Hey,
Do u guys all have the Nerds-Jocks-PopularKids Divide in your school? I used to be a popular kid and then I developed a liking for math and science..so I hung out with the nerds(I don't know why they are called so, they are all so intelligent and nice) and now I'm being called NERD and being ragged about it 24X7...:(  :(
If you even TRY not to belong to one group, like I sit with some people on one day and with some other people on another day, then people are really hurtful to me because of this and I have no friends at all...do you guys go through the same thing?
I really believe that it's not mandatory to belong only to one category but no-one else thinks so...does this happen to everyone?

[quote="7h3.D3m0n.117"]It is through my experiance that you can only really "fit in" with one group. Some of my jock friends try to "unNerd" me as they call it, and I don't really care if they think I am...eventually they accepted me for who I am and even hung around some of my nerd friends.

I would advise you to stick with one group, preferablly a group that you are comfortable with, and that you will be welcomed.[/quote]

Hmm...I Quite disagree...
What's happening with me is that I am called a MAJOR nerd and I'm proud of it, but that doesn't stop me from going and hanging out with all the un-nerds..I guess I have one thing in common with one group or the other..I don't belong to one group, but go from one group to another...one day with the so called "failures" because I discuss cricket with them, one day with the so called "cool" people to discuss music or movies...and no-one really minds coz they know that I HATE groupism...so I won't hang out with them for long!!
I btw think it's the most evil thing in school today and should be got rid of....
While you're so busy trying to fit in somewhere, you hardly notice someone who maybe sitting right next to you and badly needs your help but whom you've ignored till today coz she wasn't "Part Of The Group"
What I do is get almost the whole class to get together and play ball or eat lunch...and none of them mind coming out of their groups for my sake coz they know that I'll get up and go if they refuse...and then they end up finding out that they are actually having fun in a group of 40.
I think you should try the same...don't stick only with one group all the time..
I'd say hang out with the nerds coz ur new to school right now, wait for a while until the rest of the class notices that you're not a one sided person...but try as far as possible not to be part of this groupism...then you'll suddenly find everyone becoming friends :)

I used to be considered "cool" by my friends from middle school. Then i went to an extremely good high school, and became interested in math and studied on my free time. I went to visit my old friends, and I had a notebook with me, because I had to ride the subway for one hour to get to their neighborhood, and i wanted to study some math during thaat time. They saw me with the notebook, and they never spoke to me again, becauseif i hanged out with them i would "ruin their reputation". So i guess that they weren't real friends at all.

A good article on the subject: [url=http://www.paulgraham.com/nerds.html]Why Nerds are Unpopular[/url]

The people at our school called "preps", or people who like having a reputation for being unspectacularly popular, are really the only ones who care about who they can't and can communicate with in their daily life, or in the school spectrum. That tightly-knit group is really the only group that should care about something as miniscule as this. Preps at our school are the ones that do things to bring out adulation, such as calling hot dogs "Look it's a weiner!" and what-not. Or sticking a french fry up their nose or talking about, oh wow let's rate which one of these girls is the hottest and we will ask them out nonsense."

The rest of us are just considered normal, everyday people, and only the preps frequently refer to anyone as a nerd. We usually just don't want to hang out with gangs, badly influencing people, etc.

I don't know how it's like at other schools, but this is how it feels like to be in my school.

I've always figured people just hang out with whoever they like to, and this has an overwhelming tendency to have people develop almost-closed groups of friends, independent of some sort of clique 'classification.'

[quote="MissSmartyPants"]I'm being called NERD and being ragged about it 24X7...
[/quote]
since i come from a smaller country where the division into groups is not so strong, i'm not sure if i'll be able to grasp the situation but i'd like to know what is the thing that gets to you the most?
is it the fact that you're not popular anymore or the fact that people tease and make fun of you?
i mean, if somebody makes fun of you you should be able to return the favour ten times better(the only thing i never knew how to defend myself from was when somebody ran up to me and yelled NERD acompanied with a couple of moose/gorrilla like sounds... that's because i don't speak animal languages...)
if you just miss being popular,i'm really sorry for you but you need to understand that it really isn't the most important thing in the world. the best thing to do might be to (big clichee coming up) be yourself and if you're an ok person people will notice you for who you are. just don't worry about it to much, people who do tease you are just jealous of you because their parents keep asking them why can't they be more like you.
about not belonging to a group.
since everybody is in a group(that's what you say) maybe they don't like you trying to be different,why should you be better than them and not be in a group. again, people in puberty are selfish and jealous and try to get you down so they can feel better and more powerful,that because they're confused and in a hard period of life(just like you).they just express that confusion in a different manner...
so again,don't worry so much, don't stress out, do what you like, hang out with whoever you like and don't let anybody nag you about it. and oh yeah,this might be important. don't lose any verbal confrontation ;) that should help out...

My experience in high school goes in line with what elemennop said.  I never felt excluded from certain groups.  Rather, I had my own friends that I stayed around most of the time.  Most of the friends I made were from my classes or from cross country.  Similarly, for example, the football players would in general become good friends with each other.  So you could say they formed a clique.  However, I never felt that the clique was exclusive.

[quote="goc"]i mean, if somebody makes fun of you you should be able to return the favour ten times better(the only thing i never knew how to defend myself from was when somebody ran up to me and yelled NERD acompanied with a couple of moose/gorrilla like sounds... that's because i don't speak animal languages...)[/quote]

Haha. I got a good laugh out of this one.

And anyway, there will always be social divides in an environment. Live with it, do what you want about it, but don't obsess over it.

Groupism does not matter if you act like you want, not what other people want.
Oh to get back at the people when they say that I am something I am not, I say
"You know your voice when you said that sounded like a cow giving birth to siamese twins."

here's what i think of this:
if these people are the kind that would hate you for being smart, they probably aren't the kind of people you should be hanging out with anyways
and if they are bugging you about being a nerd, that probably means they have some academic or other troubles themselves or they are simply doing it because others are doing it and they want to follow the pack
i'm personally lucky in that there are plenty of people, popular and not, that admire intelligence.  there are also some who don't, but they for the most part don't rag on me about it, partially because they have friends who think its a good thing

You got the wrong point. When you choose your friends, your first criterion should be their mentality and warmth. I have many friends who are not sharp, but they are great as human beings. Of course the very second criterion is sharpness. Sharp people are the ones I like to hang out with. (Don't confuse with grades - they are not always strongly correlated.) I had no best friend in school either, but then nor had I any enemy. What I have now got in my old school friends is a bunch of admirers. Warmth and feelings conquers all, mate, so make that your best weapon. My advice is: do not make any foes; that will suffice to keep you happy. Nerdiness should not be a criterion at all.

The best thing you can do is to ignore tastless remarks. Paying attention to people who don't deserve your companionship and respect usually just slows you down and diverts your thoughts from what really matters. But if you're being hurt and if it's causing you to feeling bad or unhappy, then you should strike back as hard as you can. Just stick to what you enjoy doing the most and you'll be fine. :-)

MissSmarty, you are a great girl because you actually can get the truth that the person who are called nerdy actually are cool.
Well, I think the film about US high-school life (such as "American Pie", "not another teen movie", etc.)  is not true, it's too bombastic and hateful, and too much. I think America is the superpower and wonderful country because they honor their snart citizen. But, after I see the posts here, I think the film tell the truth..
I don't think that sporty boy or cheerleaders girls are cool. Meanwhile, I think, they whgo are cool in the school are they who are clever in academic lessons or at least interested..

im the smartest in my grade(evry1 sez that), and i play sports really well, but evry1 doesnt care, they still think im a nerd, and they always sarcastically congratulate me. they dont mean it, and i can skool them bad, but they dont care...
 :(  :(  :(

wear your nerdiness proudly and stick up for your friends.

Everyone in school calls me a nerd and a geek but I have a huge number of friends and they keep telling all and sundry that I'm the smartest in the grade...that doesn't stop them from not sitting with me during classes coz I concentrate and they don't...I'm quite happy with that! :)
Whatever the world says, nerds rule!!
Today's nerds are tomorrows Bosses!!!

[quote="MissSmartyPants"]Well, the other day they took us to Cambridge as part of a field trip...
One of my FRIENDS(everyone calls him a nerd) was:
a)Pushed into a pond
b)Served non vegetarian food and drinks even though he's a pure vegetarian and doesn't drink(AND it was stuffed down his throat forcing him to puke later)
c)Had his shoe pushed down one of the girls' toilets
d)Subject to repeated attempts to push his head into the dustbin
e)Due to his frail build was thrown and slammed against a lab door about ten times
f)His bag was put into an elevator and taken to the terrace from where it was dropped down
g)His glasses were thrown into the pudding
h)He had atleast one glass of juice poured on him each day
And the teachers JUST KEPT QUIET...they said IT WAS PART OF LIFE and that HE SHOULD STOP BEING SO NERDY... :o
And all this started when he volunteered a bit of information about Newton...  :mad:
And forget retaliating, they slap him hard if he even opens his mouth... :mad:
Being new to the school and wanting to pursue activities like the Science, Math, Puzzles and Astronomy Clubs, I hope you can see why I'm scared now... :(   :([/quote]

Holy...i agree with 7h3.D3m0n.117, you should sue the teachers.

Haha im lucky. wow. we have enough nerds in our grade such that we can stick together and form our own big clique.

I'm the only one in my whole class.... :(  :(
And I get called NERD, GEEK, FUNNY, BRAIN and WEIRD alternatively...wouldn't say I disagree!!
EDIT: They call me GENIUS too...but I guess that's debatable... :)

[quote="sylow_theory"]I dont fit any of those. I happen to be a very likeable person. But people dont describe me as a geek rather than "the strange one" or something like that.[/quote]

I see :|

that's strange, at my school, i guess you would call me a nerd, but i'm what you might say "moderately popular" too, so it kind of cancels out.  :|  either way, people respect me for my nerdiness.

[quote="K81o7"]
Dictionary.com definitions of geek:

1. a peculiar or otherwise dislikable person, esp. one who is perceived to be overly intellectual.
2. a computer expert or enthusiast (a term of pride as self-reference, but often considered offensive when used by outsiders.)
3. a carnival performer who performs sensationally morbid or disgusting acts, as biting off the head of a live chicken. [/quote]

if i fit into category 1, i don't know it.
don't fit into category 2
category 3, no. that's disgusting!!!  (however, nowadays nobody pays attention to category 3, people just call those kinds of people freaks or dorks or losers)

[quote="K81o7"]Dictionary.com definitions of geek:

1. a peculiar or otherwise dislikable person, esp. one who is perceived to be overly intellectual.
2. a computer expert or enthusiast (a term of pride as self-reference, but often considered offensive when used by outsiders.)
3. a carnival performer who performs sensationally morbid or disgusting acts, as biting off the head of a live chicken.

I might fit into category 1, though probably not.
I definitely don't fit into category 2.
Category 3...maybe...[/quote]

i fit 1,2,&3 :lol:

jkjkjkjkjk!!!!!!!!!!!!!!!!!!!!!

only 1&2

only one for me, but this is getting off topic...

[quote="Nerd_of_the_Ages"]but this is getting off topic...[/quote]

i agree.
i think the last page of posts was just people trying to convice others( and themselves for that matter) that they are popular... posts like
i'm the smartest one in my class/school/county but people still like me
are really starting to bother me. i don't know if i'd like to hang around with someone who can say such an egocentric sentence... and furthermore i think that most of the people here are like that(really smart), so there's no point in pointing it out :)
what i'm wondering about is
has the situation with misssmartypants changed? what happened?
transfer? legal action? massive linch...?

[quote="MissSmartyPants"]Well, the other day they took us to Cambridge as part of a field trip...
One of my FRIENDS(everyone calls him a nerd) was:
a)Pushed into a pond
b)Served non vegetarian food and drinks even though he's a pure vegetarian and doesn't drink(AND it was stuffed down his throat forcing him to puke later)
c)Had his shoe pushed down one of the girls' toilets
d)Subject to repeated attempts to push his head into the dustbin
e)Due to his frail build was thrown and slammed against a lab door about ten times
f)His bag was put into an elevator and taken to the terrace from where it was dropped down
g)His glasses were thrown into the pudding
h)He had atleast one glass of juice poured on him each day
And the teachers JUST KEPT QUIET...they said IT WAS PART OF LIFE and that HE SHOULD STOP BEING SO NERDY... :o
And all this started when he volunteered a bit of information about Newton...  :mad:
And forget retaliating, they slap him hard if he even opens his mouth... :mad:
Being new to the school and wanting to pursue activities like the Science, Math, Puzzles and Astronomy Clubs, I hope you can see why I'm scared now... :(   :([/quote]

Wow! What is wrong with your teachers and the other students at your school? Personally, if I were you, I would definitely try to be good friends with him (which you are i guess). He really needs a friend. But then again.. you don't want that to happen to you. Hmm.... those teachers suck! They should be fired.

