geometry
trapezoid
circumcircle
geometry proposed

[quote="Fang-jh"]Given a trapezoid $ ABCD$ with $ AD\parallel BC, E$ is a moving point on the side $ AB,$ let $ O_{1},O_{2}$ be the circumcenters of triangles $ AED,BEC$, respectively. Prove that the length of $ O_{1}O_{2}$ is a constant value.[/quote]

Let $ l_1$ , $ l_2$ be perp. bisectors of $ AD,BC$. let $ l_3$ , $ l_4$ be perp bisector of AE and EB. then $ O_1O_2$= distance between intersection of$ (l_1,l_3)$ and $ (l_2,l_4)$. since $ l_1,l_2$are fdixed and perp. dist.of ($ l_3,l_4$)=$ \frac{AB}{2}$,our result follows.

[quote="ith_power"][quote="Fang-jh"]Given a trapezoid $ ABCD$ with $ AD\parallel BC, E$ is a moving point on the side $ AB,$ let $ O_{1},O_{2}$ be the circumcenters of triangles $ AED,BEC$, respectively. Prove that the length of $ O_{1}O_{2}$ is a constant value.[/quote]

[quote="ith_power"]Let  ,  be perp. bisectors of . let  ,  be perp bisector of AE and EB. then = distance between intersection of and . since are fdixed and perp. dist.of ()=,our result follows.[/quote]Let $ l_1$ , $ l_2$ be perp. bisectors of $ AD,BC$. let $ l_3$ , $ l_4$ be perp bisector of AE and EB. then $ O_1O_2$= distance between intersection of$ (l_1,l_3)$ and $ (l_2,l_4)$. since $ l_1,l_2$are fdixed and perp. dist.of ($ l_3,l_4$)=$ \frac{AB}{2}$,our result follows.[/quote]

I found the distance between the circumcenters is $(AB/2).cosec(ADC)$


The lengths of two sides of a triangle are 5 and 8 units.  If the length of the third side is also an integer, what is the largest possibe length of the third side?

[hide]
I think the answer is 12, because a side on a triangle has to be smaller than the sum of the other 2.[/hide]

According to Triangle Inequality theorem, 3<x<13. So the largest possible length of the third side is 12. ;)

We learned this at school a couple weeks ago.

[hide] By the triangle inequality thm, 3<x<13. x=12. Which was basically what conandoyle said. :) [/hide]

[hide]According to triangle inequality, the length of the third side must be no larger than 5+8=13. As the third side must also be an integer, the largest possible length is 12.[/hide]

[hide="i learend this 2 weeks ago"]
as everyone else said it has to be less than 13 so it has to be 12[/hide]

[quote="biffanddoc"]Can't it be 13?
It would be degenerate though[/quote]

Yeah, a line isn't really a triangle  :D

X is the length of the 3º side.

Then, 5+8>X

13>X.

As X must be an integrer, X=12 is the biggest possible value.

[quote="biffanddoc"]But it's still a triangle :D[/quote]

What? That would mean that a square is also a line with sides n by 0?


invariant
Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

If $ R$ is a PID, $ M$ and $ N$ are $ k$ by $ k$ matrices with entries in $ R$, and the
invariant factors of $ M$ and $ N$ are $ {m_1, m_2, ..., m_k}$ and $ {n_1, n_2... n_k}$ respectively. Prove that if the invariant factors of $ P \equal{} MN$ are
$ {p_1, p_2, .., p_k}$, then $ m_i|p_i$ and $ n_i|p_i$ for all $ i \equal{} 1,2...k$.

$ N,N^T$ have the same invariant factors and so do $ N^TM^T\equal{}(MN)^T$ and $ MN$.
So it's enough to show $ m_i|p_i$ for all $ i\equal{}1,\ldots,k$.

For all $ m,n\in R,m\not\equal{}0$ we have also $ \gcd(m,n)\not\equal{}0$ (the $ \gcd$ is always taken arbitrarily)
and then $ m(R/nR)\equal{}(mR\plus{}nR)/nR\equal{}\gcd(m,n)R/nR\cong R/{n\over\gcd(m,n)}R$ as $ R$-modules.

