 33rd in the Nation back then(I screwed up bad) in MATHCOUNTS. Last year, I didn't do any math at all, so I deserved my horrible scores. On the AMC A date, we had a snow day. I took the test afterwords for practice and got a 150. Then, came the B which actually mattered, I got a 120. All very stupid mistakes. I missed the first question. Then, I got a 6 on the AIME.

I really don't need learn to much for the AMC, but how do I learn to do USAMO and AIME problems? I am quick at getting tricks(the crux move), but my arithmetic and lower level manipulation skills are low. The biggest problem though is learning new stuff such as Ceva and all that stuff. I need to boost my knowledge of logs and trig and geometry in general and algebra.

There are so many tools that I lack. I can only do problems that require 7th grade level knowledge. How do I return to the land of mathematics? I need knowledge, and I want to be skilled with it.

Well on a site that sells math classes and math books (which are really really good) you're going to get people saying buy these books or take this class. My advice to you though, since you pretty much already know everything you need is just doing problems. Additionally, when you do learn new topics, don't just learn the formula, learn how to derive the formula. This will make it much easier to remember.

Yeah.  Buy the old tests or look in the resources section.  Study solutions.  For AIME/USAMO problems, buy AOPS volume 2, ACoPS, and engel.

Thanks. It's just been so long, and it's going to be hard to start again, but after I get started I'll be okay. Thanks again.

I am also in 10th grade and I've lost my interest for math. The last time I actually studied and prepped for a competition was MATHCOUNTS over two years ago. Since then, I've slaved through my math classes with boredom, hating the subject of math more and more.

I still haven't done anything significant in math, but I do think that finding an interesting aspect of mathematics to delve into is key in getting started again. Practicing for the AMC/AIME/USAMO gets as repetitive as the busywork assigned in school math classes. Whether it be game theory, statistics, or even physics (shudder), try and do math that's [i]not[/i] competition related.

Well, obviously, don't prepare if it isn't fun. I probably end up doing 10-20 hours of math a week, but it's all fun. I recommend AoPS 2 and especially ACoPS, the best math book I've ever read.

yeah, but since he's already in 10th grade, and some of the material in AoPs vol 2 and ACoPs overlap it may be that he will skip ACoPs.

hey guys

im in a similar position as the original poster of this thread. i did mathcounts back in the day, and now i do  local math contests (i live in new mexico). i want to do something more broad, however. i made the aime last year but i didnt study well, so i didnt do well. this year i want to make usamo and maybe beyond.
however, since im new to this site, i dont know the commonplace acronyms  :)
what are acops and engel, and where can i find them?

ACoPS is The Art and Craft of Problem Solving by Paul Zeitz. Engel is Problem-Solving Strategies by Arthur Engel. I think both are available in the AoPS bookstore.

thanks. are there any other books that might be helpful?

new mexico, even though it sounds like remote, is awesome for math and science. I used to live in Los Alamos, New Mexico.
1. Do the Supercomputing Challenge (challenge.nm.org) if you're interested in programming. Even though registration is closed for this year, they may be able to squeeze you in.
2. AMC 10/12 is offered at UNM at Albuquerque
3. there's the UNM-PNM which is total awesomeness.

DUDE, ENJOY NEW MEXICO AS MUCH AS YOU CAN, I MISS IT A LOT NOW THAT I MOVED TO FLORIDA. FLORIDA SUCKS COMPARED TO THE SCIENTIFIC COMMUNITY OF LOS ALAMOS (THE NATIONAL LAB WHERE THE ATOMIC BOMB WAS MADE IS THERE MIND YOU)

[quote="simo14"]even if you are in 10th grade, dont underestimate the power of vol 1.. it can get you very far.
besides aops 2, acops, and engel, i would recommend doing aops subject books in your weak areas.
take the pretests. If you can do most (at most not being able to do 1) questions, then dont get the book. Intro is good for amc 10, 12, and easy aime. intermediate is for aime and easy olympiad.
Once you get to the point where you can solve more the half the problems on the aime (8+), maybe you should try olympiad problems. they've really helped me- both in terms of boosting aime score-and olympiads (come on, aime is not an end in itself  :wink: ).[/quote]

Also: Russian Olympiad is REALLY good for combinatorics. There are also many advanced books that focus on one particular subject (number theory, geometry, etc...). For these books, it is very important to try to do most of the problems.

um, pythag, he hasn't done much math since MC and you don't expect him to be as beast as you in a week do you?

thanks for all these suggestions guys, i'll look into them

and, if its easier for you to understand my skill lvl this way (poincare, you can probably understand this the best, since you know NM), i got first in my grade on the unm pnm last year. about a 27 out of 70 score, i think.
knowing that thats where im at, do i go aops vol 1 or straight to vol 2?

if you can afford it, definitely buy both. If not, you should do pretty well with Vol 2.

[quote="worthawholebean"][quote="Poincare"]in some cases, you may want to skip through ACOPS after doing Vol 2 and move directly to engel and come back to ACoPs if you have time.[/quote]I don't think this is a good idea; ACoPS is just such a great book in general. I really like the writing style, and it has taught me more about math than any other book.[/quote]

I agree with worthawholebean: though I really liked the AoPS series, I think Zeitz does a better job showing you how to investigate a problem. I only wish I had a solutions manual for it...

I'll step in here since most fundamental questions have been answered.

@master188: First of all, is 150/120 on AMC12 or AMC10? This makes a big difference. 6 is not a bad score but it certainly needs to be polished more. I can offer you two options that others stated but with more information.

Track 1: Do past AIME problems.
Don't worry about AoPS, ACoPS, Engel, and whatnot. Just work on problems and do all of them from 1983 to now. This is NOT an easy task but if you do ALL problems, you will immensely improve. All books are what they call "prep" books (although these books are much better than SAT prep books...). Now, here is a key part: you have to do all the problems and try on them without looking at solutions. Spend at least couple hours doing them and look at solutions at the end of day. Next day, do the problems that you missed/not answered from previous day and new set of AIME problems. Don't rush through them.

Track 2: Do some problems and go through books suggested.
I recommend AoPS most because they have solutions. I'm a strong believer of good solutions because they are as good as, if not better, than problems. AoPS books are excellent because of their phenomenal friendly lectures, problems, and solutions. Books after them matter less at this moment until you go through them and build good skills. I've heard of people who score about 9-10 in AIME with just Volume I (by getting ALL geometry right... you don't "need" to know Ceva's to get Ceva problems right) so yeah it's possible.

I offer you two tracks because you're sophomore and you'll have to take SAT, worry about your GPA, and trust me, it's going to fly. Math [i]is[/i] not everything. Don't gamble all of your time into it.

You goal should be learning how to learn. It doesn't help you much to just read how other people have figured out to learn. The most important part of becoming a good problem solver is to constantly think about how you learn and improve yourself. When you get to college, it doesn't help you much to know a bunch of math tricks. But it helps you immensely if have taught yourself how to learn on your own.

As for more concrete ideas, I think AoPS vol. 1 &2, ACoPS were helpful. They expand your mindset a lot and they give you a good overview of the necessary problem solving tools. But the majority of your studying should just be doing problems building your thinking skills through practice.

I completely stress doing problems. I learned a [hide="thing"]Find the number of distinct numbers in the list $ \lfloor \frac{1^2}{10000}\rfloor, \lfloor \frac{2^2}{10000}\rfloor, /dots , \lfloor \frac{10000^2}{10000}\rfloor$[/hide] or two on one AIME alone.

Read AoPS volume 1 for good solid ground in logarithms, exponents, complex numbers, and use AoPS volume 2 for trig.

Volume 1 is very good for a geometry review, and starting algebra. Volume 2 is very good for expanding everything in volume 1.

