ARML geometry AMC USA(J)MO USAMO search Support

Just curious, if anyone out there has a strong opinion which high schools offer the best program for excelling in competitive math?

Things I'm thinking of are an established math club program, participation in AMC and ARML contests, great selection of advanced math courses, magnet school, etc. (Dream school would use AoPS text books!!!)

I've heard great things for example about University High in Irvine, CA. I'm sure there are a lot of other excellent schools though.

There are at least four separate categories here:

1. Boarding schools that recruit nationally. The most famous one in this category is Phillips Exeter, which is New Hampshire. There are others.

2. State level (or county-level, or metropolitan area-level) magnet schools with a math/science emphasis (and presumably some level of competitive admissions). The most famous one in this category is probably "TJ" or to spell it out, "Thomas Jefferson HS for Science & Technology" in Virginia. Others in this category would include AAST in New Jersey, IMSA in Illinois, and there are others, in perhaps a third or half of all states. (Would Stuyvesant HS in NYC count? I'm not from there so I don't know.)

3. Private schools but not the ones in category 1. (Of course, there are lots of private schools. We're only talking about a few of them.)

4. Public high schools that draw a population which for whatever reasons (often related to the demographics and common employers of a particular region) happens to be particularly interested in this kind of mathematics. The one you mention - Irvine University, which happens to be in the area of the ARML team I coach, so I know quite a bit about it - falls into this category. The largest concentration of such schools in one small area is probably the "Silicon Valley" area in California (mostly Santa Clara County with some Alameda County - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=267153]this topic[/url] for more on that subject. If you checks lists of things like USAMO qualifiers or ARML winners in years past, you'll find high schools like this in various places: North Carolina, Alabama, and so on.

The list in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199100]this topic[/url], which will probably be updated in a few days to include 2009 data, shows where USAMO qualifiers largely come from. You can see each of the four categories above listed.

I think you'll find that a pretty healthy proportion of AoPS poster attend one of the schools from which multiple USAMO qualifiers come.

[quote="crazypianist1116"]Related to the question, what makes a good "competition math" school? Do they actually offer courses designed for competition math? I know in my school, everything is done mainly after school.[/quote]

I doubt you would fine direct courses offered as a school elective. I think a math culture goes along way to answer your first question. That includes having a great coach, an interested pool of math students, and supportive parents. Obviously, prior success in math tournaments helps there as well. I know some schools are able to invite and get former alumni to come back and teach the current crop of competitors.

[quote="Kent Merryfield"]2. State level (or county-level, or metropolitan area-level) magnet schools with a math/science emphasis (and presumably some level of competitive admissions). The most famous one in this category is probably "TJ" or to spell it out, "Thomas Jefferson HS for Science & Technology" in Virginia. Others in this category would include AAST in New Jersey, IMSA in Illinois, and there are others, in perhaps a third or half of all states. (Would Stuyvesant HS in NYC count? I'm not from there so I don't know.)[/quote]  Well, students in NYC can apply to any high school in NYC, so with respect to the pool of possible students Stuyvesant is no different from any other NYC high school.  But the application process at Stuyvesant consists entirely of a competitive entrance exam, and in that sense it surely belongs where you placed it.
A related question to which I don't know the answer, and which I'm not going to look up because I'm lazy: for several decades, Bronx High School of Science was probably the top public high school in the country from the math-sci perspective.  What was its admissions process like then?  Now it uses the same exam that Stuyvesant does, but has a lower cut-off score for acceptance.  That, together with a lot of unfortunate behavior by the administration of the school, seems to have made a serious negative dent in its performance in, say, math competitions.

As for crazypianist's question: try a search of the forum -- there have been several conversations about this in the past.  At Stuyvesant, math team is a class that appears on your schedule like any other (though it's graded pass-fail, or at least was when I was a student) and total enrollment is in the triple digits out of ~3000 students in the school.

I would have to disagree with Kent Merryfield on his 4th point.
The OP asked about which high schools offer the best program for excelling in competitive math. This is very different from producing more people who do math. I do not believe that the schooling environment, math programs, etc. of many public schools that do tend to have a large number of math competition participants are actually responsible for their achievements in contests.

Simply said, public schools with lots of math people are not that way because of the school itself. They are just normal high schools in areas with more people interested in competitive math. Thus it is the region that produces the people, not the schools.

Take Gunn (where I attend) for example. I would argue that attending Gunn would be detrimental to competition problem solving ability if it were not for the few people there that are interested in math. The math courses here are pretty much your typical high school math courses, even if names like "Analysis Honors" and "Trig/Analyt Honors" are tacked on. The math club here has around 10 people who show up consistently. Coming to Gunn from most high schools would bring one to a larger group of people who are interested in math, but the school itself has nothing to do with it.

I am not too sure about other high schools, but it seems like Gunn is a lot less laid back than most other schools around here and the homework load is possibly greater on average. Even if that is not true, there is a marked difference between Gunn and some other schools, for example Paly right next door. At Paly USAMO qualifiers receive an automatic A in math class so I hear, or at least are exempt from homework. There is also the saying that Paly teachers grade more leniently in general, although I don't know how true that is.

As a Gunn student the feeling I get is that I am penalized for representing the school in some contest that cuts into my personal time, and that there is little support from the school for any academic contests. There is very little awareness of math contests. Our club advisor is the only teacher on campus who is interested in contest math, and I highly doubt any other teacher would be willing to run math club and administer the contests.

[quote="JBL"]As for crazypianist's question: try a search of the forum -- there have been several conversations about this in the past.  At Stuyvesant, math team is a class that appears on your schedule like any other (though it's graded pass-fail, or at least was when I was a student) and total enrollment is in the triple digits out of ~3000 students in the school.[/quote]

Very interesting. Perhaps with all the success that Stuy has had, it's a great idea for other schools to borrow.

Right, but where's the money??

The Massachusetts House of Representatives wants to cut a small math-science school from the budget altogether. If the state Senate and Governor support the cut, the school will have to immediately shut down. Apparently the state of Massachusetts doesn't think spending 2 million on a math-science school is a good investment. They'd rather give it to the old people.

It was a great school, students could take a full load of college courses at WPI in their senior year, among many other things. It was small (~ 90 students), and the budget was under 2 million/year.

Of course, I've already drafted several letters to send out at these people, and we have a website and stuff. Watch this boil down to picketing, which I [i]will[/i] do. It was a great place, not to mention that I don't want to be an alumnus of a school that doesn't exist.

I think the idea of math team as a class is certainly borrowable and exportable, and "regularization" probably will increase interest in it.  The likelyhood (and even desirability) of getting such a huge student response in most places is quite small, of course :).  And it's worth pointing out that at least half the math team consists of freshmen, i.e., there's a lot of yearly attrition, a lot of students taking math team only because there are very few electives available to freshmen, and they've been told that it's something that will get them into college.

[quote="TZF"]They'd rather give it to the old people.[/quote]  So, this is probably not worth doing, but I'm going to be an obnoxious nit-picker.  The state of Massachusetts is currently working with a defecit in the billions of dollars (and I don't know whether that counts the hundreds of millions or billions of debt in the various transportation agencies -- it almost certainly doesn't count the MBTA, and I don't know about Turnpike and Highways).  The two million dollars isn't going to be "given" to anyone, it just doesn't exist.  So, the option isn't "take 2 million away from math education and give it to someone else," it's "take 2 million away from math education, or take 2 million away from someone else, or raise taxes by 2 million."

serialk11r makes a good point.  There are two very different ways that a high school can be a nurturing environment for math competition success.  One way is to have a good peer group.  The other is to have good support from the establishment.  It's entirely possible to have one without the other, although in many cases, these two factors feed into each other.

For example, even though Gunn may not offer much support for math competition success (at least by serialk11r's standards), it's undoubtedly a great environment for excelling at math competitions just because of the peer group.  On the opposite end of the spectrum, contrary to what JBL says, it's possible to get wide math competition participation even at not-so-hot schools.  Where I grew up, there was a pretty-good-but-not-extraordinary suburban high school with an enormous math team.  I think that essentially everyone who took honors math was on the team.  I believe that the teachers achieved this mainly by making math team sort of a fun social club that happened to go to math competitions (sort of like how some schools have huge bands in which the playing of music seems almost secondary).  There was probably extra credit involved, too.  In contrast, at my own high school (which I will contend had better math students) we literally could not find a teacher who was willing to take us to local competitions.  It was an embarrassment.  (Ah, memory lane...)

Also, I'm not sure how important money is to the general discussion.  Teachers don't get paid for running math clubs, so it just has a lot to do with the dedication of the teachers.  And I could be wrong about this, but even offering elective math classes probably does not cost a school anything "extra" as long as the classes are fully enrolled.  The money just goes to pay for X teachers in X classroom teaching Y students.  TZF is talking about funding for some sort of specialized math/science school, which is only tangentially related to the main topic.

[quote="yenlee"] it's possible to get wide math competition participation even at not-so-hot schools.  [/quote]
It's possible. I know some examples from my area. But ...

[quote] ... so it just has a lot to do with the dedication of the teachers. [/quote]
I know several long-established places in which the competition program is in the hands of a teacher now within 5 years of retirement. My confidence that the program in question will survive the retirement of that teacher is not all that high in some cases. There are two factors.

1. The principal probably won't step in to find somebody to take it over. This is generally pretty low on the priorities of principals - and the principals themselves often have shorter terms (and perhaps shorter horizons) than the long-time teachers I'm talking about.

2. Even if a principal did point a finger at a teacher and say, "You - you're in change of the math competition program," it might not work. Someone who is doing it as an assignment may never fully "buy in" to the program, and it might wither anyway.

[quote="yenlee"]On the opposite end of the spectrum, contrary to what JBL says, it's possible to get wide math competition participation even at not-so-hot schools.[/quote]  I don't think this is contrary to what I said -- I was speaking relative to the size of the Stuyvesant math team, which is (as I mentioned) in the triple digits.  I also agree with what Kent just said.

Lexington High School in Massachusetts is a public high school that consistently does well on math exams.  My sister's kids go there and they do have a strong math program, because her kids get math homework way above and beyond what my kids get.
However, the kids who make it the the AIME and beyond do way more than the school.  One kid has parents who make him do 4 hours of extra math after school.  Another kid who scores well is apparently a genius, so I don't know how much extra math he does.
You could try to find out how much the school provides for the kids, too.
There is alot of motivation on the part of the families, but obviously there's affirmation from the school.

I don't think many of the strong math students are parent-motivated, at least not primarily - my parents, at least, have left it all up to me. Everybody on this forum is already self-motivated to some degree.

[quote="serialk11r"]Take Gunn (where I attend) for example. I would argue that attending Gunn would be detrimental to competition problem solving ability if it were not for the [b]few[/b] people there that are interested in math. ... The math club here has around 10 people who show up consistently. [/quote]
serialk11r goes on to draw some comparisons with "Paly" where he implies the math contest people get better treatment.

I wouldn't know anything about either school, since I don't live in that metropolitan area. But for some perspective on what "few" means:

Palo Alto Gunn HS had more than 40 AIME qualifiers. (The exact number is a little hard to pin down because of students taking AMC exams at multiple locations.) Of those, 8 qualified for the USAMO.

As a non-resident, I'm not up on the slang, but my best guess as to the meaning of "Paly" is Palo Alto HS. Palo Alto almost exactly the same number of AIME qualifiers (a little over 40) and 6 USAMO qualifiers.

And Fremont Mission San Jose HS had almost 40 AIME qualifiers and 4 USAMO qualifiers.

Saratoga HS had only about 20 AIME qualifiers, but 9 of them were USAMO qualifiers.

And to top things off, San Jose Lynbrook had nearly 60 AIME qualifiers and 6 USAMO qualifiers, with another 15 or more AIME qualifiers and another 2 USAMO qualifiers from Miller MS, which has the same ZIP code as Lynbrook.

And there are several other schools in the region with large numbers of AIME qualifiers and several USAMO qualifiers. All told, there were 65 USAMO qualifiers from the San Francisco Bay Area, and more than 50 of them come from a relatively small arc around the south end of the bay, from Palo Alto to Fremont.

What I was intending to say was the number of people interested in math is not particularly large, it is just that there is a small group of people that scores quite well. Out of the 38 AIME qualifiers (I am pretty sure of this number) only around 25 showed up to the AIME, and there are around 12 people who attend math contests besides the AMC, which is mandatory for "higher lane" math students.
I don't think 12 people is an overwhelmingly greater number of people interested in math than most schools with some number of USAMO qualifiers. Our ARML team round participation hovers at around 6 people average (although Lynbrook has around 30, but they are a somewhat special case). The number of people qualifying for USAMO does not have a direct correlation to the number of people interested in math. At school math definitely still carries a somewhat negative connotation.
A more realistic comparison with a "special" school listed before:
At Stanford Math Tournament, Gunn brought 1 full team of 8 people, as did many other Santa Clara County schools. AAST from New Jersey flew over here to participate, alone bringing 5 full teams of 8.