Now assume $ m_i\nmid p_i$ for some $ i\equal{}1,\ldots,k$.
Then $ p_i\not\equal{}0$ and we find some prime element $ p\in R$ and $ n\in\mathbb{N}$ with $ p^n|m_i$ and $ p^n\nmid p_i$.
Then for all $ j\ge i$ we get $ p^n|m_i|m_j$ and $ p|{m_j\over\gcd(p_i,m_j)}$.
So the $ R$-module $ p_i(R/m_jR)\cong R/{m_j\over\gcd(p_i,m_j)}R$ has epimorphic image $ R/pR$ for all $ j\ge i$.

But $ R^k/MNR^k\cong R/p_1R\times \cdots\times R/p_kR$ as $ R$-modules
gives $ p_i(R^k/MNR^k)\cong R/{p_{i\plus{}1}\over p_i}R\times \cdots\times R/{p_{k}\over p_i}R$ generated by $ \le k\minus{}i$ elements over $ R$.

Then $ MNR^k\subseteq MR^k$ gives $ R^k/MR^k\cong R/m_1R\times \cdots\times R/m_kR$ as epimorphic image of $ R^k/MNR^k$.
So $ p_i(R^k/MNR^k)$ has epimorphic image $ p_i(R^k/MR^k)\cong p_i(R/m_1R)\times \cdots\times p_i(R/m_kR)$,
which has epimorphic image $ (R/pR)^{k\minus{}i\plus{}1}$.

This gives $ (R/pR)^{k\minus{}i\plus{}1}$ also generated by $ \le k\minus{}i$ elements as $ R$-module, and then also as $ R/pR$-module.
But this contradicts the fact, that $ (R/pR)^{k\minus{}i\plus{}1}$ has dimension $ k\minus{}i\plus{}1$ over the field $ F: \equal{}R/pR$.


abstract algebra
function
algebra
polynomial
calculus
derivative
complex analysis

Which of the following rings are noetherian?
(1) The ring of rational functions in ($\mathbb{C}(X)$) having no pole on $|x| = 1$.
(2) The ring of power series in $X$ $\mathbb{C}[[X]]$  with positive radius of convergence.
(3) The ring of power series in $X$ $\mathbb{C}[[X]]$  with infinite radius of convergence.
(4) The ring of polynomials in $X$ whose first $k$ parial derivatives vanish at $0$.

2. Denote by $\mathbb C\{x\}$ the ring of series with positive radius of convergence around 0.
[b]Remark[/b] It can be shown that $\mathbb C\{x\}$ is isomorphic to the ring of germs of holomorphic functions around 0. I say this only to justify the notation.
Let $f\in \mathbb C\{x\}$. On a small disc around 0, $f$ is holomorphic. If $f(0)\neq 0$, then on a smaller disc around 0, $\frac 1f$ exists as a holomorhic function, hence it defines an element $f^{-1}$ of $\mathbb C\{x\}$ such that $f\cdot f^{-1}=1$. These prove that if $f(0)\neq 0$, then $f$ is invertible in $\mathbb C\{x\}$.
Let $I$ be a nonzero ideal of $\mathbb C\{x\}$. Let $f\in I$ such that 0 is a zero for $f$ of minimal order among the elements of $I$. Then $f=x^k\cdot u$ for some $k\in\mathbb N$ and $u\in C\{x\}$ with $u(0)\neq 0$. We have proved that $u$ is invertible, hence $x^k\in I$ implying $(x)^k\subset I$. Every element of $I$ can be written as $x^l\cdot v$ with invertible $v$ and by the minimality condition in the choice of $k$, $k\leq l$. Therefore $I\subset (x)^k$.
So $I=(x^k)$, thus $\mathbb C\{x\}$ is a principal ideal domain, hence noetherian.

The third ring (let's denote it by $R$) is not noetherian, I think. It is the ring of entire functions, and it's well-known that a finitely generated ideal of entire functions is actually principal. It thus suffices to exhibit a non-principal ideal of $R$.

Now consider the ideal $I$ generated by some entire functions $f_1,f_2,\ldots$, such that the zeros of $f_i$ are $i,i+1,\ldots$, and assume that there is an entire function $g$ such that $I=(g)$. Then $g$ cannot have any zeros, because for every $z\in\mathbb C$ there is an $i$ and an entire function $h_i$ such that $f_i(z)\ne 0,\ f_i=gh_i$. This, in turn, means that $(g)=R$. This is impossible, since it's obvious that every function in $I$ has a zero.

Also for (3), inspired by a problem in Rudin, take $I_n=(g_n)$, where $g_n(z)=\prod_{i=n}^{\infty}(1-\frac{z^2}{i^2})$. This is an explicit example to grobber's post.
$I_n$ is a strictly increasing sequence of ideals.

