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I keep seeing these articles pop up that say females are not as intuitive as males in mathematics and although ethically one wants to disprove it, I seem to see many strong male math students in my school and only a handful of female students. I want to know what everyone here really thinks.

Note: I am not trying to bring up a gender war in the forum, I just want a real, clean discussion and see the opinion of the math community.

I think that this issue goes back to historical gender stereotypes.  Females were left home to clean and raise the family, and males went out and did the work and brought home the food and money.  Even as females entered the workforce, they (in most cases) worked in more humanitarian areas, as the math and science disciplines were left for males.

Even in schools today many of my friends (girls) don't like math or science because they feel that it is for boys, and english if for girls.  I also know guys who are ashamed to admit that they enjoy english more than math/science.

Maybe it does have something to do with how the brains of the different sexes work, but I think that gender stereotypes also play a large roll in the lack of females in math and science disciplines.

I agree with byrdgyrl to some extent.
It just doesn't explain why girls, these days, prefer biology and chemistry infront of math and physics. I don't know how it's over there but there are plenty of female students reading the two mentioned subjects in college.
The majority of the dispensers (from what I have seen) are girls.
And in some areas the step from chemsitry to physics isn't that big at all.

I think the main problem is motivation.

Maybe lack of motivation from oneself and/or one's relatives really prevent girls from joining math teams, and if they join math teams, lack of motivation still make them eliminated in the selection for the team. And if they are selected, well boys who are selected are still more than girls who are selected :roll:

I think the difference of the brain has to play a big part. Males' brains are more suited to mathematics than females'.

Female and male brains are different.  That fact has been proven.  But there is significant evidence that shows that this should not be stopping women from working in math and science fields.

Male and female brains mature differently.  Female brains develop language skills earlier than males, male brains develop spatial skills earlier than females.  By the age of 2.5 girls and boys are not playing together because boys have not conquered the verbal aspect, and girls are not as spatially developed.

By the time that the male and female brains have caught up to each other, in high school, males are already practiced at math and science skills, females at linguistic skills.  These prior experiences keep males and females from emerging from their "comfort zone" and trying a new field, even though each sex has caught up to each other.  Here is where we lose many females in the math field - many don't take the highest level classes because that is not their [b]original[/b] aptitude, although their brains have caught up to the visual/spatial abilities of the males.  Likewise, this can be said for males in language arts.

So brain development does play a part, but both males and females should be able to perform well in mathematics.  It is just that many females get turned off of it when they are behind the males in elementary school.

http://www.sfgate.com/cgi-bin/article.cgi?file=/chronicle/archive/2005/03/03/BAGSKBJI981.DTL

[quote="Ignite168"]I think females need to stay in the kitchen and we should go back to all boys school to teach the men professions and how to support a family and teach the women home economics.

This way, we eliminate the problem of a skewed amount of genders in terms of participation in math.[/quote]
You are wrong!!!!! I am a girl just by some things I said people have the idea i am a guy... just the other day my mom made me make some spagettios or however you spell them in cans.
I cut myself so badly and was announce unfit for doing anything womanly and why can't i be a normal girl who likes kitchens... but i LOVE maths and don't want to have to cook since i HATE it
my hands hurt i cut myself two days ago with another can and i cut myself with a knife when making sandwiches a year ago....you are so wrong!

[quote]Possibly the problem for girls with joining math team is that it is about the nerdiest thing that it is possible to do, besides chess team and comic book club.[/quote]
i think that is most likely it i get laughed at alot and called names for chatting about math other than here

[quote="Ignite168"]I think females need to stay in the kitchen and we should go back to all boys school to teach the men professions and how to support a family and teach the women home economics.

This way, we eliminate the problem of a skewed amount of genders in terms of participation in math.[/quote]
that's not very nice ignite.

[quote]You are wrong!!!!! I am a girl just by some things I said people have the idea i am a guy... just the other day my mom made me make some spagettios or however you spell them in cans.
I cut myself so badly and was announce unfit for doing anything womanly and why can't i be a normal girl who likes kitchens... but i LOVE maths and don't want to have to cook since i HATE it
my hands hurt i cut myself two days ago with another can and i cut myself with a knife when making sandwiches a year ago....you are so wrong!

[quote]Possibly the problem for girls with joining math team is that it is about the nerdiest thing that it is possible to do, besides chess team and comic book club.[/quote]
i think that is most likely it i get laughed at alot and called names for chatting about math other than here[/quote]

in latin nouns ending in "a" are femininum (except some like nauta which are masculinum). so noun "funcia" is femininum. but, whole this time i thought that you are a male. wait, you are a female, aren't you? i'm confused.

Funcia is being sarcastic.  He is a guy.

Hmm, I have noticed (at my school) that females tend to be more prevalent in honors math classes, but the people who go beyond the curriculum and are stars on the math team are almost always males.

Anyway, frankly, it could be true either way, and this is an issue for scientists to debate  :wink:

i'm a girl and i like math and science a lot! I acctually suck at languages. I niess it also depends on how you are brought up. LIke My dad wanted a son or something, so he taught me like math since i can remember and we built RC cars and other things together when i was little. On the contrary, these twins i know, their mom read them a story every night ever since they are very young, so they are super good with writing.
and another kind of general and stereotypical thing that is kinda true. Girls can do what they are caught well, but guys are the ones that takes the extra step that find and learn more in depth. Girls can do as well as guys in math classes, but when it came to math contest and stuff, we are just a step behind.

I don't mind the fact that "girls aren't as good at math and/or science" because since i do like math and science, its easier for me to get into tech school since they are always looking for more girls to balance the population of guys.

In my school, over 50% of the science club is female, but there's only like 3 out of around 25 in the math club.

The reason why girls don't join math clubs as much is because they know that they know that there aren't as many girls in the math clubs, so they won't join.

Eventually this will lead to exclusive-male math clubs.

[quote="Ignite168"][quote="deimos"][quote="Ignite168"]I think females need to stay in the kitchen and we should go back to all boys school to teach the men professions and how to support a family and teach the women home economics.

This way, we eliminate the problem of a skewed amount of genders in terms of participation in math.[/quote]
that's not very nice ignite. [/quote]

How is it not? It's been this way for thousands of years and the system's been working fine. I see no reason why we suddenly changed it recently.[/quote]

So by keeping women from getting an education, are you going to let us vote?  But why let us vote if we have no understanding of a topic that we would be voting for without schooling to begin to understand how our government works?  And if you take away our right to vote how would the United States have any sort of "equality" which it has founded itself upon?  I know that is how the US started, but it has evolved - it shouldn't revert.

I'd like to ask you how you plan on telling me and other girls that we are not allowed to go to school anymore.  Tell all the professional women out there that they cannot hold their jobs anymore, they must be housewives.  Keep girls from going to school.

What makes men so [i]superior[/i] to women that men are allowed to become educated, but women can only be housewives?  I know many women who are perfectly happy with that role, but if you were to tell them that they had to to it, no education or chance to break away from it, they would be very angry.  Very few women would take this decree that you hand out happily.  What should stop a woman from being and equal to a man.  Does this mean that once again, man [i]owns[/i] his wife and can do anything to her, including beat/abuse her?  Is this what you want?[/i]

I am taking 3 maths classes now.

Special topics in analysis: 4m, 0f
Probability Theory: 20ish m, 2ish f
Galois Theory: 20ish m, 6ish f

Math club officers: 6m, 2f

Maybe things are better in Europe than in USA. :P

[quote="byrdgyrl"]
What makes men so [i]superior[/i] to women that men are allowed to become educated, but women can only be housewives?  I know many women who are perfectly happy with that role, but if you were to tell them that they had to to it, no education or chance to break away from it, they would be very angry.  Very few women would take this decree that you hand out happily.  What should stop a woman from being and equal to a man.  Does this mean that once again, man [i]owns[/i] his wife and can do anything to her, including beat/abuse her?  Is this what you want?[/i][/quote]

Just a comment, Ignite never mentioned anything about superiority.  And besides, who said the men have a choice?  The way I see it, in Ignite's new scenario, they're being forced into a situation as much as women are.
Not at all to say I support it...you're just putting a strong bias on it and putting words in Ignite's mouth, that's all...

The mathematical world needs minds of both types.  Well, then again, I would say that the discrepancies among specific minds are greater than those between females and males.

Anyways, I don't really have anything to say on this topic apart from the fact that I, in the position of colleges, etc., wouldn't distinguish much between girls and boys in terms of admission.  Of course, I would never want to go to a 99% male college (though I'm not keen on a 99% female college either, that would definitely have its benefits  :D ).  When I see a person, I first see a person, and then I see physical characteristics.

Well to post statistics for math-leaning classes,

Math Seminar: 7m, 1f
AP Physics: 6m, 5f
Though I could swear I have 12 ppl in ap physics...

My! Well what Xevarion and fredbel say is quite helpful....i think that it actually points to how people are made so they fall in love(funcia who does not like that fact dies) and therefore girls are off chasing guys during high school and/or college instead of focusing on maths of course I might live in a strange town where everyone just is different....

about superior stuff: some of us think math is superior to language on this site...at least me...
@deimos & K81o7
The latin is "A" for most feminine and I do regret my name.
You see it starts out with me and my mental problems. I like to
talk to myself and other people using mult. personalities and all...so I made The F.U.N. of the C.I.A.. or the The Frivolous Undercover Ninnies of the Criminal Iditarod Assassinators and we are just called Funcia and that's why setna and athena have this same name on different sites....it doesn't really matter if I am a boy who makes spaghettios does it? i mean they sucked anyway..

[quote="byrdgyrl"][/quote][quote="Ignite168"][quote="deimos"][quote="Ignite168"]I think females need to stay in the kitchen and we should go back to all boys school to teach the men professions and how to support a family and teach the women home economics.

This way, we eliminate the problem of a skewed amount of genders in terms of participation in math.[/quote]
that's not very nice ignite. [/quote]

How is it not? It's been this way for thousands of years and the system's been working fine. I see no reason why we suddenly changed it recently.[/quote][quote="byrdgyrl"]

So by keeping women from getting an education, are you going to let us vote?  But why let us vote if we have no understanding of a topic that we would be voting for without schooling to begin to understand how our government works?  And if you take away our right to vote how would the United States have any sort of "equality" which it has founded itself upon?  I know that is how the US started, but it has evolved - it shouldn't revert.

I'd like to ask you how you plan on telling me and other girls that we are not allowed to go to school anymore.  Tell all the professional women out there that they cannot hold their jobs anymore, they must be housewives.  Keep girls from going to school.

What makes men so [i]superior[/i] to women that men are allowed to become educated, but women can only be housewives?  I know many women who are perfectly happy with that role, but if you were to tell them that they had to to it, no education or chance to break away from it, they would be very angry.  Very few women would take this decree that you hand out happily.  What should stop a woman from being and equal to a man.  Does this mean that once again, man [i]owns[/i] his wife and can do anything to her, including beat/abuse her?  Is this what you want?[/i][/quote]

Women should not be allowed to vote, if women weren't allowed to vote in the latest elections, we wouldn't have elected all of these Democratic presidents. Even Ann Coutler, a woman herself, and one of the most famous speakers on this subject, agrees with the fact that women should not be allowed to vote. Women still ARE getting an education, however, it's more oriented toward sociology and psychology especially dealing with kids. Please find me where in the original US Constitution does it say anything about voting being necessary for equality.

And I never said men are superior to women, did I? Both are necessary for a proper function of society. Without women, our kids cannot be raised, without man, women cannot provide for their kids. I never said a man owns a wife either, I merely said that women should be in charge of raising the kids, and the men should be in charge of getting as best a job as they can for the family.

[quote="fredbel6"][quote="Ignite168"]How is it not? It's been this way for thousands of years and the system's been working fine. I see no reason why we suddenly changed it recently.[/quote]

Babies used to die of things like diarrhea for millennia.  Yet the system did seem to work.
Lots of European countries were ruled by monarchs for centuries.  Yet the system did seem to work.

Maybe you should write "seems to work fine FOR ME".

In my last year of maths there are in

A. Pure maths :  4 men, 2 ladies
B.Mathematical physics and astrophysics : 2 men,1 lady
C. Applied math : 3 men, 4 ladies

Not really a huge difference....[/quote]

Babies still die of diarhea and I don't think you understand the difference of a system being functional and a system working fine. All your giving me is examples of functional systems.

What's wrong with monarchs?

Sorry, but are you really that ignorant¿

You give as only reason that it worked in the past, and call it working fine.
I would work fine for you, that's correct.
Is it really clear to you that you want to forget more than 100 years of human rights movement, especially equality, education and free will/decision/etc.¿

And now you also claim that women are the cause of bad elections...
No, the real reason for bad elections are ignorant people that like it better to have someone $\neq$ themself as the reason for the bad condition of the society (and for the bad elections  :roll: ).