ACoPS is good if you are persistent. If you give up on problems easily, ACoPS might not be for you.
Engel is a bit advanced, but if you can get a 6 on the AIME, then you might be almost ready for it.

And about doing problems, don't wait for them to come to you. Find them. Do old contests, read mathematical magazines(Crux Mathematicorum), and join a local math circle for practicing contests(if there is one reasonably near you).


floor function
function
combinatorics open
combinatorics

is there any closed form for
\[ \sum_{m\equal{}1}^{k}\sum_{j\equal{}1}^m(\minus{}1)^{j\minus{}1}\binom{2k\minus{}m\minus{}j}{m\minus{}j}\]

??

\[ \sum_{m \equal{} 1}^{k}\sum_{j \equal{} 1}^m( \minus{} 1)^{j \minus{} 1}\binom{2k \minus{} m \minus{} j}{m \minus{} j} \equal{} \sum_{j \equal{} 1}^k( \minus{} 1)^{j \minus{} 1} \sum_{m \equal{} j}^{k} \binom{2k \minus{} j \minus{} m}{m \minus{} j} \\
\equal{} \sum_{j \equal{} 1}^k ( \minus{} 1)^{j \minus{} 1} \sum_{m \equal{} 0}^{k \minus{} j} \binom{2k \minus{} 2j \minus{} m}{m} \equal{} \sum_{j \equal{} 1}^k ( \minus{} 1)^{j \minus{} 1} F_{2k \minus{} 2j \plus{} 1} \equal{} F_k^2.
\]

nice short proof, thank you !

here is the next one
\[ \sum_{m\equal{}1}^{K} (\minus{}1)^{m\minus{}1}3^{k\minus{}2m\plus{}1}\binom{k\minus{}m}{m\minus{}1}\]
where $ K\equal{}\left\lfloor\frac{k\plus{}1}{2}\right\rfloor$. this one should be $ F_{2k}$.
any ideas how to deal with this one ?

It is nice, but what reason to put it in Open Questions?

I didn't knew if there was a closed form for the first one

the second "looks similar", so I continued here instead of writing new topic

[quote="knoppix"]\[ \sum_{m \equal{} 1}^{K} ( \minus{} 1)^{m \minus{} 1}3^{k \minus{} 2m \plus{} 1}\binom{k \minus{} m}{m \minus{} 1}
\]
where $ K \equal{} \left\lfloor\frac {k \plus{} 1}{2}\right\rfloor$. this one should be $ F_{2k}$.
any ideas how to deal with this one ?[/quote]
There is a generating function:
\[ \sum_{t\equal{}0} {t\choose s\minus{}t} x^t \equal{} \frac{ \left(\frac{x\plus{}\sqrt{x^2\plus{}4x}}{2}\right)^{s\plus{}1} \minus{} \left(\frac{x\minus{}\sqrt{x^2\plus{}4x}}{2}\right)^{s\plus{}1} }{\sqrt{x^2\plus{}4x}}\]
Just plug in $ s\equal{}k\minus{}1$, $ x\equal{}\minus{}3^2$ and use Binet's formula.

thanks,
could you tell how did you compute this sum ?
Probably one should guess some reccurent relation for it (and next solve it), but I don't see it .......


ratio
calculus
derivative
logarithms
factorial
function
geometric series

Evaluate $ \frac{1}{2} \minus{} \frac{2}{4} \plus{} \frac{3}{8} \minus{} \frac{4}{16} \plus{} \frac{5}{32} \minus{} \ldots$

Not sure if this will be helpful, but going back to Mewto55555's third line,

$ 3S\equal{} \frac{3}{4} \minus{} \frac{1}{8}((1\minus{}\frac{1}{2} \plus{}\frac{1}{4}\minus{} \frac{1}{8} \plus{} ...)$

$ \equal{}\frac{3}{4} \minus{}\frac{1}{8}(\frac{1}{1\plus{}\frac{1}{2}})$,

since the sum to infinity of a geometric series with first term $ a$ and common ratio $ r$ is $ \frac{a}{1\minus{}r}$ and here $ a\equal{}1$ and $ r\equal{}\frac{\minus{}1}{2}$.

So $ 3S\equal{} \frac{3}{4} \minus{}\frac{1}{8}(\frac{2}{3})\equal{} \frac{3}{4} \minus{}\frac{1}{12}$

Therefore, $ S\equal{} \frac{1}{4} \minus{} \frac{1}{36} \equal{} \frac{8}{36} \equal{}\frac{2}{9}$.

[quote="007math"]How do you calculate the sum of Taylor Series?[/quote]
In this case, if we treated the series as a Taylor series, we could solve as follows:

$ \sum_{0}^{\infty }r^{n} \equal{} \frac {1}{1 \minus{} r}$

Taking the derivative of each side,
$ \sum_{1}^{\infty }nr^{n \minus{} 1} \equal{} \frac { \minus{} 1}{(1 \minus{} r)^{2}}$

If we apply the derived formula to the problem, we have:
$ \minus{} \frac {1}{2}\sum_{1}^{\infty }n \times \left (\frac {1}{2} \right )^{n \minus{} 1} \equal{} \frac {1}{2}\times \frac {1}{(1 \plus{} \frac {1}{2})^{2}} \equal{} \frac {2}{9}$

[quote="007math"]I'm asking more of a general formula/way.  Like for example:

$ 1 \plus{} 5 \ln{4} \plus{} \frac {(5 \ln{4})^2}{2} \plus{} \frac {(5 \ln{4})^3}{6} \plus{} \frac {(5 \ln{4})^4}{24} \dots$

How would you find the sum?

I'm guessing this stuff is way to advanced for me...[/quote]

Looking at the question, we see its similarity to the Taylor series expansion for $ e^{x}$.
$ e^{x} \equal{} \sum_{0}^{\infty }\frac {x^{n}}{n!} \equal{} 1 \plus{} x \plus{} \frac {x^{2}}{2!} \plus{} \frac {x^{2}}{3!} \plus{} ...$

In fact, we can merely substitute $ x \equal{} 5\ln 4$ into the formula to find that the sum equals $ e^{5\ln 4}\equal{}1024$.

To solve general problems in this form, just become familiar with all of the common Taylor/Maclaurin series expansions and try to recognize which expansion the given series resembles. After some manipulation perhaps, you should be able to find the sum of the series.

Do you know its $ e^x$ just because of the $ 5 \ln{4}$?

I had figured out that each term is $ \frac{(5\plus{}\ln4)^x}{x!}$ if $ x_0 \equal{} 1$, but I thought it had no relevance.  What are the other taylor series expansions?

[quote="007math"]Do you know its $ e^x$ just because of the $ 5 \ln{4}$?

I had figured out that each term is $ \frac {(5 \plus{} \ln4)^x}{x!}$ if $ x_0 \equal{} 1$, but I thought it had no relevance.  What are the other taylor series expansions?[/quote]

I recognized its similarity to $ e^{x}$, because of the factorial denominators and the numerators that increase by a power from term to term.  If you have not learned calculus yet, I would not worry too much about Taylor/Maclaurin series, besides maybe being familiar with some common ones...

A list of common Taylor series:
http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions


All I know about spyware is that it is a software that can either help you or be annoying. It does things on your computer without your consent. Spyware might collect your personal information and use it as identity theft (I'm not totally sure what that means). It can enter any type of computer.

Spyware can be killed though.

I just want to know if there are any free anti spyware programs that last forever.

What more is spyware?

(my worst case of spyware:not letting me install DSL to my computer)

spyware usually doesn't help you. it stays hidden while monitoring your computer to see your passwords, account numbers, stuff you buy, etc.

Spybot Search & Destroy works quite well on Windows. Or of course you can switch to Firefox, which hasn't much trouble with spyware, or all the way to Linux, and have fun going to sites that people tell you are virus-laden just to show off how immune you are to viruses.