Sounds like Silicon Valley reason has a bunch of power houses. Nice to read that the east coast has some strong contenders also. I think its great for parents to know about these schools and they should be celebrated.

Parents can help push their kids, but they can only push so far. Math coaches too. The kids that I've seen who are the best, take their parent's encouragement and run with it. That they are brilliant is a given. These kids are highly motivated and they know its about hard work.

But a lot too is having math clubs/circles and even knowing outlets such as AMC and ARML exist. There are so many schools that could be participating and do not. Maybe less so because sites like this, but still way too many.

[quote="Kent Merryfield"]Palo Alto Gunn HS had more than 40 AIME qualifiers. (The exact number is a little hard to pin down because of students taking AMC exams at multiple locations.) Of those, 8 qualified for the USAMO.

As a non-resident, I'm not up on the slang, but my best guess as to the meaning of "Paly" is Palo Alto HS. Palo Alto almost exactly the same number of AIME qualifiers (a little over 40) and 6 USAMO qualifiers.

And Fremont Mission San Jose HS had almost 40 AIME qualifiers and 4 USAMO qualifiers.

Saratoga HS had only about 20 AIME qualifiers, but 9 of them were USAMO qualifiers.

And to top things off, San Jose Lynbrook had nearly 60 AIME qualifiers and 6 USAMO qualifiers, with another 15 or more AIME qualifiers and another 2 USAMO qualifiers from Miller MS, which has the same ZIP code as Lynbrook.

And there are several other schools in the region with large numbers of AIME qualifiers and several USAMO qualifiers. All told, there were 65 USAMO qualifiers from the San Francisco Bay Area, and more than 50 of them come from a relatively small arc around the south end of the bay, from Palo Alto to Fremont.[/quote]Wow. I was the only AIME qualifier in my highschool and we only had 4 in the entire district. No one made the USAMO.

I should add a quick correction to something I said earlier: apparently math team at Stuyvesant is no longer graded pass-fail, i.e., it does not appear on your transcript.  It still counts as a class for scheduling purposes, though.

I always knew that silicon valley was strong, but seeing those numbers laid out by Kent is just crazy.  I'm not sure how typical this year was, but consider this:

Between the two high schools in Palo Alto (which are both non-selective neighborhood schools), there were more USAMO qualifiers than at Stuyvesant or in the entire state of Florida, the 4th most populous state.

In Santa Clara county alone (which includes Palo Alto, San Jose, Saratoga, and Cupertino, but not Fremont, total population 1.8 million in 2000), the number of USAMO qualifiers is larger than that of any state other than California.  The number also beats Virginia, Maryland, and DC combined.  It also seems comparable to the entire NYC metro area (which is hard to count), and so it's probably larger than every other metropolitan area of the country.  (My SoCal geography isn't good enough to easily figure out LA/OC metro area.)

I wish I lived there :(

[quote="yenlee"] (My SoCal geography isn't good enough to easily figure out LA/OC metro area.)[/quote]
Since I'm an ARML coach, I tend to think in terms of our team's region, which does not include San Diego; for the purposes of dealing with your comment it's probably best that I do include San Diego.

There were 43 USAMO qaulifiers in Southern California. By counties:

19 in Orange County
13 in San Diego County
9 in Los Angeles County
1 in Ventura County
1 in Santa Barbara County

(I'm using some information I know that wouldn't be apparent from the list downloaded from the AMC and am assigning some students to the county in which they live and attend school rather than the non-school test center that's listed on the official list.)

There are counties with pretty substantial populations - San Bernardino, Riverside - with no USAMO qualifiers, but it goes beyond that: there are [i]no AIME qualifiers[/i] from San Bernardino County and [i]one[/i] AIME qualifier from Riverside County. And from the entire Los Angeles Unified School District, which is one of the largest school districts in the country, there are 18 AIME qualifiers - 15 from North Hollywood HS, one from a middle school that feeds North Hollywood, and two from a different magnet school. An no one from any of the vast assortment of other schools in that district. Opportunity is not evenly distributed.

The largest cluster in the above data is in the city of Irvine, with University HS having 7 USAMO qualifiers, with the other three high schools of the Irvine school district having one each and a private school in Irvine having one. But competing with that cluster is the La Jolla/Torrey Pines neighborhood of coastal north San Diego, with Torrey Pines HS by itself accounting for 6 USAMO qualifiers.

Okay perhaps "few" was an incorrect description of the number of people who do math. I still stand by what I said about the 13-14 people not being an overwhelmingly greater number.
The main point is, there are schools where you can argue going to the school makes you better at math, i.e. Exeter perhaps, due to both a stronger curriculum and a substantially larger group of peers interested in math than almost any other school (just look at how Exeter floods the Mandelbrot results).

Going to Gunn, or Paly, or Saratoga would not necessarily have that effect. The group of people interested in math is "large", but not incredibly. You'd have to really be pretty close to some of the very very strong math students to be influenced to do more math.

[quote="serialk11r"]The main point is, there are schools where you can argue going to the school makes you better at math, i.e. Exeter perhaps, due to both a stronger curriculum and a substantially larger group of peers interested in math than almost any other school (just look at how Exeter floods the Mandelbrot results).[/quote]  But why would you assume that this has anything to do with the quality of Exeter as a school, rather than with the fact that Exeter recruits nationally, giving it the massive advantage of any cherry-picker?

(Side note: you've mis-used the abreviation "i.e." in the quoted sentence: "i.e." means "[i]id est[/i]," or "that is."  You wanted "e.g.," which means "for example.")

[quote="serialk11r"]I am pretty sure there is a big difference in the quality of education you would get at a prestigious private school like Exeter and the education you would get at a normal public school. The reason why they cherry pick, why they are ABLE to cherry pick, is because of that.[/quote]

I don't think that is a safe conclusion to make. If you notice, Phillips Exeter had [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=272133]virtually no USAMO qualifiers before 1996[/url], and only recently have they had an outstanding number of them. This increase is due in large part to Zuming's arrival there in 1995 and his recruiting of strong math students rather than to any significant changes in the way they teach math there. I'm not saying that the quality of teaching there isn't very good. In fact, I'm sure that it is. I'm just saying that Zuming's presense (and the offering of financial scholarships) is what has been attracting elite math students to Phillips Exeter recently.

To me, Thomas Jefferson and Montgomery Blair (and their feeder middle schools) are the most impressive schools because they draw from fairly small regions (population wise), yet they have consistently produced exceptional math students (far more than their population demographics suggest they should) over a large number of years.

[quote="serialk11r"]The problem is, number of USAMO qualifiers does not indicate how much their students improve. If Exeter draws elite math students, that is one thing. If Zuming can coach them well so that they can win everything, that is another.

I don't know how this whole magnet school thing works, since apparently we don't have those around here, but I am assuming that because they are able to selectively draw students, the quality of education at those schools must be different.[/quote]

Wait, is Harker a "magnet" school?

In the classification scheme in my post number 2, I think Harker belongs in category 3: private school, doesn't recruit nationally but does draw students from a good distance across a metropolitan area. Harvard-Westlake in my area would answer to this description; Sidwell Friends in D.C. is one I've heard of. I'm sure people who know could supply examples from all over.

I don't mean to flog a dead horse here, but I find it difficult to believe that Gunn is not in the top 5% of high schools nationally in terms of how *nurturing* it is for math students.  I say this *taking into account* the fact that the students who go there are *already* smart.  So seriously, stop whining that your high school is not as good as Exeter.  (For starters, Exeter is just a wee bit more expensive than Gunn.)  Maybe you'd be equally good at math if you lived in Riverside County, but if that were true (and I'm not saying it is or isn't), you'd probably be equally good if you went to Exeter.  Quality of school is either important to you or it's not, and if it is, you should thank your lucky stars (actually, your parents) that you live in Palo Alto.  While you're at it, thank them for all the other opportunities they've given you.

@gauss202:  There's no doubt that Thomas Jefferson and Montgomery Blair are great schools.  (No love for Stuy?)  But *small* population areas?  According to Wikipedia, in 2007 Fairfax County had a population of 1,077,000, and Montgomery County had a population of 931,000.  These are actually pretty large talent pools as far as suburban school districts go.  In contrast, the city of Palo Alto, with its 14 USAMO qualifiers, had a population of 61,000.

I'm not complaining that Gunn isn't as good as Exeter, I'm only pointing it out. I'm also pointing out that the fact that we live in Palo Alto has more to do with the good scores than the fact that our school is good.

I think in assessing the "quality" of a school's math program there are several factors that have to be taken into account. One question is, are we talking about the quality of the program for upper level kids, mid level kids, or low level kids. A school may have a program that does a good job with one set of kids, but not another. I assume we are talking about the quality of the program for its "upper" level kids (however that is defined).

Second question is, how do we define the "quality" of a program and how do we measure it? Is it based on national math contest results that test problem solving skills and base it on numbers like the number of AIME qualifiers or USAMO qualifiers? Or could it be based on the quality of the teaching of "fundamental skills", like standardized test or AP test performace? Or could you base it on something like the number or percentage of students who end up majoring in math related fields when they go to college?

The third question, which in my mind is much more important, is how much is the school itself responsible for producing the quality of the math students in relation to the quality of kids that they have coming into the school. Does the school inherit students who were already strong in math before they even get there - like from a really strong middle school or Mathcounts program? Is the school located in a demographic area where the students have a lot of access to opportunities that have nothing to do with the school (like math circles, strong peer groups from their neighborhood, parents who are mathematically trained, University classes, etc)? These are factors that are hard to really account for, but should be considered.

I still think that Thomas Jefferson High School does an outstanding job. They have excellent feeder middle schools, and are a selective school in a populous areas that is one of the wealthiest areas in the country. So they are getting a large number of students coming into the school from the outset who are already ahead of the curve. But, with all of those advantages, what is impressive to me is not just the "numbers", it is how their students show a dramatic improvement over the course of their 4 years there, and how it happens year after year. Also the fact that their math program is not dependent on the strength of just one single person as it is at so many other schools.

I think if you were to close down one single school, the one that would have the greatest impact on the quality of math education, both locally and nationally would be Thomas Jefferson.


Find the biggest $n$ that divides $a^{25}-a$ for all $a$.

[hide="Well..."] We consider the set of primes $p$ such that $\varphi(p) | 24$.  (We do not consider the prime powers because $p^{25}-p$ cannot be divisible by any powers of $p$.)

Firstly, the divisors of $24$:

$1, 2, 3, 4, 6, 8, 12, 24$

This means we have

$p = 2, 3, 5, 7, 13$

Our maximal $n$ is therefore

$2 \cdot 3 \cdot 5 \cdot 7 \cdot 13 = \boxed{ 2730 }$.

(I don't really want to prove the converse of the Totient Theorem - who wants to do it for me?) [/hide]


inequalities
logarithms
inequalities proposed

Prove that:\[ x^x\ge e^{x\minus{}1}\quad \forall x>1
\]

$ y\equal{}x\ln x$ is concave up and increases monotonously.
Consider the tangent line of $ y\equal{}x\ln x$ at $ x\equal{}1$, yielding $ y\equal{}x\minus{}1$.
$ \therefore x\ln x>x\minus{}1$.

$ f(x)\equal{} xlnx\minus{}x$ we see $ f'(x)>0$ for all x>1 so $ f(x)>f(1)\equal{}\minus{}1$ for all $ x>1$ from this it is easy to get the solution :P


Family                               Thomas         Shawn       David       Sui            Lisa

Number of Consecutive              10              5              8          6                8
Nights

The table above shows the number of consecutive nights that each of five families stayed at a certain hotel during a 14-night period. If the Sui family’s stay did not overlap with the Lisa family’s stay, which of the
14 nights could be a night on which only one of the five families stayed at the hotel?

(A) The 3rd  (B) The 5th  (C) The 6th  (D) The 8th  (E) The 10th

sorry i dont really have a solution but im pretty sure the answer is
(A) the 3rd night


A ant starts on vertex A of equilateral traingle ABC. If the probability of the ant going to the other two vertices from one vertex is the same, what is the probability the ant ends up at vertex A after 5 moves?

i would have thought that on move 5 the ant could either be on vertex A, B or C if there is an equal chance of being on any one of these then $ P=\frac13 $

Very good intuitive step. However, the probability is not 1/3, but you are right about it being around 1/3.

I smell a recursion.  ;)

do you know that the Ant WILL move each time...?

i think the answer is :
$\frac{2}{9}$ ;)

It's not 2/9.

What is the relationship between the probability the ant returns to A after a certain amount of moves and the probability the ant doesn't reach A on those number of moves?

Try a few examples, this may become clearer. Recursive sequences can be tricky. It's always been difficult for me at least.

