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Tonight I was reading through [url=http://www.edge.org/3rd_culture/debate05/debate05_index.html]a debate[/url] between Steven Pinker and Elizabeth Spelke on whether or not their are differences in abilities between men and women in regards to their cognitive abilities.

Listing to both sides, I feel like both Pinker and Spelke make many good points.

This is a subject I've thought a lot about, and I find myself coming to conclusions that are somewhat askew from both speakers:

Men and women are fundamentally different.  That doesn't mean that there is such thing as an idea that neither can grasp.  But it does mean that there could be biological features that cause one sex to gravitate more toward certain activities than the other.

The act of carrying a child to term is so fundmentally important to our species that a lot of energy is devoted to it as a biological process -- not excluding the acts of finding mates to begin with.  That alone would shift the balance of sexes in fields that require such singular dedication to achieve.

Additionally, men seem to me to be more likely than women to focus on anything single-mindedly -- something that may not be a good overall life strategy, but which may be a strategy that makes scientific achievement a more natural fit to a personality.

None of this precludes the participation of either sex in mathematics and science, though my opinion is that even if all forces of [unfair] discriminatory bias went away today (unfortunately very unlikely), there would likely still be more men in math and science than women -- though perhaps by a smaller margin than is currently the case.

It's not a matter of inferiority/superiority (maybe there are some cognitive differences, maybe there aren't): women have to deal with a priority that men don't (child birth), and that's not an insignificant priority.

It won't surprise me if the next generation of mathematicians and scientists includes more women, but I also believe that those women will be having fewer children than the average.

Any thoughts?

Interesting factoid: female mammals become smarter (they experience dendrite growth) with every gestation.

I did not year hear the audio, but based on the slides alone, if this were a boxing match then Pinker would have won a massive points victory over Spelke if not an outright KO.  (A "male" metaphor, to be sure.)

But is there any difference in IQ or general intelligence between sexes?

Most of intelligence is genetics. Based on your environment, your intelligence can bounce around in about a 10 or 15 point range. Getting good parents is more important than a small factor such as sex.

[quote="PenguinIntegral"]
Most of intelligence is genetics. Based on your environment, your intelligence can bounce around in about a 10 or 15 point range.[/quote]

What is your basis for this assertion?

[quote="PenguinIntegral"]But is there any difference in IQ or general intelligence between sexes? [/quote]

"Intelligence" is not a precise term.   IQ is well-defined but problematic as a measure.  Mathematics, music, etc  are limited enough domains that one can ask about more precisely testable abilities such as visualization and the other tasks that Pinker and (to a much lesser extent) Spelke cite studies about.

[quote="PenguinIntegral"][quote="MCrawford"][quote="PenguinIntegral"]A parent with a psychology PhD.[/quote]

I'm not interested in the person, but rather, the research it's based on.  I've read a lot of different opinions from researchers and I like to look at the basis for those opinions so that I can form my own.[/quote]
Let me restate : The brain has a genetically determined capability for the raw processing of information. This happens at the neural level, so because we never grow new brain cells, it is only going downwards, usually because of drug use.

Is that a good definition?[/quote]

However, intelligence is not determined by the number of neurons you have or the size of your brain.  In fact, for most activities, brain activity never even comes close to 100%.

[quote="dakyru"][/quote][quote="PenguinIntegral"][quote="MCrawford"][quote="PenguinIntegral"]A parent with a psychology PhD.[/quote]

I'm not interested in the person, but rather, the research it's based on.  I've read a lot of different opinions from researchers and I like to look at the basis for those opinions so that I can form my own.[/quote]
Let me restate : The brain has a genetically determined capability for the raw processing of information. This happens at the neural level, so because we never grow new brain cells, it is only going downwards, usually because of drug use.

Is that a good definition?[/quote][quote="dakyru"]

However, intelligence is not determined by the number of neurons you have or the size of your brain.  In fact, for most activities, brain activity never even comes close to 100%.[/quote]
Still, brains have a fixed capability and usable capability to process info.

1.: the statement that the brain never reaches near to 100% is useless, since it says nothing (every part of the brain has to do a special job).
2.: well, there is probably some genetic aspect, but who tells that the number of neurons etc. depend mainly on the DNA only¿ Other circumstances can also cause it.

[quote="ZetaX"]
2.: well, there is probably some genetic aspect, but who tells that the number of neurons etc. depend mainly on the DNA only¿ Other circumstances can also cause it.[/quote]

Besides when the brain is developing, which is early in life, what would change the mental ability besides negative events?

[quote="ZetaX"]Did I exclude happenings in early life¿
Indeed, I agree that the main part is probably the DNA, but where you got that $15$ points range from¿ (especially, IQ can't really be measured in my oppinion)[/quote]
The exact range is disputed, but it is somewhere between 10 and 20 IQ points.

IQ, not to mention "intelligence", is a composite of too many uncontrolled variables to lead to a meaningful discussion.   The Pinker/Spelke debate cites narrower measures such as mental rotation tasks.  Of the two, Spelke cites more generic cognitive abilities that develop by age 6, which is a huge weakness in her arguments as compared to Pinker; her data are irrelevant to the top-level "tail of the curve" group that Pinker (and Larry Summers) refers to in discussing the composition of hard-science academia.

[quote="MCrawford"]Additionally, men seem to me to be more likely than women to focus on anything single-mindedly -- something that may not be a good overall life strategy, but which may be a strategy that makes scientific achievement a more natural fit to a personality.
[/quote]

I have actually, from my own observations, found this to be the case.  So now that I've found someone else who thinks so, I agree.  Most of the weird things you see out there are done by men.

This reminds me of the following story (I can't remember the exact details):

This one man was trying to prove that women were dumber than men by taking the average weight of a lot of man brains and women brains, and he found that women's brains were slightly lighter... Ironically enough, when he died, someone found that his brain was lighter than women's brains...

On that topic, I read an interesting article in a French science magazine about how, after his death, Lenin's brain proved to be slightly below-average weight, which gave many fellow party members quite a shock.

[quote="PenguinIntegral"][quote="MCrawford"][quote="PenguinIntegral"]A parent with a psychology PhD.[/quote]

I'm not interested in the person, but rather, the research it's based on.  I've read a lot of different opinions from researchers and I like to look at the basis for those opinions so that I can form my own.[/quote]
Let me restate : The brain has a genetically determined capability for the raw processing of information. This happens at the neural level, so because we never grow new brain cells, it is only going downwards, usually because of drug use.

Is that a good definition?[/quote]

This article gives some evidence that brain cells do in fact regenerate:
http://healthlink.mcw.edu/article/926345803.html

Although I doubt it is very substantial.

And how did this guy expect to prove that brain weight is a determining factor of intelligence?

[quote="ZetaX"]Did I exclude happenings in early life¿
Indeed, I agree that the main part is probably the DNA, but where you got that $15$ points range from¿ (especially, IQ can't really be measured in my oppinion)[/quote]

But I never heard that Gauss or Newton had very smart parents...

That says nothing...
A lot of facts have to come together in the DNA to cause anything, so e.g. their mother could have had the genes needed for a higher intelligence, but only his father had the genes to activate his mothers genes.

Gauss' son was a merchant who reportedly had unusual capabilities in mental arithmetic (multiplication of 20-digit numbers, or something like that).   Newton's father is unknown, as Isaac is believed to have been a bastard child.  That he had access to schooling was the result of his mother marrying a wealthy man, and that a precondition for the marriage was that young Newton be sent away for schooling (i.e. the husband wanted nothing to do with the boy and had the money to get rid of him).

Brains shrink slightly over time, so measuring the brains of aged scholars is not a very reliable statistic.  There have been famous intellectuals with very small heads, such as Anatole France.

I was not so impressed by the Pinker/Spelke debate, myself.  That's probably because I've been semi- keeping track of the studies which look at the impact of environment on cognitive abilities.

[quote="fleeting_guest"]I did not year hear the audio, but based on the slides alone, if this were a boxing match then Pinker would have won a massive points victory over Spelke if not an outright KO. (A "male" metaphor, to be sure.)  [/quote]
Pinker may have argued more forcefully, but I found his reasoning unconvincing.

[quote="fleeting_guest"]
The Pinker/Spelke debate cites narrower measures such as mental rotation tasks. [/quote]
By the way, I don't recall whether either Pinker or Spelke mentioned this, but there are recent studies which showed that the gender gap in spacial rotation/visualization disappears if you either (a) eliminate the time requirement on the test or (b) have the test-takers practice for a couple of hours first.  The neuroscientist whom I heard speaking about (a) said that there was a theory (on the part of those conducting the study) that  when under time pressure perhaps males guess more than females, and that females are more likely to leave a question blank when they aren't certain of the answer.

[quote="MCrawford"]
This is a subject I've thought a lot about, and I find myself coming to conclusions that are somewhat askew from both speakers:

Men and women are fundamentally different.  That doesn't mean that there is such thing as an idea that neither can grasp.  But it does mean that there could be biological features that cause one sex to gravitate more toward certain activities than the other. [/quote]
Yes, there could be.  But I have yet to see any evidence that it's the case (or not the case).

[quote="MCrawford"]
Additionally, men seem to me to be more likely than women to focus on anything single-mindedly -- something that may not be a good overall life strategy, but which may be a strategy that makes scientific achievement a more natural fit to a personality. [/quote]
I agree.  However, there doesn't seem to be anything which suggests that this is a biological rather than social pattern.

[quote="MCrawford"]
None of this precludes the participation of either sex in mathematics and science, though my opinion is that even if all forces of [unfair] discriminatory bias went away today (unfortunately very unlikely), there would likely still be more men in math and science than women -- though perhaps by a smaller margin than is currently the case. [/quote]
In my view, that's a biased opinion.  We don't have enough evidence to have any idea what would happen if all forces of unfair bias evaporated.  There's a study which Virginia Valian quotes, which modelled an organizational structure which began 50/50 male/female at every level of the hierachy and added a 1% bias in favor of males (for promotion/retention decisions).  Quickly the organization shifted to being 65/35 at the upper levels.  MCrawford, if you're interested in looking at the social side of these questions, you might look at Valian's recent book.  I haven't read it myself, but the reviews indicate that it's quite interesting and Valian gave a fascinating talk at a recent Joint Mathematics Meetings.

A significant amount of unfair bias happens unintentionally, and studies (again Valian is a good source for this stuff) show that males and females apply this bias equally.  The two most famous studies are the reviews people gave a paper when it had male vs. female names, vs. a blind review, and the difference in who was selected for orchestral positions when auditions were conducted with vs. without a screen to hide the identity of the auditioners.

[quote="MCrawford"] It won't surprise me if the next generation of mathematicians and scientists includes more women, but I also believe that those women will be having fewer children than the average.
[/quote]
Yup.  I'm sure that this effect would be lessened (though not eliminated by any means) if we had a social structure which guaranteed adequate child care for professionals and supported gender-equal division of housework.

On a lighter note,

 [quote="zeb"]This one man was trying to prove that women were dumber than men by taking the average weight of a lot of man brains and women brains, and he found that women's brains were slightly lighter...[/quote]
Yes, I learned about that study when I was in elementary school (through the SRA series... is that used any more?  how much does that date me?).  It turned out that some subsequent study, using similar data, showed that higher IQs correlated to smaller/lighter brains.

There are all kinds of problems with the structure of such studies.  One study (this is described in Anne Fausto-Sterling's _Sexing the Body_) centered on measuring the volume of an organelle in the brain.  This is problematic to begin with, because it's not clear that there should be a link between the size of brain organelles and their functions... but worse, the study extrapolated from the area of a single cross-section of the organelle to its volume!  Any student of multivariable calculus will know that a single cross-section of an irregularly-shaped object is not a good indicator of the volume.

In general, there is much more in common between males and females than there are differences.  This is verified by just about every study (and if I remember correctly, Pinker/Spelke mentioned it as well).  In fact, some people question why anyone cares about such subtle differences, especially when they tend to occur three standard deviations out from the mean.  (Not me, but I did hear this expressed at a recent panel on the topic.)