I suggest to force Ignite into what he thinks a woman should only do for some months.
Especially not allowing him to post on MathLinks since this would obviously need some education (writing, math).

[quote="rainynightstarz"]you guys have very small classes[/quote]
Well we started off in our first year with about 80 people, half of which flunked.  And again half of those who passed went on to physics instead.   :wink:

[quote]
Women should not be allowed to vote, if women weren't allowed to vote in the latest elections, we wouldn't have elected all of these Democratic presidents.[/quote]
Uhm, even though participation in USA elections is voluntary (which it isn't in Belgium, unless you're not a Belgian...) it's a better representation of what the people want than when they can't vote.  Your reasoning is interesting : democratic president are not as good, women vote for them, therefore women shouldn't vote

Again : maybe we should just let you vote, because I think your opinion is what everything is about. :huh:

[quote]Please find me where in the original US Constitution does it say anything about voting being necessary for equality. [/quote]
Where in the US Constitution can I find that slavery should be illegal and that African-Americans should have the right to vote?
Why on earth should one refer to a more than 200 year old constitution?  Are the founders of the USA saints or something?
I'm happy to live in a country where people have the guts to call the founders of the nation a "bunch of highwaymen" and the second king a "megalomaniacal mass murderer". :)

[quote]What's wrong with monarchs?[/quote]
Well that and other things.

Well I for one think that women should be schooled(which if I am not that ignorant they are in most places.) And that elections ARE voluntary I mean if they weren't it would be like forcing people to be free! (sorry nationstates addicts like to talk like that)
*puts on mascara*

in my school there are many girls on math team but none of them have been good enough to make team one for the past four years (I don't know about before that).  One curious note is that on my arml team there were about 10 Indian males while there were no Indian females but for Chinese/Koreans there was a more balanced representation from the two sexes.

Another observation I have made is that although math team is one of the nerdiest activities two others need mention: compter science and dungeons and dragons.  In my schools AP Computer science class there has been one girl the past three years.  At the comp sci competitions I went to there were only one or two girls.  at least in my school computer science is not seen as a super nerdy class.  In the groups I have played dungeons and dragons girls are surprisingly common given the nerdiness of hte activity.

Our math league team was smaller this year than most.  Out of 60+ people who tried out, 22 made it (10-12 grade math league, although freshmen who are in calc must try out for this instead of the 9th grade one, so we have I think 2 freshman males this year).

But out of that group, there are 5 females, although I know for certain that more males than females also tried out.

Strangely though, out of the 9 seniors who made it, including myself, all were males.

Ok since a lot of people were giving statistics, here are mine from both past and present classes:

Math:
AP Calculus AB: 6 males, 9 females
AP Calculus BC: 11 males, 13 females
Multivariable Calculus/Linear Algebra: 14 males, 4 females
Math League Team: 9 males, 1 female
Princeton Comp. Math Team: 9 males, 1 female
HMMT projected team: 8 males
AMC participants: 13 males, 5 females
AIME qualifiers: 13 males, 2 females
USAMO qualifiers: 2 males (me and another kid)
School Winner for last three years: All male (me and another kid)
Highest AIME scores last three years: All male (me and other kid)

Science:
Honors Physics: 10 males, 10 females
AP Physics B exam: 6 males, 2 females
AP Physics C: 15 males, 2 females
Honors Chemistry: 11 males, 16 females
AP Chemistry: 14 males, 15 females
Honors Biology: 10 males, 14 females
AP Biology: 13 males, 16 females
Physics and Chemistry Club: 9 males, 4 females
Officers of Club: 5 males, 1 female
Physics USA(second round) Olympiad qualifiers: 3 males
Science League: Physics - 11 males, 1 female
                      Chemistry - 8 males
                      Biology: 5 females, 3 males

Ok. I have provided an overabundance of statistics. I noticed that females tend to outnumber males in classes that are based on mostly memorization, such as AP Biology. And that too, they only outnumber slightly.

But come down to the hardest courses such as AP Physics C, 15 males vs. 2 females. Multi-Variable Calc/Linear Algebra only has 4 females.

USAMO quialifiers from my school have always been male as well as AMC school winners. Also, AIME qualifiers: only 2 of 6 females made it through to the next round versus 13 of 13 males.

Other observations, but not necessarily conclusions, can be drawn from these statistics. Just a few numbers to ponder upon.

Just wanted to say that society probably influences greatly the composition of math classes and as is it is today, the education for boys is very different than that of girls. Maybe not that much in scadinavian countries, I don't know. I'm willing to assume small differences between females and males in this regard. Something that might show in the statistics but definitely not in the way we see it now. In the other hand I don't feel science should spend effort deciding whether it is a matter of the female brain or merely a social effect. I think proven the difference is not that big, probably balanced in other respects and unessential for things that are more important in life, so I wouldn't think too much about it.

Finally either Ignite is making a very big bad joke or he just landed from the tenth century which is nonetheless fine, I'm just too tired to start that kind of update

Ursula

Edit: I strongly support the idea that Ignite is joking, I mean, c'mon, "...we woudn't have elected all these democratic presidents..." jaja good one

Edit 2: I just read that paragraph again, to refresh my memory, oh my! it's worth gold!

2 people from my school qualified for MOSP one boy and one girl ( the only girl there)
there is accutually a big amount of girls in the math club at my school. Though its not as much as guys, but still a lot.
In science club, its about half and half.

you people have small classes
my math teacher last year, ( the calc BC teacher) have around 40 people PER class and she have 4 classes this year
and all the other classes at my school are at least 30 people
my stats class right now, is after school so there is less people, about only 15
USAMO qualifiers from my school last year was 3 girls and 5 or 6 boys , so my school is a lot more balanced than other girls

[quote="Xevarion"][b]fredbel6[/b] and I are university students. Advanced classes in universities are often small. 4 is the smallest class I've had, but last semester I had a class of 7 (Fourier Analysis, and they were all guys).

There are a fair number of girls majoring in math that I know of. But a vast majority of guys. There are also a good number of women in the graduate school for math, but still a large majority of men. So it's not absurd, but the males outnumber the females easily.

I just noticed that I refer to females my own age as "girls" and older as "women"  :| even though grad students are not that much older than I am. Weird. I wonder if I should already be calling fellow undergraduates women, and if not now, when? :|[/quote]
Oh yes, I was talking about university classes and nothing else.
In high school we only get any choice once every two years, and there just a few programmes to choose from.  I ended up (pretty obvious) taking "Science-8hours of maths".
The Flemish education system - even though offering much less choice- is known to be much better than the USA system.
When you simply have to take all that history, when you are forced to learn French, English and some German, when you have to take six years of geography... the system works pretty well. :wink:   Newer and more exotic isn't always better.

Well four people from pure mathematics stayed at the university last year to obtain a phd, three of them are "ladies".  As I lost one year, one of them is actually a month younger than me. :huh:   So what do I call HER? Lady?  Woman?  Girl?  :D

good idea but who made this to begin with??
and to point from what fredbel and Xevarion said go up to funcia's original post!!!

let x be the person that started the forum.
Take 0 2 X, and de¯ne ~X = f0g, ~¸(0) = p(0).
If 9z 2 X, z =2 ~X , extend ~¸ from ~X to the subspace generated by ~X and z,
~¸
(tz + ~x) = t~¸(z) +~¸(~x):
Suppose x1; x2 2 ~X and ® > 0; ¯ > 0.
¯¸(x1) + ®¸(x2) = ¸(¯x1 + ®x2)
= (® + ¯)¸ ³ ¯
®+¯ x1 + ®
®+¯ x2´
· (® + ¯)p ³ ¯
®+¯ x1 + ®
®+¯ x2´ = (® + ¯)p ³h ¯
®+¯ i(x1 ¡ ®z) + h ®
®+¯ i(x2 + ¯z)´ · ¯p(x1 ¡ ®z) + ®p(x2 + ¯z)
Thus,
¯ [¡p(x1 ¡ ®z) + ¸(x1)] · ® [p(x2 + ¯z) ¡ ¸(x2)]
1
®
[¡p(x1 ¡ ®z) + ¸(x1)] ·
1
¯
[p(x2 + ¯z) ¡ ¸(x2)]
sup
x12X;®>0
1
®
[¡p(x1 ¡ ®z) + ¸(x1)] · a · inf
x22X;¯>0
1
¯
[p(x2 + ¯z) ¡ ¸(x2)]
De¯ne ~¸(z) = a, then assuming t > 0,
~¸
(tz + ~x) = t~¸(z) +~¸(~x) = ta +~¸(tx)
· t ³1
t p(~x + tz) ¡ ¸(~x)
t ´+ ¸(~x)
· p(tz + ~x)
the case of t < 0 is similar. To extend for X, we will use the Zorn's lemma.
Let E be a collection of all extensions e of ¸, e(x) · p(x) for all x 2 Xe. Where
e1 Á e2 if Xe1 µ Xe2 and e1(x) = e2(x) in Xe1 . Thus, E is a partially ordered
and x=neelnal

Well neelnal is my friend I'll go PM them!!!

[quote="Xevarion"][b]fredbel6[/b] and I are university students....There are a fair number of girls majoring in math that I know of. But a vast majority of guys. There are also a good number of women in the graduate school for math, but still a large majority of men. So it's not absurd, but the males outnumber the females easily.[/quote]

Xevarion, the gender statistics at Princeton are not representative of the national scene, where 46% of mathematics majors are female.  The Princeton math department has a historically mixed reputation as regarding women, so there are plenty of people who discourage females from attending Princeton if they're intending to concentrate in mathematics.

As for what you should call your female peers... as long as it's the same word-type as the way you refer to yourself, you should be fine.  If you call yourself a boy, call them girls; if you call yourself a man, call them women.  15 years ago, you would have had to call them women, but in the last 5 years or so females under age 30 or so have frequently called themselves girls, so it's no longer clear.

[quote]The Princeton math department has a historically mixed reputation as regarding women, so there are plenty of people who discourage females from attending Princeton if they're intending to concentrate in mathematics. [/quote]

That's interesting.  I hadn't heard this about Princeton before.

Princeton certainly has some exceptional women mathematicians.  Two years ago, their Putnam team had two women on it, which is pretty unusual among schools that consistently place in the top five on that contest.  Two female Clay junior fellows have gone to Princeton for undergrad, more than any other school except Harvard, I think.  They have succeeded in attracting some very strong women to their graduate program, including Melanie Wood, who mentors their Putnam team.   A few years ago, Princeton's valedictorian was a female math major who went on to become a Rhodes scholar and is now back in their PhD program, so presumably she was not unhappy with her experience as an undergraduate math major there.

And they have some very well known female faculty--Ingrid Daubechies at the senior level and Maria Chudnovsky at the junior level.  (I know that women are still very much in the minority on the faculty, but this seems to be true at all the major research university math departments.)  EDIT:  I just noticed that Prof. Chudnovsky has left Princeton for Columbia, but she certainly did a lot of very impressive work during her time as a doctoral student and junior faculty member at Princeton.

I do know that the women's colleges (and more generally liberal arts colleges) have a very strong record of having their graduates go on to do PhDs in the sciences compared to R1 universities in general, but I hadn't heard that Princeton, in particular, was any worse an environment for women math majors than other highly ranked math departments at R1 universities.

[quote="sophia"]
Princeton certainly has some exceptional women mathematicians... A few years ago, Princeton's valedictorian was a female math major who went on to become a Rhodes scholar and is now back in their PhD program, so presumably she was not unhappy with her experience as an undergraduate math major there.

And they have some very well known female faculty--Ingrid Daubechies at the senior level and Maria Chudnovsky at the junior level.
...but I hadn't heard that Princeton, in particular, was any worse an environment for women math majors than other highly ranked math departments at R1 universities.[/quote]

Maria is at Columbia now, not Princeton.

I make no claims as to Princeton's current status as being worse or better than other R1 departments.  Some years ago, it used to be worse, on average---though as far as I know there have always been some women who have been happy at Princeton---and that's my reason for saying it historically has had a very mixed reputation.  Reputations take a while to change, which is probably one reason why the percentage of female math folks is lower there than at comparable schools. The people I know currently at Princeton have not been able to give me a clear picture of what the environment is like for women these days, though it has to be better now than (say) 15 years ago.

[quote] Maria is at Columbia now, not Princeton. [/quote]

I had actually gone back to edit my original post to correct that before you posted this--sorry for not checking in advance.  (After I posted it, I immediately thought about the fact that junior faculty move around a lot and that I should check if she's still there.  So I visited her web page and then edited my post right before your appeared.  :)  )

I think you are right that reputations are slow to change in some circles.  Princeton didn't even admit women as undergraduates at all until around 35 years ago, so 15 years ago is relatively ancient history there.