For the fourth (4) problem: The ring is the ring of polynomials in $\mathbb C[X]$ that don't have the terms $X$, $X^2,\ldots,X_k$. It can be described as $\mathbb C\oplus X^{k+1}\mathbb C[X]$.
It is easy to see (once you had the idea ;) ) that this ring is $\mathbb C[X^{k+1},X^{k+2},\ldots,X^{2k+1}]$ which is obviously noetherian.

I think, but I am not sure, that any ideal of this ring can be generated by $k$ elements.

For the $1^{st}$ and last problem :D:

The set $S$ of polynomials that do not vanish on $|z|=1$ is a multiplicatively closed subset of $\mathbb C[X]$, therefore our ring is $S^{-1}\mathbb C[X]$ which is obviously noetherian.

For (1):
We note $R$ the ring. Then, if $I$ is an ideal of $R$, $I\cap \mathbb{C}[X]$ is an ideal of $\mathbb{C}[X]$ so $I\cap\mathbb{C}[X]=(P)$. It is easy now that in fact $I=(P)$ (here this is an ideal for $R$). It's correct ?
For (3):
$($sin$X/2^{n})_{n}$ gives a strictely increasing sequence of ideals I guess.

A holomorphic function f on an open set U in C vanishes only at a discrete set of points. Conversely, given any discrete set of points D in U there is a holomorphic function vanishing exactly on D.

Let H(U) denote the holomorphic functions on U. Let D < U be discrete and denote by I(D) < H(U) the set of functions vanishing on D. I(D) is an ideal. It is finitely generated if and only if D is finite. (Don't take this on faith!)

If D1 > D2 > .... is a strictly descending chain of discrete sets in U then I(D1) < I(D2) < ... is a strictly ascending chain of ideals. Why? Because there is a function vanishing exactly on D2 and not on all of D1 etc.. So H(U) is not noetherian.

The maximal ideals in H(U) are indexed by the points: for z we have the ideal I({z}). For any proper ideal I in H(U) the set D of common zeros is a nonempty discrete set. As we pass to the stalk of the sheaf H at a point z all ideals vanish except for ones where z is a common zero. Why? If D doesn't contain z choose a neighborhood V around z disjoint from D. Then the restriction homomorphism H(U) -> H(V) extends I to the whole ring H(V), so it vanishes.

If z is a common zero of the ideal I then by moving close enough to z only z remains as a common zero and therefore this ideal becomes finitely generated. That's why the stalks of H are noetherian rings. As amfulger pointed out these are just the convergent power series.

The noetherianness of the stalks expresses the GAGA wisdom that "local analytic geometry = (global) algebraic geometry".

Here's a problem: Describe the ideals in H(U).


trigonometry
inequalities
geometry
search
geometry unsolved

Let $ ABC$ be a triangle and $ P$ an arbitrary point in the plane. Let $ \alpha, \beta, \gamma$ be interior angles of the triangle and its area is denoted by $ F.$ Prove:
\[ \ov{AP}^2 \cdot \sin 2\alpha + \ov{BP}^2 \cdot \sin 2\beta + \ov{CP}^2 \cdot \sin 2\gamma \geq 2F
\]
When does equality occur?

Dose anybody have the solution of this problem?
Obviouosly equality occurs if P is circumcenter..

Hint:[hide]I think the proposers of the question expect you to use this theorem:

If G is the centroid of the n masses  $ m_1 , m_2 , ............m_n$    at points     $ A_1 ,A_2,...........A_n$ respectively, then

$ \sum m_r.(PA_r)^2$ -  $ \sum m_r.(GA_r)^2$ =$ (m_1 \plus{} m_2 \plus{} ....... \plus{} m_n)PG^2$ .

You could also use this inequality:

If P is a point in the plane of the triangle triangle ABC and x ,y ,z are any real numbers then

$ (x \plus{} y \plus{} z)(x.PA^2 \plus{} y.PB^2 \plus{} z.PC^2)$ $ \geq$ $ (yz.a^2 \plus{} zx.b^2 \plus{} xy.c^2)$

See http://www.mathlinks.ro/Forum/viewtopic.php?search_id=691839049&t=15558[/hide]


Princeton
college
FTW
University of Chicago
function
MIT
algorithm

I just want to see what everyone's opinion of the best colleges are.