[quote="smbelcas"]Pinker may have argued more forcefully, but I found his reasoning unconvincing.  [/quote]Pinker argued more relevantly, if the slides are any indication.  Spelke ignored the essential points and cited, as though they were refutations, studies of mostly irrelevant questions.  Generic cognitive abilities in children younger than 6 are clearly not the main point in hard-science research competence, and it is ridiculous to cite grades as a measure of anything.  Pinker, for his part, can be faulted for citing SAT scores, but at least he correctly indicated their relevance, which is to solving "novel familiar problems", rather than the correlation with grades.

Any convincing will ultimately be done by the data (currently inconclusive) rather than public debates, and maybe Pinker is more practiced with large audiences, but it is striking how little Spelke managed to say.

[quote][in recent studies] the gender gap in spacial rotation/visualization disappears if you either (a) eliminate the time requirement on the test or (b) have the test-takers practice for a couple of hours first. [/quote] Both modifications would nullify the pertinence of the test to research capabilities.  If the test-takers practice and then attempt problems much harder than the practice (a real-life version of this happens every year in the mathematical olympiads), that might be apposite.

[quote] The neuroscientist whom I heard speaking about (a) said that there was a theory (on the part of those conducting the study) that  when under time pressure perhaps males guess more than females, and that females are more likely to leave a question blank when they aren't certain of the answer.  [/quote] If male-female mathematical cognitive abilities are absolutely identical that would settle one part of the controversy, but it would leave open the possibility that other intrinsic differences, such as overconfidence of the kind the neuroscientist suggests, lead to the disproportion of men among top performers.

[quote]In general, there is much more in common between males and females than there are differences.  This is verified by just about every study (and if I remember correctly, Pinker/Spelke mentioned it as well).  In fact, some people question why anyone cares about such subtle differences, especially when they tend to occur three standard deviations out from the mean.  (Not me, but I did hear this expressed at a recent panel on the topic.)[/quote] Larry Summers' comments that occasioned the debate referred to scientists he supposed were "five or more standard deviations above the mean".  Pinker and Summers address that population, Spelke did not.

So, as a side note, I'm reading [i]The Selfish Gene[/i] by Richard Dawkins, and there's a chapter in it that deals with this idea. I suggest you all read it; it's very interesting.

I'm not sure that it's possible to do a study to indicate convincingly the degree of difference (in aptitude or anything else) between men and women.  Thus, I think a combination of personal experience and logic guides most of us in our views of this issue.

Certainly gender has a component that is innate and independent from social pressures.  Since there are differences between men and women, these differences probably play some role in the aptitude of men and women in some subjects.  (If the differences didn't do this, what would they do?)  Math and science seem to be among the most likely subjects for the differences to surface, since we see there the least balance between genders.

It may be true that female scientists are less likely than the average woman to have children.  However, I don't see that the systems for child care and such would have more of an impact on scientists than on professionals in other areas (psychology professors, for example).  And what does this have to do with aptitude?

I agree with Pinker that the question of differences in aptitude should not have anything to do with future opportunities for women.  It shouldn't affect anything, really, that I can think of.  Either way we want to minimize bias in the systems for employment.

Actually, if we accept that men and women are different, there seems to be less potential for workplaces to pressure women to be more like men; women sometimes complain that they need to act like men to be powerful.  You said yourself, smbelcas, that not everyone should have to be just like you to succeed in the math community.  Similarly, men and women should not have to be the same in order to have adequate choices.

[quote="fleeting_guest"]Pinker argued more relevantly, if the slides are any indication.[/quote]

I read a transcript (and don't remember whether I saw the slides as well) which could easily account for our differing interpretations.

[quote="smbelcas"] The neuroscientist whom I heard speaking about (a) said that there was a theory (on the part of those conducting the study) that  when under time pressure perhaps males guess more than females, and that females are more likely to leave a question blank when they aren't certain of the answer.  [quote="fleeting_guest"] If male-female mathematical cognitive abilities are absolutely identical that would settle one part of the controversy, but it would leave open the possibility that other intrinsic differences, such as overconfidence of the kind the neuroscientist suggests, lead to the disproportion of men among top performers.  [/quote] [/quote]

So, to be clear: yes, it would leave open the possibility that other intrinsic differences could lead to a disproportion among top performers.  But (a) there's no reason to think that over/underconfidence is intrinsic as it's trainable and part of our gender-linked social expectations (b) of course confidence could be intrinsic but I know of no evidence to support that and (c) the notion of "top performers" is in itself social.  Let me expand on that last point.  There are lots of ways we can quantify "top peformers," for example by number and/or quality of research papers produced, or by prestige and rank of position.  But each of these measures is in some subtle ways socially influenced.  Researchers who receive funding will have more time to devote to research and will therefore be more likely to produce more results which leads in turn to more funding.  Hiring decisions are made by small groups of people, on the basis of some documents written by the candidate and some letters written by recommenders.  If the candidate isn't good at selling him/herself, but is otherwise a "top performer," this could lead to having a job in which there isn't enough time to do the research which would reveal the "top-ness" of his/her performance.  This could (and often is) a byproduct of a tough job marked as well.  Previous prestige and rank influence future hiring and promotion decisions.  I could go on and on with this, but hopefully I've conveyed my point.  (Tell me if I haven't and I'll try to be clearer.)

Moreover, it's a social value to say things like "more papers" = "top performer" as a society with different values might say "balanced life" = "top performer."  I won't argue that we [i]should[/i] have different standards for evaluating performance; I'm merely saying that the standards we use are socially based.

[quote="snorkack"] Certainly gender has a component that is innate and independent from social pressures. Since there are differences between men and women, these differences probably play some role in the aptitude of men and women in some subjects. (If the differences didn't do this, what would they do?) Math and science seem to be among the most likely subjects for the differences to surface, since we see there the least balance between genders. [/quote]

First off, I'm not sure you and I are using the word "gender" in the same way.  It used to be defined as completely separate from sex, where sex was the biological  bit and gender was the socially constructed bit.  It's now recognized that gender and sex influence each other.  In any case, the definition isn't clear... so I'm not sure that I agree that gender has an innate component.  Biological sex does have a component that is innate and independent from social pressures.

But in any case, let me answer your question: If the differences between men and women didn't play some role in the aptitude of men and women for some subjects, what would they do?  The differences between men and women are evolutionarily based, in the sense that we have gender differentiation as part of sexual reproduction.  The differences do things like let female bodies grow fetuses (is that the correct plural?).  I don't see why we would expect differences in physiology to be linked to differences in the way we "perform" in various "subjects."  I've put the word perform in quotation marks to emphasize the social construction of what we mean by it, and the word subjects is in quotation marks because education is entirely socially constructed.  We decide what separates "subjects" from each other (where's the boundary between physics and chemistry?) and our culture has developed ideas of what it's important to study. (Yes, I do think that we would expect physiological differences to be linked to traits that do not appear to be physiological, via hormone systems and such.  But the number of layers between chemical transport in the body and social people going to graduate school is so large that it's nigh impenetrable by current study.) Again, there's no reason to expect a link---and, I think, also no reason to be surprised by such a link if one does turn up.

[quote="snorkack"] It may be true that female scientists are less likely than the average woman to have children. However, I don't see that the systems for child care and such would have more of an impact on scientists than on professionals in other areas (psychology professors, for example). And what does this have to do with aptitude?[/quote]

I don't think it has anything to do with aptitude.  I was just responding to MCrawford's comment.  But systems of child care may have more impact on scientists than on others if the way we practice science is less compatible with including children in our work.  (I have no idea whether this is true or not, as young children are not part of my life.)  Some psych professors don't have labs and may be able to have their children in their offices while they work, whereas lab-based scientists pretty much can't have kids around while they work.  There's also a stereotype that scientists are expected to work longer hours than other people, but again I don't know if that has any basis in fact.

[quote="snorkack"] Actually, if we accept that men and women are different, there seems to be less potential for workplaces to pressure women to be more like men; women sometimes complain that they need to act like men to be powerful. You said yourself, smbelcas, that not everyone should have to be just like you to succeed in the math community. Similarly, men and women should not have to be the same in order to have adequate choices.[/quote]

Oh, I didn't say that; I'm not that cool.  I agree with it, but that was said [i]to[/i] me.  Long ago, I was wondering aloud in the hallway how it was that other women could reasonably complain of having trouble being in the sciences, when one of my peers said, "Not everyone should have to be like you in order to succeed in mathematics."  This changed my view of the situation completely, and I still try to keep it in mind today.

But, there're some issues with "accepting that men and women are different."  We can accept that there are biological differences, but that doesn't directly lead to acceptance of behavioral differences.  And once we accept behavioral differences, our social value system can come into play and value masculine behaviors over feminine behaviors...which leads right back to women adopting masculine behaviors in order to be equally valued.  It also ignores the women who happen to behave like men---do we have to change our behavior in order to be valued as women?  I think our goal should be to have a social structure which welcomes all kinds of reasonable ways of acting.  Loud scientists and quiet scientists should be listened to alike; pushy scientists and shy scientists should be listened to alike.  It's hard to do this in practice, of course.

[quote="smbelcas"]First off, I'm not sure you and I are using the word "gender" in the same way.  It used to be defined as completely separate from sex, where sex was the biological  bit and gender was the socially constructed bit.  It's now recognized that gender and sex influence each other.  In any case, the definition isn't clear... so I'm not sure that I agree that gender has an innate component.  Biological sex does have a component that is innate and independent from social pressures. [/quote]
All right, so can you describe more clearly what [i]you[/i] mean by "gender"?  I always sort of thought that sex had to do with the body, and gender had to do with behavior, which could be influenced by sex-linked chemicals and such.  Everyone (almost) has one of two sexes, but not everyone has one of two genders.  Maybe this isn't the conventional usage?

[quote="smbelcas"]I think our goal should be to have a social structure which welcomes all kinds of reasonable ways of acting.  Loud scientists and quiet scientists should be listened to alike; pushy scientists and shy scientists should be listened to alike.  It's hard to do this in practice, of course.[/quote]
Indeed.  I personally seem to have trouble listening to anyone, loud or quiet.

"gender" is socially constructed.

"sex" is biologically constructed. There are only two biological sexes (well, with the exception of turner's syndrom, hermaphrodites, etc)

a biological male can be gender woman and a biological female can be gender man or neither gender or both  gender.

Men and women are different.  To borrow a saying from one of my friends, "This is intuitively obvious to the most casual observer."

To answer a question posed in the discussion:  Men and women have the same IQ.  This I can state with certainty, as IQ was created to help place French school children in the correct class.  This was in about 1920, if I remember correctly.  The different results for men and women (actually boys and girls) were controlled and both averaged out at 100, with a standard deviation of 20.  People of different races answer questions differently, since they were not included in the sample that set the standard, so we have different results by race ON AVERAGE.  This has caused a great deal of pain and trouble.  None of it worthwhile, in my opinion.

I have no reason to believe that what led to the creation of this tool to sort French school children is appropriate or useful to judge the value of people’s thinking, their ideas or their intrinsic value.  For example, I can increase a reader’s IQ score (at least as measured by one test) without adding anything useful to their thinking, ideas or value.  All I would have to do would be to provide the correct answers to certain questions.  Those that adminster these tests agree not to do this.  A reader's score could jump, without changing anything except for knowing a little bit more trivia.  Why can’t the world see the rest of IQ score differences in the same light?

In the real world, we very rarely don’t have a better team by having a diversity of different thinking on the team, rather than having all folks that “know” the same thing and “think” the same way.

Anyway, people are all different, they are all made in God’s image and are all valuable.  This is one of those discussions where I’m sure we can’t change the design of people, I wouldn’t want to even if I could and we have a good chance of creating hard feelings and hurt people by pursuing proving which class of people is “best.”

Let me end with two sentences from Forest Gump:  “My mamma said, ‘Stupid is as stupid does,’” and “That’s all I have to say about that.”