(All the other top R1 math departments that I can think of have had women in their undergrad math classes for much longer:  MIT, Chicago, Berkeley, Harvard, Michigan, Wisconsin, UCLA, NYU, Columbia, (EDIT:  and Stanford, probably others I'm forgetting as well.) --though in the case of Harvard and Columbia, it was originally through some sort of cross-registration with Radcliffe and Barnard, respectively.  But Princeton had no women at all in undergrad math classes before around 1970, as far as I can tell.)

But as someone who has written letters of recommendation in recent years for a number of young women who are strong in math and were planning math majors, none of them had told me that they'd heard anything particularly negative about Princeton as opposed to other schools with comparable research math departments.   (It should be noted that none of them actually wound up attending Princeton, however.  But, as far as I know, nobody discouraged them from going there on the particular basis that it was a bad place for women math majors.)

I never heard anything about Princeton being bad for women math majors. (Although I'm a guy, so there wasn't much reason for anyone to tell me.) The only remaining possible deterrent I can think of is if some of the old professors still miss the allegedly good-ol' days back when all of the students were male, and don't like to advise the women for independent work or grade them fairly in classes. But I don't even have any reason to suspect that's true -- none of my female friends here have ever said anything to that effect.

Maybe I am overly optimistic, but my inclination is to reject that stereotype as outdated.

Also, I think the proportion of grad students is more balanced than that of undergraduates, though I'm not sure.

I completely agree with Xevarion(Nathan).

I don't think Princeton is sexist anymore so why hesitate girls? I believe it is incredibly ridiculous to say that sexism has rooted out females math students in the last 10 years.

Even with sexism, (hypothetical) it comes to the point at which a female says, "Will I be servile and docile to the social pressure?" (the weak stance) or "Will I strive forward?" (the strong stance)

Life is a fight, don't back out.

Also, there is no sexism in the AMC, AIME, USAMO, and IMO! How do you explain the lack of girls there? It is equally nerdy for boys and girls, so then why are the boys strong enough to shrug this concept of nerdiness off and actually try? The girls should also be able to do the same. It all comes down to what you think about yourself and what you want to accomplish in life. It all comes down to determination.

The struggle of life...


Let p be a prime number , a polynomial $ Q(x)\in Z[x] $ satisfying :
 $ Q(k) \equiv 0 $ (mod p) or $ Q(k) \equiv 1 $ (mod p)  for $ k \in N $
 $ Q(0) = 0 $ and $ Q(1) = 1 $
 Prove that :   deg $ Q  \geq p-1 $

If I'm not wrong then It's a Problem In IMO Shortlist.

Just use the Lagrange interpolation polynomial and show that it's leading coefficient cannot be $0$ (work in the field $\mathbb F_p$).

In order to show that the leading coefficient cannot be $0$, we use the fact that this leading coefficient is $\sum c_i\cdot u_i^{-1}$, where $c_i=f(i)$, and ${u_i=(i-0)\cdot\ldots\cdot(i-(i-1)})(i-(i+1))\cdot\ldots\cdot(i-(p-1))$. That's equal (in $\mathbb F_p$, remember) to $-\sum c_i$. It's easy to see now why this cannot equal $0$ if there is a $c_i=0$ and there is a $c_j=1$.


[quote="TALENTO"]If the sum of the individual 3-digit numbers $ \overline{abc}$ and $ \overline{bca}$ is $ 10$ times the product of the numbers $ \overline {abc}$, is the sum of the figures of $ \overline {abc}$.

Greetings  :D[/quote]
Your problem statements are generally not clear at all ...
You nearly never thank when we help you ...
I wonder why I continue to give you solutions to your problems ... :(

I suppose the problem is Let $ a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}$ such that $ \overline{abc}\plus{}\overline{bca}\equal{}10abc$ (and not $ \overline{abc}$), find $ a\plus{}b\plus{}c$.

If so :

$ 100a\plus{}10b\plus{}c\plus{}100b\plus{}10c\plus{}a\equal{}10abc$ $ \iff$ $ 101a\plus{}110b\plus{}11c\equal{}10abc$

So $ a\plus{}c\equal{}0\pmod {10}$
Hence a first solution $ a\equal{}c\equal{}0$ and so $ b\equal{}0$

If $ a\neq 0$, we get $ c\equal{}10\minus{}a$ and the equation becomes $ 9a\plus{}11b\plus{}11\equal{}ab(10\minus{}a)$ or again $ b\equal{}\frac{9a\plus{}11}{\minus{}a^2\plus{}10a\minus{}11}$

In order to have $ \minus{}a^2\plus{}10a\minus{}11>0$, we need to have $ a\in\{2,3,4,5,6,7,8\}$, so just $ 7$ tests in order to find either $ a\equal{}5$, either $ a\equal{}6$. So three solutions :

$ a\equal{}0$, $ b\equal{}0$, $ c\equal{}0$ $ \implies$ $ 000\plus{}000\equal{}10\times 0\times 0\times 0$ and $ \boxed{a\plus{}b\plus{}c\equal{}0}$

$ a\equal{}5$, $ b\equal{}4$, $ c\equal{}5$ $ \implies$ $ 545\plus{}455\equal{}1000\equal{}10\times 5\times 4\times 5$ and $ \boxed{a\plus{}b\plus{}c\equal{}14}$

$ a\equal{}6$, $ b\equal{}5$, $ c\equal{}4$ $ \implies$ $ 654\plus{}546\equal{}1200\equal{}10\times 6\times 5\times 4$ and $ \boxed{a\plus{}b\plus{}c\equal{}15}$

pco I'm grateful, I am also very grateful to everyone and I apologize because my English is not good   :)


I just read the round 2 problems (which appear to be much harder than the round 1 problems) and I have a question about the wording of problem 2. Can I ask about it here on the forum?

We don't want any public discussion of the problems, so don't post your question here.  You can send me a question, but be forewarned that you are likely to get the response 'We feel the question is clear as it stands.'

For other participants: If Antonio raises a question on which we do give a substantive answer, we will post the question and answer here.

In general, if you are uncertain about your interpretation of a question, you should state your interpretation in your solution.


inequalities
geometry
circumcircle
geometry proposed

[color=purple] In triangle $ ABC$ which have three sides length $ a,b,c$ and circumcenter $ O$, orthocenter $ H$ prove that :
  $ 2OH\ge \max \lbrace|a \minus{} b|,|b \minus{} c|,|c \minus{} a| \rbrace$
  Nice problem I think  :D [/color]

The following sharper inequality is also true :  $2OH\, \ge\, \boxed{\ 2OI\, \ge\, \max\, \{\, |a-b|,|b-c|,|c-a|\, \}\ }$ .

Taking into account the well-known distances in a triangle : $\{\begin{array}{lllll}
OH^2=9R^2-(a^2+b^2+c^2) \\\\
OI^2=R^2-2Rr\end{array}$ , the inequality $OH\, \ge\, OI$

is equivalent to: $a^2+b^2+c^2\, \le\, 8R^2+2Rr$ , which is weaker than the well-known [u][b]Gerretsen[/b][/u] i.e. $\sum\, a^2\, \le\, 8R^2+4r^2$ .

Now let's draw our attention to $\triangle\, AON$ , where $N$ represents [u][b]Nagel[/b]'s point[/u]. Since : $\|\begin{array}{ccccc}
AO+ON\, \ge\, AN \\\\
AN=\sqrt {(b-c)^2+4r^2} \\\\
ON=R-2r\ ;\ AO=R\end{array}\|$

we obtain that : $2(R-r)\, \ge\, \sqrt {(b-c)^2+4r^2}\iff 4R^2-8Rr\, \ge\, (b-c)^2\iff\boxed{\ 2OI\, \ge\, |b-c|\ }$ .

Similarly we can obtain the analogous inequalities and thus the boxed inequality is proved.


[quote="babyloverain"]Given $ a,b,c \ge 0$Prove that
$ \frac {(a \plus{} b)^2}{(a \minus{} b)^2} \plus{} \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \plus{} \frac {(c \plus{} a)^2}{(c \minus{} a)^2} \ge$$ \frac {3}{2} \plus{} \frac {b}{2(a \plus{} b)} \plus{} \frac {c}{2(b \plus{} c)} \plus{} \frac {a}{2(c \plus{} a)}$
With this problem,we can solve by AM-GM.[/quote]

isn't that obvious? :wink:
we have:
$ \frac {(a \plus{} b)^2}{(a \minus{} b)^2} \plus{} \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \plus{} \frac {(c \plus{} a)^2}{(c \minus{} a)^2} \ge 3 \ge\frac {3}{2} \plus{} \frac {b}{2(a \plus{} b)} \plus{} \frac {c}{2(b \plus{} c)} \plus{} \frac {a}{2(c \plus{} a)}$

[quote="babyloverain"]Given $ a,b,c \ge 0$Prove that
$ \frac {(a \plus{} b)^2}{(a \minus{} b)^2} \plus{} \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \plus{} \frac {(c \plus{} a)^2}{(c \minus{} a)^2} \ge$$ \frac {3}{2} \plus{} \frac {b}{2(a \plus{} b)} \plus{} \frac {c}{2(b \plus{} c)} \plus{} \frac {a}{2(c \plus{} a)}$
With this problem,we can solve by AM-GM.[/quote]
nice ineq,my friend :) so i  understood [b]sayat[/b]'s solution   :D

[quote="babyloverain"]Given $ a,b,c \ge 0$Prove that
$ \frac {(a \plus{} b)^2}{(a \minus{} b)^2} \plus{} \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \plus{} \frac {(c \plus{} a)^2}{(c \minus{} a)^2} \ge$$ \frac {3}{2} \plus{} \frac {b}{2(a \plus{} b)} \plus{} \frac {c}{2(b \plus{} c)} \plus{} \frac {a}{2(c \plus{} a)}$
With this problem,we can solve by AM-GM.[/quote]
I don't think this ineq is hard and strong. Similar the way above of SaYaT, we can prove the stronger:  :maybe:
$ \frac{(a\minus{}b)^2}{(a\plus{}b)^2} \geq\ 1 \plus{} \sum\ \frac{a}{a\plus{}b}$


inequalities
trigonometry
geometry
trig identities
Law of Sines
rearrangement inequality

Prove that in a triangle with sides a,b,c and opposite angles A,B,C (in radians):

(aA+bB+cC)/(a+b+c)  :ge:  :pi:/3

fuzzylogic wrote:Hint:
[hide]Use Chebychev's Inequality[/hide]

[hide]I was hoping you wouldn't say that.

Chebyshev makes it almost trivial [/hide]

I know neither ---- nor his inequality, but I say that [hide]the law of sines helps... sort of.  I still need a good way to explain, but it's somewhat trivial if you see what I mean.[/hide]

[quote="probability1.01"]I know neither ---- nor his inequality, ...[/quote]
See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7942]here[/url].

nr1337 wrote:[hide]Chebyshev[/hide] makes this a very fun problem!!!!

[hide]
By Chebyshev,

3(aA + bB + cC)  :ge: (A+B+C)(a+b+c)
aA+bB+cC :ge:  (a+b+c):pi: /3

(aA+bB+cC)/(a+b+c)  :ge: :pi:/3
[/hide]

Can you prove Chebyshev given the rearrangement inequality?


Every compact subspace of a Hausdorff space is closed.

Ashwath

Use the fact that if $ X$ is Hausdorff and $ A,B\subset X$ are compact and disjoint then $ A$ and $ B$ has disjoint neighborhoods.


What is the significance of divisions 1 and 2? I know what they are and how they are defined, etc., but for what purpose have they been added, I'm wondering?

Division 1 seems to be schools which draw from a larger student population, such as private/regional/magnet schools, as opposed to standard public schools which draw from much smaller populations.

I don't really see the point to split students into divisions like this because the tests are the same anyway.

[quote="nittanylion"]It also takes away the possibilty of a small high school upsetting like Exeter, e.g.[/quote]That... just won't happen. The point is to allow normal public schools to compete amongst themselves while still having a higher level of competition for the schools that are perennial powerhouses.


Let $S$ be a set of 2006 numbers. We call a subset $T$ of $S$ [i]naughty[/i] if for any two arbitrary numbers $u$, $v$ (not neccesary distinct) in $T$, $u+v$ is [i]not[/i] in $T$. Prove that

1) If $S=\{1,2,\ldots,2006\}$ every naughty subset of $S$ has at most 1003 elements;

2) If $S$ is a set of 2006 arbitrary positive integers, there exists a naughty subset of $S$ which has 669 elements.

If all members T is odd, then T is "nauhty", and $|T|\le 1003$. If "nauhty" T have even x (x maximal even), then $|T| \le (2006-x)/2+x/2-1=1002,$ because greater members are odd (<=(2006-x)/2), $(x/2) \not \in T, \ if \ y<x: y \in T, \ then \ (x-y) \not \in T$.

...but I'm not very sure that you are right! If you take $T=\{1004,1005,1006,...,2006\}$, then the set has an even member, it [b]IS[/b] naughty and $|T|=1003$.

