Putnam articles Harvard college geometry quadratics function

Hello,

I was wondering what people can say something about women in Math Olympiads in their respective countries. Unfortunatelly, math tend to be a preponderant male field. I mean, anyone who works in mathematics know that at any time (Undergrad, Grad, Faculty) there are a lot more males than females. I have just taken a course in math (after highschool) thought by a woman, and for god sake, I have taken a lot of math courses.

Also in olympiads, at least in Costa Rica, not too many women make it to the final (there are two previus rounds), even when the first round use to open with a non that few women (by the first round there are around 40% of female participant, but by the final it use to be not more than 15%, and usually less). From 1996 to 2002 I remember a few (maybe about twenty) female bronce medalist, may be three or four silver medalist, but not even one gold medalist. Just for information There are three levels for the final, with around 50-80 participant on each level, and there are 9 (5-3-1) medals for the first level, 12 (6-4-2) for the second, and 6 (3-2-1) for the third level (bronce-silver-gold). The only time that a woman has gone to a presential international math olympiad, she was the only one not getting at least a mention in the conquest.

This is, I think, very sad. I am totally convinced that there are as much math potential in women's brains than in men's brains, but for some reason, not that many women want to show it.

I know, for example, than Ana Caraiani has done it very good in olympiad. A Romanian probably may say how accurate is the information I have heard, but I think Ana Caraiani, a former Romaian IMO team's member, participated three times, one silver medal, two gold medals, including the hightest Romanian medal in 2003. She also made it quite good in the Putnam on 2003, and I suppose that also in 2004, but I haven't seen the results yet.

What makes Ana Caraiani a discontinuity. Of course, she has showed to everyone she is a very intelligent woman, out of the ordinary, but why when someone is that smart, or at least when someone has shown to be that smart, it use to be in almost all the cases (I guess, at least 90% of the times), a man?

Well, I wrote more than what I spected, my question remains, how well do women usually do in Math Olympiad in your countries?

Best,

I can confirm Ana's performances at the IMO (I was twice in the same team with her, and once an observer :D) and the Putnams' :) But she's unique (in Romania). I also remember that in 2003 there were not less than 4 girls who got gold medals in IMO Japan and a lot more with Silver and Bronze in the IMO. Let's not forget some of the IMO girls from out forums:  Valiowk (singapore team the past two years), Alison (gold medalist last year with the IMO team), and which MoPPer doesn't know first girl in the US IMO team ever,  Melanie Wood (silver in 99,00 I think).
Another topic has been discussed last year about the putnam fellow women (actually more about Ana, but well :) ): http://www.mathlinks.ro/Forum/viewtopic.php?highlight=duke&t=4751

You may be interested to read:   [url=http://www.maa.org/news/melanie-wood-interview.html]Melanie Wood Interview[/url] and also  an article [url=http://www.chronicle.duke.edu/vnews/display.v/ART/2003/03/31/3e884ee29fa4b]Having her number retired[/url]

Hope you find it interesting.
[quote]The jersey-retiring ceremony will not be in Cameron Indoor Stadium, though, but rather, the Physics Building, and Wood will be honored for something other than basketball.

Out of the thousands that competed last December in the 63rd Lowell Putnam Competition, arguably the most prestigious math competition worldwide, Wood, an A.B. Duke scholar, placed among the top five scorers. As a result, officials from the Mathematical Association of America, the organization that administers the annual Putnam Competition, named Wood a Fellow, making her the first American woman and only the second woman in the world to accomplish such a feat.[/quote]

Or in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=melanie+wood&t=5673 :D

I found the discussion in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1850]this thread[/url] very interesting. I really appreciated the ideas in [url=http://www.mathlinks.ro/Forum/weblog.php?w=1]Richard Rusczyk's web log[/url] too.

Does anybody want to the discuss the cover title of the most recent [url=http://www.time.com/time/]TIME Magazine[/url] with me ?

Orlando,

Do you want to discuss only the cover or the article itself. Unfortunatelly, they just have a few parragraph, do you happen to have the complete article (and would be able to send it to me ;) ).

The cover itself does not seem (at least to me) too controversial. So far, I think, that to ask if women are more or less intelligent than men is like asking if lefthanded people are more or less intelligent than righthanded. It is just something that is not exactly correlated.

Before you want to lynch me, read the whole idea. WOMEN AND MAN ARE DIFFERENT, and I love those differences. I am not talking only about the physical differences (that, by the way, I do love the fact that physically, my wife is very different than me  :D ) but also when we talk about usual skills and responses to different situations.

For example, one of the complains of most of the married women is that men does not listen to them carefully, and most of the married men (I include myself  :? ) is that often women don't stop talking. But analyze that. Most of the time, and in most of the western cultures (at least I have heard that from African, Europeans, and people from all the Americas) is that men use to go to the point of the issue A LOT faster than women. But also, women use to deal better with emotional problems (according to several psicologist, or at least that is what I have read in some articles, and hear in some documentals).

Given than even when we talk about hormons, men and women are quite different, I wouldn't surprise if someone proves that there really are anatomical differences between men an women. Nevertheless I think it is stupid to use it to discuss that some of the genders is smarter than the other by nature!

I remember, talking about this, the story about T. Bischoff, a prestigious anatomist on the second half of the XIX century, that during decades collected information about the weight of the brain, and discovered that the average weight of the brain of a man was 1350 grams, while the average weight of a woman is 1250, and he used this to defend his theory that men are superius than women, intellectually speaking. Nevertheless, when he died, his brain was weighted, and its weight was 1235 grams  :rotfl:

What may be true, is that, by those anatomic differences, may be males are more prone to competence than women. I am only doing speculation, but everybody agree that psicologically women are different from men, then, why not in this aspect!  :what?:

Best regards,

[quote="djimenez"]For example, one of the complains of most of the married women is that men does not listen to them carefully, and most of the married men (I include myself  :? ) is that often women don't stop talking.[/quote] :rotfl:  ... this is very funny :)

[quote="djimenez"]
What may be true, is that, by those anatomic differences, may be males are more prone to competence than women. [/quote]

How are we defining 'competence' here?  I think this would be a tough position to defend, even empirically.  'Outstanding excellence' can at least currently be empirically defended, but I'm not as convinced about mere 'competence'.  Again, this could be simply a question of different distributions - men having fatter tails on both ends (i.e. more Fields medalists, and more US Presidents).

[quote="djimenez"]Orlando,

Do you want to discuss only the cover or the article itself. Unfortunatelly, they just have a few parragraph, do you happen to have the complete article (and would be able to send it to me ;) ).

[/quote]

I wanted to discuss the article from the cover.  ;)  Yes, I have the article.

It seems true that male and female children have different preferences. Some time ago I watched the three parts of the BBC documentations "[url=http://www.dvd-galaxis.de/index.html?genre=suche&nach=Titel&suche=Kinder+unserer+Zeit]Kinder Unserer Zeit[/url]", in English: "[url=http://www.bbc.co.uk/parenting/tv_and_radio/child_of_our_time/]Child of Our Time[/url]". And they made an experiments with some parents. They always send some mother  or father in a room. There he/she is given some children in sequence and he/she always shall give him some of the toys to play. Typical boys and girls toys were available. People usually sticked to gender typical stuff. And they were confused that mostly the children did not want to have the respective toys but just very few tried to give some boys girl toys or vice versa in return. Later the parents got told that male children got the typical clothes of females ones and vice versa.

It is not clear for me what to conclude. Either there is really a fundamental difference or the male/female children were just to much used to their typical toys that they refuse other toys now. :)

[quote="rrusczyk"]How are we defining 'competence' here?  I think this would be a tough position to defend, even empirically.  'Outstanding excellence' can at least currently be empirically defended, but I'm not as convinced about mere 'competence'.  Again, this could be simply a question of different distributions - men having fatter tails on both ends (i.e. more Fields medalists, and more US Presidents).[/quote]

No, what I mean is that probably men are more prone to involve themselves in situations that are developed in a highly competitive enviroment. Of course that I know that it is a position hard to defend, specially because for centuries women didn't have any access to this kind of situations, and there are still too many preconseptions, and those, of course, play an important roll in this situations.

But look at the fact. Usually in the schools, most of the females are in a good standing, but the outstandingly good and the outstandingly bad students, use to be males!

For example, when I was in the last cycle of high school in Costa Rica (the system is a little different than here in the US) I got into the "Highscool for the Scieces of Costa Rica", that is considered the best highscool (academically) of the country, and also the most competitive. My class started with 17 males and 13 females, but ended with 13 males and 4 females, and my class was sligly over the average, some classes have just one or none female.

Best,

Ah, now I understand.  Just a tip - the word 'competence' doesn't have anything to do with competitiveness.  It means roughly 'having adequate ability'.

OK, I appologize, it was a lack of vocabulary :blush: , in Spanish they are the same word, and the sence is given by the context!  :?

Best,

[quote="Gyan"][quote] is that often [b]women don't stop talking.[/b] But analyze that. [i]Most of the time[/i], and in [i]most of the western cultures[/i] (at least I have heard that from African, Europeans, and people from all the Americas) is tha[b]t men use to go to the point of the issue A LOT faster than women.[/b] But ...[/quote]

Bold statements indeed!  Let us  take a small sample, like this thread   and  a  few posts above  .. hmmm.. who is "talking" and "talking"  here? ..And getting  to the "point a LOT faster than a women" .. well ...

To put  it mildly, IMO, these kind of statments are really  no more than nonsense. .. not much different than  some one "analyzing" why  there are more men dirvers than  women dirvers   in Saudi Arabia.)

Sad part is that similar arguments/comments do tremendous harm.  I know  of a case of a Harvard worthie (Math Prof) giving similar  words of wisdom to  a person looking/ trying to decide if Harvard was good for her undergraduate studies. She was told  by this worthie,  "Women can't do math"(or something like that)  . nevermid, this candidate was an  IMO medal winner.   (BTW she  is   a Putnam fellow now). Fortunately  other schools, and profs,  were much more encourageing and  this person is doing  very well.[/quote]

Well, may be you're right, and this is a lot of nonsence for most of the people (and may be I am not going to the point) but I am not saying that ALL women go around the point to much before going there, or that all men go inmediatelly, what I say is that in most of the cases, those are the complain of most of the men and women I know, it is:  [i]"My wife don't stop talking, and some times everything I want is a couple of minutes of silence"[/i], and [i]"I need to express myself and my husband don't pay any attention to what I say"[/i].

The statment [b]Women can't do math[/b] is false, I agree with everyone, the question that we are trying to discuss is [b]Why usually a lot more men than women do Math?[/b]. Of course that I agree that Ana Caraiani (the only female Putnam fellow currently, and gold medalist on IMO) is capable to do math, probably more capable that many males who are currently mathematicians, but why is she an exception?

By the way, I am not telling that my statments are an absolute truth, or something similar. Just some ideas in the air, and of course, what I want is to hear some one refuting them, or supporting, them in a civilized discussion. There are so many cultural aspects that people accept, makes the people of some places more prone to some  actitudes or behaviors than others. Why can't there be also some anatomical aspects that can do that.

[b]PS:[/b] An example of what I call a cultural aspect. I have a big office with other three grad students, one is Spanish, other German and the other is Chinease, and the German and Chinease complain some time that we, the hispanic part, are a little too noisy. And that is true, generally, hispanics use to be noisy, that does not mean that there are not hispanic that are quiet.

Best,

[quote]what I say is that in most of the cases, those are the complain of most of the men and women I know, it is: "My wife don't stop talking, and some times everything I want is a couple of minutes of silence", and "I need to express myself and my husband don't pay any attention to what I say". [/quote]

David, I hope you realize the nature of the audience here, and the effect your posts may have here. We are not in a graduate students’ lounge discussing this with a few friends. Many here who are young middle/high school students look up to a Post- Doc in Math  like yourself as a role model and hence you have to be more careful … . That is the reason I gave the example of that Harvard worthie   and  why I think that  it was very sad that he was talking about this (instead of encouraging her) . (BTW personwas  not Ana.  Ana   may be the third woman  Putnam  Fellow)

Remember all the reaction about the current Harvard President and his loud wonderings about all the “inherent differences and innate abilities? ? Like many I think that was very unworthy of a Harvard president to make that type of comments.

[quote="qvc122"]I looked at some of the posts in the topic of new SAT scores and found that Jennifer Lin got 2400. She is a jonior and 3 times USAMO qualifier. She just sets an example that when a female is that good with math, she is simply perfect.

Imagine, even she is so great at math (3 times USAMO qualifier), she may not choose math as her career. Instead, she may become a lawyer or an MD and earns 3-4 times more than what a typical mathematician makes. Regardless of what she chooses, you cannot say that she is less smart than a mathematician or she has less ability.