[quote="snorkack"]
All right, so can you describe more clearly what [i]you[/i] mean by "gender"?  I always sort of thought that sex had to do with the body, and gender had to do with behavior, which could be influenced by sex-linked chemicals and such.  Everyone (almost) has one of two sexes, but not everyone has one of two genders.  Maybe this isn't the conventional usage?
[/quote]

[quote="agolsme"]
"gender" is socially constructed.

"sex" is biologically constructed. There are only two biological sexes (well, with the exception of turner's syndrom, hermaphrodites, etc)
[/quote]

For a long time (before I was born) "gender" and "sex" were used almost interchangeably.  Then it became recognized that not all the trappings of masculinity and femininity are biologically based (example: if it wears pants, is it male?  if it has long hair, is it female?) and so "gender" started being used for the social construct and "sex" for the biological state of things.  Recently---I'm going to say that this started about 15 years ago---things have been getting blurry again.  Consider a person who was born with XY chromosomes and exhibits secondary sex characteristics we generally associate with females; what sex does that person have?  Here's an example of how social constructs can affect biological states: suppose someone was born XY and exhibited secondary sex characteristics associated with males and then underwent a sequence of operations which resulted in the exhibition of secondary sex characteristics associated with females.  There might be some question as to whether the person has sex type "male" or "female" but most people would for social reasons consider the person female (especially if they didn't know her genetic type).  Biological states affect social ones as well; women who have broad shoulders and narrow hips are perceived as less feminine.

So, snorkack, I'm not sure that I can be very clear about what I mean by "gender"...what I can say is that I think of gender as a social construct which is not completely distinct from the associated biology.  In attempting to answer your question, I went to my bookshelf and looked at the sources that have most clearly informed me about the changing ways in which sex and gender are discussed.  However, I found no definitions (even on the pages suggested by an index heading of "Gender, definitions of").

It may be helpful for me to give you a couple of places to look, in case you're interested.
(1) Judith Butler's essay "Contingent Foundations" has a section (pp. 15 - 16 and bits of 17 - 19) discussing how one might define the word "woman."  I can quote that here if you'd like, as the book it appears in (_Feminists Theorize the Political_) isn't easy to find outside of academic libraries.
(2) The first chapter of Anne Fausto-Sterling's book _Sexing the Body_ contains some historical discussion of sex and gender as well a summary of the nature/nurture debate in science.

So I looked at the little Amazon preview of [i]Sexing the Body[/i] and found... a Möbius band!  Yay!!!!!

However, I concluded, [b]smbelcas[/b], that you are just trying to confuse me, you confusing person.  That was your point, wasn't it?  Even so, I continue to be interested in your apparently unconventional views of gender/sex/whatever; maybe I'll contact you privately, though.

[quote="smbelcas"]It also ignores the women who happen to behave like men---do we have to change our behavior in order to be valued as women?[/quote]
This does seem to suggest that you see clear behavior differences between men and women.

You might want to check out "Gender Trouble" by Butler where she criticizes feminism's fascination with heteronormativity and tendency to ignore alternative modes of criticism(i.e. from queer view point). She defines gender in the first 50 pages or so of that book.

(I think she also criticizes the sex/gender binary, but I am not sure about that because I only bothered to read the first chapter. It's really dense.)[/quote]

[quote="wizard2378"]To answer a question posed in the discussion:  Men and women have the same IQ.  This I can state with certainty, as IQ was created to help place French school children in the correct class.  This was in about 1920, if I remember correctly.  The different results for men and women (actually boys and girls) were controlled and both averaged out at 100, with a standard deviation of 20.[/quote]

There is a conspicuous excess of men on the low, as well as the high, end of society.  Look at a jail, and you'll find a gender asymmetry larger than you'll find in modern mathematics.  This suggests that men might exhibit a higher standard deviation in [i]g[/i] than women, as Lawrence Summers (and some folks before him) proposed.  Or it might just be a consequence of men tending to be more single-minded, as MCrawford mentioned, and be completely independent of [i]g[/i].  I don't think we have enough data to resolve this question yet.  I certainly would not consider the French schoolchild study to be definitive, since that was done before we learned that much about constructing reliable and useful psychometrics.

The case isn't quite closed on average [i]g[/i] being identical between the sexes, either.  I think Helmuth Nyborg's research suggests there is a small (~3 IQ point) male advantage.  I don't think we have enough data to disprove his hypothesis yet.  (Modern IQ tests have been explicitly normalized to yield male and female means of 100.  Read what I say in the next section about IQ vs. [i]g[/i].)

[quote="wizard2378"]I have no reason to believe that what led to the creation of this tool to sort French school children is appropriate or useful to judge the value of people’s thinking, their ideas or their intrinsic value.  For example, I can increase a reader’s IQ score (at least as measured by one test) without adding anything useful to their thinking, ideas or value.  All I would have to do would be to provide the correct answers to certain questions.  Those that adminster these tests agree not to do this.  A reader's score could jump, without changing anything except for knowing a little bit more trivia.  Why can’t the world see the rest of IQ score differences in the same light?[/quote]

Because your view is based on a misunderstanding.

There is nothing sacred about an IQ test score.  As you point out, coaching or outright cheating can raise it.

The reason IQ tests are useful is because they are among the best measures we have of [url=http://en.wikipedia.org/wiki/General_intelligence_factor]Spearman's [i]g[/i] factor[/url], and all kinds of employers care about [i]g[/i].  All that coaching/cheating does is corrupt the measurement, without changing what the test is really trying to measure.  It's analogous to wearing high heels when you're getting your height measured (except that, fortunately, most employers don't care about your height).

[quote="wizard2378"]In the real world, we very rarely don’t have a better team by having a diversity of different thinking on the team, rather than having all folks that “know” the same thing and “think” the same way.[/quote]

You can set up mathematical models incorporating this principle, as well as other human resource principles, and estimate how to maximize your team's performance.  Companies spend a fair amount of money setting up such models, since their hiring decisions go a long way to determining their fate.  If they hire aggressively from a group everyone else is undervaluing, they can gain a major advantage.

What kind of racial and gender mix do we see in the most competitive companies?

[quote="wizard2378"]Anyway, people are all different, they are all made in God’s image and are all valuable.  This is one of those discussions where I’m sure we can’t change the design of people, I wouldn’t want to even if I could and we have a good chance of creating hard feelings and hurt people by pursuing proving which class of people is “best.”[/quote]

What happens when the alternative to allowing genetic engineering in your country is to doom your country to fall further and further behind the rest of the world?

I don't think it's a matter of intelligence, but different types of it. If we're talking about maths, I'd say that in general more boys appear because they can manipulate and apply ideas better. Girls may be able to understand them and so on, but are weaker in problem solving.

In general. How I see it.

Still, it's no coincidence, nor purely social, that most contenders in elite maths events are male. Looking at my own country I'd guess that at least 70-80% invited to the national olympiad are male, and I've checked that since 1981 four girls have been on the Australian IMO team. Four, as opposed to about a hundred boys?

Males and females have minds that develop very differently. You can't claim they are the same and that differences are lead on by society alone.

Whew!  The debate is getting hot again!  I have some responses:

snorkack, I was not trying to confuse you.  Also, my views on sex/gender may seem unconventional for a mathematician to hold, but are informed by the mainstream views of humanists who work on sex/gender.  (Admittedly, their mainstream is nowhere near the public mainstream.)

agolsme, I will eventually read _Gender Trouble_.  I think I'm supposed to read _Bodies That Matter_ first, though.  I like reading Butler partly because she writes so densely; it's a nice challenge.

Dog of Justice and wizard2378, there was a presentation at this past week's Joint Mathematics Meetings on IQ and gender by some guy at MIT; I read his handout on the plane home today.  The original versions of the Binet tests had females scoring on average 2 points higher than males (standard deviation 15) and the test was revised to force the male and female averages to meet.  We have no way of determining a statistical gender differential in the high IQ end of the population because the WAIS (the adult IQ test) only scores up to 155 or 160, which is 4 standard deviations above the mean.  The latest version of the WAIS scores all the way down to 40, so we are more easily able to see statistical gender differentials in the low IQ end of the population.

Dog of Justice, I think your question of what racial/gender mix we see in competitive companies which mathematically model who to hire is rather misleading.  There are all sorts of social forces which statistically affect different ethnic groups, genders, and economic classes differently; some of these forces influence the preparation of workers quite significantly.  Measuring inputs by outputs is even more misleading socially than it is mathematically.

vulgarfraction, are you saying that interest in contests correlates positively with mathematical ability?  It's definitely true that those who are more interested in contests are likelier to take them and to practice for them, and that one can't do well on contests if one doesn't take them.  I know of no evidence that various groups are equally interested in participating in contests.  In fact, there was a session at the Joint Mathematics Meetings a few years ago which focused on how to increase participation (note: not scores, just participation) in contests by females and ethnic minorities.  I also know of no evidence that females are weaker at problem solving than males... do you have some?  Anecdotal evidence doesn't count here, because of self-selection---you're not coming into contact with a random sampling of people.  Finally, I'd add that I don't know of any studies which show that contest performance is a good predictor of professional mathematical success.  Even anecdotal evidence suggests that the correlation isn't strong, though it does seem to indicate that those who succeed at high levels for contests who also succeed in professional mathematics do so at a high level.

[quote="smbelcas"]Dog of Justice and wizard2378, there was a presentation at this past week's Joint Mathematics Meetings on IQ and gender by some guy at MIT; I read his handout on the plane home today.  The original versions of the Binet tests had females scoring on average 2 points higher than males (standard deviation 15) and the test was revised to force the male and female averages to meet.  We have no way of determining a statistical gender differential in the high IQ end of the population because the WAIS (the adult IQ test) only scores up to 155 or 160, which is 4 standard deviations above the mean.  The latest version of the WAIS scores all the way down to 40, so we are more easily able to see statistical gender differentials in the low IQ end of the population.[/quote]

160-100 = 100-40.  If a lower limit of 40 is sufficient to reveal low end statistical gender differentials, an upper limit of 160 should do the same on the high end all other things being equal.  (That said, I wouldn't expect all other things to be equal.)

I don't doubt that the IQ test normalizing process has sometimes resulted in pro-male tweaks.  But the real question being discussed here is about [i]g[/i], not about imperfect measurements of it.  The question will probably be answered by biology and psychology in the next few decades.  If you forced me to bet on the answer today, I honestly would not know which way to bet.

[quote="smbelcas"]Dog of Justice, I think your question of what racial/gender mix we see in competitive companies which mathematically model who to hire is rather misleading.  There are all sorts of social forces which statistically affect different ethnic groups, genders, and economic classes differently; some of these forces influence the preparation of workers quite significantly.  Measuring inputs by outputs is even more misleading socially than it is mathematically.[/quote]

I was responding to a comment by wizard2378 about the practical issue of optimum team composition.  One of my pet peeves is uncritical acceptance of diversity.  Sometimes it's good, sometimes it's bad, and it's important to analyze situations critically and distinguish between the two cases.  (For example, I think Japan and Canada both have more rational stances toward immigration than the US at the moment.)

You are correct that my question did not point to a good way of studying the input.  That's the domain of biologists and psychologists.

[quote="Dog of Justice"][quote="smbelcas"]We have no way of determining a statistical gender differential in the high IQ end of the population because the WAIS (the adult IQ test) only scores up to 155 or 160, which is 4 standard deviations above the mean.  The latest version of the WAIS scores all the way down to 40, so we are more easily able to see statistical gender differentials in the low IQ end of the population.[/quote]

160-100 = 100-40.  If a lower limit of 40 is sufficient to reveal low end statistical gender differentials, an upper limit of 160 should do the same on the high end all other things being equal.  (That said, I wouldn't expect all other things to be equal.)
[/quote]

Ha!!  *grin* Yes, it's true that the WAIS scoring range is even.  It's also true that what I wrote didn't fully reflect what I was thinking, so it's no wonder you noticed the apparent flaw in my reasoning.