Or if you still want to retain a level of user friendliness which Linux doesn't provide, try a Mac; although, the market of malware for Macs is growing.

Here's some tips for mantinence:

If you're using Windows, Update. Regularly. http://windowsupdate.microsoft.com

Try CleanUp! http://www.stevengould.org/software/cleanup/ The first time you're running it, go to options, and make sure only these are checked: Empty Recycle Bins, Delete Cookies, Delete Prefetch files, CleanUp! All Users
This is just to save space and files on your computer and some malware hides in your Temp Internet Files so this may help flush them out.

Scan Ad-Aware along with Spybot Search & Destroy (one at a time though) regularly. Google these if you want to know where to get them.

Two very good free online virus scanning sites are Panda ActiveScan at http://www.pandasoftware.com/products/activescan.htm and TrendMicro Housecall http://housecall.trendmicro.com/
This isn't really maintenence although if you have time, it's always good to make sure your computer is clean. One of them is more than enough for maintenence.

For a free trojan remover, try ewido anti-malware. Instructions to install are here: http://wiki.castlecops.com/Malware_Removal:_Trojan_Removal_Programs#Ewido_Security_Suite_Trial

spyware doctor is what I use, and I personally think it's the best and most efficient  :wink:

[quote="solafidefarms"][quote="Ignite168"]Or if you still want to retain a level of user friendliness which Linux doesn't provide, try a Mac; although, the market of malware for Macs is growing. [/quote]

Actually, if you want a relatively seamless transition from Windows, Linux is easier than Macs to learn; as for user friendlyness, Linux is competitive with the userfriendlyness of a Mac. :) :P[/quote]

Linux is definitely not easier to learn than a Mac. I bet if you take a random IE & Windows XP user and give them a computer with Linux for an hour and Mac for an hour, he'll learn how to do a lot more things on the Mac than on Linux.

I have helped many friends fix their spyware problems.

Download Adaware SE Personal from download.com
Download Spybot S&D (Search and destroy)

Update both programs before you scan.

Download firefox, and start using it. Its better than IE anyways and it keeps out the bugs.

if you have further trouble

http://www.iamnotageek.com
Lots of spyware experts there, so ask them.

Wow, why hasn't anybody mentioned WINDOWS DEFENDER.  Sure, it's from Microsoft, but they bought the product from another company so...:lol:.

i don't use any virus protection or spyware protection. on the other hand, i don't go to many places on the internet. i go here, google, the weather website, and wikipedia. i have to be careful on google, though.

[quote="Ignite168"]Linux is definitely not easier to learn than a Mac. I bet if you take a random IE & Windows XP user and give them a computer with Linux for an hour and Mac for an hour, he'll learn how to do a lot more things on the Mac than on Linux.[/quote]
Well that depends on which Linux distro. Give them SuSE and they'll get XGL installed in 5 minutes. And Mac cannot match the awesomeness of XGL.  :P


algebra
polynomial
calculus
integration
number theory

Given arbitrary positive integer $ a$ larger than $ 1$, show that for any positive integer $ n$, there always exists a n-degree integral coefficient polynomial $ p(x)$, such that $ p(0)$, $ p(1)$, $ \cdots$, $ p(n)$ are pairwise distinct positive integers, and all have the form of $ 2a^k\plus{}3$, where $ k$ is also an integer.

similar to http://www.mathlinks.ro/viewtopic.php?t=1366

Let $ n!\equal{}p*q$, where $ p$ shares prime factors with $ a$ and $ (q,a)\equal{}1$.
Then we can find $ t>1$ such that $ q|a^t\minus{}1\equal{}b$. Also $ p|a^p$.
$ P(x)\equal{}2*a^p(1\plus{}\sum_{i\equal{}1}^n\frac{x(x\minus{}1)\cdots(x\minus{}i\plus{}1)}{i!}b^i)\plus{}3$.
Since $ i!|n!\equal{}pq|a^p*b^i$, the coefficients are all integers.
For any $ 0\le m\le n$,
$ P(m)\equal{}2*a^p(1\plus{}\sum_{i\equal{}1}^mC_m^ib^i)\plus{}3\equal{}2*a^p*(b\plus{}1)^m\plus{}3\equal{}2*a^{p\plus{}tm}\plus{}3$.

[quote="xxp2000"]similar to http://www.mathlinks.ro/viewtopic.php?t=1366

Let $ n! \equal{} p*q$, where $ p$ shares prime factors with $ a$ and $ (q,a) \equal{} 1$.
Then we can find $ t > 1$ such that $ q|a^t \minus{} 1 \equal{} b$. Also $ p|a^p$.
$ P(x) \equal{} 2*a^p(1 \plus{} \sum_{i \equal{} 1}^n\frac {x(x \minus{} 1)\cdots(x \minus{} i \plus{} 1)}{i!}b^i) \plus{} 3$.
Since $ i!|n! \equal{} pq|a^p*b^i$, the coefficients are all integers.
For any $ 0\le m\le n$,
$ P(m) \equal{} 2*a^p(1 \plus{} \sum_{i \equal{} 1}^mC_m^ib^i) \plus{} 3 \equal{} 2*a^p*(b \plus{} 1)^m \plus{} 3 \equal{} 2*a^{p \plus{} tm} \plus{} 3$.[/quote]

[quote="xxp2000"]$ q|a^t \minus{} 1 \equal{} b$[/quote]
What does this mean?  Do you mean $ qb\equal{}a^t\minus{}1$?  If so, then how does:

[quote="xxp2000"]$ 2*a^p*(b \plus{} 1)^m \plus{} 3 \equal{} 2*a^{p \plus{} tm} \plus{} 3$[/quote]??
 i.e. how does $ (b \plus{} 1)^m\equal{}a^{tm}$?  Because obviously $ b\not\equal{} a^t\minus{}1$.

$ q|a^t\minus{}1\equal{}b$ means $ a^t\minus{}1\equal{}b$ and $ b$ is multiple of $ q$.

[quote=P-H-David-Clarence]It's simple using Language's interpolation formula.Assume $$P\left( i\right) =2^{k_{i}}+3$$then$$P\left( x\right) =\sum ^{n}_{i=0}P\left( i\right) \cdot\prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$$For a big $k$,$$f\left( x\right) =\sum ^{n}_{i=0}\left( 2a^{k}+3\right) \cdot \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }=2a^{k}+3$$So we just need to make $$\sum ^{n}_{i=0}2\left( a^{k_{i}}-a^{k}\right) \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$$a integral polynomial,thus we just need to make $$n! \vline  a^{k_{i}}-a^{k}$$,it's easy just make $k$ big enough and $\varphi \left( n!\right) \vline k_{i}-k$[/quote]

 "$For a big $k$,$$f\left( x\right) =\sum ^{n}_{i=0}\left( 2a^{k}+3\right) \cdot \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }=2a^{k}+3$ So we just need to make  $\sum ^{n}_{i=0}2\left( a^{k_{i}}-a^{k}\right) \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$a integral polynomial" What do these 2 lines mean? Can you explain, please.

Also the last line $\varphi \left( n!\right) \vline k_{i}-k$. I think you use Euler theorem but if you use like that, we need $(a,n!)=1$.


What can we do? Please vote on something.  :)

I'm setting a model example by posting here in the early morning(it's 6 o'clock here).
[size=200]Please save this forum!!!  [/size]

Here is my way:

[color=red]Admin edit: snipped![/color]

I voted "bug ppl till they post more".

*poke poke* post more!

So far, 3 out of 6 voted "bug ppl". I shouldn't expect more people to vote since this is all actice members we have. :rotfl:

Who should we start with?

start some interesting topics.  don't have to be pointless... but they do have to be interesting.

Should I start posting problems for you guys to solve again?  *snicker snicker*

YOu could post problems that are realy time consuming(people would ask questions about them).

