Harvard college Putnam AMC AIME USA(J)MO USAMO

Go to any math or science camp or competition, and it is no secret that boys are overrepresented in these subjects. While we have long thought that men where more "spatially" oriented and hence better at math (and thus poorer at reading), with even the president of Harvard having to resign over such a controversy, a new study suggests that the gender gap is much more cultural than innate:

[url]http://education.guardian.co.uk/schools/story/0,,2283083,00.html[/url]

Discuss.

(Hopefully we can have an intelligent discussion on this here, but if not, move to GFF)

hmmmm that's an interesting study.

i took some online classes for math and english. the math classes r dominated by boys the most girls in a math class that i have been in is 2. me being one of them :) as for my english class. we have 4 girls and no boys.

our school math club had about 10 boys and 6 girls.

for comparison's sake, the IMO participants of the more "gender-equal societies" of Iceland, Sweden, and Norway (this is based on my best guess -- some of the pics are, well, hard to tell)

2008

Iceland 5 M, 1 F
Sweden 5 M, 1 F
Norway 6 M, 0 F

2007

Iceland 5 M, 1 F
Sweden 6 M, 0 F (I think?)
Norway 5 M, 1 F (blargh hard to tell wtf)

also for comparison's sake, the composition of our own team (in a country which does have a gender gap):

2008 6 M, 0 F (probably, haven't kept careful tabs)
2007 5 M, 1 F
2006 6 M, 0 F

the same trend would probably continue for Putnam and IMC but it's a bit harder to separate based on country there, so I won't bother looking up that data

The historical fact that these gaps have always decreased over time points to an origin in culture and discrimination. Consider the US IMO team: no girls in the first 24 years, three appearing a total of five times in the 11 years since. The gap is still there, but going from none in 100 to about 5-10% is a big change.

The Putnam wouldn't be useful for international comparison here, since it's USA and Canada only.

Some AMC statistics (2008):
$ \begin{array}{ccccc}\text{Exam}&\text{\# F}&\text{Avg F}&\text{\# M}&\text{Avg M}\\12\text{A}&29401 &60.8&39597&69.3\\12\text{B}&15074&63.6&21160&71.8\\10\text{A}&26832&53.4&31578&62.6\\ 10\text{B}&13765&60.6&15830&69.4\\ \text{AIME}&1823&4.48&6695&5.08\end{array}$

USAMO data isn't easily available; only names are available on the site. The gender data above is self-reported, and the substantial fraction who didn't report anything are not included.

What does the gap look like in the contest world? There's a participation gap, with significantly more males than females showing up at the open contests. This gap grows with age; it's larger for the AMC 12 than the AMC 10. There's also an achievement gap among those who do participate- about 9 AMC points. This gap appears at all levels of success, leading to much larger gaps in the participation numbers for selective contests.

Minor point -- many good students come from overseas to compete in the Putnam including Chinese IMOers (who always take up a couple of the fellow spots), Russians, Romanians, Moldovans, and so forth, which is why it may have some (but admittedly not a large amount of) use even in international comparison. That's why I noted the IMC, though, since yeah, American data is probably not terribly interesting as we're already pretty aware of that.

I think the study was more related to the average student, not IMO participants. There might still be a social stigma against girls in these "gender equal" countries about math competitions (though perhaps not, I'm just hypothesizing). Also, there have already been studies that try to explain why in "very high level" activities, males are overrepresented.

[url]http://www.dailybruin.ucla.edu/news/2005/jan/28/general-intelligence-equal-bet/[/url]

If I recall correctly, another study (or perhaps this same one) described that the average intelligence of males and females were exactly the same, but females had a lesser deviation from the average, whereas in males there was slightly more deviation (i.e. more smart and dumb people).

Of course, there are those that say IQ is a useless measure anyway, so the study might be completely moot. I think, however, that is irrelevant.

But again, I think the study was geared to the basic, average student, and I think the study was interesting when considering this.

Also, note that countries where there is a wide social stigma against women (e.g. Middle East and Asia), female participation is virtually nonexistent, compared to those in more "gender" equal countries (e.g. Northern and Western Europe), where there is a noticeable female participation, if still underrepresented.

[quote]Sapienza said: "Our research indicates that in more gender equal societies, girls will gain an absolute advantage relative to boys."[/quote]
Here is just some observations from my own life.

(Russia, St. Petersburg; 1974-1993; I didn't notice any cultural gender inequality in that society; also you should know that Russia does not embrace the idea of "privacy of student records", so everybody knew exactly how well everybody else was doing in class):

Low school (1-3) grade. No real difference between boys and girls.
Middle school (4-8 grade). The girls are quite uniform and doing mostly well with 1 or 2 exception; the boys get more separated with each next grade; a noticeable amount of them barely pass or just fail (some of those ended in prison just a few years after school), many are just OK and a few are what I would call brilliant.
High school (a different place; the specialized math school with very strict student selection criteria): Almost no real difference. The girls are slightly better on average but among the top 6 students in my class 4 are boys.
University: Almost all top students are boys. At the middle or the bottom level the gender difference in abilities is not really noticeable.

Conclusions? I'm inclined to agree with the statement that boys tend to deviate much more from the average than the girls, which sort of makes sense given their biological roles; it is more efficient to try the random changes in the male species. So, I believe that among the top mathematicians and in other cases when you select a small group of best performers out of a huge number, the boys will always prevail. The performance difference on the average level depends on the nature of the task. It seems like the girls are better than boys at doing several things simultaneously but the boys have an advantage when it comes to something that requires deep concentration on one task, though neither of these cannot be overcome by proper training.

So, I do not think that there is any real difference in what the boys and the girls can potentially achieve. The question is what they really want to do. The successful professional career in science is not the only thing worth living for, and men and women tend to view the world in different ways.

Some of it is certainly cultural and we should try to encourage more girls to try science (what is desperately lacking now is role models; I've heard a story quite recently that 2 rather bright female students had serious doubts about whether the academic career is for them until some top rate female mathematician came and gave a colloquium lecture at their college. They didn't understand mathematics in that lecture but they saw that she was confident, happy about what she was doing, and well respected, and that alone changed their minds and gave them necessary confidence).

But some of it may be genetic. It is not that mathematics is inaccessible for an average girl, it may just not have enough appeal for her. Everybody tries to study which gender can do better in math. but it is not telling you much without the parallel study of which gender likes math. more (I do not mean going around and asking directly: do you like math or not?; I mean something much subtler like giving a 2 year old child, which cannot possibly be spoiled by cultural stigmas, a choice of two toys one of which requires something that we usually associate with math and the other one that requires something else and seeing whether there is a clear gender difference in the preferences; well, any qualified psychologist will come with a dozen of ways of how to check what people like without letting them know what you are checking). I would be really interested in the results of such study among small children, schoolchildren, and adults. It may be much more revealing than the study of achievements on the standard tests.

Just my 2 cents :)

Some innate differences are plausible- but they're tiny compared to everything else. Success in problem solving starts with liking math, and the "Math is hard/You're no good at it" message drives people away.

Continuing my side of the thread- the gap in contests:
Also, at the elite level, the gap is shrinking. Compare MOP now to ten years ago: in my years (1998, 1999), Melanie Wood was the only girl both times (out of 24 and 24). This year, there were at least 9 (Colleen Lee, Wendy Mu, Lynnelle Ye, Carolyn Kim, Jennifer Iglesias, Katherine Rudolph, Elizabeth Synge, Shijie Zheng, Iris Xu) out of 56, and maybe some I missed with non-English names. That's still underrepresented, but it's a big change in only ten years.
One striking pattern in those nine names: they don't come alone. Six of them come from schools with two or more MOPpers, including two pairs of two girls from the same school. It's also worth noting that Melanie Wood went to the same schools as Chris Mihelich, who reached MOP in 1997; she wasn't alone either.

That's the gap we see now: three or four times as many boys as girls taking contests and problem solving seriously. It's closer to equality where the contest culture is strongest, and farther away where only a few interested kids stick out.

whoa there, jmerry, that's simply due to the invitation of the CGMO participants to MOP. Not to mention the fact that red MOP has since been added. If we look simply at blue/black MOP during the past years, the number fluctuates between zero and two, I believe.

OK, I didn't realize that difference. The 12th graders involved should have been a clue.
As for red MOP, I'd prefer to include it, and simply compare to the total size. That's not the only change anyway- there was no TST before 2000.

My other point stands.

Perhaps taking the MOs as an example is a bad idea. Perhaps it just shows that boys are naturally more competitive

Well, you can then argue that looking at the performance on the standard tests is even a worse idea, because, perhaps, it just shows that girls are more disciplined, more inclined to do what the teachers tell them to do, and are more inclined to do their homework properly (this is rather undeniable and nothing else is really required to prepare well for a standard test, while just being more competitive would not bring you far as long as MO is concerned).

The last remark is sort of pointless but once the Harvard president had to resign for just openly stating his opinion (which, in my humble opinion is a shame to the society that takes pride in its free speach rights), I do not think there will be anybody in the USA in the near future who will dare to argue that boys are any better and the studies with the conclusions like the one I quoted in my previous post will appear like mushrooms after a rain. :P

I didn't notice this in the O.P.'s post, so I'm going to quote Fedja instead:
[quote="fedja"]the Harvard president had to resign for just openly stating his opinion[/quote]  This is not a complete description of what took place.  Some substantial portions of the Harvard faculty had been at odds with Summers for academic-political reasons long before his comments.  Those comments provided a focal point for news coverage of the ensuing events, but the comments were not the reason the faculty was aligned against him.  How Summers was treated by the Harvard Faculty is a much less-good example of the touchiness of this subject than people seem to think.  (How the entire incident was treated in the news is, perhaps, a different story.)

Girls just dont seem pressured to do things like math and science. In my advanced math class their is one girl in the 22 students.... Also, people tend to do things their friends like to do. If math is all boys, girls wont want to participate and so math competitions are dominated by boys.

i agree with that. also, if they think they CAN'T do it, or they dont feel it's important, they won't try at all. i learned this from my MATHCOUNTS 6th-grade group at my school: all the girls said that they didnt really care what they got on the sprint, they didnt have a chance. then, when the 6th grade top 5 list came out, i was the only girl on it (5th place :( ). probably 'cause i was the only one who actually TRIED. but i still sucked at it (10 out of 30).

durt, I knew that you would ask me that question ("What are the social forces driving girls away from math?"), but I don't see the point.  Obviously, you yourself have heard the usual stated reasons and can easily come up with a list on your own, but you want to play a tiresome game in which I describe some such social force, and then you try to shoot it down.  I'd prefer not to play that game, because my goal (in my recent posts) is not to identify these factors, but rather to indirectly argue the general plausibility and likelihood that environmental factors are the driving force behind the "gender gap."  Or to summarize, the cause of the actual observed "gender gap" is primarily a consequence of how our *entire society* is constructed.  In any case, I'm sure it will be nearly impossible to prove (to your satisfaction) that any one factor is both (1) significant and (2) environmental rather than innate.

Against my better judgment, I'll address your question anyway (see below).  But you can also address mine:  Are you just completely skeptical of the idea that the way our society is constructed could have anything to do with the observed gender gap, or do you think it's possible but probably not the case?  And what is your criticism of my scientific argument that the hypothesis of innate superiority (or to split hairs, superiority via greater variance) of males is just flat-out illogical?

That's where I'd really like to stop this post, but since this is an issue that's important to me, I'm going to go ahead and list some social forces driving women away from mathematical fields.

Here's a list off the top of my head:

[hide]
Family pressures:  Many people (including Summers) rightly point out that a big reason why women opt out of high-stress, high-workload careers is for family reasons, and obviously they are correct.  The wrong conclusion to draw from this is that there is no problem (which is an approximation to Summers' attitude).  The right train of thought (imho) is to think about why this is so.  Society pressures women into spending far more time on childcare than men.  Sometimes this causes women to leave work entirely, and sometimes it causes women to leave elite career tracks.  While it is frequently argued that women naturally prefer this role, again I would wager that the preference that adult women actually have (whether innate or acquired) is not as strong as society requires.  For example, many intelligent, educated women would love to take time off or work part-time while their children are very young and then return to elite career tracks.  In practice, this is nearly impossible.  Also, surveys have shown that even in two-income families, women spend a lot more time on both childcare and housework than their husbands.  You might try to argue that women like spending more time with their kids, but can you really argue that they prefer spending time on housework?  The fact of the matter is that our society expects more from wives at home, whether they are working or not.

Family pressures 2:  It is far more common for women to make career sacrifices for the sake of the husband's career than vice versa.  There are many reasons for this, some of which are related to the childcare issue above, but a lot of it comes from the (conscious or unconscious) belief that the husband's career is more important.  For example, survey results show that a doctor-professor couple is more likely to think that the wife's career is "more flexible" independent of whether the wife is the doctor or the professor.