Congrats, Jennifer.[/quote]

You are may be loosing the point of the conversation. I did not know about Jennifer Lin, but I do not doubt for a second that there are [b]amazingly smart[/b] women in this world (I personally know some of them), but the women that are in the top of the fame in the academic enviroment (say universities) or in the top scores in the academic competences (example: IMO, USAMO, Putnam, etc) are very very few, even when in some cases they are really better than any of their male competitors.

It's a completely irrelevant point.  We have, if I remember correctly, a three or four time male qualifier for USAMO who also had a perfect (then 1600) score on the SAT 1.  There are probably 20-30 people in every grade who will qualify for three or four (or more) USAMOs, and I would expect that their SATs would be in the upper 1500s.  Its not as if the female portion of this segment is 'better' than the male segment just because they perform better relative to their (by gender) peers.

Obviously it's an impressive achievement; one that marks her as one of the nations brightest young individuals.  But it doesn't carry with it notions of associated perfection.

Heh, I'm not being talked about, am I?  Anyhow, I agree that individual performances are essentially irrelevant: the interesting question is one of the behaviors and successes of groups.  Any group will have exceptional individuals.  They tell you very little about the group itself.

Wildoctagon, I had a question for you: you listed 4 reasons that you suspected there was a biological difference that trumped the social influences, quoted here:
[quote="wild octagon"]I [i]think[/i] biological factors outweigh environmental ones because of
[list]1. Scientific studies that show sex hormones can affect brain function.  This [i]Scientific American[/i] [url=http://www.sciam.com/article.cfm?articleID=00018E9D-879D-1D06-8E49809EC588EEDF]article[/url], for example, cites several studies that found that men and women differ in spatial reasoning ability, an important component of math ability.[/list][list]2. Math contest results.  Maybe performance in competitions is not the best indicator of math potential but surely there's a strong connection.  This year girls made up 48% of AMC10 participants, 43% of AMC12 participants, 23% of AIME qualifiers, and only 7% of AIME high scorers (floor of 9 or above).[/list][list]3. Personal experience.  Many years ago I tried to recruit girls for my high school math team so I wouldn't be the only girl.  More recently I tried to recruit girls for the middle school math team I was coaching so my daughter wouldn't be the only girl.  The number of girls I see in math teams, math clubs and math camps is always much lower than 50%.  I guess it's possible that girls are discouraged at such a young age, they give up on math before they reach the math team/club level.[/list][list]4. AoPS.  The number of female members here is very small.[/list][/quote]
I'm curious -- why would any of those studies or indications or facts lead you to believe that the biological trumps the social?  Tests on adults are essentially useless as they've gone through a lifetime of socialization, and young children are not a good group to study if you're interested in higher mathematics, as most of them cannot add.  I find it difficult to imagine any test or series of tests that would [i]ever[/i] be able to distinguish the biological from the social, in fact.

One interesting, tangentially related comment.  I had dinner a few months ago with a group of people, primarily women in scientific fields (led by an engineer and a computer scientist, actually, so from some of the most male-heavy of the fields) and we came to one interesting conclusion: they were simply not very well equipped to figure out why women left mathematics because they themselves had never had any question about it.  The people who love math are not the people best able to understand why others don't.  Just a thought.

[quote="djimenez"]You are may be loosing the point of the conversation. I did not know about Jennifer Lin, but I do not doubt for a second that there are [b]amazingly smart[/b] women in this world (I personally know some of them), but the women that are in the top of the fame in the academic enviroment (say universities) or in the top scores in the academic competences (example: IMO, USAMO, Putnam, etc) are very very few, even when in some cases they are really better than any of their male competitors.[/quote]

Djimenez,

Well, I do not personally know Jennifer Lin at all. As you stated that women in IMO, USAMO, and Putnam are very very few, this is absolutely true. However, I am already amazed with seeing female Putnam fellows three years in a role. The 1st female Putnam fellow was in 1996. Then we had one each in 2002, 2003 and 2004. Putnam competition started in 1938 and we had zero female fellows until 1996. We are catching up fast, believe or not. I went to a state MAO convention this weekend and saw many female students competing there. It was quite common to see 2-4 female students placed in top 10 in many categories. My 9th grade daughter is a USAMO qualifier this year. But I can tell you that she has been discouraged from math for many years. We (my mathematician husband and I) do not want her to major in math for her collage and her grad school. We want her to be a medical doctor (I hope she will listen). Even with all that discouragement that female students have been encountered in math (not necessarily from school, but at their homes and communities); I still  believe that female will continue to catch up in math and science in the future, probably not as fast as in medicine and law. As long as the resource in math is limited (I heard that one state collage has travel funding of $500 for the graduate students this summer in their entire department), the number of top female students will be shunt away from making the choice of math as their major. The resource is truly a BIG factor that we have not looked into in the discussion of this topic. Generally, females care less about whether other people think that they are smart or not. Many of them are practical. Females have already taken many professional positions today. I was very surprised to hear from NPR news one late afternoon that in about 25% of the families in US today females make more income than their male counterparts, (my own family is certainly one of them. But before hearing that, my thought was somewhere less than 10%). If men are willing to loss some positions in math, we will sure take them. How's that. No offense, please.

One of the problems with what you're saying is that a large number of the high-scoring female Putnam students are not originally from the US.  It's never really been in question as much in, say, Eastern Europe whether women have historically made up a much larger proportion of the IMO teams and whatnot.

I don't think it makes any sense to say that women who make it to the USAMO/IMO/insertarbitrarymathematicalstandard are more smart than their male peers (e.g of equal qualification) which seems to be what you're getting at.  There are USAMO qualifiers - even IMO qualifiers - who come from places without an established mathematics competition tradition, and that in and of itself is a huge barrier to overcome, and probably the biggest barrier to making it to, and succeeding in, these competitions.

[quote="blahblahblah"] I don't think it makes any sense to say that women who make it to the USAMO/IMO/insertarbitrarymathematicalstandard are more smart than their male peers (e.g of equal qualification) which seems to be what you're getting at. [/quote]

No, I am not getting there.

How do you define smarter? Like what you stated that “a large number of the high-scoring female Putnam students are not originally from the US.” Why? It is basically economic reason.

In Unite States, a smart female has many more choices in her career. If they have to go out to make a living rather than to stay at home taking care of kids, most of them would choose more profitable professions. I do think that mental ability between male and female is close, but with different focuses. Top female students in math are probably less focused on math than the top male students in math. Many mathematicians’ children are not doing math as their careers, male or female, for the economic reason.

Smart can be mathematically, verbally and socially. If you are only smart mathematically, it will be very difficult for you to have a good life. Feeling good about being smart can only go that far (as far as while you are still a student and not in real life yet). Imagine, your next door neighbor, a lawyer, who is doing so much better (better house, kids in private schools…..) than you are (a successful mathematician). You would say to yourself how could that happen: the guy is not even half smart compare to you. You probably feel sick and unjustified. Your next door lawyer probably has never considered you as smart. This society does judge people as smart or not on their earnings, unfortunately.

Btw, I noticed that many top math students, male or female have high SAT scores in both math and verbal. I am very pleased to see that.

I don't think there's an economic reason towards why many of the high scoring females on the Putnam are not from the US.  We're talking about a sample size of five or six people here, most of who go to schools such as Princeton and Harvard.  Mathematicians at these schools routinely land jobs in mathematical finance - some more mathematics oriented, some more finance oriented - at big financial firms on the East Coast.  There's not a lot of economic sacrifice implicit in choosing to study mathematics at such an institution.

[quote] Top female students in math are probably less focused on math than the top male students in math. [/quote]

If you're saying that top female students in math relative to the pool of females who do math below a certain age are less focused on math than the top male students in math relative to the male pool satisfying the above conditions, then you're correct.  The biggest difference between males and females is not so much in the top 25 scores on the Putnam (there are, after all, at LEAST 3 female IMO gold medalists in North American universities) but in the next 25-75.  Or 25-200.  Or probably even 25-500.

I think a lot of it has to do with behavioral differences in gender.  Its not uncommon to see boys who really become engrossed in a topic, whether it be mathematics, computer science, or writing star wars fan fiction, regardless of what social stigmas may be associated with it.  Instead, 'highly intelligent' girls, for lack of a better term, seem to become quite well rounded and often seek social (I am referring to things like running student councils) achievements, not academic ones.

But I don't think that you can say that the few females who make academic achievements equal to those of the boys are 'simply perfect', it just means that they have put in the time necessary to achieve such things.  Yes, it is even more impressive that they have succeded with a (relative) lack of female peers, but you often see IMOers who come from areas where they don't have mathematical peers (I guess this happens to every sufficiently good young mathematician at some point in time)

[quote="JBL"] Wildoctagon, I had a question for you: you listed 4 reasons that you suspected there was a biological difference that trumped the social influences . . .  I'm curious -- why would any of those studies or indications or facts lead you to believe that the biological trumps the social?[/quote]
I think it's [i]possible[/i] that environmental factors outweigh biological ones when explaining the scarcity of females in higher level math but it's [i] unlikely[/i] given the great disparity in numbers.  Only 7% of the AIME high scorers this year were girls.  It's very difficult for me to imagine that 7% figure approaching 50% without drastic changes to the test or to the scoring.  Yes, more girls would excel in math if we made an all-out effort to train them, but that would be true for boys too.

If you're willing to concede there are biological differences in the way men and women think, in the way they behave, then you should be willing to entertain the [i]possibility[/i] that biological differences can explain why girls are less likely to study math, why few score as high as the top males.  True, we don't understand the brain processes involved in mathematical reasoning.  (Someone should use PET or fMRI to peer inside the brains of prominent mathematicians to see what happens when they're problem solving.)  When I read about studies that find physical explanations for gender differences in behavior, however, I start to wonder how much of our behavior can be explained by sex hormones.

The [i]Scientific American[/i] article I cited in a previous message, called [url=http://www.sciam.com/article.cfm?articleID=00018E9D-879D-1D06-8E49809EC588EEDF][i]Sex Differences in the Brain[/i][/url], mentions that men typically score higher than women in spatial ability tests.  You might argue that environmental influences explain the difference.  But consider the medical condition called CAH, which causes girls to be exposed to excess male hormones in the prenatal or neonatal stage.  The article states that these girls, who tend to favor male-typical activities, score higher on spatial ability tests than average girls.  In fact the optimal level of male hormone for spatial ability is in the low male range, which means those of us women who excel in spatial problems are probably more "masculine" than average and you guys who excel tend to be more "feminine" than average.

The May issue of [i]Scientific American[/i] includes an article titled [i]His Brain, Her Brain[/i].  It also cites studies that have found physical differences in how male and female brains work.  One intriguing study examined the boy-truck/girl-doll issue: "The researchers presented a group of vervet monkeys with a selection of toys, including rag dolls, trucks and some gender-neutral items such as picture books.  They found that male monkeys spent more time playing with the 'masculine'  toys than their female counterparts did, and female monkeys spent more time interacting with the playthings typically preferred by girls.  Both sexes spent equal time monkeying with the picture books and other gender-neutral toys."

Don't these studies make you question the view that cultural factors are the primary reason why few women study math and science?

[quote="JBL"]. . . they were simply not very well equipped to figure out why women left mathematics because they themselves had never had any question about it.  The people who love math are not the people best able to understand why others don't.[/quote]
You might want to read the [i]Loss of Confidence[/i] section of the 1998 NSF report on [url=http://www.nsf.gov/sbe/srs/nsf99338/access/c3/c3s1.htm#c3s1l1][i]The Undergraduate Experience in Science, Mathematics and Engineering[/i][/url].  According to the authors, women tend to leave the sciences for various reasons, including choosing a science major to please someone else and lack of attention from professors.

[quote="wild octagon"]
I think it's [i]possible[/i] that environmental factors outweigh biological ones when explaining the scarcity of females in higher level math but it's [i] unlikely[/i] given the great disparity in numbers.  Only 7% of the AIME high scorers this year were girls.  ...

If you're willing to concede there are biological differences in the way men and women think, in the way they behave, then you should be willing to entertain the [i]possibility[/i] that biological differences can explain why girls are less likely to study math, why few score as high as the top males.

... The article states that these girls, who tend to favor male-typical activities, score higher on spatial ability tests than average girls. [/quote]

I'm fascinated by the implicit assumption that there's a correlation between  females scoring high on the AIME and females in higher level mathematics.  If females are near parity in undergraduate mathematics majors, then I think that girls are not much less likely to study mathematics than boys.   As a mathematical culture, we should separate our expectations for contests (which are very culture-laden to begin with) from our expectations for mathematics in general.  Plenty of people enjoy mathematics and study mathematics but do not participate in contests.

As for spatial ability tests, a bit of training erases the gender gap.  I wish I could remember the reference for that study...