Before explaining what I _meant_ to convey, let me say that I don't place much stock in standardized tests of any form, except as statistically likely to differentiate between drastically different abilities.  (For example, I'll believe that most people with SAT verbal scores of 740 have more english-language ability than those with SAT verbal scores of 360.)

When we're talking about the very high end of ability and achievement in the sciences, we're talking about more than three standard deviations above the mean.  The IQ guideline for admission to honors classes in high school is (or at least used to be) in the 120 - 130 range.  Given the stratification which subsequently happens in the career track, my guess is that the average IQ of strong college mathematics students would be higher than that of honors students in high school,  and that this would increase again at the graduate school level.  (I'm not saying that there's a correlation between those who are smarter and those who go to graduate school, but observing an effect on the average measurement from the narrowing of the pool of people we're looking at.)  My guess is that the IQ of most people who attend the top 20 mathematics graduate schools is 145 or above.   That's the group of people likeliest to go on to careers in academia, and so that's the group we're talking about when we ask questions about differential representation at top universities (the question which started this whole debate a year ago).

On the other hand, when we think of the low end of the IQ spectrum, we're thinking of people who are 1 - 2 standard deviations below the mean.  That's the 85 - 70 range in IQ.  Either 70 or 65 (I don't remember which) is the cutoff for severe/profound mental retardation.  We don't even [i]see[/i] people with IQs below that in our society unless we work with them professionally; many of them live in care facilities.  Many years ago, I read a book on education for students in this group, and it contained several case studies.  I recall one person who, at age 10 or so, could not move a left leg together with a right arm.  Others could not recognize alphabet letters or numerals at similar ages. Another example, for those who've read _Flowers for Algernon_, is that the lead character was close to the severe/profound mental retardation line.  When we think about the jail populace, the low-intelligence people there suffer from mild mental retardation rather than severe/profound mental retardation.

So what I was trying to convey is that we think of 3 s.d. and above as the high end, but we think of 1 - 2 s.d. as the low end; the WAIS measures that low end more accurately than it measures that high end.  In other words, I don't think we're talking about the same statstical populations when we discuss these groups of people.  I'm sure that most studies have correct statistical descriptions of the groups they consider, but I think that in the rest of society there's an imbalance in our perceptions.

[quote="smbelcas"]vulgarfraction, are you saying that interest in contests correlates positively with mathematical ability?  It's definitely true that those who are more interested in contests are likelier to take them and to practice for them, and that one can't do well on contests if one doesn't take them.  I know of no evidence that various groups are equally interested in participating in contests.  In fact, there was a session at the Joint Mathematics Meetings a few years ago which focused on how to increase participation (note: not scores, just participation) in contests by females and ethnic minorities.  I also know of no evidence that females are weaker at problem solving than males... do you have some?  Anecdotal evidence doesn't count here, because of self-selection---you're not coming into contact with a random sampling of people.  Finally, I'd add that I don't know of any studies which show that contest performance is a good predictor of professional mathematical success.  Even anecdotal evidence suggests that the correlation isn't strong, though it does seem to indicate that those who succeed at high levels for contests who also succeed in professional mathematics do so at a high level.[/quote]

Yeah, you've got a good point there- you wouldn't be in the running if you're not interested, would you? But conversely, it's also difficult to be interested in something you're not good at, so that works both ways. Statistically, there's no evidence of the kind you mentioned, because looking at contest results, those are influenced by uneven participation, however those are random down to perhaps country, and the fact that they performed outstandingly in contests. Individual sampling isn't anywhere near broad enough. Encouraging interest in contests in females can help gauge that though.

On professional success in maths, well that doesn't require the very elite. It requires interest, though, and relatively ordinary people can go into that, even if the elite from the IMO and such contests would prove to do well. On the other hand, they could choose to go be a zookeeper. Or a rock star. It's down to interest. But then, I was referring to problem solving skills alone. :)

[quote="smbelcas"]Before explaining what I _meant_ to convey, let me say that I don't place much stock in standardized tests of any form, except as statistically likely to differentiate between drastically different abilities.  (For example, I'll believe that most people with SAT verbal scores of 740 have more english-language ability than those with SAT verbal scores of 360.)[/quote]

The value of standardized tests is context dependent.  Obviously it's bad news when the Army decides to flagrantly exceed its usual limit of Category IV recruits, even though Category IV is defined by nothing but a range of scores on a standardized test.  On the other hand, frequently one deals with groups of individuals that have been pre-screened to mostly fall in a narrow [i]g[/i] range (graduate students, etc.), and in that case exact standardized test scores are essentially useless compared to other forms of information.

[quote="smbelcas"]On the other hand, when we think of the low end of the IQ spectrum, we're thinking of people who are 1 - 2 standard deviations below the mean.  That's the 85 - 70 range in IQ.  Either 70 or 65 (I don't remember which) is the cutoff for severe/profound mental retardation.  We don't even [i]see[/i] people with IQs below that in our society unless we work with them professionally; many of them live in care facilities.  Many years ago, I read a book on education for students in this group, and it contained several case studies.  I recall one person who, at age 10 or so, could not move a left leg together with a right arm.  Others could not recognize alphabet letters or numerals at similar ages. Another example, for those who've read _Flowers for Algernon_, is that the lead character was close to the severe/profound mental retardation line.  When we think about the jail populace, the low-intelligence people there suffer from mild mental retardation rather than severe/profound mental retardation.[/quote]

Oops, caught me there.  It's obvious to me now that my comment about gender imbalance in jail doesn't have much relevance to the Summers hypothesis of greater male standard deviation in [i]g[/i], since other factors such as greater male aggression predominate.

In other news, there was an article published today, [url=http://biology.plosjournals.org/perlserv/?request=get-document&doi=10.1371/journal.pbio.0040019]Men, Women, and Ghosts in Science[/url], that should be of interest to those following this thread.

[quote="smbelcas"][quote="fleeting_guest"]Pinker argued more relevantly, if the slides are any indication.[/quote]

I read a transcript (and don't remember whether I saw the slides as well) which could easily account for our differing interpretations.  [/quote]

Thanks, I found and read the transcript, which is at the same web site.

The transcript confirms the impression made by the slides, which is that Spelke dodged Summers and Pinker's principal and most controversial argument: that the male/female discrepancies are consistent with, and would be largely explained by, higher variance (with or without a higher mean) in the distribution among men of whatever innate traits are relevant.   Her only counterargument was to simultaneously apply high skepticism to the SAT while placing complete uncritical confidence in college grades as a measure of performance.

Spelke gainsays the SAT as [i]potentially[/i] gender-biased in favor of "male" (e.g. geometric) problem solving strategies, to the point that no further explanation is needed of the 13-to-1 disproportion of males in the CTY study.  She states as a self-evident fact that one could reverse-bias the SAT in favor of "female" approaches to problem solving by changing the numbers of (say) linguistic versus geometric questions, but given that both types of question are already well-represented, it is a mystery how such engineering could alter the male/female ratio [b]by a factor of 169[/b].   In fact, people have tried hard to eliminate not only gender but racial biases in the mathematics SAT, and no amount of fine-tuning of the questions has led to significant changes in outcome. Spelke's claims about the super-sensitivity of the results to the proportion of this or that style of question are implausible, particularly as a blanket dismissal of the SAT.    On the other hand, she uncritically posits college grades as the gold standard of gender-neutral performance evaluation, ignoring the fact that grades and graduation rates are well-known to be higher for females, and for gender-specific reasons (whether social or innate in origin).  Alternative measures at university age, such as performance of male and female math majors from the CTY sample in mathematics competitions or academic research, are not discussed, although they would probably replicate the SAT imbalances.


group theory
abstract algebra
geometry
3D geometry
vector
symmetry
function

I think that this is a very difficult problem :

Suppose that G is a group and H a subgroup of G. H has finite index n. Show that there exists a homomorphism phi : G -> Sn with ker(phi) being a subgroup of H.
(Sn is the symmetric group of degree n.)

Think about this: a homomorphism into $ S_n$ is the same as an action on a set of order $ n$. What are there $ n$ of that $ G$ could act on?

[hide="Answer/Hint"]
There are $ n$ cosets! Act on them. How might you act on them? (In such a way that the kernel is contained in $ H$)
[/hide]

[hide="Solution"]
For a left coset $ aH$ and $ g \in G$, we define $ g(aH)$ to be the coset $ gaH$. It is fairly elementary to check that this is an action: the set $ gaH$ is another coset, and $ f(g(aH)) \equal{} f(gaH) \equal{} fgaH \equal{} (fg)(aH)$ for $ f,g \in G$.

If $ g$ is in the kernel, then $ g(H) \equal{} gH \equal{} H$, which means that $ g \in H$.

In fact, if $ N \triangleleft G$, $ N \le H$, then for $ n \in N$ and $ aH$ a left coset, we have $ n(aH) \equal{} naH \equal{} a(a^{ \minus{} 1} n a)H \equal{} aH$ since $ N$ is normal in $ G$ and contained in $ H$. Thus $ N$ is in the kernel of this homomorphism. This shows that the kernel of this homomorphism is the "largest" normal subgroup of $ G$ contained in $ H$.

[/hide]

[hide="Applications"]
Let $ |G| \equal{} p^2$ for a prime $ p$, and let $ H$ be a subgroup of order $ p$. What does it tell you when you let $ G$ act on the left cosets of $ H$?

In general, $ H \le G$ and $ |G| \nmid ([G: H])!$, what can you say about the homomorphism $ G \rightarrow S_{[G: H]}$? What does this tell you about $ G$?
[/hide]

May i ask what exactly do you mean by saying ''prove that this is an action'' ?
Thanks

[quote="http://mathworld.wolfram.com/GroupAction.html"]A group $ G$ is said to act on a set $ X$ when there is a map $ \phi: G\times X \rightarrow X$  such that the following conditions hold for all elements $ x$ in $ X$.

1. $ \phi(e,x) \equal{} x$ where $ e$ is the identity element of $ G$.

2. $ \phi(g,\phi(h,x)) \equal{} \phi(gh,x)$ for all $ g,h \in G$. [/quote]

So I'm proving that the map which takes $ g$ and $ aH$ to $ gaH$ satisfies the axioms above.

Is this a renaming of a homomorphism? I haven't seen something similar in the course, and this was a problem given to us after the end of the course. I mean, is this something i should prove in order to follow with the proof of the first problem?
Thanks

Sorry, I should have provided background on group actions. The natural examples of groups come from groups of symmetries on different object. For example, you have the group of permutations on a set of $ n$ elements, $ S_n$. You also have the group of symmetries of some geometric figure, such as a cube or a regular polygon. You can also look at the group of isometries of $ \mathbb{R}^n$ or the group of linear transformations of a vector space.

In each of these cases, an element of the group defines a bijective mapping on the elements of the object it acts on (e.g. an isometry defines a mapping on the points of the space). And in each of these cases, multiplication of group elements corresponds to composition of mappings. For example, the mapping defined by the product of two linear transformations $ A$ and $ B$ (where by "product" I mean the group operation) is simply what you get when you first apply the mapping corresponding to $ B$ and then apply that of $ A$.

One way to think of this phenomenon is as a homomorphism from our group into the group of bijective mappings on a set. As I explained, for each group element, we get a bijective mapping on that set. The property I explained above implies that it is a homomorphism. For example, for each symmetry of a regular polygon, we get a bijective mapping from the set of its vertices to itself. This is a homomorphism from the group $ D_n$ (the group of symmetries of a regular $ n$-gon) into the group of bijective mappings of $ n$ points.

Of course, you can also think of it like this: for each element $ g$ of the group $ G$ and element $ x$ of the set $ X$ we're acting on, we get another element of $ X$, i.e. $ g(x)$. So if we take a symmetry $ \sigma$ of a regular $ n$-gon  (i.e. an element of $ D_n$) and a vertex $ v$ of that $ n$-gon, we get another vertex, i.e. wherever $ \sigma$ takes $ v$. This is the definition of a "group action" I cited above. You can think of it either way: as a homomorphism from a group into a group of bijective mappings, or as the definition of "group action" I mentioned above. The key concept is that the group can be viewed as a group of mappings on some set.

