Does anyone want to suggest some good book that has lots of good challenging problems offered to solve in the area of gemoetry particularly circle/tangent etc. WHich book do smart students normally follow ?

I strongly recommend "Geometry Revisited" by Samuel Grietzer(dont trust my spelling).  It has many important theorems on euclidean geometry and many interesting practice problems (Some easy and some hard).  Just by reading it and doing the exercises makes me improve a lot on geometry (I manage to solve all three geometry problems in USAMTS this year, while before I skip all geometry problems on AIME and AMC).  you can check it out at some college library or just buy it from MAA.

[quote="PenguinIntegral"]A new one was just released free on the web!
[url=http://math.mit.edu/~kedlaya/geometryunbound]http://math.mit.edu/~kedlaya/geometryunbound[/url][/quote]
Thanks for showing us this site. It is a nice book.

[quote="PenguinIntegral"][quote="chess64"]I think that's pretty difficult... I like Challenging Problems in Geometry.[/quote]
Not really. You should see Excalabur...[/quote]

Sorry, I was talking about Geometry Revisited. Geometry Unbound is also for Olympiads anyway, though. What's Excalabur? :huh:

[quote="PenguinIntegral"][quote="chess64"]I think that's pretty difficult... I like Challenging Problems in Geometry.[/quote]
Not really. You should see Excalabur...[/quote]

The book is aimed at Olympiad level students.  I consider that to be quite a high difficulty level...especially when we're in the Intermediate Forum.

I'll have to second [i]Challenging Problems in Geometry[/i] (CPIG).  Also, as always, AoPS Volume 1 is totally awesome at AMC/AIME geometry (excluding trig which is covered in AoPS V2).  CPIG is at a higher difficulty level than AoPS V1.

If you're looking for something at the Olympiad level, [i]Geometry Unbound[/i] looks like an indispensable resource (especially considering that it is free!).


calculus
integration
number theory
prime numbers

What is the largest positive integer that is not the sum of a positive integral multiple of 42 and a positive composite integer?

[hide="Solution"]
We seek a positive integer $n$ such that no number $k \equiv n \bmod 42, k < n$ is composite.

We then want a sequence of primes congruent to each other $\bmod 42$, with the first prime necessarily less than 42, or else the smallest prime in the sequence may be expressible as $42 +$ a composite number.  The sequence obviously cannot start $p_1 = 2, 3$, so $p_1 = 5$ produces a sequence $5, 47, 89, 131, 173, 215$ until we hit a composite number, so our first solution is $n = 215$.

Any other prime $\neq 5$ will be $\equiv 1, 2, 3, 4 \bmod 5$, and so any sequence of the form $p_1 + 42k$ will be $\equiv 0 \bmod 5$ after $2, 4, 1, 3$ steps, respectively, so the only possible larger value of $n$ will start with a prime $p \equiv 2 \bmod 5$, and the prime must be $< 42$.  This gives us $p = 37, 17$.  Setting $p_1 = 37$ produces $37, 79, 121$.  Setting $p_1 = 17$ produces $17, 59, 101, 143$.

Hence, our answer is $\boxed{215}$.
[/hide]

[hide]
not so nice but we can look at every possible case:
looking at multiples of 42 plus a prime...we see that all of prime numbers except 5 produce a composite number either with 42 or 84. so we have to consider a multiple of 42 plus 5. checking the multiples, we see that 5*42+5 is the desired number.so $\boxed{215}$[/hide]

edit: looks like t0rajir0u got it before me...

[hide="DoNotClick"]We look at some number and turn it into $42k + n$, where $k, n > 0$. Now, if we make $n$ as small as possible, we want $n, n + 42, n + 84, \ldots n + 42(k-1)$ to be prime. So, we can look at all possible "small" values of $n$ - from 1 to 42. We know that $n$ must be prime so that takes out a bunch of possible $n$. The possible remaining $n$ are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41.
Moving on, we want $n + 42$ to be prime. 5 is the smallest prime not in the factorization of 42, so it would be easiest to find $n$ to make $n + 42$ composite. We see this is composite if $n$ ends in 3. The largest remaining number ending in 3 is 23, we have to make sure that there is a number greater than it which is not the sum of a positive multiple of 42 and some composite number before we can cross it off our list. For 23, the largest possible number is 65. We can see that for $n = 5$, we have that $5, 27, 89, 131, 173$ are all prime, so the largest number for $n = 5$ is 215. So, we can cross off numbers ending in 3. We get 1, 2, 5, 7, 11, 17, 19, 29, 31, 37, 41.
If $n + 84$ is prime, $n$ cannot end in 1. Testing the highest $n$ ending in 1, we see it gives 125, which is less than 215. So we cross off numbers ending in 1 to get 2, 5, 7, 17, 19, 29, 37.
If $n + 126$ is prime, $n$ cannot end in 9. For $n = 29$, we get 155, which is less than 215. So we cross off number ending in 9 to get 2, 5, 7, 17, 37.
If $n + 168$ is prime, $n$ cannot end in 7. We check $n = 37$ and get 205, which is less than 215. So we cross off numbers ending in 7 to get 2, 5
From here we just evaluate them; 2 yields 2 while 5 yields 215. So 215 is the desired answer.[/hide]

[hide] In the following table we have listed the first five multiples of $ 42$ and then added 5 to each.

42        47
84        89
126     131
168     173
210     215

This shows that any integer $ n > 215$ can be represented as the sum of a multiple of $ 42$ and a composite number. Why? If $ n$ ends in an even digit, we take the multiple of $ 42$ in the above table that ends in the same digit and add an appropriate multiple of $ 10$. Likewise, if $ n$ ends in an odd digit, we take the corresponding multiple of $ 42$ and add an appropriate multiple of $ 5$. Now it's a simple matter to check that $ 215$ cannot be represented in the desired way.
[/hide]


MATHCOUNTS email calculus AMC USA(J)MO USAMO Putnam

[Admin edit] Some other users have submitted introductions in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=20070. We don't expect those users to re-introduce themselves, so you might want to read that topic too ;)

MATHCOUNTS forum co-moderator yif man12 had the good idea of setting up a place where forum participants could introduce themselves to one another, and that's an especially good idea now soon after the AoPS-MathLinks merger. I'll start out.

I'm "tokenadult," a screen name I made up for this forum. I joined at age 45 and am now 46. I am the homeschooling father of four children and occupy myself mostly with writing about education reform. I grew up in Minnesota and Wisconsin. My wife is from Taiwan, which is where we met. I studied Chinese language as an undergraduate subject and worked for quite a few years as a language teacher, translator, or Chinese-English interpreter. I also have a degree in law and worked for a time as an immigration lawyer. I do NOT have any experience with university-level mathematics; I came to this site because of my oldest son's interest in mathematics, and because I am coach of our local homeschooling group's math team. The site management has kindly allowed me to become the moderator of the College forum here and a co-moderator of the Getting Started forum.

Just for fun, here is my AoPS-MathLinks "genealogy." I am the father of A+MATH, who joined the site soon after I did. For years I have corresponded by email (never personal email, I think, but only list email) with the mother of Alison, a recent IMO gold medalist. I am on the board of a Minnesota nonprofit organization with the paternal grandmother of Sly Si. To the best of my knowledge and belief, any connection I have with any other AoPS-MathLinks forum participant is through the forum itself.

I cheer on (from afar) the Minnesota MATHCOUNTS team every year, and the Minnesota teams in ARML. And of course I root for the United States team each year at IMO.

Please tell us about yourself. Hey, we could break a record here for thread with the most MEANINGFUL, ON-TOPIC replies. That would be cool. Every member can participate here.

joml88:  the jo comes from my first name, the m and the l are my middle and last name initials and 88 is my favorite number(8) twice and also my birthyear.  I joined at age 15 and am still 15.  I will be 16 in a few months.  I was born in Maryland.  I then moved to Florida during 3rd grade.  I stayed there until the end of middle school when I moved to South Carolina where I currently reside.  So right now I am a junior at Summerville High School in Summerville, SC.  I didn't start doing competition math until high school.  Right now I am studying for Calculus BC.  I am also training(studying) to get into the USAMO at least once in high school.

Outside of math I run a lot.  I am on the cross country and track teams at my high school.  I m not the fastest runner but that doesn't bother me too much since I know that I am doing the best that I can.  After school, running, and math I don't have much time for anything else.

As for my AoPS-Mathlinks "genealogy",  I am not really linked to too many members.  Gauss202 is is one of the coaches of the SC All-State/ARML team which I was a part of this year(thanks to him).  There are a few members of the all-state team on this forum that I know(trickdacy and bobofbobs).  Genius1601 goes to my school but he isn't a very active member.

Heya. My name is Lun Mu, I'm a fifteen year old girl from New Jersey. I go to AAST, and I know there are a lot of members on this forum (not necessarily active though.. lol) from my school.. I'm going into tenth grade, and will be taking BC+ Calculus.

Um.. well, my screenname for this forum is undoubtedly simply my name.

I was born in China, then, moved to Ottawa, Canada when I was three and a half. I stayed in Ottawa for seven years, moving to Pearl River, New York when I was ten after my father got a job transfer. Then, five months later, we moved to Ramsey, New Jersey, where I currently live.

Outside of math and school, I play the piano (but then, that's not very surprising, most Chinese kids play piano..). I also play Go (aka Wei Qi), a Oriental board game. (I've been playing since age seven). I love to read, write, umm... yea.

As for genealogy... I know the people on here who are from my school. And I know MithsApprentice pretty well now.. (met him last December...). I am still hoping to get to know other people on a first name basis too.. =]

BHorseMath: The B stands for book since I love to read and spend a lot of time doing so.  The Horse is because I am totally horse crazy.  The math well isn't it obvious?  I was 13 years old when I joined and am still 13 but I will be 14 in a few months.  I was born in Germany and moved to the U.S> when I was 8 months old.  I lived in virginia for 4 years before moving to Texas 9 1/2 years ago.  I am currently in 8th grade taking Algebra 1.

I didn't start getting into math until last year when I first joined the math team at my school I didn't do very well but I hope to get better.

Outside of school I love to read as I said and I play both piano and sax.

as for Aops Genealogy I know aalindsey100 personaly and am best friends with her I also have met rep123max.  I also know several other members who are not very active.

Real name: Feiqi Jiang

I was born in China and moved to US 3 years ago. I live in Massachusetts now. I am going to 9th grade in 2 weeks. I always liked math but I never participated in any math competition until 8th grade. My main goal next year is also qualify for USAMO. I also like chess and soccer.

I'm Peter VDD, I'm going to university this year (doing both maths and civil engineer studies). I'm too old for IMO now :( and since all the stuff like Putnam is for Americans only, I guess my problem-solving carreer is on hold for a while...

Nevertheless I am on this site all day :D mainly in the problem solving section. I came here as a real n00b somewhat a year ago, but in the mean time I've developped my skills to a more-than-decent level. (so being here really helps newbies!) ;)

Fellow math students that want to add me to their msn: vindicate69@hotmail.com

Next to solving problems I also like making problems (yes to be taken litteraly too :twisted: ) and I'm also writing a few documents on some easy subjects. (just to train my teaching skills)

My main goal for this year will be succeeding in my heavy study combination (maths+engie), I just hope my presence on this site won't have to suffer for it :) but it is very well possible :(

Oh yes and I just love abusing smileys :D :maybe: :cool:

Thanks for the credit Tokenadult.

