MIT college Princeton Stanford Harvard Columbia probability

Hi folks, I intend to pursue an M.S. in computer science from a U.S. university, but I would like to know - Is a PhD in computer science much better than an M.S. in computer science? (I mean a PhD or an M.S. from an a good/above average school)

I would like to elaborate my query, I meant- Is slogging 5 years (this is variable) for a PhD (in computer science) really worth it? Will I be able to get a job in a reputed/good software firm (for instance say Google, Microsoft, Yahoo! etc) with a PhD, or will doing a PhD confine me to only academia? (With due respect to academia, IMO,I really loathe being landed up in an academia job).

[b] Main Query [/b]: Which qualification places me in receiving a better pay- MS or PhD (in computer science)?

Will I be able to work in at least a "start-up" in California with a PhD in computer science? What are the other options that are available to a person who has a PhD other than academia?

I ask you folks to respond to these queries of mine, I am thankful to you all!

Provide replies/suggestions/advice by [b]not taking the financial crisis into account [/b]; I know that markets are plummeting and the financial crisis is snow-balling, but please provide your valuable suggestions.

P.S.: [b] A PhD in computer science (as for me) means obtaining it from a good/above-average graduate school, don't include schools such as MIT, Princeton, Stanford, Harvard, Columbia etc, or any school of such "caliber". [/b]

Mostly everyone I know getting jobs at places like Google, Apple, Microsoft and the like are doing so right out of a bachelor's degree.

I don't know what kind of people you know, but I can assure you that [i]most[/i] people with bachelors degrees in computer science do not get jobs like that. [i]Some[/i] do, but not most.

On the flip side, I would bet that [i]most[/i] people who work in upper level technical positions for high-tech companies have PhD's in their field.

[b]@gauss202 [/b]: Those who work at high tech companies in upper technical positions got (most of them) their PhDs from elite schools like MIT,Caltech etc. I am asking are there any lucrative avenues available for people who got their computer science PhDs from less reputed (good/above average schools, not the great/elite schools) schools? (by lucrative avenues I mean- [b] not in academia [/b])

I wouldn't make that assumption that all of those upper level technical positions are to people with PhD's from MIT, Caltech, etc. I bet a lot of them are from what you would call good and above average universities.

I would bet that a PhD in computer science from any good/above average school is pretty rare and in high demand. But probably more important than just the degree and the name of the school is how you perform, the type of specialty that you focus on, and the quality of your relationship with your thesis advisor.

A PhD program is not something to go into lightly though. From what I hear it is a long and grueling process, and if your heart is not in it for the love of the subject you are studying it can be even longer and more grueling.

But all of this stuff about salaries (which I think it is unwise to dwell on too much anyway) should be something you could find statistics on. Maybe by googling or talking to someone in the field?

If you think of doing of PhD as "slogging through" something, then you should definitely not pursue one, regardless of whether or not it is more lucrative.

In any case, if your goal is to make money, a PhD is not efficient.  Does a typical CS PhD have better job prospects and a higher starting salary than an MS?  Yes, but usually not enough to make up for the opportunity costs.  Very few software engineers have PhDs, because very few need them.  (I guess I disagree with gauss202 on this.)

@gauss202:  You do realize that what you wrote is completely consistent with what max_tm wrote, right?

[b] @yenlee [/b]: I never thought of pursuing or at least nurturing a drive to pursue a PhD or the very notion of completing a PhD in computer science as "slogging" (Maybe I should have used a more appropriate "word" than slogging).

So, do you mean to say that getting a PhD in computer science might not actually land up one in a lucrative job? If it's so, then what should I do to get a lucrative software job ( I mean M.S or something else)?

Actually, When someone completes a PhD in computer science, he/she have to show something "new" which they had discovered or do they have to show what they have concocted from existing ideas/theories?

[b] P.S. : [/b] I might sound like as an extremely money-minded person ( I can't concoct a better term for this, I am not a native speaker of English), but the fact is I want to earn some bucks before I pursue other aspirations of mine.

[quote="Gen8"]So, do you mean to say that getting a PhD in computer science might not actually land up one in a lucrative job?[/quote]  If there were a magic recipe that anyone could follow that would lead to the guarantee of a high-paying job, don't you think that everyone would follow this recipe?  You seem to be struck by a deep delusion about the relationship between formal credentials, skills, and job prospects, and you should knock it out of your head as quickly as possible.

A Ph. D. is a reasearch degree, and requires one to write a thesis consisting of original research.  An MS is more like a professional degree (an MD or a JD).

[b] @ JBL [/b] : But dude, There exists a relationship between good credentials, a good degree, and a high paying job. Certainly, corporate giants like Google, Microsoft, IBM won't recruit you unless you have great credentials and a good degree.

In terms of investment banks , it's even worse, D.E.Shaw explicitly mentions (at their site) that you need to be 'extra-ordinary' in order to be recruited (or have a chance of being recruited) at the firm.

[quote] You seem to be struck by a deep delusion about the relationship between formal credentials, skills, and job prospects, and you should knock it out of your head as quickly as possible. [/quote]

Well, I couldn't accurately interpret this statement of yours, do you ask me to discard the notion- that good credentials might always not lead to a high paying job and vice-versa.

[b] May I know what you are pursuing at present? I think you're from Harvard, what's your specialization? [/b]

I think what JBL is saying is that no body on this forum can guarantee you anything in the form:
if you do ____________ you can find a job like ___________
(you could fill in the second blank with a salary, benefit, desired position, anything of that sort)
There is no way to guarantee yourself a job of any sort, and gaining a degree just to meet a requirement probably isn't the right idea.
The investment banks say you need to be exceptional, but they never explicitely say something along the lines of "you need to have graduated from a top __ university with a __ degree in ________"  In many cases those things may be signs of an exceptional person, but not necessarily

Also: You probably shouldn't be pursuing anything in life with the sheer goal of making money, even if the plan is to pursue other interests later, once "stuck" at a job, you may find it hard to move on to something else and take that risk, you'll find yourself stuck in a rut forever...

P.S. I'm saying these things with very little life experience, I haven't gone to college or attempted to find a job

[quote="Quickster94"]I think what JBL is saying is that no body on this forum can guarantee you anything in the form:
if you do ____________ you can find a job like ___________
(you could fill in the second blank with a salary, benefit, desired position, anything of that sort)[/quote]  Yes, precisely: of course there is a relationship between learned skills, in-born talent, formal credentials and success on the job market, but this relationship is not deterministic and is heavily influenced by randomness and luck.  [i]Of course[/i] getting a Ph.D. in computer science does [b]not[/b] guarantee anything -- there is never a guarantee of any sort to be found in these areas.  There are probability distributions and error bars that you can shift by learning new skills or getting new credentials, but no one is going to tell you the magic recipe to ensure a lucrative career in CS because no such magic recipe exists.  Your insistence on acting as if such a recipe does exist, and that you can discover it by creating enough similar threads in this forum, is very exasperating (to me, at least).

With respect to D. E. Shaw, the word "extra-ordinary" is notable for saying nothing whatsoever about specific skills, talents, or formal credentials.

I received my bachelor's degree from Harvard -- I'm currently a Ph. D. student in mathematics at MIT studying combinatorics.  With a little bit of snooping, you can probably find my website and read a bit more about me ;)

[b] @ JBL [/b] : I heard that Harvard's math department is much more 'stronger' than MIT's math department, I can't understand how you managed to do under-grad at Harvard, and end up at MIT pursuing a PhD in Math. Anyways, I'm constantly seeking fruitful suggestions/advice to my queries/apprehensions (whatever it is!) on what to do after my under-graduation, so if this infuriates you, then maybe you should avoiding reading my threads. ( I understand your point which is valid).

I also like combinatorics.

I've actually heard it the other way around.

Well, I heard it here in AOPS and other science forums, it might be the case that the people from whom I heard might be more biased towards Harvard.

[quote="Gen8"]I heard that Harvard's math department is much more 'stronger' than MIT's math department
...
I also like combinatorics.[/quote]  I'm glad to hear that you like combinatorics :).  MIT and Harvard both fall at the top of nearly any list of the strongest math departments in the country -- for example, [url=http://grad-schools.usnews.rankingsandreviews.com/best-graduate-schools/top-mathematics-programs/rankings]US News[/url] ranks Princeton, Harvard, MIT, Berkeley, Stanford and Chicago as the top six programs (and as all essentially interchangable in overall strength), and [url=http://math.knox.edu/weblinks/top50grad.html]the National Research Council[/url] agrees on these top 6, though not in precisely the same order.  (At least, they did in 1995 -- as far as I can tell, the next update on that data is still in the process of being generated -- see [url]http://sites.nationalacademies.org/pga/Resdoc/PGA_044479[/url].)

@ Ravi B, jmerry:  :rotfl:  :lol:  :rotfl:

Stop digressing folks! There are enough experienced people in here who can provide some help to me (I'm really tensed, in a state of imbroglio), I have observed that some of the posts are actually a paraphrase of earlier posts, so don't indulge in paraphrasing someone's suggestions. Give suggestions based upon your personal experience, it would be of immense help.

Thanks!

Well, there is a bit of real advice behind the joke I posted a few minutes ago:

Calm down. Stop worrying. You don't have to perfectly optimize every decision, and you'll be happier if you don't.

[quote="Ravi B"][quote="Gen8"][b] @ JBL [/b] : I heard that Harvard's math department is much more 'stronger' than MIT's math department, I can't understand how you managed to do under-grad at Harvard, and end up at MIT pursuing a PhD in Math.[/quote]
Joel, how far you have fallen!  Are you now the black sheep of your family?

-- Ravi B (PhD, MIT)[/quote]

haha this thread is the best thread

[quote="jmerry"]Bah. Elitists.

- jmerry (UC Berkeley, Washington)[/quote]

blahblahblah: Indeed  :D

Well, I asked people who post in this thread not to digress, even if they digressed, it would be much better and helpful after giving some invaluable advice to my query.

Posting a reply that isn't fruitful to me or isn't valuable in any sense would amount to  [b] spamming [/b]. Don't indulge in it!

[quote="Gen8"]Posting a reply that isn't fruitful to me or isn't valuable in any sense would amount to  [b] spamming [/b]. Don't indulge in it![/quote]

Don't backseat moderate.  You don't get to decide what gets posted in a thread just because you started it.

Well, You and any sane human being knows the intent of this thread, and [b] I'm not backseat moderating [/b], I asked/requested people, even though they post their anecdotes or post their weird opinions on something else, it doesn't matter as long as they also give some suggestion.

I opine that if a person could give suggestions, and then post their anecdotes/ weird opinions, then his/her post won't be labeled as [b] SPAM [/b].

What do you call something that is of no use to any one: Trash, and Trash on the internet or in forums is : [b] SPAM [/b] and one who does it is a - [b] SPAMMER [/b].

Also, though it is unnecessary, It would be better if they were a much better term other than the same old cliched phrase (in AOPS forums) -" Don't backseat moderate" being used so recursively.

[quote="blahblahblah"]Don't backseat moderate. [/quote]  Awesome.  (Also, he who starts digressions shouldn't throw stones, or something.)

Anyhow, I think the original question has received a good deal of attention.  In my estimation, it is unlikely that a Ph. D. will open up a substantially larger number of job prospects than an M.S. in computer science, with the notable exception of jobs in academia.  As several other posters have noted, an M.S. is not a prerequisite for getting jobs in the field of computer science.

Edit: actually, the comments you (Gen8) are complaining about have been widely regarded as the most amusing in this entire thread.  Just because they are of no interest to [i]you[/i] doesn't mean they are of no interest to [i]anyone[/i].  (There may be a reason that you have the lowest rating of anyone participating in this conversation  :wink: )

[b] @jmerry [/b] : Well, I understand what you say by not optimizing every decision, I can't really comprehend your post - [b] bah, Elitists! [/b], what do you ask me to interpret from this particular post of yours? The post is highly ambiguous!