Alot closer. Now take that recursion one more time! And you will have the answer.

$\frac{5}{16}$

you could actually draw a tree diagram to do this.

In case people are wondering, let P(n) denote the probability of being on A after the nth move, and let X(n) denote the probability of not being on A.  X(n) = 1/2*X(n-1) + P(n-1), and P(n) = 1/2*X(n-1).  Well.... *cough*substitute*cough*

[hide="The answer is..."]
$\frac{5}{16}$
[/hide]

oops didnt see my mistake... edited up there.


geometry
3D geometry
calculus
integration
number theory
relatively prime

Prove that the only integral solutions of m^3 = n^2 + 2 are (m,n) = (3, 5) and (3, -5).

[hide]
We can use algebraic number theory. we can factor b^2+2 as (b+i\sqrt{2})(b-i\sqrt{2}). It is easily confirmed that these two are relatively prime, so b+i\sqrt{2} is a perfect cube. we have b+i\sqrt{2} = (x+yi\sqrt{2})^3. Expanding and equating the imaginary parts, we get $ 1 \equal{} 3x^2y\minus{}2y^3$. We have that y | $ 3x^2y\minus{}2y^3 \equal{} 1$, implies that y = 1 or -1. We get some cases, and then we can find what b+i\sqrt{2} is by cubing.[/hide]

[quote="pythag011"][hide]
We can use algebraic number theory. we can factor b^2+2 as $ (b\plus{}i\sqrt{2})(b\minus{}i\sqrt{2})$. It is easily confirmed that these two are relatively prime, so $ b\plus{}i\sqrt{2}$ is a perfect cube. we have $ b\plus{}i\sqrt{2} \equal{} (x\plus{}yi\sqrt{2})^3$. Expanding and equating the imaginary parts, we get $ 1 \equal{} 3x^2y \minus{} 2y^3$. We have that y | $ 3x^2y \minus{} 2y^3 \equal{} 1$, implies that y = 1 or -1. We get some cases, and then we can find what $ b\plus{}i\sqrt{2}$ is by cubing.[/hide][/quote]

Added dollar signs.

Thanks. I had a long-winded way: m and n are both even (giving a contradiction) or m and n are both odd. m =2a+1, n = 2b+1.

Then a> 0, a < b.

4a^3 + 6a^2 + 3a - (2b^2 + 2b +1) = 0.

a=1 gives us (a-1)(4a^2 + 10a + 13 - 13)=0.

a = n, n>1 gives us (a-n)(4a^2 +(n+6)a +(n^2 + 6n +3) - ((n^2 + 6n +3)/n)) = 0. n = 1, a=1, b=2, m=3 n = 5 or -5.


Prove that the sum of divisors function is multiplicative.

Note:  unlike an arbitrary function, we say that an arithmetic function $ d$ is multiplicative if $ d(ab) \equal{} d(a) d(b)$ holds whenever $ a, b$ are [i]coprime[/i] positive integers.

[hide="Hint"] What does a divisor of $ ab$ look like? [/hide]

Sum of divisors of a number $ p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ is $ (1\plus{}p_1...p_1^{e_1})... (1\plus{}p_k...p_k^{e_k})$. Therefore its multiplicative if the numbers involved are coprime.

One way to prove that formula is to argue that $ \sigma(p_1^{e_1} p_2^{e_2} ...) \equal{} \sigma(p_1^{e_1}) \sigma(p_2^{e_2}) ...$, so that's somewhat circular (or at least requires citing a statement that's stronger than what we want to prove).


Solve the system of equations:
\[ x^2+y^2+x+y=8, \]
\[ xy+x+y=5. \]

Adding two times the second equation to the first we get:
$(x+y)^2+3*(x+y)-18=0$, which factors into $(x+y+6)(x+y-3)=0$, so $x+y=-6$ or $x+y=3$.  If $x+y=-6$, then the two equations become $x^2+y^2=14$, $xy=11$, which yields $x^2-2xy+y^2=(x-y)^2=14-2*11=-8$, so there is no real solution to this possibility.  If $x+y=3$ then the system becomes $xy=2$, $x^2+y^2=5$, which clearly has the solutions $(1,2), (-1,-2), (2,1), (-2,-1)$.

Edit: Woops.  Those negative solutions don't work at all.

Well .. why not give the all (including  complex) possible solutions?

[hide](eg -3  + i :sqrt: 2   and -3 - i :sqrt: 2) .. giving x+y = -6, xy= 11, x^2+y^2=14)[/hide]

[hide]Let x+y = a, xy = b.  Then the equations become

a^2 - 2b + a = 8

a + b = 5 => b = 5 - a

a^2 - 2(5-a) + a = 8

a^2 + 3a - 18 = 0

(a+6)(a-3) = 0

a = -6, 3

b = 11, 2

Since a is the sum of the roots and b is the product of the roots, the equation whose roots are x and y is z^2 - az + b = 0

z^2 + 6z + 11 = 0

OR

z^2 - 3z + 2 = 0

The first equation has no real roots.  The second one factors to (z-2)(z-1) = 0, so z = 1 or 2.  Thus, the only answers are (x,y) = (1,2) or (2,1).[/hide]


function
algebra
polynomial
calculus
calculus computations

Determine whether the following functions have a local maximum, local minimum or saddle point at
the origin:

a.)  f(x, y) = 1 - (x^2)(y^2).
b.)  f(x, y) = (x^3)(y^2).
c.)  f(x, y) = (x^2  - sin^2x)y^2.
d.)  f(x, y) = (x^2  - tan^2x)y^2.

Hello
[hide="a"]
local max.
[/hide]

Thank u.

Hello
[hide="b"]
saddle point.
[/hide]
Thank u.

:what?: A polynomial is not differentiable? That is quite a discovery! :P.

Hello
thanks Fedja .Ur comment actually told me my mistake.And right now I learnt how to solve such function .
Thank u.

hello, $ f(x,y) \equal{} 1 \minus{} x^2y^2$ can we estimate for all real $ x$ and $ y$ as follows $ 1 \minus{} x^2y^2\le 1$, so we must have a maximum in $ x\equal{}0,y\equal{}0$.
Sonnhard.


You are given a two by three array, with each cell having room for one marble.  Prove that for any arrangement of three marbles, at most one marble must be moved so that no two marbles are in adjacent cells.  (cells diagonal from each other are not adjacent)

The arrangement in which no two marbles are adjacent involves a marble at each end of one row and one in the middle of the other, like these:
1 0 1 - - - - - - - 0 1 0
0 1 0 - - - - - - - 1 0 1
These are the two sets of cells in which the marbles can be. In one there can be 0, 1, 2, or 3 marbles in the original arrangement. This means in the other set of cells there must be 3, 2, 1, or 0 marbles, respectively. Therefore there must be at least two marbles in one of the sets. Then, the other one can be moved into that set, and no two will be adjacent.


percent
algebra
difference of squares
special factorizations

Albatross and Frankinfueter are playing each other on a game over the internet. Frankinfueter has 1000 rating points. After beating Albatross once he gets plus 5% rating points. The next game Albatross beats Frankinfueter so that Frankinfueter loses 5% of his points. How many less points does Frankinfueter have now?

[hide]997.5[/hide]

I don't feel like showing my works because I have to go see Foot Ball

GO STEELER!

Thats the right stuff, but ask what the problem asks for. This is a good way to stop stupid mistakes.

P.S. Can't wait till tommorow Vikings are playing the Eagles.

[quote="A+MATH"]How many less points does Frankinfueter have now?[/quote]

Hehehe....

[hide="I do dumb things like that all the time"]2.5[/hide]

STEELERS Won Again!!

go STEELERS!!

okay..

I made a mistake...

subtract 997.5 by 1000

the answer is 2.5

Go STEELERS

Yup. Those answers are obviously right. Sheesh the vikings stunk. Oh well I my second fav would be the Patriots.

[quote="Treething"][hide="generalization"]
If a quantity is raised by x percent and lowered by x percent, the end result will always be lower than the original quantity, provided x is not 0 (it works for negative numbers).
[/hide][/quote]
To be more specific...
[hide]Call the quantity y.
When you raise it by x%, you are taking
y(100+x)/100
When you then decrease it by x%, you are taking the above quantity and multiplying it by
(100-x)/100
So the end result is
$\frac{(100+x)(100-x)}{100^2}\cdot y$
Or, by difference of squares,
$\frac{100^2-x^2}{100^2}\cdot y$

$\left(1-\frac{x^2}{100^2}\right)\cdot y$
The quantity you are multiplying y by is clearly less than 1 unless x=0, because x^2 is never negative.
[/hide]

[hide="click here to see the hidden content"]he gets 10 points

now he has 1200 points

he loses one game so the answer is 2.5[/hide]

[quote="jli"][hide="click here to see the hidden content"]he gets 10 points

now he has 1010 points

he loses one game so the answer is 2.5[/hide][/quote]

[hide="To jli"]
I don't quite get how he only gained 10 points if he gets 5 percent more...
[/hide]

[quote="jli"][quote="DarkKnight"][quote="jli"]oops... it should be 1200(fixed it)[/quote]

[hide="to jli"]
Hint: 5% ;)
[/hide][/quote]

grrr. it should be 1020. i don't like this problem[/quote]

[hide="to jli"]
Sorry that i'm borthering you but 5% of 1000 is 50 isn't it?
1000/100=10
10=1%
10*5=50
1000+50=1050 ;)
[/hide]


[hide]Writing out the series, we get:

$\log(\log 11)+\log(\log_{11}12)+\log(\log_{12}13)+...$

Combining these using log properties, we get:

$\log((\log 11)(\log_{11}12)(\log_{12}13)...)$

From the chain rule for logs, the entire series, when represented in this manner, becomes:

$\log(\log 11)(\log_{11}12)...(\log_{99}100)) = \log(\log 100)) = \log 2$.[/hide]

[hide="What hes talking about"] $\log_{a}b \log_{b}c \log_{c}d \log_{d}e ... \log_{y}z = \log_{a}z$ [/hide] [hide="The underlying principle"] Simply use the base change formula,

$\log_{a}b = \frac{\log b}{\log a}$

And the expression becomes

$\log 11 \log_{11}12 \log_{12}13 ... = \log 11 \frac{\log 12}{\log 11}\frac{\log 13}{\log 12}... = \log 100$ [/hide]


Tamika selects two different number at random from the set
$ \{8,9,10\}$ and adds them. Carlos takes two different numbers at
random from the set $ \{3,5,6\}$ and multiplies them. What is the
probability that Tamika's result is greater than Carlos' result?  Express your answer as a common fraction.

There are only two ways Tamika's result is greater than Carlos' result. There are $ \binom{3}{2} \times \binom{3}{2}$ or $ 9$ possibilities in total. Thus our answer is $ \boxed{\frac {2}{9}}$.

EDIT: Oops, vallon's solution is correct.

are you sure the answer is $ \frac {2} {9}$?

i got $ \frac {4} {9}$

tamika could add 8 and 9, resulting in a sum of 17
to be lower than tamika, carlos must multiply 3 and 5 for a product of 15 (1 choice)

tamika could add 8 and 10, resulting in a sum of 18
to be lower than tamika, carlos must multiply 3 and 5 for a product of 15 (1 choice)

tamika could add 9 and 10, resulting in a sum of 19
to be lower than tamika, carlos could multiply 3 and 5, for a product of 15, or 3 and 6, for a product of 18 (2 choices)

adding those choices up, we have 4 ways tamika gets a higher number than carlos, and since there are 9 possibilities, our answer is $ \frac {4} {9}$


How man y positive integers less than $50$ have an odd number of positive integer divisors?

$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 11$

[hide]This basically asks for the number of perfect squares les than 50, so the answer is 7.[/hide]

There are 7 squares less than 50. $\implies \boxed{C}$


trigonometry
LaTeX
function
calculus
calculus computations

Here's a two part problem that I was wondering if you could look over my work and see if I was in the ballgame, trig has never been my strong suit!

a) find $f' of y = f(x) = (tan(x)-1)/sec(x)$

using quotient rule, let $u = tan(x)-1, v = sec (x), y = u/v$
$f' of tan(x) = sec^2(x)$
$f' of sec(x) = sec x * tan x$
so, after plugging in,
$d/dx y = {(sec[x])^3 - sec(x) * (tan[x])^2 + sec(x) * tan(x)}/(sec[x])^2$

b) find $f' of y =f(x) = [cos(x)-cot(x)]/csc(x)$

using quot rule, let $u = cos(x)-cot(x), v= csc(x), y=u/v$
$f' cos x = -sin x$
$f' cot x = -csc^2 x$
$f' csc x = -csc(x)* cot(x)$
after plugging in,

$d/dy = [(csc{x})^3 - (cot(x))^2 * csc(x) + cos(x) * cot(x) * csc(x) - csc(x) * sin(x)] / (csc[x])^2$ ????