Let x be the largest element in T,  y and x-y cannot be both in T,at least half of the int below x is not in T, so |T| <= floor((x-1)/2)+1 <= 1003

You can find a pdf file named coach.rar here : http://www.mathlinks.ro/Forum/viewtopic.php?highlight=tst+usa&t=47106 .
This is given by qxy. In the section Number Theory see problem 22 which almost the same - it's USA TST 2001 :wink:

It is clear that $1\not\in{T}$. Suppose that, $u,v\in{T}$ ,Then $mu+nv\not\in{T}$, with $n,m\in{\mathbb{N}}$
Suppose that $|T|>1003$. We choose $u,v$ such that $u+v>2006$,$(*)$(which is the worse case), for each $u,v$. Note that , any positive integer can be written as sum of other positive integers, so the$(*)$ choosing $u,v$ is the only case to get $|T|>1003$. But $T'=\{1004,1005,1006,...,2006\}$ , $|T'|=1003$,which is a contradiction.


Hi,  I am trying to fight my way through the convergency concepts so any help about this problem is greatly appreciated.

Suppose that $ \{X_{n}\}$ is an uncorrelated sequence, meaning $ Cov(X_{i}, X_{j})\equal{}0, i\neq j$.  If there exists a constant c > 0 such that $ Var(X_{n}) \leq c, \forall n\geq 1,$ then for  any $ \alpha > 1/2$ we have $ \frac{\sum_{i\equal{}1}^{n}X_{i}}{n^\alpha} \rightarrow 0$ in $ L_{2}.$

Thank you very much.

What do you know about the means of these random variables? Is $ E(X_i)\equal{}0?$ If you don't have that, than I can cause a lot of trouble. In fact, what if $ X_i\equal{}1$ with probability $ 1?$ Then $ \frac{\sum_{i\equal{}1}^nX_i}{n^{\alpha}}\equal{}n^{1\minus{}\alpha},$ and for $ \alpha<1,$ that certainly doesn't go to zero in $ L^2.$

So I'm going to try to prove your problem under the additional hypothesis that $ E(X_i)\equal{}0$ for all $ i.$ It now follows that $ \text{Cov}(X_i,X_j)\equal{}E(X_iX_j).$

Let $ U_n\equal{}\sum_{i\equal{}1}^nX_i.$ Then

$ E(U_n^2)\equal{}\sum_{i\equal{}1}^n\sum_{j\equal{}1}^nX_iX_j\equal{}
\sum_{i\equal{}1}^n\sum_{j\equal{}1}^n\text{Cov}(X_i,X_j)$

$ \equal{}\sum_{i\equal{}1}^n\text{Var}(X_i)$ by the hypothesis of uncorrelatedness.

$ \le\sum_{i\equal{}1}^nc\equal{}cn.$

Then $ E((n^{\minus{}\alpha}U_n)^2)\le cn^{1\minus{}2\alpha}.$

As long as $ \alpha>\frac12,$ this tends to zero.

So, Lionheart, what is your hypothesis about $ E(X_i)?$

Good afternoon Kent.

First of all thank you for helping me on this topic and the one that I posted a while ago.  I guess in both of those problems you asked me for additional hypothesis, but I am indeed copying these problems word by word from the book called "A Probability Path" by Sidney Resnick and there is NO word about expectations of random variables in this problem.

So it left me reading the similar techniques, but in many of them the assumptions are either iid, or means of zero.  Thus, my post here.

Regards.

[quote]there is NO word about expectations of random variables in this problem. [/quote]
In that case, the statement is very badly false.


probability
function
inequalities
calculus
probability and stats

Let $ \lbrace S_n\rbrace$ be the symmetric random walk starting at the origin and taking diagonal steps on the $ d$-dimensional integer lattice $ \mathbb{Z}^d$, i.e.
\[ S_n = \sum_{k = 1}^n X_k\]
where
\[ X_k\in \Omega_X = \lbrace(x_1,x_2,...,x_d): x_j\in \lbrace - 1,1\rbrace,1\leq j\leq d\rbrace\]
and
\[ P(X_k = x) = 2^{ - d} \text{ for each } x\in \Omega_X\]
For $ d > 1$, notice the difference from the usual random walk.  Only "diagonal" steps are allowed, and the name comes from the cases $ d = 2,3$.  Some problems:
(1) Show that the number of "returns" to the origin is almost surely infinite for $ d = 1,2$, and finite for $ d\geq 3$.  i.e., show that $ \sum_{k = 1}^\infty \chi_{\lbrace S_k = \mathbf{0} \rbrace}<\infty$ a.s. if and only if $ d\geq 3$.
(2) Let $ \rho_d$ be the probability that the random walk returns to the origin at some finite time, i.e. $ \rho_d = P(\exists \tau \in \mathbb{N} \text{ s.t. } S_{\tau} = \mathbf{0})$.  Show that, for $ d\geq 3$,
\[ \frac {\rho_d}{(1 - \rho_d)^2} < \pi ^{ - \frac {d}{2}}\zeta(\frac {d}{2})\]
where $ \zeta$ is the Riemann Zeta.

1) It is well known that, for a Markov chain with at most countable state space, the following  properties are equivalent:
(a) a state $ i$ is recurrent, i.e. $ P_i(\exists n>0: X_n\equal{}i)\equal{}1$;
(b) conditioned that $ X_0\equal{}i$,  the number of consecutive visits to $ i$ is infinite with probability 1;
(c) $ \sum_{n\equal{}1}P_i(X_n\equal{}i)\equal{}\plus{}\infty$.
In your settings, $ X_n\equal{}S_n\plus{}X_0$ and $ i\equal{}0$. It is clear that $ P_i(X_n\equal{}i)\equal{}P(S_n\equal{}0)\equal{}0$ when $ n$ is odd. For the diagonal random walk $ P(S_{2n}\equal{}0)$ can be calculated explicitly because, in fact, $ X_n$ contains $ d$ independent one-dimensional Bernoulli walks 'along the main diagonals'. Hence $ P(S_{2n}\equal{}0)\equal{}
(u_n)^d$, where $ u_n\equal{}C_{2n}^n2^{\minus{}2n}$. It is known that $ u_n\sim {1\over \sqrt{\pi n}}$ hence your statement comes from (c) and that
$ \sum_nn^{\minus{}d/2}<\plus{}\infty\Leftrightarrow d\geq 3$.

2) Are you sure about this power 2 in the denominator? For now, I can prove that \[ \frac{\rho_{d}}{(1\minus{}\rho_{d})}<\pi^{\minus{}\frac{d}{2}}\zeta\left(\frac{d}{2}\right)\]
The argument is based on the relation $ (1\minus{}F(z))U(z)\equal{}1$ between the generating functions for the sequences $ f_n: \equal{}P($the first return to $ 0$ happens at the time moment $ n$), $ n\geq 0$ ($ f_0: \equal{}0$) and $ p_n: \equal{}P(S_n\equal{}0), n\geq 0$ ($ p_0: \equal{}1$). One has $ \rho_d\equal{}F(1)$ and hence $ (1\minus{}\rho_d)^{\minus{}1}\equal{}1\plus{}\sum_{n}(u_n)^d$.
Denote $ v_n\equal{}{1\over \sqrt{\pi n}}$, then $ u_n<v_n$. ([b]Remark.[/b] Thank you for the problem! I have not realized this cute relation before  :lol: The proof: it is a straightforward computation to verify that $ r_n: \equal{}{u_n\over u_{n\plus{}1}}< q_n: \equal{}
{v_n\over v_{n\plus{}1}}$ Then $ u_n\equal{}r_nr_{n\plus{}1}\dots r_{N\minus{}1}u_N<
q_nq_{n\plus{}1}\dots q_{N\minus{}1}u_N$, and since $ u_N\sim v_N, N\to \infty$ we get $ u_n<v_n$). Finally,
$ \rho_d(1\minus{}\rho_d)^{\minus{}1}\equal{}(1\minus{}\rho_d)^{\minus{}1}\minus{}1\equal{}\sum_{n}(u_n)^d<\pi^{\minus{}d/2}\zeta(d/2)$.

Clearly, this is not a precise calculation: we have lost something replacing $ u_n$ by $ v_n$. Maybe if one makes this estimation more precise your initial formulation can be covered as well? I'll try to take look later on, when I got some more free time  :wink:

You're absolutely right, the inequality should state $ \frac {\rho_{d}}{(1 \minus{} \rho_{d})} < \pi^{ \minus{} \frac {d}{2}}\zeta\left(\frac {d}{2}\right)$, that was a miscalculation on my part.  Although I think a big part of the problem is really in proving the statements that you assert are well known (in particular that $ u_n \sim \frac {1}{\sqrt {\pi n}}$, admittedly a quick exercises, and that (b) is equivalent to (c), a nice little problem on its own), which I suppose can be left for someone less familiar with this type of problem.

Well, of course for all these statements references for various textbooks could be given, but this is not interesting  :lol: There are some tasty little problems inside your large one. For instance: the routine way to prove $ u_n\sim 1/\sqrt{\pi n}$ is to  use the Stirling formula (like it is done in the 'probabililists Bible' by W.Feller  :) ). A good exercise is to prove  this relation without such an extra tool. Note that this is a  Calculus exercise (related to the properties of  the  Gamma function).


Ya que no hay soluciones posteare la mia:
denotemos O1, O2 Y O3  a los centros de la circunferencia que tocan a los lados AB y AC, AB y BC, BC y AC, del triangulo ABC.
entonces como son del mismo radio y son tangentes a una misma recta:
O1O2 $paralela$ AB, O1O3 $paralela$ AC y O2O3 $paralela$ BC por lo que como AO1, BO2 y CO3 se intersectan en I y como O1O2O3 tienen sus lados respectivamente $paralelos$ a ABC entoces estos dos triangulos son homoteticos con un centro de homotecia en I, por lo que las circuncircunferencias de OIO2O3 Y ABC son homoteticas con un centro de homotecia en I ahora como las circunferencias son del mismo radio entonces OO1,OO2,OO3 tiene la misma longitud y porlo tanto O es el circuncentro de O1O2O3, por lo tanto el circuncentro de ABC, el incentro I de ABC y O son colineales.
jeje :lol:

Oe y eso se supone k es un resultado conocido en la homotecia???
Podrias poner la demostracion xfavor.!!!

la demostracion de eso, pues es simplemente ver la construccion para buscar los centros de homotecia de dos circunferencias.

a qui posteo otro problema de homotecia:

sea  ABC un triangulo denotamos M como el punto de tangencia del incirculo de ABC con el lado AC, denotemos por K al punto diametralmente opuesto de M con respeto al incirculo de ABC y N como el reflejado del punto M con respecto al punto medio de AC.
demostrar que  B, K y N son colineales.
posteen soluciones. jeje :lol:

[quote="conejita"]Oye, entendi toa tu solucion, excepto el final, que que argumento usas, para decir que como O es el circuncentro de O1O2O3 entonces el circuncentro de ABC, el incentro y el punto O son alineados???  Siento que no stan muy relacionados...[/quote]

Hay un error de concepto, dos circunferencias siempre son homoteticas. Sus centros de omotecia es el [i]centro de similitud de las circunferencias[/i] (son dos centros de omotecia ya que dos regiones si son homoteticas lo son de dos formas. Es facil demostrar que los centros de similitud de dos circunferencias dividen el segmento de sus centros armonicamente.

[quote="Tony2006"]eh estudiado un poco de homotecia y me parece muy interesante, me gustaria qe algunos participantes de este foro compartan algunos problemas de este tipo y que despues de un rato pusieran las soluciones, yo inicio poniendo esta IMO de 1981

Tres circulos congruentes tienen un punto en común O y estan dentro de un triangulo. cada circulo toca un par de lados del triangulo.
provar que el incentro y el circuncentro de el triangulo son colineales con el punto O.[/quote]

Aqui hay un problema, yo lo resolvi usando omotecia

Dado $\triangle ABC$ circunscrito a $\mathcal(C)(I,r)$, sea $M_{a}$ el punto medio de $\overline{BC}$y $H_{a}$ el pie de la $A-$altura. Sea $T=M_{a}I\cap AH_{a}$, demostrar que $AT=r$.

denotemos por K al punto de tangencia del incirculo de ABC con BC y W al punto diametralmente opuesto de K copn respecto al incirculo, denotemos por R a la interseccion de AW con BC por el problema anterior tenemos que Ma es el punto medio de KR. entonces IMa paralela a WR y como WK paralela a AHa entonces AWIT es paralelogramo y por lo tanto AT=WI=r

[quote="Tony2006"]A MIRA SI DOS CIRCUNFERENCIAS SON HOMOTETICAS DESDE UN PUNTO ENTONCES LOS CENTROS DE ESAS CIRCUNFERENCIAS SON COLINEALES CON O.[/quote]

Ademas dividen el segmento que une los centros armonicamente

ver [b]Circunferencias de Similitud [/b]


number theory
inequalities unsolved
inequalities

If $\frac{a}{b}<\frac{x}{y}<\frac{c}{d}$ and $cb-ad=1$, prove that $y\geq b+d$.

easy.
I suppose all numbers must be naturals, otherwise it doesn't make sense.
From $\frac{a}{b}<\frac{x}{y}$ we get $ay<bx$ thus $ay+1\le bx$.
From $\frac{x}{y}<\frac{c}{d}$ we get $dx<cy$ thus $dx+1\le cy$.
So $bcy\ge {bdx+b}\ge {ady+d+b}$, thus $y\ge\frac{b+d}{bc-ad}=b+d$. No number theory needed :)
(By the way, note that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$ if $bc-ad>0$.)