So for $ G$ acting on the left cosets of $ H$, I'm simply thinking of each element of $ G$ as a bijection from the set of left cosets of $ H$ into itself. For $ g \in G$ and $ aH$ a left coset, we get the new left coset $ gaH$. This can be thought of as a function $ G \times X \rightarrow X$, where $ X$ is the set of left cosets of $ H$. We can also think of it as giving us a permutation of the left cosets for each element of $ g$, i.e. a homomorphism from $ G$ into $ S_n$ (as there are $ n$ cosets).


If  $ F_1$ contains $ F_2$; $ F_2$ contains $ F_3$;… is a nested sequence of nonempty closed sets, then the intersection of $ F_n$ is not empty.

Let $ F_n \subset \mathbb{R} \equal{} \{ x \in \mathbb{Z} | |x| \ge n \}$.  The point is that closed doesn't imply compact.

thank you both for the example and the code! I'm a rookie.
I think it should be:
$ F_n \subset \mathbb{R} \equal{} \{x \in \mathbb{R} \mid |x| \ge n \}$

Both examples work.  I guess I wanted a "simpler" example, but it's a matter of preference.


probability
number theory
greatest common divisor
relatively prime

1. Let $ a$ and $ b$ be real numbers satisfying
$ a^4 \plus{} 8b \equal{} 4(a^3 \minus{} 1) \minus{} 16\sqrt {3}$
and
$ b^4 \plus{} 8a \equal{} 4(b^3 \minus{} 1) \plus{} 16\sqrt {3}$.
Find $ a^4 \plus{} b^4$.

2. There are three bags. One bag contains three green candies and one red candy. One bag
contains two green candies and two red candies. One bag contains one green candy and
three red candies. A child randomly selects one of the bags, randomly chooses a first
candy from that bag, and eats the candy. If the first candy had been green, the child
randomly chooses one of the other two bags and randomly selects a second candy from
that bag. If the first candy had been red, the child randomly selects a second candy
from the same bag as the first candy. If the probability that the second candy is green
is given by the fraction m/n in lowest terms, find m + n.

3. How many positive integers less than 200 are relatively prime to either 15 or 24?

15 = 3 x 5, 24 = 2^3 x 3
All multiples of 3 have GCD > 1 for both 15 and 24.
In addition, nos. having factors 2 and 5 have GCD > 1 for both 15 and 24.
So all multiples of 3 or 10 are to be excluded.
198 = 3 x 66, 190 = 10 x 19, 180 = 30 x 6
66 + 19 - 6 = 79 nos. have GCD > 1 for both 15 and 24.
Rest of the nos. are relatively prime to either 15 or 24 ( or both ).
Answer is 199 - 79 i.e. 120.

I believe GANITEE's solution is wrong because it does not take away numbers like 4, 25, etc. which are neither multiples of 3 nor 10, but are not relatively prime to 15 or 24. Therefore, the answer is much less than 120.  :!:

But it asked which are relatively prime to EITHER. 4 is relatively prime to 15, so it should be included. 25 is relatively prime to 24, so it should be included.

Ceegers, you know math ( and I think you enjoy it)...that's why you're here.

 VIPMaster, thanks for going through my solution. I try to make it more clear....

 Let A = set of nos. < 200 that are relatively prime to 15
      B = set of nos. < 200 that are relatively prime to 24
 We want n( A or B ) and we first find no. of elements in it's compliment set.
 ( A or B)' =  A' and B'
              = set of nos. not relatively prime to both 15 and 24
 Nos. such as 4 and 25 are not in this set. Only multiples of 3 or 10 are in it.[/b]


The official solution to Fermat is [url=http://www.cemc.uwaterloo.ca/english/contests/fermat.shtml]available[/url].

Look at the solution to #25. I wonder if they are going to call the problem off or retain B as the correct answer.

I sure hope they keep it, I got it right for a change!

As I remember, Dr. Rice mentioned a few weeks ago that there was something funny about #25.

When I freaked out, he assured me that 57 would be the answer.. I'm sure it won't count against you..
Argh, this is exactly the reason I don't like multiple choice contests very much.

I hope they keep B as the answer.  If they call off the problem, then so many ppl will get the perfect scroe of 142, thus they can't afford to give each one of us $200.
If this really happens....Maybe people will have to share the money, say 50 poeple got perfect, then each of us will get 4 bucks...lol :D

Hahaha...that reminds me of that insane year where like 23 people got perfecct on cayley and 21 people got perfect on fermat.  That was the year I wished I wrote the Fermat instead of Cayley, lol.

Ah, well...
I totally screwed up this year's Fermat.

Are they going to ignore question 25, or is B still the right answer (I blanked it :( )

what so you automatically get money if you score perfect on the harder contests?

I know it obviously wasn't for the Gauss because I didn't get any :(

Only regional winners of Fermat and top ~5 of Euclid get money from Waterloo.


If sec x + tan x = 22/7, find cosec x + cot x.

Well, right now, I see that cosec x + cot x = cot x/2 so I try to make a few substitutions to express the first expression in terms of cot x/2, which I can then solve for. However, I dont see how to do this. Any ideas?

[color=white] First, multiply top and bottom by (secx-tanx). Now on the top we have 1 and downstairs (secx-tanx). This is still equal to 22/7, so (secx-tanx)=7/22. Now you have two independent equations w/ two unknowns. Solve to find secx=(533/308) and tanx=(435/308).

Next, solve for sinx in the above. This yields (435/533). Now, plug in the numbers to get (533/435)+(308/435)=(29/15)

[/color]

if sec x + tan x = 22 / 7, and since

(sec x)2 - (tan x)2 = 1,

sec x - tan x = 7 / 22, therefore we have a system of two equations and can solve for sec x and tan x, which you can use to find sin x and cot x.


[size=150][color=blue]PishA Pish [/color][/size][color=green][size=200][i]Eide NoWroOz[/i][/size][/color] :clap2:   [size=150][color=blue]ro be haMeyE HAM vAtAnaNe aziZ :love:  taBrik MiGam isHallA sAle $1385=5\times277$ sale khooB bAraToOn BoOde va ishalla emsal yani sale $1386=2\times3^{2}\times7\times11$ :D  haM sAle khOoBo KhOsHkeltAri baRatOon BasHe[/color][/size]

[size=150][color=blue]baraye ma ham Doa koniiiiiiiiiiiiiiiiiiiiiiiiiiiDa yadetoon narehaaaa[/color][/size]... :lol:

Manam Sale jadido be hame tabrik migam , omidvaram sali por az moaffaghiat bashe baraye hame . natayeje marhale 1 ham ke oomad , be oonayi ke ghabool shodan tabrik migam va oonayi ham ke ghabool nashodan omidvaram salhaye badi ya too konkoor moaffagh bashan :)
be omide didar too marhale 2 :)

manam sale noro pishapish be hameye barobachehayee riazii tabrik migam ishala sali khobo khosh dashte bashin hamchenin be hameye bachehayii ke marhalle 1 ghabool shodan ishala ke hamaoon sali khobo khosh va sar shar az movafaghiayt va olamp dashte bashim:D

che ghadr baraye ham arezoohaye bahal mikonin!