Yeah we need to find some way to save this forum.

Desperate times call for deperate measures.

[quote="rem"]Yeah we need to find some way to save this forum.
Desperate times call for desperate measures.[/quote]
I agree. Lets make a topic thats Canadian to the max. :rotfl:

[quote="Singular"][quote="Sunny"][quote="hello"]Post about mathematics! Or maybe, where to find beautiful Russian girls? lol[/quote]

How come no one talks about finding canadian girls?! lol, go canada![/quote]

because russian girls are beautiful[/quote]

i like canadian girls!  :D

Here's something you can try. (My perhaps fruitless attempt to get something going)

[url]http://www.cs.uwaterloo.ca/~shallit/Talks/pmc2.ps[/url]

It's the "Ten Problems that I Can't Solve" by Jeffrey Shallit of the CS Dept at U of Waterloo. (Some of you at UW may know him.) He gave this talk at the pure math club years ago and offered money for solutions to any one of ten problems. I'm guessing that the money offer is still open. Two of these problems (problems 3,5) have been solved.

Don't believe that these problems are unsolved because they are overly difficult. But only because not enough people have looked at them.

[quote="rem"]What do I open the file with? (For some reason it does not open :( )[/quote]
Well, the file type is PS. Which (I think) means Post Script. I'm getting a program of the web right now that says it can view PS, if it works, I will post the address for the program.
Edit: I was right, it does stand for Post Script. :lol:
Anyway you don't have to download anything. To open the file, go to
http://view.samurajdata.se/
and in file view, upload the file. Then just wait a while while it uploads and renders it, and then Success! :)
In case you didn't save the file on your computer, no problem. Just type in the address provided by whisper00.

[quote="Inspired By Nature"][quote="rem"]I like Russian girls! :blush: :love:[/quote]
I don't think you guys are contributing ways to save the forum. :P  :rotfl:[/quote]

Hey, isn't talking about girls the most inspiring topic ever? (or at least it's inspiring for girls to read). Or at least it was for me (maybe because I'm Russian  :P ). Soon, this topic will have a complete classification of girls. Yay!


From a set of consecutive natural numbers one number is excluded so that the aritmetic mean of the remaining numbers is $ 50.55$. Find the initial set of numbers and the excluded number.

[hide="not vigorous"]The arithmetic mean of a sequence of natural numbers beginning with one is $ \frac {n \plus{} 1}{2}$.  For $ n \equal{} 99\rightarrow101$, the mean is very close to the mean with one missing number.  This suggests that the mean of the initial numbers is either $ 50, 50.5, 51$.  Arithmetic mean does not vary if we shorten the sequence by eliminating the front and end numbers in pairs.

Now we need to find how short to cut it.  $ 50.55$ suggests that there are $ 21$ initial terms because of the $ \frac {11}{20}$ found in the mean without one term.  This eliminates numbers with a mean of $ 50.5$ since that occurs from an even number of natural numbers.  Furthermore, the greatest difference that can result from removing one integer from a sequence is $ .5$, which eliminates $ 50$ as a possible mean.  Thus, the mean must be $ 51$.  Removing the largest number from the sequence ($ 61$) results in a shift of the mean to $ 50.5$.  Since we want a number $ \frac {1}{20}$ away from this mean, we can remove the second largest number.

The initial set of numbers was $ 41\rightarrow61$ and the removed number was $ 60$.[/hide]


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There is a square board with 4 pockets with no asymmetry of any kind. Each pocket can either contain a ball or it is empty. It is not visible from the top (or bottom) whether a pocket is empty or full. You have to insert your hand in the pocket to know its condition. There is also a whistle/buzzer in the centre of the square board which would sound if all the pockets are empty OR if all the pockets are full. For any other condition, the buzzer would rest in peace!!!!

Now, it’s a game. You have to make the buzzer sound. The regulations for the game follow.
(i) At the start of the game, the buzzer is not sounding. This implies that all the pockets are not empty neither all of them are full.

(ii)Now, you get a chance. When you get a chance, the buzzer is disconnected. If have to first verbally select any 2 pockets.

(iii)When you have selected, the two non-selected pockets are covered. And, you can modify the 2 selected pockets by inserting ball(s) (extra balls are available) and/or removing ball(s) and/or keeping them unchanged.

(iv)Then you have to declare that you are done with the modification.

(v)The buzzer would get connected and it would sound if the desired condition has been reached.

(vi)If the buzzer is not making any sound, the board would start rotating at a very high speed so that it is impossible to keep track of the earlier selected pockets.

(vii)Then, the board would stop and you get a chance again.

(viii)And, the same rule follows as described from the start.

What is the minimum number of chance in which you can make the buzzer sound? You cannot trust probability. Can you make it sound against each and every probability? Please, elaborate the method with the answerJ.  :?

-- Shamik

wouldnt that be 1? If two were filled than you could select the others by chance and win on the first try

THe question was to find out a process which is against any favourable chance :lol:

Only thing I can tell you now is that the game started with either 1 ball or 2 balls or 3 balls in the pockets. All the pockets aren't empty to start with and neither they are all filled up. Then the buzzer would start 'making noise' from the beginning.

"If you say that money can't buy happiness, it means that you haven't looked in the relevant markets :lol: "

[quote="Shamik Banerjee"]"If you say that money can't buy happiness, it means that you haven't looked in the relevant markets :lol: "[/quote]

That's incorrect. Money can't buy happiness, it can only rent happiness.

Move 1:  Change a diagonal to 11.

Board: (a,b unknown)

1a
b1

**************************************************
Move 2:  Change a diagonal.

You either get ab, which cant be 11, and win (change it to 11), or you get 11, which you change to 00.

If you win, great.  If you havent won, you know $a \ne b$.

Board:

00
10

**************************************************
Move 3:

Pick two adjacent.  You have a 1 in there (which you change to 0 and win) or you get 00, which you change to 01.  You now have either:

01
10

OR

10
10.

*********************************************
Move 4:  Choose a diagonal.

If you have 00, or 11, then flip their state and you win.
If you have 10 then leave it alone, and you know you have

10
10.

********************************
Move 5:  Choose 2 adjacent.

You have 11 or 00 which you flip to win, or 10 which you flip to 01.

You now have

01
10

**********************
Move 6:  Choose two diagonal

Flip their state, you win!

Though the process was good the minimum number of steps required to make the buzzer sound is 5 (not 6!!!).

Step 1: Choose a diagonal and change it to 1-1. If the buzzer doesn't sound then you know that we have any one of the following two arrangements.

       1   0          or       1   0
       0   1                    1   1

We have the board as    1  x    where both x's are not equal to 1.
                                       x  1

Step 2: Choose any two adjacent pockets. Two adjacent pockets can be either 1-1 or 1-0. If it is 1-1 leave it as it is. If it is 1-0 change it to 1-1. Even if the buzzer doesn't sound then we have the following arrangement in the board.

        1   1
        1   0

Step 3: Choose a diagonal. A diagonal can be either 1-0 or 1-1. If it is 1-0 then just making it 1-1 ends the game!!! If it is 1-1 just make it 1-0. Now, the board looks as shown below.

       1   0
       1   0

Step 4: Choose 2 adjacent.

You have 11 or 00 which you flip to win, or 10 which you flip to 01.

You now have

       1   0
       0   1

Step 5: Choose two diagonal

Flip their state, you win!