This thread is based on a faulty premise.

I don't think consensus will be reached in any way or form on this poll...
Everyone has a different definition of "best" for colleges that could include factors such as cost, location, atmosphere, campus, students, professors, academics, student groups, etc.
Also, "Other" forms a big category, it's hard to list just 10 or so "best" colleges.
The college many people think is the best might not be on the poll, for example: Princeton, CalTech, UPenn (though probably not on a math forum as much), a host of international schools...

Baylor by far. I predict the poll isn't even close. Everyone should vote Baylor!

There is no 'best' college. There are colleges which are strong overall by some metrics, but this doesn't make them the best. If you are looking for a college and you are going for a college because people think highly of it, you're making a mistake. Everyone wants something different out of a college.

Furthermore, just because the faculty of a college is highly ranked doesn't mean its the best place to go. Many research universities get their rankings because their professors are good at [i]research[/i] not teaching. Of course the two tend to be positively correlated, but some researchers are shitty teachers.

[quote="qntty"]Of course the two tend to be positively correlated[/quote] Data?  (My personal guess is that research success is probably very mildly anti-correlated with teaching quality, but to actually carry out this experiment would be very difficult.)

Hehe jbl. Go to every college for a year!!! I'll be old then...  :sleeping:  :moose:

I have recently visited bayor, and i must admit that it was an amazing experience!!!!

Uh oh. Forgot about Princeton!  :blush:  :oops:

and what about Uchicago and cornell?  :lol:

If you want a list of numbers, pull out a copy of US News and World Report and read down the list. Rankings are a consistent, quantitative summary of many qualitative factors, rather than an inconsistent, qualitative summary of many qualitative factors.

But while rankings are useful for average students, they're not that useful for ambitious or proactive students, who are willing to seek out the best opportunities wherever they are. A particular higher ranked school might be like bending down and picking up a quarter. A particular lower ranked school might be like having the option to bend down and pick up a dime, or to walk three feet over and pick up a dollar bill.

Consider BU, for example. It get's shot down in the rankings, but there are [i]lots[/i] of opportunities for good students. They're great with research funding, for example. That, and they have lots of merit scholarships, which attract many very bright students whose parents might be working extra-hard to earn a good salary, only to be rewarded with no financial aid at the ivy leagues.

Take another example: Cornell consists of seven affiliated undergraduate colleges, some of which draw significantly less "competetive" applicant pools than others. This lowers its ranking, but for students, it provides diversity, a broad range of academic options, and classes/faculty/research in pretty much every conceivable field of study. One of the reasons I chose it over significantly higher ranked options (aside from them having a godly applied physics program).

In summary, what you should be asking is, "Do you know of anything particularly cool or interesting about any particular college, or any good/unique opportunities that I might want to know about, that are [i]not[/i] considered in USNWR?"

You are making at least two faulty assumptions in asking this question:

- Colleges quality can be described independent of your interaction with a college as a student.

- Colleges are well-ordered by quality.

The second assumption is easier to shoot down because it's just blatantly false.  Even if you accept the first assumption, [b]college quality is a vector.[/b]

The first assumption is more subtle, and you need to appreciate it.  What college will be the most appropriate for you is not a function on the set of colleges - it's a [b]relation[/b] between the set of students and the set of colleges.  Different students have different needs and will respond differently to certain environments.  Which environment is best for you is something that it is your responsibility to discover, and it may not necessarily be the case that that corresponds with whatever someone else thinks about the "best" colleges.

Case in point:  I have a friend in computer science who is currently attending UChicago.  While he was deciding where to go, he made a point not to let the glamor of MIT and other big-name colleges distract him and he studiously graded every college he was interested in on several criteria that were important [b]to him[/b] - and MIT ended up falling short.  He's now extremely satisfied at UChicago and I have no doubt that he made the right decision [b]for him.[/b]

I'll offer another analogy:  asking what the best college is to go to is like asking who the best woman is to marry.

[quote="t0rajir0u"]Even if you accept the first assumption, [b]college quality is a vector.[/b]
...
it's a [b]relation[/b] between the set of students and the set of colleges.[/quote]  You're such a nerd  :P  :lol: .  (And I agree entirely, except that I might argue that such a thing as "college quality" is not well-defined, for example by virtue of the second point.)

I just think that mathematical terminology is a great source of precise metaphors  :P

Well, if students have college preferences and colleges have student preferences, optimizing where the students go is a mathematical problem. If enough information is shared, there's even a decent algorithm or two to do it.