Thanks again!!!!

Notation: Use more words, and don't put those words inside $\LaTeX$. Mathematics has grammar, and you shouldn't be so careless about defining what you use. Also, if you put a \ before each trig function, it looks better. The \frac code is also very useful for clarifying expressions.

(a) rephrased: Let $f(x)=\dfrac{\tan x-1}{\sec x}$. Find $f'(x)$.

The best first step is algebra: $\dfrac{\tan x-1}{\sec x}=\cos x\cdot(\dfrac{\sin x}{\cos x}-1)=\sin x-\cos x$.

(b) rephrased: Let $f(x)=\dfrac{\cos x-\cot x}{\csc x}$. Find $f'(x)$.

Again, start with algebra: $\dfrac{\cos x-\cot x}{\csc x}=\sin x\cdot\left(\cos x-\dfrac{\cos x}{\sin x}\right)=\sin x\cdot\cos x-\cos x$

In both cases, the algebra makes things easy.

Your first answer is missing a pair of parentheses, but is otherwise correct. The second is correct. Both could be greatly simplified.


I want to change the email adress of my profile. When edited my profile (I tried to change the email adress) there was still the 'old' email adress under my username (in the profile section), even when I activited it.

So what to do ?

If you cannot receive mail from this site at your new email address, first check if you have whitelisted our domain (webmaster@artofproblemsolving.com or webmaster@mathlinks.ro depending on which url you use). And then try again, checking your bulk folder for the activation email of your new email address.

Okay, is it possible (for somebody) to delete your username ?

We do not delete accounts.  This would make a mess of all the threads users have posted in, etc.

[quote="Mohamed Assalafie"]I don't understand, deleting an username => deleting all the posts of the username ?[/quote]If you want we can deactivate your account. But that's the most we can do.

[quote="Valentin Vornicu"]If you want we can deactivate your account. But that's the most we can do.[/quote]

I understand, thanks.

[quote="fedja"]I also tried to change my e-mail address and had the same problem: the address listed in the profile is still the old one. It doesn't bother me at all but it is strange.[/quote]I'm wondering why nobody reads the messages when changing an email address. You need to confirm the new email address with an activation link that is sent to the new address. If you don't click on that link there will be no change.

[quote]Profile updated. Please check your new email address for an email containing the activation link.
Without it, you cannot activate your new email address.
Give it up to an hour to arrive, and also check your SPAM/Bulk folder.[/quote]

Not quite, Valentin, not quite :) I read it, went to that link I received, got the message like "your e-mail address has been activated" or something like that (I do not remember the exact words now) and still saw no change after that.

Edit: Just to make it completely sure, I went there again. The exact message is
"Your new e-mail address has been already activated". I vaguely remember that this word "already" was there even the first time and made me suspect that something was wrong but I cannot give you a 100% warranty on that now :(.

You guys were right. There was a small typo preventing the activation. It's now fixed :)


probability
superior algebra
superior algebra unsolved

Let $\mathbb{K}=\mathbb{F}_{2^{n}}$ (the field with $2^{n}$ elements). We consider the equation (1):$x^{2}+ax+b=0$ where $a\in\mathbb{K}^{*}$ and $b\in\mathbb{K}$ are randomly chosen (the distribution is uniform in $\mathbb{K}^{*}$ or in $\mathbb{K}$).
  What is the probability so that (1) has 2 distinct roots in $\mathbb{K}$ ?

Some hints :

- You cannot have equal roots since $a\neq 0$
- You can reduce this equation to get something like $y^{2}+y+c=0$
- an equation of the latter form has a solutions (and thus two distinct solutions if and only if the "trace" equals zero)

Different hint: equations having solutions correspond to their sets of solutions. owk


Show and prove that in

$ ax^2 \plus{} bx \plus{} c \equal{} 0$

$ x\equal{}\frac{\minus{}b\pm\sqrt{b^2 \minus{} 4ac}}{2a}$

[hide="Proof"]Dividing through by $ a$, we get:
$ x^2 \plus{} \frac {b}{a}x \plus{} \frac {c}{a} \equal{} 0$
Rearranging:
$ x^2 \plus{} \frac {b}{a}x \equal{} \minus{} \frac {c}{a}$
Completing the square:
$ x^2 \plus{} \frac {b}{a}x \plus{} \frac {b^{2}}{4a^{2}} \equal{} \minus{} \frac {c}{a} \plus{} \frac {b^{2}}{4a^{2}}$
Factoring and using common denominators for the RHS:
$ (x \plus{} \frac {b}{2a})^{2} \equal{} \minus{} \frac {c}{a} \plus{} \frac {b^{2}}{4a^{2}} \equal{} \frac {b^{2} \minus{} 4ac}{4a^{2}}$
Taking the square root of both sides:
$ x \plus{} \frac {b}{2a} \equal{} \frac {\pm \sqrt {b^{2} \minus{} 4ac}}{2a}$
Rearranging:

$ \boxed{x \equal{} \frac { \minus{} b\pm\sqrt {b^2 \minus{} 4ac}}{2a}}$[/hide]

This is essentially completing the square. You may want to use the search mechanism sir.

I myself have posted this very same topic!


equation
algebra
square roots
IMO
IMO 1959

Since $\sqrt{2x-1}$ is assumed to be real, then $x \geq \frac{1}{2}$. Let
\[ \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A  \]
Squaring both sides we obtain
\[ (1) 2x+2\sqrt{x^2-(2x-1)}=A^2  \]
Since $x^2-(2x-1)=(x-1)^2$, we can write $(1)$ in the equivalent form
\[ (2) x+|x-1|=\frac{A^2}{2}  \]
$(a)$ suppose $A=\sqrt{2}$; then $\frac{A^2}{2}=1$, and we look for solutions $x \geq \frac{1}{2}$ of
\[ (3) x+|x-1|=1  \]
For $\frac{1}{2} \leq x \leq 1, x-1 \leq 0$, so $|x-1|=x-1$ and (3) becomes
\[ x+x-1=2x-1=1  \]
This equation is satisfied only if $x=1$, which contradicts the assumption that $x>1$. We conclude that $A=\sqrt{2}$ if and only if $\frac{1}{2} \leq x \leq 1$. In view of this result, we can confine attention to the range $x>1$ in solving part $(b)$ and $(c)$. Then $x+|x-1|=2x-1$.
$(b)$ If $A=1$, then $2x-1=\frac{A^2}{2}=\frac{1}{2}$, so $x=\frac{3}{4}$. Since this is not in the range $x>1$, there are no solutions for $A=1$.
$(c)$ If $A=2$, then $2x-1=\frac{A^2}{2}=2$, so $x=\frac{3}{2}$

For IMO's problem, this is very easy, I think. :)

For $a>b>0,$ use $\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b},\ \sqrt{a+b-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}$

On the condition that $2x-1\geq 0\Longleftrightarrow x\geq \frac{1}{2},$ remark that $(2x-1)+1=2x,\ (2x-1)\cdot 1=2x-1,$

we have $A=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{\frac{2x+\sqrt{2x-1}}{2}}+\sqrt{\frac{2x-\sqrt{2x-1}}{2}}$

$=\frac{\sqrt{2x-1}+1}{\sqrt{2}}+\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}=\begin{cases} \sqrt{2} & (\frac{1}{2}\leq x\leq 1) \\ \sqrt{2}\sqrt{2x-1} & (x\geq 1) \end{cases}$

Therefore sketching the graph of $A=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}$ on $x-A$ plane gives the desired answers immeadiately, that is to say, find the absissae of the intersection points for the graph of $A=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}$ and $A=the\ given\ constant.$
Therefore the desired answers are for $(a)\ \frac{1}{2}\leq x\leq 1,$ for $(b)$ there are no solutions and for $(c)\ x=\frac{3}{2}.$

kunny

I wish we could solve contemporary problems as easily as we do archaic ones. :D

So all we did is square both sides to make the left side a little friendlier?

I dont understand the different conclusions of the previous posts :oops:

Using DPopov path (adding obviously $A\ge 0$), we get that the equation is equivalent to $x\ge\frac 12$ and $x+|x-1|=\frac {A^2}2$

Which is equivalent to : ($x> 1$ and $2x-1=\frac {A^2}2$) or ($1\ge x\ge \frac 12$ and $A^2=2$)

$\iff$ ($x> 1$ and $x=\frac {A^2+2}4$) or ($1\ge x\ge \frac 12$ and $A^2=2$)

[u][b]And so [/b][/u]:
If $A>\sqrt 2$ : one solution $x=\frac {A^2+2}4$

If $A=\sqrt 2$ : infinitely many solutions : all $x\in[\frac 12,1]$

If $A<\sqrt 2$ : no solution.

[quote=DPopov]For $\frac{1}{2} \leq x \leq 1, x-1 \leq 0$, so $|x-1|=x-1$ and (3) becomes
\[ x+x-1=2x-1=1  \][/quote]

Ummm....this is a mistake, isn't it?

Although I am not good at English reading,I think the describtion is well.The question is only solving three equations,it is not worth to solve all As.

[quote=DPopov]For what real values of $x$ is \[ \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A  \]  given

a) $A=\sqrt{2}$;

b) $A=1$;

c) $A=2$,

where only non-negative real numbers are admitted for square roots?[/quote]

 Multiply both sides by $\sqrt{2}$, we get $\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=A\sqrt{2}$

$\Leftrightarrow \sqrt{(2x-1)+2\sqrt{2x-1}+1}+\sqrt{(2x-1)-2\sqrt{2x-1}+1}=A\sqrt{2}$

$\Leftrightarrow \sqrt{(\sqrt{2x-1}+1)^2}+\sqrt{(\sqrt{2x-1}-1)^2}=A\sqrt{2}$

$\Leftrightarrow |\sqrt{2x-1}+1|+|\sqrt{2x-1}-1|=A\sqrt{2}$

Squaring, we have that:
$$A^2 = 2x + 2\sqrt{x^2-2x+1} \implies A^2 = 2x+2|x-1|$$If $A=\sqrt{2}$, then, $2=2x+2|x-1| \implies 1-x=|x-1| \implies x \le 1$. Given that $\sqrt{2x-1} \in Z \implies x \ge \frac{1}{2}$, we have that $x \in \left[ \frac{1}{2},1\right]$.
If $A=1$, then $1=2x+2|x-1| \implies x = \frac{3}{4}$, which we can check to be valid.
If $A = 2$, then $4=2x+2|x-1| \implies 2-x=|x-1| \implies x = \frac{3}{2}$. However, if $x = \frac{3}{2}$, then $x-\sqrt{2x-1} < 0$, so there are no solutions.

For the inequality to hold, we must have: $x \geq \frac{1}{2}$ and $(x-1)^2 \geq 0$.
note that if we square the identity we have the following
\begin{align*}
&\Rightarrow  \left(\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}}\right)^2 \\
&\Rightarrow 2x + 2\left(\sqrt{x+\sqrt{2x-1}})(\sqrt{x-\sqrt{2x-1}}\right) \\
&\Rightarrow 2x+2\mid x-1 \mid.
\end{align*}
(The second follows by difference of squares.)
From here it is easy to see: that $2x+\mid x-1 \mid = 2$ precisely when $\frac{1}{2} \leq x \leq 1$, $4-2x = 4 \Rightarrow x = \frac{3}{2}$ and there is no soltion for $1$

Why does it seem so much that early IMO problems are way easier than IMO problems nowadays?

I guess that at that time competition math wasn't so popular as it is nowadays

[quote=DPopov]For what real values of $x$ is \[ \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A  \]  given

a) $A=\sqrt{2}$;

b) $A=1$;

c) $A=2$,

where only non-negative real numbers are admitted for square roots?[/quote]
Truly ancient problem
[hide=Solution]Squaring both sides the equation becomes $$2x+2|x-1|=A^2$$
Case 1: $A=\sqrt{2}$ then the equation becomes $x+|x-1|=1 \Leftrightarrow x \in \left[\frac{1}{2},1 \right]$
Case 2: $A=1$ then the equation becomes $x+|x-1|=\frac{1}{2}$ which clearly is extraneous as for $x \le 1$we 've a contradiction and for $x>1$ we get a value of $x <1$, which again is a contradiction
Case 3: $A=2$,then the equation becomes $x+|x-1|=2$, which clearly has no solution for $x\le 1$ , then defining $x>1$ we get a solution $x=\frac{3}{2}$ , which indeed is a solution for this case .
[/hide]

[hide=Solution](a) Squaring both sides gives $A^2=2x+2|x-1|$. Plugging in $A=\sqrt2$ gives $2=2x+2|x-1|$, or $1=x+|x-1|$. This can be rewritten as $1-x=|x-1|$ and since one is the negative of the other we have $x \leq 1$. We also must have $x \geq \frac{1}{2}$, so the solution set is $\boxed{[\frac{1}{2},1]}$.