If $ x,y,z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \left[(y\minus{}z)^2\plus{}3\sqrt{3}yz\right]\left[(z\minus{}x)^2\plus{}3\sqrt{3}zx\right]\left[(x\minus{}y)^2\plus{}3\sqrt{3}xy\right]\geq3\sqrt {3}xyz.$

[quote="Ji Chen"]If $ x,y,z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \left[(y \minus{} z)^2 \plus{} 3\sqrt {3}yz\right]\left[(z \minus{} x)^2 \plus{} 3\sqrt {3}zx\right]\left[(x \minus{} y)^2 \plus{} 3\sqrt {3}xy\right]\geq3\sqrt {3}xyz.$[/quote]
This inequality is just a corollary of Schur's Inequality for third degree
\[ \sum x(x\minus{}y)(x\minus{}z) \ge 0\]
:)

[quote="Ji Chen"]If $ x,y,z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \left[(y \minus{} z)^2 \plus{} 3\sqrt {3}yz\right]\left[(z \minus{} x)^2 \plus{} 3\sqrt {3}zx\right]\left[(x \minus{} y)^2 \plus{} 3\sqrt {3}xy\right]\geq3\sqrt {3}xyz.$[/quote]$ \left[(y \minus{} z)^2 \plus{} 3\sqrt {3}yz\right]\left[(z \minus{} x)^2 \plus{} 3\sqrt {3}zx\right]\left[(x \minus{} y)^2 \plus{} 3\sqrt {3}xy\right]$

$ \minus{} 3\sqrt {3}xyz(x \plus{} y \plus{} z)^3$

$ \equal{} \sum{(x \minus{} y)(x \minus{} z)\left[x^2(y \minus{} z)^2 \plus{} 3\left(9 \minus{} 5\sqrt {3}\right)xyz(y \plus{} z) \plus{} 3\sqrt {3}y^2z^2\right]}\geq 0$

which is true for

$ x^2(y \minus{} z)^2 \plus{} 3\left(9 \minus{} 5\sqrt {3}\right)xyz(y \plus{} z) \plus{} 3\sqrt {3}y^2z^2\geq0,$ and

$ \frac {[y^2(z \minus{} x)^2 \plus{} 3(9 \minus{} 5\sqrt {3})yzx(z \plus{} x) \plus{} 3\sqrt {3}z^2x^2] \minus{} [z^2(x \minus{} y)^2 \plus{} 3(9 \minus{} 5\sqrt {3})zxy(x \plus{} y) \plus{} 3\sqrt {3}x^2y^2]}{y \minus{} z}$

$ \equal{} \minus{} x\left[\left(3\sqrt {3} \minus{} 1\right)x(y \plus{} z) \plus{} \left(29 \minus{} 15\sqrt {3}\right)yz\right]\leq 0.$

[b]Remark[/b] For symmetric $ \sum S_{x}(x \minus{} y)(x \minus{} z) \geq 0$ :

$ S_{x}\geq 0\models S_{y}\geq 0,S_{z}\geq 0$;

$ (y \minus{} z)(S_{y} \minus{} S_{z})\leq 0\models$ the triples $ (x,y,z)$ and $ (S_x,S_y,S_z)$ are reversely sorted.

And then,  [url=http://www.mathlinks.ro/viewtopic.php?t=24555]the generalized Schur inequality[/url] works here.

This problem is record as Exercise 1.9 on page 31 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.china-pub.com/1786406]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)[/quote]

Dear jcc0107, your Schur partition is very nice.:coolspeak: In fact,

$ \prod{\left[(y \minus{} z)^2 \plus{} 3\sqrt {3}yz\right]} \minus{} 3\sqrt {3}xyz(x \plus{} y \plus{} z)^3$

$ \equal{} \left(\frac {33}{2}\sqrt {3} \minus{} 28\right)\sum{yz\left(2yz \minus{} zx \minus{} xy\right)^2} \plus{} \sum{x^3(y \minus{} z)^2\left[x \plus{} \left(55 \minus{} \frac {63}{2}\sqrt {3}\right)(y \plus{} z)\right]}$

$ \geq 0,$ which is clearly true without using Schur type inequalities.:juggle:

[quote=jcc0107][quote="can_hang2007"]This inequality is just a corollary of Schur's Inequality for third degree[/quote]$ \prod{\left[(y \minus{} z)^2 \plus{} 3\sqrt {3}yz\right]}\minus{} 3\sqrt {3}xyz(x \plus{} y \plus{} z)^3$

$ \equal{} (y \minus{} z) ^{2} (z \minus{} x) ^{2}(x\minus{} y) ^{2} \plus{} 3\sqrt {3}\sum{yz(yz\minus{}zx)(yz\minus{}xy)}$

$ \plus{} 3\left(9 \minus{} 5\sqrt {3} \right) xyz\left[\sum{x^2(y\plus{}z)}\minus{}6xyz\right]\geq0.$[/quote]

Also$:$
$$\displaystyle=  \left( 26-15\,\sqrt {3} \right)  \left( xy+zx+yz \right) \sum yz \left( x-
y \right)  \left( x-z \right) +\left( x+y+z \right) \sum yz \left( y+z \right)  \left( x-y \right)
 \left( x-z \right) +3 \left(6\sqrt {3} -10 \right) {y}^{2}{z}^{2} \left( x-y \right)
 \left( x-z \right) $$


Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side.  His preliminary drawings are shown.

[img]http://data.artofproblemsolving.com/images/2006_AMC_10A-12.GIF[/img]

Which of these arrangements gives the dog the greater area to roam, and by how many square feet?

$ \textbf{(A)}\text{ I, by } 8\pi \qquad \textbf{(B)}\text{ I, by } 6\pi \qquad \textbf{(C)}\text{ II, by } 4\pi \qquad \textbf{(D) }\text{II, by } 8\pi \qquad \textbf{(E)}\text{ II, by } 10\pi$

[hide]

I only gives us $32\pi$ (which is a semicircle)

II gives us a semicircle with radius 8 and a quatercircle with radius 4 . So the extra area are $\frac{1}{4}4^2\pi=4\pi$

$\boxed{C}$
[/hide]

[hide="Answer"]The first gives $32\pi$ for obvious reasons. The second gives the same plus one-fourth of a circle with radius $4$, so the answer is $\boxed{C}$.[/hide]


vector
linear algebra
matrix
analytic geometry
induction
invariant
combinatorics unsolved

An $n*n$ Latin square has the numbers from $1$ to $n^2$ arranged in its cells (one per cell) so that the sum of every row and column is the same. For every pair of cells in a Latin square the centers of the cells are joined by an arrow pointing to the cell with the larger number. Show that the sum of these vectors is zero.  :?:

Lets assume that the sum of one row is $S$.
First mention than for an entry $c$ of the matrix there are $n-c$ vectors going out of it and $c-1$ vectors coming to it.

Lemma : We will show that by swapping two adjacent rows the sum of vectors will not change.
Proof :
Suppose that we are swapping rows $i$ and $i+1$. Assume that we first cut the row $i$ from paper and move it over row $i+1$ (Then row $i$ and $i+1$ will be on each other for the moment) and then we cut row $i+1$ and move it up one unit so the two rows will be swapped.
In the first move how does the sum change? All vectors will only be changed by their $y$ coordinates. The vectors going out of row $i$ will increase in one unit in $y$ coordinate and the ones coming to row $i$ will decrease by one unit in $y$ coordinate. Number of outgoing vectors is $\sum_{j=1}^n(n-a_{ij})=n^2-S$ and number of incoming vectors is $\sum_{j=1}^n(a_{ij}-1)=S-n$ and so the sum of $y$ coordinates will increase by $n^2-S-S+n=n^2+n-2S$.
The same arguments hold for the second move but with opposite sign so the second move will decrease the $y$ coordinate by $n^2+n-2S$. So applying both moves will not change the $y$ coordinates and lemma is proved.