It would be much better if you found some time and elaborated your "post".

[b] @ JBL [/b] : Firstly, I post less and read more in this forum, and I am not an expert/technically brilliant in math to post solutions to problems that are beyond my level.[b] I am a jack of all trades [/b] (Meaning- I am interested in many diverse fields than being only confined to a specialized area). (These are the two logical reasons for the number of posts (mine) being less).

Secondly, a person may have low rating because he/she didn't post much in the forum, it doesn't necessarily imply he/she is posting worthless 'posts'. (This implied argument of yours clearly depicts some unsound reasoning).

Although this thread has received considerable attention courtesy of some worthless posts, some of them being a paraphrase of others; I still am devoting some considerable time to see some coherent and concise posts to my queries.

[b] P.S. [/b]: Between I didn't even post a 100 posts, so, it sounds coherent that I might have a low rating as and when compared to you people who posted more than 2000 posts.

[quote="Gen8"]
Secondly, a person may have low rating because he/she didn't post much in the forum, it doesn't necessarily imply he/she is posting worthless 'posts'. (This implied argument of yours clearly depicts some unsound reasoning).
...
[b] P.S. [/b]: Between I didn't even post a 100 posts, so, it sounds coherent that I might have a low rating as and when compared to you people who posted more than 2000 posts.[/quote]

I assume the last statement is along the lines of: "Because I don't even have 100 posts, it seems plausible that I would have a lower rating compared to someone else who has 2000 posts"

Your rating (I assume) is a function of the quality of you posts, not the quantity.  Whether you have 100 or 1000 low quality posts, you'll still end with the same bad rating.  Your rating tells me that others have deemed many of your posts "spam" so it's ironic that 1. you're reprimanding others for spamming, and 2. You call people out for "unsound reasoning" when you seem to display the most.

[quote="Ihatepie"]How do you find out someone else's rating?[/quote]

I think only moderators can see them  :ninja:

[b] @ Quickster94 [/b] : I had already cited the reasons why the number of posts of mine might be less, and also why I don't post in technical threads.(I surmise most of the ratings given to posts are based on your solutions to problems, rather than general posts)

I had seen plenty of people in here who had a similar rating (like mine), and they too, had almost the same number of posts. So, it is recommended to peruse through AOPS before making foregone conclusions about members and the quality of their posts!  :ninja:

Read what I had written properly and stop making ludicrous assumptions!

Finally, You accusing me of being guilty of some unsound reasoning is comparable to the famous idiom- The pot calling the kettle black.

[b] [color=green] So, I would like to end (since I didn't start this) this puerile debate on the quality of a person's posts and instead ask people to peruse through my queries and come up with satisfactory replies/suggestions. [/color][/b]

Thanks!

Where are all the people? I have been awaiting some response/suggestion to my queries. (I need more suggestions).

I think everyone's said all they have to say. The rest will be up to you to make a decision.

Okay, I would like to thank members who provided their suggestions/advice.  :P  I really appreciate it.

Sugarcoating with a clear history of negativity does not make others feel appreciated by a thanks.
Now you should really consider the idea of paraphrasing. If jmerry, JBL, and all of the other brilliant, older, and successful persons have seemingly paraphrased a few specific points, take the hint. It means they all share the same idea, and are NOT reading a previous post to gather thoughts as advice to give. They told you what the difference is, how the two degrees are weighted in the world.
Your posts show how quick to anger you are and people certainly don't want to deal with that, after trying to help. Remember they take their time to answer your questions, be grateful.
Maybe that will help with why they dont' respond now, though it could also be from marking exams :)

[b] @ Fermat's Little Turtle [/b] : I really can't understand what prompted you to [b] [i] fly off the handle [/i] [/b].

Between,[b] JBL [/b] himself said some of my posts were amusing.

I know, I never [i] flew off the handle [/i] in response to any of the posts, maybe it may seem like that given the "choice" of words that I might have used. As I have said, I am really thankful to all of them (I never intended to hurt/disparage any of those geniuses).

I think the discussion has ended in regards to this thread.


$ n$ coins are flipped. At least one of them came up heads. The probability that at least one other coin came up heads is $ \frac{19}{21}$. What is $ n$?

The problem is very unclear. "The probability that at least one other coin came up heads" sounds two heads came up. Even with conditional probability, it doesn't make much sense to me. Let's say there are 6 coins. Then what does the probability 19/21 mean? Can somebody clarify that for me?

It's asking, given that at least one coin came up heads, what is the probability two or more did?

We want $ \frac{\binom{n}{2}\plus{}\binom{n}{3}\plus{}....\plus{}\binom{n}{n}}{\binom{n}{1}\plus{}\binom{n}{2}\plus{}...\plus{}\binom{n}{n}}\equal{}\frac{19}{21}$

We know that

$ \binom{n}{1}\plus{}\binom{n}{2}\plus{}...\plus{}\binom{n}{n}\equal{}2^n\minus{}1$ (see [url=http://www.artofproblemsolving.com/Wiki/index.php/Combinations#Formulas.2FIdentities]here[/url])

So our fraction becomes

$ \frac{2^n\minus{}n\minus{}1}{2^n\minus{}1}\equal{}\frac{19}{21}$

Multiplying out

$ 21\cdot2^n\minus{}21n\minus{}21\equal{}19\cdot2^n\minus{}19$

$ 2^{n\plus{}1}\minus{}21n\minus{}2\equal{}0$

$ 2(2^n\minus{}1)\equal{}21n$

We need $ 2^n\minus{}1$ to be a multiple of 21. We can quickly see this occurs when $ n\equal{}6 \implies 2^n\minus{}1\equal{}63\equal{}3\cdot21$

Thus, $ n\equal{}6$

Plugging in to our first equation, we get $ \frac{57}{63}\equal{}\frac{19}{21}$ so it works.


[img]http://img522.imageshack.us/img522/5126/tourneyvs1.png[/img]
After quite a bit of work, I have come up with this structure for an AoPS FTW Tournament:
[b]It is a Single Elimination[/b] with 8 teams.
[b]Teams[/b] are 3 users, with no necessary leader.  Teams need to provide a square avatar (the exact size is 66 x 66px) and a name (duh...).
[b]Normal Rounds[/b] consist of 2 Normal games, where all 3 players from each team (total of 6) plays.  The winning team is the team with the highest sum of all their players points from both games.  At least one member of each team should post both teams' totals (2 posts total).
[b]The Countdown Round[/b] consists of 9 Countdowns.  Each member of a team plays each of the 3 members of the other team.  The team with most wins advances.  Each member should post all the wins from the games they played.

I will try to update the bracket weekly...
[b]Scheduling times to play is up to you.[/b]
No games should start until March 01, 2008.  Let's give some time for people to make teams.

If you have any questions or comments, post a reply.  If you feel there should be a change to the structure, [i]please[/i] start a new poll and post a link to it here.

Wow, cool chart.  Did you make it yourself or did it come from somewhere?

Do you have a website where you can post it so everyone is up to date (or we could put it somewhere on AOPS)?

I made it myself using The GIMP...It took a while ;).

And for the chart...I don't see the trouble in posting multiple files on ImageShack and updating them here, if that is what you're talking about.

Or, are you asking for the .xcf file?

Image shack would probably be a good place.

We are doing 2 person teams i thought
k this is our team
TEAM PLASTIC
ra5249
stevenmeow
budi713

add us to the list please, ty
nice chart btw

Well, why ae we changing the # of people?

Team CarmlSaml
CatalystOfNostalgia
perfect628

TEAM FURBIES ON FIRE

infinity4ever
mz94
aufha (if he feels like it)

Team OhioIzBallerz

[list][*]#H34N1
[*]George S Bashour
[*]xpmath[/list]

our team pic is a picture of pikachu
just use google images and take your pick...

OK, 3 person teams...w/e...And why single elimination? Seriously, why is it up to one person? I suppose if we must do it this way, we need a third an a pic cat nose.

What is the number of questions, time limit, and scoring type in the normal games?

Why don't you have both countdown and normal together each round? That would seem more fair...

NEW TEAM!!!

boyz of darkness

Verizonwires
Vallon22
emton

we will kick nerdybuttox!!!!

Which tournament should i register for now, the March or the April?
Thanks.

and this is not cool... team penguin used the avatar i have...
maybe they got it at the same place, but i request that next time they use something different.
thank you.

[quote="Ignite168"]NEW SAGGINWOLLEY TIMES:
All times are in EST
Monday: After 18:30
Tuesday: After 19:00 but not from 21:00-22:00 as Big Brother is on then
Wednesday: Call for appt, never before 17:00 or from 20:00-21:00 as Big Brother is on then
Thursday: After 22:30
Friday: NO
Saturday: NO
Sunday: From 12:00 onwards except for 20:00-21:00 as Big Brother is on then[/quote]

As in the topic, 4PM Eastern on a Sunday. Is that good with you?

[quote="lotsofmath"]Team WeWillWin

lotsofmath
fireemblem13
(undecided)

SIGN US UP![/quote]

Sign up yourself using the website.

i think we should do individual, like i said on the website, unless we are doing it....

what happened to the signup website... it says it is no longer in service... :(

The tournament was cancelled.  When I try to set up a tournament with MustItAlwaysBeMe, we will let you know.
So all tournaments for FTW are history until I get something arranged.
Anyone that would like to help out (I'll need it) can PM me or email me at aopsftw@gmail.com
Thanks,
minicon

this be the old tournament thread

the new one is elsewhere

this may be quite random, but can you explain to me how u made the chart and how u used GIMP?

[quote="ash-Pokemon"]have these games even started??????????[/quote]The last tournament that took place was the April tournament.  The next one will probably be in July.


geometry
3D geometry
absolute value
combinatorics unsolved
combinatorics

Provided $n^3$ unit cubes, $n > 1$ which are numbered from 1 to $n^3$. All these unit cubes are put together to form a cube with side length $n$. In this cube two unit cubes will be neighbors if they share at least one vertex. As distance of two neighbor unit cubes we define the absolute value of the difference of the numbers assigned to those two unit cubes. Imagine for all possible compositions of the "big" cube the biggest distance among any of these two unit cubes - all these maximal distances are written on a board. What is the minimal number on the board ?

I guess people jut didn't feel like reading the whole thing, because otherwise it would've been solved until now :D.

I think the answer is $n^2+n+1$. Let's assume there is a large cube s.t. all the distances are $\leq n^2+n$. In this case we start from the cube containing $1$ and go outwards (like ripples in the water :)) to the neighbours of $1$, then to the neighbours of those neighbours and so on. each time the value of the numbers increases with at most $n^2+n$ and we can take at most $n-1$ such steps (when $1$ is in a corner of the large cube). This means that all numbers in the cube are at most $1+(n-1)(n^2+n)=n^3-n+1<n^3$, which is false.

Conversely, we have to find a cube in which the maximal distance is $n^2+n+1$. We arrange them in the most natural way possible: $a_{i\ j\ k}=in^2+jn+k+1,\ i,j,k\in \overline {0,n-1}$. Just imagine it and you'll see it's ok.


Rhombus ABCD, with side length 6 is rolled to form a cylinder of volume 6 by taping AB to DC.  What is sin(ABC)?

[hide="not_sure,_but..."]

Firstly, let $H$ be projection of $A$ onto $BC$, making $AH$ the height of the cylinder that will be formed.

Secondly, the base of the cylinder will be a circle with circumference $6$.  Therefore the radius is $\frac{6}{2\pi}= \frac{3}{\pi}$, and it will have an area of $\pi * (\frac{3}{\pi})^{2}= \frac{9}{\pi}$.  For the volume of the cylinder to be $6$, $AH$ must be $\frac{2\pi}{3}$.