Female temperament:  It's been argued that females are less competitive than males, and that's part of the reason why there are so few females who are CEOs or big shot scientists.  It's plausible that this difference in competitiveness is innate, but as far as I know there is no scientific evidence.  But even if the difference is innate, there is no a priori reason why competitiveness should be an essential part of the scientific enterprise.  This could very well be a social artifact.  Certainly both cooperation and competition contribute to the advancement of science, and it's possible that our current system is simply overvaluing competition.

Discouragement based on "science":  For example, someone reading this very thread or about the Summers' brouhaha might decide that they are unlikely to cut it as scientist and give up that career track.  This is not as stupid as it sounds.  As it is, if you are a bright high school student who dreams of becoming a prize-winning scientist, the odds are against you.  Speaking strictly probabilistically, given the added piece of information that you are female, if the "Summers hypothesis" (about the greater variances for men) is true, then your calculated probability of success may be lower.  (This might require some more explanation.)

Discouragement based on perceived biases:  Some females might not want to constantly second-guess or deal with the issue of whether their abilities and accomplishments are either underestimated or overestimated because of gender.

Unintentional discrimination:  For example, psychological experiments show that people rating job candidates on the strength of their resumes rated resumes from males higher than identical resumes from females.  Even well-intentioned people (including myself) are capable of making these sorts of errors, which is why it's important to have policies that encourage the hiring of females.  People often think of these policies as giving an "unfair advantage" to women but the point is to combat the (admittedly difficult to quantify) unfair disadvantage.  In my mind, the problem is that too many people believe that they don't have a biased bone in their bodies and believe themselves to be 100% objective and fair, which is preposterous.

Plain old discrimination:  It's impossible to deny that there really are people who are out-and-out sexist.

The role model issue:  Without female role models in certain careers, younger females might be less likely to seriously consider those careers.

The outsider issue:  Many females might not want to be like Sunny and work in a room filled with men.

Parental influence:  Many people's parents have a large role in shaping their career paths.  Many of these parents have preconceived notions about which careers are appropriate or desirable for their daughters.  Some parents might just have lower expectations for daughters when it comes to math ability.

Parents as role models:  Most people's first conception of how a man is supposed to be comes from dad, and and their first conception of how a woman is supposed to be comes from mom.  Dads are far more likely to be mathematically inclined than moms, primarily for historical reasons.

Socialization of children through toys:  It may be true that boy toys more naturally lead to scientific curiosities than girl toys.  One might argue that boys and girls have innately different toy preferences, but that doesn't exclude socialization as a factor.

Social stigma:  The social damage of being a "math person" may be greater (in some subcultures) for females than males.

Romantic complications:  Some boys (usually not the smartest ones) may be intimidated by girls who are smarter than they are.  This is especially true for math, because as we all know, it's often really obvious when one person is better at math than another.

[/hide]

Dammit I want to punch you in the face, yenlee.

[quote="yenlee"]durt, I knew that you would ask me that question ("What are the social forces driving girls away from math?"), but I don't see the point.[/quote]

If you're going to try to counter these forces, you'd best know what they are right?

[quote="yenlee"]But you can also address mine: Are you just completely skeptical of the idea that the way our society is constructed could have anything to do with the observed gender gap...[/quote]

No.

[quote="yenlee"]... or do you think it's possible but probably not the case? And what is your criticism of my scientific argument that the hypothesis of innate superiority (or to split hairs, superiority via greater variance) of males is just flat-out illogical?[/quote]

Just because it's a difficult hypothesis to test doesn't make it illogical. You may mean "conclusion" instead of "hypothesis," in which case I'd agree that we can't draw such a conclusion (at least from these studies). So we ought to keep experimenting and observing.

[quote="yenlee"]Unintentional discrimination: For example, psychological experiments show that people rating job candidates on the strength of their resumes rated resumes from males higher than identical resumes from females. Even well-intentioned people (including myself) are capable of making these sorts of errors, which is why it's important to have policies that encourage the hiring of females. People often think of these policies as giving an "unfair advantage" to women but the point is to combat the (admittedly difficult to quantify) unfair disadvantage. In my mind, the problem is that too many people believe that they don't have a biased bone in their bodies and believe themselves to be 100% objective and fair, which is preposterous.[/quote]

In a free society, I ought to be able to hire whomever I please (assuming I'm a private entity). If my prejudices lead me to hire subpar employees, then I'll pay for that in lost revenues. I'm 100% against more government mandated bull****.

[quote="yenlee"]Certainly both cooperation and competition contribute to the advancement of science, and it's possible that our current system is simply overvaluing competition.[/quote]

Dude totally man. It's possible the moon is actually too far away. Yeah it sucks.

Anyway, your list is pretty comprehensive and stuff. But most of them I don't think need fixing or even can be fixed. Some of them maybe yeah are bad mkay. For example,

[quote="yenlee"]For example, many intelligent, educated women would love to take time off or work part-time while their children are very young and then return to elite career tracks. In practice, this is nearly impossible.[/quote]

What's the deal with this?

dude, chill

why would you even type something like that

it could not possibly serve any purpose

durt is a troll -- "not serving any purpose" is its purpose.  Ignore it and hope it goes away.

[quote="orl"][quote="fedja"] The last remark is sort of pointless but once the Harvard president had to resign for just openly stating his opinion (which, in my humble opinion is a shame to the society that takes pride in its free speach rights), I do not think there will be anybody in the USA in the near future who will dare to argue that boys are any better and the studies with the conclusions like the one I quoted in my previous post will appear like mushrooms after a rain. :P[/quote]

[url=http://www.president.harvard.edu/speeches/2005/nber.html]Here[/url] is Lawrence Summer's speech, tenured in the [url=http://www.hks.harvard.edu/about/faculty-staff-directory/lawrence-summers]economics department[/url] at age 28 and [url=http://www.deshaw.com/articles/20061019.pdf]managing director of D. E. Shaw Group[/url] who had to resign as [url=http://query.nytimes.com/gst/fullpage.html?res=9902E7D7103AF930A25750C0A9679C8B63&scp=3&sq=Harvard%27s%20New%20President%20&st=cse]Harvard's president[/url] in 2005.

I have the same impression that on average females are slightly stronger and males show a higher standard deviation in intelligence. So what would you prefer? (1) Having an average intelligence most likely, (2) or having a higher chance to be quite smart or rather quite stupid? :D  And Summer's speech was wrongly interpreted that men are inherently more capable than women in mathematics which he did not say. In the speech he worked backwards to infer implied standard deviations from observed percentages, i.e. what kind of difference in standard deviations could explain the gender difference in the tail of the distribution. He concluded men's standard deviation might be 20 percent higher than women. Then he used this ratio not to look at the top five percent of smart guys but rather the ratio of males to females in the top 1/10,000 (which seems to correspond to the pool of research scientists at leading universities). Doing some forward statistics calculation, assuming SDV difference of 20 percent, in the tail at 3.5-4 SDVs we might have five men for each women. Actually it rather looks like almost 20 to 1. This statement appeared to have got him in trouble.[/quote]

There is an interesting article [url=http://www.voxeu.org/index.php?q=node/1435]Performance Gender-Gap: Does Competition Matter?[/url] by Evren Örs that suggests that male variance in admission tests settings is higher than for females. And females seem to be better on average in A-level exams and intermediate university exams. But in very competitive settings this higher variance might be an advantage for males.

These days there are the [url=http://www.lindau-nobel.de/WebHome.AxCMS?ActiveID=1012]The Lindau Nobel Laureate Meetings for Economics.[/url] There are online video lectures to watch it live!

I recommend the article "Cross-Cultural Analysis of Students with Exceptional Talent in Mathematical Problem Solving," by Titu Andreescu, Joseph A. Gallian, Jonathan M. Kane, and Janet E. Mertz, which appeared in Notices of the American Mathematical Society, Vol. 55, No.10, pages 1248-1260, November 2008.

The participation by girls in the IMO is addressed in some detail in this article, with interesting comparisons across different countries.  The authors particularly contrast Slovakia vs. the Czech Republic:  from 1998-2008, Slovakia had 10 girls out of 66 on the IMO team, while the Czech Republic had 2 out of 66.  (These two countries were formerly united.)  In the same time frame, South Korea had 7 girls out of 66 IMO team members, while Japan had 0 out of 66.  Up to 1990 (pre-reunification), East Germany had 15 girls on the IMO teams out of 202, while West Germany had 0 out of 84.  The authors comment:  "Chi-square analysis indicated that it is virtually impossible that this observed country-to-country fluctuation was caused by random chance."
The p value is less than 0.0001.

The article makes evident the role of cultural influences on the participation by girls.  Personally, I would not actually anticipate equality of interest or success (I'm female), but I think we do not yet know what the "natural" ratio of girls to boys would be.  As long as the % of female IMO participants is rising in the U.S., I'd suspect we have not yet reached the "natural" ratio.

I don't think anyone was doubting that (except durst). Since qualification for IMO is by an (I believe) unbiased competition, I think it illustrates the problem - "recruitment" of women at the lower levels. yenlee seems to have argued a few months ago that discrimination in recruitment at the top levels is part of the issue, but I want to believe it is not the case for academia.

There may be small differences in learning ability of men and women, but I don't think that our culture has developed by chance. That is to say, I think that men take dominant roles in society more naturally than women because men are naturally physically stronger.

So one of the advantages that men have is that they create a society that is not advantageous for women.

Another hypothesis that I have is that women are more inclined to be side-tracked by child birth than men are. Of course this is true in our society, but I think it is something that is inherent in our biology.

So where does this leave us? I think that people can take the liberal point of view but it will always remain that people want to be able to claim superiority based on a group that they belong to. Many men will always be biased in the sense that they'd rather think of women as inferior because by comparison, it would make men superior. Likewise, women will want to exclaim their opposition to the male dominance.

I think that people are more likely to use group identification as a means of gaining self-esteem rather than bolstering oneself by claiming to fight the evils of society.

But I think things will generally settle into the equilibrium where men take more dominant roles because of the clear biological differences in physical strength.

So are there significant gender differences in absolute intelligence? probably not. But will a human society believe in such a difference? probably.

Here's [url=http://econ-www.mit.edu/files/4298]another new study[/url] (PDF link), along with [url=http://web.mit.edu/newsoffice/2009/math-gender.html]a story[/url] summarizing it. This study is focused on the contest data, mostly the AMC.

In reaction to the link jmerry just posted (or at least the summary article, which is all that I read):

This squares with an anecdotal observation that I made some years ago. The top female math competitors have tended to come from the places with the largest, strongest and best organized school-based competition groups. Many male competitors also came from those places - they were, after all the strongest and best organized schools - but some other top male competitors could pop up almost anywhere. For a Southern California example of this, Lauren Williams and her younger sisters all came through the Palos Verdes schools, and at that time the Palos Verdes schools were by far the strongest and best organized for contests of the region.

Every rule has its exceptions, of course. Wasn't Sherry Gong already well-established as a top competitor before she transfered to Exeter? I don't think there are any top-handful schools in Puerto Rico.

But what I wonder now is whether this whole observation is outdated. Have we moved into a time in which the schools are less important than before? That AoPS and other on-line sources, and competition-oriented weekend schools and tutoring programs, and summer camps, and all of that non-school-based contest support structure is now so important that even going to a top school isn't enough any more? And if schools and their programs are relatively less important, who is going to produce the next generation of female math competitors? Can the non-school programs do it?

Virtually all of the data used in that article is from 2007 or 2008. In fact, one of the main justifications for that assertion is the CGMO team composition- see my first-page quote:
[quote="jmerry, 6/08"]One striking pattern in those nine names: they don't come alone. Six of them come from schools with two or more MOPpers, including two pairs of two girls from the same school.[/quote]

Sherry Gong represented Puerto Rico twice at the IMO before transferring to Exeter. Her school was clearly the top contest school in Puerto Rico, with all 8 USAMO qualifiers ever from the island.
Of the other two girls to reach the IMO team, Melanie Wood went to Park Tudor (strong, but never more than two USAMO qualifiers in a year), and Alison Miller was homeschooled.
It's a small sample, but it does fit the feeling that the very top spreads out a bit more.

The two strongest relationships found in the study:
Poverty; schools with high proportions of students qualifying for free lunches had a smaller gender gap.
Contest strength; schools with lots of qualifiers had a smaller gap.

It's hard to draw conclusions from that, short of wild speculation.

One last point: the table on page 23 of the study is blatantly wrong, at least for the male half. Those students (on the IMO team) did go to schools with USAMO qualifiers.