Now using the swaps defined by lemma one can easily show that we can permute the rows arbitarily. So just reverse their order (The indices of rows will become $n,n-1,\ldots,1$).
Now by the same arguments you can permute the columns in reverse order too. So just permute them.
Now in the resulting matrix it is not hard to see each vector has the opposite sign of the corresponding vector in the first matrix.
So the sum here is just opposite of the first sum. But by lemma they are the same. So for a vector $v$ we have $v=-v$ so $v=0$.

that is a nice solution here is my solution for the problem:
ok now the sum of $x$ coordinates is equal to:
$\sum{}_{a{}_i{}_j>a{}_p{}_q}(i-p)=\sum_{x,y=1}^n \underbrace{\left((a{}_x{}_y-1)\right)}_{<a{}_x{}_y}x+\underbrace{\left((n^2-a{}_x{}_y)\right)}_{>a{}_x{}_y}(-x)=\sum_{x=1}^n \sum_{y=1}^n 2xa{}_x{}_y-(n^2+1)\sum_{x=1}^n \sum_{y=1}^n x=\sum_{x=1}^n 2x\frac{n(n^2+1)}{2}-n\frac{(n^2+1)n(n+1)}{2}=\frac{n(n^2+1)n(n+1)}{2}-\frac{n(n^2+1)n(n+1)}{2}=0$
we can use this proof to proove that the sum of $y$ coordinates are equal to $0$ to
so the sum of the vectors are equal to $0$

sorry that I didnt explain it much
if you didnt understand it please let me know and I`ll explain the details for you
 ;)

we have an $2m*n$ chessboard which half of its cells are white and the rest are black for every pair of cells (one black and one white) we join them by an edge
prove that we can choose an orientation for every edge such that the sum of these vectors will be equal to $0$(we also know that the upper left cell is black and the opposite cell is white)

I think the last problem you proposed is wrong. Consider simply the case $m=n=1$ we have one edge only.
Also for the case that $mn$ is even I have a proof that doesn't use your last assertion that the upper-left corner is black and the opposite is white.
The idea of my solution is : Show that for every even $n$ we can direct the edges of $K_{n,n}$ such that number of edges going to a vertex is equal to the numer of outgoing edges from that vertex. It is rather simple (for example take two $\frac n2$ parts from each part of $K_{n,n}$ say $X=X_1\cup X_2$ and $Y=Y_1\cup Y_2$ where $X$ and $Y$ are bipartite parts. Then direct the edges like this $X_1\rightarrow Y_1 , Y_1\rightarrow X_2 , X_2\rightarrow Y_2 , Y_2 \rightarrow X_1$.)
Then the graph is composed of some eulerian tours.
Apply this idea for the graph $K_{mn,mn}$ in the main problem. (Our parts are the white and black ones)
Now direct the edges like I told before. Then sum of vectors over each eulerian tour is $0$ and we have the desired result.

I dont know maybe your example is correct or maybe $m$ should be greater than $2$
but read my solution and you`ll find out:
[b]Lemma:we can decompose the edges of a graph into cycles if and only if $d(v_i) \equiv 0 (\bmod 2)$
proof:the proof is very simple the condition is necessary because every circle uses even number of edges of vertex $v$,...
for sufficiency just use induction on the number of edges[/b]
now as you have said at your solution we have a $K{}_m{}_n_,{}_m{}_n$ graph now we have two conditions head of us:
$i)mn \equiv 0 (\bmod 2)$ then the degree of every vertex is equal to $mn \equiv 0 (\bmod 2)$
there for we can use our lemma to decompose the graph into circles
now direct the edges of these circles clockwise then you`ll get the desired result
$ii)mn \equiv 1 (\bmod 2)$ in this condition just cut the two corners that we have talked about
now we get a $K{}_m{}_n_,{}_m{}_n$and also the degree of every vertex is even now like the last part you can prove that the sum of vectors of these cells is equal to $0$
note that in condition $ii)$ you cant just cut two arbitary cells you should cut two cells which the sum of their vectors is equal to $0$ we used that assumption here

proof2:
[b]Lemma$2$:for every $4$ points on the plane like $A,B,C,D$ such that $ACBD$ is convex we have:
$\vec{AD}-\vec{BD}+\vec{BC}-\vec{AC}=0$[/b]
now according to this lemma we can swap one white cell with one black cell without changing the whole sum
now we divide the chessboard with a line into two equal parts
now for every white cell at the upper part we have a uniqe black cell at the lower side
now swap the white cells at the upper side with the black cells in the lower side
now the upper side is completely black and the lower side is completely white
there for for every vector with head at the upper side we have a vector with head at the lower side
thus the sum of these two vectors are equal to $0$ now so on you can get the desired result

[quote="BaBaK Ghalebi"]
$ii)mn \equiv 1 (\bmod 2)$ in this condition just cut the two corners that we have talked about
now we get a $K{}_m{}_n_,{}_m{}_n$and also the degree of every vertex is even now like the last part you can prove that the sum of vectors of these cells is equal to $0$
note that in condition $ii)$ you cant just cut two arbitary cells you should cut two cells which the sum of their vectors is equal to $0$ we used that assumption here
[/quote]
I can't understand what does "you cant just cut two arbitary cells you should cut two cells which the sum of their vectors is equal to $0$ we used that assumption here" means here (sum of which vectors?)
Also you just showed how to direct the edges not passing these two cells. You didn't say the direction of the edges passing these two cells.
Now I also have a proof that simply says when $mn$ is odd we always have a counterexample (In any case there is)
Simply note that $x=-x (mod 2)$ so how ever you direct the edges sum of their $x$ coordiantes plus sum of their $y$ coordinates is invariant $mod 2$. So take this counter example :
Color the table like a chessboard (Your assertion is true here, I mean the top-left one is black and buttom-right one is white)
Then you can easily see that $x+y$ for every vector in the sum is odd. We have $(mn)^2$ edges and so the sum over all edges is also odd.

[quote="BaBaK Ghalebi"]here is another beautiful one:
we have $2000$ points on the plane such that no $8$ of them are collinear
prove that there exists $250$ segments with different length :?:[/quote]
Please post a new problem in a new topic. And I think that this problem was from mathlinks contest and I just say that the idea is a famous one, that is double counting the number of one color $K_{1,2}$s in the colored graph of the points (each color represents one distance)

hmm...
I realy dont know what to say
now its late (its about 2:05 AM) and I cant think about it sorry tomorrow I`ll send your answer and the reason that your counterexamples are wrong and why the problem is true  ;)
about the problem I think its from romania or somehow
and sorry about posting it in here
my idea is the same for this problem


1. Seven cards in a pile are numbered 1 through 7. One card is drawn. The units digit of the sum of the numbers on the remaining cards is 7. What is the number on the drawn card?
A.   1
B.   3
C.   5
D.   6
E.   7
I don't think there is any way that the sum of the rest of the cards is 7?

2. How many four-digit numbers have only even digits?
I just want to ask when we say:first digit, is this the digit from the left or the right?

Thanks a lot.

1. A

2. (from the left)

$ 4\times5\times5\times5\equal{} 500$

lol, can't count

1. The sum of all the numbers: $ 1\plus{}2\plus{}\ldots\plus{}6\plus{}7 \equal{} 28$. Only if you remove card #1 will the sum of the remaining cards end in 7.

2. The question asks for the number of 4 digit numbers where all digits are even. The first digit has 4 possibilities (2,4,6,8) while the other three digits have 5 possibilities (0,2,4,6,8), so the answer is 4*5*5*5 = $ \boxed{500}$


[hide="Hint."]What is the cardinality of the span of $k$ independent vectors?[/hide]

[quote="Omid Hatami"]Find the number of matrices in $GL_n(\mathbb F_p)$[/quote]

I had this in my final exam today  :P and $SL_n$ as well.

nice corollary: $n!$ divides a nice produttory of $n$ terms...
not obvious at all :)

I got that

$\textrm{GL}_n \left( \mathbb F_p \right) \, : \, \prod_{k=0}^{n-1} \left( p^n - p^k \right)$

$\textrm{SL}_n \left( \mathbb F_p \right) \, : \, \frac1{p-1} \cdot \prod_{k=0}^{n-1} \left( p^n - p^k \right)$

Is that correct? Also, what's with $n!$ dividing something?