Finally, $sin{ABC}= \frac{AH}{AC}= \frac{\frac{2\pi}{3}}{6}= \frac{\pi}{9}$.

[/hide]

thank you i am pretty sure that that is right!


number theory
Diophantine equation
number theory unsolved

Find all four positive integer numbers $ a,b,c,d$ such that $ a>b>c>d ; b+c+d> a$ and:
        $ 16(a+b+c+d)^{2}= (-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)$

This Diophantine equation has no solution according to my proof which is in the attached pdf-file below.  However, under the condition that $a$, $b$, $c$, $d$ are not distinct, there are solutions to this equation, for example $(a,b,c,d)=(8,5,5,2)$ and $(a,b,c,d)=(14,6,5,5)$.


geometry
3D geometry
induction
irrational number
algebra
polynomial
Rational Root Theorem

Prove that the cube root of any irrational number is an irrational number.

Let $ u_n \equal{} 5^{1/ \left(3^n \right)}$. Given that $ \sqrt[3]{5}$ is an irrational number, prove by induction that $ u_n$ is an irrational number for every positive integer $ n$.

Hence, or otherwise, give an example of an infinite sequence of irrational numbers which converges to a given integer $ m$.

Thought for a: If $ p\notin\mathbb{Q}$, then $ p\neq\frac{a}{b}\Rightarrow p^{3}\neq\frac{a^{3}}{b^{3}}\equal{}\left(\frac{a}{b}\right)^{3}$, which is of course a rational number.  Hence $ p^{3}$ is not rational.

[quote="JRav"]Thought for a: If $ p\notin\mathbb{Q}$, then $ p\neq\frac {a}{b}\Rightarrow p^{3}\neq\frac {a^{3}}{b^{3}} \equal{} \left(\frac {a}{b}\right)^{3}$, which is of course a rational number.  Hence $ p^{3}$ is not rational.[/quote]Not quite correct. If $ p \equal{} \frac1{\sqrt [3]{2}}\not\in\mathbb{Q}$ but $ p^3 \equal{} \frac12$...  :huuh: BTW, it's asking about cube root of any irrational number, not cubing ;)

[hide="Discussion"] The cube of a rational number is rational.  Think [url=http://en.wikipedia.org/wiki/Contraposition]contrapositives[/url].

It's not hard to prove that the cube root of an integer is either irrational or an integer:  apply the Rational Root Theorem to $ x^3 \minus{} n$ for any integer $ n$.  This lets us prove that $ \sqrt [3]{5}$ is irrational.

Note that we can replace $ 3$ with any positive integer greater than $ 1$.  This immediately proves the second result without the need for induction.

As for the third question, let $ r_1, r_2, ... r_n$ be the sequence of [url=http://mathworld.wolfram.com/ContinuedFraction.html]convergents[/url] of some irrational number $ \alpha$ and consider the sequence $ u_n \equal{} m \plus{} \alpha \minus{} r_n$.  If $ \alpha$ is algebraic, we get the additional property that the degree of $ \mathbb{Q}[u_n] : \mathbb{Q}$ is bounded.  In the given construction, it is $ 3^n$.  [/hide]

Oh sorry, I misread the question. :oops:

Thanks for the insights  :) So by quoting contrapositives, we can use it directly?

You can state without proof that a statement is logically equivalent to its contrapositive, if that's what you mean.

A nice way to handle problems such as this, which handle irrationality, transcendental numbers,... is use proof by contradiction.  Suppose what you want to prove is not true.  Then do the math and show that in the end, you wind up with a contradiction, hence the initial assumption must have been wrong.


Let $ABC$ be a triangle with circumcircle $\omega$. The angle bisectors of $A,B,C$ intersect $\omega$ again in $A', B', C'$ respectively. Prove that the area of $\Delta ABC$ is less or equal to the area of $\Delta A'B'C'$.

Denote: $R$ be the circumradius of triangle $ABC.$ we have:
$\frac{S(ABC)}{S(A'B'C')}=\frac{2R^{2}.\prod\sin A}{2R^{2}.\prod\sin A'}=\frac{\prod\sin A}{\prod\sin\frac{B+C}{2}}=8\prod\sin\frac{A}{2}\le 1\quad\mathbb{QED}$


:rotfl:  :rotfl:  :rotfl: [url]http://www.kongregate.com/games/ArmorGames/achievement-unlocked[/url]

499 sec

95% completed... this game is genius.

EDIT: 96%. And although I clicked "hint," I didn't read it.

407 Seconds :rotfl:

To change colors, press the numpad. to jump for a LONG time, go onto one of the jets and hold up.

1196 seconds....w00t
 :wallbash_red:

yet i looked at the hint

lol had to use hints on 13, 49, 77, 78, and ultimately 99.

Others i figured out on my own.

Ima try for high score!

Yay!  Sub 400

396:20

Pointless, yet fun game.

I completed it in 1767 seconds. I did not look at the hints, but I did click it (of course).

Second time: 333

Wait, how do you do it in under 300 seconds if one of the achievements is to play for 300 seconds?

256:93 (of course I had already played for the first time before) :)

Finished (no hints)! I did it again, and got 395 seconds.

[quote="eggylv999"][quote="33286"]Now 218.6. The 300 seconds must have been a misreading.[/quote]

 :o how do you do graveyard so fast???[/quote]
graveyard is easy. do you mean century and one more? they are really time consuming.

I would have shortened my time to 240, had the game accepted the Contra code earlier. Anything special about this achievement that I don't know...

EDIT: Oh, of course. I didn't know the code  :) Finally got under 200 - 196:33.

267 Seconds. I am spending too much time on Pi, (2*125)+3, THIS IS SPARTA, We Have Liftoff, I CAN COUNT TO FIVE, 3*(30/(15-13)), and Meaning Of Life.

It's strange how you automatically get the "No Cheese For You" achievement when you click "Restart" after beating the game.

Also, the link on the main menu counts as an Armor Games link for You Are El.

Also, Programmer's Credits also gets you Shocking Pink.

I CAN COUNT TO FIVE also gets you 3*(30/(15-13)).

Also, you can get Get Off Your Seats by holding "up" on a launcher.

No Cheese for You = click on the play area = click on the "restart" button

so its logical.

Am currently stuck on the color problem.

91% at 223 seconds.

231 seconds.

no, actually 214.40!

:)

EDIT: Sub-3 minutes: 168:86

TIP: Get a laggy computer.

Guide:

You should divide gameplay into the following "Phases":

[hide]0) Pregame Phase:
Bandwidth exploiter, You are El, Clink Clash Clink, Play the Game, Get to Level 1, Menu Explorer [6]

1) Keypressing Phase
QUICKLY TYPE/CLICK:
1,jmtb02,3,4 [color change+Developer's Credit]
up/down, left/right, Space [Confusions]
u-u-d-d-l-r-l-r-b-a [Konami Code (Contra)]
Click: Hint?, Quality, Mute
left, right, up
Click gameplay screen
Scroll the scrollbar
Adds up to 17 achievements.

2) Exploration Phase
Get all the spikes. Get all the jets.
Get the 3 secret alcoves.
Get the Dev's favorite spot.
Get the 5 secret locations.
lefty/righty
Adds up to 23 achievements.

3) Location Phase.
Hit the locations, in order:
5-4-3-2-1-0-1-2-3-4-5-4-2-5-3-0-0
You may die in between, however you may want to receive the longevity badges here...
7 achievements.

4) Survivalism Phase.
Jump for 10 seconds, then wait for 20 seconds [You can couple this with Phase 3]
3 achievments.

5) Dying Phase.
Use the moving platform for "Die in a Flash", "Cemetery"+"Graveyard", "Half-Century"+"Century"+"And one More"
Use the pit for 1/3/10, 2/3 simul
11 Achievements.
And also: "Death at 0", "1-o-clock death", potentially "airwalk death"
so a total of 14 achievments.[/hide]

I missed a couple (20 or so) achievements.

On my "speedy" computer I got 300.0 seconds. Exactly.

[hide="modified guide"]0) Pregame Phase:
Bandwidth exploiter, You are El, Clink Clash Clink, Play the Game, Get to Level 1, Menu Explorer [6]

1) Keypressing Phase
QUICKLY TYPE/CLICK:
1,jmtb02,3,4 [color change+Developer's Credit]
up/down, left/right, Space [Confusions]
u-u-d-d-l-r-l-r-b-a [Konami Code (Contra)]
Click: Hint?, Quality, Mute
left, right, up
space
l+r, u+d
Click gameplay screen
Scroll the scrollbar
Adds up to 20 achievements.

2) Exploration Phase
Get all the spikes. Get all the jets.
Get the 3 secret alcoves.
Get the Dev's favorite spot.
Get the 5 secret locations.
lefty/righty
paint 25, 50, 75, 100, masterpiece, extreme makeover
stalagmite
stalactite
speedy downfall
Adds up to 32 achievements.

3) Location Phase.
Hit the locations, in order:
5-4-3-2-1-0-1-2-3-4-5-4-2-5-3-0-0
You may die in between, however you may want to receive the longevity badges here...
7 achievements.

4) Survivalism Phase.
Jump for 10 seconds, then wait for 20 seconds [You can couple this with Phase 3]
3 achievments.

5) Dying Phase.
Use the moving platform for "Die in a Flash", "Cemetery"+"Graveyard", "Half-Century"+"Century"+"And one More"
Use the pit for 1/3/10, 2/3 simul
11 Achievements.
And also: "Death at 0", "1-o-clock death", potentially "airwalk death"
so a total of 14 achievments.
[/hide]

missing 21

481 seconds  :D

P.S. That's the most weird game I've ever played  :D

[hide="completed guild"]0) Pregame Phase:
Bandwidth exploiter, You are El, Clink Clash Clink, Play the Game, Get to Level 1, Menu Explorer [80, 81, 82, 85, 86, 88]
and also "Don't move a muscle" [0]
totally 7

1) Keypressing Phase
QUICKLY TYPE/CLICK:
1,jmtb02,3,4 [color change+Developer's Credit] [72-75, 77]
up/down, left/right, Space [Confusions] [47-49]
u-u-d-d-l-r-l-r-b-a [Konami Code (Contra)] [78]
Click: Hint?, Quality, Mute [76, 83, 87]
left, right, up [1-3]
Click gameplay screen [79]
Scroll the scrollbar [84]
Adds up to 17 achievements.

2) Exploration Phase
Get all the spikes. Get all the jets. [4, 5, 65-70]
Get the 3 secret alcoves. [27-29]
Get the Dev's favorite spot. [50]
Get the 6 secret locations. [14-20]
lefty/righty [6, 7]
paint 25, 50, 75, 100, masterpiece, extreme makeover [8, 9, 59-64, 71]
stalagmite [12]
stalactite [13]
speedy downfall [10]
escape velocity [11]
Adds up to 32 achievements.

3) Location Phase.
Hit the locations, in order:
5-4-3-2-1-0-1-2-3-4-5-3-1-4-2-5-3-0-0
You may die in between, however you may want to receive the longevity badges here...
7 achievements. [22, 23, 54-58]

4) Survivalism Phase.
Jump for 10 seconds, stay in a launcher for 10 seconds, then wait for 10 seconds [You can couple this with Phase 3] [24, 32, 38, 39]
4 achievments.

5) Dying Phase.
Use the moving platform for "Die in a Flash", "Cemetery"+"Graveyard", "Half-Century"+"Century"+"And one More"
Use the pit for 1/3/10, 2/3 simul
11 Achievements.
And also: "Death at 0", "1-o-clock death", potentially "airwalk death"
so a total of 14 achievments.
and also: "Wrecking Ball", "Hellbound", "Heavenbound"
[21, 25, 26, 30, 31, 33-37, 40-46]
totally 17.