*adjusts microphone*

Um, hi. I'm yif man12. The first part of my login name here is just the first 3 letters of my name. The "man" part is, well, because I'm male. 12 is my lucky number, so I guess I put that in as well. I'm 14 now, 13 when I joined. I live in Connecticut. I was born in China but moved to the US when I was 5. Since then, I've moved a ton but my family has settled down for the last 5 years. I really started taking math seriously at the end of 7th grade. That was mainly because I qualified for Mathcounts Nationals. I had participated in some math contests before then, like some Math Olympiad stuff and the AMC 8, but hadn't really cared too much about how I did. This site has really gotten me more interested in math. I'll be going into 9th grade in the fall. Some of the things I also enjoy doing are playing baseball, watching TV, playing video games, reading, running and joining in some little sports games with my friends. (Like wiffleball, touch football, basketball, etc.) Some of the things I don't especially enjoy but still do often are getting yellled at, making stupid mistakes, breaking or ruining things in my house and losing bets to JSRosen3. Some people also say I'm not serious enough, but I think they're just plain crazy.

i'm Peter
i'm a freshman in high school, taking Trig/Analysis (trig and precalc)
live in CA, in Bay Area
play cello, like reading, playing games...
i have met some people on this forum in person, but won't list them because i don't know all their usernames

Hi! My name is Emily, adn I joined AoPS when I was 13, and am now 14. I live in Indiana, and have all my life. The whole 'mathfiend' thing was my dad's idea, cuz I absolutely LOVE maths.

I am a freshman in high school now (woohoo!!!), and hope to participate in ARML this year. I will be helping coach the MathCounts team at the nearby middle school, where I attended last year. I found AoPS through MathCounts which I officially participated in in 8th grade, but was a sub in 7th.

Outside of maths, I enjoy playing soccer (o yea!!!). Our season started just this past week, and we opened with a win! yay! I also enjoy reading, and spending time on the internet chatting, and singing.

I don't personally kno anyone from the AoPS/MathLinks site, but have made some fairly close friendships via Instant Messenger.  :lol:

*grabs microphone away from yif man12*

Hey! I'm Jeff Rosen, and in 19 days I will be a 10th grade student at Choate Rosemary Hall, a high school in Connecticut which yif man12, a good friend of mine, will be also be attending. My nicknames include "Jeff the Chef", "Frozen Rosen", "El Jefe", and "JeFrozen", of which you can take your pick. I'm currently taking AP Calculus BC, which I enjoy, but my real love is problem solving. I competed in National Mathcounts twice, once on the same team as yif man12 and Yasha. Sadly, I missed qualifying for USAMO (and freshman MOP) by one question last year, so I'm working extra hard to make it this year. My hobbies include sports, piano (which I've played for 8 years), card games, Bush-bashing, reading, inequalities, and beating yif man12 in all sorts of bets. My goals for next year are many: first and foremost, to qualify for USAMO; second, to answer at least one question on it; third, to convince at least three Republicans that I know to support Kerry; fourth, to tackle Rachmaninov's Prelude in G minor; fifth, to win more Chinese candy from yif man12; and sixth, to live through a daunting courseload that I've set up for myself. Oh, and winning the IMO wouldn't be bad either.

Hmm...This was a good idea!

Hola!  My name is Danny and I just joined this site fairly recently...and it's been fun hanging out with people obsessed with math as much as me.   I used to live in Mass. until I was four and then I came to Pawtucket, RI.  Fun times.  My s/n. SingTheSorrow, comes from the title of my favorite band's recent album....and that band would be Un Fuego Adentro (That's A Fire Inside, for all you non-spanish speakers...long for AFI)

Anyway, I'm fourteen (15 soon) and I'm a sophomore...taking Geometry and hopefully, if my family isn't poor, Pre-Calculus too at my local college.  Um, I began my psycho math-craze about January of this year, and annoyed my teacher for awhile after school.    (Oh yeah, my school is horrible.  We only have ONE AP course, and that's AP Calc)

Um, for non-math interests...I like listening to music ...anything really.  Um, I'm on my school's swim/math/golf (since my school is *ghetto* ...there is absolutely NO interest in golf, so me and my friends just decided to be on the golf team for laughs....and for something on our college application.  You know the drill.  I'm serious.  We ALL get one or two below DOUBLE par.  it's really sad.  We have fun though)/key club, and I might do mock trial.....but that's like uh, crazy for me anyway.  I really like to chat on AIM you know (IM me at KillingInSleep)  and I really like Scrabble.  Before math, I was ALL about Scrabble.  Oh yeah, I love Spanish.  I have a really bad accent, pero lo que sea.

Oh, and I talked to a few people on AIM (Yeah, I'm sure they love me annoying them with stupid math questions) and I've met Julie at three rock concerts!  Oh yeah, fun times!
 :P

I'm not a very interesting person, so I guess this post will be relatively short.

As you can tell from my signature, my name is Simon Rubinstein-Salzedo. I am a math major (surprise, surprise) at the University of California at Santa Barbara's College for Creative Studies. (I will be starting my second year soon.) I also plan to get a minor in music.

I like to do math a lot. I am most interested in analysis and number theory, and I will probably study analysis in graduate school. Unfortunately, I'm not very good at it, so that's something that will have to be fixed soon.

The second most important thing in my life is music. My parents are professional musicians, so they encouraged me to play music when I was young. So I play the piano and the cello. Unfortunately, I'm not very good at music either.

My third hobby is playing chess. I'm rated 1669 USCF, and if I ever want that to increase, I'll have to get to another tournament some day.

I think I used to be smart at one time, but then I went to high school, and now I'm just an ordinary not-very-smart person who likes to study advanced mathematics.

*shakes head*

People are too modest these days.

HI. My name is Jarey (Pronouced Jerry - but spelled differently - don't ask) Wang. I joined this forum when I was 13, and now I am 14 - just turned 14 a few days ago. My lucky number is also 88 because my birthday is August 8th. I am taking Algebra 2 next year in 9th.

I enjoy music, and I play the alto sax. I like tennis, but I don't dislike any particular sport. I used to play soccer, which I still like very much. I go to an obscure little high school, which nobody will ever hear of because we accomplish so little some years. But, life's good overall, and this forum is a great place to learn and pick up knowledge.

I was born in china. I moved to Edmonton Alberta when I aws 4, went to Vancoucer when I was 7, moved back to China for 1/2 year, moved to Madison Wisconsin when I was 8, moved to Wilmington DE when I was 9, and moved to West Chester, PA when I was 12, yup... I can't believe I stayed this long because my family was thinking about moving to North Carolina during 7th.

Hi!

My name is Fred Sala. I am 16 years old and I am in 11th grade in high school. I currently live in Troy, Michigan. I was born in Bucharest, Romania, where I lived until I was 7; afterwards, I lived in Toledo, Ohio for two years, followed by another year in Bucharest. After this I moved to my current residence.

Math is one of my favorite activities, and I have been involved in math competitions for the last few years. Otherwise, I play a lot of video/computer games, play piano, albeit not very well, and play football and basketball. I also enjoy reading. Various driving lessons also take up some of my time.

My goals for the next year simply are to do well in the various classes I'm taking and the competitions I will be involved in. I also hope to enjoy the two years of high school I have remaining without stressing out about the college admissions process.

As for genealogy, this summer I went to Mathcamp and in the process met about 25-30 people that post a lot on AoPS. Naming them would take a long long time!

Greetings, Earthlings.  I'm Keone, formerly from Nevada, now from Hawaii, and soon to be from New Hampshire.  I joined AoPS in the summer of '03, back before there was LaTeX, spoiler, or online classes.  Or, to put it another way, I'm old enough to remember jojobird.

My username, mathfanatic, was coined one summer day when I couldn't think of anything else.  Since then, it's seemed to stick.  I like math, perhaps to the point of fanaticism, and while people laugh about me taking math books to the beach, I don't mind.  Math is my hobby.

In two weeks, I'll head off to New Hampshire to attend Phillips Exeter Academy.  I'll be a lower (sophomore) there.  I like Spanish too, but don't speak it nearly as well as I ought to.

I enjoy music; my favorite bands are Dave Matthews Band and Counting Crows, although I also enjoy the Beatles, the Police, James Taylor, and other such old fare.  I used to play the piano, but stopped; I want to learn to play the drums now.  My other non-math interests are skiing, soccer, running, and waterskiing/windsurfing.

My other accomplishments include developing a [url=http://goodlymaths.0catch.com]math website[/url] and then abandoning it in favor of AP exams; almost being on a Cody Bowl team, and taking the USAMO as an 8th (but not 9th) grader.  I enjoy the AoPS site and am thankful for the community - AoPSers have constantly motivated me and helped me to have fun.  (Thanks, Matt, Neal, Simon, Hann, Adam, Julie, Tom, Mr. R., Mr. C., and all the others.)

I'm...uh...well look 2 ur left and u will see ...
So im a girl who lives in arizona, which has 110 degree weather, so as soon as u step outside--even in a bikini or bathing suit--u sweat. Buckets. Gallons. Olympic pools. Get the point? anyways, i'm 11 turnin 12 next month and going to 7th grade at a middle school with over 600 kids. (HUGE!)
I was seriously born 2 play music. if u knew me when i was 3, u would see me banging on keyboards everywhere. EVERYWHERE. so i've been playing piano for over 6 years. but i've started percussion last year and i luv it =]
but i love 2 listen 2 my ipod, play on my DS, swim, read, watch tv..............and much more.

Hi, everyone!  I'm alexhhmun, which comes directly from my name.  I GREATLY dislike writing about myself, due to the fact that I don't want to appear arrogant, nor as someone untalented.  Well, here I go...(/me is unenthusiastic)

Several of my friends have called me extremely mature for a 14 year old, that is, when I'm not making a complete and total idiot of myself, which I do a [size=150][b]lot[/b][/size] (meenamathgirl, TheHobo, and 1=2 can contest to this).  I GREATLY dislike having to repeat myself and when people lack common courtesy to respond to you when you say "Hi!", even when they're a total stranger.  My biggest pet peeve is when people disrupt me when I'm working.  I know the 3 aforementioned people via national MATHCOUNTS 2007-2008 (this year), as well as many others also through National MC, or For the Win.

Err.. I don't have many hobbies.  I guess I spend my time online, mainly on AoPS, or working on math.  I have a VERY bad self-image, but will be greatly insulted if someone adds on to one of my self-insults.  I like using words most people don't, like "rather", "pet peeve", and "aforementioned" in conversation, but I am constantly shaming myself by misusing them.  I guess that's how you learn, though.  I also like saying "Hi!" and smiling to total strangers.

I made myself sound like a rather dull person, but that's the kind of mood I'm in right now.  There's still 2 weeks until AMSP...    :(

Thanks.  That's my bad self-image shining through there.   :lol:

[quote="myyellowducky82"]anyways, i'm 11 turnin 12 next month and going to 7th grade at a middle school with over 600 kids. (HUGE!)
[/quote]

Umm... my middle school has over 1000 kids, that's not very big...

[quote="tinytim"]Hi, I'm tinytim. My name is not Tim. I live in the United States. I live in a state that is located in the United States. I live in a city that is located in the state I live in. I live in a house. I enjoy wasting my time in the Games&Fun Factory Forum on this website named artofproblemsolving.com. You should visit it sometime, its a really really cool.[/quote]

You should also say that you and I are teletubby geeks :D

Anyways, I wanna change the way I introduce myself.

HI EVERYBODY! MY NAME IS GAUSS1181 AND I PHAILED STATES THIS YEAR. I AM IN SEVENTH GRADE. I LIKE THE TELETUBBIES!!!!! THAT'S ALL I HAVE TO SAY ABOUT MYSELF. THIS HAS BEEN A PUBLIC SHAMING OF GAUSS1181, THANK YOU!!!!! :P

hahahaha :|

Edit: And everyone likes my small monkey mouth. :|
And I can do the Rubik's Cube in 18 seconds!

Hi I'm Angie. My username is: xoangieexo so if you split it: xo angiee xo
Please don't call me xoan or xo because that's not my name.
I'm 13 and I live in NY. My hobbies are: tennis violin, reading, math, writing, and going to partiess of course. I waste a lot of time on AoPS =P

Join the AoPS Family!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211650

That's all I have to say.  :)

[quote="chess64"][size=75]hi im chess64 im 9yrs old and i like math.[/size][/quote]

1. How was MOP?
2. People expect other people to tell the truth on this forum.

у тебя есть сыр?  nice.

1. Massive revival
2. It was something that you could have PMed to me.
3. We should keep this going anyways.

[quote="chess64"]i'm just wondering.  why do you have that in your signature?[/quote]
probably because у него есть сыр?

It says "This is my sig", now lets stay on topic.