As far as I can tell, that table I objected to in the last post is comparing students in both 2007 and 2008 to 2008 statistics excluding those students, being somewhat inconsistent in how that exclusion is handled, and comparing to a rather unhelpful sample of all AMC schools.
Here's my version of the table, strictly looking at USAMO qualification numbers:
[hide="Table (very big)"]\begin{tabular}{llrl}IMO 2007&School&Number&\%ile\\ \hline Sherry Gong&Exeter&16&96-100\\Eric Larson&South Eugene&1&00-42\\Brian Lawrence&Montgomery Blair&9&89-91\\Tedrick Leung&North Hollywood&2&42-64\\Arnav Tripathy&East Chapel Hill&2&42-64\\Alex Zhai&Univ. Lab (Urbana)&1&00-42\\ \hline IMO 2008&School&Number&\%ile\\ \hline Paul Christiano&Harker&5&71-79\\Shaunak Kishore&Unionville&1&00-41\\Evan O'Dorney&Homeschool&-&-\\Colin Sandon&Essex&1&00-41\\Krishanu Sankar&Horace Mann&1&00-41\\Alex Zhai&Univ. Lab (Urbana)&1&00-41\\ \hline IMO 2009&School&Number&\%ile\\ \hline John Berman&Hoggard&1&00-43\\Wenyu Cao&Andover&8&81-86\\Eric Larson&South Eugene&3&61-68\\Delong Meng&Baton Rouge M.&2&43-61\\Evan O'Dorney&Homeschool&-&-\\Qinxuan Pan&TS Wootton&2&43-61\\ \hline CGMO 2007&School&Number&\%ile\\ \hline Marianna Mao&Mission San Jose&3&64-74\\Wendy Mu&Saratoga&4&74-82\\Colleen Lee&Palo Alto&3&64-74\\Patricia Li&Lynbrook&5&82-85\\Sway Chen&Lexington&4&74-82\\Jennifer Iglesias&IMSA&2&42-64\\Wendy Hou&Hillsborough&1&00-42\\Sherry Gong&Exeter&16&96-100\\ \hline CGMO 2008&School&Number&\%ile\\ \hline Wendy Mu&Saratoga&8&88-90\\Colleen Lee&Palo Alto&7&80-88\\Jennifer Iglesias&IMSA&5&71-79\\Carolyn Kim&Lawton Chiles&2&41-57\\Jenny (Ning) Jin&Taft School&3&57-65\\Joy (Shijie) Zheng&Lakeside&5&71-79\\Lynnelle Ye&Palo Alto&7&80-88\\Inyoung Cho&Exeter&17&96-100\\ \hline CGMO 2009&School&Number&\%ile\\ \hline Carolyn Kim&Lawton Chiles&1&00-43\\Joy (Shijie) Zheng&Exeter&16&93-96\\Elizabeth Synge&Boston U. Acad&2&43-61&Cynthia Day&Lynbrook&6&74-78\\Patricia Li&Lynbrook&6&74-78\\Shiyu Li&Cupertino&2&43-61\\Ramya Rangan&Harker&5&70-74\end{tabular}

Interpretation: A line such as "43-61" in the percentile column means that 43% of USAMO qualifiers (excluding homeschoolers and non-USA residents) went to schools with strictly fewer USAMO qualifiers than that student did, and 61% went to schools with that many or fewer.
[/hide]

If I average those percentile numbers, I can create a "cluster index", in which a random sample of USAMO qualifiers would score about 50. Cluster index by year:
\begin{tabular}{rcc}Year&IMO&CGMO\\ \hline 2007&56&68\\2008&31&77\\2009&55&64\\Total&48&70\end{tabular}
It's a dramatic result; the boys on the IMO teams are distributed almost exactly like USAMO qualifiers overall, while the girls on the CGMO teams are heavily concentrated in schools with lots of qualifiers.

[quote="JBL"]Altheman, it sounds to me like you're trying to extrapolate from your personal experience to human universals in a way that's unlikely to be valid.  The United States is only middling when it comes to political participation by women, for example: even Pakistan, of all places, does better.  ([url=http://www.feministing.com/archives/018807.html]link[/url])  I think, for example, that one lesson to be learned from Sweden and other Nordic countries is that you can (largely through modest changes to public policy) create a society in which [url=http://yglesias.thinkprogress.org/archives/2009/10/the-daddy-factor.php]men do their fair share of child-care[/url] even while [url=http://fistfulofeuros.net/afoe/economics-and-demography/the-myrdals-and-feminist-natalism/]birth rates remain relatively high[/url].  It's not at all clear why, in light of examples like this, we should assume that the gender distribution in various different kinds of employment (e.g., politics or mathematics) is at some biologically-determined equilibrium.[/quote]
How can this possibly constitute a valid argument? The so called modest changes to public policy ensure that unless you are rich (after taxes), the family really has no choice but to split the parental leave at least 75/25 between the mother and the father. So, it really says nothing about the validity of anything.

Modestly different regulatory regimes have substantially different outcomes, ergo human nature is not a stronger constraint on division of labor than some rather small details of public policy.  In other words, it demonstrates that the American division of child-tending labor is not the only one consistent with human nature and therefore disproves the notion that the current division of labor is the "natural" one.


geometry
rectangle
combinatorics unsolved
combinatorics

arrange 100 dots in a 10*10 grid and color each dot either white or black (randomly).

prove that these dots contain a monochromatic rectangle (means that the 4 corners of rectangles are either all white or all black).

does this result hold for any n*n grid?

You can guarantee there is a monochromatic rectangle iff n > 4. It is easy to construct a 4 x 4 grid without a monochrmatic rectangle. We show that a 5 x 5 grid must contain a monochromatic rectangle, and any larger grid contains a 5 x 5 grid.

In each column, there must be either at least 3 black dots or 3 white dots. There must therefore be three columns that all have either 3 black dots or 3 white dots. If we superimpose these three rows, such that two dots lie untop of each other if they are in the same column, we have 9 dots of one color in five places, so there must at least 4 overlaps. But, since there are only 3 distinct pairs of rows from which two dots can come, at least one pair of overlaps must come from the same pair of rows--in other words, there must be four dots with the property that two are in each of two columns, and two are in each of two rows, and so which form a rectangle.

Sorry for my icky writing, but thats the idea.

Mmmm this problem has become inexplicable popular!, The funniest is that I composed an identical problem for our country olympiad!, well, good then!


In the figure, what is the ratio of the gray squares to the area of white squares?

$ \textbf{(A)}\ 3: 10\qquad
\textbf{(B)}\ 3: 8\qquad
\textbf{(C)}\ 3: 7\qquad
\textbf{(D)}\ 3: 5\qquad
\textbf{(E)}\ 1: 1$

[hide]

There are $ 6$ gray squares and $ 10$ white ones to get $ 6: 10$.  We can simplify that to $ \boxed{\textbf{(D)}\ 3: 5}$[/hide]


Newton $(x+a)(x+a^{2})(x+b)(x+b^{2})$ on $2^{2}\ge 1\cdot3$.
It is
$\left(\frac{(a+a^{2}+b+b^{2})^{2}}{4}\right) \ge \left(\frac{1}{1}\right)\left(\frac{ab+a^{2}b^{2}+(a+b)(a^{2}+b^{2})}{6}\right)$

Key is to rearrange the LHS a little (as you did) and then notice the RHS is products of cross terms of those. Of course the expand and apply AM-GM strategy seems to be quite effective as well, even if not beautiful. ;)


We contruct the sequence $\{x_{n}\}$ as follow: $x_{1}\in R$,$x_{n+1}=\frac{1}{n}\sqrt{1+x_{n}^{2}}\forall n \geq 1$.
Find $lim_{n\to \infty}(nx_{n})^{\frac{1}{x_{n}}}$


A circle is inscribed in a square. Prove that the probability that a point inside the square lies outside the circle is [tex]\displaystyle \frac{4- \pi}{4}.[/tex]

Something I can do!

[hide]The square has side length x, so the area of the square is x^2.  The circle has area equal to (x/2)^2*pi = x^2*pi/4.  Thus, the area of the region outside the circle = area of square - area circle = x^2 - x^2 * pi/4.  The probability is this divided by the area of the entire region, the area of the square:

x^2 - x^2 *pi/4

-------------------

x^2

x^2 ( 1-pi/4)

-------------

x^2

1 - pi/4 = (4-pi) / 4[/hide]

Okay, Julie, I think you got the right answer. But how did you get [hide]1- :pi:/4=4-:pi:/4[/hide].

I think it's easier this way:[hide]

Let the radius of the circle be r. Then the side of the square is 2r. The area of the circle is :pi:r:^2: and the area of the square is (2r):^2: =4r :^2: . So the area not in the circle, but within the square  is (4r:^2:)-(:pi:r:^2:). Then that is over the total area of the square so that it's probability.

so [(4r :^2:)-(:pi:r :^2: ]/(4r :^2:)  Divide both sides by r :^2:  and you have:

(4-:pi:)/4.[/hide]

That's a good proof I recommend that everyone try it.

BobofBobs wrote:Okay, Julie, I think you got the right answer. But how did you get [hide]1- :pi:/4=4-:pi:/4[/hide].

[tex]

\displaystyle1 - \frac{\pi}{4} = \frac{4}{4} - \frac{\pi}{4} = \frac{4-\pi}{4}

[/tex]

BobofBobs, in your solution, you need a pair of parentheses in a couple of places.  For example, when you wrote 4 - :pi:/4, you mean (4 - :pi:) / 4.


modular arithmetic
quadratics
Diophantine equation
algebra
number theory unsolved
number theory

Find all positiv integers n for which n^3+3^3 is a perfect square

this is not a complete solution, just some ideas maybe leading to a solution...

it's easy to prove that $\gcd(n+3,n^{2}-3n+9)|27$, which by a little reasoning implies WLOG that there are two cases:

(i) $n+3=x^{2}$ and $n^{2}-3n+9=y^{2}$, $x,y\in \mathbb{N}$. multiplying the latter equation by 4, we get $4n^{2}-12n+36=4y^{2}$; $(2n-3)^{2}+27=(2y)^{2}$; $(2y-2n+3)(2y+2n-3)=27$; and it is easy to show that this produces no solutions.

(ii) $n+3=3x^{2}$ and $n^{2}-3n+9=3y^{2}$, $x,y\in \mathbb{N}$.

it's clear in this case that $n=3k$ for some $k\in \mathbb{N}$. making this substitution in the above equations gives

$k+1=x^{2}$ and $3k^{2}-3k+3=y^{2}$ (*)

now it's clear that $y=3Y$ for some $Y\in \mathbb{N}$.

making this substitution in the second equation in (*) gives

$k^{2}-k+1=3Y^{2}$ (**)

now for (**) to hold we must have $k\equiv 2\pmod{3}$. then $k+1=x^{2}\equiv 0 \pmod{3}$, so we must have $x\equiv 0 \pmod{3}$. letting $k=3l+2$ and $x=3X$, $l, X\in \mathbb{N}$, in (*) and (**), we have

$l+1=3X^{2}$ and $3l^{2}+3l+1=Y^{2}$.

multiplying the latter equation by 4, we get

$12l^{2}+12l+4=4Y^{2}$,

$3(4l^{2}+4l+1)+1=4Y^{2}$,

$(2Y)^{2}-3(2l+1)^{2}=1$.

now this is a pell equation with smallest solution $(2Y,2l+1)=(2,1)$, so all solutions $(2Y,2l+1)$are given by $2Y+(2l+1)\sqrt{3}=(2+\sqrt{3})^{k}$,$k\in \mathbb{N}$. the solutions to our original problem will be given by those $(2Y,2l+1)$ for which $2l+1$ has the form $6X^{2}-1$.

[b][size=134]THIS IS ALL WRONG, I MADE A MISTAKE SO READ THIS ONLY IF YOU WANT TO LAUGH[/b][/size]

This, on the other hand, is a complete solution :)

As I said in here http://www.mathlinks.ro/Forum/viewtopic.php?t=113283  , so $d=gcd(n+1,n^{2}-n+1)|3$.

$1^{o}$ If $d=3$ then $3|n+1$ so $n^{3}+27 \equiv_{9}-1$ which is not a
quadratic residue $mod$ 9.

$2^{o}$ If $d=1$ then numbers $n+1$ and $n^{2}-n+1$ are perfect
squares. $n=1$ isn't a solution, and for $n>1$ it stands $(n-1)^{2}< n^{2}-n+1 < n^{2}$ so number $n^{2}-n+1$ can't be a perfect
square, thus there is no solutions for this equation.

Bye[/b]

I don't understand, which problem i solved?

OOOOOOps :)

Well, that's what you get when you write the solution right after you read the problem, without thinking...

I will delet the post

The Pely Equation is obtained exactly that way(as in the second post)
but there is no way to go from there:(
Interesting...