Gauss
number theory
least common multiple
floor function

[hide]Integers that are multiples of both 3 and 5 are divisible by 15. Those numbers between 1 and 200, inclusive, are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, and 195. Of those, 7 of them aren't divisible by 4 or 7.[/hide]

[quote="JRav"]Use the Principle of Inclusion-Exclusion.[/quote]
What's that??? :maybe:

[quote="JRav"]The only problem with a brute force method is that it takes too long under test conditions.[/quote]

Gauss isn't brute forcing anything.

PIE is the best method but I think during a contest, I would be using Gauss1181's methodbecause i'm lazy

[hide]
just make a bit simpler, 3 and 5 have an LCM of 15, so there is 13 of them, since the highest integer below 200/15 is 13. So, 4 intersects with 15 each 60, so there is 200/60 so 3. 7 intersects each 105, so it happens once,

so 13-3-1=9[/hide]

The answer isn't 7 gauss.

Sorry never mind...Using brute force can lead to a careless error... :oops:

[quote]Gauss isn't brute forcing anything.[/quote]

He even admitted that he brute-forced it.

[quote]What's that??? [/quote]

Principle of Inclusion-Exclusion is basically a nifty little counting trick.  We basically overcount and then subtract out everything that we counted twice.  Problems that ask things like "How many numbers between 1 and 1000 have 3 and 4 but not 5 as a factor?" are often primed for PIE.

[hide]First, know this simple formula:  The number of multiples of $ d$ between $ 1$ and $ n$ is $ \lfloor \frac {n}{d} \rfloor,$ where the brackets mean the largest integer less than or equal to the number in brackets.

Second, notice that multiples of "3 and 5" are the same as multiples of 15.

In our case, the number of multiples of $ 15$ between $ 1$ and $ 200$ is $ \lfloor \frac {200}{15} \rfloor \equal{} 13,$

the number of multiples of $ 15\times 4 \equal{} 60$ between $ 1$ and $ 200$ is $ \lfloor \frac {200}{60} \rfloor \equal{} 3,$

the number of multiples of $ 15\times 7 \equal{} 105$ between $ 1$ and $ 200$ is $ \lfloor \frac {200}{105} \rfloor \equal{} 1,$ and

the number of multiples of $ 15\times 4\times 7 \equal{} 420$ between $ 1$ and $ 200$ is $ \lfloor \frac {200}{420} \rfloor \equal{} 0.$

With just two or three multiples to worry about, I would draw a Venn diagram, using this information to fill in the sections.  It's only a little more work to completely fill in the Venn diagram.  (Compute the number of multiples of 4, 7, and 28.)

In the following diagram, the full 360 degree yellow circle represents all multiples of 15.  The full cyan circle represents all multiples of 4.  And the full magenta circle represents all multiples of 7.  The red region, where yellow and magenta intersect, represents multiples of 105.  The white region, where all three circles intersect, represents multiples of 420.  [What do the green and blue regions represent?]  A black and white pencil Venn diagram works just as well, but it's harder to describe the regions.

Finally, decide which section(s) includes the multiples you want and excludes the multiples you don't want.  What's the final answer?

BTW, a Venn diagram, which is easy to draw for up to three sets, is equivalent to PIE.   Once you can visualize why the PIE rules work for up to three sets, it is often easier to make the leap to understand why PIE rules work for higher numbers of sets.  More than three sets are rarely seen in MATHCOUNTS.[/hide]

Bonus problem:  How many positive integers less than or equal to 200 are multiples of 4 or 7, but not 15?  [The answer should be available from the following Venn diagram.]


Let $ A,B,C,D$ be four points on a circle in this order. Prove that there are four points $ M_{1},M_{2},M_{3},M_{4}$ on the circle which form a quadrilateral with perpendicular diagonals such that for each $ i\in\{1,2,3,4\}$, $ \frac{AM_{i}}{BM_{i}}\equal{}\frac{DM_{i}}{CM_{i}}$


vector
linear algebra
matrix
conics
geometry
search
algebra

We have the formula for the distance between two straight lines in ${\mathbb R}^{2}$ and ${\mathbb R}^{3}$ and formula for the distance between one straight line and a plane in ${\mathbb R}^{3}$. How about the general formula for the distance between a plane of dimension $p<n$ and another plane of dimension $m<n$ in ${\mathbb R}^{n}$?

actually i dont remember where i saw it, but a nice shot.

using vectors we can generalize the formula for distance from a point to a line to a formula to a n-plane to a m-plane

we can try it. but imagine high dimension is not for me :)

but, use vectors is the actual solution

EDIT:

http://ltcconline.net/greenl/courses/107/Vectors/LINPLNE.HTM

Let n be a normal vector to the hiper-plane.

but what is the definition of normal for more than 3 dimension?! we will need more than 2 vectors to definie ne n-plane.

what i said, i cant think in more than 5 dimensions. hehe :)

Let $U,V$ be affine subspaces of $\mathbb{R}^{n}$ of dimension $m,p$ and $\delta$ be the distance between $U$ and $V$. Assume that $m+p<n$ (otherwise $\delta=0$) and that $U$ and $V$ are in general position. $U=\{x|Mx=\alpha\}$ where $rank(M)=n-m$; $\{a_{1},\cdots,a_{m}\}$ is a $ker(M)$ basis. $V=\{y|Py=\beta\}$ where $rank(P)=n-p$; $\{b_{1},\cdots,b_{p}\}$ is a $ker(P)$ basis.
a) $\delta=\parallel b-a\parallel $ where $(a,b)\in{U.V}$ and $b-a$ is orthogonal to $U$ and $V$.
Proof of the second part (classical by Lagrange method): we seek $inf\parallel x-y\parallel ^{2}$ when $(x,y)\in{U.V}$; for the variable $x$ there exists a row $\lambda$ s.t. $\forall{h}\in\mathbb{R}^{n}$ $2(a-b|h)+\lambda{Mh}=0$ that is equivalent to $Mh=0$ implies ${(a-b|h)=0}$ or $a-b$ is orthogonal to $U$. With the variable $y$, $a-b$ is orthogonal to $V$.
b) Let  \[N=\begin{pmatrix}a_{1}^{T}\\\cdots\\a_{m}^{T}\end{pmatrix},Q=\begin{pmatrix}b_{1}^{T}\\\cdots\\b_{p}^{T}\end{pmatrix}\]
Thus $N(b-a)=0$ and $Q(b-a)=0$. We deduce the matricial system:
\[\begin{pmatrix}M&0\\0&P\\N&-N\\Q&-Q\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}\alpha\\\beta\\0\\0\end{pmatrix}\]
If $U,V$ are in general position, then this $(2n,2n)$ system admits an unique solution $(a,b)$.

I have deduced a general formula via Lagrange Multiplier but rather complicated.

I found it not possible to deduce a general formula for the distance between two conics in higher dimension by the same method.

I wonder if we have a formula for the minimum distance between a straight line and a conic in High school geometry?  :(

i) if you search common normals to 2 conics in the plane (which don't intersect)  you must solve a polynomial of degree 12 with 8 real roots (for example).
ii) If you search common normals to a straight line and a conic in the plane (which don't intersect) then you must solve a polynomial of degree 2 with 2 real roots (for example). Thus, in this case, you can write a formula for the distance as the minimum of 2 values.


Define \[\begin{cases}d(n, 0)=d(n, n)=1&(n \ge 0),\\ md(n, m)=md(n-1, m)+(2n-m)d(n-1,m-1)&(0<m<n).\end{cases}\] Prove that $d(n, m)$ are integers for all $m, n \in \mathbb{N}$.

[quote="Peter"]Define \[ \begin{cases}d(n, 0)=d(n, n)=1&(n \ge 0),\\ md(n, m)=md(n-1, m)+(2n-m)d(n-1,m-1)&(0<m<n).\end{cases}\] Prove that $ d(n, m)$ are integers for all $ m, n \in \mathbb{N}$.[/quote]

Very easy: First we are going to prove the identity

[color=green][b](1)[/b][/color] $ m\binom{n}{m}^{2}=m\binom{n-1}{m}^{2}+\left(2n-m\right)\binom{n-1}{m-1}^{2}$

for any integer $ n$ and any positive integer $ m$.

In fact, for $ n=0$, proving [color=green][b](1)[/b][/color] is easy (exercise to the reader), while for $ n\neq 0$, we have

$ \binom{n-1}{m}=\frac{\left(n-1\right)\left(n-2\right)...\left(n-m+1\right)\left(n-m\right)}{m!}=\frac{n-m}{n}\cdot\frac{n\left(n-1\right)\left(n-2\right)...\left(n-m+1\right)}{m!}$
$ =\frac{n-m}{n}\cdot\binom{n}{m}$

and

$ \binom{n-1}{m-1}=\frac{\left(n-1\right)\left(n-2\right)...\left(n-m+1\right)}{\left(m-1\right)!}=\frac{m}{n}\cdot\frac{n\left(n-1\right)\left(n-2\right)...\left(n-m+1\right)}{m\cdot\left(m-1\right)!}$
$ =\frac{m}{n}\cdot\frac{n\left(n-1\right)\left(n-2\right)...\left(n-m+1\right)}{m!}=\frac{m}{n}\cdot\binom{n}{m}$,

so that

$ m\binom{n-1}{m}^{2}+\left(2n-m\right)\binom{n-1}{m-1}^{2}=m\left(\frac{n-m}{n}\cdot\binom{n}{m}\right)^{2}+\left(2n-m\right)\left(\frac{m}{n}\cdot\binom{n}{m}\right)^{2}$
$ =\left(m\left(\frac{n-m}{n}\right)^{2}+\left(2n-m\right)\cdot\left(\frac{m}{n}\right)^{2}\right)\cdot\binom{n}{m}^{2}=m\cdot\binom{n}{m}^{2}$,

and [color=green][b](1)[/b][/color] is proved.

The rest of the solution is trivial, but formally explaining it is some work. We claim that

[color=green][b](2)[/b][/color] $ d\left(n,m\right)=\binom{n}{m}^{2}$

for any nonnegative integers $ n$ and $ m$ satisfying $ m\leq n$. Of course, [color=green][b](2)[/b][/color] immediately will yield that $ d\left(n,m\right)$ is integer for any nonnegative integers $ n$ and $ m$ satisfying $ m\leq n$. Thus, in order to solve the problem, it is enough to prove [color=green][b](2)[/b][/color].

We will prove [color=green][b](2)[/b][/color] by induction over $ n$:

The induction base, $ n=0$, is trivial (in fact, if we have $ n=0$, then $ m\leq n$ becomes $ m\leq 0$, so that $ m=0$ (because $ m$ must be nonnegative), so that $ d\left(n,m\right)=d\left(0,0\right)=1=1^{2}=\binom{0}{0}^{2}=\binom{n}{m}^{2}$).

The induction step is straightforward, but long to write up:

Let $ u$ be a positive integer. Assume that [color=green][b](2)[/b][/color] is proven for all nonnegative integers $ n$ and $ m$ satisfying $ m\leq n$ and $ n<u$. We want to show that [color=green][b](2)[/b][/color] also holds for all nonnegative integers $ n$ and $ m$ satisfying $ m\leq n$ and $ n=u$.

In fact, let $ n$ and $ m$ be nonnegative integers satisfying $ m\leq n$ and $ n=u$. Then, $ n-1<u$. Since $ m$ is nonnegative and satisfies $ m\leq n$, one of the three cases $ m=0$, $ m=n$ and $ 0<m<n$ must hold. We distinguish between these three cases:

[i]First case.[/i] Assume that $ m=0$. Then, $ d\left(n,m\right)=d\left(n,0\right)=1=1^{2}=\binom{n}{0}^{2}=\binom{n}{m}^{2}$.

[i]Second case.[/i] Assume that $ m=n$. Then, $ d\left(n,m\right)=d\left(n,n\right)=1=1^{2}=\binom{n}{n}^{2}=\binom{n}{m}^{2}$.