(b) If $A=1$, we have $1=2x+2|x-1|$, which has no solutions after casework on whether $x-1$ is negative or positive.

(c) If $A=2$, we have $4=2x+2|x-1|$, or $2=x+|x-1|$. After casework on whether $x-1$ is negative or positive, we get $x=\boxed{\frac{3}{2}}$.[/hide]

We know that $2x-1\ge0$ so $x\ge\frac12$.

We can square both sides of the equation $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A$ to get
\begin{align*}
x+\sqrt{2x-1}+2\sqrt{(x-\sqrt{2x-1})(x+\sqrt{2x-1})}+x-\sqrt{2x-1}&=A^2\\
2x+2\sqrt{x^2-2x+1}&=A^2\\
2x+2|x-1|&=A^2\\
x+|x-1|&=\frac{A^2}2
\end{align*}

We can do casework on the equation $x+|x-1|=\frac{A^2}2$.

We know that if $x\ge1$ then $x+|x-1|=x+x-1=2x-1$ so $2x-1=\frac{A^2}2$ and $x=\frac{A^2+2}4$.

We also know that if $x\le1$ then $x+|x-1|=x+1-x=1$ so $1=\frac{A^2}2$ and $A=\sqrt2$.

We can now solve for when $A=\sqrt2$, $A=1$, and $A=2$.

(a) $A=\sqrt2$

We already figured out that when $A=\sqrt2$ that $x\le1$. We combine this with $x\ge\frac12$ to get $\frac12\le x\le1$.

(b) $A=1$

We know that if $x\le1$ then $A=\sqrt2$ which is false. We know that if $x\ge1$ then $x=\frac{A^2+2}4=\frac34$ but $\frac34<1$ so this is also not a solution. Thus, there are no solutions.

(c) $A=2$

We know that if $x\le1$ then $A=\sqrt2$ which is false. We know that if $x\ge1$ then $x=\frac{A^2+2}4=\frac32$. Thus, $x=\frac32$.

We have that $2x-1=0\iff x\geq\tfrac{1}{2}$ and $x+\sqrt{2x-1},x-\sqrt{2x-1}\geq0\iff x\geq0$. Henceforth assume that $x\geq\tfrac{1}{2}$. Since $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}$ and $A$ are both nonnegative, we may square both sides of the given equation to obtain an equivalent formulation:
$$(x+\sqrt{2x-1})+(x-\sqrt{2x-1})+2\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=(x+x)+2\sqrt{x^2-2x+1}=2x+2|x-1|=A^2,$$
or, equivalently, $x+|x-1|=\tfrac{A^2}{2}$. For (a) $A=\sqrt{2}$, when $x\leq1$, we have $x+(1-x)=1=\tfrac{(\sqrt{2})^2}{2},$ always true; and when $x\geq1$, we have $x+(x-1)=1\iff x=1$. Hence, the solutions to (a) are given by $\boxed{x\in[\tfrac{1}{2},1]}$.

For (b) and (c), we have $x+(1-x)=1=\tfrac{A^2}{2}$ when $x\leq1$, never true, so it suffices to consider $x\geq1$. For (b),
$$x+(x-1)=\frac{1^2}{2}=\frac{1}{2}\iff x=\frac{3}{4}\not\geq1,$$
so $\boxed{\text{no solutions}}$; and for (c),
$$x+(x-1)=\frac{2^2}{2}=2\iff\boxed{x=\frac{3}{2}}\geq1.$$
$\blacksquare$

Note that $x\geq 1/2.$ Squaring it we get $2x+2|x-1|=A^2.$ Easily --
a) $\implies x\in [1/2,1]$
b) $\implies \not\exists x$
b) $\implies x=3/2$

We can square this expressed to get $2x+|x-1|=A^2/2.$ we can now substitute values of a and solve

$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}}$, where $c^2=a^2-b$

Plug it in and trivializes

[hide=solution] Squaring both sides, we get $$2x + 2|1-x|=A^2.$$ Now, plugging in the values of $A$ and solving for x, we find that if $A$ is $\sqrt{2}$, then $\boxed{x=1},$ if $A=1,$ then $\boxed{x \in \emptyset},$ and if $A=2,$ then $\boxed{x = \frac{3}{2}}.$ $\square$[/hide]

[hide=sol]
$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A $
squaring both sides:

$x + \sqrt{2x-1} + x - \sqrt{2x-1} + 2\sqrt{x^2 - 2x + 1} = A^2$
$2x + 2|x-1| = A^2$

Now plugging in $A = \sqrt{2}$, we get $2x + 2|x-1| = 2 \implies 1 - x = |x - 1| \implies x \le 1$. Finally, since $2x-1 \ge 0$, we have $x \ge \frac{1}{2}$. This means we obtain $\boxed{(a) \text{ } \left[\frac{1}{2}, 1\right]}$.

For (b), we have $2x + 2|x -1| = 1 \implies \frac{1}{2} - x = |x - 1| \implies x < \frac{1}{2} \rightarrow \bot$. Therefore there are no solutions for (b).

Finally, for (c), we have $2x + 2|x - 1| = 4 \implies 2 - x = |x - 1| \implies x = \frac{3}{2}$.

So our final answers are $\boxed{(a) \text{ } \left[ \frac{1}{2}, 1 \right], (b) \text{ No Solutions}, (c) \text{ } \frac{3}{2}}$. $\blacksquare$


How do I find all solutions of x|0<=x<360

$\sin (\frac{ax}{b}) = y$

Take the inverse sine of y.  There are 2 different degree values for a given inverse sine.  Just remember that $\sin{x}=\sin{180-x}$.  Then multiply by b/a.

If you found that z was a solution to the equation, you can't say z + 360n is also a solution correct? Because if you plug the same value with 360n added, it wouldn't fit the equation?

Yes, but the problem's requirements set the angle between {0,360).  Therefore, x+360n where n is anything but 0 isn't considered valid due to that requirement.


I already searched through the forum about this, and the instance that I found was not answered.

I wanted a fraction in my exponent, so I typed $ 3^{\frac{2 \cdot 3^{n}}{3}}$, which is ugly.

How can I reduce the size of the exponent, or otherwise beautify it?

Thanks :)

EDIT: It looks nice on the forum, but not on  the .dvi file...

[quote="Aryth"]EDIT: It looks nice on the forum, but not on  the .dvi file...[/quote]
The forum uses \dfrac instead of \frac which makes the fraction larger (for which you need \usepackage{amsmath} in the preamble), but in a document that looks worse to me. It may have something to do with the fonts being used. Could you give us a screenshot of the 'ugly' fraction, along with a minimal document you used to produce it (ie only the preamble plus the fraction) so we can see what is happening.

Using \frac seems to work.
Weird...
Thanks :)

Here's my minidoc:

%%%%%%%%%%%%%%%%%%%%

\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document}

$ 3^{\frac{2 \cdot 3^{n}}{3}}$

\end{document}

%%%%%%%%%%%%%%%%%%%%

EDIT: Whenever I try to say \dfrac  in the forum, it automatically reverts to \frac. But my original document has a \dfrac.

So are you happy with what you have in your document now?


Problem 1
A point $P$ lies in $\triangle ABC$. The lines $BP,CP$ meet $AC,AB$ at $Q,R$ respectively. Given that $AR=RB=CP, CQ=PQ$, find $\angle BRC$.

Problem 2
We have two distinct positive integers $a,b$ with $a|b$. Each of $a,b$ consists of $2n$ decimal digits. The first $n$ digits of $a$ are identical to the last $n$ digits of $b$, and vice versa. Determine $a,b$.

Problem 3
Find the greatest real number $k$ such that, for any positive $a,b,c$ with $a^{2}>bc$, $(a^{2}-bc)^{2}>k(b^{2}-ca)(c^{2}-ab)$.

Problem 4
Let $p,q\geq 2$ be coprime integers. A list of integers $(r,a_{1},a_{2},...,a_{n})$ with $|a_{i}|\geq 2$ for all $i$ is said to be an expansion of $p/q$ if
$\frac{p}{q}=r+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{...+\frac{1}{a_{n}}}}}$.
Now define the weight of an expansion $(r,a_{1},a_{2},...,a_{n})$ to be the product
$(|a_{1}|-1)(|a_{2}|-1)...(|a_{n}|-1)$.
Show that the sum of the weights of all expansions of $p/q$ is $q$.

Problem 5
Find the greatest possible integer $n$ such that one can place $n$ points in a plane with no three on a line, and color each of them either red, green, or yellow so that:
(i) inside each triangle with all vertices red there is a green point.
(ii) inside each triangle with all vertices green there is a yellow point.
(iii) inside each triangle with all vertices yellow there is a red point.

http://www.mathlinks.ro/Forum/viewtopic.php?p=820074#820074
http://www.mathlinks.ro/Forum/viewtopic.php?p=820075#820075
http://www.mathlinks.ro/Forum/viewtopic.php?p=820076#820076
http://www.mathlinks.ro/Forum/viewtopic.php?p=820077#820077
http://www.mathlinks.ro/Forum/viewtopic.php?p=820078#820078

Thanks for the submission, keep 'em coming http://www.mathlinks.ro/Forum/resources.php?c=87&cid=55&year=2003


Follow this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160783]link.[/url]

Okay, I'll do it for $ 7$ because it gets a little messy with your denominator by the idea is generalizable.

$ \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{6\pi}{7}$

Now, $ \cos \frac{4\pi}{7}=-\cos \frac{3\pi}{7}$ and $ \cos \frac{6\pi}{7}=-\cos \frac{\pi}{7}$.

So we have,
$ X=\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{3\pi}{7}$.
So,
$ X^{2}= \cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{3\pi}{7}\cos \frac{3\pi}{7}\cos \frac{2\pi}{7}\cos \frac{\pi}{7}$
Reduce (by $ \cos (\pi-x) =-\cos x$),
$ X^{2}= \cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{3\pi}{7}\cos \frac{4\pi}{7}\cos \frac{5\pi}{7}\cos \frac{6\pi}{7}$
Now follow the link that has this similar problem.
Finally extract square roots.

I don't really understand that link, and there are so many posts. Can you rewrite the solution?

[hide]Using a catalyst of $ \frac{\sin{2\pi}}{1999}$, I'm getting $ ({\frac{1}{2}})^\frac{999}{2}$.[/hide]


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\neq0.$ Prove that
\[ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{a^2 \plus{} b^2 \plus{} c^2} \geq\ \sqrt {a^2 \minus{} ab \plus{} b^2} \plus{} \sqrt {b^2 \minus{} bc \plus{} c^2} \plus{} \sqrt {c^2 \minus{} ca \plus{} a^2}
\]

Note that

$ 3\sqrt {\frac {2(a\plus{}b\plus{}c)^2\minus{}5(ab\plus{}bc\plus{}ca)}{3}} \ge \sum \sqrt {a^2\minus{}ab\plus{}b^2}$

So we want to prove

$ 3(a^3\plus{}b^3\plus{}c^3)^2 \ge (a^2\plus{}b^2\plus{}c^2)^2(2(a\plus{}b\plus{}c)^2\minus{}5(ab\plus{}bc\plus{}ca))$

But also $ (a^3\plus{}b^3\plus{}c^3)(a\plus{}b\plus{}c) \ge (a^2\plus{}b^2\plus{}c^2)^2$

So, we have to show
$ 3(a^3\plus{}b^3\plus{}c^3) \ge 2(a\plus{}b\plus{}c)^3 \minus{} 5(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$
$ \Leftrightarrow 3(a^3\plus{}b^3\plus{}c^3) \ge 2(a\plus{}b\plus{}c)^3\minus{}\frac 53 ((a\plus{}b\plus{}c)^3\minus{}(a^3\plus{}b^3\plus{}c^3)\plus{}3abc)$
$ \Leftrightarrow 4(a^3\plus{}b^3\plus{}c^3)\plus{}15abc \ge (a\plus{}b\plus{}c)^3$
$ \Leftrightarrow (a^3\plus{}b^3\plus{}c^3)\plus{}5abc \ge (a\plus{}b)(b\plus{}c)(c\plus{}a)$

which is Schur's inequality
 :)

All right, hell_ever!
What about the following inequality?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228646


AMC AIME probability ratio Gauss ARML USA(J)MO

1. a, b, and n are positive integers, and three consecutive terms of the standard binomial expansion of (a + b)^n are 9375, 11,250, and 6750. Find the next term in the expansion.

2. A standard, fair 6 sided dice is rolled 10 times successively. Find the probability that the number 6 is rolled 6 or more times in a row.

[hide="1."]

Let $a^{n-k}b^k\binom{n}{k} = 9375$.