[quote="qvc122"]We (my mathematician husband and I) do not want her to major in math for her collage and her grad school. We want her to be a medical doctor (I hope she will listen).[/quote]

I hope you will allow your daughter to follow her heart.  As a professor, I have encountered many students whose talents and/or passions lie in areas other than those their parents would prefer, and it's painful to see students either squelching their dreams or having conflict with their parents because they refuse to squelch their dreams.

I might add that financial incentives are far greater to pursue mathematics---grad school is generally free plus a stipend---than medicine or law, which are both costly.  So I doubt the disparity is economic in nature.

[quote="qvc122"]This society does judge people as smart or not on their earnings, unfortunately.[/quote]

I disagree here as well.  Society sees mathematicians as quite intelligent, and teachers/professors in general as quite intelligent, regardless of their income levels.  Perhaps the socioeconomic stratum composed mainly of doctors and lawyers differs, but that's not society.  Housepainters are routinely impressed by mathematics majors.

[quote="smbelcas"]I'm fascinated by the implicit assumption that there's a correlation between  females scoring high on the AIME and females in higher level mathematics.  If females are near parity in undergraduate mathematics majors, then I think that girls are not much less likely to study mathematics than boys.   As a mathematical culture, we should separate our expectations for contests (which are very culture-laden to begin with) from our expectations for mathematics in general.  Plenty of people enjoy mathematics and study mathematics but do not participate in contests.[/quote]
That's an interesting point.  How strongly does math competition performance correlate with future math success?  I was assuming a close connection between the two but perhaps that's incorrect.  To me the AMC/AIME problems seem to test basic math skills and knowledge that would make up any well-rounded math education.  Does the ability to whip off these easy problems at a young age not give one a great advantage when doing math research later?

I agree that it is not necessary to excel in contests in order to be a successful mathematician, who typically works on a single problem in a single topic for weeks, months, years at a time.  There are also math olympians who end up in careers other than math -- perhaps because research in math is too different from contest math?  But if these contests are not a strong predictor of future math success, should schools like MIT be asking for scores on their admissions applications?

[quote="wild octagon"]How strongly does math competition performance correlate with future math success?... Does the ability to whip off these easy problems at a young age not give one a great advantage when doing math research later?[/quote]

First, I don't know of any studies done in this vein; second, it would be quite difficult to do such a study because one would not only need to track which mathematicians had earlier had success in contests but also what the various success levels in mathematics contest participants reached... and somehow try to control for interest factors (people who could be successful in contests but aren't interested in taking them and people who are successful in contests but are not interested in mathematics careers).

That said, my sense is (based on anecdotal evidence, of course) that there's little correlation between math competition performance and future math success, where for the former I'm thinking of AIME/USAMO qualification and for the latter I'm thinking of completing an intended mathematics major in college.

It's certainly true that many mathematical luminaries were successful in either high school or college mathematics contests; and, many mathematical luminaries were unsuccessful or did not participate in such contests.  It is also true that some students who are successful at contests can't prove their way out of a paper bag (so to speak) and end up not being able to hack upper-level collegiate mathematics, and that plenty of students are successful both in contests and in collegiate mathematics.

I wouldn't characterize AMC/AIME problems as 'easy' myself.  Still, I don't sense any advantage (certainly not a great one) in doing mathematics research in having that type of facility.  In order to get to doing math research, one must go through lots of other mathematical education (to acquire background) and for some of that background algebraic/numerical facility is an advantage, but most successful people have some of that.  Those who succeed at contests have LOTS of facility with algebraic/numerical manipulations, and that does help.  (So I guess my conclusion is that it's an advantage, but not a great one.)

[quote="wild octagon"]But if these contests are not a strong predictor of future math success, should schools like MIT be asking for scores on their admissions applications?[/quote]

Yes, absolutely!  Every school should be interested in all the facets of their applicants' academic and extracurricular lives.  Even if we're not predicting future mathematics success via contest scores, we are seeing evidence of other things like persistence and interest and focus, all of which are essential for college success.

Hello,

In my personal opinion, I know many many cases of people who considered to study (and ended up studying) mathematics after they participate on a math competition, even if they didn't do really well on such competition. I can think right know in fifteen cases right now, and all but two didn't finish their bachelor in Math (both went to computer sciences in some moment). And many of us are either studying for our Ph.D in Math, or have already completed it.

I honestly consider that if I had never participated on a Math Competition, probably I would have never discovered that I like math that much, and would probably study some other thing. I can tell that either me or any of the cases I have in mind are highly successful mathematicians, mainly because those who already finished their doctorates finished less than two years ago, and who knows if some will do really good some time.

At least my point is that those competitions for sure are attracting people to study mathematics as a carrier, and making some kids to really discover their skills, interest and potential.

[quote="zeb"]Here is something that is happening to me:

My sister is quite good at math, all of her teachers tell her that she has great potential, she is in a fairly good school and is practically the top in her class (intelligence wise.. i don't know about her grades), etc. And yet she says that she doesn't see the point in devoting any time to it. She had the oppurtunity to skip class and take the amc-8, for free, and she didn't want to.

This may not mean anything.[/quote]

Well, as smbelcas suggest, many people that enjoy studying mathematics never participate in such contest, and many of them just because they are not interested!, There are thousands of mathematicians (I would say that more than 90% of the total, at least) that do research but never participate in such contest, even when they are young enough to have participate in them (well, in Hungary, that "young enough" means that they were born after 1880 :P ).

On the other hand, one of the most mathematically gifted minds that I have seen (a kid that with 15 years had already tought himself a lot of complex analysis, topology and similar) when finally got into college decided to study biology, and now live happy doing research about whatnot in the jungles of all Central America.

She may have the gift, and you may suggest the path to follow, but at the end is her decision what to do with that gift!

I've known a few female math nuts, but they were few and far between.  I think, in Zeb's case, his sister might be a little intimidated by his level of achievement-  I am, and I'm twenty years older than him and pretty versed, myself.

On top of that, there's the old cultural argument that women's skill sets are determined from an early age, the way boy's sets are, and that this is largely a sex-role assignment, no matter how liberal the parents are (and we were pretty friggin' liberal).

The question of why ANYONE can be interested and good at math and science is answered by Polanyi in "Personal Knowledge" after the section on articulation.  This is a good book on many fronts, but his explanation of the importance of the perception of truth as being beautiful in terms of why it fascinates scientists or mathematicians to first sight some new truth-  it's akin to mountain climbing.   Because it's there, because it's challenging, because it's something new.

Why some are interested in it more than others, and can dedicate themselves to it-  I think that's too broad a question.  My experience is that math fascinates most people, until they hit some area that they don't understand well, at which point they either bugger off, or decide to come back to it later, and never do.  Those who do well with it are either personally motivated to overcome their lack of understanding very strongly, or they have someone or someones to talk to who do understand the subjects and who are passionate about them to keep them interested.  And a lot of it has to do with how questions are answered when you are young.  If your parents or teachers tended to answer questions with "because", rather than with real answers, interest is probably blunted.  In my case, I always tried to answer every question to the very best of my ability, and to make sure that books were available or problems were created that were within the scope of my children's interest (before they were severed from me).

Hope this gives thoughts!


[quote="keta"]Prove that no three odd integers greater than 3 are all prime numbers.

I can't figure out which method to use to prove this statement (contradiction), contrapositive, etc... without providing a counterexample.

Here's my work:

For contradiction, assume there does exist three odd integers greater than 3 that are all prime numbers. Then, all three must be relatively prime. Then I can't think of what to write next. Should I mention the that the only factors of primes are 1 and the number itself?

Can someone help? Thanks.[/quote]
im making the assumption youi mean consecutive odd integers in your original statement; otherwise, we can have 5, 7, and 11.
Think about modulos and which one could be used best here

[hide]
$n>3$ and $n\in N$
If n is a prime number, we want to prove that $n+2$ or $n+4$ is not a prime number.

Because $n$ is an odd prime number greater than $3$, it is not divisible by $2$ or $3$.
So that tells us that $n+1$ or $n+2$ is divisible by 3.
If $n+1$ is divisible by 3, then $n+4$ is divisible by $3$ also.
So if $n$ is a prime number, $n+2$ or $n+4$ is divisible by $3$, and is not a prime number. So no $3$ consecutive odd integers greater than $3$ can all be prime numbers.
QED[/hide]

Multiples of 3.  As Jan said, one of n, n+2 and n+4 must be a multiple of 3.  You know this because they are all different mod 3.  And if each is >3, then none can be 3.

I posed this problem a while back on the forums :D  You don't really have to specify that they are odd in the problem statement, btw, but obviously three primes each two apart can't be even...

You can use contradiction.  One way would be as follows:

[hide]
Assume that there exists an $n>3$ such that $n$, $n+2$, and $n+4$ are all prime.  Therefore, since $n>3$, $n$ cannot be divisible by 3 since it is prime.  So $n\equiv \pm 1\mod 3$.

If $n\equiv 1\mod 3$ then $n+2\equiv 0 \mod 3$ which contradicts the assumption that $n+2$ is prime.

Similarly if $n\equiv -1\mod 3$ then $n+4\equiv 0 \mod 3$ which contradicts the assumption that $n+4$ is prime.

Thus our assumption wrong and there does not exist an $n>3$ such that $n$, $n+2$, and $n+4$ are all prime. QED[/hide]

Whoa, never considered using divisiblity or mods. I guess from this view, you can prove the statement for a general case. I initially tried finding a counterexample but that doesn't constitute  a full proof. Thanks everyone.

[quote="Palytoxin"][quote="joml88"]
[hide]
Assume that there exists an $n>3$ such that $n$, $n+2$, and $n+4$ are all prime.  Therefore, since $n>3$, $n$ cannot be divisible by 3 since it is prime.  So $n\equiv \pm 1\mod 3$.

If $n\equiv 1\mod 3$ then $n+2\equiv 0 \mod 3$ which contradicts the assumption that $n+2$ is prime.

Similarly if $n\equiv -1\mod 3$ then $n+4\equiv 0 \mod 3$ which contradicts the assumption that $n+4$ is prime.

Thus our assumption wrong and there does not exist an $n>3$ such that $n$, $n+2$, and $n+4$ are all prime. QED[/hide][/quote]
hmm? this is the same idea as Jan's solution...[/quote]

I am quite aware of that.  The purpose of my post was to write a proof that clearly used contradiction.

A counterexample wouldn't work.  The only way it would work is if you found three numbers $n$, $n+2$ and $n+4$ ($n>3$) that were all prime thus proving the original statement in the problem wrong.  Or if the problem were worded differently as saying that prove for [i]all[/i] $n$ that $n$, $n+2$, and $n+4$ are always prime.


A set $P$ of $2002$ persons is given. The family of subsets of $P$ containing exactly $1001$ persons has the property that the number of acquaintance pairs in each such subset is the same. (It is assumed that the acquaintance relation is symmetric). Find the best lower estimation of the acquaintance pairs in the set $P$.

Mmmh.... I'm afraid there has been a worsening in my brain desease.
Number the vertices of the associated graph from $1$ to $2002$ (set which I will call: $\mathbb{N}_{2002}$) in such a way that their corresponding degrees satisfy $d_{1}\leq d_{2}\leq \ldots\leq d_{2002}$. Suppose $d_{i}\neq d_{j}$ for some $1\leq i<j\leq 2002$; then if given a set of vertices $X\subset \mathbb{N}_{2002}$ we call $E(X)$ the set of all edges $\{ i,j \}\in X$  (giving for granted $i\neq j$), we get:
$|E(\{ i/ \;1\leq i\leq 1001 \}|=\frac{\sum_{i=1}^{1001}d_{i}-|\{ \; \{i,j\}/\;1\leq i\leq 1001\;\;1002\leq j\leq 2002 \}|}{2}<$
$<\frac{\sum_{i=1002}^{2002}d_{i}-|\{ \; \{i,j\}/\;1\leq i\leq 1001\;\;1002\leq j\leq 2002 \}|}{2}=$
$=|E(\{ i/ \;1002\leq i\leq 2002 \}|$;
which contradicts the hypothesis conditions. So $d_{i}=k\;\;\forall i\in \mathbb{N}_{2002}$ for some $k\in \mathbb{N}_{2001}$.
Now take any two vertices $i\neq j$; wheather they are joined or not, each one must be joined with the same number of vertices outside the pair, so suppose there are $a$ other vertices joined with both, $b$ vertices joined with none, and for each of the two, $c$ vertices wich are joined just with it. If we have $c>0$ then it is possible to choose a set $Y$, $|Y|=1001$ of vertices distinct from $i,j$ so that $|\{ \;\{ i,k \}/\;k\in Y\}|\neq |\{ \;\{ j,k \}/\;k\in Y\}|$ which means that given a certain $l\in Y$ there exists a $m\in \{ i,j \}$ so that by interchanging $l$ and $m$ in $Y$ we alter $|E(Y)|$, which, again contraddicts the hypothesis. So $c=0$ for every $i\neq j$.
Now take, $i',j'$ so they are joined (at least one such a pair must exist, if we want to avoid the trivial case where all people are mutual strangers) and suppose for this pair $b>0$. If that is the case we can pick a vertex $v$ which is not joined with $i'$ nor $j'$. Now, because of the previous general result $\forall w\in \mathbb{N}_{2002},w\not \in \{i',j',v \}\;\;[\{ v,w \}\in E(\mathbb{N}_{2002})\leftrightarrow \{ i',w \}\in E(\mathbb{N}_{2002})]$ and, since $\{ i',j'\}\in E(\mathbb{N}_{2002})$, $k=a+1$, but, however $k=d_{v}=a$: a contraddiction. So $c=b=0$, which implies all vertices are joined to both $i'$ and $j'$, $k=2001$, $|E(\mathbb{N}_{2002})|$=$\binom{2002}{2}$.

so person is vertex and acquaintance is edge. show graph is regular, then complete...or empty


In the finite sequence $ 7,8,6,5,10,\dots, 0$, the odd-numbered terms decrease by 1, and the even numbered terms increase by 1. The last term in the sequence is 0. What is the sum of the terms?