If it keeps up like that... he may try to commit suicide.. it really does get that serious.

This will make my post too long so I'll hide it:

[hide]One day, when I was a freshman in high school, I saw a kid from my class walking home from school. His name was Kyle. It looked like he was carrying all of his books. I thought to myself, "Why would anyone bring home all his books on a Friday? He must really be a nerd." I had quite a weekend planned (parties and a football game with my friend the following afternoon), so I shrugged my shoulders and went on.

As I was walking, I saw a bunch of kids running toward him. They ran at him, knocking all his books out of his arms and tripping him so he landed in the dirt. His glasses went flying, and I saw them land in the grass about ten feet from him. He looked up and I saw this terrible sadness in his eyes.

My heart went out to him. So, I jogged over to him, and as he crawled around looking for his glasses, I saw a tear in his eye. I handed him his glasses and said, "Those guys are jerks. They really should get lives."

He looked at me and said, "Hey, thanks!" There was a big smile on his face. It was one of those smiles that showed real gratitude. I helped him pick up his books, and asked him where he lived. It turned out he lived near me, so I asked him why I had never seen him before. He said he had gone to private school before coming to this school.
I would have never hung out with a private school kid before. We talked all the way home, and I carried his books. He turned out to be a pretty cool kid. I asked him if he wanted to play football on Saturday with me and my friends. He said yes. We hung all weekend and the more I got to know Kyle, the more I liked him. And my friends thought the same of him. Monday morning came, and there was Kyle with the huge stack of books again. I stopped him and said, "Damn boy, you are gonna really build some serious muscles with this pile of books everyday!". He just laughed and handed me half the books. Over the next four years, Kyle and I became best friends.

When we were seniors, we began to think about college. Kyle decided on Georgetown, and I was going to Duke. I knew that we would always be friends, that the miles would never be a problem. He was going to be a doctor, and I was going for business on a football scholarship. Kyle was valedictorian of our class.

I teased him all the time about being a nerd. He had to prepare a speech for graduation. I was so glad it wasn't me having to get up there and speak.

Graduation day arrived - I saw Kyle and he looked great. He was one of those guys that really found himself during high school. He filled out and actually looked good in glasses. He had more dates than me and all the girls loved him! Boy, sometimes I was jealous. Today was one of those days. I could see that he was nervous about his speech. So, I smacked him on the back and said, "Hey, big guy, you'll be great!"

He looked at me with one of those looks (the really grateful one) and smiled. "Thanks," he said. As he started his speech, he cleared his throat, and began. "Graduation is a time to thank those who helped you make it through those tough years. Your parents, your teachers, your siblings, maybe a coach... but mostly your friends. I am here to tell all of you that being a friend to someone is the best gift you can give them. I am going to tell you a story."

I stared at my friend in disbelief as he told the story of the first day we met. He had planned to kill himself over the weekend. He talked of how he had cleaned out his locker so his Mom wouldn't have to do it later and was carrying his stuff home. He looked hard at me and gave me a little smile. "Thankfully, I was saved. My friend saved me from doing the unspeakable."

I heard the gasp go through the crowd as this handsome, popular boy told us all about his weakest moment. I saw his Mom and dad looking at me and smiling that same grateful smile. Not until that moment did I realize its depth. Never underestimate the power of your actions. With one small gesture you can change a person's life. For better or for worse.

- Author unknown -

from the site:http://inspiringwords.blogspot.com/[/hide]

[quote="MissSmartyPants"]Well, the other day they took us to Cambridge as part of a field trip...
One of my FRIENDS(everyone calls him a nerd) was:
a)Pushed into a pond
b)Served non vegetarian food and drinks even though he's a pure vegetarian and doesn't drink(AND it was stuffed down his throat forcing him to puke later)
c)Had his shoe pushed down one of the girls' toilets
d)Subject to repeated attempts to push his head into the dustbin
e)Due to his frail build was thrown and slammed against a lab door about ten times
f)His bag was put into an elevator and taken to the terrace from where it was dropped down
g)His glasses were thrown into the pudding
h)He had atleast one glass of juice poured on him each day
And the teachers JUST KEPT QUIET...they said IT WAS PART OF LIFE and that HE SHOULD STOP BEING SO NERDY... :o
And all this started when he volunteered a bit of information about Newton...  :mad:
And forget retaliating, they slap him hard if he even opens his mouth... :mad:
Being new to the school and wanting to pursue activities like the Science, Math, Puzzles and Astronomy Clubs, I hope you can see why I'm scared now... :(   :([/quote]

that was seriously mean. the teachers should get fired for that kind of stuff. people should learn to appreciate nerds, b/c someday, the people who made fun of them will work for them. bill gates also had similar problems in his childhood, but he became the richest man in the world.


geometry
inequalities
analytic geometry
Princeton
college
FTW
trigonometry

This is the collection of funny quotes from this year's IMO training camp. Most of them are from Peng, and we call them PCC, Peng's Common Comments. Enjoy!

[size=150]Peng[/size]
This is what he said right after he read a problem:
[b]"So strong (one second later) obvious (another second later) not really"[/b] - June 28

Peng always made tons of mistakes while bashing. This one shows how bad it was:
[b]"I hope I made an even number of mistakes"[/b] - June 29
Another one involving his mistakes:
[b]"(I) always show that 1=2 when I make mistakes"[/b] - June 30
Another horrible mistake:
[b]"What I thought the problem is and what the problem actually is, are off by a factor of 64"[/b] - June 31
Yet another one:
[b]"(talking to himself) Don't say I proved the problem is wrong (pause) Just proved that 1=-1"[/b] - June 31

Richard was presenting a solution, and two of the variables were P and Q. Even though the solution was understood by most of us, Peng couldn't understand it. Later he said:
[b]"Oh! It's not queue. It's Q"[/b] - June 30

During a discussion about whether combinatoric geometry is grungable:
[b]"Combinatorics is not grunge. Geometry is mass grunge. Therefore, combinatoric geometry is null set"[/b] - June 31

Peng was explaining analogy between a math problem and a maze, the logo of IMO 2006:
[b]"My grunge is like a tractor, going through walls. Menelaus, Jensen, and everything nice are like spikes in front of the tractor, making it more strong"[/b] - July 5
Later he added:
[b]"My bulldozer has machine guns"[/b] - some time after IMO

We met the British team on the excussion day. First thing they said to us was "We heard that you guys solve all geometry problems with coordinates", and Peng answered:
[b]"No, we sometimes use complex numbers"[/b] - July 5

Discussing ways of cheating in IMO:
[b]"No, that's useless. We should stuff the moose with Yufei"[/b] - July 6

Presenting his solution for an inequality problem:
[b]"Happier this side gets, closer we are to solving the problem. (later) This is ideal happiness"[/b] - July 6

Commenting on Yufei's insane solution:
[b]"This is maximum craziness you can get"[/b] - July 7

Describing bashing method in programming:
[b]"If you only do math, you don't know what bash is"[/b] - July 8

I'll add quotes from other people in few days(Probably Monday since I will be away till then).

This.

Thread.

Rocks.

Peng is awesome.  Why didn't you come to Waterloo, you BASTARD?!?!?   :P

Ah hahahahaha.  I definitely love the 'stuff the moose with yufei' one.   :D

This is hilarious. :rotfl:
"I hope I made an even number of mistakes"

btw I think this is the best thread in the past few months!!! :)

haha this thread is way too funny. I couldn't believe a CS guy like Peng makes that many mistakes in talking.

[quote="jpark"]You missed one...

"I learned this in grade 5!! In PUBLIC SCHOOL!!"  :D[/quote]

"grade 5!!  [b]Canadian[/b] public school system!!"[/b]

[quote="Singular"]Grunge like a tractor FTW

ps.  anyone remember "taking brute force to the next level" ?[/quote]

Yeah, do you still have the picture?   :D

Wow, an entire page devoted to the BS that comes out of my mouth?! I feel honoured.

I voted for "No, we use complex numbers", prolly cuz i'm too dumb to get the joke.

These are gold. I don't get this one: "Oh! It's not queue. It's Q", doesn't queue and Q sound the same?

[quote="weileaf"]It's Q", doesn't queue and Q sound the same?[/quote]
I think that is the funny part.

I'm back. Below are more quotes from other people.

[size=150]Farzin[/size]
Farzin praising ugliness of Peng's solution:
[b]"I acknowledge my friend and brother Peng, because his solutions make mine so elegant!"[/b] - June 30

Unfortunately, I wasn't there when he said this:
[b]"Peng is like a computer - Pengtium 5"[/b] - July 4

Everyone was saying 'crap' all the time by the end of the camp. After complaining that we should watch our language, he said:
[b]"Crap is like an oyster"[/b] - July 7

[size=150]Victoria[/size]
Victoria saying how obvious a step of her solution is:
[b]"officially obvious"[/b] - July 4

During a presentation:
[b]"A 3*1 square"[/b] - July 7

I forgot why she said this:
[b]"Will you miss them with fibres of your soul?"[/b] - July 7

[size=150]David[/size]
Richard said, "Sign language is working great so far", I replied:
[b]"Have you tried cosine language, too?"[/b] - July 4

During a conversation on inversion:
[b]"What would happen if we invert Farzin?"[/b] - July 5

[size=150]Yufei[/size]
Yufei was trying to say that the light is broken:
[b]"the fluorescent light is malfunctioning"[/b] - July 4

[size=150]Robert[/size]
After writing $\frac1{n!}+\frac1{(n+1)!}+\frac1{(n+2)!}+\cdots$ on the blackboard, our leader said:
[b]"rather small denominators..."[/b] - July 4

[size=150]Joel[/size]
Looking at a problem handwritten by our deputy leader, this Swedish read:
[b]"Stew that..."[/b](supposed to be 'show that') - July 5

[quote="lightrhee"]

[size=150]Farzin[/size]
Farzin praising ugliness of Peng's solution:
[b]"I acknowledge my friend and brother Peng, because his solutions make mine so elegant!"[/b] - June 30

Unfortunately, I wasn't there when he said this:
[b]"Peng is like a computer - Pengtium 5"[/b] - July 4

[size=150]Victoria[/size]
Victoria saying how obvious a step of her solution is:
[b]"officially obvious"[/b] - July 4

During a presentation:
[b]"A 3*1 square"[/b] - July 7

[/quote]

These are great :!:   :rotfl:  :rotfl:  :rotfl:
This topic is very very very amusing!!!
Thanks for these pills of wisdom [b]lightrhee[/b]  :lol:

My favourite Peng quote was about premier Ralph Klein wanting to make Alberta the centre of the new intergalactic civilization.  There was another great quote about a ditch at the Halifax citadel, but you had to be there.


geometry
circumcircle
perpendicular bisector
geometry proposed

Let $ D$ be an interior point of the side $ BC$ of a triangle $ ABC$, and let $ O_1$ and $ O_2$ be the circumcenters of triangles $ ABD$ and $ ADC$. The perpendicular bisector of the side $ AC$ meets the line $ AO_1$ at $ E$, and the perpendicular bisector of the side $ AB$ meets the line $ AO_2$ at $ F$. Prove that the bisectors of the angles $ \angle O_1EO_2$ and $ \angle O_1FO_2$ are orthogonal.

A rotational homotethy which takes O2 to O1 takes angle AO2H to HO1E (try to prove why). Hence AO1HO2 is inscribed (H is intersection of EO2 and FO1). Now we use well known lemma to prove that bisectors of theese angles are orthogonal QED


inequalities
number theory unsolved
number theory

Two people, $A$ and $B$, play the following game: $A$ start choosing a positive integrer number and then, each player in it's turn, say a number due to the following rule:

If the last number said was odd, the player add $7$ to this number;
If the last number said was even, the player divide it by $2$.

The winner is the player that repeats the first number said. Find all numbers that $A$ can choose in order to win. Justify your answer.

For starting values $>21$, this value is never reached again (proof by simple inequalities on the size of the number after two steps).
So checking gives that only $1,2,4,7,8,14$ work.