[i]Third case.[/i] Assume that $ 0<m<n$. Then, $ n-1$ and $ m$ are nonnegative integers satisfying $ m\leq n-1$ and $ n-1<u$, so that, by the induction assumption, these integers satisfy [color=green][b](2)[/b][/color], i. e. we have $ d\left(n-1,m\right)=\binom{n-1}{m}^{2}$. Also, $ n-1$ and $ m-1$ are nonnegative integers satisfying $ m-1\leq n-1$ and $ n-1<u$, so that, by the induction assumption, these integers satisfy [color=green][b](2)[/b][/color], i. e. we have $ d\left(n-1,m-1\right)=\binom{n-1}{m-1}^{2}$. Finally, since $ 0<m<n$, we have
$ md\left(n,m\right)=md\left(n-1,m\right)+\left(2n-m\right)d\left(n-1,m-1\right)$,
so that
$ md\left(n,m\right)=m\binom{n-1}{m}^{2}+\left(2n-m\right)\binom{n-1}{m-1}^{2}$.
Comparing this with [color=green][b](1)[/b][/color], we get $ md\left(n,m\right)=m\binom{n}{m}^{2}$, so that $ d\left(n,m\right)=\binom{n}{m}^{2}$ (since $ m>0$).

Thus, in all three cases, we get $ d\left(n,m\right)=\binom{n}{m}^{2}$, so that [color=green][b](2)[/b][/color] holds for our two integers $ n$ and $ m$. Hence, the induction step is complete, and [color=green][b](2)[/b][/color] is proven. As said above, this solves the problem.

  Darij


Compute
S[size=75]1[/size] = F[size=75]0[/size]:^2:/1 + F[size=75]1[/size]:^2:/c + F[size=75]2[/size]:^2:/c:^2: + ...

and
S[size=75]2[/size] = F[size=75]0[/size]F[size=75]1[/size]/1 + F[size=75]1[/size]F[size=75]2[/size]/c + F[size=75]2[/size]F[size=75]3[/size]/c:^2: + ...

where F[size=75]0[/size] = F[size=75]1[/size] = 1 and c is an arbitrary constant such that
|c| > (3 + :rt5:)/2,  and F[size=75]n+1[/size] = F[size=75]n[/size] + F[size=75]n - 1[/size] for n :ge: 1.

all i have is that for S2, [hide] S2 -(S2/c)=1+S1...i think you could find the first sum with binets formula, but ill work that out in a bit.[/hide]

That's an excellent start.  You could use Binet, but I wouldn't recommend it, although it may turn out nice than my personal preferred method.  Think about manipulations you can do to the first sequence.

actually the real way to solve it is to add [hide] cS1 to S1 and with some slick factoring you get what you want. then i think you get (c+1)S1=c+2+S1-2S2-(5/c), and you also know that S2-(S2/c)=1+S1

lets solve! from eq 1 we get c^2S1=c^2+2c-5-2cS2, and from eq 2 we get (cS2-S2-c)/c=S1, putting back in eq 1 we get (c+1)(cS2-S2-c)=c^3+2c^2-5c-2(c^2)S2. oh how fun. expanding LHS:

(c^2)S2-cS2-c^2+cS2-S2-c-->(c^2)S2-S2-c^2-c, bring the S2's on LHS-

2(c^2)S2+(c^2)S2-S2=c^2-c+c^3+2c^2-5c-->c^3+3c^2-6c

S2(3c^2-1)=c^3+3c^2-6c

S2=(c^3+3c^2-6c)/(3c^2-1)?

[/hide]

thats just S2, and it did have cubes...

Well, you've gone wrong somewhere, but I honestly don't know where.  My guess would be that you made some sort of error in the very first step, but I'm not positive.  Anyhow, we can tell your answer can't be right, because if we make c very very large, our answer should approach to 1 (since all the terms after the first will shrink away), but your answer approaches infinity.

yes i was, and S2 had a c^3 term in it, and it was nasty. its all in my second post in this thread. yeah i kept making errors in the computation which really delayed it, so i'll have to do the problem over again.


$\text {What does a+b+c+d+e equals?}\\
abcde+abce+abcd+abde+acde+bcde+abc+abe+abd+ace+acd+bce+bcd+ade+bde+cde+ab+ac+bc+ae+ad+be+bd+ce+cd+de+a+b+c+d+e=101385660$

This is pretty nasty, even I made it. :lol:

[hide]If you write f(x)=(x-a)(x-b)(x-c)(x-d)(x-e) and expand it (you don't have to expand it one by one if you know this trick) it looks like:
$x^5-(a+b+c+d+e)x^4+(ab+bc+cd+de+...[all_combinations_of 2])x^3-(all_combinations_of_3)x^2+(all_combinations_of_4)x^1-abcde$

Next, consider f(-1) which = -1-(a+b+c+d+e)-(all comb. of 2)-(all comb. of 3)-(all comb. of 4)-(abcde).  The equation given is equivalent to saying:
$-f(-1)-1=101385660  \rightarrow  -f(-1)=101385661$
Now the search is to find a+b+c+d+e.  Factoring 101385661 into it's prime factor decomposition gives 13*37*41*53*97.  If the numbers a,b,c,d,e are only to be integers, then they must be those 5 numbers and so a+b+c+d+e=241[/hide]

PS - Nice problem.

[quote]PS - Nice problem.[/quote]

Thank you. The problem was basically in form of:

$(a+1)(b+1)(c+1)(d+1)(e+1) = 101385661$

 :lol:

What I saw reminded me of expanding a polynomial in (x-a)(x-b)... form, so I went from there.

You must have been really bored to factor that.


Να βρεθούν οι γωνίες του τριγώνου ΑΒC αν είναι γνωστό ότι ισχύουν οι σχέσεις:
$ (cosA\plus{}cosB)^3\equal{}2cosC$

$ (cosB\plus{}cosC)^3\equal{}2cosA$

$ (cosC\plus{}cosA)^3\equal{}2cosB$

έστω $ x\equal{}cosA, y\equal{}cosB, z\equal{}cosC$ τότε ($ x,y,z, \in [0,1]$)

$ (x\plus{}y)^{\frac{1}{3}}(y\plus{}z)^{\frac{1}{3}}(z\plus{}x)^{\frac{1}{3}} >\equal{} (xyz)^{\frac{1}{3}}\plus{}(xyz)^{\frac{1}{3}} \Leftrightarrow$

$ (2x2y2z)^{\frac{1}{3}}>\equal{}2(xyz)^{\frac{1}{3}} \Leftrightarrow 8xyz\equal{}8xyz$

άρα στην ανισότητα ισχύει η ισότητα, δηλαδή:

$ (x,y,z)\equal{}k(z,x,y)$ που τελικά δίνει $ x\equal{}y\equal{}z$ και άρα το τρίγωνο που ψάχνουμε είναι το ισόπλευρο

(πράγματι, αντικαθιστώντας $ x\equal{}y\equal{}z$ είναι $ 8x^2\equal{}2 \Leftrightarrow (x>\equal{}0) x\equal{}\frac{1}{2}\equal{}cos(\frac{\pi}{3})$)

P.S. Επειδή έχω να λύσω επαρκώς competitive ανισότητα πολύ καιρό, αν έχω εφαρμόσει λάθος την ανισότητα (η οποία παρεπιπτόντως δε θυμάμαι πώς λέγεται) πείτε μου.

Καταρχήν σε ευχαριστώ για την ανταπόκριση και την ωραία λύση.

Μια πιο στοιχειώδης προσέγγιση μπορεί να είναι η εξής:
Όπως και στη λύση του maniopas θέτουμε $ x \equal{} cosA,y \equal{} cosB,z \equal{} cosC.$ Χωρίς βλάβη υποθέτουμε ότι $ x\leq y\leq z.$
Τότε $ 2z \equal{} (x \plus{} y)^3\leq (y \plus{} z)^3 \equal{} 2x.$ Άρα $ z\leq x\leq y\leq z$ οπότε $ x \equal{} y \equal{}z.$


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

This was picked straight from the source. I know it's not really mathematics, still it should have something mathematical in it. Strangely, the first 38 questions are really easy but the last one is surprisingly hard.

I think it should be A, but I'm not sure. Should have something to do with moving rows/columns.

after 8 months still no one? I still haven't figured this out... :(

it is E.

observe that a "Table" has exactly 4 elements in common with the previous one ("common" = same symbol in the same row and column)

PS: Sorry if I am wrong.

no. in the lowest row, nr1 and nr2 only have 1 element in common...

but still a very smart remark :)

Going from one row to the next, the boxes are 90 degree rotations, but other than that I don't see any obvious patterns. Why do you think it's A?

I do not have a theory for a, just on my "intelligent" estimate I thought A was most "likely" to fill in. (due to the placement of the things... but it's not entirely closing :))

Anyway that's a nice pattern you discovered with the rotation :) that may help a lot!

It even brings me an idea: within a row, we rotate the board counterclockwise, and then do some switches :)
(looking at the triangles)

but still it is hard to find a pattern... we may have to switch rows/columns, do permutations, switch symbols, ...

square 1 column 1 -> square 2 column 2  -> square 3 column 3 (+some shape switching)
square 4 column 1 -> square 5 column 2  -> square 6 column 3
square 7 column 1 -> square 8 column 2  -> square 9 column 3

square 1 column 3 -> square 2 column 1  -> square 3 column 2 (+some shape switching)
square 4 column 3 -> square 5 column 1  -> square 6 column 2
square 7 column 3 -> square 8 column 1  -> square 9 column 2

[quote="perfect_radio"]huh??? what does that mean?[/quote]

Make the change x to o, o to triangle and triangle to x. Then, in each row, you can obtain the second column of the second square from the first column of the first square and the third column of the third square from the second column of the second square.

The second claim is less clear (you have the right symbols but in different order). If this approach is correct, the answer is B.

[quote="fedja"]Make the change x to o, o to triangle and triangle to x. Then, in each row, you can obtain the second column of the second square from the first column of the first square and the third column of the third square from the second column of the second square.

The second claim is less clear (you have the right symbols but in different order). If this approach is correct, the answer is B.[/quote]

This is only a partial solution --- it limits us to picking either A or B as the answer, but does not provide enough information to distinguish which one is correct.

Here is the full answer (as well as enough information to continue the pattern for an indefinite number of iterations):

[list]Replace each X with an O, each O with a triangle, and each triangle with an X, as fedja has suggested.
Slide all the symbols to the right by one column (column one becomes column two, column two becomes column three, and column three becomes column one).
Slide the symbols of the newly formed column one down one row (col. one row one becomes col. one row two, col. one row two becomes col. one row three, col. one row three becomes col. one row one).[/list]

In this way, it is possible to create the second grid from the first grid, and likewise, the third grid from the second grid (see attached image, below).

As it has been pointed out by pilot, to produce the first grid of a new row of grids, take grid number 3 from the previous row, and rotate it clockwise 90 degrees.  Thus, the pattern can be continued for any number of rows of grids.


According to a random bureau, the number of bank tellers in 2010 is expected to be 440, 000 this is a decline of 12% from 2000-2010. Assuming this continued, how many bank tellers will there be in 2030?

i think it is 334400 in 2030 :) [/quote]


Pretty cool:

The following equations have the same roots $a$ and $b$ for some constants $P$ and $Q$.  Find $\frac{1}{a} + \frac{1}{b}$.

$(\log_2{x})^2 + \log_2{x^2} + P = 0$
$4^{2x} + 4^{x+Q} + 64 = 0$

[hide="Solution"]The first equation is quadratic in $\log_2 x$, hence $\log_2(ab)=\log_2a+\log_2b=-2$ (Vieta's formulas). Therefore $ab=2^{-2}={1\over 4}$

The second equation is quadratic in $4^x$, hence $4^{a+b}=4^a4^b=64$ (Vieta's formulas). Therefore $a+b=\log_464=3$.

Now we have ${1\over a}+{1\over b}={{a+b}\over{ab}}=12.$[/hide]


I think you meant,

[quote="manlio"]For $a,b,c,d,e$ positive reals such that $abcde=1$ prove that

$\frac{(a+b+c)^2}{2a+3b+4c}+\frac{(b+c+d)^2}{2b+3c+4d}+\frac{(c+d+e)^2}{2c+3d+4e}+\frac{(d+e+a)^2}{2d+3e+4a}+\frac{(e+a+b)^2}{2e+3a+4b} \geq 5$[/quote]

which is trivial from Cauchy and AM-GM.

$\sum_{cyc} \frac{(a+b+c)^2}{2a+3b+4c} \geq \frac{9(a+b+c+d+e)^2}{9(a+b+c+d+e)} = a+b+c+d+e \geq 5\sqrt[5]{abcde} = 5$.

Manlio, these tivial problems are not from GMA!

Gazette... Mathematik... A..?