The consensus of the general public in the United States a few years ago, according to the Gallup Poll organization:

http://www.gallup.com/poll/9109/Harvard-Number-One-University-Eyes-Public.aspx

[quote="worthawholebean"]College quality is not only a vector; it is a vector function on the student.[/quote]
It's not even a vector function, it's just something multi-faceted that can only be measured subjectively and changes for each student.

[quote="t0rajir0u"]
What college will be the most appropriate for you is not a function on the set of colleges - it's a [b]relation[/b] between the set of students and the set of colleges.
[/quote]
Not sure what you mean.  Perhaps a function of several variables.

[quote="t0rajir0u"]
I'll offer another analogy:  asking what the best college is to go to is like asking who the best woman is to marry.[/quote]
That's a good one.  Only on this site would that just be an aside from the mathematical content of a post explaining real-life decisions.


Putnam Harvard college AMC USA(J)MO USAMO Princeton

the putnam results came back to my school today, in case you are curious:

This year, 3615 students competed in the exam.  The median
score was 2 out of 120, and students getting 8 points or
more (essentially getting most of one problem correct)
placed in the top third.

I got 26 and ranked 264th nationally. My team ranked 31st. Amazingly, even my inexcusibly low score put me first in my college. Our coach seemed very disappointed by our results.

[quote]Amazingly, even my inexcusibly low score put me first in my college.[/quote]

Hehe, so you beat that Jonathan Lewd (sp?) kid you were talking about a long while ago, huh? ;)

Yes, I beat him by quite a lot actually. 1/1 against him in college so far. (Actually his last name is Lowd.)

Jonathan Lewd ... hehehe

I want to get my results back!  Waa!

Okay, those were my forays into severe immaturity for the day.  I'll edit this whenever I find out how I did.

By the way, good job Simon!  You'll have quite a reputation by the time you're a senior at this rate, if you don't have it already :)

Thanks. What reputation will I have? People here don't make a big deal about the Putnam, and it's not really open for anyone to take it, so I doubt many people will know that I did ``well.'' Now I have to keep it up for four years.

[quote]I want to get my results back! Waa!

Okay, those were my forays into severe immaturity for the day. [/quote]

Geez, that's it for a day? Gosh Joel, you really gotta take after these fun-loving F&G'ers a tad bit more.  :mrgreen:

According to Joe Gallian, three perfect scores have been attained on the Putnam.

Okay, so I'll be getting my scores tomorrow, hopefully.  Just waiting for them to come in my e-mail box ... I do know I didn't get any special designation, though.  I'll have to practice for next year :)

[quote="tAsDoOm"]has anyone ever gotten a perfect score on this??[/quote]

No one in the last 10 years has gotten a perfect score. Someone came close last year with a score of 116 out of 120 (I presume it was either Barton or Caroll - but I don't know for sure). The highest score usually ranges from 85 to 105. But scores even coming close to perfect are very very rare. Even Ciprian Manolescu, who scored perfect (42) on the IMO three times, only scored over 100 once on the Putnam.

Joel, how do you get your score emailed to you? Is that just the way things work at Harvard? By the way, if you see David Simmonsduffin, tell him congratulations from me. Hopefully it won't scare him too much.

The scores of Putnam fellows this year were apparently 110, 96, 95, 92, and 82. One of {Gabe, Reid} did much better than the other one it seems.

There are ~100 people who sign up for the Putnam here and ~70 who take it, so Svetlana (in charge of these kinds of things) sends out and e-mail and you reply asking for your score.  I got a 41, which means I should go back and figure out which question I thought I got that I didn't, or something like that.  Rank 106.5.  Any idea how close I was to being Honorable Mention?

I'm afraid I don't know that David, sorry.

I think Honorable mention is top 60 or so. Congratulations on your 41! That's halfway to being a fellow.

hehe, that's exactly what I've been saying.  Thanks.

Hon. Mention this year was top 69, which means I would have needed 51 points -- one more problem.  So now, I just have to not fall out of practice on that type of math by next winter :)

[quote="ComplexZeta"]By the way, if you see David Simmonsduffin, tell him congratulations from me. Hopefully it won't scare him too much.[/quote]

How do you know David?  He's a roommate of a close friend of mine, so I can pass the congratulations on if you wish.