Err?? WRONG PROBLEM? Doesn't make sense man.

I know it doesn't make sense :P

From original:

[img]http://img187.imageshack.us/img187/4884/1997statesprint15ak8.gif[/img]

[quote="cognos599"]was this photoshopped, cuz i remember doing this problem and it was like 7,8,6,9,5,10.....[/quote]
If it's like that, then it's simple to brute force :P


geometry
incenter
cyclic quadrilateral
geometry proposed

$ABCD$ is a cyclic quadrilateral with $BC=CD$. The diagonals $AC$ and $BD$ intersect at $E$. Let $X,Y,Z,W$ be the incenters of triangles $ABE,ADE,ABC,ADC$ respectively. Show that $X,Y,Z,W$ are concyclic iff $AB=AD$.

Since $\triangle AXB\sim\triangle AWC$ and $\triangle AYD\sim\triangle AZC$, we have respectively $\frac{AX}{AW}=\frac{AB}{AC}$ and $\frac{AY}{AZ}=\frac{AD}{AC}$. So
$X, X, Z, W \text{are concyclic}$
$\Leftrightarrow AX\cdot AZ=AY\cdot AW$
$\Leftrightarrow \frac{AX}{AW}=\frac{AY}{AZ}$
$\Leftrightarrow \frac{AB}{AC}=\frac{AD}{AC}$
$\Leftrightarrow AB=AD$


Any reason why there is a rule for consecutive shouts?

I find it's a problem in less-viewed blogs and I can't leave the message I want because no one has shouted other than me. If it was meant to reduce flooding/spam, then I would suggest a temporal aspect?

Or disable/enable consecutive shouts, based on user opinion

I can add it to the wishlist, though it's not going to happen any time soon. My schedule is full.


Try to use the fact that $\prod_{p prime}{\frac{1}{1-p^{-k}}}=zeta(k)$ ou $zeta(k)=\sum_{n\geq 1}{1/n^k}$. Then use the fact that $zeta(2)=\frac{\pi^2}{6}$ and $zeta(4)=\frac{\pi^4}{90}$. Next, compute $\frac{zeta(2)^2}{zeta(4)}$ in two different ways.


MATHCOUNTS
search
blogs
geometry
3D geometry
pyramid
algebra

What is the coefficient of the term $ x^3y^5z^2$ in the expansion of $ (x \plus{} y \plus{} z)^{10}$.

Is there anything like the trinomial theorem to solve this, or do you just have to bash?

The answer is [hide="answer"]2520[/hide] btw

For $ x^3y^5z^2$, count how many ways there are to choose $ 3,5,2$ of $ x,y,z$ respectively from $ 10$ total. Now try to find the coefficient of $ x^ay^bz^c$ from $ (x\plus{}y\plus{}z)^n$, and then generalize to more variables.

... whence the [color=red]multinomial [/color] coefficients $ \genfrac {(} {)} {0pt} {} {n} {a_1,a_2,\ldots,a_k} \equal{} \frac {n!} {a_1!a_2!\cdots a_k!}$, with $ a_i$ non-negative integers having $ \sum_{i\equal{}1}^k a_i \equal{} n$.
So $ (x_1 \plus{} x_2 \plus{} \cdots \plus{} x_k)^n \equal{} \sum_{a_1\plus{}a_2\plus{}\cdots\plus{}a_k\equal{}n} \frac {n!} {a_1!a_2!\cdots a_k!} x_1^{a_1}x_2^{a_2}\cdots x_k^{a_k}$.

[quote="math154"]For $ x^3y^5z^2$, count how many ways there are to choose $ 3,5,2$ of $ x,y,z$ respectively from $ 10$ total. Now try to find the coefficient of $ x^ay^bz^c$ from $ (x \plus{} y \plus{} z)^n$, and then generalize to more variables.[/quote]

Darnit Im terrible at deciphering math-related clues. Could you please expand on that?

And I have no idea what mavropnevma said, but judging by all the $ \sum$'s, this doesn't look Mathcounts level. Ah well.

The multinomial theorem is derived the same way as the binomial theorem is, and once you understand it, it's really not that complicated.

In $ (x\plus{}y)^n$, the coefficient of $ x^ay^b$ where $ a\plus{}b\equal{}n$ is $ \binom na\equal{}\binom nb\equal{}\frac{n!}{a!b!}$. Why? Because each of the $ n$ factors $ x\plus{}y$ of $ (x\plus{}y)^n$ either contribute an $ x$ or a $ y$ to the product. Hence it is equivalent to arranging $ a$ x's and $ b$ y's in a row: an $ x$ being in the $ k^{th}$ position means that the $ k^{th}$ factor $ x\plus{}y$ in $ (x\plus{}y)^n$ contributes an $ x$ to the product and a $ y$ being in the $ i^{th}$ position means that the $ i^{th}$ factor $ x\plus{}y$ in $ (x\plus{}y)^n$ contributes a $ y$ to the product. The number of ways to arrange these x's and y's is simply $ \frac{n!}{a!b!}$.

Using essentially the same logic, finding the coefficient of $ x^3y^5z^2$ in $ (x\plus{}y\plus{}z)^n$ is equivalent to arranging $ 3$ x's, $ 5$ y's, and $ 2$ z's in a line. How many ways are there to do this?

Now it should be clear how to generalize.

Ohhhhh okay thanks math154 I understand it now :lol:

and now I get what mavropnevma was saying.

EDIT: so I get his part of the summation which is what math154 was saying and could generalize $ (x_1\plus{}x_2\plus{}\dots \plus{}x_y)^n$: $ \boxed{\sum_{a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{k} \equal{} n}\frac {n!}{a_{1}!a_{2}!\cdots a_{k}!}}{x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{k}^{a_{k}}}$ but what does this part mean:$ \sum_{a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{k} \equal{} n}\frac {n!}{a_{1}!a_{2}!\cdots a_{k}!}\boxed{x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{k}^{a_{k}}}$

Hm? $ \frac{n!}{a_1!a_2!\cdots a_k!}$ is the coefficient of $ x_1^{a_1}x_2^{a_2}\cdots x_k^{a_k}$ so each term is of the form $ \frac{n!}{a_1!a_2!\cdots a_k!}x_1^{a_1}x_2^{a_2}\cdots x_k^{a_k}$ where $ a_1\plus{}a_2\plus{}\cdots\plus{}a_k\equal{}n$.

[quote="AlphaBetaTheta"]there's no such thing as trinomial theorem[/quote]

There is sort of a trinomial theorem, although much less known than the Binomial Theorem. Both are generated from Pascal's Triangle or variant.

But most of the time, this is way beyond MATHCOUNTS, so don't worry too much about this.:)

I made a random google search, and came across
[url=http://www.staff.washington.edu/~znwu/tritheo/theorem.htm]this[/url]
i'm not sure if it's any help, though

Yep thats basically what math154 was saying.

I posted a [i]blog[/i] entry on my [i]blog[/i] if anyone would like to visit my [i] blog[/i] to find out more about this topic. </shameless plug>

But most of the time, this is way beyond MATHCOUNTS, so don't worry too much about this.:)[/quote]

Yeah I think I said that to myself last year. Oops!

[quote="PowerOfPi"]Also try [url=http://en.wikipedia.org/wiki/Pascal%27s_pyramid]Pascal's Pyramid[/url]. Similar to how Pascal's triangle relates to the Binomial Theorem, Pascal's Pyramid helps with trinomials.[/quote]

That's what I meant earlier :P but is there a general formula for the expansion of $ (a_1\plus{}a_2\plus{}a_3\plus{}\cdots\plus{}a_n)^{k}$ in terms of the coefficients and $ k$


Hi, I need help with this problem http://acm.timus.ru/problem.aspx?space=1&num=1428. To summarize the description of the problem:
We are given 1<=A,B,C<=32 such that A % (C-1) = B % (C-1) = 0 and we are searching for 1<=X,Y,Z<=10^50, such that X^A + Y^B = Z^C thanks in advance for any help.

How about ${(X,Y,Z)=(2^\frac{C-1}{A},2^\frac{C-1}{B},2})$ :)
(Oh, and you had the mod constraints the wrong way around.)


modular arithmetic
number theory unsolved
number theory

Given a fixed $ N\equal{}pq$ and a known $ (a,b)$ and $ (c,d)$ pairs s.t.

$ g^{ab}\equiv g\pmod{pq}$ and $ g^{cd}\equiv g\pmod{pq}$

and $ ab\equiv 1\pmod{(p\minus{}1)(q\minus{}1)}$ and $ cd\equiv 1\pmod{(p\minus{}1)(q\minus{}1)}$ with the value $ (p\minus{}1)(q\minus{}1)$ unknown.

How do we use $ a,b,c,$ and $ d$ to find out the value of $ (p\minus{}1)(q\minus{}1)$ or $ p\plus{}q$?

For it you must factored $ N \equal{} pq$ (or find $ p,q$). But period is $ lcm(p \minus{} 1,q \minus{} 1) < (p \minus{} 1)(q \minus{} 1)$.

i don't understand... actually, i'm trying to determine the value of p+q to factor N=pq


Θα μπορουσε καποιος να μου εξηγησει τον μηχανισμο αποδηξης των παρακατω ασκησεων????
Nα δειξετε οτι
$ 1) \binom{n}{0}^2\plus{}\binom{n}{1}^2\plus{}...\plus{}\binom{n}{n}^2\equal{}\binom{2n}{n}$
Χρεισημοπιωντας την ιδια μεθοδο με τo $ \boxed 1$ να δειξετε οτι εαν $ p,r \in N, p \leq m,p \leq n,r\leq n$
$ 2) \binom{n}{0} \cdot \binom {m}{p}\plus{}\binom{n}{1} \cdot \binom{m}{p\minus{}1}\plus{}...\plus{}\binom{n}{p} \cdot \binom {m}{0}\equal{}\binom{m\plus{}n}{p}$
$ 3) \binom{n}{o} \cdot \binom{n}{r}\plus{} \binom{n}{1} \cdot \binom{n}{r\plus{}1}\plus{}...\plus{}\binom {n}{n\minus{}r} \cdot \binom{n}{n}\equal{}\frac{(2n)!}{(n\minus{}r)!(n\plus{}r)!}$
Πειμενω μια κατατοπιστικη απαντηση :)

[quote="Dimitris X"]Θα μπορουσε καποιος να μου εξηγησει τον μηχανισμο αποδηξης των παρακατω ασκησεων????
Nα δειξετε οτι
$ 1) \binom{n}{0}^2 \plus{} \binom{n}{1}^2 \plus{} ... \plus{} \binom{n}{n}^2 \equal{} \binom{2n}{n}$
Χρεισημοπιωντας την ιδια μεθοδο με τo $ \boxed 1$ να δειξετε οτι εαν $ p,r \in N, p \leq m,p \leq n,r\leq n$
$ 2) \binom{n}{0} \cdot \binom {m}{p} \plus{} \binom{n}{1} \cdot \binom{m}{p \minus{} 1} \plus{} ... \plus{} \binom{n}{p} \cdot \binom {m}{0} \equal{} \binom{m \plus{} n}{p}$
$ 3) \binom{n}{o} \cdot \binom{n}{r} \plus{} \binom{n}{1} \cdot \binom{n}{r \plus{} 1} \plus{} ... \plus{} \binom {n}{n \minus{} r} \cdot \binom{n}{n} \equal{} \frac {(2n)!}{(n \minus{} r)!(n \plus{} r)!}$
Πειμενω μια κατατοπιστικη απαντηση :)[/quote]