. If the first number is 1:
then the sequence is 1(A), 8(B), 4(A), 2(B), 1(A), A is the winner.
. If the first number is 2:
then the sequence is 2(A), 1(B), 8(A), 4(B), 2(A), A is the winner.
. If the first number is 3:
then the sequence is 3(A), 10(B), 5(A), 12(B), 6(A), 3(B), B is the winner.
. If the first number is 4:
the sequence is: 4(A), 2(B), 1(A), 8(B), 4(A), A is the winner.
. If the first number is 5:
the sequence is 5(A), 12(B), 6(A), 3(B), 10(A), 5(B), B is the winner.
. If the first number is 6:
the sequence is 6(A), 3(B), 10(A), 5(B), 12(A), 6(B), B is the winner.
. If the first number is 7:
the sequence is 7(A), 14(B), 7(A), A is the winner

. If the first number > 7:
In the sequence, after an odd, an even follows, so the sequence has the following form: (First number), E, ..., E, O_1, E, ... , E, O_2, E, ... , E, O_3, ...
O_i > (O_i +7)/2 if only if O_i >7, then, If O_i >7, then O_(i+1) < O_i because  O_(i+1) is less than or equal to (O_i +7)/2 < O_i.
It's clear that if an even of the sequence is less than first number then the next term is also less than first number. So the first number won't be in the sequence a second time if all numbers of the sequence are > 7, then there exists a number N =< 7 in the sequence. In the first cases (where the first number was 1,2,3,4,5,6,7) we saw what numbers can follow in the sequence after N=< 7, so the first number can only be 8, 10, 12, 14 since these are the only numbers >7 that are after some number less than or equal to 7.
*) the first number is 8:
the sequence is: 8(A), 4(B), 2(A), 1(B), 8(A), A is the winner
*) the first number is 10:
the sequence is: 10(A), 5(B), 12(A), 6(B), 3(A), 10(B), B is the winner
*) the first number is 12:
the secuence is:  12(A), 6(B), 3(A), 10(B), 5(A), 12(B), B is the winner
*) the first number is 14:
the sequence is: 14(A), 7(B), 14(A), A is the winner

So A will win if only if A chooses 1, 2, 4, 7, 8 or 14.


We have a sequence of numbers $ a_i$. For a number $ a_i$ from the sequence we define previous number $ a_j$, such that $ j$ is maximum possible, $ j<i$ and $ a_j<a_i$.
Calculate how many numbers from the sequence  are not previous for any other number from the sequence????
The time is limited, so this cannot be solve by a banal algorithm. I need something more optimised....

$ a_j$ is previous to $ a_{j\plus{}1}$ if and only if $ a_j < a_{j\plus{}1}$. $ a_j$ is previous to [i]anything[/i] only if $ a_{j\plus{}1}$ doesn't supersede it by being smaller.
So, a one-pass algorithm: count the number of values that are $ <$ whatever comes next.

OOOhhhh... THat's really easy.... thanks a lot!!


Find the perimeter of square ABCD when only the diagonal length is given to be the square root of 32.


Urmatoarea problema este probabil cunoscuta(?), eu o propun totusi FORUMO-nautilor spre rezolvare:

"Sa se arate ca doua triunghiuri ABC si A1B1C1 care sunt simetrice fata de o dreapta d, sunt ortologice!"

DEFINITIE:
Doua triunghiuri ABC si A1B1C1 sunt ortologice, atunci cand ele au una dintre urmatoarele 3 proprietati echivalente:
(i). perpendicularele duse din varfurile A1, B1 si C1 pe dreptele BC, CA si respectiv AB sunt concurente intr-un acelasi punct P1;
(ii). perpendicularele duse din varfurile A, B si C pe dreptele B1C1, C1A1 si respectiv A1B1 sunt deasemenea concurente intr-un
alt punct P;
(iii). are loc relatia: |AB1|^2+|BC1|^2+|CA1|^2=|A1B|^2+|B1C|^2+|C1A|^2.
In aceasta situatie, punctele P si P1 se numesc centrele de ortologie ale celor doua triunghiuri.

Pentru demonstratia echivalentei proprietatilor (i), (ii) si (iii), v. de exemplu cartea noastra:
Miculita&Branzei: ANALOGII TRIUNGHI-TETRAEDRU, pag.124, teorema 45.

INDICATIE(pentru rezolvarea problemei propuse): Folositi proprietatile patrulaterelor inscriptibile! (nu e grea!)

ALTA SOLUTIE[folosind condifia 3), din definitia anterioara]:
AA1B1B-trapez isoscel---->|AB1|=|A1B|---->|AB1|^2=|A1B|^2                  (1);
BB1C1C-trapez isoscel---->|BC1|=|B1C|---->|BC1|^2=|B1C|^2                (2)
si
AA1C1C-trapez isoscel---->|CA1|=|C1A|---->|CA1|^2=|C1A|^2                 (3).
In fine, adunand relatiile (1), (2) si (3), memru cu membru, ajungem la relatia (iii).

Cu privire la metoda pe care ati indicat-o, cu patrulatere inscriptibile, intr-adevar e usor de observat ca, notand cu $ K,L,M$ picioarele perpendicularelor  din $ A,B,C$ pe laturile $ B_{1}C_{1}, A_{1}C_{1}, A_{1}B_{1}$, avem niste patrulatere inscriptibile... Notam $ P_{1}$ intersectia dreptelor $ BL$ si $ AK$ si cu $ P_{2}$ intersectia dreptelor $ AK$ ci $ CM$.
Deci, avem patrulaterele inscriptibile $ P_{2}MB_{1}K$ si $ P_{1}KC_{1}L$. De aici rezulta $ \angle{MP_{2}K}\equal{}\angle{A_{1}B_{1}C_{1}}\equal{}\angle{ABC}$
$ \angle{KPL}\equal{}\angle{A_{1}C_{1}B_{1}}\equal{}\angle{ACB}$. De aici rezulta $ ABP_{2}C$ si $ ABP_{1}C$ sunt inscriptibile, de unde se vede ca $ P_{1}\equal{}P_{2}$. Deci sunt concurente.
Si rezolvarea folosind iii) mi se pare frumoasa, cum vad eu pct. iii) se deduce din lema lui Carnot, probabil.


[hide]For 2, just use synthetic division to factor out $(x-1)$ and then just factor that to find the roots.[/hide]

[quote="lotrgreengrapes7926"][quote="lyra"]um i don't think what you are saying for 1 works...

lyra[/quote]
[hide="It works"] $\sqrt{2-x}+2=\sqrt{3x-2}$.
$2=\sqrt{3x-2}-\sqrt{2-x}$.
$4=3x-2-2\sqrt{(3x-2)(2-x)}+2-x$.
$4=2x-2\sqrt{-3x^2+8x-4}$.
$2\sqrt{-3x^2+8x-4}=2x-4$.
$4(-3x^2+8x-4)=4x^2-16x+16$.
$16x^2-48x+32=0$.
$x^2-3x+2=0$.
$(x-2)(x-1)=0$.
$x=2$ or $x=1$.
Squaring can lead to extraneous solutions, so check the answers in the original problem. And we see that only $x=\boxed{2}$ works. [/hide][/quote]

oh, thank you!  I get it now I was forgetting the middle term.

Smiles,
lyra

[quote="ch1n353ch3s54a1l"][quote="lyra"]HI I have a few questions and I was wondering if you know how to solve them?

to start: $sqrt(2-x)+2=sqrt(3x-2)$

$6x^3-11x^2+6x-1$ if one of the zeros is 1 what are the others?

$x^3-4x^2-x+4$

Thanks, also, if you solve one of them the same way as the other just tell me and I will solve, thanks for your help,
lyra[/quote]
[hide="1"]$\sqrt{2-x}+2=\sqrt{3x-2}$
Just move $\sqrt{2-x}$ to the other side the start squaring and simplifing
[/hide]
[hide="2"]Um...just divide by $(x-1)$?
You get $6x^2-5x+1=(3x-1)(2x-1)$ and yeah.[/hide]
[hide="3"]Find possible rational roots: $\pm 1, \pm2, \pm4$
$f(1)=0$ so divide by $(x-1)$ to get $x^2-3x-4$, factor and find roots. [/hide][/quote]

1. see post below, $x=2$
2. $x=1/3$ or $x=1/2$
3. $(x-4)(x+1)$ $x=4$ or $x=-1$
    and 4 works so divide by (x-4) to get $x^2-x$ um, what do you do with this?

THanks, lyra

For 3, we can factorize $x^3-4x^2-x+4$ by grouping.
[hide]
$x^3-4x^2-x+4$
$=x^2(x-4)-(x-4)$
$=(x^2-1)(x-4)$
$=(x+1)(x-1)(x-4)$

Hence the roots are $-1,1,4$[/hide]

[quote="ch1n353ch3s54a1l"][hide="1"]$\sqrt{2-x}+2=\sqrt{3x-2}$
Just move $\sqrt{2-x}$ to the other side the start squaring and simplifing
[/hide][/quote]
Yeah, that works (as lotrgreengrapes7926 posted), but you don't need to. In fact, if you just square both sides of the original equation, the calculation is a bit easier ;)

[quote="ch1n353ch3s54a1l"][quote="lyra"]HI I have a few questions and I was wondering if you know how to solve them?

to start: $sqrt(2-x)+2=sqrt(3x-2)$

$6x^3-11x^2+6x-1$ if one of the zeros is 1 what are the others?

$x^3-4x^2-x+4$

Thanks, also, if you solve one of them the same way as the other just tell me and I will solve, thanks for your help,
lyra[/quote]
[hide="1"]$\sqrt{2-x}+2=\sqrt{3x-2}$
Just move $\sqrt{2-x}$ to the other side the start squaring and simplifing
[/hide]
[hide="2"]Um...just divide by $(x-1)$?
You get $6x^2-5x+1=(3x-1)(2x-1)$ and yeah.[/hide]
[hide="3"]Find possible rational roots: $\pm 1, \pm2, \pm4$
$f(1)=0$ so divide by $(x-1)$ to get $x^2-3x-4$, factor and find roots. [/hide][/quote]

No need to rational root #3, just group and factor and you get your roots.

oh, yeah, that works, thanks, one more question on this topic,
How would I find zeros of $x^3+2x^2-18x+48$
factors of 48 are (plus/minus all): 48,1,24,2,16,3,12,4,8,6
factors of 1 are (plus/minus all):
 1
So zeros are what none of them seem to work???

Smiles,
lyra

It is possible that there are no rational roots to the equation, but there has to be at least one real root.

Also, was that just a random polynomial that you created.  Not all polynomials have a rational root.

So, I don't think there are any rational roots to the equation, just roots.

For 3 I suggest you just factor by grouping rather than finding all possible 0's and plugging in. For example if you have $x^3-4x^2-x+4$ you can easily factor out an $x^2$ out of $x^3-4x^2$ and factor out a -1 out of $-x+4$ giving you $x^2(x-4)-1(x-4)$. That gives you $(x-1)(x+1)(x-4)=0$ hence your zero's are 1, -1 and -4.

[quote="lkryptonicl"]For 3 I suggest you just factor by grouping rather than finding all possible 0's and plugging in. For example if you have $x^3-4x^2-x+4$ you can easily factor out an x out of $x^3-4x^2$ and factor out a -1 out of $-x+4$ giving you $x(x-4)-1(x-4)$. That gives you $(x-1)(x-4)=0$ hence your zero's are 1 and -4.[/quote]It's $x^3-4x^2-x+4=x^2(x-4)-1(x-4)=(x^2-1)(x-4)=(x+1)(x-1)(x-4)=0$, so the zeros are $-1, 1, 4$.


geometry
inradius
algebra
polynomial
MIT
college
Galois Theory

It's well-known problem of which I know solution, but I need to present it to ordinary high school students which seems to be a little difficult to me...

Prove that it's impossible to construct (following only the rules of Euclidean construction) a triangle given by lengths of its two sides and its inradius (=radius of its inscribed circle).

I use this well-known formula:
$r =$ :sqrt:$[(s-a).(s-b).(s-c)/s] $
where $r$ is inradius, $a, b, c$ are lengths of sides of our triangle and $s = (a+b+c)/2$
We can adjust this formula to the cubic equation with variable $c$ or (after substitution $c = 2s - (a+b)$) $s$ (it's an easier form of demanded cubic equation). After a little counting we'll state that this cubic polynom has no rational root which will imply its irreducibility in rational numbers. And as a final step we'll use Wantzel Theorem which will show that no root of considered polynomal is a constructible number (following Eucleides). But is there anything clearer and easier...

Thanks for any help.

I just found this problem discussed in

[url=http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D243885]P. Buchner, [i]Eine Aufgabe, die mit Zirkel und Lineal nicht lösbar ist[/i], Elemente der Mathematik 2, p. 14-16[/url].

I myself have never studied Galois theory, so I don't know whether there is a considerably simpler proof...

  dg


Here hucht again  :roll:, You know hucht work a lot in order you increase your knowledges, here you have the exam of [b]OIM 1989[/b] wich was taken in La Habana Cuba  :D

PD. Lo mejor que tiene cuba es el arroz a la Cubana, lo han probado??
PD2. Luego posteo la preparacion del arroz a la cubana
PD3. Vayan alistando huevos, arroz y platano y si trabajan un poco de hot dog

hucht:
Creo que seria mejor que no pongas IMO pon Iberoamerican Olympiad o puede ser Ibmo, para evitar confusiones... a y corrige algunas cosas en el pdf... por ejemplo la fecha... y por ahi vi que decia sucha en vez de such... bueno... errores minimos

$Tipe$


I just counted from triangles that have 1 triangle in them, 2 trignales and so on. I got 22? Any higher answers?