6) others
[51-53]
played for 1 minute
[89-99]
get when all other achievements are done.
totally 14
[/hide]
done. :)

[quote]#56: Meaning of Life = Hit 4-2 in that order[/quote]

This part with the meaning of life is extremely funny.  :lol:

Programmer's Credit: what is it?
ArmorGames is blocked on my computer so I can't access hints  :(

New time: 232.43 on fast computer

On the "lagging" computer I had warcraft III running in the background, plus about 7 other tabs in Firefox.

Question: Does everybody else have to suicide about 30 times at the end?

Played about 5 times

fastest (and most recent) 204.6!

[hide="hints"]
#0: Don't Move a Muscle = Keep the elephant from moving
#1: It's a Jump to the Left = Find a way to move westward
#2: And it's a step to the Ri-i-ight! = Strain yourself to move right
#3: Fight Gravity = Figure out how to jump upwards
#4: Elephant Skewers = Hit a spike strip
#5: Ascension = Be lifted into the sky
#6: Lefty = Show love to the left wall.
#7: Righty = Show love to the right wall.
#8: Purple Ceiling = Hit the top of the hardest level
#9: Rock Bottom = Find the bottom of the easiest level
#10: Speedy Downfall = Fall from the top to bottom without stopping
#11: Escape Velocity = Go from bottom to top without stopping
#12: Stalactite = Skewer yourself at high speed on downward facing spikes
#13: Stalagmite = Skewer yourself at high speed on upward facing spikes
#14: Found Point 0 = Uncovered secret point 0
#15: Found Point 1 = Claimed secret point 1
#16: Found Point 2 = Excavated secret point 2
#17: Found Point 3 = Popped open secret point 3
#18: Found Point 4 = Tripped over secret point 4
#19: Found Point 5 = Visited secret point 5
#20: Bloodhound = Uncovered all points on level
#21: Floating Corpse = Get a skeleton stuck in a launcher
#22: I Can Count to 5! = Visit each point in order
#23: We Have Liftoff = Visit each point in reverse order
#24: Airwalking = Stay off the ground for 10 seconds
#25: Hellbound = Launch corpse to bottom
#26: Heavenbound = Launch corpse to ceiling
#27: Secret Alcove = Get surrounded by three sides (1)
#28: Secret Cave = Get surrounded by three sides (2)
#29: Secret Hideout = Get surrounded by three sides (3)
#30: Death at 0 = Fire a corpse at Point 0
#31: 1 o'clock murder = Fire a corpse at Point 1
#32: Get Off Your Seats = Jump for 10 seconds nonstop
#33: Cemetary = Have 5 live corpses on the screen
#34: Graveyard = Have 7 live corpses on the screen
#35: Half-Century = Reach 50 elephant deaths
#36: Century = Reach 100 elephant deaths
#37: And One More... = Hit 101 elephant deaths
#38: Stayin' Alive = Stay alive for 15 seconds
#39: Longevity = Stay alive for 30 seconds
#40: Gone in a Flash = Die in less 0.23 seconds
#41: Wrecking Ball = Hit the center spiked platform
#42: In the Pit = Land a corpse in the pit
#43: Pit Hat Trick = Land 3 corpses in the pit
#44: 10 Pin Pit = Land 10 corpses in the pit
#45: Double Play Pit = Have two corpses in the pit at once
#46: Triple Play Pit = Have three corpses in the pit at once
#47: Horizontal Confusion = Hold both the left and right keys
#48: Vertical Confusion = Hold both the up and down keys
#49: Total Confusion = Space bar has no purpose, dummy
#50: My Favourite Spot = Find the developer's favourite spot
#51: 60 Seconds = Play for one minute
#52: 180 Seconds = Play for three minutes
#53: 300 Seconds = Play for five minutes
#54: Pi = Hit points 3-1-4 in that order
#55: THIS IS SPARTA = Hit 3-0-0 in that order
#56: Meaning of Life = Hit 4-2 in that order
#57: (2*125)+3 = Hit 2-5-3 in that order
#58: 3*(30/(15-13)) = Hit 4-5 in that order
#59: Paint 10 = Touch 10 blocks
#60: Paint 25 = Touch 25 blocks
#61: Paint 50 = Touch 50 blocks
#62: Paint 75 = Touch 75 blocks
#63: Paint 100 = Touch 100 blocks
#64: Masterpiece = Touch every touchable block
#65: Jet 5 = Hit 5 launchers
#66: Jet 10 = Hit 10 launchers
#67: Master Jetter = Hit every launcher
#68: 3 Spikes = Hit 3 stationary spikes
#69: 7 Spikes = Hit 7 stationary spikes
#70: Spikeful = Hit all stationary spikes
#71: Extreme Makeover = Recolour all areas
#72: True Blue = Turn your elephant blue
#73: Shocking Pink = Turn your elephant Pink
#74: Sweet Orange = Turn your elephant orange
#75: Stealth Black = Turn your elephant black
#76: Cheater = Ask for a Hint
#77: Programmers Credit = Enter jmtb02 into level
#78: Too Much Contra = Enter Contra level select code
#79: No Cheese for You = Click the level area
#80: Bandwidth Exploiter = Preload the game
#81: You Are El = Click the Armor Games link
#82: Menu Explorer = Find the main menu
#83: Quality Snob = Adjust the quality
#84: Scrolliastic = Scroll the achievements section
#85: Play the Game = Start the game
#86: Get to Level 1 = Play Level 1
#87: Not Listening = Mute the sound
#88: Clink Clash Clink = Found the sponsor screen
#89: Best Player Ever = Earned an achievement
#90: 10 Percent Done = Earn 10 achievements
#91: 20 Percent Done = Earn 20 achievements
#92: 30 Percent Done = Earn 30 achievements
#93: 40 Percent Done = Earn 40 achievements
#94: 50 Percent Done = Earn 50 achievements
#95: 60 Percent Done = Earn 60 achievements
#96: 70 Percent Done = Earn 70 achievements
#97: 80 Percent Done = Earn 80 achievements
#98: 90 Percent Done = Earn 90 achievements
#99: Too Much Free Time = Earn 99 achievements
[/hide]

For 72-75: press 1-2-3-4 on keyboard.  For #78, sometimes, I can just type "contra" and it counts, but u-u-d-d-l-r-l-r-b-a always works.

"contra" only works after you typed u-u-d-d-l-r-l-r and then typed "jmtb02", in which case the a is the only letter missing.

I got 98%, then couldn't figure out total confusion, until I decided to press every key on the keyboard.  :P

So, I tried again for below 200 seconds, but miserably failed.

I got Century and And One More and then realized I hadn't gotten Longevity yet. So yeah, I had to wait for 60 more seconds. I ended up getting 253 seconds...


function
real analysis
real analysis unsolved

Let $f: R\to\ R$ be a function( :D  ) and  $M=\{|f(x)-f(y)| \mid x,y\in R,x\neq y\}$.
   Prove that $\inf M=0$.

If $\inf M>0$, around every point $z\in f(\mathbb{R})$, we can draw an open ball $B_z$ such that $B_x\cap B_y=0$ for every $x\not=y$ in $f(\mathbb{R})$.

There can only be countably many such balls, whereas $f(\mathbb{R})$ is uncountable, which is absurd.

What`s a ball?My solution also invovles the uncountability of $f(R)$.

The ball of radius $r$ around $x$, $B(x,r)$ is defined by
\[ B(x,r)=\{y: |x-y|<r\}  \]

actually the idea is that if we presume the contrary we get that f is injective and we eventually get a bijective correspondence between an uncountable and a countable set, in this direction. Contradiction so that's it :roll:

see http://www.mathlinks.ro/Forum/viewtopic.php?highlight=dilatation&t=45121 for a related problem and a very nice solution.

[quote="blahblahblah"]If $\inf M>0$, around every point $z\in(\mathbb{R})$, we can draw an open ball $B_z$ such that $B_x\cap B_y=0$ for every $x\not=y$ in $f(\mathbb{R})$.

There can only be countably many such balls, whereas $f(\mathbb{R})$ is uncountable, which is absurd.[/quote]

Why is $f(\mathbb{R})$ uncountable?

Furthermore, what do you mean by " around every point $z\in(\mathbb{R})$, we can draw an open ball $B_z$ such that $B_x\cap B_y=0$ for every $x\not=y$ in $f(\mathbb{R})$ " ???

Suppose $\inf M =\alpha > 0$ then there exist a rational number $q$ such that $\alpha>q$ which means that in each interval of the form $[(k+1)q,kq]$ there exist at most a si.ngle value of $f$ in this interval and since $f$ is injective denote that value  by $x_k$ . Since Q in numerable so is $x_k$. So this makes $f$ an injective function from $R$ to $Q$. Which is imposible because $R$ is not numerable while $Q$ is.


geometry
inequalities
inradius
trigonometry
geometry solved

Let K be a point on the side BC of the triangle ABC. The incircles of the triangles ABK and ACK touch BC at points M and N, respectively. Show that [tex]BM\cdot CN>KM \cdot KN[/tex].

[quote="Sailor"]Denote by p1 the semiperimeter of ABK, and by p2 the semiperimeter of AKC. We have to prove that (p1-AK)(p2-AK)>(p1-c)(p2-b) after opening the paranthesis the conclusion follows.[/quote]

Hmm, that took me some time to understand... In order to make your solution a bit clearer for beginners, I will write a few more details: We have $BM=\frac{AB+BK-AK}{2}$, $CN=\frac{AC+CK-AK}{2}$, $KM=\frac{AK+BK-AB}{2}$, $KN=\frac{AK+CK-AC}{2}$; thus, proving the inequality $BM\cdot CN > KM\cdot KN$ reduces to proving $\left(AB+BK-AK\right)\cdot\left(AC+CK-AK\right)>\left(AK+BK-AB\right)\cdot\left(AK+CK-AC\right)$. Now, after opening the brackets, this simplfies to $AC\cdot BK + AB\cdot CK > AK\cdot BK+AK\cdot CK$, or, equivalently, $AC\cdot BK + AB\cdot CK > AK\cdot BC$. But this follows from the Ptolemy inequality, applied to the quadrilateral ABKC (equality cannot occur since the quadrilateral ABKC can never be cyclic).

Anyway, here is my solution of the problem:

Let $s=\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Let r be the inradius and $r_a$, $r_b$, $r_c$ be the exradii of triangle ABC. Then, it is well-known that $\cot\frac{B}{2}=\frac{r_c}{s-a}$ and $\cot\frac{C}{2}=\frac{s}{r_c}$; thus, $\cot\frac{B}{2}\cdot\cot\frac{C}{2}=\frac{r_c}{s-a}\cdot\frac{s}{r_c}=\frac{s}{s-a}$. Since, obviously, s > s - a, we have $\frac{s}{s-a}>1$, and thus,

$\cot\frac{B}{2}\cdot\cot\frac{C}{2}>1$.

Now, the point M was defined as the point where the incircle of triangle ABK touches the line BC. Thus, if X is the center of this incircle, we have $XM\perp BC$, and consequently, the triangles BMX and KMX are right-angled. It follows that ${BM=XM\cdot\cot\measuredangle XBM}$ and $KM=XM\cdot\cot\measuredangle XKM$. Thus,

$\frac{BM}{KM}=\frac{\cot\measuredangle XBM}{\cot\measuredangle XKM}$.

But since the point X is the center of the incircle of triangle ABK, the lines KX and BX are the angle bisectors of the angles BKA and ABK. Thus, $\measuredangle XKM=\frac{\measuredangle BKA}{2}$ and $\measuredangle XBM=\frac{\measuredangle ABK}{2}$. The latter of these equations rewrites as $\measuredangle XBM=\frac{B}{2}$. Thus,

$\frac{BM}{KM}=\frac{\cot\measuredangle XBM}{\cot\measuredangle XKM}=\frac{\cot\frac{B}{2}}{\cot\frac{\measuredangle BKA}{2}}$.