[quote="chess64"]it actually doesn't say anything.  i guess you tried to write у меня есть сиг, which means "i have a whitefish" (not sig).  i believe "this is my signature" would be "это мой подпись". :P

а Erken, ты из России?)[/quote]

it should be "это моя подпись" or "это моя сигнатура",but not "это мои подпись".
Nah,I am not from Russia,I just know russian...
I guess you're not from russia,according to what you wrote,but anyway,how do you know these russian words?
Please don't response "by the use of google-dictionary" :)


AMC USA(J)MO USAMO geometry AIME search USAMTS

Here is something that I handed in for my English class that readers of this forum might find amusing.

lmao
arnav going to your school and getting a 9
you wish

Dang. You freaked me out for a second there. I was like, what? Arnav? Nuh-uh. And then I read the beginning again, and I was like ok.

omg, "THE NUMBER DEVIL!"
that book is soooooooooo awesome. when i first read it, i didn't really understand it
but when i read it when i was in 6th grade, everything came together and i was stunned by its content.

[quote="Altheman"]so when i try and download it tries to download the .php file...[/quote]

Ditto.

Check your popup blockers (if any). Have you tried to right click on the Download link and do Save as ... ?

Right click and save as tries to save viewtopic.php.  Clicking with Control to override any pop-up blockers gives viewtopic.php as well.

I haven't had any problems downloading attachments before.

I thought the other person at your school got a 17?

[quote="solafidefarms"]Have you tried saving it as memoir.doc and opening the resulting file with Word?[/quote]That will not work, the problem (s)he is describing essentially means that the downloading script is not being accessed by the browser, and I'm 95% sure it's because of a web-blocker of sorts (pop-up blocker, or firewall, or an all-in-one antivirus program).

Wow...my math experiences (so far) match up with yours [i]really [/i]closely, except I used a different book (Algebra the Easy Way) to do algebra in elementary school.

That was really well written, and inspiring. You should consider being a writer as well as a mathematician.
And I LOLLED at the Arnav thing, too.

[quote="Valentin Vornicu"]It's clearly on your side, as I've just tried it on several different computers, with different browsers and on all of them it works. What browser are you using?[/quote]

Sorry, got busy and didn't get back to here until just now.  IE 6.  I don't know if there are any secondary pop up blockers installed.  I don't think so.  I always get popups.....

I've never had a problem downloading before this one.  I have noticed some funkiness since the new rate posts script stuff went in.  I always hit "back" to go back to the topics list from a thread, and I now have to hit it twice.  Perhaps it's related somehow, though I've no idea how.

[quote="K81o7"][quote="The Herald of Doom"]dude no I heard that apparently decision of the team will be announced sometime during the first week of MOP[/quote]

It's been decided.[/quote]

Ummm ok... but this does not contradict what The Herald of Doom has said.

Well I suppose the "dude no" part contradicts it.

Who else beside Yichen Wu would use my name as his aops spamming username? (he even left his email in the profile)

But you aren't delong you are neal wu right?  :o  :wink:

[quote="K81o7"]Either way, I never said that my post contradicted that of Herald of Doom: I was simply giving information because, based on the circumstances, I thought it would be more likely that it would be unknown to Herald of Doom.[/quote]

OK, well either way, people should definitely keep in mind that the race is very close.

Huh what that is not what I heard . . . team selection to be announced during MOP?

wait are you delong because I'm pretty sure he just posts stuff to make it seem like he didn't do well

BUT DUDE I AM PRETTY SURE THE TEAM HASN'T BEEN ANNOUNCED YET SO WTF

DUDE I RECANT MY STATEMENT THAT THE RACE IS OVER IN THE AMC FORUM

LIKE THIS IS TOP SECRET INFORMATION

THE RACE IS VERY CLOSE

ARGHHHHHHHH

I'm locking this thread, as it has gotten way off topic and has degenerated into spam.

As of this moment, the US IMO team has not been publicly announced.  Until it is, please stop asking for, speculating about, or divulging information that is not yet public.


number theory
greatest common divisor
algebra
polynomial
invariant
geometry
geometric transformation

If $ p$ is prime, show that $ x^p\minus{}x\minus{}1$ is irreducible over $ Q$.

[hide="Hint"] If $ \alpha$ is a root of $ x^p \minus{} x \minus{} 1$ over an algebraic closure of $ \mathbb{F}_p$, then so is $ \alpha \plus{} k, k \in \mathbb{F}_p$.  [/hide]

i think that for all p>1  its works
however  if this polynome is reductible then we have  a rationnal  x such that x is a solution  whith x=p/q and  gcd (p,q) = 1 then we conclud a contradiction

[quote="ouboujlal"]however  if this polynome is reductible then we have  a rationnal  x such that x is a solution  whith x=p/q and  gcd (p,q) = 1 then we conclud a contradiction[/quote]Nope. It has to have a polynomial with rational coefficients divide it; that polynomial isn't necessarily linear.

Is there a way to do this without field theory?  The furthest I can get is that there are no rational linear factors of $ x^p\minus{}x\minus{}1$, which is trivial from the rational root theorem.  Eisenstein's Criterion can't be used either.  Any ideas?

If you're willing to use Eisenstein's criterion, which is based on showing that a polynomial is irreducible modulo a prime, you should be willing to think about more general ways to show that a polynomial is irreducible modulo a prime.  Anyway, here's a solution which avoids explicit use of field theory.

[hide="Solution"] Suppose $ x^p \minus{} x \minus{} 1 \equiv f(x) g(x) \bmod p$ with $ f$ irreducible.  Because the LHS is invariant under translation, it also follows that

$ x^p \minus{} x \minus{} 1 \equiv f(x \plus{} k) g(x \plus{} k) \bmod p$

for every $ k \in \mathbb{F}_p$.  Since $ f(x \plus{} k)$ is also irreducible for every $ k$ and cannot be linear by the rational root theorem, the polynomials

$ f(x), f(x \plus{} 1), ... f(x \plus{} (p\minus{}1))$

cannot all be distinct $ \bmod p$, so $ f(x) \equal{} f(x \plus{} k)$ for some $ k$.  Since $ p$ is prime, this gives $ f(x) \equal{} f(x \plus{} k), k \equal{} 0, 1, ... p\minus{}1$, so $ f(x) \minus{} f(0)$ must have degree at least $ p$; contradiction. [/hide]


MATHCOUNTS
AMC
AMC 8
AMC 10
AMC 12
LaTeX

Just curious to see what grade ya'll are in.

6th here

I'm in 8th grade, but I skipped a grade...it stinks that my Mathcounts years are over before I realized my full potential. Hmm...maybe I should fail English on purpose to stay back? :D

8th my last year in mathcounts, i wish i competed last year and the year before

6th grader but didn't go to Mathcounts. Practicing a lot though.

:O lots more 6ths than I thought there would be...

8th here...
didn't know bout MathCounts till a new math teacher came this year

woah, a sixth grader with 800+ posts and lots of posts in the higher level forum... wow

well i am in 8th grade but i don't know if my grade should count or not because im still only 12. so um... ya. so maybe u wanna take my vote off cause i sorta count as a 7th grader.... i dunno

8th and I wish our twon had 6th grade Mathcounts (6th is in a separate school) then... they just introduced it this year... I think Mathcounts is way better than any other math competition that I've done.. including the Math Olympiads, AMC 8/10... etc..

Our team of 6 graders placed 2 in state no one went to nationals but we were the only team with all 4 in the top 20 and we're from utah
do u think this is pretty good

Yes... our team of 6th graders made states in their first year... so we're proud of just that. In terms of ranking, it depends on the difficulty of state (pls don't discuss your scores, I'm not asking you to) but our state-winning team last year had people in the following places: 3, 24, 46, 49 and that was good enough... of course CT states have over 300 ppl but still it's pretty good no matter what if EVERYONE is in the top 20... sorry if i'm rambling...

[quote="led14pa2"]i agree that mathcounts is the best and hardest math contest that ive done. it is way harder than the CML which u only miss a maximum of 1 out of the 30 questions per year, the AMC8 doesnt really have a purpose in my opinion, unlike the AMC10 and AMC12, and math olympiads is fairly easy too, the best part about it is that u get a ton of awards at the end like a gold pin, trophy, patch, and certificate.[/quote]
But for us, none but Mathcounts (and Olympiad, but I did that in 5th adn 6th) has that competitive feel, and the Countdown round is that makes Mathcounts special...

Now I'm in seven grade, being the first year I am exposed to Mathcounts related problems. I should've joined in 6th grade, and now I'm getting old...

how is seventh grade old? In our school, mathcounts doesn't even start til seventh grade.... feeling kinda young with the eighth graders....  :lol:

Hmm... I wanted to go to MathCounts but they only let one team go... I could've used some practice at least. :P

[quote="pascal_1623"]5th[/quote]

5th? Wow... I joined in 5th grade but... i didn't go to the mathcounts section. :P TOo advanced for me back then but that's just me.

[quote="led14pa2"]i agree the countdown round is very origional, and mathcounts is much more competitive, i like how it is divided into 4 levels, school chapter state and national. good luck at states hockeyadi i hope we both make nationals but it will be tough, CT isnt a very easy state to do well in[/quote]
That would be cool... I just wish that Jason hadn't locked up first already (barring no one gets a perfect score)... b/c then it turns into a tiebreaker-fest here in CT... but I'm aiming in the forties to 39 for a baseline...

yea i wish that jason isnt pretty much gaurenteed to get first but jason is my teammate so i still want him to do well so my team can win, my goal is about the same as yours, and ive been practicing extra hard for sprint for that exact reason, tiebreakers.

maybe i dont have right definition, but cant a 7th or 8th grader be an alumnist?

By alumnist we mean that you can't participate in Mathcounts anymore

I'm in [color=red]8th [/color]grade, and its over for me :(
I can't be in MATHCOUNTS anymore :(
Oh well too bad...  :(

Then I guess you qualify for Mathcounts Alum lightspeed ;) Anyways I hope you have better luck at the AMCs...

aw, i'm sorry lightspeed, actually, you shouldn't be sad, i mean, Mathcount is just another competition among all the others..... I don't see anything special... the only thing is that you can compete with people older than you..... just think, at least you made it to state.... my friends would say that you are  weird because they didn't even make it to states and you are just trying to  make them feel bad!

this is totally random but:
[hide]I just posted my first post on this forum using latex! I know that's not much but it took me a while! hahaha, feeling proud![/hide] :D  :D  :D

So, it seems so far that there are more eigth graders here in AOPS than any other grade right? Surveys aren't always right

7th grade (i wonder if a 7th grader has ever won at nationals)?

Yes.. because many 8th graders found this forum recently and are preparing for their last states (like me), although I've found the AoPS forums extrememly helpful in strengthening my Mathematical skills...


Find all $f: \mathbb N \to \mathbb N$ such that
\[f(n\plus{}1)\plus{}f(n\plus{}3)\equal{}f(n\plus{}5)f(n\plus{}7)\minus{}1375\]
for all positive integers $n$.

[quote="stvs_f"]find all $ f: N \to N$ that:
$ f(n \plus{} 1) \plus{} f(n \plus{} 3) \equal{} f(n \plus{} 5)f(n \plus{} 7) \minus{} 1375$[/quote]

Let $ P(n)$ be the assertion $ f(n\plus{}1)\plus{}f(n\plus{}3)\equal{}f(n\plus{}5)f(n\plus{}7)\minus{}1375$

First notice that we can study solutions $ f(n)$ for $ n$ odd and translate them immediately for $ n$ even.