We can examine the ratios to find $\displaystyle \frac{b}{a}\left(\frac{n-k}{k+1}\right) = \frac{11250}{9375}$ and $\displaystyle \frac{b}{a}\left(\frac{n-k-1}{k+2}\right) = \frac{11250}{9375}$. Looking at the factors, $9375 = (3)(5^5)$, $11250 = (2)(3^2)(5^4)$, and $6750 = (2)(3^3)(5^3)$, it's easy to tell that $a=5$ and $b=3$.

Then we go back to our ratios and find that $\displaystyle \frac{n-k-1}{k+2} = 2$ and $\displaystyle \frac{n-k}{k+1} = 1$ so we solve to get $n=5$ and $k=1$.

The term we want to find is $(5)(3^4)\binom{5}{4} = 2025$.[/hide]

2.
[hide]The probability is just $\frac {5}{6^6}-\frac {4}{6^7}$.  I'm just sticking this here in case anyone wants an answer later.  I finished this problem in the shower, without any paper, which makes me happy.[/hide]

Gauss, it's good to see you back!

[quote="d343seven"][hide]
for 2,
if you put $6^{10}$ in the denominator and then for each case of in a row 6's, you find the number and sum on top, would that work?
if you had 6 6's in a row, that means that there are 4 numbers left that should not be 6's, therefore, 5^4, multiplied by 5 (the sixes can be blocked in 5 places) and then you do it for 7, 8, 9, 10.

is this right?  where are any errors because the answer doesn't come out ot be te same as blahblahblah's.[/hide][/quote]

Overcounting is the devil in this problem, and my answer is correct.  Try to analyse your thought process again.

I derived my solution by considering ALL of the sequences of length 6 of sixes starting at place $n$, $1\leq n\leq 5$.  This is of course a gross overcounting, but bear with me.  The probability of this is $\frac {5}{6^6}$.  Now, see what happens if we subtract all of the sequences of length 7 of sixes in a row, starting at place $n$, $1\leq n\leq 4$ (having probability $\frac {4}{6^7}$).

We claim that every sequence of sixes with length $n\geq 6$ is counted exactly once and every sequence of length $n<6$ is counted zero times.  The last part is trivial, so consider the first.  For example, say that we have a sequence with $6$s in the $2-9$th place.  This is counted $3$ times by the first part ($2-7$,$3-8$,$4-9$) and twice by the second part ($2-8$,$3-9$), so it is counted exactly once, as desired.  If you guys want, you can prove rigorously that each sequence with the desired property is counted exactly once, which is what we wanted in the first place (it's not that hard)

i understand
that's a good solution,  seems very hard to come up with
but i don't see what i overcounted? :?
this is what i had
$\frac{5^4*5+5^3*4+5^2*3+5*2+1}{6^{10}}$

Problem 2 is interesting because I've seen the exact same problem on my local ARML tryouts...  That makes sense, though, as they probably take problems from the AIME, which are justthe right difficulty...

JB

There are AIME [i]type[/i] problems.  ARML writes problems theirself.  So I would assume that gauss202 got these from an old ARML. (and a lot of ARML problems are around AIME difficulty, btw).

[quote="d343seven"][hide]
for 2,
if you put $6^{10}$ in the denominator and then for each case of in a row 6's, you find the number and sum on top, would that work?
if you had 6 6's in a row, that means that there are 4 numbers left that should not be 6's, therefore, 5^4, multiplied by 5 (the sixes can be blocked in 5 places) and then you do it for 7, 8, 9, 10.

is this right?  where are any errors because the answer doesn't come out ot be te same as blahblahblah's.[/hide][/quote]

Why is it 5^4? That wouldn't allow any of the other numbers to be 6, I think for each case you only want to restrict the number(s) next to your string of 6's.

I am probably going to the ARML.  So, the problems are around AIME difficulty?

ARML has a bunch of different competitions:  individual, team, power, 3 person relay, and super relay (am I forgetting any?).  There are 8 individual questions.  You get them in pairs of two with each new pair being increasingly more difficult.  For each set you have 10 minutes.  I think these problems are usually around the first 10 on the AIME level (with variation of course).  The team round has 10 questions.  These problems are pretty difficult and are probably about AIME level also.  The power round consists of 10 problems (some can have multiple parts though) each one usually requires proof of your answer.  These are in increasing difficulty.  Since these are proofs they are not like AIME.  The last 2 or 3 problems might compare to USAMO level.  Finally, both the relays are fairly easy problems, which should be expected of a speed based event.  These are about AMC level.

[hide="Attempt_at_No._2"]I enjoy probability-type questions; what I would likely do for 2 (And it's horribly inefficient, I think) is write out all possible combinations of the 10 rolls resulting in 6 consectutive 6's, or, at the least, figure out how many different permutations of each series of ten rolls that result in a string of 6 6's there are for a given number of total 6's greater than 6.

For example:

With 6 6's, You have 666666XXXX, X666666XXX, XX666666XX, XXX666666X, and XXXX666666. These five rolls have probability (1/6^6)((5/6)^4) of each occuring, so the total probability for these rolls is 5 times that product. 9 and 10 are just as easy (Two ways to make 9, only one to make 10). 7 and 8 would probably require some form of combinations to find the total of, but the total sum is:

(5)(1/6^6)((5/6)^4)+(# of 7-combintations)(1/6^7)((5/6)^3)+(# of 8-Combinations)(1/6^8)((5/6)^2)+(2)(1/6^9)(5/6)+(1)(1/6^10)

If my counting is correct, there's, I think, 16 ways to make a row of 10 rolls with 7 6's, at least 6 in a line, and 18 ways to make a row with 8 6's, at least 6 in a line (One or both of these is likely wrong. Our school is not big into teaching anyone how to do these types of problems without trial and error.)

So, I get:

((5*5^4)+(16*5^3)+(18*5^2)+(2*5)+1)/(6^10), which is about .00009238. I probably messed up something.[/hide]

[quote="d343seven"]for 2, for 7 6's, does that mean xxx6666666, xx6666666x, x6666666xx, 6666666xxx....?
because that's what i thought the question meant.[/quote]

A sequence like "666666XX6X" also works, although there's only 6 6's in a row. So does one like "X6666666XX" with more than 6 6's in a row.

pilot's right
its good to see it in picture...

you could have

6666662142

but those last four are only restricted to 1 through 5.

why can't you have 6666662164?


I have two nice solutions, but impossible to find. I want another nice approach.

Does the inequality
$(x+y)(y+z)(z+x) /geq frac{8}{9} (xy+yz+zx)(x+y+z)$ applied to the LHS help us then some Schur?
I don't have time to check I will return back to this idea sorry..

[quote="iura"]Does the inequality
$(x+y)(y+z)(z+x) \geq \frac{8}{9} (xy+yz+zx)(x+y+z)$ applied to the LHS help us then some Schur?[/quote]

Hmm... not directly, i. e. not if you use it in the form $(x^2+y^2)(y^2+z^2)(z^2+x^2) \geq \frac89 (x^2 y^2 + y^2 z^2 + z^2 x^2)(x^2 + y^2 + z^2)$. At least, if you do this, then you have to use the inequality $\left( x^{2}+y^{2}+z^{2}\right) \left( x+y+z\right) ^{2}\geq 9\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\right)$ later, but this inequality generally doesn't hold (for instance, take x = 1, y = 12, z = 13).

A really tough (and yes, surely nice) problem...

I have no proof of the inequality, but if you are interested in a geometric equivalent, here is one:

Let ABC be an acute triangle with the altitudes AA', BB', CC' (the points A', B', C' being their respective feet). The circle with diameter BC intersects the altitude AA' at two points; let X be one of them (which one doesn't matter). Similarly define two other points Y and Z using the circles with diameters CA and AB and the altitudes BB' and CC', respectively. Then one can show that AY = AZ, BZ = BX, CX = CY. If we call x = AY = AZ, y = BZ = BX, z = CX = CY, then the inequality

  $x+y+z\geq \sqrt{2}\cdot \left( B^{\prime }C^{\prime }+C^{\prime }A^{\prime }+A^{\prime }B^{\prime }\right) $

holds.

When someone finds a proof of this (I hope!), I will explain why it is equivalent to Harazi's inequality.

  Darij

[quote="harazi"]Prove that if $ a,b,c\geq 0$ and $ a+b+c=1 $ then $ (a^2+b^2)(b^2+c^2)(c^2+a^2)\geq 8(a^2b^2+b^2c^2+c^2a^2)^2 $.[/quote]
Dear harazi, i know this one is old
but now i am very suprise when i see the same ineq with this one
$ (a^2+b^2)(b^2+c^2)(c^2+a^2)\geq 8a^2b^2c^2(a^2+b^2+c^2)^2/(ab+ac+bc)^2$[/tex]
Can you compare them ;)

Oh, sorry, I did not know that it s an old inequality. Well, it certainly can happen to everyone. Can you give me more details about the first source where you found it?

The inequality
   $ (x^2+y^2)(y^2+z^2)(z^2+x^2)(xy+yz+zx)^2 \ge 8x^2y^2z^2(x^2+y^2+z^2) (1)$
is of course equivalent to Harazi's one
   $ (x^2+y^2)(y^2+z^2)(z^2+x^2)(x+y+z)^2 \ge 8(x^2y^2+y^2z^2+z^2x^2)^2 (2)$ by putting $x --> 1/x, y --> 1/y, z --> 1/z$

I used (1) to prove another ineq, posted by Harazi long time ago:
      $ \sum \sqrt{\frac{abc}{b+c-a}} \ge a + b + c $

In this article, I proved (1) by indentity:
   $LHS - RHS = x^6(y-z)^2(y^2+4yz+z^2) + y^6(x-z)^2(x^2+4xz+z^2) + z^6(y-x)^2(y^2+4yx+x^2) + 2xyz(x^2y^2+y^2z^2+z^2x^2)(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y-6xyz) $

It was effective, but not easy to find. Recently, my student suggested one nice solution for (1), using geometry. And I also found nice sln for (2), using mixing variables.

Let try first, I will come soon.

I am very interested in seeing your solutions, Namdung.

Thank you very much.

Nobody tries the general case that I also posted not long time ago on this forum? Is this also a very old and known problem, dreammath?

Hmm, maybe I've found a very simple way to prove this inequality. Is it all right :o?
If we fix $a+b$ as constant then we can see that $a^{2}+b^{2}$ is smallest when $a=b$ (for example from AM-QM) and $ab$ is largest also when $a=b$ (from AM-GM).
Above argument implies that if we take $f(a, b, c) = LHS-RHS$ then $f(a, b, c) \geq f(t, t, b)$

So we will just assume that $2a+b=1$ and we will prove that:
$(a^{2}+b^{2})^{2}(2a^{2}) \geq 8(2a^{2}b^{2}+a^{4})^{2}$
which is equivalent to
$a^{2}+b^{2}\geq 4ab^{2}+2a^{3}$
or
$(a^{2}+b^{2})(2a+b) \geq 4ab^{2}+2a^{3}$
which after cancelling common terms and dividing by $b$ becomes:
$a^{2}+b^{2}\geq 2ab$

I don't understand how you are arguing $f(a,b,c) \ge f(t,t,c)$. Don't forget about terms $a^{2}+c^{2}$ etc that also involve a or b.

[quote=harazi]Prove that if $ a,b,c\geq 0$ and $ a+b+c=1 $ then $ (a^2+b^2)(b^2+c^2)(c^2+a^2)\geq 8(a^2b^2+b^2c^2+c^2a^2)^2 $.[/quote]

[img]http://s9.sinaimg.cn/middle/006ptkjAzy77mhUSpAIf8&690[/img]


hello.  how do we prove that

$ tan^{2}  \  \frac{\pi}{7} \plus{} tan^{2}  \  \frac{3\pi}{7} \plus{} tan^{2}  \  \frac{5\pi}{7} \equal{} 21$

do we prove it using some polynomial?   :(

hello, see here
http://mathworld.wolfram.com/TrigonometryAnglesPi7.html
Sonnhard.