Gia tin 1 pare $ (1\plus{}x)^{n}\cdot (1\plus{}x)^{n}$

[quote="Dimitris X"][quote="Γιώργος"][quote="Dimitris X"]Θα μπορουσε καποιος να μου εξηγησει τον μηχανισμο αποδηξης των παρακατω ασκησεων????
Nα δειξετε οτι
$ 1) \binom{n}{0}^2 \plus{} \binom{n}{1}^2 \plus{} ... \plus{} \binom{n}{n}^2 \equal{} \binom{2n}{n}$
Χρεισημοπιωντας την ιδια μεθοδο με τo $ \boxed 1$ να δειξετε οτι εαν $ p,r \in N, p \leq m,p \leq n,r\leq n$
$ 2) \binom{n}{0} \cdot \binom {m}{p} \plus{} \binom{n}{1} \cdot \binom{m}{p \minus{} 1} \plus{} ... \plus{} \binom{n}{p} \cdot \binom {m}{0} \equal{} \binom{m \plus{} n}{p}$
$ 3) \binom{n}{o} \cdot \binom{n}{r} \plus{} \binom{n}{1} \cdot \binom{n}{r \plus{} 1} \plus{} ... \plus{} \binom {n}{n \minus{} r} \cdot \binom{n}{n} \equal{} \frac {(2n)!}{(n \minus{} r)!(n \plus{} r)!}$
Πειμενω μια κατατοπιστικη απαντηση :)[/quote]

Gia tin 1 pare $ (1 \plus{} x)^{n}\cdot (1 \plus{} x)^{n}$[/quote]
Μπορεις να γραψεις αναλυτικα την λυση για το 1???[/quote]

Εχεις $ (1 \plus{} x)^{2n} \equal{} x^2n \plus{} \binom{2n}{1}x^{2n \minus{} 1} \plus{} ... \plus{} 1 \equal{} P(x)$
Και
$ (1 \plus{} x)^{2n} \equal{} (1 \plus{} x)^n(x \plus{} 1)^n \equal{} (x^n \plus{} \binom{n}{1}x^{n \minus{} 1} \plus{} ... \plus{} 1)(x^n \plus{} \binom{n}{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} 1) \equal{} Q(x)$
Κανεις τις πραξεις στο $ Q(x)$ και αφου $ P(x) \equal{} Q(x)$ για καθε $ x$, εξισωνεις τον συντελεστη του $ x^n$ στο $ P$ με το συντελεστη του $ x^n$ στο $ Q$ και παιρνεις τη ζητουμενη σχεση

Για το (2) μπορείς είτε να κοιτάξεις το $ (1\plus{}x)^m(1\plus{}x)^n$ ακριβώς όπως εξηγούν ο Γιώργος και ο Νίκος είτε να δουλέψεις με τον ακόλουθο τρόπο:

Έχουμε m αγόρια και n κορίτσια και θέλουμε να φτιάξουμε μια ομάδα με ακριβώς p άτομα. Αυτό γίνεται μα ακριβώς $ \binom{m\plus{}n}{p}$ τρόπους. Η ομάδα θα έχει είτε 0 αγόρια και p κορίτσια, είτε 1 αγόρι και p-1 κορίτσια, είτε ... Αυτό γίνεται με $ \binom{m}{0}\binom{n}{p} \plus{} \binom{m}{1}\binom{n}{p\minus{}1} \plus{} \cdots$


I have the file "graph01.bmp" in the same folder as the .tex file with the following code:

[quote]\documentclass{article}
\usepackage{graphicx}
\begin{document}
Testing BMP inclusion...\\
\includegraphics[width=2in,height=2in]{graph01.bmp}\\
\end{document}[/quote]

When I run latex, the error message:

[quote]LaTex Error:  File "graph01.bb" not found[/quote]

Could anyone help me to solve the problem? The funny thing is, the error message ask for a "graph01.bb" file which is not of the same extension.
I did try another file "myimage.png" but the error message ask for a "myimage.bb" file?

You cannot use bmp files, only png, jpg or gif, so convert it first to one of these formats. And you must be compiling to pdf not dvi.

LaTeX is looking for a bounding box which isn't available in graph01.bmp or myimage.png (it is there in some pngs). So it looks for the details in a .bb file which it can't find. The solution is to give the dimension of the bounding box when including the file. For example
\includegraphics[bb=0 0 100 100]{myimage.png}

Could it be a problem of my version of  WinEdt?

I have made the changes but the error message comes up:

[quote]Paclage graphics warning:  pdftex option does not work with standard TEX[/quote]

[quote]Package pdftex.def warning:  Option 'bb' does not make sense[/quote]

 :?:

Yes, sorry! In fact it says what to do in the line after the warning and that is to use viewport (which uses relative coordinates). So try
\includegraphics[viewport=0 0 100 100]{myimage.png}
You will need to experiment with the dimensions.

If you are using MikTeX you can use the bb.exe provided with pdflatex, you will find it in [b]..\texmf\miktex\bin\ebb.exe[/b] this small soft create the bounding box file *.bb  in the same folder,  to do this write in your command prompt:
[code] ebb filename[/code]

You can also [url=http://www.imagemagick.org]imagemagik[/url] converter to convert to *.eps, with this syntax:
[code]convert filename  newfilename.xxx[/code]
When xxx= jpg, eps,.....


Find the largest positive integer n with the following property: One can
choose $ 2007$ distinct integers from the interval $ [2\cdot10^{n\minus{}1}, 10^{n})$ such that,
for any $ i, j$ with $ 0 < i < j < n\plus{}1$, among the selected numbers there is one has a decimal represantation
$ a_{1}a_{2}. . . a_{n}$ with $ a_{j}< a_{i}\plus{}3$.

For $ n\equal{}4$ there exist 2007 numbers each of which violates the required property:
namely, it is enough to take numbers $ a_1a_2a_3a_4$ with $ a_1\plus{}3\leq a_4$ (there are 1500 such numbers, and numbers with $ a_1\in\{7,8,9\}$ and $ a_2 \plus{}3\leq a_4$ (there are 630 of them).
On the other hand, for $ n\equal{}3$ the total number of numbers in the interval $ [2\cdot10^{n\minus{}1}, 10^{n})$ is 800 and 2007 numbers cannot be selected. Therefore, the required $ n$ does not exist.


vector
calculus
geometry
trigonometry
analytic geometry
graphing lines
slope

vecor calculus doesn't belong in this forum. it should be in college playground.

I do not think that this topic is appropriate for the Getting Started Forum.

Neither do I. Moved from Getting Started.

This is actually a vector geometry question, not a calculus question. When the boats meet, boat A and boat B must be equally far from their starting points. That's an isosceles triangle with known base and known base angle, which is enough information to construct it. Once you find the third vertex, you can calculate the distance and the time.

[quote="JesusFreak197"]This problem can't be solved. There are two variables - the time $t$ that they meet, and the velocity of boat $B$ - but we need to know one to find the other.[/quote]
Thats what my teacher was saying as he was trying to solve it.

As for the thing that it isn't calculus, well I'm not too sure myself as it came under the heading 'vector calculus' in my math textbook.

Sorry for posting this on getting started, I didn't know the level this was supposed to be at :(

[quote="chillypacman"]I did this one a while ago, my teacher said the question was quite 'interesting' which is his way of saying a little bit harder than the other questions. But it is quite a cool question in my opinion.
My teacher done the question on the board, but I never saw the way he actualy done it. THe working out I have is nothing but a whole heap of scribbles here and there that somehow end up with the correct answer...[/quote]

With boat A starting from (1,3) and boat B starting from (2,7) moving at the same speed, $|3\hat{i}+5\hat{j}|$, a) How long until the boats meet? b) What is the velocity of boat B?

a) The vector difference in the boats's initial positions is (2-1,7-3) = (1,4) = $\vec{A}$.

Since the boats are moving at the same speed, $v=\sqrt{34}$, the difference in their initial positions, and their velocity x time vectors form an isosoles triangle.

$A^2 +(vt)^2-2Avt \cos \theta =(vt)^2$

$A = 2vt \cos \theta$

$t = \frac{A}{2v \cos \theta}$

$\arctan \frac{1}{4}+\theta +\arctan \frac{5}{3} = 90^o$

$\theta = 16.93^o \Rightarrow \cos \theta = 0.9567$

$\boxed{t = \frac{\sqrt{17}}{2\sqrt{34} (0.9567)} = 0.3696 \text{ hrs.}}$

Boat A's final position = $ 1\hat{i}+3\hat{j} + (3\hat{i}+5\hat{j})(0.3696)= 2.1087 \hat{i} +4.8478 \hat{j}$

Boat A's final position = Boat B's final position.

$\boxed{\vec{v}_B= \frac{(2.1087 -2) \hat{i} + (4.8478 -7) \hat{j}}{0.3696}= 0.2941 \hat{i} -5.8235 \hat{j} \text{ kph}}$

[quote="jmerry"]That's not wrong, but almost everything should be rational.[/quote]

Why? When dealing in triangles one encounters radicals more often than not, and radicals are seldom rational.

[quote="jmerry"]The reason everything should be rational is that every line involved has rational slope, and passes through rational points. Since $90^\circ$ rotations and reflections across rational lines have rational coefficients, every key point is rational. Finding the intersection should be done by solving a system of linear equations. You don't even need to invoke distance formulas in the last step; even the time comes out rational.

This wasn't rigged to be nice, so you get a denominator of 46 in the key point's coordinates, and in the time taken.[/quote]

That's clearly not the case. I solved it using the law of cosines. The distance between the initial positions of the boats is $\textstyle\sqrt{17}$, not a rational number. The speed of either boat is $\textstyle\sqrt{34}$. When one divides this into $\textstyle\sqrt{17}$, as one must there is a factor of $\textstyle\frac{1}{\sqrt{2}}$ floating around in the mix; again not a rational number.

The upshot is unless I analyzed the problem completely wrong - something not beyond the realm of possibility I do admit - the answers are not rational numbers.

The time taken is precisely $\frac{17}{46}$, as you can verify yourself.


[quote="quykhtn-qa1"]Let$ a,b,c$ be the sides of a triangle.Prove that:
$ \frac {\sqrt {b \plus{} c \minus{} a}}{\sqrt {b} \plus{} \sqrt {c} \minus{} \sqrt {a}} \plus{} \frac {\sqrt {c \plus{} a \minus{} b}}{\sqrt {c} \plus{} \sqrt {a} \minus{} \sqrt {b}} \plus{} \frac {\sqrt {a \plus{} b \minus{} c}}{\sqrt {a} \plus{} \sqrt {b} \minus{} \sqrt {c}} \le 3$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130812


calculus
inequalities
inequalities proposed

Given $ a, b, c$ are the [b]real[/b] numbers satisfy $ a\plus{}b\plus{}c\equal{}3$. Prove that:
\[ 3\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \ge ab^2\plus{}bc^2\plus{}ca^2\plus{}ab\plus{}bc\plus{}ca\]
 :)

[quote="nguoivn"]Given $ a, b, c$ are the [b]real[/b] numbers satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ 3 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \ge ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} ab \plus{} bc \plus{} ca\]
:)[/quote]

It is equivalent to :

$ \sum a^4 \plus{}18 \sum a^2b^2 \plus{}\sum a^3b\minus{}8 \sum ab^3 \minus{}12 \sum a^2bc \geq 0$

Which is true :)

[quote="Materazzi"][quote="nguoivn"]Given $ a, b, c$ are the [b]real[/b] numbers satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ 3 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \ge ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} ab \plus{} bc \plus{} ca\]
:)[/quote]

It is equivalent to :

$ \sum a^4 \plus{} 18 \sum a^2b^2 \plus{} \sum a^3b \minus{} 8 \sum ab^3 \minus{} 12 \sum a^2bc \geq 0$

Which is true :)[/quote]
Yes, we can complete the proof by Can_hang's lenma.
Besides, the original problem also exists an easy proof by Am-Gm (not to use any calculus)  :)

[quote="nguoivn"]Given $ a, b, c$ are the [b]real[/b] numbers satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ 3 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \ge ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} ab \plus{} bc \plus{} ca\]
:)[/quote]
The inequality is equivalent to
$ 9\minus{}3\sum ab \plus{}3\sum a^2b^2\minus{} 9abc \ge 3\sum ab^2\minus{}9abc.$
Note that
$ 9\minus{}3\sum ab \equal{}(a\plus{}b\plus{}c)^2\minus{}3\sum ab\equal{} \frac{1}{2}\sum (a\minus{}b)^2,$
$ 3\sum a^2b^2\minus{}9abc\equal{}3\sum a^2b^2\minus{}3abc\sum a\equal{}\frac{1}{2}\sum (2ca\minus{}ab\minus{}bc)^2.$
Therefore, by applying the AM-GM Inequality, we can see that the left hand side of the above inequality is not less than
$ \sum (a\minus{}b)(2ca\minus{}ab\minus{}bc)\equal{}3\sum ca(a\minus{}b)\minus{}(ab\plus{}bc\plus{}ca)\sum (a\minus{}b)\equal{}3\sum ca(a\minus{}b)\equal{}3\sum ab^2\minus{}9abc,$
which is equal to the the right hand side.
The proof is completed.