I guess A is just the class since you got A and B, B being the student's level...
(or it's beginner and advanced :D)


Se considera un cub de latura n care se imparte in $n^{3}$ cuburi de latura 1.
Cele $n^{3}$ laturi se coloreaza in roz si vernil(x roz, y vernil - logic x+y=n). Fie k numarul de fete care unesc un patrat roz de unul vernil. Sa se afle
 $min(k)$ in functie de x si y

[quote="luke999"]-logic x+y=n
 Sa se afle
 $min(x+y)$ in functie de k[/quote]
??? Nu intelg...poti sa explici mai clar te rog :)

Laturile sau cuburile se coloreaza? Si cum x+y=n??? nu e n^3? Adik nu coloram x cubulete alb si y cubulete negru si cum sunt n^3 cubulete atunci x+y=n^3?

Cuburile se coloreaza si sunt $x+y=n^{3}$...Scuze

enuntul asta e al naibii de neclar, cel putin eu nu inteleg nimic  :oops:

KK....O sa dau un caz particular de la care a pornit problema

Am un cub cu latura `10.Si cu 1000 de cuburi de latura unu in interior.Colorez cuburile in rosu si verde(500-rosii si 500-verde).Sa se arate ca exista cel putin 500 de fete de cuburi mici care sa uneasca doua cuburi de culori diferite.


If $x_1$,$x_2$ are roots of $x^2-2bx+b^2-c^2 = 0$, find the equation which have roots $x_1^3$ and $x_2^3$


[quote="anirudh"]Help. What is the line X=Y?[/quote]
[hide]
It's a diagonal line that goes through all points in which the x-coordinate is equal to the y-coordinate. I call it the "perfect diagonal" It goes through (1,1)(2,2)(3,3)(4,4)(5,5). Can you imagine the line now?[/hide]

Drunken_Math is right, it also goes through the origin.


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Solving instructions: this puzzle is made in the style of MIT mystery hunt. The puzzle consists of a title, some flavourtext (intended to help with solving the puzzle), and the puzzle itself. No instruction is given as to how to proceed.

[url=http://web.mit.edu/puzzle/www/]MIT mystery hunt[/url] is a yearly event held in January on MIT campus; teams compete to find a "coin" hidden somewhere on the campus by solving a series of challenging puzzles. The winning team wins the right to design the next mystery hunt.

The title of this particular puzzle isn't very helpful; however it is an in-joke for those who attended mathcamp '03.

Please don't hesitate to PM me to
- verify your answer
- ask for a hint

However I cannot give hints that give the puzzle away. Read the flavourtext carefully.

[b]Note: the answer is not a number.[/b]

---

[b]How to Beat Children <sub>(at their own games)</sub>[/b]

Some say life is just a game. I concur. By the time you figure out the rules, it's half over!

I may have seen something similarish. I've printed it out and I'll have a scribble soon. May take a while.

Heh, cool. The answer:(which gives nothing at ALL away, so don't worry if you accidentally read it) is [hide]9[/hide].

That's not quite it yet - you haven't used the word grid.

An AoPS MIT Mystery Hunt team would be cool. I bet there would be a lot of interested people. You could do something like what the Mathcamp did to allow people to participate online. I personally will be there on the Mathcamp team.

I don't think we have a sizable presence at Cambridge though... It's a great idea nonetheless.

People could participate online, though. You could easily create a temporary forum.

Bumps... do people have done some work on it (please show) or should I start giving out more explicit hints?

I'll attach a GMail invite to this puzzle. Whoever solves this first (either here or on my Livejournal) [and does not have a GMail account] gets an invite.

Would the flavourtext be "Some say life is just a game. I concur. By the time you figure out the rules, it's half over!" ? If so I can't seem to figure out how it helps, if at all...

You can't expect it to be OBVIOUS, can ya? :)

Heh. I'll get it soon.

[quote]do people have done some work on it (please show) or should I start giving out more explicit hints?[/quote]
I got the obvious answer posted by TripleM, can't seem to do much more yet.

Edit: I assume one can't use Mathematica on this - that would be almost too easy. How about a TI-84+ calculator?

You can use whatever tools are available to you, though I doubt it'd be of much help.

For those of you who have made progress on this puzzle, why don't you talk about what you have gotten so far?

Well, you can get the value "9" out of the numbers, pixel-puzzle style. Now as for the letters, I thought maybe either some sort of caesar cipher relating to the number 9 on either all the letters, or the shaded in or shaded out letters, or perhaps every 9th letter of the same two options, or .. but none of these have provided anything legible.

Hmm. This is like a 100% totally random guess, but would the wine (or grape) 'Tokay' have anything to do with this?

Congratulations! PM me with the email address that I should send the invite to. :D

TripleM wrote:WAHOO!
[hide]
The infinite bit put me off. I was playing with actual edges meaning edges. You created the puzzle yourself! Thats brilliant
Oh, the answer: John Conway. Funny that.
[/hide]

Yay, a gmail account

How did you get that?

I will PM you in case other people want to keep trying.

TripleM, can you PM me too?

I'm interested in knowing the solution.

woah this has been inactive for 13 years

[quote=KiwiYum]-______- y u bump this.

There' threads that have been inactive longer.

And I think users have been inactive longer.[/quote]

NOT FF THREADS

"RT-2PM2 Topol-M Silo-based intercontinental ballistic missile hits streak"


function
induction
inequalities
algebra
functional equation
algebra proposed

Here another functional equation
Determine all functions $f$ satisfying $f(f(n))+f(n)=2n+6$ for all $n \in N_0$

nice one, i guess the method i recently used for some similar thing will work here as well, let's see:

let $g(x)=|f(x)-x-2|$ for simplicity. we want to show that $g(x)=0$.

let $M_n=\{g(x): x\leq n\}$. let $x\leq n$ be s.t. $g(x)$ is maximal in $M_n$.

the functional equation implies $g(f(x))=2g(x)$. therefore it suffices to show that $g$ is bounded since by repeated application of $g(f(x))=2g(x)$, we then see that in fact $g(x)=0$. we will show that $g(x)\leq 3$. thus assume that we had $g(x)\geq 4$.

if we had $f(x)\leq x$, we would have $g(f(x))>g(x)$ for some $f(x)\leq n$, contradiction to our choice of $x$. therefore we have $f(x)>x$ and since $g(x)\geq 4$ this even implies that $f(x)\geq x+6$. this implies by the functional equation that $f(f(x))\leq x$.

now, we get the clue: $g(x)\geq g(f(f(x)))=2g(f(x))$ (the first inequality comes by the choice of $x$). but this contradicts $g(f(x))>g(x)$. therefore $g(x)\leq 3$; we may do this for all $n$, therefore $g(x)\leq 3$ in fact for all $x\in \mathbb{N}_0$. this again implies by $g(f(x))=g(x)$ that $g(x)=0$. but it is obvious that $f(x)=x+2$ is a solution.

every time i give that proof, some immediately gives a proof by induction. i guess they want to show how ugly my approach is :D

Poor Peter  :( .
Thank you to both of you for the nice proofs.


[hide]3^1=3
3^2=9
3^3=27
3^4=81
3^5=243

the last digit repeats 3, 9, 7, 1 so.... divide 33 by 4 and u get 8 r. 1 so the last digit is a [b]3[/b][/hide]

[quote="DPopov"]What is the units digit of $3^{33} \displaystyle{?}$[/quote]
[hide="big hint"]
The units digit of 3^(0 mod 4) is 1.
The units digit of 3^(1 mod 4) is 3.
The units digit of 3^(2 mod 4) is 9.
The units digit of 3^(3 mod 4) is 7.
[/hide]

wow....

I just freaked out when I saw the problem...

I guess there are more ingenious ways to solve this problem...

since I saw those people's hints, I won't answer it...

I'll just say the actual number I typed in to my calculator..

[hide="the actual number"]5559060566555523[/hide]

[hide]you don't need to worry about the whole number... you just need the units digit
3^1 = [u]3[/u]
3^2 = [u]9[/u]
3^3 = 2[u]7[/u]
3^4 = 8[u]1[/u]
3^5 = 24[u]3[/u]

as you can clearly see, the powers of 3's units digits go in a cycle of 3,9,7,1,3,9,7,1... divide 33 by 4, we get a remainder of 1. [b]3[/b][/hide]

good, what's the units digit of $3^{3^5}$

[hide]I'm not really sure... but wouldn't 3^{3^5}  be the same as 3^15?
3^nth power can end in 3, 9,7,1 and 15/4 = 3R3... so I think it's 7

------------------------------ or ---------------------------------------
or you can you 3^3 = 27 and take the unit digit, and find 7^5
7^nth power can end in 7, 9, 3, 1   and 5/4 = 1R1 and so it end in 7[/hide]

Lucy,

The order of operations for exponents starts at the top of the exponential chain.

While $(3^3)^5 = 3^{15}$, this is because the parantheses dictate that we work inside them first.

On the other hand, $3^{3^5} = 3^{243}$ because $3^5 = 243$.

[hide]powers of 3 go in a cycle of 4
3,9,7,1,3
33/4 has a remainder of 1 so the answer is [hide]3[/hide][/hide]

Lucy -  Be careful .. Be very careful..

$ (3^3)^5  $   is   same as  $3^{15} $  but it is NOT same as $3^{3^5} $  which is   $3^{243}$

Edited later: Okay - .. I see that MCrawford has already explained it nicely but I will still leave this post - Just to let one know that it is very  important to see the difference.. Hope this helps.

For pespective $3^{15}$ is  about 14 million  about number of people in world's largest city (say Mexico  City) (or about   what some  parents  spend on a  kid's education  measured in pennies :))   while the other number  $3^{243} $ is a number so huge that it is greater ..much greater .. than the number of electrons in the whole known universe....  more than a trillion  googol..

o, sorry, didn't know that, thank you!  :D  :blush:

[hide]We can easily find that powers of 3 go in cycle of 4, so that 3^32 ends with 1. Hence 3^33 ends with 3[/hide]

DPopov is going to like this one: what is the units digit of $3^{3^{3^5}}$?  :D

PS: this one is going to be very hard, just one hard challenge for those who wants to 'take a step further'

Eek...I didn't understand a word Mamat said.  :blush:

But this is the way I did it:

[hide]
3^3^3^5
=3^27^5
=3^14348907

Cycle of units digit of powers of 3:
3,9,7,1,3

14348907/3=goes in evenly because its digits add up to 36, which is divisible by 3

Since there's no remainder, the units digit must be the last number in the cycle.

So the units digit of 3^3^3^5 is 1.

Just wondering, is there a way to do this problem without using a calculator or "mod" at all? Thanks!
[/hide]

Edit: Whoops, sorry, lots of mistakes. Thanks 3X for pointing them out. I'll fix them later.

[quote="SoccerBrainy40"]Eek...I didn't understand a word Mamat said.  :blush:

But this is the way I did it:

[hide]
3^3^3^5
=3^27^5
=3^14348907

Cycle of units digit of powers of 3:
3,9,7,1,3

14348907/3=goes in evenly because its digits add up to 36, which is divisible by 3

Since there's no remainder, the units digit must be the last number in the cycle.

So the units digit of 3^3^3^5 is 1.

Just wondering, is there a way to do this problem without using a calculator or "mod" at all? Thanks!
[/hide][/quote]
[hide=""Grave error Please read SoccerBrainy40""]$3^{3^{3^5}}\neq3^{27^5}$ you can't do that, OK? Because the order of operation, or hierarchy, begins from the top most exponent.
Also note that the 'cycle' of units digit of powers of 3 is not 3 but 4, you wrote it out yourself.[/hide]
To answer your last question: Yes, my suggested problem CAN be done [u]without[/u] calculator. Without "mod" can also be done but it takes more writing, "mod" is created to simplify calculations.


Do you mean $ 0.\overline{9}\equal{}1$ ? Yes it is correct!


[quote="ali_ms2004"]ببینید بچه ها من داشتم فکر می کردم که این جا چه کاربرد خوبی می تونه داشته باشه یعتی رو هوا نچرخه بعد به یک نتیجه خیلی خوب رسیدم !!
...این طرحی که الان می خوام بگم قبلا تست شده و جواب هم گرفته و مسلما به پیشرفت همه کمک می کنه حالا بسته به هر آدمی یکی برا المپیاد یکی برا علاقه و
ایده هم اینه
چهارتا تاپیک درست کنیم : ترکیبیات نظریه عداد هندسه و جبر
بعد به جای این که هر دفعه یک تاپیک جدید درست کنیم سوالهای هر بخش رو تو همون تاپیک بزاریم
بعد هر سوالی رو هم توی دوتا پست بفرستیم
پست اول خود سوال
پست دوم پیش نیازهای سوال که اگه کسی خواست یا نتونست حل کنه اون جا یه کمک خوبی براش باشه
نظرتون چیه ؟؟؟؟[/quote]
che ideye jaalebi :roll: (ghaabele tavajohe aaghaaye hatami)

benazar manam fekre galebiye ( ah! injoori ke nemeeshe farsi nevesht!!!!)

manam be nibeye khodam movafegham.va azat kamale tahsakoro daram. ;)

ali joon man vaght nakardam ke posteto bekhoonam.vali fahmidam manzooret chie.

bebin man ba idat movafegham,vali...