Find all three digits numbers which is  25 times as large as sum his digits

All three digits numbers which is 25 times as large as sum his digits are 150; 225; 375;

Check the answer:
1) 150
150 = 25(1+5+0)
150 = 25(6)
150 = 150 [b]True[/b]

2) 225
225 = 25(2+2+5)
225 = 25(9)
225 = 225 [b]True[/b]

3) 375
375 = 25(3+7+5)
375 = 25(15)
375 = 375 [b]True[/b]

To learn more on "Number, Number Sense and Operations" click below:
http://www.winpossible.com/Grade_Four_Math.aspx

[hide="Solution"]
We can define the number as $ \overline{abc}$ where $ a,b,c$ are digits and $ a\neq0.$ Thus, we have the equation
\[ 100a \plus{} 10b \plus{} c \equal{} 25(a \plus{} b \plus{} c)\Rightarrow75a \equal{} 15b \plus{} 24c.
\]
[hide="Most important step!"]Dividing both sides by $ 15$ gives
\[ 5a \equal{} b \plus{} \frac {8c}{5}.
\]
Since $ a,b,c$ are digits, we must also have $ \frac {8c}{5}$ be an integer, specifically $ c$ is a multiple of $ 5.$ In other words, $ c$ can only be $ 0$ or $ 5.$[/hide]
[hide="Cases"]Now we just have to tackle 2 cases.
[u]When $ c \equal{} 0.$[/u]
We have $ 5a \equal{} b.$ The only solution (with $ a\neq0$) is $ (1,5,0).$
[u]When $ c \equal{} 5.$[/u]
We have $ 5a \equal{} b \plus{} 8.$ That means $ b\equiv2\pmod{5}.$ This restricts $ b$ to being $ 2$ or $ 7,$ and they give the solutions $ (2,2,5)$ and $ (3,7,5).$[/hide]
[hide="Checks"]
We want to know whether or not $ 150 \equal{} 25(1 \plus{} 5 \plus{} 0).$ Simplifying, $ 150 \equal{} 25(6)$ which is true, so $ 150$ works.
We want to know whether or not $ 225 \equal{} 25(2 \plus{} 2 \plus{} 5).$ Simplifying, $ 225 \equal{} 25(9)$ which is true, so $ 225$ works.
We want to know whether or not $ 375 \equal{} 25(3 \plus{} 7 \plus{} 5).$ Simplifying, $ 375 \equal{} 25(15)$ which is true, so $ 375$ works.[/hide]
[hide="Answers"]
The answers are $ \boxed{150, 225, \text{and }375.}$[/hide][/hide]

[quote="small_unicorn"][quote="RoFlLoLcOpT"][hide="Solution"]
We can define the number as $ \overline{abc}$ where $ a,b,c$ are digits and $ a\neq0.$ Thus, we have the equation
\[ 100a \plus{} 10b \plus{} c \equal{} 25(a \plus{} b \plus{} c)\Rightarrow75a \equal{} 15b \plus{} 24c.
\]
[hide="Most important step!"]Dividing both sides by $ 15$ gives
\[ 5a \equal{} b \plus{} \frac {8c}{5}.
\]
Since $ a,b,c$ are digits, we must also have $ \frac {8c}{5}$ be an integer, specifically $ c$ is a multiple of $ 5.$ In other words, $ c$ can only be $ 0$ or $ 5.$[/hide]
[hide="Cases"]Now we just have to tackle 2 cases.
[u]When $ c \equal{} 0.$[/u]
We have $ 5a \equal{} b.$ The only solution (with $ a\neq0$) is $ (1,5,0).$
[u]When $ c \equal{} 5.$[/u]
We have $ 5a \equal{} b \plus{} 8.$ That means $ b\equiv3\pmod{2}.$ This restricts $ b$ to being $ 2$ or $ 7,$ and they give the solutions $ (2,2,5)$ and $ (3,7,5).$[/hide]
[hide="Checks"]
We want to know whether or not $ 150 \equal{} 25(1 \plus{} 5 \plus{} 0).$ Simplifying, $ 150 \equal{} 25(6)$ which is true, so $ 150$ works.
We want to know whether or not $ 225 \equal{} 25(2 \plus{} 2 \plus{} 5).$ Simplifying, $ 225 \equal{} 25(9)$ which is true, so $ 225$ works.
We want to know whether or not $ 375 \equal{} 25(3 \plus{} 7 \plus{} 5).$ Simplifying, $ 375 \equal{} 25(15)$ which is true, so $ 375$ works.[/hide]
[hide="Answers"]
The answers are $ \boxed{150, 225, \text{and }375.}$[/hide][/hide][/quote]
What is $ b\equiv3\pmod{2}$ ?[/quote]

[url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction#Examples]Modular Arithmetic[/url] (Read the part that says "Congruence")

Yeah sorry about that, made a typo. It should read $ b\equiv2\pmod{5}.$ That will solve the problem.


A triangulation of a polygon is a partitioning of the polygon into triangles all of whose vertices are vertices of the original polygon. Given an $n-$side convex polygon, determine the number of different possible triangulations of it.

If $a_{n}$ is the corresponding number for $n-$gon, I've found out that
$a_{n}= \sum_{i=2}^{n-1}a_{k}\cdot a_{n+1-k}$ for $n \geq 4$
with $a_{2}= a_{3}= 1$. Maybe this can help.  :)

[quote="RDeepMath91"]A triangulation of a polygon is a partitioning of the polygon into triangles all of whose vertices are vertices of the original polygon. Given an $n-$side convex polygon, determine the number of different possible triangulations of it.
[/quote]

The desired number for the $(n+2)$-gon is well-known to be the $n^{th}$-Catalan's number $C_{n}= \frac{1}{n+1}\binom{2n}{n}.$

Pierre.


PLease help.

Let $A \in C^{n \times n}$ be a defective matrix. Use Schur's theorem to show that for every $\epsilon > 0$, there is a semisimple matrix $A_{\epsilon} \in C^{n \times n}$ such that $\Vert A - A_\epsilon \Vert_2 < \epsilon$.

Schur's Theorem:
$A \in C^{n \times n}$ there exists a unitary matrix $U \in C^{n \times n}$ and an upper-triangular $T \in C^{n \times n}$ such that $T=U^*AU$

How would u use Schur's Theorem to prove... anyone?

what's your definition of defective ? semisimple ?


Gauss
linear algebra
matrix
parameterization

Determine the rank of the following matrix
\[ \begin{array}{|cccc|}
   4&3&2&1\\
   1&0&a\minus{}2&1\\
   5&2\plus{}a&1&2a\\
   5&2&\minus{}a&3
\end{array}\]
 in dependence to the real parameter $ a$.


[quote="math_99"]How do you find the cosine of a line in cartesian plane given its equation? I can only do tangent for it. thanks for help.[/quote]So the problem is equivalent to finding $ \cos\theta$ given $ \tan\theta$. Make use of the secant-tangent identity, ie $ \sec^2\theta\equal{}\tan^2\theta\plus{}1$, and work from there.

[quote="t0rajir0u"]What is the "cosine of a line"?[/quote]

Maybe express the equation of a line given $ Ax \plus{} By \equal{} C$ in terms of $ \cos$?

I mean the cosine of the angle formed by the line and x-axis counterclockwise. I used that identity, but when you take the square root, is it positive or negative?

In the range of 0-180 degrees, cosine and tangent are of the same sign.


If m+n>50 and m-n :le: 10, what is the smallest possible integer value of n?

50 > n+m
-m+m >= -10

adding and manipulating,
30>m so m at most 29, wherein n is its minimum at 20.

[quote="4everwise"][hide]20 is the minimum.[/hide][/quote]
no its not, and where's your solution?

The minimum is 21.
The sum must be greater than fifty.  Making m 30, or 31, it doesn't matter, but n has to be 21 for the minimum value

[quote="Syntax Error"]If $m+n>50$ and $m-n \leq 10$, what is the smallest possible integer value of n?[/quote]

$m>-n+50$, and $m \leq 10+n$. Because $m \leq 10+n$, $m<9+n \Rightarrow -m > -9-n$.

\[-n+50 = -9-n+2m\]
\[59 = 2m\]
\[\frac{59}{2} = m\]
\[\frac{59}{2}-n \leq 10\]
\[n \geq \frac{39}{2}\]
\[n \geq 20\]

That was definitely not the best way...

Edit: nat mc is right, I don't know what i did wrong

[quote="chess64"][quote="Syntax Error"]If $m+n>50$ and $m-n \leq 10$, what is the smallest possible integer value of n?[/quote]

$m>-n+50$, and $m \leq 10+n$. [size=150][[/size]Because $m \leq 10+n$, $m<9+n$ [size=150]][/size]$\Rightarrow -m > -9-n$.

\[-n+50 = -9-n+2m\]
\[59 = 2m\]
\[\frac{59}{2} = m\]
\[\frac{59}{2}-n \leq 10\]
\[n \geq \frac{39}{2}\]
\[n \geq 20\]

That was definitely not the best way...[/quote]
How do you get that $\uparrow$ in the brackets
That part doesn't make sense

You should get $n>20$
$m+n>50$
$m \leq 10+n$
Since the last inequality was less than or [b]equal to[/b], we set $m$ to its maximum, $10+n$ and plug it in to the other equation:
$10+n+n>50$
$2n>40$
$n>20$
$min(n)=21$


Is this one a well known inequality?
\[
\frac{a}{-a+b+c}+\frac{b}{a-b+c}+\frac{c}{a+b-c}
\]
provided that $a,b,c$ are the sides of a triangle.

Well known,well known.I have lots of solutions for this one.One of them is very simple.Just apply AM-GM inequality
 twice and it is over. ;)

oldman,prove this sharper inequality:

     $\frac{a}{-a+b+c}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq\frac{s\sqrt3}{3r}$ ;)

 Well,I think you know that $s$ is the semiperimeter of the triangle given.

Could you post the most elegant solution to this ineq., please   :)

Thanks

[quote="oldman"]sorry

\[
\frac{a}{-a+b+c}+\frac{b}{a-b+c}+\frac{c}{a+b-c}\geq 3\][/quote]

Put $x=a+b-c$ and so on.  This gives

$\sum \frac {a}{b+c-a}=\frac {1}{2}\sum \frac {x+y}{z}\geq \frac {6}{2}=3$, where the middle inequality follows from AM-GM.


As we discussed in another thread, it is an open question whether there exists an integer polynomial $f \in \mathbb{Z}[x]$ such that $f(1) \neq -1$ and $f(x)+x^n$ is reducible for all $n$. Here is a 'converse' problem:

Show that for every $d \in \mathbb{N}$ there exists a [b]monic[/b] polynomial $f \in \mathbb{Z}[x]$ of degree $d$ such that $f(x) + x^n$ is irreducible for every $n \in \mathbb{N}$.

Hmmm... Strange... what about $f(x)=1+Mx+x^d$, where $M>3$ is an arbitrary natural, + Perron criteria.

But how do you apply Perron criteria here? The coefficient of $x^{n-1}$ is zero and you might have some problems if $n=d$. Or am I missing something?  :?

[quote="nalpaction"][quote="vess"]As we discussed in another thread, it is an open question whether there exists an integer polynomial $f \in \mathbb{Z}[x]$ such that $f(1) \neq -1$ and $f(x)+x^n$ is reducible for all $n$[/quote]

How about $f(x)=x$?May i missunderstand something :? ?[/quote]
There was a slightly different statement. See http://www.mathlinks.ro/Forum/viewtopic.php?t=28770


The number of square inches in the areas of rectangles A, B and C are shown in the diagram. How many square inches are in the area ot rectangle D?

[asy]draw((0,0)--(15,0)--(15,12)--(0,12)--cycle);
draw((0,7)--(15,7)); draw((6,0)--(6,12));
label("A",(0,12),SE); label("B",(6,12),SE); label("C",(0,7),SE); label("D",(6,7),SE);
label("30",(3,9),SE); label("45",(11,9),SE); label("63",(11,3),SE);[/asy]

The side lengths of A are 6 and 5, the side lengths of B are 5 and 9, and the side lengths of D are 9 and 7. Thus C has an area of $ 6\cdot7\equal{}\boxed{42}$. And yes, there is a typo in the question or the answer would be $ 63$.


calculus
derivative
algebra
polynomial
symmetry
algebra proposed

The phrase "This should be easy" is a JOKE and should never have been paired next to a problem.  Nothing in the solver's set-up hints at
the subjective "easy."  And  any other methods to solve this that require these levels of math techniques pulled together, OR slightly less in difficultly but more steps, just go to show the insincerity of the baiting of the author.

Even shorter  and more direct  method of solving it (than presented by jin) can lead you to the same result .
Using the theory of [u]symmetric[/u] polynomials  I don't find difficult and is quite natural here.
Anyway,nice solution jin!

It is reasonable to me looking at the coefficients in the equations that some sort of other method based on symmetry exists, and can be shorter, and EASIER (note the relative word usage).

What benefit does it do the reader to know that you don't find doing
symmetric polynomials difficult?  To know you're arrogant?

But it remains that you were wrong to use "easy."  Drop a subjective word like this.

Although I normally disagree with A cute angle, I agree that the title is bad:
Not because it's subjective, but because it says nothing about the problem. Something like "find x^5+y^5+z^5" should be better (bad topic naming is one reason why it is so hard to find some threads again).

[quote="A cute angle"]It is reasonable to me looking at the coefficients in the equations that some sort of other method based on symmetry exists, and can be shorter, and EASIER (note the relative word usage).