I've seen people talk about "Simon's Favorite Factoring Trick".

What is it?? :huh:

http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick

[quote]The general statement of SFFT is: ${xy}+{xk}+{yj}+{jk}=(x+j)(y+k)$. Two special cases appear most commonly: $ xy+x+y+1 = (x+1)(y+1)$ and $ xy-x-y+1 = (x-1)(y-1)$.[/quote]

Videos such as

[url]http://www.artofproblemsolving.com/Movies/simon1.php[/url]

and

[url]http://www.artofproblemsolving.com/Movies/simon2.php[/url]

help visually


Solve the system equations $ x(x\plus{}y)^{2}\equal{}9$ and $ x(y^{3}\minus{}x^{3})\equal{}7$

From the first eq. $ y\equal{}(3/\sqrt{x}\minus{}x)$(x>0) and the second $ y\equal{}(7/x\plus{}x^{3})^{1/3}$ then
$ (3/\sqrt{x}\minus{}x)^{3}\equal{}7/x\plus{}x^{3}\implies{f(x)\equal{}2x^{9/2}\minus{}9x^{3}\plus{}27x^{3/2}\plus{}7x^{1/2}\minus{}27\equal{}0}$ f(x) is increasing and so has one root for example in the intervall [0,1] with $ x\equal{}1$.

its easily to see that $ x > 0$ and $ x < y$ ,now let $ f(x) =\frac{3}{\sqrt{x}}-x$and $ g(x) = (\frac{7}{x}+x^{3})^{\frac{1}{3}}$.
for $ 0 < x < 1$ we have $ f(x) >\frac{3}{\sqrt{x}}-\frac{1}{\sqrt{x}}=\frac{2}{\sqrt{x}}$ ang $ g(x) < ({\frac{7}{x}+\frac{1}{x})^{1/3}=\frac{2}{x^{1/3}}}$ so that $ g(x) < f(x)$ since $ \sqrt{x}< x^{\frac{1}{3}}$.
if$ x > 1$ we have $ g(x) > f(x)$.hence the only solution is $ x = 1$.


function
topology
real analysis
real analysis unsolved

In a metric space lets have two closed subsets A and B , such that A cross B is connected.
Take a continuous function f  on A , and a continuous funtion g on B.
Is it possible to construct a continuous function , say h , on (A  union B) such that its testriction to A and B are f and g respectivly.

I suppose that $ f \equal{} g$ on $ A\cap B$?

Unless I misunderstand the problem, there is nothing to "construct" here: $ h$ is already defined on $ A\cup B$. Is it continuous? Consider a sequence $ x_{n}\to x$ in $ A\cup B$. Split it into two subsequences: one in $ A$ and one in $ B$. Using the assumption that $ A$ and $ B$ are closed, conclude that $ f(x_{n})\to f(x)$.

I don't know why you asked for $ A\cap B$ to be connected.

to be more specific:f is a fuction from A to {0,1} ,g is a function from B to {0,1} ;both are continuous. A CROSS B is connected.A and B are closed.
Can we construct a functin h on A union B such that h equals f when restricted to A and g when restricted to B.
I am using the fact that a set S is connected is the same as saying tht every continuous function from S to {0,1} is constant ;this is to solve the following probem: if A and B are closed , AcrossB and AunionB are connected then so are A and B;
[/code]

This doesn't make sense. There's nothing to construct- you already have a definition, with no choices left.

[quote]this is to solve the following probem: if $ A$ and $ B$ are closed, $ A\cap B$ and $ A\cup B$ are connected then so are $ A$ and $ B$[/quote]
Now this makes sense.


Tow circles intersect at $A$ and $B$. Through $A$ draw a line perpendicular to $AB$ and meets the two circles at $C$ and $D$. Through $A$ draw another line which meets the two circles at $P$ and $Q$. Show that $PQ \leq CD$.

Let $\angle{PQB}=\alpha, \angle{BCD}=\beta, \angle{QAB}= \gamma$ then $\angle{PAB}=180^{o}-\gamma$
We have $\angle{PQB}= \angle{BCD}=\alpha$, $\angle{QPB}=\angle{CDB}=\beta$. Then $\angle{QBP}=\angle{CBD}=180^{o}-(\alpha+\beta)$
By Sine Law in $\triangle{QAB}$, get:
$QB=\frac{ABsin{\gamma}}{sin{\alpha}}$
By Sine Law in $\triangle{BAP}$, get:
$PB=\frac{ABsin(180^{o}-\gamma)}{sin{\beta}}= \frac{ABsin{\gamma}}{sin{\beta}}$
By Sine Law in $\triangle{CAB}$, get:
$CB=\frac{ABsin{90^{o}}}{sin{\alpha}}$
By Sine Law in $\triangle{DAB}$, get:
$DB=\frac{ABsin{90^{o}}}{sin{\beta}}$
By Cosine Law in $\triangle{QPB}$, get:
$QP=\sqrt{QB^{2}+PB^{2}-2QB(PB)cos(PBQ)}$
$=\sqrt{\frac{AB^{2}sin^{2}{\gamma}}{sin^{2}{\alpha}}+\frac{AB^{2}sin^{2}{\gamma}}{sin^{2}{\beta}}+\frac{2 cos(180^{o}-(\alpha+\beta))AB^{2}sin^{2}{\gamma}}{ sin{\alpha}sin{\beta }}}$
$=\sqrt{sin^{2}{\gamma}*(\frac{AB^{2}}{sin^{2}{\alpha}}+\frac{AB^{2}}{sin^{2}{\beta}}+\frac{2cos(180^{o}-(\alpha+\beta)) AB^{2}}{sin{\alpha}sin{\beta}}}$
By Cosine Law in $\triangle{CDB}$ get:
$CD=\sqrt{CB^{2}+DB^{2}-2CD(DB)cos(CBD)}$
$CD=\sqrt{\frac{AB^{2}sin^{2}{90^{o}}}{sin^{2}{\alpha}}+\frac{AB^{2}sin^{2}{90^{o}}}{sin^{2}{\alpha}}+\frac{2 cos(180^{o}-(\alpha+\beta))AB^{2}sin^{2}{90^{o}}}{sin{\alpha}sin{\beta }}}$
$CD=\sqrt{sin^{2}{90^{o}}*(\frac{AB^{2}}{sin^{2}{\alpha}}+\frac{AB^{2}}{sin^{2}{\beta}}+\frac{2cos(180^{o}-(\alpha+\beta)) AB^{2}}{sin{\alpha}sin{\beta}}}$
Because $\frac{CD}{QP}=\frac{sin{90^{o}}}{sin{\gamma}}$ and $sin{90^{o}}\ge{sin{\gamma}}$, the result follows.

If PQ is perpendicular to AB then PQ=CD
And if PQ is not perpendicular to AB then add B & P ;B & Q;
Suppose the centers of the two circles are repectively M and N
So add B & M ; C & M; B & N; D& N
As the circular angle CAB= 90 degree
So the central angle BMC= 180 degree
So B,M and C are collinear and therefore BC is the diameter of the Circle of center M
Similarly B,N and D are collinear and therefore BD is the diameter of the circle of center N
Now  in triangle BCD and BPQ:
angle BCD= angle BPQ
angle BDC= angle BQP
So the triangles are similar
So PQ/CD= PB/PC

But PB<BC [ PB is the chord and BC is the diameter of the circle]
So PB/BC<1
So PQ/CD<1
So PQ<CD

So PQ<=CD


How many numbers are in the list $ 2008, 2003, 1998, \ldots, 8, 3?$

To find the number of members in an arithmetic sequence, we subtract the largest term and the smallest term, divide by the common difference, and add one.

So, for this problem: The largest term is $ 2008$, the smallest term is $ 3$, and the common ratio is $ 5$.

$ \frac{2008\minus{}3}{5}\plus{}1\equal{}\frac{2005}{5}\plus{}1\equal{}401\plus{}1\equal{}\boxed{402}$ terms.

WELL, you see that the difference between each term is 5. So then you add 2 to every term, and each term becomes a multiple of 5!!!!!

!!
0

So then you divide everything by five:

$\frac{2010, 2005, 2000,..., 10, 5}{5}$

and you get $402,401,400,...,2,1$

$\boxed{402}$


topology
complex analysis
complex analysis unsolved

Let $ G_1$ and $ G_2$ be simply connected regions neither of which is the whole plane. Let $ f$ be a one-one analytic mapping of $ G_1$ onto $ G_2$. Let $ a\in G_1$ and $ \alpha\equal{}f(a)$. Prove that for any one-one analytic map $ h$ of $ G_1$ into $ G_2$ with $ h(a)\equal{}\alpha$ it follows that $ |h'(a)|\le|f'(a)|$. Suppose $ h$ is not assumed to be one-one, what can be said?

[hide="Hint"]Consider the map $ j\equal{}g\circ h\circ f^{\minus{}1}\circ g^{\minus{}1}$, where $ g$ is a homeomorphism from $ G_2$ onto $ \mathbb{D}$ and apply Schwartz's Lemma.[/hide]


Ok that's fine but is it a unique solution?

Let $p_k=x_1^k+...x_n^k$.
We have $p_0=n$ and $p_k=a^k$ for $k=0,1,...,n$.

Let $P(t)=(t-x_1)(t-x_2)...(t-x_n)= c_nt^n+c_{n-1}t^{n-1}+...+c_1t+c_0$.
We have $c_n=1$ and $c_{n-1} = -p_1=-a$.

The well-known Newton's relations state that for $k=1,...,n-1$, we have :
$p_k + c_{n-1}p_{k-1}+...+c_{n-k+1}p_1+kc_{n-k}=0$.

For $k=2$ it gives that $c_{n-2}=0$. Thus, using these relations again, an easy induction leads to $c_{n-k} = 0$ for $k=2,...,n-1$.

Then $P(t)=t^n - at^{n-1} + c_0$.
Moreover, since $P(x_i)=0$ for each $i$, and summing these relations, we then obtain that $p_n-ap_{n-1}+nc_0=0$ so that $c_0=0$.
Thus $P(t)=t^n-at^{n-1}=t^{n-1}(t-a)$, which has roots $a$ and $0$, the latter with multiplicity $n-1$.
Then $(x_1,...,x_n)$ is some permutation of $(a,0,...,0)$.

Pierre.


calculus
integration
function
trigonometry
calculus computations

A function $ f(x)$ is defined as follows for $ x\geq 0$.

$ f(x) \equal{} \sin \left(\frac {n\pi}{4}\right)\sin x\ (n\pi \leq x < (n \plus{} 1)\pi)\ (n \equal{} 0,\ 1,\ 2,\ \cdots)$.

Evaluate $ \int_0^{100\pi} f(x)\ dx$.

[hide="brute force"]$ \int_0^{100\pi} f(x) dx \equal{} \sum_{n\equal{}1}^{100} \sin \left( \frac{n\pi }{4} \right)\int_{(n\minus{}1)\pi}^{n\pi}  \sin x dx$

$ \equal{} \sum_{n\equal{}1}^{100} \sin \left( \frac{n\pi }{4} \right) [\cos (n\minus{}1)\pi \minus{} \cos n\pi ]\equal{}\sum_{n\equal{}1}^{100} 2\sin \left( \frac{n\pi }{4} \right) (\minus{}1)^{n\minus{}1}$

$ \equal{} 2\sum_{k\equal{}1}^{25} \sin \left( \frac{(4k\minus{}3)\pi }{4} \right) \minus{}\sin \left( \frac{(4k\minus{}2)\pi }{4} \right) \plus{}\sin \left( \frac{(4k\minus{}1)\pi }{4} \right) \minus{}\sin \left( \frac{4k\pi }{4} \right)$

$ \equal{}2\sum_{k\equal{}1}^{25} (\minus{}1)^{k\minus{}1} [ \sqrt 2 \minus{} 1 ]\equal{}2 [ \sqrt 2 \minus{} 1 ]$[/hide]

[quote="ppt"]By graph, we can see the cycle of the function is $ 8\pi$. Every cycle is divided into two parts which has positive and negative values and the sum of integrals in it is zero.
So we only calculate the integral from $ 97\pi$ up to $ 100\pi$. We have:
$ \int_0^{100\pi}f(x)dx \equal{} \int_{97\pi}^{98\pi}\frac {\sqrt {2}}{2}sinxdx \plus{} \int_{98\pi}^{99\pi}sinxdx \plus{} \int_{99\pi}^{100\pi}\frac {\sqrt {2}}{2}sinxdx$
$ \equal{} \minus{} \sqrt {2} \plus{} 2 \minus{} \sqrt {2} \equal{} 2 \minus{} 2\sqrt {2} \equal{} 2(1 \minus{} \sqrt {2})$.[/quote]

That's the correct answer.