Similarly,

$\frac{CN}{KN}=\frac{\cot\frac{C}{2}}{\cot\frac{\measuredangle CKA}{2}}$.

Hence,

$\frac{BM\cdot CN}{KM\cdot KN}=\frac{BM}{KM}\cdot\frac{CN}{KN}=\frac{\cot\frac{B}{2}}{\cot\frac{\measuredangle BKA}{2}}\cdot\frac{\cot\frac{C}{2}}{\cot\frac{\measuredangle CKA}{2}}=\frac{\cot\frac{B}{2}\cdot\cot\frac{C}{2}}{\cot\frac{\measuredangle BKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}}$.

Since < BKA + < CKA = 180°, we have $\frac{\measuredangle BKA}{2}+\frac{\measuredangle CKA}{2}=\frac{\measuredangle BKA+\measuredangle CKA}{2}=\frac{180^{\circ}}{2}=90^{\circ}$. Hence, $\cot\frac{\measuredangle BKA}{2}=\tan\frac{\measuredangle CKA}{2}$. Thus,

$\cot\frac{\measuredangle BKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}=\tan\frac{\measuredangle CKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}$
$=\tan\frac{\measuredangle CKA}{2}\cdot\frac{1}{\tan\frac{\measuredangle CKA}{2}}=1$.

And consequently,

$\frac{BM\cdot CN}{KM\cdot KN}=\frac{\cot\frac{B}{2}\cdot\cot\frac{C}{2}}{\cot\frac{\measuredangle BKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}}=\frac{\cot\frac{B}{2}\cdot\cot\frac{C}{2}}{1}=\cot\frac{B}{2}\cdot\cot\frac{C}{2}$.

But we know that $\cot\frac{B}{2}\cdot\cot\frac{C}{2}>1$; thus, $\frac{BM\cdot CN}{KM\cdot KN}>1$, so that $BM\cdot CN>KM\cdot KN$. Proof complete.

  darij


Διάβαζα από τα βιβλία του κύριου Στεργίου πριν λίγο και προσπάθησα να λύσω ένα θέμα Ολυμπιάδας (ΙΜΟ) που βρήκα σε ένα Γαλλικό βιβλίο και με άξονα αυτό το θέμα έχω να κάνω τρεις ερωτήσεις:

1) Όταν μία ανισότητα είναι ομογενής μπορώ να θέσω πάνω από ένα περιορισμό? Για παράδειγμα α+β+c=3 και abc=1?

2) Όταν μία ανισότητα είναι κυκλική αλλά όχι συμμετρική μπορώ να θέσω α>=β>=γ ή μόνο να υποθέσω ότι ένα από αυτά είναι το μέγιστο ή το ελάχιστο?

3) Μπορεί κανείς να λύσει (αν γίνεται με ομογενοποίηση αφού η ανισότητα είναι από μόνη της ομογενής επειδή το προσπάθησα μάταια) την:

α^2(β+γ-α)+β^2(γ+α-β)+γ^2(α+β-γ)<=3αβγ, όπου α,β,γ πλευρές τριγώνου.

Protonie...isos dn ime o armodios gi ana s apantiso..(malon o k.Stergiu)parolayata an 8es mporo an su po merika...
Parolo pu dn mporesa na vro pu akrivos..ala ex ksanagini mia paromia sizitisi me aformi tis anisotites+exun ipo8i paromia pragmata..
Gia to
1) Genika..(+leo genika gt mpori an iparxi kapia poli eksidikeumeni periptosi pu na min gnorizo..)dn mporis..+to skeptiko ine aplo...(me mia mono sxesi..dn periorizis tis metavlites...eno me dio i perisoteres...sximatizis sxesis metaksis tus...pragma to opio dn dinete apo tin ekfonisi +ine en geni la8os.. (episis dn ine anagki na 8esis a+b+c i abc al a mporis +a^2+b^2+c^2 opos kai ales parastasis..)

2)Dn exo ayti ti stigmi mprosta mu to sigekrimeno vivlio..ala 8imame oti se kapio simio to dieykrinizi ayto kseka8arizontas tis dio periptosis....+lei kseka8ara ti isxii+se pies periptosis ...yaksto+nomizo oti 8a to vris..

3)Mia prosopiki simvuli..(an mu epitrepis..,gt dn exoume +pola xronia diaforas+ine ligo anusia i leksi<simvuli..>)Min kinise se sigekrimena plesia..!!Min epidi ides oti i anisotita ine omogenis prospa8is anagastika na ti lisis m ayto ton tropo...I anisotites ine toso sixna xrisimopiumenes se ma8imatikus diagonismus gt ka8e mia ine ksexoristi( sto va8mo pu ayto ginete..)+apetun tin protovulia tu diagonizomenu...na kani ekino na kani to alo...Den gnorizo ti anisotites kses....ala su ipa to parapano  gt to proto pu paratirisa ine oti i anisotita pu les..ine i anisotita schur..gia n=1,..xoris na xrisimopiisis ka8olu tin omogenia ... :wink: Dn iparxun panakia...pu8ena...+se oti iparxi..spania 8a dis na pefti kati tetio se sovaro diagonismo...gt poli apla dn zitite apo tus diagonizomenus na parun enan tipo..<+na kanun praksis> pu leme..i na  efarmosun enan tiflosurti gia na kataliksun stin  apodiksi...Xriazete efeuretikotita,(+pu +pu na kitas tin anisotita apo apostasi megaliteris tu misu metru ....emena sixna mu rxete tote i idea..gt mu 8imizi kapia ali pu exo di..i paratiro kalitera tin omogenia i tin simetria+vlepo ti voi8ai na metasximatiso......)

Btw...dn ksero an mpori na li8i..eukola ekmetaleuomeni tin omogenia..ala sigura to oti ine pleyres trigonu...se..<prokali> na dokimasis ton klasiko metasxomatismo se teties periptosis..

Filika..(+elpizo na se voi8isa)...

 :D

δικιο εχεις .ομογενης ανισοτητα σημαινει οτι αν στη θεση του $a_{i}$ θεσεις το $ka_{i}$,αρα μπορεις να θεσεις σε μια ανισοτητα με μεταβλητες $a_{1},a_{2},...,a_{n}$ :

$a_{1}+a_{2}+...+a_{n}=u$

$a_{1}a_{2}...a_{n}=u$ και αλλες συναρτησεις των μεταβλητων  οτι σε συμφερει βεβαια μια συνθηκη μονο .

Στο πρώτο σου ερώτημα, νομίζω ότι η απάντηση είναι αρνητική. Διότι αν βάλεις τους περιορισμούς  που ανέφερες, τότε προκύπτει $a=b=c=1.$
Στο δεύτερο ερώτημα, αν  η ανισότητα είναι κυκλική, τότε μπορείς να υποθέσεις ότι $a=max\{a,b,c\}$  ή  $a=min\{a,b,c\}.$ Αυτό  διότι: Εφόσον η ανισότητα $f(a,b,c)\geq0,$ είναι κυκλική, τότε ισοδυναμεί με
$f(b,c,a)\geq0$ και με την $f(c,b,a)\geq0.$ Άρα αν πχ είναι $b=max\{a,b,c\}$, τότε λύνουμε την $f(b,c,a)\geq0,$ η οποία είναι ισοδύναμη με την αρχική!.
 Αν όμως η ανισότητα $f(a,b,c)\geq0$ είναι συμμετρική, τότε ισοδυναμεί με τις:
$f(a,c,b)\geq0, f(b,c,a)\geq0, f(b,a,c)\geq0, f(c,b,a)\geq0, f(c,a,b)\geq0.$
Άρα  όποια σχέση διάταξης και να έχουν οι $a,b,c$ (πχ $b\geq a\geq c),$ λύνουμε την $f(b,a,c)\geq0,$ η οποία ισοδυναμεί με την αρχική! Για αυτό το λόγο δεν έχει σημασία πώς θα ορίσουμε τη διάταξη των $a,b,c.$ Είναι όλες ισοδύναμες.
Στο τρίτο ερώτημα, είναι μετά τις πράξεις η γνωστή ανισότητα Shcur, και δεν χρειάζεσαι καν τρίγωνο.

Επειδή είναι κυκλική θεωρούμε $a=max\{a,b,c\}.$ Οπότε προκύπτουν δύο περιπτώσεις: $b\geq c$ ή $c\geq b.$ Η λύση που έχεις αναφέρεται στην πρώτη περίπτωση και όμοια  η δεύτερη.

[quote="Protonios"]Ευχαριστώ πολύ για τις πληροφορίες. RAKAR το ξέρω ότι δεν πρέπει να εμμένω σε μια συγκεκριμένη ιδέα και σε ευχαριστώ που μου το θύμησες :wink: αλλά επειδή δεν είδα ότι μπορεί να γίνει SCHUR και ούτε ο μετασχηματισμός με οδήγησε πουθενά και ούτε τη λύση βρήκα στο βιβλίο :wink: σκέφτηκα μήπως η ομογενοποίηση δώσει κάτι καλύτερο.

Επειδή έχω μπερδευτεί όμως:

α^3+β^3+γ^3>=α^2*β+β^2*γ+γ^2*α

Το παραπάνω λύνεται με αναδιάταξη αλλά το δεύτερο μέλος είναι κυκλικό και όχι συμμετρικό γιατί αν εναλλάξουμε το β με το α δίνει β^2*α+α^2*γ+γ^2*β που είναι διαφορετικό. Άρα δεν μπορούμε να θεωρήσουμε α>=β>=γ. Πού κάνω το λάθος?[/quote]

 Η αναδιάταξη δεν έχει σχέση με κυκλικότηα ή συμμετρικότητα.Το μόνο που θέλει είναι να ισχύουν οι προυποθέσεις για την εφαρμογή της. Για παράδειγμα , αφού οι τριάδες $a,b,c$  $a^{k}, b^{l}, c^{m}$ έχουν για  περιττες δυνάμεις  των εκθετών την ίδια διάταξη , φτιάχνεις όσες ανισότητες θέλεις , συμμετρικές και μη.Άρα τις αποδεικνύεις με αναδιάταξη χωρίς να σε ενδιαφέρει η μορφή τους. Για αυτό η αναδιάταξη είναι τρομερή ανισότητα ! Διότι είναι πολύ γενική και θέλει μόνο τη διάταξη των αρχικών ν-άδων.
  Ως προς τα άλλα να ακολουθήσεις τα σχόλια του Nickolas !

                 Μπάμπης

[quote="Protonios"]Ευχαριστώ πολύ για τις συμβουλές κύριε Στεργίου απλά διάβασα σε ένα ξενόγλωσσο βιβλίο ότι αν δεν είναι συμμετρική μία ανισότητα δεν μπορούμε να υποθέσουμε α>=β>=γ δίχως βλάβη της γενικότητας ή τουλάχιστον έτσι κατάλαβα. Αληθεύει? Γιατί αν αληθεύει δεν μπορούμε να εφαρμόσουμε την ανισότητα της αναδιάταξης σε αυτήν την περίπτωση παραπάνω. :huh:[/quote]

Έτσι ακριβώς είναι όπως το διάβασες και όπως το λέμε και εγώ με το Σκομπρή ! Αλλά μην το μπερδεύεις : Στην αναδιάταξη δεν διατάσουμε για να αποδείξουμε την ανισότητα βασισμένοι στην διάταξη αυτή , αλλά θεωρούμε [color=red]δύο διατεταγμένες τριάδες [/color]με την ίδια ή αντίθετη διάταξη και εφαρμόζουμε την ανισότητα της αναδιαταξης. Είναι μια λεπτή αλλά σημαντική διαφορά ! Αν δεν το καταλαβαίνεις , συνέχισε τη μελέτη σου και επανέρχεσαι μόλις εμβαθύνεις περισσότερο. Μην ανησυχείς όμως , διότι : στις συμμετρικές δεν έχεις πρόβλημα και στις κυκλικές: υπάρχουν τόσοι τρόποι.Αν βρεις δύο τριάδες με την ίδια ή αντίθετη διάταξη , που να μην εξαρτάται όμως από επιλεκτική διάταξη των μεταβλητών της πρώτης τριάδας , τότε με σιγουριά εφαρμόζεις αναδιάταξη και δεν υπάρχει περίπτωση να πέσεις έξω.