Subtracting $ P(n)$ from $ P(n\plus{}2)$, we get new assertion $ Q(n)$ : $ f(n\plus{}5)\minus{}f(n\plus{}1)\equal{}f(n\plus{}7)(f(n\plus{}9)\minus{}f(n\plus{}5)$

$ Q(1)$ $ \implies$ $ f(6)\minus{}f(2)\equal{}f(8)(f(10)\minus{}f(6)$ and so $ f(6)\minus{}f(2)\equal{}f(8)f(12)f(16)(f(18)\minus{}f(14)$ ... and so on. So :

either $ f(6\plus{}4n\plus{}4)\equal{}f(6\plus{}4n)$ for some $ n$ and so $ f(6)\equal{}f(2)$ and so $ f(4n\plus{}6)\equal{}f(4n\plus{}2)$ $ \forall n\ge 0$
either $ f(4n\plus{}6)\ne f(4n\plus{}2)$ $ \forall n\ge 0$ and so $ f(4n)\equal{}1$ $ \forall n$ great enough. But then :
$ Q(3)$ $ \implies$ $ f(8)\minus{}f(4)\equal{}f(10)(f(12)\minus{}f(8)$ and so $ f(4n)\equal{}1$ $ \forall n>0$
Then $ P(4n\plus{}1)$ $ \implies$ $ f(4n\plus{}2)\plus{}1\equal{}f(4n\plus{}6)\minus{}1375$ and so $ f(4n\plus{}6)\equal{}f(4n\plus{}2)\plus{}1376$

As a first conclusion, we got :
Either $ f(4n\plus{}2)\equal{}f(2)$ $ \forall n$
Either $ f(4n)\equal{}1$ $ \forall n$ and $ f(4n\plus{}2)\equal{}1376n\plus{}f(2)$ $ \forall n$

Using the same approach starting from $ Q(3)$ instead of $ Q(1)$, we get
Either $ f(4n)\equal{}f(4)$ $ \forall n$
Either $ f(4n\plus{}2)\equal{}1$ $ \forall n$ and $ f(4n\plus{}4)\equal{}1376n\plus{}f(4)$ $ \forall n$

Combining these two conclusions, we get three possibilities :
1) $ f(4n)\equal{}a$ and $ f(4n\plus{}2)\equal{}b$ $ \forall n$ and checking back in the equation, this gives $ a\plus{}b\equal{}ab\minus{}1375$ $ \iff$ $ (a\minus{}1)(b\minus{}1)\equal{}1376\equal{}2^543$

2) $ f(4n\plus{}2)\equal{}1$ $ \forall n$ and $ f(4n\plus{}4)\equal{}1376n\plus{}f(4)$ $ \forall n$ and it is easy to check that this indeed is a solution

3) $ f(4n)\equal{}1$ $ \forall n$ and $ f(4n\plus{}2)\equal{}1376n\plus{}f(2)$ $ \forall n$ and it is easy to check that this indeed is a solution

Hence the $ 14$ solutions for even arguments :
$ 2,1377,2,1377,2,1377, ....$
$ 3,689,3,689,3,689,....$
$ 5,345,5,345,5,345,....$
$ 9,173,9,173,9,173,...$
$ 17,87,17,87,17,87,....$
$ 33,44,33,44,33,44,....$
$ 1377,2,1377,2,1377,2 ...$
$ 689,3,689,3,689,3....$
$ 345,5,345,5,345,5....$
$ 173,9,173,9,173,9...$
$ 87,17,87,17,87,17....$
$ 44,33,44,33,44,33....$
$ 1,u,1,u\plus{}1376,1,u\plus{}2\cdot 1376, ...$
$ u,1,u\plus{}1376,1,u\plus{}2\cdot 1376,1 ...$

And we obviously have the $ 14$ same solutions for odd arguments, leading to $ 196$ solutions.
For example :

$ 9,17,173,87,9,17,173,87,9,17,173,87,...$
$ u,2,1,1377,u\plus{}1376,2,1,1377,u\plus{}2\cdot 1376, ...$
$ u,1,1,v,u\plus{}1376,1,1,v\plus{}1376, u\plus{}2\cdot 1376,1,1  ...$
...


inequalities
integration
trigonometry
limit
calculus
real analysis
real analysis unsolved

Prove that:
1) $ \int_0^{\sqrt{2\pi}}\sin x^2dx>0$
2) $ \lim_{n\rightarrow\plus{}\infty}\int_{n}^{n\plus{}1}\sin x^2dx\equal{}0.$

1) $ x^2\equal{}t$ yields $ \int_0^\pi {\sin t\over 2\sqrt t}dt  \plus{} \int_\pi^{2\pi} {\sin t\over 2\sqrt t}dt \doteq I_1\plus{}I_2$
$ I_1>0$ and $ I_1>\vert I_2\vert$ since $ \sin t$ is periodic but $ 1/\sqrt t$ decreases.

2) The change $ x^\gamma \equal{}t$ togheter with  an integration by parts implies the convergence of
$ \int_0^{\plus{}\infty} \sin x^\gamma dx$ for $ \gamma>1$ and  $ \lim_{n\rightarrow \plus{} \infty}\int_{n}^{n \plus{} 1}\sin x^2dx \equal{} 0.$ is
the necessary condition for the convergence of $ \sum_{n\equal{}0}^\infty \int_{n}^{n \plus{} 1}\sin x^2dx$ which is
the integral over $ [0,\plus{}\infty)$

P.S. Very nice your inequalities as well as your proofs.


The button for editing my posts seems to have disappeared; what's going on?
I'd like to be able to fix my typos.

You are able to edit your posts. Just click the edit button then edit them. It's as simple as that.

[quote="jmerry"]That would work- if there was an edit button. It appears for some posts, but not for others.
I've edited many times before with no problems; this just happened today.

The button appears for my post in this thread, but not for my posts in the Putnam A5 thread.

(Edit: now there's no button for my first post in this thread! I really don't understand this)[/quote]Is it just your first post? ... I am not able to reproduce the bug with a normal user...  :?

When I returned to the thread after editing that post, that post's edit button was gone. This is weird.

Very strange ... does it appear only in certain forums?!

I've just looked at my posts in many different places, and none of them have an edit button right now.

(Edit)
Now I'm getting somewhere. I switched browsers, and the missing buttons (edit, quote, and delete for last posts on a thread) reappeared. I was using Netscape when the problems happened.

And merely by leaving this thread it broke again.

(Edit) Switching browsers seemed to fix it for now. I posted the correction I wanted to, and I'm going to leave for a while so this doesn't drive me quite so crazy.

It's happening to me as well, on a different machine (but also a Mac). Using Netscape, the edit buttons on all of my old posts aren't there. This is the browser I've always used; something changed today.

[Edit] Well, now it's back. [/edit]

Spoke too soon. No edit button elsewhere, and now no edit button here.

[quote="darij grinberg"]completely idiot thought: maybe the edit buttons are there but you don't see them because the pictures don't load? that's the first thing I think of when i hear about different browsers...

  dg[/quote]Yeah, unfortunately we don't have MACs to test thing on, however it might seem like an loading image problem?

The site is being tested for full-compatibility with IE 6, Opera 7, Firefox 1 (thus Mozzila included). Since our number of visitators using Netscape or Konqueror is extremely small (compared with the others above) we don't tess ALL the features in the above mentioned browsers. I would recommend you use one of the 3 browsers above :)

I can switch browsers if I need to, of course- probably Firefox. Until now, Netscape has worked very well.

Incidentally, Safari (standard on Macs) displays this forum badly in general, with large blocks of empty space at the end of posts.

It wouldn't be a problem loading images- Netscape displays a broken image graphic for those, and links are still accessible. I'm used to clicking before the images load anyway.


function
algebra
polynomial
algebra unsolved

Prove that there is no strictly monoton functions $f: R \to R$ that
$f(x+f(y))=f(x)+y^n$ for all $n>1,n \in N$

We have two cases:

Case 1: n  is even

Suppose that a strictly monotonic function f exists.

Assume WLOG that $f(-1) \ne 0$
$x = y = 1 \Rightarrow f(1+f(1)) = f(1)+1$
$x = 1, \ y = -1 \Rightarrow f(1+f(-1)) = f(1) + 1$

which is a contradiction since we assume that f is monotonic.

Case 2: n is odd:

Set y as y + b where $b \ne 0$.

Put the equation
$f(x+f(y+b)) = f(x) + (y+b)^n = f(x) +y^n = f(x+f(y))$

$y^n = (y+b)^n$

Because n>1 and n is odd, this polynomial equation with variable b has a nontrivial real solution for appropriately chosen y, so for some x, y, and b,

$f(x+f(y+b))=f(x+f(y))$ which implies noninjectivity, a contradiction.

Actually, I retract my proof for odd n. I do not know if the problem statement even holds for such n.

$f(x+f(y)) = f(x) + y^n$
$x = 0 \Rightarrow f(f(y)) = f(0) = y^n$ so any suitable function must be surjective.

Revisiting the equation

$y^n = (y+b)^n$

has only b = 0 for its real solution, so $f(x+f(y)) = f(x + f(y+b))$ only if b = 0. So...???


MATHCOUNTS
geometry
search
function
articles
calculus
AMC 10

:ninja: I need advice for the upcoming year in Mathcounts. I made it to state in seventh grade and I want to make it further. If anybody has advice could u let me know. :gathering:

Well, what have you done to practice? What math are you currently doing? Art of Problem Solving holds a [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=5]Mathcounts Problem Series[/url] class, which I found extremely helpful when I did Mathcounts. The [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts_Intro.php]Introduction[/url] books are very good and thorough books to  enhance individual subject areas you lack in. Or, if you feel that you need work in all areas of Mathcounts, [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=100]AoPS Volume 1[/url] is a good overview. AoPS Volume 1 depends more on your innate problem-solving ability, and it doesn't go as in-depth in any subject areas, but it does improve all areas. The Introduction books are relatively easy, straightforward, conversational, and in-depth.

You also might be interested in the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=2000]All-Time Greatest MATHCOUNTS Problems[/url] for more MATHCOUNTS problem practice.

Above all, welcome to the AoPS Mathcounts forum! I hope you stay around for a long time :).

Billy, your friendly MATHCOUNTS forum moderator :).

make sure to do all of the state/nats problems you can do (since you made it to states already, don't with the chapter problems).

So ask your coach is he or she has any of the national problems (it would be good practice) or any of the state mc problems that are so old that you can't find them online anymore. I can't remember where you can find the mathcount problems for 2000-2008, so ask other people for the sites.

The Mathcounts book that your coach gets every year (with all the warm-ups/work-outs) is actually pretty good. The warm-up/work-outs at the end of the book are around state-level.

Aops Volume Book #1 would help as well.

If you run out of problems to do and you have already done everything in the Aops Volume Book #1, consider doing some of the amc 10 problems. You can find them in the contests section of the website under the USA category. There are some things in there that you probably have never seen before, but you'll be able to do the rest of the problems for good practice.

Besides that, if you're someone who makes a lot of careless mistakes, practice on math problems where careless mistakes are easy to make. Increase your speed in math by either just doing more math or maybe even try playing "For the Win" (don't spend to much time on For the Win though).

Good luck doing well on the states next year.

My Suggestions (This is what I did; I made it to Nats):

Took MATHCOUNTS Problem Series Class over the summer
Got AoPS books Vol. 1 and 2
Got The All-Time Greatest MATHCOUNTS Problems Book

as posted in another topic:

There is no golden bullet, and you can't win or do well without having lots of intelligence, but there are some strategies to help you do better.

1. Do lots and lots of old tests. Before the school test, do all of your available school rounds in ascending order of year, with on the day before doing an easy chapter. Before the chapter test, do all of your available chapter rounds in ascending order of year, with an easy state on the day before. Before the state test, do all of the state tests in ascending order of year, with an easy nats on the day before. Before nats, do all of the tests in ascending order, with one test on the day before.
2. If you don't have a team, do the team rounds by your self also. Try to allot a maximum of 80 minutes Maximum for the round.
3. If you don't have enough tests, look on the net, you can usually find school, chapter, and state tests going back to 2001. In addition, tons of people on AOPS have old Mathcounts National Tests. Ask them.
4. There are plenty of mock tests on AOPS. While they are usually too easy or too hard, they are good practice.
5. After doing a test, figure out how to do all of the problems that you didn't get. If you still can't understand, post it here, in this forum, of use the search function to see if any1 has posted in before.
6. Practice every day, whether it be on this site, an actual mc test, amc problems, or this site: http://mathcounts.saab.org/mc.cgi
7. Get LOTS of sleep before the test.
8. At target, when checking, read every single word and read the question very slowly to make sure that you read it right.
9. At sprint, try to do it really fast, and then check it over. If you are stuck on a problem, or it is gonna take too long, skip it and come back to it later.