I hear that the cut off for gold was 27 out of 40. Last year 11 students got perfect score.
Congratulations to Anri Rrembeci the Albanian guy who recieved a gold medal.

did anyone take perfect score?
what was the 4th problem?

THe cut offs are
Gold 25
Silver 20
Bronze 15

the 4th problem was:
find all 3 digit numbers [b]abc[/b]=abc(a+b+c)

[quote="Sasha"]Congrats to MOL2! (I guess it's [b]freemind[/b] :? )[/quote]
yeah congratulations
i think he was the first with 33 points ain't it?

the fourth problem was quite difficult for the JBMO it was more long and with calculations than with nice tricks
the 3-rd problem was i think :
(i)prove that we can find 5 points which form 8 right triangles
(ii)prove that we can find 64 points which form more than 2005 right triangles
i'm very sorry for my friend who went to the JBMO and took 10 points :roll:  :roll:

thx sasha for the congratulations :)
and to Albanian Eagle :)
i was the first but not with 33 but 31.
good luck to sasha at imo


videos
geometry
geometric transformation

Just wondering about if AoPSers watch a lot of TV, and what do you watch?

Here's some stuff I've been hooked on lately:

HEROES!!!!
Grey's Anatomy
Lost
Desparate Housewives
Bleach

And I'm also making my way through Firefly.  I used to watch Family Guy as well...but I got bored of it pretty quickly.

I don't like watching TV...

The only thing I watch is a chinese show on sundays, and the only reason I watch that is because it has a magic competition on part of it.

NUMB3RS
House
Flashpoint
Home Improvement
Stargate SG-1
Stargate Atlantis

All the ones I can think of at the moment.

Heroes
Avatar
BSG
Stargate
Smallville
Word Girl
The Office
Arrested Development

uhhh.. I don't watch any shows. I usually watch movies with my family once or twice a week, though.

Numb3rs
The Simpsons
The Mole
I Survived a Japanese Gameshow
Family Guy
American Dad
Mythbusters
Cash Cab
CSI: NY

Thats about it...

Jeopardy
The Mole
I survived a Japanese Game Show
Wipeout
America's Got Talent
Simpsons (i watch with jeopardy. when jeopardy's on commercial, i watch it)

i also watch American Idol (Archuletta!!! :D) but that's not on now.

I also used to watch Extreme Makeover: Home Edition, but that's not on anymore.

I watch mainly anime, and it's not really on TV (on computer), but does it count? If it does, then
Bleach
Naruto
Naruto Shipudden
Sekerei (kinda pervy, but storyline's kinda good)
Yeah, that's basically all. Oh, but just don't even look at Sekerei's manga. It's sick :(

Jeopardy's a great show to have on in the background while I'm making dinner I find =P
Mythbusters is AWESOME!  I also occasionally watch House and So you think you can Dance...(I know, I'm strange)

Brainiak, Mythbuster, Stunt junkies, HOUSE!!!

Simpsons, Friends, Jeopardy, News, Numb3rs, ocassionally: Flashpoint, House, other random shows...

absolutely nothing except a few things on the discovery channel that i get a glance at b/c my brother watches them.

I don't watch very often, but when I do, I watch Mythbusters, Fresh Prince, and Wheel of Fortune.

actually, i watch animes on the comp, but...

shaman king
fullmetal alchemist
rurouni kenshin
kamichama karin
shugo chara
full moon wo sagashite
bamboo blade
sakura no ichiban
ayashi no ceres
.hack//legend of the twilight
inuyasha
azumanga daioh

ok so maybe i finished most of them already... actually make that all of them...

Oh yeah, I forgot, I watched American Idol and I've been usually watching America's Got Talent, but only cuz someone I knew was on there.

I watch too much TV...to extend my extensive list
Kyle XY (next season is next year  :( )
Smash Lab
Man vs Wild
King of the Hill
Wipeout
CSI
CSI: Miami

[quote="dynamo729"]THE OFFICE!!!!!!
House
Monk
Psych
John and Kate plus 8
My Name Is Earl
30 Rock
Ninja Warriors (occasionally)
Law and Order sometimes (its a new thing for me)
The closer
Saving grace (newly watching)
Lost
Avatar whenever I can
American Idol
Jeopardy (in the summertime only)
The local news

You probably think I'm a freak but these are in and out of season at different times so...[/quote]

Hahaha, that's a lot of TV to watch!  But yeah, there's like nothing on during the summer though!

[quote="jmerry"]I watch... nothing. I don't have a TV, and I don't miss it.[/quote]
Wow, nice. I don't watch TV (except maybe watching a basket ball game along with my dad and every other year, the olympics) either. I just use my TV for video games.  :D

[quote="jmerry"]I watch... nothing. I don't have a TV, and I don't miss it.[/quote]

Ha, you clearly have a computer though.  I watch probably around 75% of the TV I watch on my computer.  That said, I was very little TV - it's only when I'm on break.  During the semester, it's [i]extremely[/i] rare for me to watch TV.  I've been watching some this summer though:

The Office
Arrested Development
The Simpsons
Seinfeld
Chappelle Show (and Comedy Central in general)
Major sporting events (e.g. super bowl, NBA finals, world cup, olympics, tour de france, etc.)
Family Guy
Malcolm in the Middle
Mythbusters
Survivorman
and more...

lol...

i watch ALL my tv online. especially since parents are so restrictive ;)

I watch only chinese dramas/shows. so heh...:-D

[quote="splatyango"]UPDATE:
im getting bored these days, so im starting to watch tv at night. i just watch whats on the discovery channel at night depending on what day it is/ what time i watch (because sometimes im lazy and stay in the shower for half an hour :blush: )[/quote]
Ya...I'm her brother, and we just got cable recently. Before that all we had was a satellite dish that received signals from China :(  because my grandparents (maternal, my paternal grandparents never got a visa, i don't know why, maybe because my grandpa fought for the Communists during WWII and the Chinese Civil War)come over a lot from china and my parents didn't really care (they thought we didn't care either, but i guess they learned :) )
Now we got another Chinese network called KyLin TV, and my grandpa loves to watch it, I don't know why, so the TV is constantly on (except on weekends when they go to an apartment nearby that my parents rented)
And I don't watch TV online because my computer (the one I'm typing on now), even though it is dual-core and only a year old, is paired with a monitor that has been in this family for at least 5 years
And we have a 36" HD TV

I watch:
- CSI (Miami, Vegas)
- Cold Case
- Big Bang Theory
- Man v Food
- Bizarre Foods with Andrew Zimmern
- Monk
- NCIS
- The Forgotten

please do not revive. It is quite annoying. [b]Start a new thread.[/b]

MythBusters, Time Warp, Destroyed in Seconds.

Mostly just College football.

Inuyasha <3
Fruits Basket (best translation ever)
Ouran HSHC
Skip Beat
Numb3rs
The Office
Clannad
To Heart
Ah, My Goddess

Yeah, mostly anime.

The Simpsons
Family Guy
The Cleaveland Show
House
Chuck
The Biggest Loser
Fullmetal Alchemist: Brotherhood

Teh Biggest Loser
Brain Surge
Table for Twelve
18 Kids and Counting
I used to watch Jon and Kate + 8 but then it got cancelled.

Hmmm...

- The Colbert Report
- The Daily Show
- NCIS
- NCIS: LA
- A little L&O
- A little CSI
- South Park
- The Simpsons
- Family Guy

... did I forget any?

r u joking no one else watches disney channel? :|

I don't watch too much TV during winter break since I have more freedom with my computer.  But normally:

The Simpsons
Family Guy(whenever my parents aren't home :P)
Total Drama Action(DON'T LAUGH)
The news(anything but fox :P)
The Price is Right
Let's Make a Deal

Other stuff.......

Ironically enough, my parents give me plenty of freedom with movies(I watched my first r movie at 9), but TV is a different story.......

Regularly?

GLEE!, 24, Heroes, Gossip Girl, Grey's and most of the shows on Bravo. Oh, and I love Tabitha's Salon Makeover. Can't get enough!
Also, I watch the Food Network all the time. I think I might have a crush on Bobby Flay.


geometry inequalities conics trigonometry China

A circle meets the three sides $BC,CA,AB$ of a triangle $ABC$ at points $D_1,D_2;E_1,E_2; F_1,F_2$ respectively. Furthermore, line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, line segments $E_1F_1$ and $E_2D_2$ intersect at point $M$, line segments $F_1D_1$ and $F_2E_2$ intersect at point $N$. Prove that the lines $AL,BM,CN$ are concurrent.

hi hui jack, soarer and "me"!!! it's so great that you can online in the mainland!!! what are your result today?  :D  :D  :D

I'm pretty busy now, so I usually don't open a thread unless it is in the "geometry" or in the "inequalities" forum. Please post problems into separate threads in respective forums.

[quote="mecrazywong"]A circle meets three sides $BC,CA,AB$ of triangle $ABC$ at points $D_1,D_2;E_1,E_2;,F_1,F_2$ in turn. Then, line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, line segments $E_1F_1$ and $E_2D_2$ intersect at point $M$, line segments $F_1D_1$ and $F_2E_2$ intersect at point $N$. Prove that the lines $AL,BM,CN$ are concurrent.[/quote]

This one is pretty simple:

The Desargues theorem, applied to triangles ABC and LMN, shows that instead of proving that the lines AL, BM, CN are concurrent, it is enough to prove that the points $X = MN\cap BC$, $Y = NL\cap CA$, $Z = LM\cap AB$ are collinear.

Applying the Pascal theorem to the cyclic hexagon $F_1D_1D_2E_2F_2E_1$, we see that the points $F_1D_1\cap E_2F_2$, $D_1D_2\cap F_2E_1$ and $D_2E_2\cap E_1F_1$ are collinear. But $F_1D_1\cap E_2F_2 = N$, $D_1D_2\cap F_2E_1 = BC\cap F_2E_1$ and $D_2E_2\cap E_1F_1 = M$; hence, it follows that the points N, $BC\cap F_2E_1$ and M are collinear. In other words, the point $BC\cap F_2E_1$ lies on the line MN. Equivalently, the lines BC, $F_2E_1$ and MN concur; in other words, the point $X = MN\cap BC$ can be also considered as the point $X = BC\cap F_2E_1$. In other words, $X = D_1D_2\cap F_2E_1$. Similarly, $Y = E_1E_2\cap D_2F_1$ and $Z = F_1F_2\cap E_2D_1$. But applying the Pascal theorem to the cyclic hexagon $D_1D_2F_1F_2E_1E_2$, we see that the points $D_1D_2\cap F_2E_1$, $D_2F_1\cap E_1E_2$ and $F_1F_2\cap E_2D_1$ are collinear; in other words, the points X, Y, Z are collinear. $\blacksquare$

[b]EDIT:[/b] Looks like I have forgotten that I already solved exactly the same problem some time ago - see [url=http://groups.yahoo.com/group/Hyacinthos/message/9873?expand=1]Hyacinthos message #9873[/url]!

  Darij

[quote="darij grinberg"]I'm pretty busy now, so I usually don't open a thread unless it is in the "geometry" or in the "inequalities" forum. Please post problems into separate threads in respective forums.

[quote="mecrazywong"]A circle meets three sides $BC,CA,AB$ of triangle $ABC$ at points $D_1,D_2;E_1,E_2;,F_1,F_2$ in turn. Then, line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, line segments $E_1F_1$ and $E_2D_2$ intersect at point $M$, line segments $F_1D_1$ and $F_2E_2$ intersect at point $N$. Prove that the lines $AL,BM,CN$ are concurrent.[/quote]

This one is pretty simple:

The Desargues theorem, applied to triangles ABC and LMN, shows that instead of proving that the lines AL, BM, CN are concurrent, it is enough to prove that the points $X = MN\cap BC$, $Y = NL\cap CA$, $Z = LM\cap AB$ are collinear.