Someone I know made USAMO in 9th grade with this:

130something AMC 10
7 on AIME

It's below cutoff on AIME level (8 pts) and the index is only 203.5 or so... what?

According to Dunbar-Steve, cutoff was 217... right?

I think you are missing something. I don't expect AMC to make this mistake or It could be that he got a 9  in AIMe and he qualified. Please check with your friend once more.

[quote="mathdool"]maybe he lived in one of the states with no other respresentatives[/quote]
Yes. Check the Wyoming qualifier for instance.

maybe...he miss wrote one of his answers and actually bubbled the right answer.

Maybe he guessed something right and didn't realize it  ;)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ A$ and $ B$ be two subgroups of $ G$ such that $ G : A$ and $ G : B$ are finite. Prove that: \[ G : (A\cap B)\] is also finite.

$ G: (A \cap B) \equal{} (G: A)(A: A \cap B) \le (G: A)(G: B) < \infty$.

Lemma: $ (A : A \cap B) \le (G : B)$
Proof: If $ ag \equal{} a'$ for some $ g \in B$, then $ g \equal{} a'a^{\minus{}1} \in A$, so $ g \in A \cap B$. Therefore the number of orbits by the $ A \cap B$ action of $ A$ is at most the number of orbits by the $ B$ action of $ G$.


group theory
abstract algebra
algebra solved
algebra

Prove that the expression x*2^0.5+y*3^0.5+z*5^0.5, where x,y,z are integers which are not all zero, can be made arbitrarily close to zero.

The set E = {x \sqrt 2 + y \sqrt 3 + z \sqrt 5/ x,y,z integers} is an additive subgroup of |R. It is well_known that additive subgroups of are dense in |R or of the form mZ for some real m.
Since, it is not difficult to prove that  \sqrt 2 ,  \sqrt 3 ,  \sqrt 5 are linearly independant over Z, then E is not of the form mZ. Thus E is dense in |R and we are done...

Pierre.

An elementary solution :

A =  \sqrt 2 + \sqrt 3 + \sqrt 5
Let B > 0.

for x,y,z in [ 0 .. B-1], 0<= f(x,y,z) = x*2^0.5+y*3^0.5+z*5^0.5 < A*B

There are only B^3-1 intervals [(k-1)A*B/(B^3-1) .. k*A*B/(B^3-1)], 0 <= k <= B^3-1

So there are two numbers f(a,b,c), f(x,y,z) in the same interval. f(a-x,b-y,c-z) can be rendered as small as we want by increasing B.


calculus
integration
derivative
real analysis
real analysis unsolved

Hi guys.. i have a problem and i do not know where to start.

I'd really appreciate it if somebody could show me how to solve this problem. I know the basics of integration, but when powers are involved i get lost.

I have posted the problem on this webpage:

[url]http://putfile.com/pic.php?pic=main/8/21515512088.jpg&s=f10[/url]

Just substitute $u = x^{4}+2$, so that $dx= \frac1{4x^{3}}du$ :$\int_{0}^{1}2x^{3}(x^{4}+2)^{5}dx = \int_{2}^{3}2x^{3}u^{5}\frac1{4x^{3}}du= \frac12 \int_{2}^{3}u^{5}du$
I suspect You should be able to take it from here..

[quote="Ottem"]Just substitute $u = x^{4}+1$, so that $dx= \frac1{4x^{3}}du$ :$\int_{0}^{1}2x^{3}(x^{4}+1)^{5}dx = \int_{1}^{2}2x^{3}u^{5}\frac1{4x^{3}}du= \frac12 \int_{1}^{2}u^{5}du$
I suspect You should be able to take it from here..[/quote]
That's $x^{4}+2$ Ottem.  :)

It certianly is!  it's ok now tho.. :lol:

Thanks, much appreciated.

Could you please briefly explain the final part of the problem.

I understand the U=x^4 and du=4^3dx

I then get a little lost, what happens to the ^5 surrounding the brackets? Do you dis-regard that in the first step?

no I just substituded in for $u$. The thing is that we want to get rid of the $x^{4}+2$ term, and the substitution $u=x^{4}+2$ turns $\int 2x^{3}(x^{4}+2)^{5}dx$ into $\int u^{5}$ which is a nicer integral.

i'm still a little lost, could you briefly explain what's going on when you show the answer?

Its standard stubstitution method. The derivative of $x^{4}+2$, $4x^{3}$ alows us to cancel the $2x^{3}$ in the integral, leaving only $\int u^{5}du$. You might wanna read more about it here:

http://www.sosmath.com/calculus/integration/substitution/substitution.html

i appreciate any help, it wasn't a good idea for our college to let a media teacher try to teach us math!

So from that, how do i go on to get an answer to 3 decimal places?

As amcavoy showed, the integral $\int_{2}^{3}u^{5}du$ evaluates to $\frac{665}{12}$, approximately $55.416$

Thanks guys, just one more question the integral

how did the
 1
[size=150][b]S[/b][/size]
 0

become

 3
[size=150][b]S[/b][/size]
 2

Remeber to change the integration limits; The variable $x$ is integrated from $0$ to $1$, hence when you substitute $u=x^{4}+2$, u should integrate from $0^{4}+2=2$ to $1^{4}+2=3$.


If $ y^2\equal{} 36$, what is the greatest value of $ y^3$?

just find the sqrt of 36 and the answer is 6 so the answer wants you to find the answer y^3 but y=6 so plug in 6 to Y and get 6^3=216

[hide="THIS IS YOUR ANSWER"]216 IS THE ANSWER YOU SEEK[/hide][/quote]

[size=200][color=#FF80BF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH IT IS GREATLY APPRECIATED THANK YOU!!!![/u][/i][/b][/color][/size]

[size=50][color=#0080FF][b][i][u]THIS IS SO SMALL JUST LIKE THE SONE ITSY BITSY SPIDER : )[/u][/i][/b][/color][/size]

THE FOLLOWING STATEMENT WILL BE IN WHITE

[color=#FFFFFF][b][i]Pure mathematics is, in its way, the poetry of logical ideas.  ~Albert Einstein

[/i][/b][/color]


trigonometry
geometry
circumcircle
perpendicular bisector
geometry solved

Let $ABC$ be a triangle such that $\angle B>\angle C$. And let $H$, $L$, $D$ be the foot of altitude, bisector, median respectively belongs to $A$.  Suppose that $\angle HAL=\angle DAL$  prove that $\angle BAC=90^\circ$.

[i]Edited by Myth[/i]

I agree, it is very easy.
We see that $AH$ is a simedian. So it is well know $BH:HC=c^2:b^2$, in the other hand $BH:HC=\cot \beta:\cot \gamma$. Using sinus theorem we immediately obtain $\sin\beta\cos\beta=\sin\gamma\cos\gamma$, or $\sin 2\beta=\sin 2\gamma$. Since $\beta\neq \gamma$ we conclude $2\beta+2\gamma=\pi$.

That's all.

I think, there is a synthetic solution.

You used the fact that $AH$ is the isogonal conjugate of the median, but we can sue the fact that the median is the isogonal conjugate of the altitude instead :). The isogonal conjugate of the altitude is the line connecting $A$ to the circumcenter, so what this problem tells us is that the circumcenter is on the median through $A$, and $AB\ne BC$. I think the conclusion is clear now.

yes i know this solution that $AH$ is simedian... but i ask more elegant solution...synthetic solution...