 Μπάμπης

Νομίζω πως κατατοπίστηκα πλήρως. Ευχαριστώ πολύ. Ούτως ή άλλως είχα αρχίσει να υποψιάζομαι κάτι διότι στις περισσότερες ασκήσεις που έθετε α>=β>=γ δίχως βλάβη τις γενικότητας, το έκανε για να φτάσει σε μια παραγοντοποιημένη μορφή μεγαλύτερη ίση (ή μικρότερη ίση) με το μηδέν στο τέλος όπου η διάταξη θα του χρειαζόταν.
Στην αναδιάταξη δηλαδή θεωρούμε δύο τριάδες που ό,τι και να γίνει (όπως και να βάλουμε τα μεγαλύτερα μικρότερα ίσα) θα έχουν την ίδια ή αντίθετη σειρά (διάταξη) οπότε και εφαρμόζουμε τα θεωρήματα (τσεμπίτσεφ ή καθαρή αναδιάταξη).
Το κατάλαβα καλά???

[quote="Protonios"]Νομίζω πως κατατοπίστηκα πλήρως. Ευχαριστώ πολύ. Ούτως ή άλλως είχα αρχίσει να υποψιάζομαι κάτι διότι στις περισσότερες ασκήσεις που έθετε α>=β>=γ δίχως βλάβη τις γενικότητας, το έκανε για να φτάσει σε μια παραγοντοποιημένη μορφή μεγαλύτερη ίση (ή μικρότερη ίση) με το μηδέν στο τέλος όπου η διάταξη θα του χρειαζόταν.
Στην αναδιάταξη δηλαδή θεωρούμε δύο τριάδες που ό,τι και να γίνει (όπως και να βάλουμε τα μεγαλύτερα μικρότερα ίσα) θα έχουν την ίδια ή αντίθετη σειρά (διάταξη) οπότε και εφαρμόζουμε τα θεωρήματα (τσεμπίτσεφ ή καθαρή αναδιάταξη).
[color=blue]Το κατάλαβα καλά???[/color] [/quote]

[size=150][color=red] Είμαι σίγουρος πως ΝΑΙ !!![/color][/size]


calculus
integration
function
real analysis
real analysis unsolved

Find $\inf\int\int_{x^2+y^2\leq 1} \( (\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2 \)dxdy$

for $C^{\infty}$ functions $u$ that vanish at $0$ and are equal to $1$ on $x^2+y^2=1$

Click [hide="here"] The answer is $0$. It is enough to consider radial functions only. Then the problem reduces to finding the infimum of $\int_0^1 r f'(r)^2\,dr$ under the restrictions $f(0)=0$, $f(1)=1$. [/hide] for a hint.


geometry
probability
rectangle
analytic geometry

[quote="#H34N1"]A point is randomly selected within the rectangle with vertices at  (0,0), (2,0), (2,5)  and  (0,5) . What's the probability that the x-coordinate of the point is less than the y-coordinate? Express your answer as a common fraction in lowest term.[/quote]
[hide]This would be the area of the rectangle "above" x=y. (See picture)

The area of the rectangle is $2\times5=10$ and the black part is $\frac{2\times2}{2}=2$ so then the probability is $\frac{2}{10}=\boxed{\frac{1}{5}}$[/hide]

[quote="usaha"]wouldnt it be 1-1/5=4/5?[/quote]
No, $x<y$ so the graph is shaded above the line of $x=y$.

[quote="usaha"]ur graph is wrong..[/quote]

yeah usaha is right.  it should be more like

----------
[
[
[
[
[
[
----------

like tall, not wide.

[quote="usaha"]ur graph is wrong..[/quote]
It's not precise.

Here, now it's labelled.

Edit: Oh yeah, I'm stupid...

look at the question and then ur coordinates


How difficult are they compared to AoPS 1 and 2?  I know that they cover their respective subjects more in depth, but is there one (or more) book(s) in particular that you would really suggest as necessary and going beyond?  Thanks!

None of the introductory books go beyond the material found in volume 2 (they don't even really touch on a lot of the material), but they will give you a much stronger base in any given topic you might struggle with in volume 2. Wait for the intermediate books to come out, those are probably closer to what you're looking for.


If $ a \plus{} 4b \equal{} 33$ and $ 6a \plus{} 3b \equal{} 51$, what is the value of $ a \plus{} b$?

if we add the two equations together, you get

$ 7a \plus{} 7b \equal{} 84$

we divide both sides by 7 to get
$ a \plus{} b \equal{} 12$

ans is $ \boxed {12}$


Hi, I was having a little trouble with one of the problems on the NT book.

Well ill just write the part i dont get,...........

If  -n=6(mod 2000) , they say that n= 2000x  + 6  for some positive integer x :o

shouldnt it be that -n= 2000x  + 6  for some integer x??????

 :huh:


does the bouyancy force increase as you increase the depth?

ha ha

it depends. In the real world, yes, since the density of the displaced liquid will go up as you go down in depth. In the realm of physics problems, where density is magically uniform, it does not since bouyant force only depends on the density and volume of the displaced liquid.


AMC
AIME
USA(J)MO
USAMO
email
AMC 10
inequalities

I have two questions:

1) Until the stats come out, how many people got a perfect score? Post here if you somehow know or if you got a perfect score (and which test).
2) Why is the AIME perfect score not as recognized as either the USAMO or the AMC perfect score? We could hypothesize about the dependency on stupid mistakes or whatever, but to put it simply, it's harder to get a perfect score on the AIME than on the AMC. Why do AMC perfect scorers get a page with pictures and AIME perfect scorers get... what do they get anyway?

I'll have to admit, I am a bit biased; I got a perfect score on the AIME II this year (and also AMC 12B for a 300 index!).

Not to brag or anything, but who else got a 15?

Just so you know, this thread does indeed qualify as bragging. But a 15 AIME gets you lots of bragging rights, so it's alright. :P
I believe fourierseries got a perfect index as well, on the AIME I if i recall correctly.

An interesting point. I guess AIME high-scorers should be recognized somehow.

maybe not just perfect, but like 13+ or something, because the number of ppl who get 15's are very few, i believe that there have been like 10 in history of AIME. SamE, you are a genius!

[quote="CatalystOfNostalgia"][quote="cognos599"]maybe not just perfect, but like 13+ or something, because the number of ppl who get 15's are very few, [b]i believe that there have been like 10 in history of AIME.[/b] SamE, you are a genius![/quote]

Where did you come to that conclusion? According to the Community Awards page (which may not exist anymore), there were like six AoPSers with 15 on the AIME in 2005. And there were ~4 last year.[/quote]

I believe that of those four, only one was American (Bohua Zhan). Anyway, there were definitely a whole bunch this year though... Haitao got a perfect too. This year's AIMEs weren't as... um... computation-heavy as last year's AIME 1.

[quote="CatalystOfNostalgia"][quote="cognos599"]maybe not just perfect, but like 13+ or something, because the number of ppl who get 15's are very few, [b]i believe that there have been like 10 in history of AIME.[/b] SamE, you are a genius![/quote]

Where did you come to that conclusion? According to the Community Awards page (which may not exist anymore), there were like six AoPSers with 15 on the AIME in 2005. And there were ~4 last year.[/quote]

alot of those ppl who got perfects were from like taiwan and other countries, as aspiringchesslete said, maybe there were more than 10, there aren't very many.

There are probably even fewer perfect scores on the USAMO, but those get their own little section.

Congrats on your perfect score. I definitely think you're right about the recognition thing.

[quote="SamE"]
2) Why is the AIME perfect score not as recognized as either the USAMO or the AMC perfect score? We could hypothesize about the dependency on stupid mistakes or whatever, but to put it simply, it's harder to get a perfect score on the AIME than on the AMC. Why do AMC perfect scorers get a page with pictures and AIME perfect scorers get... what do they get anyway?

I'll have to admit, I am a bit biased; I got a perfect score on the AIME II this year (and also AMC 12B for a 300 index!).

Not to brag or anything, but who else got a 15?[/quote]

I did, but I didn't even know for sure I had a 15 on the AIME until I got my USAMO winner call and was told then.  (This was back in 1989.)  So, things have at least improved on this front a little.  I never received official notification or a pin or anything (I would have preferred the 1000 dollar scholarships they gave for a year or two), but I did see it in the results book.  And I beat SamV (he was greedy on [url=http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems#Problem_13]this problem[/url], so he got a 14), which is all that really mattered.

Shoot me an email in September and I'll bring this up at the next CAMC meeting.

To the best of my knowledge, this is the first time anyone has ever raised the suggestion about national/regional/top-score awards for the AIME.

I suspect the reason has to do with the original philosophy and intention of the contests.

Ever since the math contests went national in 1952, there has been some kind of national recognition of top scores, but I would have to research how far back we have been recognizing state and regional top scores etc.  This award structure is consistent with the original nature of the Math Contest (now called the AMC 12) as a national contest.  Of course, there is national recognition of the top scores on the USAMO, instituted in 1974.  But I think that since the AIME was created as a "filter" between the AHSME and the USAMO in 1983, it was viewed as a filter, not as a contest with its own reward structure.  Richard's experience is consistent with that viewpoint, I would say.

I (with Richard's help) will make sure this gets on the CAMC agenda in the Fall.  But in the meantime, what would people here view as a proper award system for the AIME.  High score in each state?  High score in the region?  Top x% of scores among all AIMEs?  Ideas and suggestions welcome.

Steve Dunbar
MAA Director, American Mathematics Competitions

I think the top scorers in each region is a good way to award success on the AIME. Each state is not good because there are those states where the number of test takers is really small.

The AIME was considered a filter? I always viewed the AMC10 and 12's and filters since one really doesn't need that much creativity on these if they are mildly experienced and all you need is speed and lack of silly mistakes. I was always surprised that the AMC results were even included in USAMO selection because they are definitely different degrees of difficulty (aliteration not intended :P). I think school AMC prizes/certificates are important because they encourage those new to problem solving to continue. The AIME, I think, is the much more serious contest for those, in say, the top 2000 or so of qualifiers for it. I think perfect scorers should get a "grand" prize, such as a neat looking trophy, and those above or at scores of 12 or 13 a "first" prize with a small medal perhaps. I don't think small "encouraging" prizes are needed because most people who atleast qualify for AIME are strong in math and will continue.

I see that the AIME statistics are now posted at http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2008-aa/08-AIMEstats.shtml. This year, there were eight perfect scores (wow!), seven male and one female.

I do have one question, however--are these statistics official? I was looking through the AMC and AIME stats, and there seem to be some inconsistencies.

[quote]are these statistics official? I was looking at the AMC and AIME stats for my state, and there seem to be some inconsistencies.[/quote]

Since they appear on our website, compiled using our item-analysis programs, with the complete, unabridged AIME data, I would call them official.

Can you be specific about the inconsistencies?

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="AMCDirector"]Can you be specific about the inconsistencies?[/quote]
I've sent a private message. I guess specifics would in fact help. Thanks.