Applying the Pascal theorem to the cyclic hexagon $F_1D_1D_2E_2F_2E_1$, we see that the points $F_1D_1\cap E_2F_2$, $D_1D_2\cap F_2E_1$ and $D_2E_2\cap E_1F_1$ are collinear. But $F_1D_1\cap E_2F_2 = N$, $D_1D_2\cap F_2E_1 = BC\cap F_2E_1$ and $D_2E_2\cap E_1F_1 = M$; hence, it follows that the points N, $BC\cap F_2E_1$ and M are collinear. In other words, the point $BC\cap F_2E_1$ lies on the line MN. Equivalently, the lines BC, $F_2E_1$ and MN concur; in other words, the point $X = MN\cap BC$ can be also considered as the point $X = BC\cap F_2E_1$. In other words, $X = D_1D_2\cap F_2E_1$. Similarly, $Y = E_1E_2\cap D_2F_1$ and $Z = F_1F_2\cap E_2D_1$. But applying the Pascal theorem to the cyclic hexagon $D_1D_2F_1F_2E_1E_2$, we see that the points $D_1D_2\cap F_2E_1$, $D_2F_1\cap E_1E_2$ and $F_1F_2\cap E_2D_1$ are collinear; in other words, the points X, Y, Z are collinear. $\blacksquare$

  Darij[/quote]
Darij, nice proof!
In fact, all the contestants who got this problem in cmo used the trigo form of Ceva's Theorem(including me), but except 2, they worked out the same solutions as you, which is nice and simple ;)

There are two theorems related to this theme.

[b]Steiner's Theorem[/b]
Let $ABCDEF$ is an arbitrary inscribed hexagon. Then Pascal's lines of $ABCDEF$, $ADECF$, $ADCFEB$ are concurrent.

[b]Kirkman's theorem[/b]
Let $ABCDEF$ is an arbitrary inscribed hexagon. Then Pascal's lines of $ABFDCE$, $AEFBDC$, $ABDFEC$ are concurrent.

There are 60 Pascal's lines for $ABCDEF$. Each such line is included into one Steiner's triple, and it is inluded into three Kirkman's triples.

There is another reduction to Pascal's theorem by taking the points dual to
the lines AL, BM, CN  with respect to the circle (or conic) through the 6 points
on the sides.  Are the CMO contestants are expected to know projective geometry?
If yes, the problem can be solved in a few minutes, but if not then it could be a difficult
and time consuming problem (at least, I did not see how to solve it without projective geometry).

Question:  is the problem statement true if we replace the 3 sides of the triangle by
a smooth cubic curve?  (Like the proof of Pascal using addition law on an elliptic curve.)

Believe or not but I have solved this similar problem before:
$D_1E_2$ meets $D_2F_1$ at P.
$D_1E_2$ meets $E_1F_2$ at Q.
$E_1F_2$ meets $D_2F_1$ at R.
AP,BQ,CR are also concurrent.
And this problem is also solved by similar proof (use the trigo form of Ceva's Theorem).

[quote="minhkhoa"]Believe or not but I have solved this similar problem before:
$D_1E_2$ meets $D_2F_1$ at P.
$D_1E_2$ meets $E_1F_2$ at Q.
$E_1F_2$ meets $D_2F_1$ at R.
AP,BQ,CR are also concurrent.[/quote]

This problem is actually much simpler:

In order to prove that the lines AP, BQ and CR are concurrent, by the Desargues theorem, it is enough to show that the points $QR\cap BC$, $RP\cap CA$ and $PQ\cap AB$ are collinear. But we have $QR\cap BC = F_2E_1\cap D_1D_2$, $RP\cap CA = E_1E_2\cap D_2F_1$ and $PQ\cap AB = E_2D_1\cap F_1F_2$, and the collinearity of the points $F_2E_1\cap D_1D_2$, $E_1E_2\cap D_2F_1$ and $E_2D_1\cap F_1F_2$ follows from the Pascal theorem applied to the cyclic hexagon $F_2E_1E_2D_1D_2F_1$. This solves the problem.

  Darij

eh this is also seem to be right for the pappus lines.
A,B,C on d1
D,E,F on d2
we denote (ABCDEF) the papus line through AE :cap: BD BF :cap: CE and AF :cap: CD
then (ABCDEF) (ABCEFD) (ABCFDE) are concurent.
Does anyone know any theorems about this. These kind of problems are really beautiful and interesting.

i didn't knew that desargue's theorem works in both directions.Could S.O. please explain?

[quote="indybar"]How to do it using trig form of Ceva?[/quote]
Trig Ceva proof:

Construct $LL_1 \perp AB$, $LL_2 \perp AC$ where $L_1, L_2 \in AB, AC$. Thus we have:
\[ \frac{\sin{\angle{BAL}}}{\sin{\angle{CAL}}}=\frac{LL_1}{LL_2}=\frac{LF_2 \cdot \sin{\angle{LF_2L_1}}}{LE_1 \cdot \sin{\angle{LE_1L_2}}} \]
Since $\triangle{LF_2D_1} \sim \triangle{LE_1D_2}$, so:
\[ \frac{LF_2}{LE_1}=\frac{F_2D_1}{E_1D_2} \]
Connect $D_2F_1$ and $D_1E_2$, we can see that:
\[ \frac{D_2F_1}{D_1E_2}= \frac{2R \cdot \sin{\angle{LF_2L_1}}}{2R \cdot \sin{\angle{LE_1L_2}}}=\frac{\sin{\angle{LF_2L_1}}}{\sin{\angle{LE_1L_2}}} \]
Put it back into the first equation we got:
\[ \frac{\sin{\angle{BAL}}}{\sin{\angle{CAL}}}=\frac{D_1F_2}{D_2E_1}\cdot\frac{D_2F_1}{D_1E_2} \]
Similarly we can get other two trig relations, multiply them togather and finishes the problem.

Sorry for bumping an 11 old thread, but is this problem is purely projective?


The cards in a deck marked with consecutive positive integers are equally spaced around a circle. If the card marked 5 is directly opposite the card marked 18, how many cards are in the deck?

if 5 is the 5th number going clockwise, then 18 is the 5th number going counterclockwise, so u count starting with 18.   18, 19, 20, 21, 22.  the answer is 22.

I'm pretty sure you messed up somewhere.

You need to add five more, not add just four more.

So, 23 is our answer.

How many cards are on the right side, including the 18 and 5? 18-5+1=14.

What about the left side, excluding the 18 and 5? 18-5+1-2=12.

14+12=$\boxed{26}.$

In general, with $m$ and $n,$ notice that you have $m-n+1+m-n-1=2(m-n).$


please prove this  :huh:  : {p^{p}-1} over p-1 is composite (p=4k+1,p is a prime number)

http://www.mathlinks.ro/Forum/viewtopic.php?t=169135  :wink:

May I ask where you got the problem from¿


Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(f(x)y+x)=xf(y)+f(x)\]
for all $x,y \in \mathbb R$.

[quote="nguyenvuthanhha"]Excellent solution :first:[/quote]

Thanks :)

Hi There,

  My solution is similar to PCO's but differs after (2). I am only including the difference.

From PCO's solution, we have

$ f(f(x)y \plus{} x) \equal{} xf(y) \plus{} f(x) \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (1)$

1)$ f(x) \equal{} 0 \iff x \equal{} 0$
==================
$ P(1,0) \implies f(1) \equal{} f(1) \plus{} f(0)$ and so $ f(0) \equal{} 0$
Suppose $ \exists u$ such that $ f(u) \equal{} 0$. Then $ P(u,c) \implies 0 \equal{} uf(c) \implies u \equal{} 0$
Q.E.D.

2) $ f( \minus{} 1) \equal{} \minus{} 1$ and $ f(1) \equal{} 1$ and $ f( \minus{} x) \equal{} \minus{} f(x) \forall x$
=======================================
$ P( \minus{} 1, \minus{} 1) \implies f( \minus{} f( \minus{} 1) \minus{} 1) \equal{} 0 \implies f( \minus{} 1) \equal{} \minus{} 1$
$ P(1, \minus{} 1) \implies f(1 \minus{} f(1)) \equal{} f(1) \minus{} 1$
$ P(1 \minus{} f(1),1) \implies 0 \equal{} \minus{} (1 \minus{} f(1))^2$ and so $ f(1) \equal{} 1$

$ P( \minus{} 1,x \minus{} 1) \implies f( \minus{} x) \equal{} \minus{} f(x \minus{} 1) \minus{} 1$
$ P(1,x \minus{} 1) \implies f(x) \equal{} f(x \minus{} 1) \plus{} 1$
And so $ f( \minus{} x) \equal{} \minus{} f(x)$
************************************************
MY SOLUTION STARTS HERE:

We have $ f(x \plus{} 1) \equal{} f(x) \plus{} 1 \Rightarrow f(n) \equal{} n \;\;\forall \;n \in \mathbb{Z}$
Set $ x \equal{} 2$ in $ (1)$.

$ f(2y \plus{} 2) \equal{} 2f(y) \plus{} 2$

$ \Rightarrow f(2y) \plus{} 2 \equal{} 2f(y) \plus{} 2$

$ \Rightarrow f(2y) \equal{} 2f(y) \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (2)$

Set $ x \equal{} \minus{} x, y \equal{} \minus{} y$ in $ (1)$ and use $ f( \minus{} x) \equal{} \minus{} f(x)$.

$ f(f(x)y \minus{} x) \equal{} xf(y) \minus{} f(x)$

$ \Rightarrow f(f(x)y \minus{} x) \plus{} 2f(x) \equal{} xf(y) \plus{} f(x)$

$ \Rightarrow f(f(x)y \minus{} x) \plus{} f(2x) \equal{} f(f(x)y \plus{} x) \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (3)$

We note that $ f(x)y \minus{} x$ can assume any real value as long as $ x\neq 0$.
We let $ u \equal{} f(x)y \minus{} x$ and $ v \equal{} 2x$ and $ x\neq 0$ in $ (3)$.

$ f(u) \plus{} f(v) \equal{} f(u \plus{} v) \;\;\forall u,v \in \mathbb{R}\setminus\{0\} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (4)$

We note that the above equation holds even when $ u$ or $ v$ or both are 0.
Hence, we have

$ f(u) \plus{} f(v) \equal{} f(u \plus{} v) \;\;\forall u,v \in \mathbb{R} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (5)$

$ \implies f(q) \equal{} qf(1) \equal{} q \;\;\forall q \in \mathbb{Q}$

Using $ (5)$ in $ (1)$, we get

$ f(f(x)y) \plus{} f(x) \equal{} xf(y) \plus{} f(x) \implies f(f(x)y) \equal{} xf(y) \minus{} \minus{} \minus{} (6)$

$ \implies f(f(x)) \equal{} x \;\;(\;\text{ set } y \equal{} 1 \text{ in } (6))$

$ \implies f(xy) \equal{} f(x)f(y) \;\;(\;\text{ set } x \equal{} f(x) \text { in } (6))$.

$ \implies f(x^2) \equal{} (f(x))^2 \geq 0 \;\;(\;\text{ set } x \equal{} y \text { in the above equation})$.

Setting $ v > 0$ in $ (5)$ gives us $ f(u \plus{} v) > f(u)$. We note that $ f$ is an increasing function and $ f(q) \equal{} q \;\;\forall q \in \mathbb{Q}$.

Hence, $ f(x) \equal{} x\;\;\forall x \in \mathbb{R}$.


Can anyone give me a hint on this?

Find in degrees the smallest positive angle $x$ that $\sin3x=\cos7x$.  (Mandelbrot #3)
X is the angle.
Does $\cos7x = \sin(90^{\circ}-7x)=\sin(630^{\circ}-7x)$

How do you make a degree sign?
Thanks!

[size=75][color=red]Also, use \sin for sin, etc
b-flat[/color][/size]

[size=59]Grammar Ninja edits post like swinging tree in breeze.[/size]

[quote="myc"]can anyone give me a hint on this

Find in degrees the smallest positive angle $x$ that $sin3x =cos7x$.  (Mandelbrot #3)
x is the angle
Does $cos7x = sin(90[size=9: d3a9a50ef1]degree[/size: d3a9a50ef1]-x)7=sin(630[size=9: d3a9a50ef1]degree[/size: d3a9a50ef1]-7x)$  ?

Thanks[/quote]
Uh... something's wrong with your $\LaTeX$.
And about the problem, you should use the fact that $\sin a^\circ=\cos (90-a)^\circ$.

To do a degree sign, put x^\circ within dollar signs
It should show this, $x^\circ$

ok the latex looks better now  Thanks   :)

To answer your question:

Yes, $\sin(90-7x)^\circ=\cos7x^\circ$

However, $\sin(90-7x)^\circ\neq\sin(630-7x)^\circ$, they are opposites in parity.

Is this all you need?

the answer to your problem can be found easily with some number theory. we have $\sin(3x)=\sin(90-7x)$. What must be true of the relationship between the arguments (3x, 90-7x) in terms of mods, remainders, etc.