Easy one
you easily get 27*x^2-9*x+1 should be square for some positive integer x

then it equals to the (3x-1)^2+3x(6x-1)=y^2
so 3x(6x-1) is the sum of consecutive odd integers beginning from 6x-3=2*(3x-1)-1; assume that there are m terms

you write for m the quadratic equation and in dicriminant you get 108x^2-48x+something it doesn't matter

multiply first thing by foru and the difference should be also sum of consecutive odds, we know where it begins, so iif it has more then one term you get contradiction easily(try)...
and one term gives only solution x=0

so here is the final for this problem!


integration
inequalities
calculus
real analysis
real analysis unsolved

im reading from here
http://www.indiana.edu/~mathwz/PRbook.pdf
page 170 , theorem 170.1

$ \int_{X}|f|^{p}d\mu \leq \left(\parallel f\parallel _{\infty}^{p}\right) \int_{X}1d\mu$

where $\parallel \parallel _{\infty}$ is the norm used in $\textbf{Holder}$ inequality
i figgure because actually the norm
means $\parallel f\parallel _{\infty}= inf\{M;|f|\leq M \mu\ a.e.\ X \}$
now thats ok , but where does the $\int_{X}1d\mu$ come from ?
if the $\int_{X}1d\mu$ would be greater than one i guess
there'd be no problem.
im not managing understand this
please help

It's more or less clear; an integral can't be any larger than it's (a.e) maximum value, multiplied by the measure of the set you're integrating over (since $\int_{X}d\mu = \mu(X)$).

im sorry its not clear
why cant it be higher than that ?
maybe $\mu(X) \in (0,1)$.
what about that ?

EDIT: added some thoughts

or am i using
$|f|^{p}= |f|^{p}\cdot 1 = |f|^{p}\cdot 1 \leq |f|^{p}\cdot |1| \leq \parallel f\parallel _\infty \cdot |1|$
now integrating the last inequality the result follows ?


Is there an extremely easy, fast way to do this problem?

How many rectangles have sides determined by the grid lines below? How many of them are squares?

If it was normal, there is a quick way of doing it.

I don't know about abnormal.

Well, I get 72 squares.. if tahts right, I can post an easy solution to this

Wait, can you include the hole.. as a square or part of a square?If you can forget my answer...

if you can include the hole as a square or part of a square, I think it would be 91-
amount of 1X1 squares-8^2-4^2=48
amount of 2X2 squares-7^2-5^2=24
amount of 3x3 squares-none
amount of 4x4 squares-none
amount of 5x5 squares-4
amount of 6x6 squares-3^2=9
amount of 7x7 squares-2^2=4
amount of 8x8 squares-1^2=1
plus the hole as a square
so 91

but this is probalby not right..so dont trust this answer..
rectangles-im too lazy to count

[hide="rectangles"]I think this counts all exactly once-but
I'm not 100% sure. First, the # in the 2x8s on the sides-$4\binom{3}{2}\binom{9}{2}=432$. Subtracting the squares in the corners' rectangles, that is $432-4\binom{3}{2}\binom{3}{2}=396$. Now, for the rectangles w/the inner square, there are ${3}^{4}$, or $81$, for a total of $\boxed{477}$ rectangles. Tell me if I missed anything.[/hide]

Can you maybe use that formula for rectangles in an array for th ebig one and then subtract the omitted ones in the blank center?

Yes, but you would have to account for any rectangle that uses any part of that center.  Then, you would have to add back the rectangles that use the entire center.

[quote="funmath"]I got the first two parts but i dont get how u got the last one with the (3C2)4 can u plz explain and tell me how to get choose and superscripts on this thing?[/quote]

I used $\binom{3}{2}^{4}$ because I made a mistake. It's actually $3^{4}$, but it works out the same. That's b/c there arfe 3 choices for each side.

[code]\binom{m}{n}[/code] is $\binom{m}{n}$.

[code]{a}^{b}[/code] is ${a}^{b}$.

For future reference, just hover over other peoples latex, and you can see how it is made.

What do you mean by 3 choices per side?
(By the way, thanks a lot!)

You have to use the big square in the middle. But, there are choices for how far to extend each side. On each side, there are 3 different #s of squares to extend the side. (0, 1, or 2) So, there are $3^{4}$ ways to make rectangles using the big square. Is that clear?

[hide="solution"]We have somme x*y squares, where

$1/leq x /leq 2$

$1 /leq y /leq 2$

So 1*1's:

$8^{2}-4^{2}=48$

For 1*2's:

Now it's getting complicated. Let's label the columns 1-8.

There are 7 in column 1 and 2 and 7 and 8. In columns 3-6, there are 2.

So 28+8=36.

We multiply by 2 to account for the rows to get 72.

We do 2*2's.

By simply counting(which isn't hard now) we get

7*4-4(we're overcounting) =24. So there are 48+36+24=108 rectangles and 48+24=72 squares.[/hide]

Am I not getting something, or did I miss that someone already got this answer???

Well, madskillz said the answer would be 72 initially, but you and him both forgot to take in consideration the squares consisting of the big hole-so you didnt count that as a square, nor did you count the number of 5X5, 6X6, 7X7, or 8X8 squres (there are no 3X3 squres)


geometry
incenter
circumcircle
Asymptote
angle bisector

$ \triangle ABC$ is isosceles with $ AC\equal{}BC$. Let I be the incenter and let D be the point where the angle bisector of $ \angle A$ meets $ BC$. Let O be the circumcenter of $ \triangle ABD$. N is the midpoint of $ AD$ while M is the midpoint of $ BD$. P is the intersection of $ OD$ and $ BI$. Prove the N, M, and P are collinear.

It's time this problem is tried. Now if I remember how Tong Tonginated this... Any good ideas are his, and any inaccuracies are my own.

[hide]We will prove that $ N$, $ M$ and $ P$ are collinear by proving $ \angle{OMN}\equal{}\angle{OMP}$.

Since $ \angle{OND}\equal{}\angle{OMB}\equal{}90^{\circ}$, $ NOMD$ is a cyclic quadrilateral. Therefore, $ \angle{OMN}\equal{}\angle{ODN}$. Since $ O$ is the circumcenter of $ ABD$, $ \angle{ODN}\equal{}\angle{OAN}$.

Now, we will show that $ OPMB$ is cyclic. We see $ \angle{DOB}\equal{}\frac{1}{2}\angle{DAB}$. Therefore, $ \angle{DOM}\equal{}\angle{DAB}$. Also, $ \angle{IBD}\equal{}\angle{IBA}$ as $ IB$ bisects $ \angle{DBA}$. Since $ ABC$ is isosceles, $ \angle{IBA}\equal{}\angle{IAB}\equal{}\angle{DAB}$. Therefore, $ OPMB$ is cyclic.

We see $ \angle{OMP}\equal{}\angle{OBP}$ as $ OPMB$ is cyclic. Finally, $ \angle{OBP}\equal{}\angle{ABP}\minus{}\angle{ABO}\equal{}\angle{DAB}\minus{}\angle{BAO}$ (by symmetry). And $ \angle{DAB}\minus{}\angle{BAO}\equal{}\angle{DAO}$. Therefore, $ \angle{OMN}\equal{}\angle{OMP}\equal{}\angle{OAD}$. And we're done.[/hide]

Heh, I had nearly forgotten about this... I guess I will post my own solution.

[hide]
[asy]import cse5;
import olympiad;

pair B=(0,0),C=(250,0),A,D=(86,0),O,I,M,N,P;
A=IP(CR(B,130),CR(C,250));
O=circumcenter(A,B,D);
I=incenter(A,B,C);
D(MP("A",D(A),NE)--MP("B",D(B),SW)--MP("C",D(C),SE)--A);
D(A--MP("D",D(D),S));
D(B--MP("I",D(I),NE));
D(B--MP("O",D(O),NW)--D);
M=midpoint(B--D);
N=midpoint(A--D);
D(MP("M",D(M),S)--O--MP("N",D(N),NE));
unitsize(1);
D(rightanglemark(O,M,D),red);
D(rightanglemark(O,N,D),red);
P=IP(B--I,O--D);
D(MP("P",D(P),S));[/asy]

\begin{align*}
\angle PBM+\angle PDB&=\frac{1}{2}\angle ABC+90-\angle DOM\\
&=\frac{1}{2}\angle ABC+90-\angle BAD\\
&=\frac{1}{2}\angle ABC+90-\frac{1}{2}\angle BAC\\
&=90
\end{align*}

Then, $ \angle BPD=90$.

$ \angle BOD=2\angle BAD$
\begin{align*}
\angle BID&=180-\angle IBD-\angle IDB\\
&=180-\angle IBD-(180-\angle BAD-\angle ABD)\\
&=\angle BAD+\angle ABD-\angle IBD\\
&=\angle BAD+\angle ABI\\
&=2\angle BAD
\end{align*}

Therefore $ \angle BOD=\angle BID$ and $ OIDB$ is cyclic. Then $ O$ lies on the circumcircle of $ \triangle BID$. Also, $ N$ is the perpendicular from $ O$ to $ ID$, $ M$ is the perpendicular from $ O$ to $ BD$, and $ P$ is the perpendicular from $ O$ to $ BI$. Hence, by Simson Line Theorem, $ M, N,$ and $ P$ are collinear.
[/hide]

Not too difficult.

[hide="My Solution, a little different"]Let $ \omega$ be the circumcircle of $ \triangle ABD$, and let $ \angle IBA \equal{} \angle IDB \equal{} \angle DAB \equal{} \alpha$.  Extend $ DO$ to meet  $ AB$ at $ J$ and $ \omega$ at $ K$.  Since $ DAKB$ is cyclic, $ \angle DKB \equal{} \angle DAB \equal{} \alpha$.  Since $ KD$ is a diameter, we have that $ \angle DBK \equal{} \pi/2$.  Therefore $ \angle KDB \equal{} \pi \minus{} \pi/2 \minus{} \alpha \equal{} \pi/2 \minus{} \alpha$.  Now $ \angle DJB \equal{} \pi \minus{} (\pi/2 \minus{} \alpha) \minus{} 2\alpha \equal{} \pi/2 \minus{} \alpha \equal{} \angle JDB$.  Therefore $ \triangle BJD$ is isoceles and and $ P$ is the midpoint of $ DJ$.  Since $ A, J,$ and $ B$ are collinear and $ N, P,$ and $ M$ are the midpoints of $ DA$, $ DJ$, and $ DB$, respectively, we have that $ N, P,$ and $ M$ are collinear.[/hide]


linear algebra
matrix
search
linear algebra unsolved

$ A, B$ square matrix with integer entries, $ detA \equal{} 1$, $ detB\neq 0$. Prove that: there is positive integer n, such that $ BA^nB^ { \minus{} 1}$ has integer entries.


algebra
polynomial
limit
algebra unsolved

Use http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html with the polynomial $ x^{n\minus{}1}$ and see coeficients of $ x^k$ in both sides.

[quote="palazi"]I have problem with this:
$ f(x) = \prod_{i = 1}^n(x - x_i)$
Prove that:
$ \sum_{i = 1}^n \frac {x_i^k}{f'(x_i)} = \{\begin{array}{cc}0 \mbox{ for } 0 \leq k \leq n - 2 \\
1 \mbox{ for } k = n - 1 \end{array}$[/quote]
It is true only for case, when $ x_i$ are distinct.
I give one solution before.
Second solution. Let $ g(x)$ is polinom with $ degr(g(x)<n$, then $ f'(x_i)=\prod_{j\not =i}(x_i-x_j)$ and
\[ g(x)=\sum_i g(x_i)\frac{\prod_{j\not =i}(x-x_j)}{\prod_{j\not =i}(x_i-x_j)}.\]
Therefore ${ \frac{xg(x)}{f(x)}}=\sum_i \frac{g(x_i)}{f'(x_i)}\frac{x}{x-x_i}$.
It give $ \sum_i\frac{g(x_i)}{f'(x_i)}=\lim_{x\to \infty}\frac{xg(x)}{f(x)}$.

[quote="Rust"]
It is true only for case, when $ x_i$ are distinct. [/quote]
Of course it's true, i forgot write about this  :blush:

I have one question, because I'm not confident about this: [quote="Rust"] $ g(x) \equal{} \sum_i g(x_i)\frac {\prod_{j\not \equal{} i}(x \minus{} x_j)}{\prod_{j\not \equal{} i}(x_i \minus{} x_j)}$[/quote]. Isn't it only accuracy of polinom $ g(x)$ ? Can we write equals?
If answer is yes, i understand your proof :) Thanks a lot guys.

[quote="palazi"][quote="Rust"]
$ g(x)\equiv  \sum_i g(x_i)\frac {\prod_{j\not \equal{} i}(x \minus{} x_j)}{\prod_{j\not \equal{} i}(x_i \minus{} x_j)}$[/quote]. Isn't it only accuracy of polinom $ g(x)$ ? Can we write equals?
If answer is yes, i understand your proof :) Thanks a lot guys.[/quote]
It is true only for polinoms with $ degr(g(x))<n$ and $ x_1,x_2,...,x_n$ are distinct.