I wonder why there is such a thing as pre-calc. To learn calc you only need to know algerba 1.

[quote="Poincare"][quote="mewto55555"]I'm doing calc  :D  Yet I still uber-failed  :([/quote
What grade are you in dude? there is some article on AoPs about the Calculus Trap, but even if you do read it i just think its crud cause i am taking a university course for precalc(high school won't let me take it) and i am in 5th grade :D[/quote][/quote]

I'm also doing that, but AoPS 1 is teaching me more than UofDelaware has taught me this week

And I'm taking Geometry next year, so geometry

[quote="Urc"][quote="Poincare"][quote="mewto55555"]I'm doing calc  :D  Yet I still uber-failed  :([/quote
What grade are you in dude? there is some article on AoPs about the Calculus Trap, but even if you do read it i just think its crud cause i am taking a university course for precalc(high school won't let me take it) and i am in 5th grade :D[/quote]

I'm also doing that, but AoPS 1 is teaching me more than UofDelaware has taught me this week

And I'm taking Geometry next year, so geometry[/quote][/quote]
man you are so lucky , i am going to UNM- LA which is right in the middle of no where and doesn't even offer geometry ( :censored:  :cursing:  :furious: )

Matt,

There is also another post by Urc, called "how to do well at mc" or something like that. Check that out. There's tons of advice and I posted what I think should be really helpful.

[quote="Poincare"][/quote][quote="Urc"][quote="Poincare"][quote="mewto55555"]I'm doing calc  :D  Yet I still uber-failed  :([/quote
What grade are you in dude? there is some article on AoPs about the Calculus Trap, but even if you do read it i just think its crud cause i am taking a university course for precalc(high school won't let me take it) and i am in 5th grade :D[/quote]

I'm also doing that, but AoPS 1 is teaching me more than UofDelaware has taught me this week

And I'm taking Geometry next year, so geometry[/quote]
man you are so lucky , i am going to UNM- LA which is right in the middle of no where and doesn't even offer geometry ( :censored:  :cursing:  :furious: )[/quote][quote="Poincare"][/quote]

I meant my school's geometry. SOrry for the confusion...

This is great. thanks for all the advice. I have already reviewed the TAOPS volumes 1 aand 2 but mathcounts greatist hits sounds cool so I will try that out. I am taking the thursday night summer class thats starts tonight. See some of you there.
 :ninja: NINJAS ARE BEAST :ninja:

[quote="Poincare"][quote="mewto55555"]I'm doing calc  :D  Yet I still uber-failed  :([/quote
What grade are you in dude? there is some article on AoPs about the Calculus Trap, but even if you do read it i just think its crud cause i am taking a university course for precalc(high school won't let me take it) and i am in 5th grade :D[/quote][/quote]

i was doing fractions in 5th grade  :lol:


Find the limits:

1 )$ \lim_{x\to\ 0}\frac{3^x\minus{}1}{x}$
and
2) $ \lim_{x\to\ 0} \frac{ln(1\plus{}x)}{x}$

Not sure whether these are appropriate for High School Intermediate Topics.

1)

L'Hopital's rule gives

$ \lim_{x \to\  0} \frac {3^{x}\minus{}1}{x} \equal{} \lim_{x \to \ 0} \frac{3^{x}\ln 3}{1} \equal{} \ln 3$.

Alternatively, using the Taylor series expansion, $ 1\plus{}x\ln 3 \plus{} \frac{1}{2} x^{2} (\ln 3)^{2} ...$, we also get $ \ln 3$.

2)

L'Hopital's rule gives

$ \lim_{x \to\ 0}\frac{\ln(1\plus{}x)}{x} \equal{} \lim_{x \to\ 0}\frac {\frac{1}{1\plus{}x}}{1} \equal{} 1$.

Alternatively, using the Taylor series expansion, $ x\minus{}\frac{x^{2}}{2} \plus{} \frac{x^{3}}{3} ...$, we also get $ 1$.

I can't think of any other ways of doing these that might be more appropriate and am looking forward to seeing other methods.

[hide="2"]2

$ \lim_{x\to\ 0}\frac {ln(1 + x)}{x}$

$ \lim_{x\to\ 0}ln(1 + x)^{\frac {1}{x}}$

Since $ ln$ is a continuous function in this range,

$ ln(\lim_{x\to\ 0}(1 + x)^{\frac {1}{x}}$$ )$

$ lne = 1$[/hide]

\[ \boxed{\lim_{x\rightarrow 0} \frac{a^x\minus{}1}{x}\equal{}\ln a}\]


calculus
integration
analytic geometry
geometry
3D geometry
sphere
geometric transformation

Calculate $ \iiint_{V} \sqrt{ (x\minus{}a)^2 \plus{} (y\minus{}b)^2 \plus{} (z\minus{}c)^2 } dx dy dz$ where $ V \equal{} \{ (x,y,z) \mid x^2 \plus{} y^2 \plus{} z^2 \le R^2 \}$, $ R$ is some positive real number.

If you integrate one variable at a time, it's actually not as hard as you think.

What do you mean? this is a triple integral, not an iterated integral (although some variable change may make it one. I tried spherical coordinates, but had no luck with it).

Why isn't this an iterated integral?  Unless multivariable calculus changed since I last took (last semseter), when you do a triple integral, you go in order.  So start with $ x$ (or any of the other two.  Then, you have
$ \sqrt{(x\minus{}a)^{2}\plus{}\text{constants}}$
which you could find or just look up in a table.

[quote="JRav"]Why isn't this an iterated integral?  Unless multivariable calculus changed since I last took (last semseter), when you do a triple integral, you go in order.  So start with $ x$ (or any of the other two.  Then, you have
$ \sqrt {(x \minus{} a)^{2} \plus{} \text{constants}}$
which you could find or just look up in a table.[/quote]

I think maybe the original poster was hung up on the bounds of integration; the region $ V$ that you integrate over in this problem is a sphere, so doing this in Cartesian coordinates is messy, if it's even possible.

$ V$ does not change under rotation, so if you let $ l = \sqrt{a^2 + b^2 + c^2}$, you'll obtain more simple form $ I = \int_V \sqrt{x^2 + y^2 + (z-l)^2} \, dxdydz.$

Now apply the cylinderical coordinate to obtain $ I = \int_{-R}^{R} \int_{0}^{2\pi} \int_{0}^{\sqrt{R^2 - z^2}} \sqrt{r^2 + (z-l)^2} \, r \,  dr d\theta dz.$ Then

\begin{eqnarray*}I
& = & \frac{2\pi}{3} \int_{-R}^{R} \left[ ( r^2 + (z-l)^2 )^{\frac{3}{2}} \right]_{0}^{\sqrt{R^2 - z^2}}  \,  dz\\
& = & \frac{2\pi}{3} \int_{-R}^{R} \left[ ( R^2 + l^2 -2lz )^{\frac{3}{2}} - | z-l |^{3}\right]  \,  dz.
\end{eqnarray*}

The remaining calculation is now just a one-dimensional integration.

Alternatively, you can use the observation that if you define $ u(l) \equal{} \int_V{\sqrt {x^2 \plus{} y^2 \plus{} (z \minus{} l)^2}dxdydz}$ then
\[ \frac {d}{dl}u(l) \equal{} \int_V{\frac {\partial}{\partial z}\sqrt {x^2 \plus{} y^2 \plus{} (z \minus{} l)^2}dxdydz}
\]
Then integrating by parts gives a relatively painless expression for $ u'(l)$ as an integral over $ \partial V$.  Since $ u(0)$ is easy to find, you can then use the Fundamental Theorem of Calculus
\[ u(l) \equal{} u(0) \plus{} \int_0^l{u'(s)ds}
\]
Though in the end it amounts to evaluating essentially the same integrals as suggested by sos440.  Incidentally, the answers come out as:
\[ \frac {4\pi\left(5l^2R^3 \plus{} R^5\right)}{15l} \text{ for } l > R
\]

\[ \frac {\pi(15R^4 \plus{} 10l^2R^2 \minus{} l^4)}{15} \text{ for } l\leq R
\]

Or we could do it in Cartesian, which I find more interesting:
Divide out by the non-$ x$ terms, so we get
$ \iiint\sqrt{(y\minus{}b)^{2}\plus{}(z\minus{}c)^{2}}\sqrt{\frac{(x\minus{}a)^{2}}{(y\minus{}b)^{2}\plus{}(z\minus{}c)^{2}}\plus{}1}\ dx\ dy\ dz$
which has the form of a trig substitution.  Now let
$ \tan{\theta}\equal{}\frac{(x\minus{}a)}{\sqrt{(y\minus{}b)^{2}\plus{}(z\minus{}c)^{2}}}$
and
$ \sqrt{(y\minus{}b)^{2}\plus{}(z\minus{}c)^{2}}\sec^{2}{\theta}\ d\theta\equal{}dx$
Now make the substitution, yielding,
$ \left((y\minus{}b)^{2}\plus{}(z\minus{}c)^{2}\right)\iiint\sec^{3}{\theta}\ d{\theta}\ dy\ dz$
This is a well-known integral.  Then you wind up with a double integral.


what do you do to prepare?

If this has already been posted, delete this

aznwithabrain wrote a good guide thing...

[quote="diophantient"]I'd suggest going through all the past tests you have first, then posting the problems too hard for you on this website.
Then, go through the book Art of Problem Solving Volume 1.[/quote]
Yes i think that going through the past tests is extremely important as many concepts keep repeating in Mathcounts. Also I don't think you should do all of AOPS 1 as some of it (like trig is a waste of time if you are gonna just start), just do the stuff that seems like MC level. Good luck to everyone going to nats this year!


I was wondering, after JS1527's post, what programming languages have you learned? I have learned Visual Basic and Java, and both were very fun to learn. Which prog. language do you think is the most popular? I only listed some of the many high-level programming languages - but if you have learned some other language not listed, please post too.

I can make flash animations, if that counts as a language

Let's see...

BASIC, VB, C, C++, a tiny bit of Visual C++, Java, PHP; will read about Perl soon.

I know C and Java mainly, but can kind of program in C++.  But I do know a itsy bity tenny weeny part of pascal, basic, visual basic and logo.

I used basic for a while.

Where's calculator?

Basic, vB, C, Java. But nowadays, I write mainly on paper and let someone else code, and I write entirely in pseudo-C.

I have programming courses in JAVA, and I'm working on this project that is impossibly hard. I'm working on MASTERMIND, a graphical version. This may seem easy, but updating the number of colors matched is extremely difficult.

Hey, white_horse_king88, have you tried the training pages for the USACO at ace.delos.com/usacogate.?

I do JAVA, a bit of c++, JavaScript, POV-ray, HTML, and MMF, StageCast, ROBIX, RCX Code.
(The last 2 are Robot languages, and MMF and STagecast are basic(not laguage) visual languages.
[url]http://www.stagecast.com[/url]
[url]http://www.clickteam.com/English/multimedia_fusion.htm[/url]
[url]http://www.robix.com/default.html[/url]
[url]http://www.legomindstorms.com[/url]

I never learned C++, so I wouldn't know, but didn't they create JAVA just to simplify it? After all, Java is more object oriented.

TI-Basic, Batch (if that counts), GW-Basic, Q-Basic, , Visual Basic, C, C++, portions of the STL for C++, and varying amounts of the SDL, Win32 and DirectX APIs for C++. Those are listed very roughly in the order I learned them. I can also read code in most languages.

[quote="Magnara"]

Where's calculator?[/quote]

yea.  i agree... where's calculator? and does HTML count as a language? i think/hope i'm taking java next year, but i have no clue what they could have done to my schedule for next year... stupid budget cuts

[quote="h_s_potter2002"][quote="churchilljrhigh"]RRR! I can't believe College Board is switching to Java now. C++ was so awesome.[/quote]

C++ is a lot harder to learn and more complicated in my opinion, since you can compile C programs in it.  Wit Java, everything has to be OOP, making it a lot easier for people who have never programmed before to use.[/quote]

C++ is still cooler :), and it has pointers.