[/quote]
$s_{2}={p_{1}}^2-2p_{2}=7$
$s_{3}={p_{1}}^3-3p_{1}p_{2}+3p_{3}=15$
$s_{4}={p_{1}}^4-4{p_{1}}^2{p_{2}}^2+4p_{1}p_{3}=35$
[b]_______________________________[/b]
According to the terms of the problem the satisfying solutions of this system are:$p_{1}=3,p_{2}=1,p_{3}=-1$

These give the wanted result as:
$x^5+y^5+z^5=s_{5}={p_{1}}^5-5{p_{1}}^3p_{2}+5p_{1}{p_{2}}^2+5{p_{1}}^2p_{3}-5{p_{2}}^3p_{3}=\boxed{83}$

$s_{n},p_{n}$ denote fundamental Newton's and symmetric polynomials respectively.
Maybe the problem wasn't so easy for those who didn't heard of them.
My apology to all of them! :)

The solution, then esti, is not to use those troublesome words.  In fact, you used the phrase "so easy" in your recent post which is relative compared to just "easy."

I like your solution to the posed math problem, esti.


modular arithmetic
algebra unsolved
algebra

Let

$ \frac{p}{q}\equal{}1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}\plus{}...\minus{}\frac{1}{1318}\plus{}\frac{1}{1319}$

Prove that p is divisible by 1319.

Really?
$ p \equiv 164 \pmod{1319}$
[url=http://www.wolframalpha.com/input/?i=Mod%28Numerator%28sum+power%28-1%2C+k-1%29%2Fk+k%3D1+to+1319%29%2C+1319%29]Mod(Numerator(sum power(-1, k-1)/k k=1 to 1319), 1319)[/url] - Wolfram Alpha

I post so bad, dunno why. We need to prove that 1979 divises p. Sorry for mistake


function
absolute value
advanced fields
advanced fields unsolved

Prove that $ f(z) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {1}{n^z}$ does not converge uniformly on A={z|Re z > 1}

[quote="fedja"]Your definition of $ A$ doesn't make sense as written.[/quote]

You're right, I don't even know what I'm typing anymore. I know if you take a closed disk B that is inside A and $ \delta$ away from the line Re z=1 and use the analytic convergence theorem, we have to find an $ M_n(\delta)$ using the Weierstrass M-theorem. The function $ f(z)\equal{}\frac{1}{n^z}\equal{}n^{\minus{}z}$, which is the Riemann $ \zeta$ function. If we take the absolute value of the function: $ |n^{\minus{}z}|\equal{}|e^{\minus{}z*log(n)}|\equal{}e^{\minus{}x*log(n)}\equal{}n^{\minus{}x}$. But $ x \geq 1\plus{} \delta$ if $ z \in B$ so $ |n^{\minus{}z}| \leq n^{\minus{}(1\plus{}\delta)}$ so by choosing $ M_n\equal{}n^{\minus{}(1\plus{}\delta)}$ using the Weierstrass M-theorem, the function converges uniformly on B. How would we show it does not converge uniformly on A?

If it converged uniformly, you could extend it continuously to the line $ \text{Re}(z)\equal{}1$. That's impossible, because it goes to $ \infty$ at $ z\equal{}1$.


calculus
integration
LaTeX
function
calculus computations

Hi can someone help me with the double integration:
Sorry I dont know how to use the LATEX code.
it is a double integration
The first integration goes [b]from 1/2 to1[/b]. the second integration goes from 0[b] to sqrt(1-x^2)[/b]. there is no function after ,just[b] dy dx[/b].
I am asked to calculate the integral using polar coordinates.
I dont know how to find the bounds of integration corresponding for r.
Thank you

B

[quote="sangre"]The integral:
$\int_{1/2}^{1}\int_{0}^{\sqrt{1-x^2}}dydx$
(LaTeX isn't hard, trust me. Mostly you just put computations between 2 $\$$)

The line $x=\frac{1}{2}$ is $r cos\theta=\frac{1}{2}$

Hope that helps!  :D[/quote]

All righty!
It makes sense now

B


Ring Theory
calculus
integration
algebra
function
domain
complex numbers

Hi everyone,
  Here's the problem taken straight from the assignment:

  Let R = (S, +, *) be a ring which is not a field. Is it possible for R to have a subring which is a field? Either prove that this is not possible or give an example of a ring with divisors of zero which has a subring which is a field.

  The first source of confusion is this: why must the counterexample have divisors of zero in the first place? Integral domains have none and are also not fields. My grasp on the material is tenuous at best, but I think a proof of the impossibility of this situation would go something like this:

  Let (X, +, *) be a field.
    - all elements of X are units
    - has no zero divisors
  Let X also be a subset of S.
    - all elements of X have additive inverses
    - let S have a pair of elements such that a * b = 0 (zero divisors)

  Show that the introduction of these zero divisors breaks the closure of S under multiplication.

  I have no idea how to do this, or even if it's the correct approach. Would appreciate any help / clarification / explanation, as I'm still new to this subject. Thanks.

  Ari.

There are counterexamples with zero-divisors and there are counterexamples without zero-divisors. Just construct one.

How can I construct a non-field supergroup from a field?

This is a conceptual problem for me, I'm stuck in (R, +, *). The only other sets I can imagine are modulus and complex numbers, and even those are hazy...

Take for example $\mathbb{Z}_2^2$ or even every 'power' of a field.

I know what $\mathbb{Z}_2$ means, but the $\mathbb{Z}_2^2$ or even $\mathbb{Z}^x$ is completely foreign notation to me. What does that mean in terms of rings? If the explanation is too tedious, where can I find the information online?

Ari.

The example ZetaX is thinking of: Let $F$ be your favorite field. Define $+$ and $\cdot$ on $F\times F$ by $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)\cdot(c,d)=(a\cdot c,b\cdot d)$. The subrings $\{(a,a): a\in F\}$, $\{(a,0): a\in F\}$, and $\{(0,a): a\in F\}$ are each obviously isomorphic to $F.$

The exponent notation denotes the Cartesian product: $A^k$ is the set of ordered $k$-tuples of elements of $A.$

Thanks for all your help folks. That's a brilliant example and it actually helped me understand the nature of rings a bit better. Very gratefully,

  Ari.

$\{(a,0)|a\in F\}$ doesn't work if you use definitions for rings, fields, subrings that strongly involve preserving $1$.

$(a,a)$ however works for me too.

I want to present some other example: Let $K$ be a field. $K[X]/(X^2)$ is a ring that has zero divisors (take $X$ for example) and has $K$ as a subring which is a field.


geometry
perimeter
ratio
trigonometry
area of a triangle
trig identities
Law of Cosines

What is the number of square centimeters in the area of a triangle whose sides measure 8 cm, 12 cm and 16 cm? Express your answer in simplest radical form.

Use Heron's formula.  The semi-perimeter is $ s\equal{}\frac{8\plus{}12\plus{}16}{2}\equal{}18$.  So
$ A\equal{}\sqrt{18(18\minus{}8)(18\minus{}12)(18\minus{}16)}\equal{}\sqrt{18(10)(6)(2)}\equal{}\sqrt{2160}\equal{}12\sqrt{15}$

Is Heron's formula the only way to approach this, as if this were a sprint or AMC question it would be harder to calculate.

uh i think heron's is best.
you could factor out a 2 from the side lengths, and multiply the resulting heron-bash area by 4 (with similar triangle area ratio stuff)
i know i didn't make much sense but w/e

Ok thanks, I guess I need some more practice using Heron's fast lol.

The other way I think of is still pretty bashful.
Law of Cosines to find the cosine of one angle: $C$.
$\frac{ab\sin C}{2}$ for the area.


geometry
projective geometry
geometry proposed

Actually this is just another way of stating the following well-known fact (see Theorem 1 in [url=http://www.cut-the-knot.org/Curriculum/Geometry/TwoQuadris.shtml]Alexander Bogomolny's note about the "Two quadrilaterals"[/url]):

[i]If XYZW is a circumscribed quadrilateral whose incircle touches its sides XY, YZ, ZW, WX at the points X', Y', Z', W', respectively, then the lines X'Y' and Z'W' meet on the line XZ.[/i]

In your case, we call F' the point of intersection of the tangents to k at the points B and D, and we call G' the point of intersection of the tangents to k at the points C and E. Then, the quadrilateral FF'GG' does not exactly have an incircle, but it has an excircle - namely the circle k - which touches its sidelines FF', F'G, GG', G'F at the points B, D, C, E, respectively. Whether a circle is an incircle or an excircle doesn't matter for the application of the above fact, so we can apply it to our quadrilateral FF'GG' with its excircle k and get that the lines BD and CE meet on the line FG. But the lines BD and CE meet in the point A; hence, the point A lies on the line FG, i. e. the points F, G, A are collinear.

  Darij

thanks, a nice proof. theres a simpler one with the same idea:
Lemma 1 (maybe wellknown): Let ABCD be a quadrilaterl with inscribed circle. Let tangency points at AB, BC, CD and DA be A',B',C' and D'. then the four lines are concurrent: A'C', B'D', BD, AC.
Proof: this is trivial. First apply Brianchon's theorom to A,B,C,D, A',C' to see the the intersection of AC and BD lies on A'C'. THen apply Brianchon to A,B,C,D, D',B' to see the intersection of AC and BD lies on B'D'.
Now apply Lemma 1 to quadrilaterl BEDC (note ordering of vertices). That tells us that the intersection of tangents from E and B, the intersection of tangents from D and C, and the point where BD intersects EC are collinear, as required. [i know its a bit confusing.].

A (perhaps) different way is to observe that, as XZ, YW, X'Z' and Y'W' concurr (a well known fact) , then, by Desargues, the point of intersection of (WW'.YX')=X, (YY'.WZ')=Z and (W'Z'.X'Y') are collinear.

[quote="vinoth_90_2004"]Let ABC be a triangle. Let k be a circle passing through vertices B and C of the triangle, intersecting the segments AB and AC at D and E respectively. Let the tangent to k at E intersect the tangent to k at B at a point F. Le the tangent to k at D intersect the tangent to k at C at a point G. Prove F, G and A are collinear.

[Note: The original problem, slightly easier, had k being the circle with diameter BC, but that isnt necessary.][/quote]
Maybe another approach:
Let $ \{X\}\equiv DE\cap BC$, $ \{Y\}\equiv DC\cap BE$, hence, $ XY$ is the pole of $ A$ wrt $ \mathcal K$. But $ Y$ is the pole of $ EF$ wrt $ \mathcal K$. According to [b]Lahire[/b] theorem, we conclude that $ A\in EF$. The result is lead as follow.


I am going to do an independent study on Group Theory next semester, and I've been looking for a decent text.  I've had a semester of Number Theory.  Any suggestions?

I recommend [u]Contemporary Abstract Algebra[/u] by Joseph A. Gallian.
Of course, I'm a bit biased since he's my advisor for an extracurricular project on group theory, but it [i]is[/i] the best text I've seen on the subject.


How do you switch the order of things in a header? For example,

" How to manipulate a Header                                                  0 of 10 ".

Then on the next page:

" 1 of 10                                                  How to Manipulate a Header ".

Also, can someone please describe how to have the section title appear in the header? Thank you in advance.

[code]\documentclass[11pt]{book}
\usepackage{fancyhdr}

\pdfpagewidth 8.5in
\pdfpageheight 11in

\pagestyle{fancy}
\headheight 35pt

\fancyhead{} % clear all fields
\fancyhead[CE,CO]{\bfseries Headers}
\fancyhead[LE,RO]{\thepage}
\renewcommand{\headrulewidth}{0.4pt}
\renewcommand{\footrulewidth}{0.4pt}
\cfoot{}[/code]

It works, except all the even-numbered pages are shifted to the right and you can't see the text. Here is my preamble (maybe its because of the margins I set):

 [code]
\documentclass[20pt]{book}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{color}
\linespread{1.6}
\topmargin=0in
\oddsidemargin=0in
\textwidth=6.5in
\pdfpagewidth 8.5in
\pdfpageheight 11in
\usepackage[pdftex]{graphicx}
\usepackage{fancyhdr}
\pagestyle{fancy}
% with this we ensure that the chapter and section
% headings are in lowercase.
\renewcommand{\chaptermark}[1]{\markboth{#1}{}}
\renewcommand{\sectionmark}[1]{\markright{\thesection\ #1}}
\fancyhf{} % delete current setting for header and footer
\fancyhead[LE,RO]{\bfseries\thepage}
\fancyhead[LO]{\bfseries\rightmark}
\fancyhead[RE]{\bfseries\leftmark}
\renewcommand{\headrulewidth}{0.5pt}
\renewcommand{\footrulewidth}{0pt}
\addtolength{\headheight}{0.5pt} % make space for the rule
\fancypagestyle{plain}{%
\fancyhead{} % get rid of headers on plain pages
\renewcommand{\headrulewidth}{0pt} % and the line
}
[/code]

It works when I take away the margins I set, but I want to keep the margins I set. Any ideas?

for more information about headers look in the forum for a post from orl about latex... there are some very usefull links :D