[color=blue]SALE NO MOBARAK,[/color]    [color=green]SALE NO MOBARAK, [/color]     [color=yellow]SALE NO MOBARAK, [/color]
~~~~~~~~~~~~~~~:roll:  :roll:  :roll:  :roll: ~~~~~~~~~~~~~~~~

PS. man be soorate khalagh, tabrik goftam!
PS. tafrih yadetoon nare,  :surf:

salam. manam eyde nowrooza be hameye farsi zabanha too har kojaye donya tabrik migam. :clap2:  :clap:  :thumbup:

بوی باران,بوی سبزه, بوی خاک
شاخه های شسته, باران خورده, پاک
آسمان آبی وابر سپید
برگ های سبز بید
عطر نرگس, رقص باد
نغمه وبانگ پرستوهای شاد
خلوت گرم کبوترهای مست
 نرم نرمک می رسداینک بهار
خوش به حال روزگار

فریدون مشیری

Man ham Eid ro be hameye azizan tabrik migam :)

sayeye hagh,salame eshgh,sa'adate rooh,salamate tan,sarmastie bahar,sokoote 2a,sorure javedane"inast haftsine ma ,pishkeshe shoma"
eydetoon mobarak,omidvaram hamegi sali por az shadi o sarshar az omid o moaffaghiat dashte bashim :)

Eide norooz ba hame ham vatanan mobarak.
Har roozetan norooz, noroozetan pirooz.
[img]http://upload.wikimedia.org/wikipedia/commons/7/7c/Haft_Seen.jpg[/img]


This is the only problem on pre-TST 2003 that is worth attention.
However I do like it so much!

Consider n,N natural numbers and n numbers a_1..a_n smaller than N with LCM(a_i,a_j)>N
Prove 1/a_1+..+1/a_n<2

i solve the problem:
given a_1,a_2,..,a_k are positive integer,a_i<=n(with any i)
and lcm(a_i,a_j)>nwith any i different j
show that:sum(1/a_i)<2

with any i=1,..,k ,define b_i=[n/(2*a_i)]+1=1,c_i=[n/a_i]  ([*]:the integarl part of *)
b_i>n/2a_i,c_i<=n/a_i
we have: q*a_i in [[n/2]+1,n] (with any i=1,..,k),b_i<=q<=c_i
all the  above number  are distinct because if assume exist q*a_i=q'*a_j=l ==>
l is divisible by a_i and a_j==> l is divisible by lcm(a_i,a_j) ==>l>n(contract)
 we have  sum    (sum(1/qa_i))          >sum  ((c_i-b_i+1)/c_i)*1/a_i
               i=1,..,k       q=b_i,..,c_i          i=1,..,k
>1/2*sum(1/a_i)
         i=1,..,k
then, we prove that:
sum    (sum(1/qa_i))          <1/([n/2]+1) +...+1/n
i=1,..,k       q=b_i,..,c_i
so we have sum(1/a_i)<2(1/([n/2]+1) +...+1/n)<2
sorry because i don't know how to type Latex(and i'm lazy to learn)


geometry
circumcircle
geometric transformation
reflection
homothety
cyclic quadrilateral
Inversion

Let $\Gamma$ be the circumcircle of triangle $ABC.$ Denote the circle which touches to the sides $AB, \ AC$ and touches to $\Gamma$ internally at $P$ by $\Gamma_{A},$ and the circle which touches to the sides $AB, \ BC$ and touches to $\Gamma$ internally at $Q$ by $\Gamma_{B},$ and the circle which touches to the sides $AC, \ BC$ and touches to $\Gamma$ internally at $R$ by $\Gamma_{C}.$ Prove that the lines $AP,\ BQ,\ CR$ are concurrent.

Let $f_{a}$ is the inversion having $A$ for pole and $AB.AC$ for constant.
$f_{a}(B) = B' , f_{a}(C)=C'$ so $\Delta AB'C'$ is reflect $\Delta ABC$ about the bisector of $\hat{BAC}$
$f_{a}(\Gamma_{A})$ is the excircle opposite $A$ of $\Delta AB'C'$. So $f_{a}(\Gamma_{A})$ touches $B'C'$ at $P'$ - which is $f(P)$
Let excircle opposite $A$ of $\Delta ABC$ touches $BC$ at $A_{1}$. Easy to prove $A_{1}$ is reflect $P'$ about the bisector of $\hat{BAC}$. It's means that $AP$ is reflect $AA_{1}$ about the bisector of $\hat{BAC}$.
Proving $BQ$ is reflect $BB_{1}$ about the bisector of $\hat{ABC}$ and $CR$ is reflect $CC_{1}$ about the bisector of $\hat{BCA}$ are same.
Note that $AA_{1},BB_{1},CC_{1}$ is conccurency so use the ceva's theorem ( the trigonometric form ) we have $AP,BQ,CR$ is conccurency, too.
It's to difficult to say... because my English is very bad...

From your nice solution one can say that the point of concurrency is the isogonal conjugate of the Nagel point.

Another solution:
It is a known fact that, given three circumference, the 3 (3 or 1 of them are external) centres of similitude of the circumference taken two by two lie on a line. It follows from Menalaus' theroem.

Now: the centre of similitude of $\Gamma$ and $\Gamma_{A}$ is obviously $P$. The external centre of similitude of the incircle and $\Gamma_{A}$ is $A$. Then, the external centre of similitude of the incircle and the circumcircle lies on AP!
$\Rightarrow$ Ap,BQ,CR concur in the external centre of similitude of the  incentre and circumcentre of ABC.

Combining my and PDatK40SP's proof we get the nice result:
The external centre of the similitude of the incircle and the circumcircle is  the isogonal conjugate of Nagel's point.

It's also true that the internal centre of similitude of the incircle and the circumcircle is the isogonal conjugate of Gergonne's point  :)

[quote="edriv"]
It is a known fact that, given three circumference, the 3 (3 or 1 of them are external) centres of similitude of the circumference taken two by two lie on a line. It follows from Menalaus' theroem.
[/quote]

Can you show the proof please, because I can't!

Given two circles $(O_{1},R_{1}),(O_{2},R_{2})$. $S$ is the external similitude center is equivalent with $\frac{\bar{SO_{1}}}{\bar{SO_{2}}}= \frac{R_{1}}{R_{2}}$. Use this and Menelaus' theorem, you will prove the lemma easily  :wink:
With [b]edriv's solution[/b], I think $\Gamma$ needn't to be the circumcircle of triangle $ABC$. It can be an arbitrary circle which contains $A,B,C$  :)

The triangle formed by the centers of the three circles!

[quote="kunny"][color=darkred][b][u]Enunciation.[/u][/b] Let $w$ be the circumcircle of triangle $ABC$. Denote the circle $w_{1}$ which touches to the sides $AB, \ AC$ and touches to $w$ [b]internally[/b] at $T_{1}$. Define analogously the internal tangent points $T_{2}$ and $T_{3}$ of the circles $w_{2}$ (tangent to the lines $BC$, $BA$), $w_{3}$ (tangent to the lines $CA$, $CB$) respectively with the circle $w$ . Prove that the lines $AT_{1},\ BT_{2},\ CT_{3}$ are concurrently in [b][u]the isogonal conjugate of the Nagel's point for the triangle $ABC$[/u][/b].[/color][/quote]
[color=darkblue]I"ll present three well-known useful lemmas.

[b][u]Lemma[/u][/b] $1\blacktriangleright$ Let $w$ be a given circle and let $\alpha$ be a circle which is tangent internally (externally) to the circle $w$ in the point $T$.
For a point $M\in \alpha$ define the tangent d to the circle $\alpha$ and the intersections $\{X,Y\}=w\cap d$.
Then the ray $[TM$ is a internal (external) bisector of the angle $\widehat{XTY}$.

[b][u]Lemma[/u][/b] $2\blacktriangleright$ Let $ABCD$ be a convex cyclic quadrilateral for which denote $S\in AC\cap BD$. Then there are the following relations :
$\frac{SA}{AB\cdot AD}=\frac{SB}{BA\cdot BC}=\frac{SC}{CB\cdot CD}=\frac{SD}{DC\cdot DA}$.

[b][u]Lemma[/u][/b] $3\blacktriangleright$ The circle $w_{1}$ touches the sides $[AB]$, $[AC]$ in the points $X$, $Y$ respectively and touches internally the circle $w$ in the point $T_{1}$.
Then the incenter $I\in XY$ (apply the Pascal's theorem to the cyclic hexagon $VT_{1}UCAB$, where $U\in T_{1}X\cap w$, $V\in T_{1}Y\cap w$) and
$AX=AY=\frac{bc}{p}$, where $2p=a+b+c$ and $\frac{XA}{XB}=\frac{b}{p-b}\ \ ,\ \ \frac{YA}{YC}=\frac{c}{p-c}$.

[b][u]Proof of the proposed problem.[/u][/b] From the first lemma and the third lemma obtain that the rays $[T_{1}X$, $[T_{1}Y$ are bisectors of the angles $\widehat{AT_{1}B}$, $\widehat{AT_{1}C}$, i.e. $\frac{T_{1}B}{T_{1}A}=\frac{XB}{XA}=\frac{p-b}{b}$ and $\frac{T_{1}C}{T_{1}A}=\frac{YC}{YA}=\frac{p-c}{c}$. Thus, $\frac{T_{1}B}{T_{1}C}=\frac{c(p-b)}{b(p-c)}$. Denote the intersection $S_{1}\in AT_{1}\cap BC$. From the second lemma applied in the cyclic quadrilateral $ABT_{1}C$ obtain $\frac{S_{1}B}{S_{1}C}=\frac{BA\cdot BT_{1}}{CA\cdot CT_{1}}$, i.e. $\boxed{\frac{S_{1}B}{S_{1}C}=\frac{c^{2}(p-b)}{b^{2}(p-c)}}$. If for the point $N\in BC$, the ray $[AN$ is the $A$- Nagel's line, then $NB=p-c$, $NC=p-b$ and $\frac{S_{1}B}{S_{1}C}\cdot\frac{NB}{NC}=\left(\frac{c}{b}\right)^{2}$, i.e. rays $[AN$, $[AS_{1}$ are isogonal conjugate in the angle $\widehat{BAC}$ a.s.o. In conclusion, the lines $AT_{1}$, $BT_{2}$, $CT_{3}$, i.e. $AS_{1}$, $BS_{2}$, $CS_{3}$ are concurrently in the isogonal conjugate of the Nagel's point for the triangle $ABC$.

[b][u]Remark 1.[/u][/b] Denote $Z_{1}\in XY\cap BC$ and $Z_{2}\in T_{1}T_{1}\cap BC$. Then $\frac{Z_{1}B}{Z_{1}C}=\frac{T_{1}B}{T_{1}C}$ and $\frac{Z_{2}B}{Z_{2}C}=\left(\frac{T_{1}B}{T_{1}C}\right)^{2}$, i.e. the ray $[T_{1}Z_{1}$ is the external bisector of the angle $\widehat{BT_{1}C}$ and the point $Z_{2}$ is the middlepoint of the segment $[Z_{1}Z]$, where the ray $[T_{1}Z$ is the internal bisector of the angle $\widehat{BT_{1}C}$.

[b][u]Remark 2.[/u][/b] Prove analogously the following dual problem :[/color]

[quote="Virgil Nicula"][color=darkred] [b][u]Enunciation.[/u][/b] Let $w$ be the circumcircle of triangle $ABC$. Denote the circle $w'_{1}$ which touches to the sides $[AB]$, $[AC]$ and touches to $w'_{1}$ [b]externally[/b] at $T'_{1}$. Define analogously the external tangent points $T'_{2}$ and $T'_{3}$ of the circles $w'_{2}$ (tangent to the lines $BC$, $BA$), $w'_{3}$ (tangent to the lines $CA$, $CB$) respectively with the circle $w$ . Prove that the lines $AT'_{1}$, $BT'_{2}$, $CT'_{3}$ are concurrently in [b][u]the isogonal conjugate of the Gergonne's point for the triangle $ABC$[/u][/b].[/color][/quote]

[quote="Virgil Nicula"]

and the point $Z_{2}$ is the middlepoint of the segment $[Z_{1}Z]$, [/quote]

How do you know it is the middlepoint? I mean, I know the middlepoint has this ratio with respect to $B,C$ (from your topic about harmonic division, I think). But there's supposed to be another point with the same ratio. How do you know this one is the middlepoint?

And in case you are too lazy to prove it on an exam - it's called Monge's Theorem, :D

[quote="PDatK40SP"]Easy to prove $A_{1}$ is reflect $P'$ about the bisector of $\hat{BAC}$.[/quote]

Sorry, how do you prove that $A_{1}$ is the reflection of $P'$ about the bisector of $\hat{BAC}$?

This is the easy question.
These circles are ther mixtilinear circles that and by homothecy it's so easily

Dear Mathlinkers,
allso at

http://jl.ayme.pagesperso-orange.fr/Docs/Mixtilinear1.pdf  p. 23-25

Sincerely
Jean-Louis

Proof:
By the three homotheties theorem (or by Monge d'Alembert's theorem), since the center of homothety mapping the "cycle" incircle to "cycle" $\Gamma_A$ is $A$, "cycle" $\Gamma_A$ to "cycle" $O$ is $P$, the line $AP$ passes through the center of homothety mapping "cycle" incircle to the "cycle" $O$. Here I have used the word cycle for a directed circle, oriented anticlockwise by convention.