Joshua is holding a party at his house. Everyone going to the party brought food or drinks but not both. When everyone came, 2/3 of the girls bought food while 10 boys brought drinks. If there were 35 people ar the party and 3/7 of them were girls. How many people brought food?

[hide="Solution"]

$ \frac {3}{7}$ of $ 35$ gives $ 15$ girls and so $ 20$ boys, $ 10$ of whom brought drinks, so the other $ 10$ boys brought food.

Then $ \frac {2}{3}$ of the $ 15$ girls brought food, that is $ 10$ girls, giving $ 10 \plus{} 10 \equal{} 20$ people who brought food (and the other $ 5$ girls brought drinks).

[/hide]


modular arithmetic
number theory
relatively prime

Not much of a challenge...

If $m$  is  a  prime number and $a,b$ are two distinct numbers less than $m$,prove that

$a^{m-2}+a^{m-3}b+a^{m-4}b^2+.......b^{m-2}$

is a multiple of $m$

Are you sure? try  a=b=1 and m=5, for example...

This is about as far as I got.

[hide]$a^{m-2}+a^{m-3}b+a^{m-4}b^2+ \cdots +b^{m-2} = a^{m-2}\left( \frac{(\frac ba)^{m-1} -1}{\frac ba - 1} \right) = \frac{b^{m-1}-a^{m-1}}{b-a} = (b^{\frac{m-1}{2}} + a^{\frac{m-1}{2}})(b^{\frac{m-1}{4}}+a^{\frac{m-1}{4}})\cdots(b+a)$

Thus if exactly one of $a,b$ with $a \neq b$, is a quadradic residue $\mod m$, the statement is true.[/hide]

[hide]$a^{m-1}-b^{m-1}$ factors into $(a-b)S$ where $S$ is the original sum. Since $a$ and $b$ are (positive integers) less than $m$ and $m$ is prime, they must be relatively prime to $m$, so by Fermat's Little Theorem, $a^{m-1}\equiv b^{m-1}\equiv 1\pmod{m}\Rightarrow m|(a^{m-1}-b^{m-1})$.

Since $m$ is prime and $m$ divides a product, $m$ must divide one of the factors of the product. Then if $m\not| (a-b)$, $m|S$ and the statement is true. $|a-b|<m$, so if $m|(a-b)$, then $a-b$ must equal 0. So the statement is true as long as $a\neq b$.

But if $a=b$ then the statement is false, for $S=(m-1)a^{m-2}$ which is clearly not divisible by $m$.[/hide]

[quote]
If $m$  is  a  prime number and $a,b$ are two [b] distinct [/b]numbers less than $m$,prove that

$a^{m-2}+a^{m-3}b+a^{m-4}b^2+.......b^{m-2}$

is a multiple of $m$[/quote]

Shadysaysurspammed: you may like to edit the original question, as shown above,  so that the problem is correct. (see my  earlier post). HTH

Let $\bar a$ be the multiplicative inverse of $a$, mod $m$.

$a^{m-2} (1 + (b\bar a) + (b\bar a)^2 + \ldots{} + (b\bar a)^{m-2})$ (mod $m$)

$= a^{m-2} \left (\frac{{(b\bar a)}^{m-1} - 1}{b\bar a - 1} \right ) \equiv a^{m-2} \left (\frac{0}{b\bar a} \right ) \equiv 0$ (mod $m$)

as long as $b\bar a \ncong 1$ (mod $m$), or $a \ncong b$ (mod $m$)

But since $a,b<m$, this means that the theorem is true unless $a \equiv b$ (mod $m$)

Otherwise, it is clearly not true.

;)hai solve it for me
sum of the digits of 2^50


Let $ a$, $ b$, $ c$, $ d$ and $ e$ be whole numbers. The sum of the four numbers on each of the five longest line segments of the star is the same. If $ a \equal{} 18$, what is the value of $ b$?

think there should be a picture for that prob...

There should. It's a five-pointed star with $ a$,$ b$,$ c$,$ d$, and $ e$ being numbers written in circles at the vertices, going clockwise with $ b$ at the top. In addition, the inner pentagon, clockwise from top left, has $ 1$, $ 4$, $ 2$, $ 5$, and $ 3$.
[hide]
So
$ b\plus{}4\plus{}2\plus{}d\equal{}d\plus{}3\plus{}5\plus{}a$

$ b\equal{}\fbox{20}$[/hide]

btw, they should list question numbers along with the warmup set...


30. The activation energy of a certain reaction is $ 87 kJ/mol$.  What is the ratio of the rate constants for this reaction when the temperature is decreased from $ 37^\circ C$ to $ 15^\circ C$?

$ (A) 5/1 (B) 8.3/1 (C) 13/1 (D) 24/1$

[hide="Solution?"]

One equation that relates the rate constant and activation energy is the [b]Arrhenius Equation[/b], which states as follows:

\[ k \equal{} Ae^{\minus{}E_a/RT}
\]

where $ k$ is the rate constant, $ A$ is the frequency constant, $ E_a$ is the activation energy, $ R$ is the gas constant, and $ T$ is the temperature in Kelvin.

Assuming $ A$ is constant, we have

$ \dfrac{k_{310}}{k_{288}} \equal{} \dfrac{Ae^{\minus{}87000/310R}}{Ae^{\minus{}87000/288R}} \equal{} 13.17 \approx \boxed{13}$

Therefore, [b](C)[/b] is your answer.
[/hide]

[hide="hint"]for this, you can use one of the equations given on page 2 (the formula/periodic table sheet)   $ \minus{} \frac {E_a}{R} \equal{} \frac {lnk_2 \minus{} lnk_1}{\frac {1}{T_2} \minus{} \frac {1}{T_1}}$ just plug in the numbers (T is temp in Kelvins, $ E_a$ is activation energy and k is rate constant)(and don't forget to multiply E_a by 1000 to compensate for given kJ)
This is basically the same as the thing n0vad3m0n posted[/hide]


How many two-digit primes have a ones digit of 1?

If I'm not mistaken, $ 41$ and $ 61$ are also primes, so our answer is actually $ \boxed{5}$

And once again you correct me.  :blush:  Thanks.


Hello,
Here is the problem:
The situation is as in the picture below.
We apply a force $T_2$ on one end of a rope.
Let $\mu$ be the coefficient of static friction between the rope and the pole.
Then according to the following paper, http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000045000002000185000001&idtype=cvips&gifs=yes
the rope will not slip as long as the inequality $T_2 \leq T_1\exp(\mu\theta)$ is obeyed.

Does anyone know how to prove this?
Thanks,

Consider a tiny part of rope which touches the cylinder. Two forces of tension act on this, one from the left ($\vec T_i$) and one from the right ($\vec T_j$); the angle between these two vectors is $\text d\phi$. In the tangential direction an additional force of friction $F$ acts on this part of the rope; the critical condition for slipping is $F = T_i - T_j \cos\text d\phi \approx T_i - T_j$. If $N$ is the reaction of the pole on the part, the equilibrium in its direction gives us $N = T_j\sin\text d\phi \approx T_j\text d\phi$. Now, the force of friction is $F = \mu N = \mu T_j\text d\phi$, so plugging this back into the condition for slipping gives us $T_i = T_j (\mu\text d\phi + 1)$.

The number of such tiny parts is $n = \theta/\text d\phi$, so plugging this back into the above equation and applying the obtained equation for every part of the rope we get \[ T_2 = T_1 \left(1+\frac{\mu\theta}{n}\right)^n.  \] Taking the limit $n\rightarrow \infty$, this equation reduces to \[ T_2 = T_1 \exp(\mu\theta).  \]

Hi again,
Later in the same paper he says that when a rope is squeezed against the pole by another segment of rope, the tension in the lower segment differ on the two sides of the crossing: the limiting value of this difference is proportional to the tension in the upper segment of the rope.
How to prove this?

Thanks again,


calculus
derivative
calculus computations

If $ z \equal{} f(x^2 \minus{} y^2)$, show that $ y\frac{dz}{dx} \plus{} x\frac{dz}{dy}\equal{}0$.

I'm clueless  :( .

Pretty much the same thing (showing the chain rule a bit more).

$ y \frac{dz}{dx} \equal{} y \frac{d}{dx}f(x^2\minus{}y^2) \equal{} y f^`(x^2\minus{}y^2) \cdot \frac{d}{dx}(x^2\minus{}y^2) \equal{} y f^`(x^2\minus{}y^2) \cdot 2x.$

Also,

$ x \frac{dz}{dy} \equal{} x \frac{d}{dy}f(x^2\minus{}y^2) \equal{} y f^`(x^2\minus{}y^2) \cdot \frac{d}{dy}(x^2\minus{}y^2) \equal{} x f^`(x^2\minus{}y^2) \cdot (\minus{}2y).$

We get

$ y\frac {dz}{dx} \plus{} x\frac {dz}{dy} \equal{} \left (2xy f^`(x^2\minus{}y^2)\right) \plus{} \left(\minus{}2xy f^`(x^2\minus{}y^2) \right) \equal{} 0.$


wen a positive 3-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole-number quotient with a remainder of 1. How many distinct values are possible for this three digit dividend?

[hide="hint"]How many three digit numbers with units digit 1 have a factor less than 10 that isn't 1.[/hide]

Isn't anyone going to attempt this beastly problem?

[hide] I used bpms's hint of how many 3 digit number with a units digit 1 had a factor less than 10.  This is the same thing as finding the 3 digit numbers with units digit 2 that have a 1 digit divisor remainder 1.  So using the easier method, I eliminated several divisors.  2,4,6, and 8 are all out since they are even.  Composite numbers are out because their prime factors count everything before them.  5 is out because all multiples of 5 end in a 0 or 5.  So the only 2 divisors left are 3 and 7.  Using PIE, I got 30+12-4 or 38.[/hide]

I can do it with guess and check but i don't know any other way.

[quote="espark52"]:jump:   :jump:[/quote]
i wouldn't make posts that are 100% smilies.  :wink:

This is kind of irrelevant, but I still don't know how to post math formulas like x^2+x+1 in those bold "things"

And oh yeah

김치 is so good

[hide]all the three digit numbers with unit digit of 1... and can be factored by either 3 or 7.

3: the sum of the digits must be a multiple of 3
111, 141, 171, 201, 231, 261, 291... so on to...981
30 total

7: hmm... tricky       only 7 times 3 equals something unit 1
7*23=161  ...  7*133=931
13-2+1=12
 Then their is 33, 63, 93, 123 divisible by 3
 12-4=8

30+8=[b]38[/b][/hide]

I did it :lol:

[quote="Andrew Kim"]This is kind of irrelevant, but I still don't know how to post math formulas like x^2+x+1 in those bold "things"[/quote]
You mean $\LaTeX$? Just put dollar signs around the text. x^2+x+1 becomes $x^{2}+x+1$.

[quote="Andrew Kim"]Thanks

Now I can post [b]nice [/b]things  :evilgrin:

$x^{2}/25+y^{2}/9=1$[/quote]
for fractions, use the \frac function like this:

$\$$\frac{x^{2}}{25}+\frac{y^{2}}{9}= 1$\$$
which comes out to look like:
$\frac{x^{2}}{25}+\frac{y^{2}}{9}= 1$


Given $\triangle ABC$, let $D$, $E$, $F$ be the points on $AB$, $BC$, and $CA$, respectively, such that $AD: DB=BE: EC=CF: FA=2: 1$. Next, take points $X$, $Y$, and $Z$ on $DE$, $EF$, and $FD$, respectively, such that $DX: XE=EY: YF=FZ: ZD=2: 1$. Prove that $\triangle ABC$ and $\triangle XYZ$ are similar.


Let $ P(x)$ be a polynomial of degree $ 16$ with integer coefficients.

a) Knowing that $ P(x)$ admits odd values for $ x\equal{}0$ and $ x\equal{}1$, show that $ P(x)$ does not have integer roots.

b) Knowing that $ P(x)\equal{}7$ for $ 4$ distinct integer values of $ x$ , for how many integer values of $ x$ does $ P(x)$ admit the value $ 14$ ?

[hide="Solution"]

a) $ P(0)$ is odd, so the constant term is odd
$ P(1)$ is odd, so the sum of all coefficients is odd

Now for even $ x$, parity of the polynomial depends on the constant term - it is odd (every term before is even)
For odd $ x$, parity depends on the sum of all coefficients (number multiplied with odd number doesn't change product's parity)

We conclude that $ P(x)$ is always odd, so it hasn't got any zeros.

b) Because of the above, none (14 is even)

But, fact b) can be expanded to every polynomial:

We know that $ P(x)\minus{}7\equal{}m(x\minus{}a)(x\minus{}b)(x\minus{}c)(x\minus{}d)Q(x)$ where all the numbers are integers
If $ P(x)\equal{}14$, then the RHS of the expression above is 7 and its number of divisors isn't enough (remember, $ a,b,c,d$ are different!), so no value of $ x$ satisfies $ P(x)\equal{}14$

[/hide]

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=354820815&t=309455

And it wouldn't surprise me if some of your other questions were in http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=354820815&t=309443

Do a search before you flood a forum!