I like pointers. They are amazing.

but PHP is. and it's not in the list. :(

I can do some Java.  Does anyone learn Fortran anymore?  I noticed it had 0%.  My dad actually did learn it, but that was in the seventies.

I've learned:

C++ - Pointers are cool once you get used to them  :D
C
BASIC, though I have forgotten all of it
Cocoa, which isn't on the list
Visual Basic, which used to be my best language (now C++), which I forgot once I got a mac
A [i]very[/i] small bit of PHP, if that counts
A [i]very[/i] small  bit of java

Java isn't very fun.  Too bad they had to switch to it for AP.

[quote="Osiris"]HTML is not a programming language. It stands for HyperText Markup Language, which is a subset of the SGML (Standardised General Markup Language), which is used to describe a multitude of documents. HTML belongs in the same category as TeX.[/quote]

i know it's not a programming language, but because i know some of it and everyone here knows like five languages, it look better to al least know html than not know anything..

I'm really struggling to do USACO. I was in the middle of a bulk-read and Tensorflow algorithm when all of a sudden, I decided to be smart and fall asleep. I still finished the test and made it past silver, but that was dumb.

Why can't we select multiple options :mad:
If we could I'd choose Python and C++

Python, Java, Html/Css/Javascript, and some swift i guess...

Super basic Javascript.(not mentioned in poll)


AMC
AMC 12
AIME
USA(J)MO
USAMO
MATHCOUNTS
ARML

The AMC says that it has 700,000 annual participants, which must be true since the AMC doesn't lie.

However, there were 150,000 AMC 8 takers, 100,000 AMC 10 takers, and 150,000 AMC 12 takers this year (very roughly).

Summing these gives 400,000 participants.  Where are the other 300,000?

Perhaps you didn't take into account tests A and B?

I think he did. Just go to general stats for AMC 10/12.

Hmm...I was all but sure the AMC12 number approaches a quarter million...*mutters*

[url=http://www.unl.edu/amc/whatswhat.html]it says 413,000 students here[/url]

check the bottom of the AMC page w/ all the color and what not.  It also says "Over 700,000 annual participants."

Maybe it really says "700.000" participants :).  Well the AMC page is a lot of times unreliable.  You can't expect the AMC guys to make such a large site 100% up-to-date and accurate while dealing with all the other stuff they face, after all.  Anyways, a little bending of figures never hurt anybody :D

Could it include people that use the material but don't take the tests?

[quote="probability1.01"]Anyways, a little bending of figures never hurt anybody :D[/quote]

Famous last words of Enron.

The claim of 700,000 annual participants was an old (I believe exaggerated) claim.  I now routinely (under)estimate about 400,000 annually, with about 160,000 in AMC 8, about 240,000 in AMC 10/AMC 12 and of course about 12,500 in AIME and about 250 in USAMO.

(The old exaggerated claim was probably based on the number of contests shipped or even printed, since until we started scanning answers sheets in the late 80's or early 90's there was no reliable way to accurately count the number of actual participants.)

I will see that the exaggerated claim is removed.

Steve Dunbar
AMC Director

Wait! Do not see to it that the exaggerated claim is removed unless you absolutely must!!!

No one, except a loser with nothing better to do like me, is going to check to see if that 700,000 figure is ccorrect.  You could probably change it to over one million annual participants.  If people do find out, they'll just assume the other participants are hidden somewhere.  It's good to have a large number, even if it's not quite true.

Of course, if you feel that you need to change it, go right ahead.  I'm sure there are ethical and other reasons why you would want to change it.  But don't change it because I pointed it out on this forum--the last thing I want to do is hurt the AMC.  You could order moderator deletion of the entire thread if you want to keep it a secret.  I want the best for the AMC, and if that means exaggerating your figures, go right ahead.  Say over 10000000000000000 annual participants for all I care.

There's reasons beyond simple moral values that might require such figures be changed. Prof. Dunbar is obviously doing the right thing by seeing to it that the estimate is corrected.

of course.  I know that.  I just wanted his reason to be "Morally I had to, plus I want accurate information on the AMC website" not "gee, some kid named bubala created this thread on AoPS and unless I change it, the AMC will be the next Enron."

Let's all tell our friends and neighbors to participate and make the higher number true.

My reason is because I want honest and accurate information on our website, plain and simple.

I acknowledge the frequent complaints about accuracy of information on our website.  Our website orginated in the second half of the 1990's as a simple HTML information page.  Obviously the site has grown since then, but what has not grown with it is a modern Web content management system.  Our (part-time!) webmaster is this very moring learning about exactly such a system.  In time, that should help minimize inaccuracies.   In the meantime, gentle reminders will help us to make corrections.

Over the same time that we have been adding information to the Website, we have also added a second AIME date, the B contests, and doubled the participation in the USAMO and MOSP.  Our attention may have been slightly diverted from the fine print on our pages.

Steve Dunbar
AMC Director

So less people participate in the AMCs than MathCounts? *sigh* They should have a MathCounts-like competition in high school.

[quote="Phelpedo"]They should have a MathCounts-like competition in high school.[/quote]

ARML anyone?

And really, the AMCs probably aren't too far off. The problem is that by the time people get to high school, too many of them will be able to score perfects on MathCounts...

Not if it is harder. The point is that MathCounts is a middle-school contest, while AMC is like middle school and high school.

Haha, I just noticed that new edition to your sig Miths. Or "miss perfect"

Mathcounts is good for middle schoolers as prep for higher competitions. The only thing that distinguishes mathcounts is the countdown round. That might be cool for high schools, but it might not work either.

[quote="Phelpedo"]So less people participate in the AMCs than MathCounts? *sigh* .[/quote]

I don't think so.  The MathCounts numbers come from how many students are exposed to MathCounts materials.

[quote="MithsApprentice"][quote="yif man12"]Haha, I just noticed that new edition to your sig Miths. Or "miss perfect"[/quote]

Yes, it's quite wonderful isn't it. I must say that it was designed by custom order. (i.e. I have an awesome best friend.)

Anyways, as for the countdown round, don't they do something like that for the top scorers of ARML?[/quote]

I don't even know what the ARML is, except that it's American Regional Math League (?)

[quote="paladin8"]Go here: [url=http://www.arml.com/]http://www.arml.com/[/url][/quote]Of course, if your town doesn't [i]have[/i] an ARML...
Online stuff is great, but I really wish there was some way of broadening ARML (and Math Meet)-style competition to a greater geographic area.  Alas, the hopeless conclusion is that I must move somewhere bigger or mather after all my studying.

[quote="fedwinri"]
Of course, if your town doesn't [i]have[/i] an ARML...
Online stuff is great, but I really wish there was some way of broadening ARML (and Math Meet)-style competition to a greater geographic area.  Alas, the hopeless conclusion is that I must move somewhere bigger or mather after all my studying.[/quote]

Where do you live? The great thing about the ARML competition is that teams can be organized by anyone in just about any kind of way - either by a teacher, a student, or just any old local citizen who likes math.

If you live in an area without a team, and you think you can get together 15 people who love math enough to travel to one of the sites, then you could maybe organize a team. All you need is someone not on the team to serve as your "coach" or sponsor.

Well, actually, you need two proctors. But if you're not already in the geographic area of an existing team, there are probably lots of places around you that aren't either, so you could try to organize a team.

In terms of ARML teams, I just wanted to add that some states have ARML teams composed of students from across the state---that is, there isn't a geographic concentration of students on the team. It may be that no two students are from the same school/city.  So even if you're the only person you know who might be interested, that shouldn't stop you.  I'd start by asking one of your teachers if s/he is a member of your state's chapter of the NCTM (National Council of Teachers of Mathematics) because then s/he can use that network to find other interested teachers/students across the state.

[quote="gauss202"]Where do you live? The great thing about the ARML competition is that teams can be organized by anyone in just about any kind of way - either by a teacher, a student, or just any old local citizen who likes math.

If you live in an area without a team, and you think you can get together 15 people who love math enough to travel to one of the sites, then you could maybe organize a team. All you need is someone not on the team to serve as your "coach" or sponsor.[/quote]I live in Utah--and the town isn't much of a helpful aspect either, but even in Salt Lake, I doubt if I could get more than 25 bites tops.  Trouble is--flying on site to a competition no one (except a few) has even heard of requires rounding up a lot of rich kids Or a lot of funding.  Either way, it could be done...
[quote="smbelcas"]I'd start by asking one of your teachers if s/he is a member of your state's chapter of the NCTM (National Council of Teachers of Mathematics) because then s/he can use that network to find other interested teachers/students across the state.[/quote]I'll ask about the NCTM; I think so.  There is a problem too in the prospect of winning.  Nobody excactly wants to lose.  From the competitions you've all been to, is it worth going (over a distance) if you don't win anything?  I guess the math is good any day, but is it worth it?  (By the way, this is the kid from Mathcamp talking, so it's not me you're talking into this thing, it's the other fifteen ppl needed for the team.)

 [hide="Eddy"][Edit:}Yay, I'm a Hodge Conjecture!  I need the somewhat useless reminder to mark my passage from newbie-dom (as here defined).  'Better l[url=http://www.claymath.org/millennium/Hodge_Conjecture/]o[/url]ok [hide="that"][i][size=125]H[/size][size=117]odge [/size][size=125]C[/size][size=117]onjecture [/size] [/i] In the twentieth century mathematicians discovered powerful ways [size=67]to [/size][size=50]inv[/size][size=42]es[/size][size=34]t[/size][size=17]ig[/size][size=9]ate [/size] [size=25]This[/size][size=59] tech[/size][size=67]nique[/size][size=75] turned out [/size][size=84]to[/size][size=100] be so useful t[/size][size=84]hat i[/size][size=67]t go[/size][size=50]t general[/size][size=42]ize[/size][size=34]d in m[/size][size=25]any di[/size][size=17]ff[/size][size=9]eren[/size][size=17]t[/size][size=25] wa[/size][size=34]ys[/size][size=42], ev[/size][size=50]en[/size][size=59]tu[/size][size=67]al[/size][size=75]ly [/size][size=84]le[/size][size=92]ad[/size][size=100]ing to [/size][size=117]powerful tools[/size][size=100] t[/size][size=92]ha[/size][size=75]t[/size][size=59] e[/size][size=42]n[/size][size=25]a[/size][size=9]bled[/size][size=42] mathematicians [/size][size=34]to[/size][size=17] m[/size][size=9]ake [/size][size=42]The [/size][size=50]Hodge conjecture [/size][size=59]asserts [/size][size=67]that for [/size][size=75]particularly [/size][size=84]nice types of spaces [/size][size=92]called [/size][size=100]projective algebraic varieties, the pieces called Hodge cycles are actually [/size][size=75](rational linear) [/size][size=100]combinations of geometric pieces called algebraic [/size][size=92]cycles.[/size][/hide] up...[/hide]

Utah actually used to have an ARML team back in the 90s.  I'm not sure when they stopped coming, but it may have been when the western site was moved from UNLV to San Jose State in 2001. I imagine San Jose is a pretty long way from Utah. :oops:

But its certainly possible to organize a team. There are probably more students interested than you might think, and raising funds to get there just requires some time and some persistence on the part of either the coordinators or the students or both.

The main point of ARML, or any math competition for that matter, though, isn't winning. There's something like 90 teams that go to ARML, and only a few of them can win awards. So the main point is just the camaraderie between the teammates, and the chance to get to see some good math problems.

I went to ARML last year as a freshman and had a great time.  The only thing I wish could have happened was that we would have actually had to travel rather than just going the 5 miles or so to the U of I.  If you can organize it somehow I would highly recommend going.


calculus
integration
calculus computations

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=19273[/img]

is anyone interested in solving those! i solved most of them, but don't know if i got it right or not. checking for solution.