In this problem $\Gamma$ can be any circle


analytic geometry
symmetry
complex numbers
algebra unsolved
algebra

Let $ a,b,c$ be distinct complex numbers such that
$ (a\minus{}b)^{7}\plus{}(b\minus{}c)^{7}\plus{}(c\minus{}a)^{7}\equal{}0$
Prove that $ a,b,c$ are the complex coordinates of the vertices of an equilateral triangle

[b]Solution.[/b] Let $ u\equal{}a\minus{}b,\ v\equal{}b\minus{}c,\ w\equal{}c\minus{}a$.  Clearly, $ u\plus{}v\plus{}w\equal{}0\implies (u\plus{}v)^{7}\minus{}u^{7}\minus{}v^{7}\equal{}u^{7}\plus{}v^{7}\plus{}w^{7}\equal{}0$.  Equivalently, $ uv(u\plus{}v)(u^{2}\plus{}uv\plus{}v^{2})^{2}\equal{}0$.  Since $ a\neq b\neq c$, $ uvw\neq 0\neq u\plus{}v$.  It follows that $ u^{2}\plus{}uv\plus{}v^{2}\equal{}0\implies u^{3}\equal{}v^{3}$.  Hence, by symmetry, $ u^{3}\equal{}v^{3}\equal{}w^{3}\implies |u|\equal{}|v|\equal{}|c|$.

And the conclusion follows. $ \Box$


[quote="cnyd"]$ 4(3a \plus{} b \plus{} c)(b^{2} \plus{} c^{2} \minus{} 2bc\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(b \plus{} c)}) \equal{} a^{2}(3a \plus{} b \plus{} c) \minus{} 4a(a \plus{} b)(a \plus{} c) \plus{} 4a(a \plus{} b)(a \plus{} c)(\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(b \plus{} c)})^{2}$[/quote]

What a great Olympiad problem !

Btw, it's false : Choose $ a\equal{}c\equal{}0$ and $ b\ne 0$ and $ RHS\equal{}0$ while $ LHS\equal{}4b^3$

[quote="cnyd"]ahh sorry
$ 4(3a \plus{} b \plus{} c)(b^{2} \plus{} c^{2} \minus{} 2bc\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} c)}) \equal{} a^{2}(3a \plus{} b \plus{} c) \minus{} 4a(a \plus{} b)(a \plus{} c) \plus{} 4a(a \plus{} b)(a \plus{} c)(\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} c)})^{2}$[/quote]

Are you sure it's not $ 4(3a \plus{} b \plus{} c)(b^{2} \plus{} c^{2}$ $ \minus{} 2bc\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} c)}) \plus{}$ $ 4(3a \plus{} 7b \minus{} c)(b^{2} \plus{} c^{2}$ $ \minus{} 2bc\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} b\plus{}c)}) \equal{} a^{3}(3a \plus{} b \plus{} c) \minus{}$ $ 4a(a \minus{} b)(a \plus{} c) \plus{}$ $ 4a(a \plus{} b)(a \plus{} c)(\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} c)})^{2}$ $ \minus{} a^{2}(3a \plus{} b \plus{} 2c) \minus{} 4a(\minus{}a \plus{} b)(a \plus{} c) \plus{} 4a(a \plus{} b)(a \plus{} c)(\frac {(a \plus{} b)^{2} \plus{} (b \minus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} c)})^{2}$ ?

Check again before we deal with ...

[quote="cnyd"]ahh sorry
$ 4(3a \plus{} b \plus{} c)(b^{2} \plus{} c^{2} \minus{} 2bc\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} c)}) \equal{} a^{2}(3a \plus{} b \plus{} c) \minus{} 4a(a \plus{} b)(a \plus{} c) \plus{} 4a(a \plus{} b)(a \plus{} c)(\frac {(a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \minus{} a^{2}}{2(a \plus{} b)(a \plus{} c)})^{2}$[/quote]
Since you said nothing, I think you checked and that's now the good great olympiad problem.

It's still false : choose $ a \equal{} 0$ and $ RHS \equal{} 0$ while $ LHS \equal{} \minus{} 4b(b \plus{} c)(b \plus{} 2c)$


Let $O$ be the circumcenter of a triangle $ABC$, $P$ be the midpoint of $AO$, and $Q$ be the midpoint of $BC$. If $\measuredangle ABC = 4\measuredangle OPQ$ and $\measuredangle ACB = 6\measuredangle OPQ$, compute $\measuredangle OPQ$.

What? If $O$ is circumcenter, then $A, P, O, Q$ are collinear. Please review the statement.

I don't get you. Why should they be collinear?  :huh:

Because that was my regular daily moment of insanity... Forget it, I'm sorry :)

And here is the solution too. See the attached diagram. Let $R=PQ\cap OC,x=\angle OPQ$.

Angle chasing:

\begin{eqnarray*}\angle ABC=4x &\implies& \angle ROP=8x\\ &\implies& \angle CRQ=180^\circ-9x\end{eqnarray*}

\begin{eqnarray*}\angle AOC=8x &\implies& \angle ACO=\frac{180^\circ-8x}{2}=90^\circ-4x\\ &\implies& \angle RCQ=6x-(90^\circ-4x)=10x-90^\circ\end{eqnarray*}

Now $\angle RQC=180^\circ-\angle CRQ-\angle RCQ=180^\circ-(x-90^\circ)=90^\circ-x$. Since $\angle OQC=90^\circ$, we get $\angle OQR=x$. Therefore $\triangle OQP$ is isosceles, giving $OA=2OQ$. But $OC=OA$, hence $\triangle OCQ$ is 30-60-90, yielding $10x-90^\circ=30^\circ\iff x=12^\circ$

A possibly long winded approach:

[hide]Let $\measuredangle OPQ = \theta$. Draw the segment $OC$ to form the isosceles triangle $AOC$

Since $\measuredangle ABC = 4\theta$, $\measuredangle AOC = 2\measuredangle ABC = 8\theta$ and $\measuredangle ACO = 90^{o}-4\theta$

$\measuredangle ACB = 6\theta$, so $\measuredangle OCB = 6\theta-(90^{o}-4\theta) = 10\theta-90^{o}$.

$\angle OQC$ is a right angle so $\measuredangle COQ = 90^{o}-\measuredangle OCQ = 180^{o}-10\theta$.

$\measuredangle OQP = 180^{o}-(\theta+8\theta+(180^{o}-10\theta)) = \theta$. Therefore triangle $POQ$ is isosceles and $OP = OQ = r/2$.

$OC = r$, so triangle $OCQ$ is a $30^{o}-60^{o}-90^{o}$ triangle. This makes $\measuredangle OCQ = 10\theta-90^{o}= 30^{o}$. So $\theta = 12^{o}$.[/hide]

NOTE: I've appended my solution to my previous post. And it seems to be exactly the same as gauss202's.


inequalities
function
analytic geometry
calculus
derivative
vector
graphing lines

Um... what about $a=b=c=0.1, d=3.7$?
In that case the LHS works out at about 0.8.

Sorry, the sign is reversed.

If $a,b,c,d$ are non-negative numbers such that $a+b+c+d=4$, then
$\frac{1}{5-abc}+\frac{1}{5-bcd}+\frac{1}{5-cda}+\frac{1}{5-dab}\leq1$.

I think it is better start solving this problem with the following one,
which is case for n = 3.
Let a,b,c be positive integers and a+b+c=3. Prove that:

1/(4a-abc)+ 1/(4b-abc)+ 1/(4c-abc) ≤ 1/abc

Sorry, it was a bad attempt. :(
Today in the class we realize that this inequality is wrong putting a = 3/2, b goes to 3/2, c goes to 0.
So we have to prove for n = 4.
1/(5a-abcd) + 1/(5b-abcd)+1/(5c-abcd)+1/(5d-abcd) ≤ 1/abcd.

OK, I think that it would be a nice thingh to have a help or a hint by Vasc?

Aren't  you agree with me, Prowler?  ;)

Thank you. ;)

Dear Vasc ,is it possible to have your proof, please, or is it secret? ;)

Of course, it is a great secret. :ninja:

[quote="Vasc"]If $a,b,c,d$ a non-negative numbers such that $a+b+c+d=4$, then
$\frac{1}{5-abc}+\frac{1}{5-bcd}+\frac{1}{5-cda}+\frac{1}{5-dab}\leq1$.[/quote]
Why?
Assume that $a\ge b$
$t=\frac{a+b}{2}$
$u=\frac{a-b}{2}$
$f(u)=\frac{1}{5-cd(t-u)}+\frac{1}{5-cd(t+u)}+\frac{1}{5-c(t^2-u^2)}+\frac{1}{5-d(t^2-u^2)}$
$f'(u)=\frac{-cd}{(5-cd(t-u))^2}+\frac{cd}{(5-cd(t+u)^2}-\frac{2cu}{(5-c(t^2-u^2))^2}-\frac{2du}{(5-d(t^2-u^2))^2}$
And now $f'(u) \le 0$
???....
If this ineq is really difficult,I sure that I wrong!

Can you please post your solution, Vasc?

Sorry Manlio, I cannot post for some time my solution to this inequality.
But the following more general inequality holds:
If $a_i$ are non-negative numbers such that $a_1+a_2+...+a_n=n$ ($n\ge4$), then
$\frac{1}{n+1-b_1}+\frac{1}{n+1-b_2}+...+\frac{1}{n+1-b_n} \le 1$,

where $b_i=\prod_{k<>i}{a_k}$ for all $i$.

Hmm... I see it again and again.
Of course, it is only my own opinion, but I consider such behaviour as a very unpleasant one.

Here is just my idea.

Let $(x-abc)(x-bcd)(x-cda)(x-dab)=x^4-px^3+qx^2-rx+s\ \cdots\ [E]$,

where $p=abc+bcd+cda+dab,\ q=abcd(ab+ac+ad+bc+bd+cd)$,
$r=4(abcd)^2,\ s=(abcd)^3$.

Then taking natural logariythm of both sides of equation of [E] , we have

$\ln |x-abc|+\ln |x-bcd|+\ln |x-cda|+\ln |x-dab|=\ln |x^4-px^3+qx^2-rx+s|$.

Differentiating  both sides of equation of $[E]$ ,with respect to $x$, we obtain

$\frac{1}{x-abc}+\frac{1}{x-bcd}+\frac{1}{x-cda}+\frac{1}{x-dab}=\frac{4x^3-3px^2+2qx-r}{x^4-px^3+qx^2-rx+s}$.

Plugging $x=5$ into this equation, we get

$\frac{1}{5-abc}+\frac{1}{5-bcd}+\frac{1}{5-cda}+\frac{1}{5-dab}=\frac{500-75p+10q-r}{625-125p+25q-5r+s}$

OK, let's finish it off. We proved that $\sum_{\text{cyc}}abc+\sum_{\text{cyc}}(abc)^2\leq 8$. So, it remains to prove that the condition $x+y+z+t+x^2+y^2+z^2+t^2\leq 8$ implies $\frac 1{5-x}+\frac 1{5-y}+\frac 1{5-z}+\frac 1{5-t}\leq 1$. ($x,\dots,t\geq 0$). Let us start with the case $\max\{x,y,z,t\}\leq 2$. Then we may use the inequality $\frac 1{4-w}=\frac 14+\frac w{16}+\frac{w^2}{64(1-\frac w4)}\leq \frac14+\frac w{16}+\frac {w^2}{48}$ for $-1\leq w\leq 1$ and write $\frac 1{5-x}\leq \frac14+\frac1{16}(x-1)+\frac1{48}(x-1)^2=\frac14+\frac1{48}(x+x^2)-\frac1{24}$ and similarly for 3 other variables. Adding up, we get
\[
\sum_{\text{cyc}}\frac1{5-x}\leq 1+\frac1{48}\left(\sum_{\text{cyc}}(x+x^2)-8\right)\leq 1
\]
Now consider the general case. Let $X=x+x^2$, $Y=y+y^2$ and so on. Consider $\frac {1}{5-x}$ as a function of $X$. Implicit differentiation gives $\frac d{dX}\frac1{5-x}=\frac{1}{(5-x)^2}\frac1{2x+1}$ and \[\frac {d^2}{dX^2}\frac1{5-x}= \frac{2}{(5-x)^3}\frac1{(2x+1)^2}-\frac{2}{(5-x)^2}\frac1{(2x+1)^3} =\frac{2(3x-4)}{(5-x)^3(2x+1)^2}\], so the function is concave for $x\leq 4/3$ and convex for $x>4/3$. Fixing the sum $X+Y+Z+T\leq 8$, we see that it is enough to consider the case when $x=y=z<t$ (I do not know whether it is exactly the famous Myth's IPT, but it is something close to it). Also, it is enough to deal with the case $t>2$ because the case $t\leq 2$ has been covered already.
Thus, we have to show that the conditions $3(x+x^2)+t+t^2\leq 8$ and $\frac 3{5-x}+\frac1{5-t}>1$ are incompatible. The latter condition is equivalent to $(2-x)(4-t)<3$. Now comes the boring part. Suppose $x>0$ and $t>2$ satisfy both inequalities. Look at the inequality $3(x+x^2)+t+t^2\leq 8$. Since $t>2$, we must have $x<\frac12$. Let $t=2+\delta$. Since $(t+t^2)'=1+2t> 5$ for $t>2$ and $3(x+x^2)'=3(1+2x)<6$ for $x<\frac12$, we have $x<\frac12-\frac56\delta$. Thus $3>(4-t)(2-x)>(2-\delta)\left(\frac 32+ \frac 56\delta\right)=3+\frac 16\delta-\frac56\delta^2$, whence $\delta>\frac 15$ and  $t>\frac{11}5$. But then $t+t^2>\frac{176}{25}>7$ and $x+x^2<\frac 13$ from where we conclude that $x<\frac 27$. Now let $t=\frac{11}5+\delta$. Since the derivative $3(x+x^2)'=3(1+2x)<\frac{33}7$ for $x<\frac27$, we have $x<\frac 27-\frac{35}{33}\delta<\frac27-\delta$. Thus $3>(4-t)(2-x)>\left(\frac 95-\delta\right)\left(\frac {12}7+ \delta\right)=\frac{108}{35}+\frac3{35}\delta-\delta^2$ whence $\delta^2>\frac 3{35}>\frac1{25}$, i.e., $\delta>\frac15$. But then $t>\frac{12}5$ and $t+t^2>\frac{204}{25}>8$. The end.     :lol:

In your solution, when does the equality hold?

Complete solution?Yeah,we can prove that the inequality:

$\sum{abc}+\sum{a^2b^2c^2} \le 8$

tronger than Vasc problem.which this one is not difficult!

Indeed,let $x=abc,y=bcd,z=cda,t=dab$

the problem is:

$\sum\frac{1}{5-x} \le 1$
$<->\sum\frac{x-1}{5-x} \le 0$
$<->\sum\frac{(x-1)(x+2)}{(5-x)(x+2)} \le 0$

Because $(x-1)(x+2)$ is inreasing and $(5-x)(x+2)$ is increasing ,too.

By Trebusep:

$\sum\frac{(x-1)(x+2)}{(5-x)(x+2)} \le\sum{(x-1)(x+2)}\frac{1}{(5-x)(x+2)}$

And the rest is prove:

$\sum{(x-1)(x+2)} \le 0$

Which this one is:

$\sum{abc+a^2b^2c^2}\le 8$

[quote="hardsoul"][quote="hungkhtn"]
By Trebusep:

$\sum\frac{(x-1)(x+2)}{(5-x)(x+2)} \le\sum{(x-1)(x+2)}\frac{1}{(5-x)(x+2)}$

[/quote]

What is Trebusep? :?  Is it Chebyshev's Inequality?But then can you use it for negative terms?[/quote]

I think he means chebyshev.
I think both rearrangement and chebyshev do not require terms to be positive.

[quote="hungkhtn"]Because (x-1)(x+2) is inreasing and (5-x)(x+2) is increasing too. [/quote] I believe that the first one is increasing, but  $(5-x)(x+2)=3x-x^2+10$ is increasing for $x<\frac32$ and decreasing for $x>\frac32$  :?

No,you can check $x+y \le 3$
Then $(5-x)(x+2) \ge (5-y)(y+2)$ if $x \ge y$

[quote="manlio"]This inequality is really devastating.

However we have not yet a simple non calculus solution. ;)[/quote]

I don't think so,Manlio.

Prove $\sum{abc}+\sum{a^2b^2c^2} \le 8$ not use more calculus.
The calculus is only used incase three of $a,b,c,d$ equality!

For $n=5$ we have to prove the following inequality:

If $a,b,c,d,e$ are non-negative numbers such that $a+b+c+d+e=5$, then
${\frac{1}{6-abcd}+\frac{1}{6-bcde}+\frac{1}{6-cdea}+\frac{1}{6-deab}}+\frac{1}{6-eabc}\leq1$.

It is routine now: using $(n-1)$EV, we see that it is enough to prove the inequality $\frac 4{6-x^3y}+\frac 1 {6-x^4}\leqslant 1$ under the conditions $x,y\geqslant 0$,  $4x+y=5$. Substituting $y=5-4x$, multiplying by the denominator, combining similar terms, and factoring, we finally obtain the inequality
\[
(1-x)^2[6+12x+18x^2-x^3-2x^4-3x^5-4x^6]\geqslant 0
\]
to prove. Dividing by $x^2$, we see that it is enough to check that the expression in brackets is non-negative for $x=\frac54$, which can be verified directly.

Actually, it seems like this approach will work for any given $n$, though I didn't try to verify it  :D.

Vasc could you pos your solution for the first inequality you posted or a hint.
I think it passed so much time. Thanks

Vasc spring 2006 passed a long time ago, could you post your solution for the 1st one?

This inequality and its generalization are proved by AC-Theorem in Algebraic Inequality, page 273.
Besides, see also
http://reflections.awesomemath.org/2006_3.html


online version of truth or dare,  :D

choose truth or dare, and the person below you dare you or ask u something, and the dare has to be able to be done online, and you need to be able to prove that you have done it.

i'll start, i choose dare.

[quote="tinytim"]what does rickroll mean??[/quote]

uhhhh....someone show him the ARML video

1=2:
How many times have you gone to the bathroom today? :rotfl:

[quote="tinytim"]what does rickroll mean??[/quote]
[url=http://www.youtube.com/watch?v=eBGIQ7ZuuiU]This.[/url]

Dare would be either waaay boring or waaay offensive on the internet.

oh come on Rickrolling rrusczyk boring? :P

he hasn't posted proof yet... eh I should prod (^_^) and myyellowducky to post.

hav u ever made an online profile that stated u were Mr. LaBob?

no, I dont even know what Mr.LaBob is...

Dare

i dare u to... get a picture. 'cause everyone should have a pic. yay pretty blue flowers!

[quote="tinytim"]what does rickroll mean??[/quote]

go search internetisseriousbusiness.com

Whoa, truth or dare could get really offensive.

[quote="pianogirl"]We can't get a pic...we are bogged down with bills...so: Promise to Repeat.

Truth[/quote]

Has 1=2 ever been to your "location" as stated in your aops profile?

Truth

No...anyhow, if he tried to go to sleep in his computer, he would either smash it, electrocute himself, or both.

Dare


1. When does an even number have a unique even-prime factorization?

2. Prove that a number is perfect when it is in the form

$P=2^{p-1}q$, where $q=2^{p}-1$ (a Mersenne prime).

[quote="brady00n"]2. Prove that a number is perfect when it is in the form

$P=2^{p-1}q$, where $q=2^{p}-1$ (a Mersenne prime).[/quote][hide] A number is perfect when the sum of its factors is twice the number. The sum of the factors of $2^{p-1}(2^{p}-1)$ is $(1+2^{1}+2^{2}+...+2^{p-1})(1+2^{p}-1)=(2^{p}-1)2^{p}$. $\frac{2^{p}(2^{p}-1)}{2}=2^{p-1}(2^{p}-1)$, the original number, therefore, $P$ is perfect. [/hide]

[quote="matt276eagles"][quote="t0rajir0u"][quote="brady00n"]1. When does an even number have a unique even-prime factorization?[/quote]

I don't understand what an even prime is in this context.[/quote]

I think it's an even number that cannot be expressed as the product of two even numbers.[/quote]

Well, in that case, isn't the number just a number that contains only 1 two in its unique prime factorization?


Does anyone know if isometry is an absolutely necessary condition for an isomorphism between metric spaces with different metrics to preserve metric properties?  In particular, are there more general conditions for an isomorphism to preserve completeness, or is isometry necessary?  I have not found much information to answer this question so far.

Of course, you do not need to exactly preserve the metric to preserve completeness: distorting it not more than twice, say, will do just as well. Isometry allows you to preserve [b]all[/b] metric properties you can possibly think of. For each particular property, it may be an overkill.


4. To make one pound of honey, a worker bee must visit 25 million flowers. An average visit at a flower, including flight time to the flower is 30 seconds. How many days will it take to the nearest day for a swarm of 10,000 worker bees to continually visit enough flowers to make 5 pounds of honey?

[hide="Answer"]4 days[/hide]

[quote="The_Scintillator"]Okay. I'm not entirely sure if this is right, but this is what I got:
Since a ONE bee must go to 25,000,000 flowers, taking a total of 30 seconds for each trip, that's  750,000,000 seconds for ONE bee to make ONE pound of honey.
However, you have 10,000 bees splitting the workload.
So then you divide 750,000,000 by 10,000--giving you 75,000. Note this is still in seconds.
Now you take that and divide it by 60, to get the number of minutes, then by 60 again to get the number of hours. (Or just divide by 3,600.)
You should get 20+5/6. This is the time it takes 10,000 bees to make ONE pound of honey.
Now you multiply by 5 to get 104+1/6. Since this is still in hours, and you want days, you divide this by 24 to get 4+49/144. Round that down to 4 and there's your answer.
(This entire thing is probably flawed, but it's the best I could think of. Let me know if I'm wrong! :))[/quote]
It looks good to me. I forgot to multiply 5 to get 5 pounds..hehe.. but Thanks.


Ενα συναρτησοειδές J:V->IR, όπου V γραμμικός χώρος με νόρμα, είναι συνεχής στο yo που ανήκει στο V αν για κάθε ε>0 υπάρχει δ>0 τέτοιο ώστε $ |J(y)-J(yo)|<ε$ε όταν $ \parallel{}y-yo\parallel{}<ε$δ

α)  Έστω $ V=C^1[a,b]$. (όπου C1 το σύνολο των συνεχών συναρτήσεων των ποποίων η πρώτη παράγωγος είναι
     επίσης συνεχής)
     Να αποδειχθεί ότι αν το J είναι συνεχές ως προς την ισχυρή νόρμα, τότε είναι συνεχές και ως προς την ασθενή νόρμα.
     ( ισχυρή νόρμα :  $ \parallel{}y\parallel{}_M=max|y(x)| , a<x<b$
    (ασθενής νόρμα: $ \parallel{}y\parallel{}_w=max|y(x)|+max|y'(x)| , a<x<b$
β) Να δοθεί ένα παράδειγμα που να δείχνει ότι η συνέχεια ως προς την ασθενή νόρμα δεν συνεπάγεται συνέχεια ως προς την
    ισχυρή νόρμα .

Ενα συναρτησοειδές J:V->IR, όπου V γραμμικός χώρος με νόρμα, είναι συνεχής στο yo που ανήκει στο V αν για κάθε ε>0 υπάρχει δ>0 τέτοιο ώστε $ |J(y)\minus{}J(yo)|<$ε όταν $ \parallel{}y\minus{}yo\parallel{}<$δ

α)  Έστω $ V\equal{}C^1[a,b]$. (όπου C1 το σύνολο των συνεχών συναρτήσεων των οποίων η πρώτη παράγωγος είναι
     επίσης συνεχής)
     Να αποδειχθεί ότι αν το J είναι συνεχές ως προς την ισχυρή νόρμα, τότε είναι συνεχές και ως προς την ασθενή νόρμα.
     ( ισχυρή νόρμα :  $ \parallel{}y\parallel{}_M\equal{}max|y(x)| , a<x<b$
    (ασθενής νόρμα: $ \parallel{}y\parallel{}_w\equal{}max|y(x)|\plus{}max|y'(x)| , a<x<b$
β) Να δοθεί ένα παράδειγμα που να δείχνει ότι η συνέχεια ως προς την ασθενή νόρμα δεν συνεπάγεται συνέχεια ως προς την
    ισχυρή νόρμα .

Το α) είναι τετριμμένο. Για το β) πάρε το συναρτησοειδές:

\[ f\mapsto f'(1/2)\]

όπου $ [a,b]\equal{}[0,1]$.

β) Να δοθεί ένα παράδειγμα που να δείχνει ότι η συνέχεια ως προς την ασθενή νόρμα δεν συνεπάγεται συνέχεια ως προς την ισχυρή νόρμα .

Mας ζητάει εδώ η ασκηση να αποδείξουμε ότι αυτό θα ισχύει για παράδειγμα για το συναρτησοειδές   'μήκους τόξου'
(      $ J(y)\equal{}\int_a^b \sqrt{1\plus{}[y'(x)]^2} dx$     )

Αγαπητέ Jondrette μήπως θα μπορούσες να μου εξηγήσεις
πως θα αποδείξουμε την συνεχεια (ή την μη συνέχεια) ως προς ένα συναρτησοειδές
ενώ δηλαδή είναι συνεχής ως προς την ασθενή νόρμα θα δείξουμε πως δεν είναι συνεχής ως προς την ισχυρή νόρμα


I'm trying to enroll in the Olympiad Problem Solving course, but when I get to the checkout page and click "I Agree To These Terms" nothing happens. Is it just me or is something wrong with the site   :?

Fixed.  Give it a try now, and then see you in class tomorrow.


[hide="by induction"]
For $n=1$ , by bernoulli we have $m^m=(m+0)^m>m^2$ Since $m\geq 3$ ,
$m^2-2m-1=(m-1)^2-2>2>0$ .Hence it is true for $n=1$

Suppose $n=k$ is true , then

$f_{k+2}(m)=m^{f_{k+1}(m)}>m^{2f_k(m+1)}$

Also $\frac{m^{2f_k(m+1)}}{(m+1)^{f_k(m+1)}}>\frac{m^2}{m+1}\cdot f_k(m+1)>\frac{m^2}{m+1}$ since $f_k(m+1)>1$ . It remains to prove $\frac{m^2}{m+1}>2$ which is true since for $m\geq 3$ .
[/hide]

Ascertain $\lim_{x\searrow 0} f_n(x)$, where $f_0(x)=1$ and $\left(\forall \right) n\in N,\ f_{n+1} (x)=x^{f_n (x)}.$