[quote="AMCDirector"]But I think that since the AIME was created as a "filter" between the AHSME and the USAMO in 1983, it was viewed as a filter, not as a contest with its own reward structure. [/quote]

I always consider AMC10/12 is a filter for AIME, and AIME is "real thing",  the training ground for the USAMO. it's like a hurdle before one gets into the more serious stuffs like functional equations, inequalities, etc.

I would agree that the top score in each region would be a good thing to give awards for--or perhaps the top male and female score in each region, as was done with the scholarships.  Perfect scores should also be publicized in some way, at a minimum with pictures on the website like AMC10/12 perfect scorers get.

Also, one inconsistency I've noticed on the website stats is that the overall stats give one perfect score by a 10th grader, but the stats for international students give two.  Do the "overall" stats include these international scores?

the top score in our state was 14 =( (obviously not me. It was jb05)

[quote="cnola"][quote="Anaxerzia"]I think perfect scorers should get a "grand" prize, such as a neat looking trophy.[/quote]
good idea.. keep the trophy small (15 cm?), but with distinctive mathematical feature, probably klein-bottle,  icosahedron, or a bust of one of the prominent mathematicians.[/quote]

Exactly my thoughts: not flashy, but memorable and worthy :)

[quote="Anaxerzia"]Exactly my thoughts: not flashy, but memorable and worthy :)[/quote]Flashy trophies are generally a result of event organizers trying to inflate the importance of their event. They are unnecessary.

and they could also give out plaques to maybe 2nd in the state?
and then a medal for 3rd

[quote="hello.world"]This may not be as applicable to the AIME as to the AMC, but what about top performing schools?[/quote]

I believe they do get prizes. (was it a book?)

[quote]Flashy trophies are generally a result of event organizers trying to inflate the importance of their event. They are unnecessary.[/quote]

When I read this, I immediately though of State Mu Alpha Theta.  They gave out at least 500 trophies to approximately 1000 students from all levels of high school math, and they were definitely what you could call flashy.  Also, the sheer number of trophies awarded devalued each one individually.

I think that only giving out trophies for 12+ AIME scores would be a good idea, because it would keep the value of actually having a trophy high.  Perhaps, a plaque on the trophy could list the person's AIME score (12,13,14, or 15) so that a 15 would be more significant than a 12.

Also, having pictures of AIME perfect scores just like for the AMC 10 and 12 seems like a good idea.

[quote="cnola"][quote="Anaxerzia"]I think perfect scorers should get a "grand" prize, such as a neat looking trophy.[/quote]
good idea.. keep the trophy small (15 cm?), but with distinctive mathematical feature, probably klein-bottle,  icosahedron, or a bust of one of the prominent mathematicians.[/quote]

Definitely a bust of one of the prominent mathematicians.

[quote="cognos599"]is the 8 perfects. just US or Inernational?[/quote]
well I found 6 scorers on the official website...3 internationals..1 from ohio..1 from colorado....1 from virginia...the other 2 I couldn't open were Texas and Virgin Islands...so I guess the other 2 are from there.... :lol:

The Texas link was wrong (just replace other state abbreviations with TX), but they don't have any perfect scorers. I found one from Ontario, but there's still one missing.

What is more confusing is that there are 2 sophomores who got perfect scores listed in the international section, but only one overall. If we throw out internationals, then we're missing 4, and we still need to find a sophomore.

This has been more interesting than I thought it would be...

[quote="SamE"]
2) Why is the AIME perfect score not as recognized as either the USAMO or the AMC perfect score? We could hypothesize about the dependency on stupid mistakes or whatever, but to put it simply, it's harder to get a perfect score on the AIME than on the AMC. Why do AMC perfect scorers get a page with pictures and AIME perfect scorers get... what do they get anyway?
[/quote]

Because to put it somewhat bluntly, getting a perfect score on the AIME is the ultimate "that's really nice and really hard, but uhh, who cares".

I think recognizing AMC perfect scorers is a nice thing. It is definitely a reachable goal for most talented math students who work hard to get a perfect score on the AMC. Also, since everybody theoretically is allowed to take the AMC, it's kind of a nice intermediate "ultimate goal".

On the other hand, if you're getting 15 on the AIME, presumably you'll be doing well on the USAMO. If not ... you should get better at USAMO since you clearly have the ability. I don't think it's worthwhile to be drawing attention to 15s on the AIME when people who are good enough to get 10+ should really just be putting the AIME aside and focusing on the next level.

Also, I think it's reasonable to refine your AMC test taking ability to the point where you can expect to get a perfect, so it's not a terribly huge waste of time (though I do think it's a waste of time, and this is not pure sour grapes as I have actually gotten a perfect AMC12, albeit only once). On the other hand, I think it's pretty hard to get to the point where you can expect 15 on the AIME. There's a lot of variables involved there that put together are very hard to prepare for, and the overall potential range of difficulty is just high enough that the incremental cost of going from being able to get 13+ every time to being able to get 15 almost every time (or even like 75% of the time) is really really steep.

Lastly, ironically, a perfect score on the AIME is possibly the least good indicator of skill at that test of the three perfect scores. Kind of like what I said above, you can get to a point where you can expect to get 150 on the AMC, and it's a pretty good indicator that you are quite good at math up to that level. A 150 is way better than a 138 on the AMC ... it's not even close. Obviously there's no such thing as a lucky 42 on the USAMO and a 42 is way better than a 35, and a 35 is way way better than a 28. But with the AIME, a 15 really isn't way better than a 13, at least not in my opinion. There's a reason why the USAMO winners almost always do really well on the AMC but you frequently get strange cases who never even get above 11 on the AIME though they still manage to make top 12 every year.

AOPS definitely takes it seriously. They always say whether an instructor for an online course has gotten a perfect AIME.


Are there infinitely many pairs $ (p, n) \in \mathbf{N}^2$ such that both $ p$ and $ p^n \plus{} n^p$ are prime?

I first posted the problem in "unsolved" board, thinking it's a "regular" problem, but now I know it's an open one. Moderators can delete the old topic ( http://www.mathlinks.ro/viewtopic.php?p=1334906 )

May be useful: (from the old topic)

[quote="Akashnil"]
Here are solutions for $ p < 10, n < 100$ from Mathematica

$ (2,\{1,3,9,15,21,33\}),(3,\{2,56\}),(5,24),(7,54)$[/quote]

[quote="djuro"]Are there infinitely many pairs $ (p, n) \in \mathbf{N}^2$ such that both $ p$ and $ p^n \plus{} n^p$ are prime?[/quote]
FYI, you can extract the list of known primes of the above form from the [url=http://www.leyland.vispa.com/numth/primes/xyyx.htm]list of known prime Leyland numbers[/url].


algebra
polynomial
calculus
calculus computations

Find $ e^{tA}$ where
$ A\equal{}$\[ \left(
\begin{array}{ccc}
1 & 1\\
2 & 2 \end{array}
\right)\]

Also Find $ e^{tA}$ where
$ A\equal{}$\[ \left(
\begin{array}{ccc}
\lambda & 1\\
0 & \lambda \end{array}
\right)\]

[hide="the first one"]
The characteristic polynomial for $ A$ is $ \lambda(\lambda \minus{}3)$. $ A$ squashes $ (1,\minus{}1)$ and scales $ (1,2)$ by 3. So under $ A^n$ ($ n$ positive), $ (1,\minus{}1)\rightarrow (0,0)$ and $ (1,2)\rightarrow (3^n,2\cdot 3^n)$, so that $ (1,0)\rightarrow (3^{n\minus{}1},2\cdot 3^{n\minus{}1})$ and $ (0,1)\rightarrow (3^{n\minus{}1},2\cdot 3^{n\minus{}1})$. So
\[ A^n\equal{}\left(
\begin{array}{cc}
3^{n\minus{}1}&3^{n\minus{}1}\\
2\cdot 3^{n\minus{}1}&2\cdot 3^{n\minus{}1}\\
\end{array}\right)\]
Raising $ tA$ to the $ n$th power multiplies the entries by $ t^n$. Now plug into the exponential series to get
\[ e^{tA}\equal{}I\plus{}\sum_{n\equal{}1}^\infty \frac{(tA)^n}{n!}\]
Now by comparing entries in $ (tA)^n$ to terms in the exponential series for real numbers, we get
\[ \sum_{n\equal{}1}^\infty \frac{3(tA)^n}{n!}\equal{}\left(
\begin{array}{cc}
e^{3t}\minus{}1&e^{3t}\minus{}1\\
2e^{3t}\minus{}2&2e^{3t}\minus{}2\\
\end{array}\right)\]
\[ \Rightarrow e^{tA}\equal{}\left(
\begin{array}{cc}
\frac{e^{3t}\plus{}2}{3}&\frac{e^{3t}\minus{}1}{3}\\
\ &\ \\
\frac{2e^{3t}\minus{}2}{3}&\frac{2e^{3t}\plus{}1}{3}\\
\end{array}\right)\]
[/hide]

Countdownmath, do you read the answers to your own [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177357]previous[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177355]topics[/url]?

o sorry, i thought it was taken out or something... Thank you very much tho..


geometry
rectangle
AoPSwiki
area of a triangle
Heron\u0027s formula

find the area of the triangle formed by the points (7,8) (12,5) (-4,3)

Uhh, did you post this at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=291581??

ya i accidently posted it in middle school. i meant to post it here... sorry

[hide="hint"]shoelace formula is your friend.[/hide]

Or if you don't know shoelace, this will help you understand:

[hide="Hint 1"]Plot the 3 points. Calculating the area directly won't help much. Find another way.[/hide]
[hide="Hint 2"]Find the smallest rectangle (with sides that are horizontal and vertical) such that all three of these points lie on the sides of the rectangle.[/hide]
[hide="Hint 3"]Subtract the extra regions in the rectangle that is not part of the triangle.[/hide]

Alternatively, you can AoPSWiki: Shoelace Theorem

could draw perpendiculars from each of the points to find 3 right angled triangles outside the shape, then use Pythagoras's Theorem to find the 3 side lengths. after that, you could just use heron's formula and that will give you the area.
Alternatively, could draw the perpendiculars, then find the area of the large rectangle and subtract the three triangle lengths.


We define a number 44...48...89 like this: First, we insert the number 48 between the two digits of 49. Next, 48 is inserted between the digits 4 and 8 of the resulting number. Finally, we repeat this second step over and over again. Is it true that every number obtained this way is a perfect square?


MATHCOUNTS calculus probability AMC AIME trigonometry USA(J)MO

I have this question about Mathcounts 2009. If Neal Wu won the competition won, then does he get to compete again. Even though Bobby Shen got second, does he still get to compete in 2009?

I think so... as long as you're still a middle schooler.

Don't you mean Darryl Wu? Neal Wu won in 2005.

I think not. One lives in Louisiana and the other lives in Washington.

well, don't get discouraged like that. After someone wins a competition they normally slow down on progress. If you increase your rate you can always beat anyone you want :)

[quote="Wickedestjr"][quote="AIME15"][quote="Wickedestjr"]That stinks, I probably have no chance of winning nat's now, because I am in the same grade as Neal Wu and as long as I am in middle school, he will be too.[/quote]

:P

There's still a TON of people that are really good out there, like 6th and 7th graders who finished in top 12 at nats last year.

Anyway, only a few of us here have a chance of winning nats, and I'm not included :P[/quote]

I have a slight chance. I still have two more years, and I almost made it to Nationals last year. I got 7th, but that is also because the people I competed against are doing college topics and high school topics now.[/quote]

college topics and high school topics don't help in MATHCOUNTS. :wink:

[quote="tinytim"][/quote][quote="Wickedestjr"][quote="AIME15"][quote="Wickedestjr"]That stinks, I probably have no chance of winning nat's now, because I am in the same grade as Neal Wu and as long as I am in middle school, he will be too.[/quote]

:P

There's still a TON of people that are really good out there, like 6th and 7th graders who finished in top 12 at nats last year.