Apparently the correct answer is 24, but I don't know how to arrive to that answer  :blush:

I only got 22, does anybody have a solution for 24?

I counted by 'size', the number of divisions of a triangle as divided by other lines.

size, number of triangles of that size
1, 8
2, 5
3, 6
4, 2
5, 2
6, 0
7, 0
8, 1
-----
sums to 24 triangles

Double checking, I counted by triangles having a particular vertex, but not one of the target vertices previously searched.

Target vertex, new triangles

Top, 12
bottom left, 6
bottom right, 3
mid left, 2
mid right, 1
bottom center, 0
center, 0

sums to a total of 24


I'm sorry, it should have been $ 6^6\minus{}5^6$, not $ 6^6\minus{}5^5$. I fixed this in the original post.

Im pretty sure that this does generalize.

Try expanding: $ a^x \minus{} (a \minus{} 1)^x$ and you should get $ a^{x \minus{} 1} \plus{} a^{x \minus{} 2}(a \minus{} 1) \plus{} a^{x \minus{} 3}(a \minus{} 1)^2 \plus{} \dots \plus{} a(a \minus{} 1)^{x \minus{} 2} \plus{} (a \minus{} 1)^{x \minus{} 1}$

Or more specifically, expanding $ a^6 \minus{} (a \minus{} 1)^6$ should get you $ a^5 \plus{} a^4(a \minus{} 1) \plus{} a^3(a \minus{} 1)^2 \plus{} a^2(a \minus{} 1)^3 \plus{} a(a \minus{} 1)^4 \plus{} (a \minus{} 1)^5$

My clue for trying that method was that this closely resembles the binomial theorem.

Can someone please verify this/give an alternate way to show the generalization?

[quote="PowerOfPi"]I noticed that $ 6^6 \minus{} 5^6 \equal{} 6^5 \cdot 5^0 \plus{} 6^4 \cdot 5^1 \plus{} 6^3 \cdot 5^2 \plus{} 6^2 \cdot 5^3 \plus{} 6^1 \cdot 5^4 \plus{} 6^0 \cdot 5^5$. Can this be generalized?[/quote]
In general if you have $ a^{n}\minus{}b^{n}$, then it factors to $ (a\minus{}b)(a^{n\minus{}1}\plus{}a^{n\minus{}2}b\plus{}a^{n\minus{}3}b^{2}\plus{}...\plus{}b^{n\minus{}1})$. It just so happens that in this particular case $ a\minus{}b\equal{}1$, and that's why what you have is that.


A guidance counselor is planning schedules for 30 students.  Sixteen students say they want to take French, 16 want to take Spanish, and 11 want to take Latin.  Five say they want to take both French and Latin, and of these, 3 wanted to take Spanish as well.  Five want only Latin, and 8 want only Spanish.  How many students want French only?

[hide="Draw your own Venn diagram."]$\boxed{7}$.[/hide]

[hide="PIE"]From PIE,
$30=(F \cap S \cap L)-(F \cap S)-(S \cap L)-(L \cap F)+F+S+L$

From the given,
$F \cap S \cap L = 3$
$L \cap F = 5$
$L=5$
$S=8$
$F\cap S+L\cap F-F\cap S\cap L+F= 16$
$F\cap S+S\cap L-F\cap S\cap L+S= 16$
$S\cap L+L\cap F-F\cap S\cap L+L = 11$

From the last equation, we get that: $S\cap L = 4$. From the second equation, we get that: $F\cap S = 7$. Now, from the first equation, we get that:

$7+5-3+F= 16$

So $F=7$. The answer is $\boxed{7}$.[/hide]


geometry
email
analytic geometry
national olympiad

mm... but they are all shortlist. which means i am not allowed to post them as yet?... all the way until the IMO itself?...

As far as I know, the German TSTs have had 3 problems from ISL 2004 (perhaps there were some more which I have forgotten):

- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24078]Some more ugly circle geometry[/url]
- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30903]Golden matrices[/url] (for Moldova's TST too)
- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30865]ab + bc + ca = 1[/url]

So I guess you can post up the other shortlist questions as they are slowly but surely being used for TSTs.

Since the problems are already used in a selection, I think that it is almost impossible to contain them (ie users can find them through email or other such methods). In plus you have not seen the shortlist so you cannot be sure if they are or not from the shortlist. So you can just simply post them if you really want as British TST 2005.

Anyway how did you do? :)

annoyingly, they didn't tell me my marks. Just that the final 8 are chosen as such:

Tom Eccles
Saul Glasman
Nathan Kettle
Andre Kueh (-'03, S'04) (me!)
Matthew Lee (R'04)
Martin Orr (R'02, S'03, B'04)
Jack Shotton
Alexander Smith (R'04)
(R stands for Reserve. 2 people here will get R'05)

To be exact, this was a FST (First Selection Test) and there will be a NST (Next Selection Test) apparantly.

In any case, we were told not to spread the questions, so it will be unlikely the questions will leak unless I leak them. So I guess I will not post them until a little later when -most- TSTs have finished.

I probably got 4 questions in 6.

I get R04, but what does S04 and B04 mean? :)

oh. and the ugly circle geometry came up in our FST, day 2, qn 1. i -almost- didn't get it, panicked, started to coordinate it, failed, and realised the synthetic method works easily 20 minutes from time.

I suspect that only Andre would try to use coordinate geometry on such a question. :) (Side note to Andre: Well, Joel probably would have done that too, once upon a time. ;) )

well i heard a trigo soln - which he confessed was almost coordinate because of the ready-made axis.

I didn't think the circle geometry was ugly... I failed to get it in the exam, but I was kicking myself afterwards. I managed two questions out of the six, I'm still very surprised that I made the final 8.

[quote="Valentin Vornicu"]When's the final part taking place? :)[/quote]

Considering my country flag I should have more information.  :D


linear algebra
matrix
abstract algebra
superior algebra
superior algebra unsolved

By using formula (so called Scots formula) Show that the group $ Sp(6,k)$ can't be constructed by two elements $ x,y$ such as $ x^{2}\equal{}y^{2}\equal{}(xy)^{7}\equal{}1$

if group is generated by two elements $ x,y$ such as $ x^{2}\equal{}y^{3}\equal{}(xy)^{7}\equal{}1$ then what is the order of this group? :maybe:

[quote="ZetaX"]If you have no restriction except $ x^2 \equal{} y^3 \equal{} (xy)^7 \equal{} e$, then the group can even be infinite.
And I don't know what $ Sp(6,k)$ is.[/quote]
it's so called Symplectic group. http://en.wikipedia.org/wiki/Symplectic_group

also Why can be infinity?  it's not so obviusly for me..
for some particular case for example $ Sp(6,31)$ I can prove by this way:

Let's try to compute the order of Symplectic group $ \equal{}Sp(2n,q)$.
Direct calculation shows that order is equal to $ |Sp(2n,q)|\equal{}\prod_{i\equal{}1}^{n}(q^{2i}\minus{}1)q^{2i\minus{}1}\equal{}q^{n^{2}}\prod(q^{2i}\minus{}1)$, therefore in our case
we have $ |Sp(6,q)|\equal{}q^{9}(q^{2}\minus{}1)(q^{4}\minus{}1)(q^{6}\minus{}1)$  I note that ($ n\equal{}3$)
Suppose our group can be generated by two elements $ x,y$ such as $ x^{2}\equal{}y^{3}\equal{}(xy)^7\equal{}1$. I also note that $ (xy)\in Sp(6,q)$ because $ x\in Sp(6,q)$
, $ y\in Sp(6,q)$ and by definitions our group is with the group operation that of matrix multiplication. That's mean that order of element $ (xy)$ is equal to
$ 7$. By Lagrange's theorem, the order of each element divides the order of the group. But for case $ q\equal{}31$ we have contradiction. Hence Our statement proved.


O, jak fajnie, nareszcie. Pozdrawiam wszystkich z Polski i nie tylko ;-)

No cześć cześć, czekamy na el_doope, filipa, jalowca, yrcha i reszte..moze kiedys przyjda...w kazdym razie poczujcie sie jak u siebie w domu ;)

[Edit] uwazajcie tylko na oszustow ktorzy moga sie pytac o zadania z trwajacej OM...potrzebny jest Megus do moderowania, bo ja sie nie podejmuje

Witam wszystkich  :) Mam prosbe - jakby ktos zauwazyl ze pojawilo sie zadanie z OM to prosze o PM, bo moze sie zdarzyc ze nie zauwaze

To i ja w swoim pierwszym poście powiem "cześć wszystkim"  :)

mnie tu juz w sumie dawno nie bylo, ale dzien dobry.

To moze by tak jakis sensowny temat zaczac  :D jak juz takie community powstalo to jakos mozna by to wykorzystac

Megus lepiej sie tlumacz, czemu na "org" sie nie pokazujesz :P.

Startujesz w tym roku w OM?

Startuje, startuje  :)

A na .org (ani na .pl) sie nie pokazuje (choc od czasu do czasu wejde) bo jakos mathlinks mnie bardziej urzekl  :P

BTW, oszustow Olimpiado - Matematycznych...
Wydaje sie ze Polacy sa w tej dziedzinie bardzo aktywni. :(
(widzialem juz kilka osob na mathlinks, a w zeszlym roku gosciu na sci.math dostal kompletne rozwiazanie...)

na ".org" jak narazie widzialem 2, ktorzy chcieli dostac rozwiazania. Jeden byl sprytny i zmienil smiesznie tresc 4-ego (problem pozostawal ten sam).

Ciekawe tylko po co im to ? Przeciez drugi etap i tak nic nie daje. A jesli nie potrafia zrobic tylu zadan, zeby sie do niego dostac, to tym bardziej sie nie dostana do finalu..

Witam wszystkich :)

Risk, w niektórych szkołach może i daje... Np. w mojej daje już wyższą ocenę 2 etap z fizyki. To nie to samo, ale pewnie w niektórch szkołach tak zachęcają uczniów i do OM.

ehhe..., witajcie wszyscy  :lol:  nieźli czarodzieje na tym forum!

cześć wszystkim :) też postanowiłem się do Was przyłączyć , choć patrząc na to jakie cuda się tu dzieją to niewiele wniosę , ale co tam :) pozdro

No witam kolegów :)

Teraz się trzeba troszke podszkolić z anglosaskiego i bedziemy tu wymiatać :P

również Witam wszystkich kolegów  i pozdrawiam ;)

Witam wszysktich w swoim pierwszym poście! ;)

Witam również  :)  przywitać się należy. Choć na tym forum już jestem nie od dziś. :roll:

Pozdrawiam

Nie powiem, calkiem fajny ten slowniczek  :)

Witam, piszę głównie dlatego, że w słowniczku zabrakło mi semiperimeter-pół obwodu  :)

Witam wszystkich serdecznie.

Widzę, że po prawie trzech latach wyciagam temat; ) Swoją drogą sam troche przypadkiem tutaj trafiłem. Zna ktoś jakieś inne duże miedzynarodowe fora matematyczne?

Z tych, na które do tej pory trafiłem to mathlinks.ro jest jednak zdecydowanie największe. Moze chińczycy mają jakieś większe?; p

thats more time that passed than how old I am lol


A , B , C are angles of the triangle ABC, prove that:
cosC(sinA + sinB)=sinCcos(A-B)

Well sin C=sin(A+B) (since they add up to 180) and cos C=-cos(A+B), I think it's an identity from there.
Actually, the problem is wrong. Take A=60,B=30,C=90.


inequalities
trigonometry
inequalities solved

Let $x,y,z$ be real positive numbers such that $x+y+z=1$, prove that
\[ \frac{xy}{\sqrt{\tfrac13+z^2}}+\frac{zx}{\sqrt{\tfrac13+y^2}}+\frac{zy}{\sqrt{\tfrac13+x^2}}\leq\frac12.  \]

[quote="M4RI0"][quote="pvthuan"]Let $x,y,z$ be real positive numbers such that $x+y+z=1$, prove that
\[ \frac{xy}{\sqrt{\frac13+z^2}}+\frac{zx}{\sqrt{\frac13+y^2}}+\frac{zy}{\sqrt{\frac13+x^2}}\leq\frac12.  \][/quote]

You shouldn't have posted it in the solved problems :P[/quote]

I solved it already. So I put it here.