At Irv's Cycle Rental Shop, Irv rents all kids of cycles:  unicycles, tandem bikes, regular bikes, and even tricycles for little kids.  He parks all the cycles in the front of his shop with a helmet for each rider strapped to the cycles.  This morning Irv counted 57 helmets and 115 wheels parked in front of his store.  He knows he has an equal number of unicycles and tandem bikes.  he also knows that he has 32 regular bikes.  How many Unicycles, tandem bikes, and tricycles does Irv have?

I've been at this quetion for a week and I still dont get it. If someone could help me out I would greatly apprishiate(sp?) it.

Thanks.

How many wheels does a tandem bike have?

We know that there are 32 regular bikes so that is 64 wheels and 32 helmets.
So there is 25 helmets for other bikes and 51 wheels.
$b$=tandem bikes
$t$=tricycles
$u$=unicycles

$b-u=0$
$b+t+u=25$
$2b+3t+u=51$

You can solve this system however you want.
There are $8$ tandem bikes, $9$ tricycles, and $8$ unicycles

[hide]u=# of unicycles
x= # of tandem bikes
t= # of tricycles

u=x
u+2x+t=25
u+2x+3t=51

1.) Substitute the x's with u's:
u+2u+t=25
u+2u+3t=51

2.) Subtract the two equations:
 u+2u+3t=51
[u]-u+2u+1t=25[/u]
 2t=26, so t=13

3.) If t=13, then 3u=12, so u=4

4.) If u=4, then x=4

5.) There are 13 tricycles, 4 unicycles, and 4 tandem bikes[/hide]

biffanddoc, I got a different answer from you because tandem bikes have two wheels, but they also have seats for two riders, so each tandem bike has 2 helmets, not 1.


Does anyone have a link to that one website where you type in four terms of a series and it tells you which one it is?


Can the Euclidean minimum spanning tree (MST) of n points in R^d be computed in time close to the lower bound of  Omega(n log n) ? :?

Azat123:

In the first place, this isn't complex analysis.

Secondly: dumping a very large number of problems, many of them famous unsolved conjectures, into these forums all at once is simply not helpful behavior.

Can you rationally discuss them yourself? Offer partial approaches? Say anything helpful?

I suspect that nearly all of your posts will go unanswered, and that the usual heavy users of the site will start treating anything that you post as something that doesn't require their attention.

You have to offer some reason for a person to want to reply to your posts. Posting in huge volume is not a means to that end.


$ x,y,z$ are positive real numbers satisfying $ xyz\equal{}1$. Prove that
$ \frac{1}{1\plus{}2x\plus{}2xy}\plus{}\frac{1}{1\plus{}2y\plus{}2yz}\plus{}\frac{1}{1\plus{}2z\plus{}2zx}\geq\frac{3}{5}$

Let $ x \equal{}\frac{a}{b}, y \equal{}\frac{c}{a}, z \equal{}\frac{b}{c}$ where $ a,b,c$ are any positive reals.
Then the inequality becomes:
$ \sum\frac{1}{1\plus{}\frac{2a}{b}\plus{}\frac{2c}{b}}\geq\frac{3}{5}$
$ \Leftrightarrow\sum\frac{b}{2a\plus{}2c\plus{}b}\geq\frac{3}{5}$
Wolog $ a\plus{}b\plus{}c \equal{}\frac{1}{2}$
Then the inequality becomes $ \sum\frac{a}{1\minus{}a}\geq\frac{3}{5}$
However, $ \frac{a}{1\minus{}a}\geq\frac{36a\minus{}1}{25}\Leftrightarrow 25a\geq\minus{}36a^{2}\plus{}37a\minus{}1\Leftrightarrow (6a\minus{}1)^{2}\geq 0$ which is true.
So, $ \sum\frac{b}{2a\plus{}2c\plus{}b}\geq\sum\frac{36b\minus{}1}{25}\equal{}\frac{36(a\plus{}b\plus{}c)\minus{}3}{25}\equal{}\frac{18\minus{}3}{25}\equal{}\frac{3}{5}$ as $ a\plus{}b\plus{}c \equal{}\frac{1}{2}$
QED

The there another way to kill it  :D
[quote="rem"]Let $ x \equal{}\frac{a}{b}, y \equal{}\frac{c}{a}, z \equal{}\frac{b}{c}$ where $ a,b,c$ are any positive reals.
Then the inequality becomes:
$ \sum\frac{1}{1\plus{}\frac{2a}{b}\plus{}\frac{2c}{b}}\geq\frac{3}{5}$
$ \Leftrightarrow\sum\frac{b}{2a\plus{}2c\plus{}b}\geq\frac{3}{5}$
[/quote]

By Cauchy Swartz we have $ \sum\frac{b}{2a\plus{}2c\plus{}b}\geq\frac{(a\plus{}b\plus{}c)^{2}}{(a\plus{}b\plus{}c)^{2}\plus{}2ab\plus{}2bc\plus{}2ca}$
So it suffices to prove that
$ \frac{(a\plus{}b\plus{}c)^{2}}{(a\plus{}b\plus{}c)^{2}\plus{}2ab\plus{}2bc\plus{}2ca}\geq\frac{3}{5}$  which is equivalent to
$ (a\plus{}b\plus{}c)^{2}\geq 3(ab\plus{}bc\plus{}ca)$ which is .... true  :)

[quote="quangpbc"][quote="patham"]Great solution!!! :lol:
But how did you come to think of $ \frac{36a\minus{}1}{25}$ ?[/quote]

We can find $ k,l$ so that :

$ \frac{a}{1\minus{}a}\ge ka\plus{}l$

Try $ a \equal{}\frac{1}{6}$ then :

$ \frac{1}{5}\ge\frac{k}{6}\plus{}l$ we choose $ l \equal{}\frac{1}{5}\minus{}\frac{k}{6}$

Now we find $ k$ so that :

$ \frac{a}{1\minus{}a}\ge ka\plus{}\frac{1}{5}\minus{}\frac{k}{6}\Leftrightarrow (6a\minus{}1)(\frac{1}{1\minus{}a}\minus{}\frac{k}{6})\ge 0$

In the expression: $ \frac{1}{1\minus{}a}\minus{}\frac{k}{6}\ge 0$ put $ a \equal{}\frac{1}{6}$ again  we have $ k\le\frac{36}{5}$, choose $ k \equal{}\frac{36}{5}$ :

Last we must prove the inequality :

$ \frac{a}{1\minus{}a}\ge\frac{36a\minus{}1}{5}$  :wink:[/quote]
We can use Fermat law
Let $ f(x)$ is continuous and on $ [a,b]$ it has extram value at $ x_{0}$ then $ f'(x_{0})\equal{}0$
Use it.

My proof is UCT theorem, it is a useful method to kill some problem :D .

Example :

Let $ a,b,c>0$ and $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}3$ .Find the minimum value of

$ \text{P}\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{4(a\plus{}b\plus{}c)}{3}$  :wink:

Dear quangpbc,what do you mean by UCT method???
Can you explain it to me please??
Btw,your problem is hard to solve. :(

No stephen38, it is not to hard  :D .

Solution :

We prove the inequality :

$ \frac{1}{a}\plus{}\frac{4a}{3}\ge\frac{a^{2}}{6}\plus{}\frac{13}{6}(1)$

$ (1)\Leftrightarrow (a\minus{}1)^{2}(a\minus{}6)\le 0$.

Similary :

$ \frac{1}{b}\plus{}\frac{4b}{3}\ge\frac{b^{2}}{6}\plus{}\frac{13}{6}$

$ \frac{1}{c}\plus{}\frac{4c}{3}\ge\frac{c^{2}}{6}\plus{}\frac{13}{6}$

We're done  :)


induction
number theory proposed
number theory

$p_{1},p_{2},p_{3}....$ sequence of primes such that holds below condition. If ${n}\ge{3}$ then $p_{n}$ is largest prime divisor of$p_{n-1}+p_{n-2}+2000$. Prove that sequence $p_{1},p_{2},..$ is bounded.

We can find 3000 consecutive composite positive integers(we can do it using Chinese remainder theorem or just take (3000+1)! + 2, (3000 + 1)! + 2 and so on).
Let this integers are a,a+1,...a+2999. We may consider $a>p_{1}$ and $a>p_{2}$
We claim that $a>p_{k}$ for all k. The proof uses induction on k.
For k = 1,2. it is true by the above asumption.
Suppose for all <k is true.
Case 1. $p_{k-1}$and $p_{k-2}$ are odd. Then $p_{k}$ is divisor of $p_{k-2}+p_{k-1}+2000$
But this number is even. WE have $\frac{p_{k-2}+p_{k-1}+2000}{2}\leq a+2999$ by induction asumption. From here and this number is composite(or smaller than a) so $p_{k}\leq \frac{p_{k-2}+p_{k-1}+2000}{4}\leq a$.
Case 2. One of $p_{k-1}$ and $p_{k-2}$ is 2. Then $p_{k}$ is divisor of $p+2002\leq a+2999$ which is either smaller than a or composite. In either case $p_{k}\leq a$
It remains the trivial case when both $p_{k-1}$ and $p_{k-2}$ are 2. But this is really trivial.


probability
AMC
AIME
analytic geometry
counting
distinguishability
number theory

[hide="solution"]
There probably is a nicer way, but if you count all of the possible arrangements and let $S_n = 1+...+n$, then the total number of arrangements is $(1) S_6 + (2) S_5 + (3)S_4 + (4)S_3 + (5)S_2 + (6)S_1 = \sum_{n=1}^{6} \frac{(7-n) n (n+1)}{2} = 7 \frac{6 \cdot 7}{4} + 6 \frac{6 \cdot 7 \cdot 13}{12} - \frac{6^2 \cdot 7^2}{8}$

The total number of possibilities for rolling the two dice is $6^4$.

The probability is then: $\frac{6 \cdot 7 \cdot 3}{6^4} = \frac{7}{6^2 \cdot 2} =\frac{m}{n}$

Therefore $m+n = 7 + 72 = 79$

Answer: $\boxed{079}$[/hide]

[hide]The dice rolls can be in the form

ABCD
AABC
AABB
AAAB
AAAA

where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.)

Now, for the first case, there are $\binom{6}{4} = 15$ ways for this. We do not have to consider the order because the combination counts only one of the permutations; we can say that it counts the correct (ascending order) permutation.

Second case: $\binom{6}{3} = 20$ ways to pick 3 numbers, $\binom{3}{1}$ ways to pick 1 of those 3 to duplicate. A total of 60 for this case.

Third case: $\binom{6}{2} = 15$ ways to pick 2 numbers. We will duplicate both, so nothing else in this case matters.

Fourth case: $\binom{6}{2} = 15$ ways to pick 2 numbers. We pick one to duplicate with $\binom{2}{1} = 2$, so there are a total of 30 in this case.

Fifth case: $\binom{6}{1} = 6$; all get duplicated so nothing else matters.

There are a total of $6^4$ possible dice rolls.

Thus,

$\displaystyle \frac{m}{n} = \frac{15 + 60 + 15 + 30 + 6}{6^4} = \frac{126}{6^4} = \frac{7}{72}$

So, $m + n = 7 + 72 = 79$.[/hide]

[quote="joml88"]A fair die is rolled four times.  The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$[/quote]

[hide="What is wrong with my solution..."]
The complement of this is the probability that each of the final three rolls is less than the roll preceding it, or the numbers are strictly decreasing. The number of ways to pick 4 numbers from $\{ 1, 2, 3, 4, 5, 6\}$ that are strictly decreasing is $\binom{6}{4}=15$ since you can just pick 4 distinct numbers and arrange them. Thus our answer is $1-\frac{15}{6^4}$. :(
[/hide]

Chess64,

Let $A$ be the probability we are looking for and $A'$ be the complement of what we are looking for.  Of course, $A=P(\text{4 dice are increasing}).$  Now, what is $A'$?  Well, it's everything else that can happen.  All of the dice decreasing is one part of it, but not all.  What if they are assorted randomly, i.e. neither decreasing nor increasing like 1,4,3,2?  That's what you forgot about.

chesspro: what do you mean by duplicate?

I mean that if we roll a 4, it is "duplicated" if we roll another 4.