Anyway, only a few of us here have a chance of winning nats, and I'm not included :P[/quote]

I have a slight chance. I still have two more years, and I almost made it to Nationals last year. I got 7th, but that is also because the people I competed against are doing college topics and high school topics now.[/quote][quote="tinytim"]

college topics and high school topics don't help in MATHCOUNTS. :wink:[/quote]

Yeah they do, if you master high school Algebra, then you will do really good at the algebra questions in MC.

[quote="isabella2296"]All the Algebra questions I've seen in MathCounts only require the stuff you learn in an Algebra I course. MathCounts is meant for middle school topics. I'm learning Calculus, but this sure as heck won't help me in MathCounts.[/quote]
lol it sometimes helped ME with the optimization problems (maximum/minimum)

My MC coach let me take a look at his copy of Introduction to Counting and Probability, and the hard problems in it helped me with the easy questions, but I guess that people have different opinions.

Let's put it this way. In most AIME problems, which are harder than MC, each question requires MC tricks, but they are just a little bit harder, so you learn how to solve harder problems, and then the easier ones are just simple.

[quote="gauss1181"][quote="AIME_is_hard"]Neal Wu's brother will pwn this year...hes in 6th at my school..[/quote]
6th in just SCHOOL? Really? Why do you think he'd be that hardcore? Or is it just that your school is very strong at math?[/quote]

I think AIME_is_hard means that he is in 6th grade at his school... not that he is 6th place in the school.

I used to do MC with absoulutely no background except for very basic stuff. After I learned trig and harder algebra i could the problems much much faster (speed was one of the main reasons i didn't make nats and the other was i wasn't in middle school :P)

I studied trig and advanced number theory this summer, and on practice tests my scores were about 4 points better.  Unfortunately, I'm a freshman now so I can't do MC anymore.  Time to beef up the USAMO stuff...

[quote="Wickedestjr"]I didn't have time to do any math over the summer, so I am the same old dumb person that I used to be.  :rotfl:[/quote]
Well but I did, and that was to my advantage. Right now my math time is less...

Neal Wu got top 4 in 2006... if you have not competed for at least three years AND are in middle schooll you can compete.

So basically you just have to be a middle schooler.

No, if you compete more than three times(hold back a grade is TRUE), then I think your team is banned for life.

[quote="AIME15"][quote="Wickedestjr"]That stinks, I probably have no chance of winning nat's now, because I am in the same grade as Neal Wu and as long as I am in middle school, he will be too.[/quote]

:P

There's still a TON of people that are really good out there, like 6th and 7th graders who finished in top 12 at nats last year.

Anyway, only a few of us here have a chance of winning nats, and I'm not included :P[/quote]
lol aime, our chapter division is not very good :) my school was new to mc but we got 2nd LOL

wait hongsquare where do you live?  what school?

Would if you were somebody that just got accepted to a school that does MC, and you were an eighth grader. Would you be able to compete three years, and keep getting held back, I mean, that seems like a possibility, but it also seems unfair.

There is in fact, I believe, an age limit for IMO... like 20 or something

Well if somebody at age 1 could read what you just said, I will give that one-year-old HUGE APPLAUSE.

Anyways, that (hypothetical) 1-year-old might not understand a thing said in that TV show.

[quote="math154"]Well, all you have to do is become Terence Tao. So pretty much watch Sesame Street when you're 1 and recognize numbers.

EDIT:
Source: Wikipedia on "Terence Tao"[/quote]

Oh my god, that guy got 760 on SAT math when he was 8 years old...got bronze at IMO at age 10, silver at 11 and gold at 12....that guy is SCARYY

[quote="AIME15"]I don't think people would flunk intentionally just for MATHCOUNTS...

Going that way, people could keep getting held back to make red MOSP every year...or go to IMO like 12 times :P[/quote]

Think about it though. If you held back in the eighth grade twice in a row, then you would have plenty of time for the IMO.


modular arithmetic
number theory
relatively prime

Consider the triangular array of numbers with 0,1,2,3... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row.  Rows 1 through 6 are shown.
                                0
                             1    1
                          2    2    2
                       3    4     4   3
                     4   7     8    7   4
                   5  11   15   15  11  5
Let $f(n)$ denote the sum of the numbers in row $n$.  What is the remainder when $f(100)$ is divided by 100?
The numbers are supposed to be in a triangular form, kind of like pascals triangle.
You can post your answer if you want, but please hide it. Also which chapter in aops book1 or perhaps book2  should I read for this kind of problem?

Thanks  :)

[hide]Find a pattern for the sum of the numbers in each row, and then try to find a way of using Fermat's Little Theorem.[/hide]

[hide="Try it yourself before you see the solution"]After some experimentation, I found that $f(n)+2=2^{n+1}$. So $f(100)=2^{101}-2$. Were looking at things in $\pmod{100}$, so:
$f(100) = 2^{101}-2$
$f(100)\equiv 2^{101}-2\pmod{100}$
From Fermat's little Theorem with Euler's generalization,
$2^{\phi(25)}\equiv 1\pmod{25}$
$2^{20}\equiv 1 \pmod{25}$.
$2^{100}\equiv 1\pmod{25}$
$2^{99}\equiv 13\pmod{25}$
This implies tha $2^{99}= 13+25j$ for some integer $j$. Multiplying this equation by 4 and modding out everything in $\pmod{100}$,
$2^{101}\equiv 52\pmod{100}$
$2^{101}-2\equiv 50\pmod{100}$
Substituting this to our other equation,
$f(100)\equiv 2^{101}-2\equiv 50\pmod{100}$

So the remainder is [b]50[/b].[/hide]

I'd suggest AoPS Volume 2, Chapter 23: Number Theory.

YOur solution has a flaw, vishalarul.

[hide="Reason"]You can only use Fermat's Little Theorem, if a^p is congruent to a(modp) iff (a,p)=1.

Also, it makes no sense that $2^{40}$ is congruent to 1(mod 100)[/hide]

[hide="Hint"]Try using mod 25, instead of mod 100[/hide]

Yeah, I guess you are right. For the theorem to work, 2 needs to be relatively prime to 100.  :oops: I'll try mod 25, then, and edit my solution.

vishalarul, your solution still has a flaw, I believe.

[hide]The $\phi{25}\neq24$.

Try to find what it is, and then you will be in better shape.

You should also combine in a way that 2^100 is congruent to 0(mod4), then you may figure it out.[/hide]

Your mod work is wrong.  You obtained $2^{102}-8\equiv 96\pmod{100}$  :wink:

[hide]By experimentation, we see that $f(n)=2^{n}-2$.  Hence $f(100)=2^{100}-2$

We know that

$2^{\phi{25}}=2^{20}\equiv1\pmod{25}$.

Also, this means that $2^{100}\equiv1\pmod{25}$.

Plus, we know that $2^{100}\equiv0\pmod{4}$

So combining these two, we get that

$2^{100}\equiv76\pmod{100}$.

Hence $f(100)=2^{100}-2\equiv74\pmod{100}$

$74$[/hide]

[quote="myc"]Is there a way to figure out the answer without knowing what  $\phi{25}$ means.  :)[/quote]

If you wanted to calculate it out, you could verify [b]experimentally[/b] that

$2^{20}\equiv 1 \bmod 25$

Do you want to?   :D

Of course not!  That is why we have the $\varphi$ function.  Basically, $\varphi(n)$ gives the number of positive integers less than and relatively prime to $n$.  For example, $\varphi(4) = 2$ because $1, 3 < 4$ are relatively prime to $4$, and $\varphi(p) = p-1$ for primes $p$ since every integer $1, 2, 3, ... p-1$ is relatively prime $p$.

There is a theorem which states that, for relatively prime integers $a, m$, we have

$a^{ \varphi(m) }\equiv 1 \bmod m$

Which is the theorem people are using in their solutions here.

Edit:  Actually, it is not so difficult experimentally.

$2^{10}= 1024 \equiv-1 \bmod 25$.  :)

[quote="squashed"][quote="boxedexe"][hide="Equation"]$f(n)=2^{n}-2$  :wink: [/hide]
Might as well post solution:
[hide="Solution"]$2^{100}-2\equiv (24)^{10}-2\equiv 24^{4}-2\equiv 74\pmod{100}$
[/hide][/quote]

Can you explain how you went from $2^{100}$ to $24^{10}$?[/quote]
$2^{100}=(2^{10})^{10}=1024^{10}\equiv24^{10}\pmod{100}$

[quote="myc"]can you explain how you get from $24^{10}-2=24^{4}-2(mod 100)$   ?
also how would you calculate 24^4 without a calculator?[/quote]

Note that $24 \equiv-1 \bmod 25$ and $24 \equiv 0 \bmod 4$.  Hence

$24^{10}\equiv 1 \bmod 25$
$24^{10}\equiv 0 \bmod 4$

Ergo,

$24^{10}\equiv 76 \bmod 100$

so if $n\equiv b \bmod c$ then  $n^{a}\equiv b^{a}\bmod c$ .
and since -1*-1=1 it all works out.   I get it now!  yay thanks a lot everyone.   :P

[quote="myc"]so if $n\equiv b \bmod c$ then  $n^{a}\equiv b^{a}\bmod c$ .[/quote]

Yes.  More generally, if $a \equiv b \bmod m$ and $c \equiv d \bmod m$ then $ac \equiv bd \bmod m$.


Let A be an invertible nxn mx. Use the Gram-Schmidt process to show that there is an nxn upper triangular mx B and an orthogonal nxn mx U with AB=U.

This result is basically a description of what the Gram-Schmidt process does. We start with a independent set (the columns of the invertible matrix $A$.) We're going to produce an orthonormal set; the orthonormal set will be the columns of the matrix $U.$ Saying $U$ has orthonormal columns is precisely equivalent to saying $U$ is an orthogonal matrix.

The first element of the orthogonal set is simply some scalar multiple of the first element of the original independent set. The way to pick out a scalar multiple of the left column of $A$ is to right-multiply $A$ by a column vector whose only nozero entry is the top entry.

Then the second vector of the orthogonal set (the second column of $U.$) is obtained, ultimately, as a linear combination of the first two columns of $A.$ We get that by right multiplying $A$ by a column for which only the first two entries is nonzero. An the third column of $U$ is a linear combination of the first three columns of $A,$ and so on.

If we simply assemble together in order all the column vectors that we right multiplied $A$ by to get the orthonormal set that is the columns of $U,$ then we have an upper triangular matrix. We then proceed to call that matrix $B.$

That make perfert sence. Thanks!

[quote="Kent Merryfield"]This result is basically a description of what the Gram-Schmidt process does. We start with a independent set (the columns of the invertible matrix $A$.) We're going to produce an orthonormal set; the orthonormal set will be the columns of the matrix $U.$ Saying $U$ has orthonormal columns is precisely equivalent to saying $U$ is an orthogonal matrix.

The first element of the orthogonal set is simply some scalar multiple of the first element of the original independent set. The way to pick out a scalar multiple of the left column of $A$ is to right-multiply $A$ by a column vector whose only nozero entry is the top two entries.

Then the second vector of the orthogonal set (the second column of $U.$) is obtained, ultimately, as a linear combination of the first two columns of $A.$ We get that by right multiplying $A$ by a column for which only the first two entries is nonzero. An the third column of $U$ is a linear combination of the first three columns of $A,$ and so on.

If we simply assemble together in order all the column vectors that we right multiplied $A$ by to get the orthonormal set that is the columns of $U,$ then we have an upper triangular matrix. We then proceed to call that matrix $B.$[/quote]