Hello
[hide="1"]
put $ e^{2x} \equal{} t$
$ \int\frac {e^{2x} dx}{1 \plus{} e^{4x}} \equal{} \int\frac {dt}{2(1 \plus{} t^2)} \equal{} \frac {tan^{ \minus{} 1}e^{2x}}{2} \plus{} C$

[/hide]
[hide="2"]
put $ sec\theta \equal{} t$
so ur integral will be
$ \int^{\sqrt2}_{1} (t^2 \minus{} 1)^2 t^2 dt$
=$ \int^{\sqrt2}_{1} (t^6 \minus{} 2t^4 \plus{} t^2) dt$
[/hide]

[hide="3"]
put $ 1\plus{}ln x\equal{}t^2$
$ \int \frac{2t^2 dt}{t^2\minus{}1}$
=$ 2 \int (1\minus{}\frac{1}{t^2\minus{}1})dt$
=$ 2\sqrt{1\plus{}ln x}\minus{}log(\frac{\sqrt{1\plus{}ln x}\plus{}1}{\sqrt{1\plus{}ln x}\minus{}1})\plus{}C$
[/hide]
[hide="4"]
$ \frac{x^3\plus{}1}{x^3\minus{}x^2}\equal{}1\minus{}\frac{x^2\minus{}1}{x^3\minus{}x^2}\equal{}1\plus{}\frac{2}{x\minus{}1}\minus{}\frac{1}{x}\minus{}\frac{1}{x^2}$
[/hide]

[hide="9"]
put $ x\equal{} tan y$
so $ sin^{\minus{}1}(\frac{2x}{1\plus{}x^2})\equal{}2 tan^{\minus{}1}x$
now use integration by parts.
Answer =$ 2 x tan^{\minus{}1}x \minus{} log(x^2\plus{}1)\plus{}C$
[/hide]
[hide="8"]
$ \int \sqrt{3\minus{}2x\minus{}x^2} dx$
$ \equal{}\int \sqrt{4\minus{}(x\plus{}1)^2} dx$
$ \equal{}\frac{x(\sqrt{3\minus{}2x\minus{}x^2})}{2}\plus{}2 sin^{\minus{}1}(\frac{x\plus{}1}{2})\plus{}C$
[/hide]

use partial frac for vi.
sorry buddy
  :| My sir will kill me if i will not leave now.So i will try to post others questions after hours only when i will online again.

anyone else wants to solve these integrals? thx..

hello, in (v) we can set $ x\plus{}1\equal{}t$, then we get $ dx\equal{}dt$ and our integral is
$ \int\frac{(t\minus{}1)^3}{t}\,dt\equal{}\int3\minus{}\frac{1}{t}\minus{}3t\plus{}t^2\,dt$.
Sonnhard.

hello, in (vi) we can write the integrand as
$ \frac{1}{(x\minus{}2)(x^2\plus{}4)}\equal{}\minus{}\frac{1}{8}\frac{x}{x^2\plus{}4}\minus{}\frac{1}{4}\frac{1}{x^2\plus{}4}\plus{}\frac{1}{8(x\minus{}2)}$
Sonnhard.

hello, in (vii) we can write the denominator as
$ \sin(x)^2\plus{}\cos(x)^4\equal{}\frac{1}{2}(1\plus{}\cos(2x)^2)$ and our integral is
$ \int\frac{\sin(2x)}{1\plus{}\cos(2x)^2}\,dx$ and you can substitute $ t\equal{}\cos(2x)$.
Sonnhard.


inequalities
integration
real analysis
function
calculus
absolute value
real analysis unsolved

I appreciate if someone can help me out with that because I think it can be true because of the inequality

\[ \left| \int f \right| \leq \int |f|,\]

but I don't know why I feel it is not that easy.  Maybe, the key is about the measurability of $ f$.

Thank you in advance

In the Lebesgue version, this is very close to the definition of integrability. The positive part $ f^\plus{}(x)\equal{}\begin{cases}f(x)&f(x)>0\\0&\text{otherwise}\end{cases}$ and negative part $ f^\minus{}(x)\equal{}\begin{cases}\minus{}f(x)&f(x)<0\\0&\text{otherwise}\end{cases}$ are nonnegative functions, with integrals in $ [0,\infty]$. If both integrals are finite, we define $ \int f\equal{}\int f^\plus{}\minus{}\int f^\minus{}$.

The integral of the absolute value is of course $ \int f^\plus{}\plus{}\int f^\minus{}$, which is finite when both integrals are finite. That's exactly the same condition.

Of course, the full statement is that if $ f$ is measurable, then $ |f|$ being integrable implies that $ f$ is integrable.

If you don't separately assert the measurability of $ f,$ then you're in trouble. For example: consider an nonmeasurable set $ E\subset [0,1].$ Let $ f(x)\equal{}1$ for $ x\in E$ and $ f(x)\equal{}\minus{}1$ for $ x\in [0,1]\setminus E.$ (And let $ f$ be zero outside of $ [0,1].$ Then $ |f|$ is integrable, but we can't do anything with $ f$ itself, since it isn't measurable.

[quote="Kent Merryfield"]Of course, the full statement is that if $ f$ is measurable, then $ |f|$ being integrable implies that $ f$ is integrable.

If you don't separately assert the measurability of $ f,$ then you're in trouble. For example: consider an nonmeasurable set $ E\subset [0,1].$ Let $ f(x) \equal{} 1$ for $ x\in E$ and $ f(x) \equal{} \minus{} 1$ for $ x\in [0,1]\setminus E.$ (And let $ f$ be zero outside of $ [0,1].$ Then $ |f|$ is integrable, but we can't do anything with $ f$ itself, since it isn't measurable.[/quote]

But in your example, if I use the definition of the integral of a simple function (in this case $ f$), then the integral should be
\[ \int f \equal{} m^* (E) \minus{} m^*([0,1]\setminus E)
\]
which is just a number since the outer measure is well-defined for [b]any[/b] real set.   The definition of the integral of a simple function is what makes me confused since I can perfectly define the integral of $ f$ to be the value which is obtained as above.  In fact, why do we need to assume measurability for the sets in order to define the integral of a simple function (or more general in any function)?

Regards

[quote="HilbertThm90"]"Definition of the integral of a simple function"

I'm pretty sure all texts define the integral for a 'measurable simple function'. So to get the integral of $ f$ you need to, by definition, take a sequence of measurable simple functions that converges up to $ f$. Here we have a problem. This can't be done.

On the other hand, what exactly is $ m^*(E)$?  :wink:[/quote]

It is what is :)

$ m^*(E) \equal{} \inf \{\sum_{j \in J} lenght (I_j): J$ is countable, $ \{I_j\}$ is a collection of intervals and $ E \subset \bigcup_{j \in J} I_j \}$

So, my question was what important property about INTEGRABILITY fails when I consider the outer measure?

thanks

Measurability is really "recombinability" - we can tear a measureable set apart into pieces and and then be able to properly put the pieces back together again.

Any attempt to base a theory of integration on nonmeasurable sets would result in very many basic theorems utterly failing. For instance, we would not be able to say that $ \int(f\plus{}g)\equal{}\int f\plus{}\int g,$ even for $ f$ and $ g$ the characteristic functions of disjoint sets.

Since the very idea of integral is rooted in sums (down to the fact that the symbol $ \int$ is really just the letter "S" which stands for "Sum") I would not accept as a theory of integration anything which denied basic linearity.

[quote="Kent Merryfield"]Measurability is really "recombinability" - we can tear a measureable set apart into pieces and and then be able to properly put the pieces back together again.

Any attempt to base a theory of integration on nonmeasurable sets would result in very many basic theorems utterly failing. For instance, we would not be able to say that $ \int(f \plus{} g) \equal{} \int f \plus{} \int g,$ even for $ f$ and $ g$ the characteristic functions of disjoint sets.

Since the very idea of integral is rooted in sums (down to the fact that the symbol $ \int$ is really just the letter "S" which stands for "Sum") I would not accept as a theory of integration anything which denied basic linearity.[/quote]

Perfect! After posting I realized that integrals are based on sums, so the key point is the countable additiviy.   Thanks everyone who answered.


Suppose $a, b, c, A,B,C$are real numbers such that for all x,
$|ax^2 + bx + c| \leq |Ax^2 + Bx + C|$.
Prove that:
$|b^2 - 4ac| \leq |B^2 - 4AC|$.

Here is an incomplete (and akward) proof:

Let $y_1=|ax^2+bx+c|$ and $y_2=|Ax^2+Bx+C|$

1) Consider $b^2-4ac\ge0$, meaning that $y_1$ has at least one x-intercept.

If $y_2$ has no x-intercepts, then............INCOMPLETE

If $y_2$ has x-intercept(s), they must be where $y_1$'s intercept(s) is(are).  If it were otherwise, $y_2$ would be zero where $y_1$ is nonzero.  Then the distance between $y_1$'s x-intercepts must eqaul the distance between $y_2$'s.   Namely, $\left|\frac{\sqrt{b^2-4ac}}{2a}\right|=\left|\frac{\sqrt{B^2-4AC}}{2A}\right|$.
As $x$ increases without bound, $y_1\to|a|x^2$ and $y_2\to|A|x^2$, meaning that $|A|\ge|a|$.
So, $\left|\sqrt{b^2-4ac}\right|=\left|\sqrt{B^2-4AC}\right|$, implying $|b^2-4ac|=|B^2-4AC|$

2) Consider $b^2-4ac<0$, meaning that $y_1$ has no x-intercepts.  ($y_2$ must also have no intercepts; otherwise, $y_2$ would be zero when $y_1$ is nonzero.)

So, $ax^2+bx+c$ and $Ax^2+Bx+C$ are each wholly on one side of the x-axis, meaning that $y_1$ and $y_2$ are each a parabola.

$y_2$ must be above and "inside" $y_1$, implying that the y-coordinate of $y_2$'s vertex be least that of $y_1$'s.

Namely, $\left|\frac{B^2-4AC}{-4A}\right|\ge\left|\frac{b^2-4ac}{-4a}\right|$.  By using $|A|\ge|a|$,
$|B^2-4AC|\ge|b^2-4ac|$.


summer program
MathZoom
counting
distinguishability
complementary counting

A PE class has 12 students, 6 girls and 6 boys. The coach has 4 jerseys in each of 3 colors to mark 3 teams for a soccer tournament. If the coach wants at least one girl and at least one boy on each team, how many ways can he give out the jerseys? (Jerseys of the same color are indistinguishable.)

This problem requires a lot of casework. I'm stuck on it.

I tried complimentary counting and still didn't get the right answer.

The official answer is 29700, but I got 26100. However, I think my solution is correct:
Regardless of the restriction there are $\binom{12}{4}\binom{8}{4}=34650$ ways to distribute the jerseys.

If 4 boys are on a team, then for the other two teams, we would either have 1B 3G + 1B 3G or 4G + 2B 2G.

In the first case there are $2(3!)\binom64\binom63\cdot2=7200$ - 2 for either no girl or no boy on a team, 3! for 3 different colors of jerseys/ 3 different teams, 6-choose-4 to choose the 4 same-gender people on the team, 6-choose-3 to divide the girls into two groups of 3's, and 2 for the last two boys to join either groups of 3 girls.

If it is 4B + 4G + 2B 2G then it is simply $\binom64^2\cdot3!=1350$

13650-7200-1350=26100

Could someone please tell me if 26100 is indeed the correct answer, and the official answer is incorrect?

EDIT: Nevermind, in the first case there shouldn't be an extra factor of 2 in the end... so 13650-3600-1350=29700 is correct... (thanks to gianlino for spotting that error)

This was what I did.  I didn't use complementary counting, though.  In the other forum, I had everything right except for the $21600 + 8100$ part; I put $29600$ instead of $\boxed{29700}$ which is the answer.  :rotfl:

[hide="My Work"]
Let's name the teams.  I'll name the teams Black, Blue, and Red (after the groups of MathZoom :P ).  We know that there have to be at least one girl and at least one boy on each team, so:

Black:  B G _ _
Blue:  B G _ _
Red:  B G _ _

Then, if we let the girls and boys be indistinguishable, we have 7 cases:

Black:  BGBB
Blue:  BGBG
Red:  BGGG

Black:  BGBB
Blue:  BGGG
Red:  BGBG

Black:  BGBG
Blue:  BGBB
Red:  BGGG

Black:  BGBG
Blue:  BGGG
Red:  BGBB

Black:  BGGG
Blue:  BGBB
Red:  BGBG

Black:  BGGG
Blue:  BGBG
Red:  BGBB

Black:  BGBG
Blue:  BGBG
Red:  BGBG

Now, we let the groups have distinguishable people.