I found a good solution now. :D

[hide]
If we randomly pick 4 integers from 1, 2, 3, 4, 5, 6, then there is a 1-1 correspondence between the sequence we pick and the numbers on the 4 rolls. So we can just solve the problem: "How many ways can we distribute 4 objects among 6 boxes," which has solutions that can be represented by $x_1+x_2+x_3+x_4+x_5+x_6 = 4$. The number of solutions of this is $\binom{9}{4}=126$. Thus our answer is $126/6^4=7/72 \implies \boxed {79}$.
[/hide]

i don't see how does this problem relates to the 1 to 1 correspondence?

[[2001 AIME I Problems/Problem 6]]; each set of increasing dice rolls can be matched with one four-element subset of $ S \equal{} \{1,2,3,4,5,6\}$ (think about it as, given any four-element subset of $ S$, we can only arrange them in one way to form a strictly increasing set of four dice rolls). Counting the number of four-element subsets can then be done by the ball-and-urn argument.

[quote="hrithikguy"]Sorry for reviving, but what is wrong with this solution:

ABCD
AABC
AABB
AAAB
AAAA

for the ABCD case, this is 6C4 = 15, but the probability that after the four numbers have been correctly ordered is 1/24, so the probability here is 15/24 = 5/8.

And then we do the total number of possible multiplied by the probability that it is correctly ordered for each case. However, this yields a probability greater than 1. What is going wrong?[/quote]

$\frac{1}{24}$ doesn't account for all the events (the $6^4$ events)- plus, ${6 \choose 4}$ already orders each of the permutations, so dividing by $24$ is unnecessary.

This is very hard for me and I cannot solve it. :bruce: :bruce: :starwars:

Recast the problem entirely as a block-walking problem.  Call the respective dice $a, b, c, d$.  In the diagram below, the lowest $y$-coordinate at each of $a$, $b$, $c$, and $d$ corresponds to the value of the roll.

[[File:AIME01IN6.png]]

The red path corresponds to the sequence of rolls $2, 3, 5, 5$.  This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths.

The solution to this problem is therefore $\dfrac{\binom{9}{4}}{6^4} = \boxed{\dfrac{7}{72}}$.

Sorry, but I had a question about the solution on the AoPS Wiki.

[hide]If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from $\{1,2,3,4,5,6\}$. We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of ${1,2,3,4,5,6}$), or dividing them amongst 5 separators. Thus, there are $\binom94 = 126$ outcomes of four dice. The solution is therefore $\frac{126}{6^4} = \frac{7}{72}$, and $7 + 72 = \boxed{079}$.[/hide]

Why is the number of ways to pick four numbers out of $6$ equal to $\binom94$ and not $\dfrac{6^4}{4!}$?

@droid347 The solution is basically trying to find the number of results of the rolls, so that the rolls are indistinguishable and taken as a group. For example, a roll of {4,5,2,4} is the same as {2,4,4,5}. Basically, we are trying to solve for the diophantine $x_1+x_2+x_3+x_4+x_5+x_6 = 4$, where $x_1$ represents the number of rolls which show 1, etc. Then, we can apply stars and bars. In essence, this is drawing a one-to-one between the possible rolls and arrangements of D|D|D||D|, where | is like a "divider". In this case, we would have $x_1 = 1, x_2 = 1, x_3 = 1, x_4 =0, x_5 =0, x_6 = 1$. This is a useful concept.

Thank you for the detailed explanation.  I now know how that method works, but why did mine fail?

Because the four dice are interchangeable, I thought.

In a roll like {2,4,4,5}, dividing by $4!$ wouldn't be correct, so we can use division by $4!$ for all cases.

Oh, so we could take cases for how many numbers are repeated?

[quote="BobaFett101"]If we were to correct for overcount in every cases, it would be quite tedious. When we pick 4 numbers from the set 1-6, we can pick numbers in the form AAAA, AAAB, AABB, AABC, and ABCD, where A, B, C, and D represent different numbers. (for example, picking the numbers 2, 4, 5, 2 would be classified as AABC: 2=A, 4=B, 5=C)

For the case AAAA, there is no overcount at all, while for ABCD, we must divide by 4 factorial. Basically, there are a lot of different cases to consider if you try casework. It is possible, but if I did it that way, I would likely make many calculation errors.[/quote]

This, basically. You could casework if you were patient and didn't make mistakes, but in an AMC-like shorter test I wouldn't recommend it.

Sorry for revive, but is there a generating functions solution to this?

Let $D_i$ be the outcome of the $i$th roll. We let $D_i$ take any value in $\{1,2,3,4,5,6\}$ with equal probability. We seek the probability that the $4$-tuple $(D_1,D_2,D_3,D_4)$ is such that $D_1\leq D_2\leq D_3\leq D_4$. Clearly, there are $6^4$ total $4$-tuples, and they are equally likely.

It remains to count the $4$-tuples meeting the condition. We can draw a bijection between the set $X$ of $4$-tuples satisfying the inequality chain and the set $Y$ of multisets of cardinality $4$ whose elements are from $\{1,2,3,4,5,6\}$. Map the $4$-tuple $(D_1,D_2,D_3,D_4)$ to the multiset $\{D_1,D_2,D_3,D_4\}$. Since the elements of any element of $Y$ can only be ordered as an increasing chain in one way, this map is injective. Further, any given element of $Y$ consists of $4$ numbers that can form an increasing chain, so there is an element in $X$ that maps to that given element, thus the map is surjective. We can compute $|Y|$ using the well-known stars and bars bijection. We find that $$|Y|=\binom94=126.$$ Therefore, the probability is $$\frac{|X|}{6^4}=\frac{|Y|}{6^4}=\frac{126}{6^4}=\frac{7}{72},$$ and the answer is $7+72=\boxed{079}$.

As we know the total sample space for the dice is 6^4 so p(e)=favourable outcomes/total outcomes
For the favourable outcomes
We can fill the first space in any of the 6 ways and let the last three space be a b c such that 1<=a<=b<=c<=6 then there can be two cases when a<b<c which can be done is 6c3 ways
And one more way when they are equal is 1 so p(e)=6(6c3+1)/6^4 which is 126/1296 simplifying to get 7/72  then m+n=79


logarithms
LaTeX
number theory solved
number theory

I'm sorry if this has appeared anywhere. Please help me with this problem.
$S(k)$ is the set of numbers $n$ for which $\ 2^n$ and $\ 5^n$ have the same first $k$ digits.
Find the content of these sets.
I've found some simple rules and have checked them for a great number of cases but I can't prove any of them. I can't come up with any general rules either.

Acutally we can find a pretty neat solution by multiply $\ 2^{n}$ and $\ 5^{n}$ at first.
Assume the first $\ k$ digits of both two numbers are $\ \alpha$, then $\ 2^{n} = \alpha * 10^{deg(2^{n})-k} +$ some number and $\ 5^{n} = \alpha * 10^{deg(5^{n})-k} +$ some number
which leads to $\ (\frac{\alpha ^{2}}{10^{k-1}}) ^{2} < 10 < (\frac{\alpha ^{2}}{10^{k-1}}+1) ^{2}$ I'm not so sure if this can be easily proved
so that the value of $\ \alpha$ is approximately $\ \sqrt{10}$ (a little bit less than $\ \sqrt{10}$, in fact)
then we'll need to find any values of $\ 2^{n}$ and $\ 5^{n}$ which has a starting as $\ 3.16...$

I could be wrong, but taking some $\epsilon$ to describ how near we have to be to $\sqrt{10}$ and then using the method from the problem in http://www.mathlinks.ro/Forum/viewtopic.php?highlight=20012001&t=50019 should do.

From what I learn from that link, we can prove that there really exist $n$s in all set $S(k)$.
Kalimdor's idea is great, and I think you mean
$(\frac a{10^{k-1}})^2 < 10 < (\frac{a+1}{10^{k-1}})^2$

But I still can't find a way to prove it. Could you please help me?

Oh, i think the way to prove this is to make a regular multiplication,
$\ 2^n * 5^ n$ can be written in this way:
which is     $\ \ \ \ \ 316...................$ $\ \rightarrow$value of$2^n$
$\ \times \ 316......................$ $\ \rightarrow$value of$5^n$
equals
$\ 1000000000000000000000000000000000000000000000...$
If the first $k$ digits do not start with $\ 3.16...$ assume it's something else, then square it, you can say that's impossible for you to get a $\ 100000$ at the begining of the result

And for the suggestion from $ZeTaX$, can you please give us your solution in using  this kinda idea: $10^s = 10^{a+b \cdot log_{10}(2000)} = 10^a \cdot 2000^b$?

Choose an $\epsilon >0$, just depending on the $k$ you have (the exact $\epsilon$ you need is some boring calculation).
Then the set of all $s \in \mathbb{N}$ with $\frac{1}{2}-\epsilon < \{s \cdot \log_{10}(2)\} < \frac{1}{2}+\epsilon$ lies in $S(k)$, and when choosen $\epsilon$ exactly, also only then.

I got it now. Thanks a lot everyone.
I really want to post the solution here but I don't know how to type it in LATex. Could anyone help me please?

See Valentin's introduction into $\LaTeX$: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261


Determine the number of integers n such that 1<=n<=1000 and n^(12)-1 is divisible by 73.

I might be mistaken (number theory is most definitely not my speciality), but...

From Fermat's little theorem $n^{p-1}\equiv1$ (mod p).  73 is prime, so we want $n$ to be a sixth power (so we have $n^{72}= n^{73-1}$).  There are 3 sixth powers less than 1000 (1, 64, 729), so the answer is 3.

You've missed an awful lot of solutions, and here's why:  it's only necessary that $n$ be [b]congruent[/b] to a sixth power $\bmod 73$.  There are $12$ such remainders.

[hide]
\[n^{12}-1=(n^{6}-1)(n^{6}+1)=(n^{3}-1)(n^{3}+1)(n^{2}+1)(n^{4}-n^{2}+1)\]
\[=(n-1)(n+1)(n^{2}-n+1)(n^{2}+1)(n^{2}+n+1)(n^{4}-n^{2}+1)\]
One of these factors must be divisible by 73. $n+1$ and $n-1$ are trivial. $n^{2}-n+1$ and $n^{2}+n+1$ are also surprisingly easy, since $9^{2}-9+1=8^{2}+8+1=73$. We get $n\equiv 9,-8$ and $n\equiv 8,-9$ in each of these cases.

For the $n^{2}+1$ factor, use $73=8^{2}+3^{2}$ and $8(-9)\equiv 1$. We find that $n\equiv \pm 27$ works here.

Finally, the quartic remains. We find $n^{2}\equiv 9,-8$. The case $n^{2}\equiv 9$ is easy. For $n^{2}\equiv-8$, we have $73=1+8\cdot 3^{2}$ and $3(-24)\equiv 1$. This gives $n\equiv \pm 24$.

So all solutions $n$ are given by
\[\boxed{n\equiv \pm 1,\pm 3,\pm 8,\pm 9,\pm 24,\pm 27}\]
I'm not writing all solutions up to 1000 out. :P

If you want to find the number, use $1000=73\cdot 13+51$. This means 13 groups of 12 solutions each, plus the solutions up to 51, or -22. There are 8 of these. This makes the number $13\cdot 12+8=\boxed{164}$.
[/hide]

Alternately, once you've found a primitive root $p$ of $73$, the elements of order $12$ are given by $p^{6k}, k = 0, 1, 2, ... 11$.


کمک!!     F(X)F(YF(X)_1)=X^2F(Y)_F(X)
 F AZ R BE R

soaletun ro lotfan ba $ \text{\LaTeX}$ benevisid(chon soaletun ro nemishe fahmid chie...) ,in ye pdf ast ke omidvaram be shoma komak kone...


Find all $n$ that $X$ and $F$ with the following conditions exist:
Suppose $|X|=n$ and $F \subset P(X)$
i)$\forall a,b \in X \exists ! A \in F:a,b \in A$
ii)$\{ a,b,x \} , \{ a,c,y \} , \{b,c,z \} \in F \Longrightarrow \{ x,y,z \} \in F$

oh nice linear algebra ,it is from IMC2003 not from iran

Well send your linear algebra solution.But I have non-linear algebra solution.
It was in one of our exams for being ready.

Can somebody explain a little the problem, i dont understand many things, what is P(X)? what are the conditions? i simply dont understand and i want to try it.

I think Omidforgotto mention that $F$ shouldonly contain sets with $3$ elements.

Oh yes all sets in $F$ have 3 elements
Also $P(X)$ is set of all subsets of $X$

what does that E! means, anctually the "E" is reversed...

$\exists!..$ means there exists a unique..


function
inequalities proposed
inequalities

[quote="billzhao"]Easy. The function is linear in every variable, so we only need to check the case where $x_k \in \{0, 1\}$ for all $k$.[/quote]

may i know why?

Because linear functions are convex.  It's worth trying to prove for yourself that a convex function in $\mathbb{R}$ does indeed assume its maximum at one of its endpoints.