Umm... Did you read the one I wrote (actually, I didn't write all of it, but the rest is clear)? It uses the fact that $O,H$ are isogonal conjugates (a quick angle chase), thereby showing that $O$ lies on the median from $A$. This, together with the fact that $AB\ne AC$, so the median is not the same as the perpendicular bisector of $BC$, gives the result. This is pretty synthetic, isn't it? :?

Ok, I will not write solutions for sinankaral53's problems next time.  :mad:

ok ok i am satisfied with your answer ...


Determine all positive integers that are expressible in the form \[a^{2}+b^{2}+c^{2}+c,\] where $a$, $b$, $c$ are integers.

Every non-negative integer can be expressed that way:
We write $ 4n\plus{}1$ as sum of three squares (which is possible by the Three-Squares-Theorem as $ 4n\plus{}1\not\equiv 0,4,7\mod 8$): $ 4n\plus{}1 \equal{} x^{2}\plus{}y^{2}\plus{}z^{2}$.
Looking $ \mod 4$ (the squares $ \mod 4$ are only $ 0$ and $ 1$), we see that exactly two of $ x,y,z$ are even, w.l.o.g. $ x\equal{}2a$, $ y\equal{}2b$ and $ z\equal{}2c\plus{}1$ for some integers $ a,b,c$.
We now have $ 4n\plus{}1 \equal{} (2a)^{2}\plus{}(2b)^{2}\plus{}(2c\plus{}1)^{2}\equal{} 4a^{2}\plus{}4b^{2}\plus{}4c^{2}\plus{}4c\plus{}1$, equivalent to $ n\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}c$.


Casino offers a simple game:
You toss a coin, heads you win 1 dollar, tails you lose 1 dollar.

John, an immortal, enters the casino with 1000 dollars in his pocket and plans to toss a coin until he makes 100 dollars, then grab the cash and leave.

What is the probability that his plan will be successful?

thanks.  although the wikipedia article didn't explain.

matt276eagles gave me the solution.
[hide="Solution"]
Let $P_{n}$ be the probability that John reaches $\$$1100 before he goes bankrupt.  We have the recursive equation $P_{n}=\frac{1}{2}P_{n-1}+\frac{1}{2}P_{n+1}$.  Rearranging, we get $P_{n+1}-2P_{n}+P_{n-1}=0$.  This recursion has characteristic equation $x^{2}-2x+1$, which has a double root at $x=1$.  Therefore, we have $P_{n}=(a+bn)1^{n}=a+bn$, for constants $a$ and $b$.  Since $P_{0}=0$, we have $a=0$, and since $P_{1100}=1$, we have $b=\frac{1}{1100}$.  Therefore, $P_{n}=\frac{n}{1100}$, and $P_{1000}=\boxed{\frac{10}{11}}$.
[/hide]

cool stuff.


Find the formula for the area of a regular n-gon with side length x. :D

There are probably lots of solutions. Please post your solutions.

[quote="surge"][hide]$\frac{1}{2} \cdot \text{inradius} \cdot {xz}$

Where z represents the number of sides of the n-gon. (isn't this a common formula in high school geometry?).

For regular polygons, the inradius is also known as the apothem, I believe.[/hide][/quote] We don't know the inradius though.

[hide="If hiding is necessary"]In an n-gon, each angle is $\frac{180(n-2)}{n}$

You can also break this polygon into $n$ congruent isosceles triangle each with base angle $\frac{90(n-2)}{n}$.

To find the area of this triangle: find the height which is $\frac{x}{2}\tan \frac{90(n-2)}{n}$

Then multiply by $\frac{x}{2}$ to get $\frac{x^2}{4}\tan \frac{90(n-2)}{n}$

Then finally multiply by $n$ to get $\boxed{\frac{nx^2}{4}\tan \frac{90(n-2)}{n}}$

Please tell me if I did anything wrong. :) [/hide]


I recently got a test back and I'm unsure about the correct solution to a one of the problems. I'm not that great at physics, but I'm still curious as to what the science/math community says.

5kg, 10kg, and 15kg boxes sit next to each other on a frictionless surface in that order. Anything more than 200N will crush the middle box (10kg). What is the minimum force that must be applied to the 5kg box to crush the middle box?

[quote="Yongyi781"][quote="le00327146"]Anything more than 200.N will crush the middle box (100kg).[/quote]

This is confusing.

(1) What do you mean by the dot between 200 and N? I would've dismissed it as a simple typo, except...
(2) Is the middle box 10kg or 100kg??[/quote]

Sorry about that. The middle box is 10kg. I've edited the original post.

[quote="rituraj007"]$ 240N$??[/quote]

That's what the teacher's solution had, but I'm a little skeptical.

just make the free body diagrams of the three separately and u will understand
if not do tell me :P

Yeah, it's 240N if you consider only the force of the 5kg box on the 10kg box. But what about the force of the 15kg box on the 10kg box (from Newton's third law)?

1st of all we hav to calculate the minimum force req

and also,action and reaction forces act on different bodies
write ur calculation,i ll tell u where u took the interaction force bw 10 and 15 into account

Well to get 240N,

A sixth of the total force is used to move the 5kg box, so the force between the 5kg and 10kg boxes is 5/6 of the total force. Since the minimum is 200N, the total force is 240N. I don't get where the 15kg-10kg force is taken into account though.

You are absolutely correct, le00327146.  Your teacher apparently didn't take into account the 10-15 force, and in doing so, you would get an answer of 600 N.  I am interpreting the question thus: the middle box will be crushed if the [i]net[/i] force on it exceeds 200 N.

BTW, le00327146, I'm responding in this thread because I actually know you in person.  You might not know who I am, though.

EDIT: In case you want even more, I shall proceed to justify my answer to death.

First justification: Assume the net force on the 10 kg box is 200 N, to the right.  Then the acceleration is 20 m/s^2.  Assuming the boxes stay in contact while the force is being applied, the acceleration of all three boxes, as a body, must also be 20 m/s^2.  This means F_min = (5 kg + 10 kg + 15 kg)(20 m/s^2) = 600 N.

Second justification (the dumb way, i.e., drawing a free-body diagram): We have three forces: $ F_{min}$, $ F_1$ (the 5-10 force), and $ F_2$ (the 10-15 force).  Call $ a$ the acceleration of all three boxes.  Thus we have the conditions:
\[ \frac {F_{min} \minus{} F_1}{5} \equal{} a; \frac {F_1 \minus{} F_2}{10} \equal{} a; \frac {F_2}{15} \equal{} a.
\]
Now set $ F_1 \minus{} F_2 \equal{} 200$ and solve.

EDIT 2: Please tell me what happens when you prove your teacher utterly incorrect.

O_o Interesting...I have a guess. You're a fan of Yale University? =P

I can't remember the exact wording of the problem, but I believe it was total force, not net force. Either way, you still have to consider the 15kg-to-10kg force right? I get 150N this way.'

Here's another one that a lot of people were unsure about. I'm less skeptical on this one, but still a little so.

Again, a frictionless system. There's two blocks, 50kg and 20kg. What is the minimum force that must be applied to the 50kg block to keep the 20kg block from falling? (Diagram in the attachment)

So as I told you during the ARML meeting today, le00327146, I agree with your answer of 150 N under that interpretation.

And as we also discussed, the answer to the second problem is $ 70g\tan 40$, so your teacher was actually right on that one.

well thinking at the micromolecular level,even if u get an interaction force of 200N from 1 side on the block,it shud crumble before it imparts a force on 15 kg block :oops:


I'm has a problem .
For $ p\in N$ and $ p>1$
If exist an infinite sequence of natural number has property
$ mn|m^p \plus{} n^p \plus{} k$
I has solve for k=1 but the general is quite hard.

What is the relationship between $ m,n$ and the sequence in question? Should $ m,n$ be consecutive elements of the sequence or what?
Also, what is $ k$, is it a constant term (similarly to $ p$)?