How beautiful it is! I think that.. :lol:

[quote="pvthuan"]We have $\frac13+x^2=\frac13(x+y+z)^2+x^2\geq xy+yz+zx+x^2=(x+y)(x+z)$. Then applying AM-GM inequality, we are done.[/quote]
How can you prove this by AM-GM:
$\sum\frac{yz}{(x+y+z)\sqrt{(x+y)(x+z)}}\le \frac{1}{2}$   :huh:

Oh, I understand it.That's really a beautiful solution. :roll:

[quote="pvthuan"]Thank you, Jin. It is just a hint, and I believe that most Mathlinkers can do it themselves. I could not solve the following inequality of similar form. Prove that if $x,y,z$ are positive real numbers such that $x+y+z=1$, then
\[ \frac{xy}{\sqrt{1+z^2}}+\frac{zx}{\sqrt{1+y^2}}+\frac{yz}{\sqrt{1+x^2}}\leq\frac1{\sqrt{10}}.  \] :roll:[/quote]

hmm... look proof like some trig substitution or the thesis $x+y+z=1$ maybe useful $(\sqrt{1+z^2})$, but $\frac{1}{\sqrt{10}}$ is not really a good number (here I don't mean this is false).

$\sqrt{1+z^2}=\sqrt{9\time(\frac{1}{3})^2+z^2}\geq\frac{1}{\sqrt{10}}(3+z)$
So we just need to prove
$\sum\frac{xy}{3+z}\leq\frac{1}{10}$
Assume $a=3+x,b=3+y,c=3+z,a+b+c=10$
$\sum\frac{xy}{3+z}=\sum\frac{(a-3)(b-3)}{c}=\sum\frac{ab-3(a+b)+9}{c}=\sum\frac{ab+3c-21}{c}$
So we just need to prove
$\sum\frac{ab-21}{c}\leq-\frac{89}{10}$
By a series of caculated:
$\Leftrightarrow 21\sum a^3(b+c)+16\sum a^2bc\geq58\sum a^2b^2$
$\Leftrightarrow \sum(21ab-8c^2)(a-b)^2\geq0$
Since $x,y,z\in(0,1),a,b,c\in(3,4)$
So $21ab-8c^2\geq21\times3\times3-8\times4\times4=61>0$
So it's correct.

That is to say that my inequality is not sharp. Thank you Jin.

Yes, but I can't prove the original one with this method.

how can i prove (xy)^2 + (yz)^2 + (zx)^2 <= 1/27 ?

then you are done by cauchy schwarz.

If $x,y,z$ be nonnegative real numbers such that $x+y+z=1$, then
$\sqrt{10}\sum\frac{yz}{\sqrt{1+x^{2}}}\leq2\sqrt{3}\sum\frac{yz}{\sqrt{1+3x^{2}}}\leq\frac{\sqrt{3}}{2}\sum\sqrt{xyz+x^{2}}\leq 1.$


If $a+b+c=1$ and $a,b,c>0$ prove that:
$\frac{a+b^2}{c}+\frac{b+c^2}{a}+\frac{c+a^2}{b}\geq4$

The best way to deal with these simple inequalities in this forum is to solve them straight away I guess... (I think this belongs to the Pre-Olympiad forum, Gohan)

\[\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq 3\] by AM GM and \[\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq a + b + c = 1\] by rearrangement (or Cauchy, if you want). Adding gives the result.

[quote="ehku"]If $a^4+b^4+c^4=1$ and $a,b,c>0$ prove that:
$\frac{a+b^2}{c}+\frac{b+c^2}{a}+\frac{c+a^2}{b}\ge 6$[/quote]

The last one can not be true!

[quote="mathpk"]I can't do it.Please help me  :) .[/quote]
See the Arne's post. :wink:  :D

mathpk, you note that if $ a^4\plus{}b^4\plus{}c^4\equal{}3$ then $ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\geq 3$,adding give the result


If the angle bisectors of a triangle are equal. Prove that the triangle is isosceles.

 This question can be done using trig but its in the euceledian geometry section of my book.

Does anyone know a simple solution?

[i]p.s sorry if this question doesn't fit here. I'm new so .. [/i][/i]

[hide="hint"]
Assume the contrary, then use the law of sines.[/hide]

This is a famous problem. There is a very nice way of doing it using areas and contradiction.

Is it possible to do the q without using trig (i.e only eucledian geometry)

[img]http://i5.photobucket.com/albums/y194/jelly119/05062302.jpg[/img]
Assume without loss of generality that $\overline{AB}>\overline{AC}$,then $\angle{ECD}>\angle{EBD}$.

Construct $D'$ on $\overline{PD}$ such that $\angle{ECD'}=\angle{EBD}$, then $BED'C$ is cyclic.

Since $\widehat{ED'}=\widehat{D'C}<\widehat{BE}\Rightarrow\widehat{BED'}>\widehat{ED'C}\Rightarrow\overline{BD'}>\overline{CE}\Rightarrow\overline{BD}>\overline{CE}$,

which leads to a contradiction that $\overline{BD}=\overline{CE}$, and we are done!

Is there a way to prove it without assuming the contrary?


What is the least positive integer $ n$ such that $ \frac{16}n < n$?

Multiply n on both sides.
Thus, you get 16<n^2.
Since 4^2=16, the least positive integer is 5.

Algebraically, in this step, you would take the square root of both sides, because if $ a \equal{} b$, then $ a^2 \equal{} b^2$.

$ 16 < n^2$

$ 4 < n$

The least possible integer value is $ \boxed{5}$.


calculus
derivative
inequalities
absolute value

Could someone please explain to me Taylor's Theorem with Remainder, as it is called in my Calculus textbook.

It states:

If f has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I,

f(x) = f(a) + f'(a)(x- a) + f''(a)/2!(x - a)^2 +... + f^(n)(a)/n!(x - a)^n + Rn(x)

where Rn(x) = f^(n+1)(c)/(n + 1)!(x - a)^(n + 1)

for some c between a and x.

This makes no sense to me. I was expecting that Rn(x) = f^(n+1)(a)/(n + 1)!(x - a)^(n + 1) + ... (basically the rest of the series). Where did that random c come from?

The textbook has an example of proving convergence of a maclaurin series where they talk about how f^(n+1)(c) has some upper bound, and they use this to illustrate the Remainder Estimation Theorem, which has inequalities and absolute value in it. Where did all of this come from???

[quote="rlee"]The c is an arbitrary value between a and x. Makes sense if you think about it a little. Basically, the purpose is to choose c such that R(n) is maximized.

If you want an easy way to think of Taylor Series, I don't know of any. The proof is extremely difficult, and if you really feel like torturing yourself, check it out at mathworld or google it. You can try thinking of it as an extension of the mean-value theorem.[/quote]

The c E [a,x] is a consequence of the fact that the Taylor Series is known as the extended MVT.  A fairly simple proof is given at mathworld (repeated integrations).

Perhaps you should first begin with the study of the Rolle's theorem (see http://mathworld.wolfram.com/RollesTheorem.html) and then of the mean value theorem (see http://mathworld.wolfram.com/Mean-ValueTheorem.html).

You should see then what this 'c' means (and where it comes from).


A hole 6 inches deep is drilled all the way through the center of a sphere. Find the volume of the resulting object.

Note that drilling a hole through a sphere removes the "spherical caps" as well as the cylindrical section that's 6 inches in height.

I hope my wording isn't really bad.

Er...don't you have to know how wide the hole is?

Well, I guess the width of the hole would be dependent on the radius of the sphere... there would be a wider hole in a larger sphere.


[quote="Valentin Vornicu"][quote="dts"]You might be an AoPSer if...

... you know that the next Olympic games are in Mexico City.[/quote]Hmm, I think we also should save this for ml description, and change Mexico City to Mérida, Yucatán, México :)[/quote]

Errr... hehe... got the city wrong, I guess.

does that mean you are NOT an aops'er?

oh wait. faulty logic...

[quote="Valentin Vornicu"][quote="dts"]You might be an AoPSer if...

... you know that the next Olympic games are in Mexico City.[/quote]Hmm, I think we also should save this for ml description, and change Mexico City to Mérida, Yucatán, México :)[/quote]

Why would it be better for ML rather than AoPS?

[quote="MithsApprentice"][quote="theone853"]more ML people are IMO'ers than AoPS'ers[/quote]

Very debatable and completely irrelevant.[/quote]I think the correct affirmation is: more old-ML people were IMO'ers than old-AoPS'ers (because of the different traits of the two site, MathLinks was concentrated on IMO Training, thus almost all old-members were expected to know what about IMO, and where the next IMO should be, etc. and the joke would make sense).

At this point it's really hard to call someone AoPS'er and not a MathLinker and vice-versa  :lol:  Maybe you guys can crop something up :P
About the description I thought of it as a better phrase to put on a MathLinks t-shirt (ie with the MathLinks logo on it) than on an AoPS t-shirt (wearing a t-shirt with that at say a Mathcounts meeting it not really a good idea, because nobody will catch the joke).

At this point we have two interfaces to one site: the AoPS interface is designed to help students from getting started, amc, aime leading up to usamo, and it's optimized for that task (ie you see the olympiad cat closed, menu in left with the most used features, etc.) the MathLinks interface is designed to help an advanced user (such as one preparing for usamo, mop, imo) with direct links to some sections, etc. which can be easily interchanged. That's why I don't know how to call one user: a MathLinker or an AoPSer :D

BUMPITY BUMP BUMP!  :rotfl:
________________________________
BUMPIY BUMP BUMP :rotfl:


algebra
polynomial
linear algebra
linear algebra unsolved

Please give an example,such that:

$ A$, $ B$ $ \in$ $ M_{n}(\mathbb{F})$

and $ AB$ has the same EigenPolynomial as $ BA$

but they don't have the same minimal polynomial :roll:

Post #2 of http://www.mathlinks.ro/viewtopic.php?t=308765 could help you...
Note that:
- the characteristic polynomials (I think you call them "eigenpolynomials") of $ AB$ and $ BA$ are always equal to each other (as long as the matrices $ A$ and $ B$ are $ n\times n$ matrices), as I proved in http://www.mathlinks.ro/viewtopic.php?t=275576 Corollary 2 (click on the words "By the way" in my post to see this corollary),
- the minimal polynomials of $ AB$ and $ BA$ can differ only in the power of $ X$ which they contain (this was proven by Kent Merryfield in http://www.mathlinks.ro/viewtopic.php?t=299895 ).

  darij


Give the exact value of sin(3pi/2)[/code]

As the first response I would like to state that this board is [i]NOT[/i] friendly to those who post Homework problems on a regular basis.

[hide="Your problem"]$  \sin\left(\frac{3\pi}{2}\right) = 0$
Note that $ \sin\theta$ is correspondent to the Y-axis, and that $ \frac{\pi}{2}$ is equivalent to $ 90^{\circ}$.[/hide]

I am not doing homework... I am going through my dad's trig book just trying to solve some problems to brush up on math before school starts back up in August.

My answer choices wer

a. -1
b. 0
c. 1/2
d 1

do you say the answer was -1??

:blush: oops man yeah its -1.

i was thinking cosine. I could use some brush up myself, it appears. I took precalc last semester. Already forgot about all the identities. I can barely do the trig problem I posted, go check it out.


請問LaTeX有否中文版 (可輸入漢字)?

可以用CJK宏包
[code]\usepackage{CJK}[/code]
后面在CJK环境中加入文字即可, 中日韩均可:
[code]\begin{CJK*}{GBK}{song}
blah blah blah
\end{CJK*}[/code]


Find the largest positive integer $n$ such that $n$ is divisible by all the positive integers less than $\sqrt[3]{n}$.

Could you give the more specific solution?I can't understand  the above solution

a more elementary solution. Let $ x$ be the largest integer less than $ \sqrt [3]{n}$. Than $ (x \plus{} 1)^3\geq n$. We have $ x|n$, $ x \minus{} 1|n$, $ x \minus{} 2|n$ and $ x \minus{} 3|n$. But than $ lcm(x,x \minus{} 1,x \minus{} 2,x \minus{} 3)|n$. But therefore:
\[ \frac {x(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)}{6}\leq lcm(x,x \minus{} 1,x \minus{} 2,x \minus{} 3)\leq n\leq (x \plus{} 1)^3\]
which is false for $ x\geq13$. so $ x\leq 12$.

Now denote $ a_n \equal{} lcm(1,2...,n)$. We have:
$ a_{12} \equal{} a_{11} \equal{} 2^3\cdot 3^2 \cdot 5 \cdot 7\cdot 11 \equal{} (2^2\cdot3^2)(5\cdot 7)(2\cdot 11) > 13^3$
$ a_9 \equal{} a_{10} \equal{} 2^3\cdot 3^2 \cdot 5 \cdot 7 \equal{} (2^2\cdot 3)(3\cdot 5)(2\cdot 7) > 11^3$
$ a_8 \equal{} 2^3\cdot 3\cdot 5 \cdot 7 \equal{} 840 > 9^3$

Therefore $ x\leq7$ and we have $ 420|n$ ($ a_7 \equal{} 420$) and $ n < 8^3 \equal{} 512$ so the solution is $ n \equal{} 420$


analytic geometry
geometry
graphing lines
slope
LaTeX
area of a triangle

[quote="daermon"][hide]using the pythagorean theorem we get $\sqrt{64+225}=17$[/hide][/quote]

[hide]But the distance from P to A itself is 8, so 17 is incorrect; however, I know 8 is wrong but don't know how to do the question.[/hide]

[quote="espark52"]The endpoints of segment AB are A(0,8) and B(15,0). Wat is the shortest distance in units from P(0,0) to segment AB? Express answer as a common fraction.[/quote]
[hide]Draw the shortest segment from the origin and label the intersection C.
$CB=x$
$CA=17-x$
$PC=\sqrt{17x-x^{2}}$
pytagorean gives us
$17x-x^{2}+x^{2}=15$
$x=\frac{15}{17}$
$PC=\frac{\sqrt{4110}}{17}$[/hide]
This seems wrong but I don't see any mistakes.