Realy easy!!!
We'll prove that:
$\prod\limits_{k=1}^{n-1}(1-x_k) + \prod\limits_{k=1}^{n-1}(1+x_k) \le \prod\limits_{k=1}^{n}(1-x_k) + \prod\limits_{k=1}^{n}(1+x_k) $
$\Leftrightarrow \prod\limits_{k=1}^{n-1}(1-x_k)x_n \le \prod\limits_{k=1}^{n-1}(1+x_k)x_n$
Now it is trival because
$1-x_k \le 1+x_k$
Hence
$\prod\limits_{k=1}^{n}(1-x_k) + \prod\limits_{k=1}^{n}(1+x_k) \ge (1-x_1) + (1+x_1) =2 $


calculus
derivative
number theory
arithmetic sequence
number theory theorems

I've searched some books but didn't find any theory about this, so I would like to ask you if you can show an e-book or teach me about this here in this topic.

It's like, a certain type of sequence, when you have, for example

$(4, 11, 22, 37)$

If you make the difference between $a_{n}$ and $a{}_{n}{}_{-}{}_{1}$, you will get:

$(7, 11, 15)$ , which is an arithmethic progression.

Now, if you have $(2, 9, 28, 64, 126)$ , if you use the difference between $a_{n}$ and $a{}_{n}{}_{-}{}_{1}$, you will get $(7, 19, 37, 61)$, but, proceeding with $a_{n}-a{}_{n}{}_{-}{}_{1}$ again, you will get $(12, 18, 24)$, which is another arithmethic progression. The first sequence of this paragraph, $(2, 9, 28, 64, 126)$ , seems to follow the formula $a_{n}=n^{3}+1$.

I would like to know some theory about these kind of progressions, like it's name, how to calculate the general therm, it's sum, and everything else I could learn. Can you help me?

hm...

Interesting.

Did you read That book that Says about that man called Eff Der Thuns?!

He give the solution for this problem.

Maybe this is what you want http://en.wikipedia.org/wiki/Difference_equations or http://en.wikipedia.org/wiki/Recurrence_relation

hmm... Thanks, Tuan, but, in fact, I don't know if that wiki page shows how to operate them simply. Recurrences are OK to do that, but seems too much complex for these simply sequences. I am searching for specific ways to solve them, like, I figured out that the more you use $a_{n}-a{}_{n}{}_{-}{}_{1}$, the more elevated is the expoent of the polynom that gives you the general therm. My examples on the post above show that.

err... By the way, sorry, but, I just saw I've created this topic at Number Theory section. Should it have been better to create it at Algebra section? I meant to do this, I don't know how my topic appeared here  :lol:

I think that I know what You want:

${F_{1}(n)=F_{0}(n)-F_{0}(n-1)}$
${F_{2}(n)=F_{1}(n)-F_{1}(n-1)}$
${\ldots }$
${F_{k}(n)=F_{k-1}(n)-F_{k-1}(n-1)=c_{1}}$

(The Last is because in some $k$ we find a arithmetic sequence)

Remember the formulae for Derivatives

${f(n)=\frac{F(n)-F(n-\epsilon )}{\epsilon }}$

If $\epsilon =1$

So, we can think that $F_{k}$ Is the $k$ quasi-derivative of $F_{0}$
Using
http://www.xuru.org/fc/Intro.asp we can conclude that
$(h \to 1, n \to k, x \to n\)$

${F_{k}(n)=\sum_{m=0}^{k}\left((-1)^{m}\left( \begin{array}{c}k \\ m \end{array}\right) F_{0}(k-m+n)\right)=c_{1}}$

And I Will not try to solve this because I'm not crazy!!!

[hide]If you want to solve lets the tip

In the $k+1$ quasi-deritivative, the diference is zero.

and you can assume that the answer is exponential $F_{0}(x) = \alpha. \beta^{x}$

See wikipedia.[/hide]

Look up difference operators or even $\delta$-operators of umbral calculus, they show the connections between finite differences and derivation.

There are methods to solve all recurrences of type $a_{n}= c_{1}a_{n-1}+...+c_{k}a_{n-k}$ for all constants $c_{i}$ and fixed $k$. And even if your reccurence is different, there are good chances that some "solution" exists.

And@Fabiomr1: doing that equations by the most simple way shows you often nothing about their structure, doing the theoretical way is much more helpfull on the longer term.


Notice the cevian can be at any angle, and x will remain the same (AAS). So, we choose the cevian to be an altitude, dividing the angle into 30 and 45 degree parts. We can now use special triangle formulas. The altitude is (1/2)(sqrt3) of the hypotenuse of the 30-60-90, and x is (sqrt2) of the altitude.

[(1/2)(sqrt3)(2)] * (sqrt2)
sqrt3 * sqrt2
$ \sqrt6$


I got a 4/5 on Round 2 Problem 4 because "you need to show that $n<k^{2}+k-2$ also doesn't work." Yet my solution was the only featured solution (see [url]http://www.usamts.org/Solutions/Solution4_2_18.pdf[/url]) for that problem. Moreover, the edited solution still doesn't account for this.

I know we're not allowed to debate one point, and that featured solutions and commendations don't really matter, but does that strike anyone as a conflict of interests?

Not a 'conflict of interests' (not really sure what you could be referring to with that phrase), but rather, a probable mistake on our part.  We'll sort it out when we're back in the office on Monday.

Yeah, basically I meant to say that whoever made my solution the official solution must not have agreed with whoever graded my solution in the first place. On the Round 3 Math Jam (see [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=189[/url]), I used my score on this problem to argue that we weren't done with Problem 4.

Now, given that my solution became the official solution, could we get around the "more than one point" rule with a point and a commendation? Given that I don't know what happened with another similar case ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135172[/url]) and that the rubric ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125937[/url]) for the problem is unclear on the meaning of "significant gap in the argument," should I repost this in the Grading Issues section?


I actually solved this in a class at MathPath with about 10000 hints. Anyway,

If $a$ is a non-negative real and $x$ is a real number satisfying
$(x+1)^{2}\ge a(a+1)$, show that $x^{2}\ge a(a-1)$.

[quote="t0rajir0u"][hide] This is for the case $x$ non-negative.  I'm too lazy to work out the complete solution  :|

$a(a+1) = a(a-1)+2a$

The first inequality becomes

$x^{2}+2(x-a)+1 \ge a(a-1)$

Now, we have $x+1 = \sqrt{a(a+1)}$.  From any of several different arguments (for example, convexity, or simply examining $(a+\frac{1}{2})^{2}= a^{2}+a+\frac{1}{4}$) we can conclude that $a < x+1 < a+\frac{1}{2}\implies (x-a) <-\frac{1}{2}$.  Then

$2(x-1)+a < 0$

From which it follows

$x^{2}> x^{2}+2(x-a)+1 \ge a(a-1)$.  QED. [/hide][/quote]

For negative x, $x^{2}> (x+1)^{2}$ :lol:

:surrender:

I figured it was something obvious like that.   :P


Let  $ABCD$ a parallellogram with  $AB>AD$ .Take points $L,K$ at $AB,AD$  respectively such that $LB=KD$ .Let  $P$ the point of intersection of the lines $BK,DL$
Prove that the parallel from $P$ to the bisector of the angle $DAB$  passes through $C$ .

Call $l$ the line through $P$ and parallel to the bissector of $\angle A$

Draw perpendiculars from $P$ to $BC$ and $DC$ and call them $H$ and $I$ respectively.
\[ PH=PB.\sin{PBH}=\frac{\sin{PLB}.\sin{PBH}.BL}{\sin{LPB}} \]
\[ PI=PD.\sin{PDI}=\frac{\sin{DKP}.\sin{PDI}.DK}{\sin{KPD}}=PH \]
So $P$ lies on the angle bissector of $\angle C$. Therefor, $PC = l$
So $C$ lies on $l$.


function
complex analysis
complex analysis unsolved

Obtain the Taylor series $ e^z\equal{}e \cdot \sum_{n\equal{}0}^{\infty} \frac{(z\minus{}1)^n}{n!}$ ($ |z \minus{} 1 | < \infty$) for the function $ f(z)\equal{}e^z$ by using $ f^{(n)}(1)$ where $ (n\equal{}1, 2, \ldots)$.

I do not see how to do this.  I was able to prove that by writing $ e^z\equal{}e^{z\minus{}1}e$ but I don't see how to do it that way.  Thanks in advance for help.

The Taylor series around a point $ a$ is defined as:

$ f(z) \equal{} \sum ^\infty _ {n \equal{} 0} \frac{f^{(n)}(a)}{n!}(z\minus{}a)^n$

Therefor, if you have $ a \equal{} 1$ then from $ f^{(n)}(a)\equal{}\left(e^z\right)^{(n)}\Big|_{z\equal{}a} \equal{} e^z|_{z\equal{}a}\equal{}e^a$ you get:

$ e^z \equal{}  \sum ^\infty _ {n \equal{} 0} \frac{e}{n!}(z\minus{}1)^n$


How many odd numbers between $ 100$ and $ 999$ have distinct digits?

Four cases here:

1. First digit odd, second odd. There are three choices for the final digit, yielding 5*4*3=60 possiblities.
2. First digit odd, second even. There are four choices for the final digit, yielding 5*5*4=100 possiblities.
3. First digit even, second odd. Same as Case 2. 100 possibilites.
4. First digit even, second even. There are 5 choices for the final digit, yielding 5*4*5=100 possibilities.

$ 60 \plus{} 100 \plus{}100 \plus{} 100 \equal{} \boxed{360}$

No.

Units Digit: 1, 3, 5, 7, or 9
Hundreds Digit: (1 to 9) - 1
Tens Digit: (0 to 10) - 2

So we have $8\times8\times5 = \boxed{320}$


geometry
combinatorial geometry
combinatorics unsolved
combinatorics

Given $ n$ points in the plane. Prove that there exist three points of them which form an angle at most $ \frac{\pi }{n}$

If three of the given points are collinear the desired result is trivial, thus we assume that no three points are collinear.
The convex hull of the $ n$ points is a $ c$-gon, with $ 3 \leq c \leq n$.

The sum of the interior angles of the convex hull is $ (c\minus{}2)\pi$, thus one of them, say the angle at the vertex $ A$, is at most $ \frac {c\minus{}2}{c} \pi$.
Now, draw each ray $ AM$ where $ M$ is any of the $ n\minus{}1$ other points. These rays divides the angle at $ A$ into $ n\minus{}2$ parts. One of them is then not greater than $ \frac {c\minus{}2}{c(n\minus{}2)} \pi$.
Now, using that $ c \leq n$, it is straighforward to verify that $ \frac {c\minus{}2}{c(n\minus{}2)} \pi \leq \frac {\pi}{n}$, hence done.

Pierre.


geometry
analytic geometry
algebra
system of equations

this problem must be pretty hard for 17+ people to view it and no one to bother a response.

A pentagon P has vertices at A=(0,0),B=(7,0), C=(13,8), D=(5,14), E=(0,14). Line L passes through the origin and divides P into two quadrilaterals with equal perimeters. What is the sum of the point F where L meets segment CD? Express your answer as a decimal to the nearest tenth.

If you make a diagram and draw in where line L goes you will divide the pentagon into two quads each of which has an area of 67 which is 1/2 of the area of the entire pentagon which is found by using the sholace method.  Using the shoelace method again on each quad yeilds the system of equations:

14x-5y=64 and -8x+13y=78  Solving this yeilds y=11.295+ and x=8.605+

therefore the sum of the coordinates is approximately 19.9

If you need more detail or help on the shoelace method let me know and I will try and add it to this solution.

Anyone else have a better or easier solution?  Assuming that I did not make any Mickey Mouse mistakes!

Mr. Boyd, I believe you are incorrect. Drawing out the pentagon, gives us side lengths of 7, 10, 10, 5 and 14. Because we are dividing the side of 10 up, we wish to divide it in a way such that one part, added to 7 and 10, equals the other part added to 5 and 14. This comes when we have FC equal 6. Thus, F is 6/10 of the way from (13,8) to (5,14), giving us (8.2, 11.6) giving for a sum of 19.8.

It would be rather awkward if that is incorrect.

JBoyd:  Thanks for joining us as a problem solver!  As you have just discovered, if you make a mistake on AoPS, you can be pretty sure someone will spot it.  [This one only took ten minutes.]  I think it's good that students check other members' work, as we all make plenty of mistakes.

Your post seems to have gotten this thread on the move.  And I think your post would make a great followup question.

Attached is a picture, drawn to scale for solving the original problem, but certainly suitable for solving the followup problem.

Are there any other original or unique solutions to the original or the followup?

My mistake,  I mesread the problem.  I did the problem as equal areas and not equal perimeters.

thanks for catching my mistake!!!