Case 1:  $\binom{6}{3} \cdot \binom{6}{1} \cdot \binom{3}{2} \cdot \binom{5}{2}$ which simplifies to $3600$.  The other people have to be on the Red Team, what other team could they possibly be on?
Cases 2 through 6 are basically the same as Case 1; the same arrangements just called different things (teams).  YAY work saved  :)
Case 7:  $\binom{6}{2} \cdot \binom{6}{2} \cdot \binom{4}{2} \cdot \binom{4}{2} = 15 \cdot 15 \cdot 6 \cdot 6 = 8100$.

So $3600 \cdot 6 + 8100 = \boxed{29700}$.
[/hide]

I got the answer wrong, but I don't know where I went wrong in my work

My method:
[hide]We'll call the teams red, yellow, and blue.

First, we have to choose the boy to be on the red team. There are 6 ways to do that. Then, we need to pick the girl for that team, and there are 6 ways to do that, so there are $6\times6=36$ ways to pick the first two people. For the yellow team, there are 5 ways to pick the boy, 5 ways to pick the girl. For the red team, there are 4 ways to pick the boy, and 4 ways to pick the girl. Therefore, the first six people can be selected in $36\times25\times16=14400$

Now to pick the remaining members. There are $\binom{6}{2}$ ways to pick the remaining people on the red team, $\binom{4}{2}$ ways to pick the remaining people on the yellow team, and $\binom{2}{2}$ ways to pick the remaining people on the blue team. Therefore, there are $\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}=15\times6\times1=90$

The total number of ways is $14400\times90=\boxed{1296000}$[/hide]

I know this answer is wrong because the total number of ways to pick the teams with no restrictions is $\binom{12}{4}\binom{8}{4}=34650$ and 1296000 is way bigger than 34650. Please let me know where I went wrong.

I did the exact same thing ^ could anybody explain?

[hide=better solution?] I found that there are two ways to make the teams based on gender. The first way: 1 boy 3 girls, 2 boys 2 girls, 1 girl three boys. Hence, there are $3!*\binom61*\binom63*\binom52*\binom32=21,600$ cases ($3!$ because of the different colors).The second way: 2 boys 2 girls, 2 boys 2 girls, 2 boys 2 girls. Hence, there are $\binom62 *\binom62*\binom42*\binom42=8,100$ At last, the number of total cases is $21,600+8,100=29,700$.

[quote=Swonica2718]I got 1296000 by the same reasoning as the others above. Why is this incorrect?[/quote]

Um, I got the EXACT same answer as what everyone else did. Someone please explain why $1{,}296{,}000$ is wrong.

I don't get this part
[hide]"However, this procedure double-counts the assignments that make one team all girls and another all boys. There are 3 choices for the girls team and then 2 choices for the boys team, and $\binom64^2=225$ ways to choose the teams, for a total of $2\cdot3\cdot225=1350$ double-counted arrangements"[/hide]
Where do the 3 and 2 choices come from, and why do we square 6 choose 2?

the complementary counting solution is somewhat well-known
like by that i mean this type of problem is p common just count total then subtract the number that have all girls and all boys
a version on this problem was in 2010 amc10b
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22
BUT i see that 1296000 is a common trap answer and to check my work i actually got this using combinations and multiplication but then by the reasoning of #6 i deduced that it could not be the answer
it def overcounts, not sure where

I think I have a theory on why 1296000 is wrong (I think the exact same cases are counted more than twice):

Number the girls $G1, G2, \cdots, G6$ and the boys $B1, B2, B3, \cdots, B6$. Using the 1296000 argument, we pick one girl and one boy for the first team, and WLOG pick $G1$ and $B1$. Then for team 2 we pick $G2$ and $B2$, and for team 3 $G3$ and $B3$. As for the remaining 6 students we put $G4$ and $B4$ in team 1, $G5$ and $B5$ in team 2 and $G6$ and $B6$ in team 3. That would count as one case.

Now let's pick another case. We pick $G4$ and $B4$ on team 1, $G5$ and $B5$ on team 2 and $G6$ and $B6$ on team 3. Then we put $G1$ and $B1$ on team 1, $G2$ and $B2$ on team 2 and $G3$ and $B3$ on team 3. The teams are the exact same, but the 1296000 approach counts them as 2 different cases!

I decided to just select the remaining students with order mattering and divide by $4!$ three times because order doesn't matter on a team, and I got 750 which was wrong for some reason.

[hide=Solution]Notice that there is either one team that contains $2$ boys and $2$ girls or all three teams contain $2$ boys and $2$ girls.

[b]Case 1:[/b] One team contains $2$ boys and $2$ girls.

We must multiply $3!$ since there are $3$ distinct teams. Then, we want to choose which of the $6$ boys join the team that contains $2$ boys. Then, we want to choose which of the $6$ girls join the team that contains $2$ boys. Then, we want to choose which of the remaining $4$ boys join the team that contains $3$ boys. Notice that the remaining boy joins the final team. Then, we want to choose which of the remaining $4$ girls join the team that contains $3$ girls. Notice that the remaining girl joins the final team.

$$3! \cdot \binom{6}{2} \cdot \binom{6}{2} \cdot \binom{4}{3} \cdot \binom{4}{3}=21600$$

[b]Case 2:[/b] Three teams contain $2$ boys and $2$ girls.

Firstly, we have to choose which of the $6$ boys join one team. Then, we have to choose which of the remaining $4$ boys join the second team. Notice that the remaining $2$ boys join the last team. Then, we have to choose which of the $6$ girls join one team. Then, we have to choose which of the remaining $4$ girls join the second team. Notice that the remaining $2$ girls join the last team.

$$\binom{6}{2} \cdot \binom{6}{2} \cdot \binom{4}{2} \cdot \binom{4}{2}=8100$$

Thus, the final answer is $$21600+8100=\boxed{29700}$$


Given 7 lines on the plane, prove that two of them form an angle less than 26 degrees.

You could have seven parallel lines, in which case they don't even form any angles.

[quote="diophantient"]Assuming that no two lines are parallel.[/quote]

...or define the angle between parallel lines as 0 since they intersect at the point at infinity

[hide="an idea?"]Do you just shift the lines so that they are all concurrent because shifting lines do not change the angle between them...?  Then, the largest that the smallest angle can be is $ \frac {360}{14} < 26$.[/hide]


Hello Everybody,
I am interested in:

1) The Art and Craft of Problem Solving by Paul Zeitz.
2) Problem Solving Strategies by Arthur Engel.
3) Geometry Revisited by H. S. M. Coxeter and S. L. Greitzer.
4)Principles and Techniques in Combinatorics by Chen Chuan-Chong
   and Koh Khee-Meng
5)Mathematics of Choice by Iven Niven

So, if you have the above mentioned books and you want to sell them for any reason whatsoever, and you are in and around the GTA, or any chance to swing by in the near future, start giving me your bids.  :D

how is this exchange going to take place?  :huh:

Well, I believe the exchange should take place at a mutually decided and agreed upon public place, like a major shopping mall, etc. Or if you want, we can have the exchange at a Coffee Time, etc. Btw, what books are you interested in selling to me?

I know, I was just answering your question.

Ok. I have added two new titles to my list of books that I need to get. Please check the list again to see if you have any of them and  pthen if you want to make some money, contact me.

Oh man, I can't believe this. No one has these books!!! No one wants to make money!!! Please people, prove me wrong, start giving me your offers :lol:

Maybe if you thought of an easier way for the exchange to take place, there would be an increase in interest.  Perhaps you and the seller send the money and book to a trusted third party?  btw, personally I AM LOOKING FOR A COPY OF GEOMETRY REVISTED :!:

Well, I am thinking about easier ways for the exchange to take place. In the meantime, if you find Geometry Revisited or one of the other above mentioned titles at a fair price, please do let me know. We are in this(book search) together. :D


Let $ n\equal{}p_1^{\alpha_1}...p_s^{\alpha_s}\geq 2$. If for any $ \alpha\in N$, $ p_i\minus{}1$ does not divide $ \alpha$, where $ i\equal{}1,2,...,s$. Prove that $ n|\sum_{\alpha \in Z'_n}a^{\alpha}$, where $ Z'_n\equal{}\{a\in Z_n: (a,n)\equal{}1\}$

you might want to recheck the problem statement. if you're gonna copy problems from wherever it is you get them, at least do it right.


A straight, vertical  43-inch wick was lit from the top and burned downward at the constant rate of 1.5 inches per hour. Three hours later, the bottom of the wick also was lit, and it burned upward at a constant rate of 4.0 inches per hour. How many total inches had the top of the wick burned down when the burning ends met.

Answer & Thoroughly explain, plz :-)

[hide="Solution"]In three hours the top of the wick loses $ 1.5\times {3} \equal{} 4.5$ inches. So only $ 43 \minus{} 4.5 \equal{} 38.5$ inches of the candle is left. Then we are told that the bottom is also lit so the total number of inches the candle loses in hour is now $ 1.5 \plus{} 4 \equal{} 5.5$ We divide $ 38.5$ by $ 5.5$ to get the number of hours it will take the candle to completely burn out after the bottom was lit. $ \frac {38.5}{5.5} \equal{} 7$. Thus, the top of the wick burned for a total of $ 3 \plus{} 7 \equal{} 10$ hours at $ 1.5$ inches an hour so the total number of inches the top of the wick burned down is $ 10\times{1.5} \equal{} \boxed{15}$[/hide]


there are no ideas for this problem?!! that's very strange..!

Lets rewrite this inequlity
for $1 > x_{i}\ge 0$ prove that
$(\frac{\sum^{n}_{i=1}x_{i}^{x_{i}}}{n})^{n}\le (\frac{\sum^{n}_{i=1}x_{i}}{n})^{\sum^{n}_{i=1}x_{i}}$
lets put $s=\sum^{n}_{i=1}\frac{x_{i}}{n})$
so we have
$\frac{\sum^{n}_{i=1}x_{i}^{x_{i}}}{n}\le s^{s}$
$(x^{x})''<0$ on $(0;e)$ so lust inequality is inequality Iensena.
also it is right for $e \ge x_{i}\ge 0$

[quote="derezin"]By $\{x\}$ you mean $x-[x]$?[/quote]

Waiting for answer..... :maybe:


Hi all!

I really wanted to ask how you guys study for the AP Chemistry tests. I'm having some difficulties with the class, particularly since I don't have a very solid base in Chemistry. Are there any practice problems that you usually do? Are there any particular strategies that you follow while studing?

If you have any, please post some links to practice problems and/or past exams (AP non-AP, whatever, as long as they are in the AP material range), especially for Equilibrium and Acid-Base dissociation, acid classifications, etc.

Thank you very much!

-Lena

1. Be a whiz at stoic.  It helps a lot.
2. Study the theory really well, and practice using a systematic approach (the one in your textbook, if you have one)
3. If you don't have a textbook, borrow or buy one.  I know Zumdahl and Brown-lemay pretty much lays out for you step by step how to approach each type of problems.
4. Make sure you attempt the hardest problems of each chapter.  If you don't, you won't really know if you actually know the material or not.

Hope it helps.

1. Make sure you understand stoich theory to the fullest extent. You'll save yourself tons of time if you know how to do the math behind chemistry.

2. With number 1 in mind, you'd only have to study theory behind each test. Most math-type chemistry problems are monotonous step-by-step playbook questions, meaning each calculation method is pretty much the same from problem to problem. So understand how to do the types of problems you should expect (Zumdahl does a nice job of showing you the different possibilities) but you should focus on theory.

3. Understand the formulas you're using. Hopefully the textbook you're using or the proffessor you have derives the formulas you use.

4. Link the concepts! Don't forget you can pull stuff from other units. Collision theory can not only be applied to the gas laws, but to kinetics also!