:oops_sign:
ur right

the answer is [hide]$\frac{17}{2}$ because the median to the hypotneuse of a right trangle is half the hypotneuse[/hide]

[quote="daermon"]:oops_sign:
ur right

the answer is [hide]8$\frac{17}{2}$ because the median to the hypotneuse of a right trangle is half the hypotneuse[/hide][/quote]
The median isn't the shortest distance ;).

:blush:  :wallbash_red:

[hide]$\sqrt{64-x^{2}}=\sqrt{225-(17-x)^{2}}\\ 34x=0\\x=0$ so the closest point on $AB$ is $A$, so $8$[/hide]

this seems wrong too :(

neither is the sidelength. ;)
[hide="click to see what is shortest"]the altitude to the hypotenuse[/hide]

ok [i]now[/i] I think I have the right answer  :lol:
[hide]the area of a triangle is $\frac{base\times height}{2}$, so $60=\frac{17x}{2}$ and $x=\frac{120}{17}\approx 7.06$[/hide]

[quote="daermon"]ok [i]now[/i] I think I have the right answer  :lol:
[hide]the area of a triangle is base$\times$height, so $60=\frac{17x}{2}$ and $x=\frac{120}{17}\approx 7.06$[/hide][/quote]
That makes more sense then my way so I think my answer is wrong :(.

[quote="daermon"][quote="bpms"]$PC=\sqrt{17x-x^{2}}$
[/quote]
how did you get that?  :huh:[/quote]
The geometric mean of $x$ and $17-x$

[quote="daermon"]what is a geometric mean?  :maybe:[/quote]
The geometric mean of $a$ and $b$ is $\sqrt{ab}$. When you draw the altitude to the hypotenuse, it splits the hypotenuse into 2 segments, let's call them length $x$ and $y$. The altitude is the geometric mean of the 2 segments, or $\sqrt{xy}$.

oh so thats a good way to find altitudes  :D

[quote="lotrgreengrapes7926"][quote="bpms"]$17x-x^{2}+x^{2}=15$[/quote]It's $15^{2}$, not $15$.  :wink:[/quote]
Thank you for pointing that out.//\\

[hide]The shortest distance is the distance that is equal to the altitude to the hypotenuse.

This is just

$\frac{8*15}{\sqrt{8^{2}+15^{2}}}=\frac{120}{17}$

$\frac{120}{17}$[/hide]

[quote="pascal_1623"][hide]
The shortest distance would be a line perpendicular to the line segment.
So, the slope of the line segment is $-\frac{15}{8}$.
Then, the slope of the line parallel to it is $\frac{8}{15}$.

For the first line segment, the equation is $y=-\frac{15}{8}x+8$
The second one is $y=\frac{8}{15}x$.
$=-\frac{15}{8}x+8=\frac{8}{15}x$
Solving for x, we get $x=\frac{960}{289}$
Plugging $x=\frac{960}{289}$ into one of the equations, we get $y=\frac{512}{289}$

Using the distance formula, we get $\sqrt{\frac{960}{289}^{2}+\frac{512}{289}^{2}}=\boxed{64}{17}$[/hide]

I don't think that's right...
Where did I go wrong?[/quote] Do you have a latex error?

[quote="espark52"]The endpoints of segment AB are A(0,8) and B(15,0). Wat is the shortest distance in units from P(0,0) to segment AB? Express answer as a common fraction.[/quote]

[hide="My Solution that I think is correct"]
Please refer to the image below:
[img]http://mediamax.streamload.com/adasarathy/Hosted/trianglehyp.JPG[/img]
$\\  \text{The smallest distance to the hypotenuse is the altitude to the hypotenuse. The area of a triangle is }\frac{\text{Base}\cdot\text{Height}}{2}.\\ \text{Next, you can find out that the area of the triangle is 60, using the base as B and the height as A.}\\ \text{Then, by using C as the base, you get the area formula as: }\\ \frac{17x}{2}=60\\ 17x=120\\ x=\frac{120}{17}\\ \text{The final answer is }\boxed{\frac{120}{17}}$[/hide]

[quote="lotrgreengrapes7926"][quote="pascal_1623"][hide]
The shortest distance would be a line perpendicular to the line segment.
So, the slope of the line segment is $-\frac{15}{8}$.
Then, the slope of the line parallel to it is $\frac{8}{15}$.

For the first line segment, the equation is $y=-\frac{15}{8}x+8$
The second one is $y=\frac{8}{15}x$.
$=-\frac{15}{8}x+8=\frac{8}{15}x$
Solving for x, we get $x=\frac{960}{289}$
Plugging $x=\frac{960}{289}$ into one of the equations, we get $y=\frac{512}{289}$

Using the distance formula, we get $\sqrt{\frac{960}{289}^{2}+\frac{512}{289}^{2}}=\boxed{64}{17}$[/hide]

I don't think that's right...
Where did I go wrong?[/quote]

You messed up the slope.  :P It's $-\frac{8}{15}$ and $\frac{15}{8}$. :D[/quote]

i think he also has a latex error


[quote="Totally Zealous"]Let $n$ be a positive integer such that $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}$ is an integer. Which of the following statements is not true?
(A) 2 divides n
(B) 3 divides n
(C) 6 divides n
(D) 7 divides n
(E) n>84[/quote]

[hide]n=42, E[/hide]

1/2 + 1/3 + 1/7 + 1/n = 41/42 + 1/n
[hide="Therefore..."]
It's obvious that n=42. So A,B,C, and D are all true, but E is FALSE, since n<42. THE ANSWER IS E.[/hide]


trigonometry
inequalities
inequalities proposed

Let ABC be a triangle with $d_a,d_b,d_c$ are the length of the bisectors from A,B,C respectively, and $2p=a+b+c$. Prove that:
$d_a\cos (A/2)+d_b\cos (B/2)+d_c\cos (C/2)\ge p(\cos A+\cos B+\cos C)$

I have a solution but I think I must not post it (Because of M&Y)

 (You can use the inequality of length triangle for this problem)

  :)  :)  :)


A number of spheres with radius $ 1$ are being placed in the form of a square pyramid. First, there is a layer in the form of a square with $ n^2$ spheres. On top of that layer comes the next layer with $ (n\minus{}1)^2$ spheres, and so on. The top layer consists of only one sphere. Compute the height of the pyramid.

[hide]take the cross section of the plane perpendicular to the base that passes through 2 opposite verticies. The cross section is triangle of circles stacked on top of each other. The hieght can be computed by adding the height of the equilateral triangle side 2n-2 formed by the centers of the outermost circles, and adding the 1 unit above and below the triangle to it. Thus, the height is $ (n\minus{}1)\sqrt{3} \plus{}2$

[/hide]

Unfortunately, that is wrong. [hide="actual answer is .."] $ \sqrt{2}(n\minus{}1)\plus{}2$[/hide]

[quote="chrischris"]I agree with azjps. The height of a square pyramid with all sides equal to $ 2$ is $ \sqrt2$.[/quote]

Oh, oops, its an iscoselese right triangle, not an equilateral one.


ARML
geometry
geometric transformation
rotation

Problem: Given square ABCD, semicircles are constructed on sides AB and BC. Points P and K on semicricles AB and BC are chosen so that PBK is a line, with PB 23 and BK 7. Find the distantce between the center of the square and a corner.

[hide="My problem with the solution"]The sol assumes that a circle with diameter PK will form a circle with the center of the square on it. Proof of this is needed. I might be overlooking something, I just sat down to do some math for the first time in months, and always sucked at geometry.[/hide]

Solution with hopefully more clear proof:

[hide]
First, let us prove that PK is a diameter of a circle that passes through the origin.

Consider an extension of both semi-circles into their full circles. We see that the circle with P is just a congruent 90-degree rotation of the circle with K. Any such circles will have two points of intersection, at B and at the center of the square.

Let the center of the square be point S. Consider any isocoles right triangle with legs SK and another equivalent leg, SK`. This leg is a 90-degree rotation of K about S. Now consider the sides of the square that are diameter of the circle capturing K. Consider a 90-degree rotation of every point on that line about S, the side AB. This overlaps the side of the square that is the diameter of the circle encompassing K`. Because both K and the line AB are rotated 90 degrees about the same point S, the intersection of AB and AB` must be intersected by KK`. Thus KK` passes through B, and K`=P.

Therefore, line KBP is a diameter of a circle containing S. From there we easily can show that SB is $\sqrt{274}$.

I'm not amazing at geometry, but I think that does it.[/hide]

[quote="10000th User"]I just consulted my cousin's copy and you read the solution wrong. Your claim that a circle with diameter PK passes through the center of the square is not what the book says. The book says the circle with diameter AB (which is a side of the square) passes through the center. However, the book also, but indirectly, proved that a circle with diameter PK also passes through center if you read the book's solution again :)[/quote]
 :blush:

I apoligize for the stupidity. I missread AB as AE. :(. So much for claiming I don't need glasses.

[quote="Gyan"]Or think of triangles APB and  BKC, while  thinking  point B is pivot (and is fixed) while BC rotates 90 degree so that Point C now coincides with A.
And now look at  APBK,  Angle PBC is now right angle  (Because  the BC and hence the triangle is rotated through 90 degrees), and so are angles APB (and angle BKC ... (Both are in semi-circle) hence APBK is rectangle...  so $AB^2 = 23^2+7^2$ (You half it and then take a square-root to find your answer)[/quote]
I had the exact same thing, I just wanted to see how circles were involved.

Circles=cool.


Can someone explain what the '3-connected' means in "3-connected simple planar graph?"

Also, in a convex polyhedron, is it correct that each face of the polyhedron can have at most one edge in common with any other face?

3-connected I'll think about.  As for the second part, it is in fact true -- if two faces had 2 edges in common, you should be able to prove without much trouble that they are in the same plane.  This isn't possible for a convex polyhedron.

3-connected means that no matter which two (or fewer) vertices (and their edges) you remove, the remaining graph is connected.


Let there be a convex polygon. $ABCDE$ such that the areas of each of the triangles $ABC$,$BCD$,$CDE$,$DEA$,$EAB$  is equal to one unit. Then compute the value of the area of the polygon.

If area of the triangles $ABC,BCD,CDE,DEA,EAB$ is $S_{1},S_{2},S_{3},S_{4},S_{5}$ then the area $S$ of pentagon can be calculated by $S=\frac{1}{2}(\sum_{i=1}^{n}S_{i}+\sqrt{(\sum_{i=1}^{n}S_{i})^{2}-(\sum_{i=1}^{n}S_{i}S_{i+1})})$
Thus $S=\frac{5}{2}$ but is this problem of you ashwinkrishnan.  :D


trigonometry
quadratics
function
algebra
quadratic formula

Given the equation $16x^4+1=16x^2$

a) Show that $\cos\frac\pi{12}$ is a solution of the equation
b) Find the three other solutions

[hide]

A)
$cos \displaystyle \frac {\pi}{12} \Rightarrow cos \displaystyle \frac {30}{2}  = \sqrt {\frac {1+cos30}{2}}$
$cos \displaystyle \frac {30}{2} = \sqrt {\frac {1}{2} + \frac {\sqrt 3}{4}}$

to find the roots of $16x^4 - 16x^2 + 1 = 0$, you can use the quadratic formula for $x^2$

$x^2 = \displaystyle \frac {16 \pm \sqrt {192}}{32} \Rightarrow \displaystyle \frac {1}{2} \pm \frac {\sqrt 3}{4}$

so $x = \pm \sqrt \displaystyle \frac {1}{2} \pm \frac {\sqrt 3}{4}$

looking at the part inside and outside the square root and taking the positive part of both of those, you can see that one of the solutions equals $cos \displaystyle \frac {\pi}{12}$

B) other roots are positive outside of square root but negative inside, the negative outside but positive inside, and the negative outside and inside

[/hide]

$16x^4+1=16x^2\Longleftrightarrow (4x^2)^2-4(4x^2)+1=0\Longleftrightarrow x^2=\frac{2\pm\sqrt{3}}{4}$

Since $2\pm\sqrt{3}=\frac{4\pm 2\sqrt{3}}{2}=\left(\frac{\sqrt{3}\pm 1}{\sqrt{2}}\right)^2$,

The root of the given quartic equation is $x=\frac{\sqrt{6}+\sqrt{2}}{4},\frac{\sqrt{6}-\sqrt{2}}{4},-\frac{\sqrt{6}+\sqrt{2}}{4},-\frac{\sqrt{6}-\sqrt{2}}{4}$.