1. A quadrilateral ABCD, AD=BC, AC meets BD at O, the bisectors of the angles DAB and CBA meets at I.
Prove that the midpoints of AB, CD, OI are colinear.

2.
Let P is an arbitrary point in a quadrilateral $ ABCD$ such that  $ \widehat{PAB}, \widehat{PBA}, \widehat{PBC}, \widehat{PCB}, \widehat{PCD}, \widehat{PDC}, \widehat{PAD}, \widehat{PDA}$
are all acute angles. Let $ M,N,K,L$ are feets of the altitude from $ P$ on $ AB, BC, CD, DA$.
Find the smallest value of $ \frac {AB}{PM} \plus{} \frac {BC}{PN} \plus{} \frac {CD}{PK} \plus{} \frac {DA}{PL}$

3.
Let $ ABCA'B'C'$ is a triangular prism, $ M,N$ is two points, chosen on $ AA', CB$ such that  $ \frac {AM}{AA'} \equal{} \frac {CN}{CB}$. Prove that $ MN$ is no less than the distance from $ A'$ to $ C'B$

4. Let $ p > 2007$ is a prime number and $ n$ is an integer exceeding $ 2006p$, Prove that
$ C^{2006p}_{n} \minus{} C^{2006p}_{k}$ is a multiple of p, where $ k \equal{} [\frac {n}{p}]$ is the largest integer which is not larger than $ \frac {n}{p}$

5.
Consider 3 positive real variables $ x, y, z$. Prove that:
$ \sum\limits_{cyc} \frac {x(y \plus{} z)^2}{2x \plus{} y \plus{} z} \geq \sqrt {3(x \plus{} y \plus{} z)xyz}$

1) Can we discuss it now?

2) Does the TST contain only 5 problems? (Does it has a link?)

[quote="JBL"]Lock, please (ongoing problems).[/quote]
I don't understand.

Jumbler, you have been fooled - none of this problems is from Singapore TST 2007, and all of these have been at least attempted to discuss before. These are Maths&Youth problems, and their deadline is over.

1: http://www.mathlinks.ro/Forum/viewtopic.php?t=167953 (deadline over?)
2: http://www.mathlinks.ro/Forum/viewtopic.php?t=164083
3: http://www.mathlinks.ro/Forum/viewtopic.php?t=164085
4: http://www.mathlinks.ro/Forum/viewtopic.php?t=190735 (I kept your topic, but locked it)
5: http://www.mathlinks.ro/Forum/viewtopic.php?t=168941

I have deleted most of your topics because the problems have been already posted long ago.

PS. This topic can be deleted in future. The problems don't have much to do with each other; they were not even all from the same M&Y issue.

  Darij

Well, darij grinberg, I am terribly sorry about posting these problems. I have no idea where they are from.

Actually I tried a search for the Singapore TST 2007 in Google, but couldn't find it.  :D


1. [url=http://www.mathlinks.ro/viewtopic.php?t=3547&search_id=745820270]You can see the solution here.[/url]

I think, $ 2$ is wrong. Try $ a\equal{}1$ and $ b\equal{}0.5.$ :wink:


Solve the system of equations \[ \left\{\begin{array}{lll} x^3\plus{}xy^2&\equal{}&40y\\ y^3\plus{}x^2y&\equal{}&10x\end{array}\right.\]

[hide]$ x(x^2 \plus{} y^2) \equal{} 40y$
$ y(x^2 \plus{} y^2) \equal{} 10 x$
Multiply them together.
$ xy(x^2 \plus{} y^2)^2 \equal{} 400xy$
Thus, $ x^2 \plus{} y^2 \equal{} 20$.  EDIT:  oops, we divided by $ xy$, which is illegal if one of $ x$ or $ y$ is zero.  However, if $ x$ is zero, then $ y$ must be zero as well, and vice versa, so $ x\equal{}y\equal{}0$ is another solution
So, $ x(20) \equal{} 40y$ and
$ y(20) \equal{} 10x$, which both give us $ x \equal{} 2y$.  Now we just plug those into $ x^2 \plus{} y^2 \equal{} 20$ to get our solutions.[/hide]

[quote="April"]Solve the system of equations
\[ \left\{\begin{array}{lll} x^3 \plus{} xy^2 & \equal{} & 40y \\
y^3 \plus{} x^2y & \equal{} & 10x\end{array}\right.
\]
[/quote]
[hide="Solution"]
Notice that this means that $ x(x^2\plus{}y^2)\equal{}40y$ and $ y(x^2\plus{}y^2)\equal{}10x$.  Divide them so $ \frac{x}{y}\equal{}4\cdot \frac{y}{x}\implies (\frac{x}{y})^2\equal{}\frac{1}{4}$, and we notice that $ x\equal{}\pm 2y$.  Plug $ 2y\equal{}x$ into the first equation, so $ x(x^2\plus{}y^2)\equal{}40y\equal{}20x$.  Thus, either $ x\equal{}y\equal{}0$ or $ (2y)^2\plus{}y^2\equal{}20\implies y^2\equal{}4\implies y\equal{}\pm 2$, so $ x\equal{}\pm 4$.  Now, if $ x\equal{}\minus{}2y$, then $ x(x^2\plus{}y^2)\equal{}40y\equal{}\minus{}20x$.  This gives that either $ x\equal{}y\equal{}0$ or $ x^2\plus{}y^2\equal{}\minus{}20\implies (2y)^2\plus{}y^2\equal{}\minus{}20\implies y^2\equal{}\minus{}4\implies y\equal{}\pm 2i$, so $ x\equal{}\pm 4i$.  Thus, the solutions are $ \boxed{(x,y)\equal{}(4, 2), (\minus{}4, \minus{}2), (4i, 2i), (\minus{}4i, \minus{}2i), (0,0)}$. [/hide]

The substitution $ x\equal{}m sin A$, $ y\equal{}m cos A$ works very well. The two equations simplify to
$ m^3 sin A\equal{}40m cos A$
$ m^3 cos A\equal{}10m sin A$
So either $ m\equal{}0$, in which case $ (x,y)\equal{}(0,0)$, or we can multiply them and get $ m^4\equal{}400$, in which case $ m\equal{} \pm \sqrt{20}$ or $ \pm i \sqrt{20}$. If $ m\equal{} \pm \sqrt{20}$, we find $ A\equal{} tan^{\minus{}1}2$, and if $ A\equal{} \pm i \sqrt{20}$, $ A\equal{}tan^{\minus{}1}\minus{}2$, from which we get all the solutions.

My solution:
$ x^4\plus{}x^2y^2\equal{}40xy$
$ y^4\plus{}x^2y^2\equal{}10xy$

So $ x^4\minus{}y^4\equal{}30xy \Leftrightarrow (x\minus{}y)(x^3\plus{}x^2y\plus{}xy^2\plus{}y^3)\equal{}30xy$
But $ (x^3\plus{}x^2y\plus{}xy^2\plus{}y^3)\equal{}40y\plus{}10x$
So $ 10(x\minus{}y)(4y\plus{}x)\equal{}30xy \Leftrightarrow 4xy\plus{}x^2\minus{}4y^2\minus{}xy\equal{}3xy \Leftrightarrow
x^2\equal{}(2y)^2$
So $ x\equal{}2y$ or $ x\equal{}\minus{}2y.$
We substitute and we take the solutions :)


are you allowed to put that a LCM is negative say the LCM of 2 and 3 is -6?

no i believe LCMs must be greater than 0 (positive)


[quote="asintota"]
If
\[ \log_{3}n \equal{} \log_{6}m \equal{} \log_{12}m \plus{} n
\]
find $ \frac {m}{n}$.
[/quote]

[hide="solution"]$ \log_{3}n \equal{} \log_{6}m \equal{} \log_{12}m \plus{} n \equal{} x$

$ 3^x \equal{} n$

$ 6^x \equal{} m$

$ 12^x \equal{} m \plus{} n$

$ 3^x(1 \plus{} 2^x) \equal{} 3^x*4^x$

$ 4^x \equal{} 1 \plus{} 2^x$

$ 2^x \equal{} y$

$ y^2 \equal{} 1 \plus{} y$

$ y \equal{} \frac {1\pm\sqrt {5}}{2}$

$ \frac {1\pm\sqrt {5}}{2}*n \equal{} m$

$ \frac {m}{n} \equal{} \frac {1\pm\sqrt {5}}{2}$

But if it's negative, then one of m or n must be negative, and we can't have that.

$ \frac{m}{n}\equal{}\boxed{\frac{1\plus{}\sqrt{5}}{2}}$[/hide]


On AMC and AIME, there are a lot of time pressure and I make a lot of stupid mistakes when I am under time pressure.
Also there are a lot of exam pressure because a lot of us consider AMC and AIME one of the most important exam in the year. We could be very nervous and can't think clearly on the exam. What should we do?

I think the most important part is to have fun! Problem solving is always great.

Just realize that this competition is a lot harder for everyone else. :D

On tests like the AIME, which is also an endurance test, I find it's great just to use 5 mintues every hour to completely zone out.  I'll start thinking about the weather, or that history test I have next week, or whatever.  It helps me hyperfocus again once I come back.

Yeah, Masa, me and my friend were discussing the AIME and sort of what the best way to solve stuff is.  He scored the highest in the school, and he said that's because he doesn't mind just doing brute computation.  I said that that ruins the fun, because the point isn't to get it right, it's how you get it right.

I agree that it's important to realize that the competition should be a fun and learning experience.  But to give you some solid advice, I find it helps to read every AMC problem twice - this helps to eliminate really stupid mistakes.  For example, on the AMC-->10A this year, at least half of the kids at my school (myself included) missed the "you and" part, and divided by 5 instead of 6.  Had we read the question a second time, we probably would have caught that.

just remember, it's only one little test. if you don't do well, it's not the end of the world. after making AIME last year, i did horrible on the AMC10 and didn't make AIME this year, but my life isn't over. anyways, there's always next year, and there's always more opportunities.


Έστω $ p(n)\equal{}a_{n}z^{n}\plus{}a_{n\minus{}1}z^{n\minus{}1}\plus{}...\plus{}a_{1}z\plus{}a_{0}$ με $ a_{0},a_{1},...,a_{n\minus{}1},a_{n} \epsilon C,a_{n} \neq 0$,πολυώνυμο βαθμού $ n$.
Να αποδειχθέι ότι

$ \frac{1}{2} |a_{n}\parallel{}z|^{n} \leq |p(z)| \leq \frac{3}{2} |a_{n}\parallel{}z|^{n},$
για κάθε $ |z| \geq R, R\equal{} max \{1, \frac{2}{|a_{n}|}  \sum_{k\equal{}0}^{n\minus{}1}   |a_{k}|     \}$

Η συγκεκριμένη άσκηση δεν είναι ιδιαίτερα ενδιαφέρουσα από μόνη της (στο κάτω κάτω είναι απλά εφαρμογή της τριγωνικής ανισότητας, και δε λες και κάτι το φοβερό, απλά εξακριβώνεις το γνωστό $ P(x) \in \Theta(x^n)$ αν $ \deg P \equal{} n$), αλλά έχει μια αρκετά... ενδιαφέρουσα non-trivial εφαρμογή: το Θεμελιώδες Θεώρημα της Άλγεβρας :)

WLOG θεώρησε ότι $ a_n \equal{} 1$. Τότε για κάθε $ z \in \mathbb{C}$ με $ |z| \geq R \equal{} \max\left\{1, 2\sum_{j\equal{}0}^{n\minus{}1} |a_j|\right\}$

$ |z|^n \minus{} \sum_{j\equal{}0}^{n\minus{}1} |a_j| |z|^j \leq |P(z)| \leq |z|^n \plus{} \sum_{j\equal{}0}^{n\minus{}1} |a_j| |z|^j$

$ |z|^n \left(1 \minus{} \sum_{j\equal{}0}^{n\minus{}1} |a_j| |z|^{j\minus{}n} \right) \leq |P(z)| \leq |z|^n \left( 1 \plus{} \sum_{j\equal{}0}^{n\minus{}1} |a_j| |z|^{j\minus{}n} \right)$

Οπότε:

$ |z|^n \left(1 \minus{} \sum_{j\equal{}0}^{n\minus{}1} |a_j| |z|^{\minus{}1} \right) \leq |P(z)| \leq |z|^n \left( 1 \plus{} \sum_{j\equal{}0}^{n\minus{}1} |a_j| |z|^{\minus{}1} \right)$

Αλλά προφανώς $ \sum_{j\equal{}0}^{n\minus{}1} |a_j| |z|^{\minus{}1} \leq \frac{1}{2} \cdot |z| \cdot |z|^{\minus{}1} \equal{} \frac{1}{2}$, και παίρνεις αυτό που θέλεις.