Especially $\frac{\sqrt{6}+\sqrt{2}}{4}=\cos \frac{\pi}{12}$

[quote="kunny"]The root of the given quartic equation is $x=\frac{\sqrt{6}+\sqrt{2}}{4},\frac{\sqrt{6}-\sqrt{2}}{4},-\frac{\sqrt{6}+\sqrt{2}}{4},-\frac{\sqrt{6}-\sqrt{2}}{4}$.
[/quote]

or just: $x=\cos\frac\pi{12},\cos\frac{5\pi}{12},\cos\frac{7\pi}{12},\cos\frac{11\pi}{12}$ :)

[hide="Using Just Trig"]

First, we plug $x=\cos \frac{\pi}{12}$ into $16x^4+1=16x^2$

\begin{eqnarray*}
16\cos^4 \frac{\pi}{12}+1 &=& 16\cos^2 \frac{\pi}{12}\\
\Rightarrow 16\cos^2 \frac{\pi}{12}(1-\cos^2\frac{\pi}{12}) &=& 1\\
16\cos^2\frac{\pi}{12}\cdot\sin^2\frac{\pi}{12} &=& 1\\
4(2\sin\frac{\pi}{12}\cos\frac{\pi}{12})^2 &=& 1\\
4\sin^2\frac{\pi}{6} &=& 1\\
4\left(\frac 12\right)^2=1
\end{eqnarray*}

And of course the last equation is true and thus $x=\cos\frac{\pi}{12}$ is a root.  Thus, since the equation is even, $x=-\cos \frac{\pi}{12}=\cos \frac{11\pi}{12}$ is also a root.  Similarily we show that $x=\sin\frac{\pi}{12}=\cos\frac{5\pi}{12}$ and $x=-\sin \frac{\pi}{12}=\cos\frac{7\pi}{12}$ are roots.[/hide]

how about this one.. i couldnt even understand the solution that was given, so hopefully someone can solve it:

Show that $4\cos^2(\pi/14)$ is the greatest root of the equation

$x^3-7x^2+14x-7=0$.

hmmmm $\frac\pi{7}$ cannot be expressed in terms of finite sums, products and roots of real rational numbers :?
which makes it quite hard to work with  :(

i guess there is a special way to prove this, so i'ld be glad to see someone solve this... :)

i got stuck at showing that $2\cos\frac\pi7-2\cos\frac{2\pi}7+2\cos\frac{3\pi}7-1=0$  :|

Believe it or not, that very equality has been verified on the Pre-Olympiad forum. So, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43118]there![/url] It's post #7 by Yetti<----who rocks by the way at trig and geom.
Also, about $\frac{\pi}{7}$, we have an equation that helps us.
$\frac{1}{\sin x} = \frac{1}{\sin 2x} + \frac{1}{\sin 3x}$. We must solve it 2 different ways. :lol:.
(1) $\frac{1}{\sin x} - \frac{1}{\sin 3x} = \frac{1}{\sin 2x}$

$\frac{1}{\sin x} - \frac{1}{\sin 3x} = \frac{\sin 3x - \sin x}{\sin x \sin 3x} = \frac{2\sin x \cos 2x}{\sin x \sin 3x}$

$\Rightarrow \frac{2\sin x \cos 2x}{\sin x \sin 3x} = \frac{1}{\sin 2x} \Rightarrow 2\sin 2x \cos 2x = \sin 3x$

$\sin 4x = \sin 3x$.

Now, to solve that, we know $\sin a = \sin b$ if $a = b$ or if $a + b = ...,-3\pi, -\pi ,\pi , 3\pi, 5\pi, ... , (2n-1) \pi, ...$. If $a = b$, $4x = 3x$ and $x = 0$. Since $\sin 0 = 0$, by the original problem, $0$ is an extraneous root, since a division by zero would occur. But, for the latter, since $a = 3x$ and $b = 4x$, we get $a + b = 7x$ and it quickly follows that $x = \frac{(2k-1)\pi}{7}$ for any integer $k$.

Now, for the other part, we must algebracalize the problem :lol:. First, we know from Demoive's that $e^{i \theta} = \cos \theta + i \sin \theta$. Squaring and cubing $e^{i \theta}$ we get the trig functions for $2 \theta$ and $3 \theta$(I will leave as an excercise to the reader.) We find, doing the aforementioned step, that $\sin 2x = 2 \sin x \cos x$ and $\sin 3x = 3\cos^2 x \sin x - \sin^3 x = \sin x (3 \cos^2 x - \sin^2 x) = \sin x(4 \cos^2 x - 1)$. So, plugging these into our original equation, we get:

$\frac{1}{\sin x} = \frac{1}{2 \sin x \cos x} + \frac{1}{\sin x(4 \cos^2 x - 1)} \Rightarrow 1 = frac{1}{2 \cos x} + \frac{1}{4 \cos^2 x - 1}$

$\Rightarrow (4 \cos^2 x - 1)(2 \cos x) = (4 \cos^2 x - 1) + (2 \cos x)$

$\Rightarrow  8 \cos^3 x - 4 \cos^2 x - 4 \cos x + 1 = 0$.

Thus, we know $\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}$ and $\cos \frac{5\pi}{7}$ are solutions to the polynomial.


For each $ n \in N^*$ and an odd prime $ p$,denote by $ v_p(n)$ the unique integer satisfying $ p^{v_p(n)}\parallel{} n$.
Denote by $ ord_p(a,b)$ the least positive integer $ x$ satisfying $ p|a^x\minus{}b^x$.($ gcd(a,b)\equal{}1,a,b\in Z$).
Then we have the theorem:
For odd prime $ p$,if $ p|a^n\minus{}b^n$,$ x\equal{}ord_p(a,b)$,then
$ v_p(a^n\minus{}b^n)\equal{}v_p(a^x\minus{}b^x)\plus{}v_p(\frac{n}{x})$

Could you please help me to prove this?

You are probably talking about [url=http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf]Lifting the Exponent[/url].


[quote="daoldman"]Sorry I posted this in the wrong section...I apologize. Hello, I have a feeling the answer to this is going to be easy. I am helping my stepdaughter with her 5th grade homework and I cannot figure this out or even how to go about it.         math question.........Twenty four chairs will be set up for the school orchestra.How many different ways can the chairs be arranged? Tell what strategy you used to solve the problem.   she wants me to write a letter to the teacher saying we can't figure it out but I NEED to figure this out now. It is driving me crazy!!!!!!!!   PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!![/quote]

I think you just factorialize it. $24!$

Or unless you arrange it in different shapes.
$1*24, 2*12, 3*8, 4*6, 6*4, 8*3, 12*2, 24*1$
$8$ ways

Wait, are the chairs the same(as in shape and size)?
I'm getting confused.

The problem doesn't really give enough information; the way it's written it could be an infinite number of ways.

That was the problem we were having. The question to me could be interpreted many ways. Thank you for the help!!!!!!!!!!!

[quote="math92"]since its 5th grade i would just go with 24! :)[/quote]

If it's 5th grade, would you really expect them to know factorial? I'd go with how many rectangles with an area of 24.

i agree, they are probably looking for equal rows, so in rectangles...


For which positive integers n you can divide the positive integer divisors of $ 30^n$ into three disjoint nonempty sets such that the product of the numbers in each set is the same for the three sets?

As the product of divisors is 30^{n*(n+1)^3/2}. Each set should have product 30^{n*(n+1)^3/6}
n can be either 2mod3 or 0mod3

[quote="t0rajir0u"]A necessary condition, not a sufficient one.  Do you have a proof that such a division is possible?[/quote]

Divide them into ordered pairs and then the proof will be easy

also, i think that should be $ 30^{\frac{n(n\plus{}1)^3}{2}}$.

[quote="pleurestique"]also, i think that should be $ 30^{\frac {n(n \plus{} 1)^3}{2}}$.[/quote]
yepp I corrected it :)


For each fixed $k$ is it possible to find an integer $x_{0}$
such that the followings hold:
[b] 1 [/b] $P(x_{0})\not=x_{0}$.
[b] 2 [/b] $P^{k}(x_{0})\not=x_{0}$.
but [b]3 [/b] $P^{k+1}(x_{0})=x_{0}$
where $P^{k}(y)$ means $P(P(..........P(P(y))..........))$ where $P$ appears k times.

Are they same?(or equivalent)
And I forgot to note that $P$ has integer coefficients.


geometry
MATHCOUNTS
trigonometry
Pythagorean Theorem
area of a triangle
Heron\u0027s formula
angle bisector

Could somebody give me a list of all geometric formulas, big and small? I know basic ones such as Pythagorean Theorem, Heron's Formula, Angle Bisector, etc. Basically, if it is not mentioned in AoPS Introduction to Geometry, I don't know it. Thanks.

[url]http://en.wikipedia.org/wiki/Heron%27s_formula[/url]

Let $ s$ be the semiperimeter = $ \frac{a\plus{}b\plus{}c}{2}$, where $ a$, $ b$, and $ c$ are the side lengths.

Then (let $ A$ be the area of the triangle),

$ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$

Gogators, to me, your request seems absurd. There are infinitely many formulas. Out of those, there are infinitely many geometry formulas. Furthermore, there are several of those without a name. Next, some of those include tools of which you are not familiar of, etc. You get the point.

My point is this: you should narrow down your question a bit. Perhaps you may include whether it is for MATHCOUNTS Nationals, or maybe trigonometry formulas, or something like that. That way, the people of AoPS will be able to better answer your questions. Thank you.


Hi,

Sometimes at basketball we have 16 people. We play teams of 5. Obviously we can make 3 teams but that leaves out 1 person. How can we deal with that extra person? What would be the optimum way to manage 16 people?

What do you mean by manage 16 people. can you phrase your question in a better way?

Yes. We are 16 people. When we are 15 people, we make 3 teams of 5. The winning team stays on. But if you win twice in a row, you get off. To prevent playing too much so that everyone gets equal playing time. WHen we are 14, we play three teams and the team to come on is missing one person, they get that one from the team coming off.
When we we are 16 people, we don't know what the best way to proceed is so that everyone gets equal playing time.

:huh: At any point you have 10 people playing, 6 people waiting in line.  Then 5 people stop playing and go to the back of the line, while the first 5 people in line go to play  :maybe:

This is very easy dude! Put in more problems like this!


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $ f(x)$ be a nonegative function, continuous on $ [a,b]$ and differentiable on $ (a,b)$. If
\[ \int_x^b {f\left( t \right)dt} \leqslant \int_x^b {\left( {t \minus{} a} \right)dt}
\]
for all $ x \in [a,b]$ and $ f'(x) \geq 1$ for all $ x \in [a,b]$ then $ f(x) \leq x \minus{} a$.

This assertion can be proved by considering the function
\[ G\left( x \right) \equal{} \left( {\int_x^b {\left( {f\left( t \right) \minus{} t \minus{} a} \right)dt} } \right)\left( {x \minus{} a \minus{} f\left( x \right)} \right).
\]
If we denote
\[ F\left( x \right) : \equal{} \int_x^b {f\left( t \right)dt} \leqslant \int_x^b {\left( {t \minus{} a} \right)dt} \equal{} : G\left( x \right)
\]
then
\[ F''\left( x \right) \leq \minus{} 1 \equal{} G''\left( x \right)
\]
and the assertion can be rewritten as
\[ F'\left( x \right) \geqslant G'\left( x \right)
\]
for all $ x \in [a,b]$. My question is that in general does the assertion hold true for every $ F$ and $ G$?

\[ \boxed{\begin{gathered}
  F\left( x \right) \leqslant G\left( x \right) \hfill \\
  F''\left( x \right) \leqslant G''\left( x \right) \hfill \\
\end{gathered} } \longrightarrow \boxed{F'\left( x \right) \geqslant G'\left( x \right)} \]

[quote="bookworm_vn"]My question is that in general does the assertion hold true for every $ F$ and $ G$?
\[ \boxed{\begin{gathered} F\left( x \right) \leqslant G\left( x \right) \hfill \\
F''\left( x \right) \leqslant G''\left( x \right) \hfill \\
\end{gathered} } \longrightarrow \boxed{F'\left( x \right) \geqslant G'\left( x \right)}
\]
[/quote]No. Take $ F(x) \equal{} 0$ and $ G(x) \equal{} x^2$ on $ [\minus{}1,1]$.

You should add the condition $ F(b) \equal{} G(b)$.

If $ F\equal{}0$ (no loss of generality here), then the assertion is: if $ G$ is a nonnegative convex function such that $ G(b)\equal{}0$, then $ G$ is nonincreasing for $ x\le b$. Which is true.

That's right. This argument looks like the following context.

$ \int_a^x f(t) dt\equal{}g(x)\Longrightarrow f(x)\equal{}g'(x)$, but the reverse doesn't hold.

[quote="mlok"]You should add the condition $ F(b) \equal{} G(b)$.

If $ F \equal{} 0$ (no loss of generality here), then the assertion is: if $ G$ is a nonnegative convex function such that $ G(b) \equal{} 0$, then $ G$ is nonincreasing for $ x\le b$. Which is true.[/quote]

Oh, I see

[img]http://img165.imageshack.us/img165/8696/clipboard01hv2.jpg[/img]

Any ideas to improve the problem?


Let $ m,n$ be positive integers. Prove that the number $ 5^n\plus{}5^m$ can be represented as sum of two perfect squares if and only if $ n\minus{}m$ is even.

[i]Vasile Zidaru[/i]

[hide]
Assume m<=n

This can be rewritten as $ 5^m(5^{n \minus{} m} \plus{} 1)$. Multiplying by a power of prime congruent to 1 mod 4 can only increase the number of ways to write a number as a sum of two squares if one already exists (see http://mathworld.wolfram.com/SumofSquaresFunction.html), so the $ 5^m$ term does not affect our answer, so we can just focus on the $ 5^{n \minus{} m} \plus{} 1$ term.

Quite obviously, if n-m is even, then that number can be written as $ 1^2 \plus{} (5^{(n \minus{} m)/2})^2$, so n-m being even is a sufficient condition. Any odd power of 5 is congruent to 5 mod 24, making $ 5^{n \minus{} m} \plus{} 1$ congruent to 6, which is not a sum of squares mod 24, so n-m being even is also a necessary condition.

(Also, the 2 can be factored out of the 6 without causing any sum of squares to disappear, leaving us with 3 mod 24, which means the term is 3 mod 4, which means it contains at least one odd power of a prime congruent to 3 mod 4, which really means it cannot be represented as a sum of squares.)

[/hide]

Supose that the number $ 5^n\plus{}5^m$ can be writen as a two sqares sum and prove that $ n\minus{}m$ is even.Supose that $ m$ and $ n$ are not in the same class(odd or even).Let $ m\equal{}2k\plus{}1$ and $ n\equal{}2p$, $ k,p \in N$. Observe that $ 25\equal{}8\cdot 3\plus{}1$ and $ 25^l\equal{}M_8\plus{}1$ for any $ l\in N$.Then $ 5^m\plus{}5^n\equal{}M_8\plus{}6$.contradiction because a perfect square can be $ M_8, M_8\plus{}1 or M_8\plus{}4$ therefore a sum of squares can't have the form $ M_8\plus{}6$

[hide="if"]If $ n\minus{}m$ is even, then $ m$ and $ n$ have the same parity. The first case is when they are both even. This case is trivial. The second case is when $ m$ and $ n$ are odd.

$ 5^m\plus{}5^n\equal{}5(5^{m\minus{}1}\plus{}5^{n\minus{}1})\equal{}(2^2\plus{}1^2)(5^{m\minus{}1}\plus{}5^{n\minus{}1})$

[hide="Lemma 1"]The product of two sums of two squares is a sum of two squares.

=== Proof ===
$ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}a^2c^2\plus{}a^2d^2\plus{}b^2c^2\plus{}b^2d^2$
$ (ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2\equal{}a^2c^2\plus{}b^2d^2\plus{}a^2d^2\plus{}b^2c^2\plus{}2abcd\minus{}2abcd\equal{}a^2c^2\plus{}a^2d^2\plus{}b^2c^2\plus{}b^2d^2\equal{}(a^2\plus{}b^2)(c^2\plus{}d^2)$
=== End of Proof ===[/hide]

Since $ 5^{m\minus{}1}\plus{}5^{n\minus{}1}$ is a sum of squares, $ 5^m\plus{}5^n$ is a product of two sums of two squares, and it is thus a sum of two squares.[/hide]

[hide="Only if"]We must prove that if $ 5^n\plus{}5^m$ is a sum of squares, then $ n$ and $ m$ have the same parity. Thus we proceed by contradiction: we assume that one is even and one is odd and that $ 5^n\plus{}5^m$ is a sum of squares, and try to find a contradiction.
Let us take $ 5^m\plus{}5^n\bmod{6}$. Squares can only be $ 0,1,3$ or $ 4$ mod 6, and since one of $ m$ or $ n$ is odd and the other is even, the sum mod 6 is 0. If the sum of two squares is 0 mod 6, then they must both be divisible by 3 but not 6, or they are both divisible by 6. But neither of them are divisible by 3 or 6, and thus we have found our contradiction. Thus $ m$ and $ n$ have the same parity, and we are done.[/hide]


inequalities
trigonometry
LaTeX
geometry proposed
geometry

Let ABC be an acute triangle. Prove that
   [tex]\left( \sum \sin^2A\right)\left( \sum \cos A\right) \leq \frac 12 \left( \sum \sin A\right)^2.[/tex]

Um,I think the inequality is still right for all kind of triangle{If i don't make a mistake}.I have had a solution for it but it is quite long so I will post it later. :maybe:

This is just the direction,but I hope it helps
       [tex] \left(\sum \sin^2A\right)\left( \sum \cos A\right) \geq \frac 12 \left(\sum \sin A\right)^2 \Leftrightarrow [/tex] [tex] \left(\sum a^2\right)\sum \frac {b^2+c^2- a^2}{bc} \geq \left(\sum a\right )^2 [/tex]
[tex] \Leftrightarrow \left(\sum  \frac {b^2+c^2- a^2}{bc}\right)  \geq { \left(\sum a\right)^2 \over \sum a^2 } [/tex]

So we must prove this inequality with a,b,c are three sides of a triangle.This one is difficult but really interesting :)

[color=red][Admin Edit: Made LaTeX compliant][/color]

Anh Cuong, can you please post your complete solution about the last hard inequality?

Thank you very much.

I assume that you mean the inequality:
  [tex]\sum\frac {(b^2+c^2-a^2)} {bc}\geq\frac {(\sum a)^2}{\sum a^2} [/tex]. I put [tex] a=y+z, b=z+x, c=x+y [/tex] and I obtained after lots of computations (which I hope are good) the equivalent form:
  If [tex] x,y,z>0[/tex] then [tex] 1+\frac{\sum x^2}{\sum xy}\geq\frac{\prod (x+y)}{4xyz} [/tex]. But something tells me that this is not good. What can you do when x tends to 0?

Damn, in fact we must prove that [tex] 1+\frac{\sum x^2}{\sum xy}\leq\frac{\prod (x+y)}{4xyz}[/tex]. I looked at Anh Cuong post and I didn't see that in fact the sign is reversed.

I finally managed to solve treegoner's problem. If the triangle is acute, then we can write [tex] a^2=y+z, b^2=z+x, c^2=x+y [/tex] with x,y,z non-negative. Then the inequality can be written:
      [tex] 4\cdot \sum \frac {x(x+y+z)}{\sqrt{(x+y)(x+z)}} \leq (\sum{\sqrt{x+y}})^2 [/tex]. But this follows from treegoner inequality [tex]\sum 4ab\sin^2(C)\leq(\sum a)^2[/tex] applyied for  [tex] a=y+z, b=z+x, c=x+y [/tex] and the triangle having angles [tex] sinA=\sqrt{\frac{x(x+y+z)}{(x+y)(x+z)}}[/tex]

Hi everybody,here is my solution(Hope that i don't make any mistake)
[tex]\left(\sum a\right )^2  \geq \left(\sum a^2\right)\sum \frac {b^2+c^2- a^2}{bc}[/tex]
[tex]\leftrightarrow\left\sum\ (a-b)^2 [\right\left a+b-c\over\ 2abc \right  - \left 1/(a^2+b^2+c^2) ]\right \geq\ \left 0 \right [/tex]

Now,let x=p-a;y=p-b;z=p-c;A1= (a+b-c)/2abc - 1/a^2+b^2+c^2; A2=...;A3=...
We can see:
abc(a^2+b^2+c^2)A1=[(x+y)^2+(y+z)^2+(x+z)^2]z-(x+y)(y+z)(z+x)=x^2z+y^2z+yz^2+xz^2+2z^2-x^2y-xy^2 .
 So,A1+A2>0;A2+A3>0;A1+A3>0 .
I think it is easy now,so let try for the rest. :)

My first solution is somewhat like Harazi's.
  But when I see  Anh Cuong said that it's also right for obtuse triangle, I have another solution to this one
    By replacing sum(cosA)=1+r/R, we have
       2(r/R)*( Sum (sinA)^2)<=2(Sum sinA*sinB) - Sum (sinC)^2
<=> p^2<=4R^2 +5Rr +r^2
    (a^2+b^2+c^2 = 2(p^2-r^2-4Rr) )
     This one can be easily proved by this following inequality
                sum sinA*sinB<=(sum cosA)^2.
                  <=> p^2 <=4R^2+4Rr+3r^2.  :)

Anh Cuong, can you please complete your solution with the missing steps?

Thank you very much.


On the enroll page, it says that there are 3 levels of classes, but it lists two levels: "Introductory" and "Intermediate."

Yes, it's Art of Problem Solving, where we don't know how to count  :wink:  :rotfl:

(No offense, the classes are awesome  :D , but haha counting error when they're listing a counting class)

[quote="now a ranger"]I think the classes are probably pretty useful, considering how little I can count  :lol:[/quote]

spam?

This is because they got rid of the Olympiad classes (they used to have geometry and inequalities), and probably never edited that.

[quote="now a ranger"]What does INTERMEDIATE mean? For example, in chess intermediate can mean a very good chess player, even though the name may not sound like it, whereas beginners have ratings up to 1250, although they aren't technically beginners and play a lot.[/quote]
beginner<intermediate<advanced< or =professional

[quote="funcia"]beginner<intermediate<advanced=WOOT< or =professional
if you need WOOT in there but that isn't what you asked I think...[/quote]
I meant woot as the third class level.

Introduction-Mathcounts/AMC 10 level
Intermediate-AMC 12/AIME level
Olympiad (WOOT)-Olympiad level


group theory
abstract algebra
symmetry
geometry
geometric transformation
rotation
superior algebra

Let $A,B$ in $O(2)$ and suppose $\det A = 1$, $\det B =-1$. Show that $B^2=I$ ($I$ is identitiy) and $BAB^{-1}=A^{-1}$. Deduced that every discrete subgroup of $O(2)$ is either cyclic or dihedral.

I can prove the two identities. I need help on the structure of discrete subgroup of $O(2)$. Why are there at most two generators?

For the first part: If $\det B=-1$, then $B$ is a symmetry with respect to a line passing through the origin of $\mathbb R^2$. Then clearly $B^2=I$.
If $\det A=-\det B=1$, then there is a nice geometric method to prove $BAB=A^{-1}$. Another method is to know that if $\det A=1$, then $A$ is a rotation and any roation is obtained by composing two symmetries i.e. $A=CD$ for some $C,D\in O(2)$, $\det C=\det D=-1$. Then $BAB^{-1}=A^{-1}\Leftrightarrow BCDB=(CD)^{-1}\Leftrightarrow BCDB=DC\Leftrightarrow (BCD)^2=I$. The last holds because $\det(BCD)=-1$.

For the second part of the problem:
Let $G$ be a discrete subgroup of $O(2)$.
$O(2)$ is a compact group, so $G$ must be finite.
$\det: G\to(\pm 1,\cdot)$ is a group homomorphism. Let $R=\ker\det$.
$R$ is the set of rotations in $O(2)$ and it identifies to a subgroup of $(\mathbb S^1,\cdot)$, where $\mathbb S^1=\{z\in\mathbb C||z|=1\}$. $R$ is a finite group, and it is known that finite subgroups of $(\mathbb C^*,\cdot)$ are cyclic. Therefore $R$ is cyclic.

If $R=G$, then $G$ is cyclic and we are done.
The alternative is $G=R\cup Ry$ for some $y\in G\setminus R$. $y\not\in R\Rightarrow \det y=-1\Rightarrow y^2=1$.
Let $R=<x>$ for some $x\in R$ with $x^n=1$ for some $n$.

It is easy to see that $G$ is generated by $x$ and $y$. We have the relations $x^n=y^2=1$ and $yxy=x^{-1}=x^{n-1}$.
This guarantees the existence of a surjective homomorphism $D_n\to G$, where $D_n$ is the dihedral group. Since $D_n$ and $G$ both have $2n$ elements, this map must be an isomorphism.