His mother and my mother were friends in college. The summer before seventh grade (for me; he is one year older), my family went to visit his family, and I was pleasantly surprised to find someone in that family who was also interested in math. So that was how I first met him. He came to California for a visit once a few years later, and then I ran into him at the USAMO central event in 2002. It's a small world, isn't it?

[quote="texas137"]Are there a lot of people who take the Putnam in high school? I have no doubt that there are high schoolers who could do well on it, but I thought people were limited to only taking it 4 times. Seems like they would want to save their eligibility for the college they are actually going to attend.[/quote]

I am not sure whether you know this, but the people on this thread who have taken it are all in college already (in case that was the reason you were asking this question).

I think it's pretty rare for high-school students to take the Putnam, for the eligibility reason you mentioned.  Strangely, the last year I took the test (20 years ago), our team consisted of two high-school kids and me.  So at age 18, I was the oldest.  One of the high-school kids (Brian Hunt) was top 10 that year, and the next year (after I left) was one of the Putnam Fellows.  Thanks mostly to him, our team ended up finishing 3rd my junior year and 5th my senior year.  I think Brian ended up jumping straight from high school to graduate school, so eligibility wasn't an issue for him.

[quote]I am not sure whether you know this, but the people on this thread who have taken it are all in college already (in case that was the reason you were asking this question).[/quote]

Thanks, that's exactly why I was asking. Someone's post made me think "high school" instead of "school"=college. And I know that sometimes high schoolers (like Gabe Carroll) take this. Thanks!

This has been a peev of mine for some time now - why do they only count the scores of the people officially named by the University for the Team Score, instead of just counting the 3 highest individuals from the school!!? Would it really be that hard to do??

It's time for them to change that. It drastically affects the order of the Universities in the Team Rankings, and the persons who scored the highest from each school should get the money.

Princeton didn't have its two highest scores (or maybe its first and third highest scorers) on the team this year, including one Fellow.  I think she may be only the second female Putnam Fellow, unless there's been someone else in the past 5 or 6 years.  You would think someone good enough to be a Putnam Fellow would have made it on the team, though -- especially over people who didn't make honorable mention.

Gauss, I think the present rule is designed to give small colleges a chance.  If we changed rules as you suggested, then colleges with lots of good students would have an advantage over colleges with say just 3 good students.

I'm not advocating one system over the other.  I'm just stating the rationale I have heard for the present rules.

Things like that happen every year under this current system. It's an embarrassment that mathematicians would allow a system as flawed as this to continue for so long.

Princeton doesn't usually take the Putnam exam very seriously. I don't think they do much in the way of any organized preparation, and the official members aren't chosen in any strategic sort of way. They're usually Juniors or Seniors who are known in the department or who did well in previous years. Both Stefen and Mihai are former Putnam Fellows.

Ana (though she was a 3 time IMO medalist) is only a freshman. That's probably why she was not chosen as an official member. Knowing how the Princton Math Department works, it's not unlikely that the person who chose the official team members had no idea who she was until now.

Joel, I think there have been 3 female Putnam Fellows:

Ioana Dumitriu (NYU), 1996
Melanie Wood (Duke), 2002
Ana Caraiani (Princeton), 2003

I remember when Ioana won.  She got a lot of press attention (such as in the New York Times).  We at NYU were quite proud of her.

Yes, that's right -- I should have remembered Melanie.

I like the way Putnam teams are formed. How would small schools have any chance at all against big giants like Harvard, MIT, etc. if the team was just the three highest people? A team with a few good people wouldn't be able to do well against a team with a lot of people, but with the teams ranked as they are, the two teams could compete equally. I like it much better that way. I wonder what would happen if they used that ranking method for the AMC12. Perhaps TJ and AAST and Stuy and Exeter and IMSA wouldn't always win.

It may help smaller schools, but it doesn't do so in a reasonble way. I think the fact that they only count 3 scores, should in itself give smaller schools a reasonable chance.

Right now, by ignoring some of the higher scores, the Team Score isn't an accurate reflection of how the school did. Should a school really want to win just because they picked the right three members to count, or because their top three were really higher than the top 3 from another school? Schools should also be rewarded for having high participation levels - which increases your chances of having 3 good performances.

How do you all know eachother? You talk about these fellows and other people as if you have grown up with them. Is there some sort of math social club that you are all a part of?? :) Anyways, where do you all go to college? Good job on your scores guys.

Scm -- I think the Putnam Fellows are a group of the (five?) top scorers on the exam.  By the way -- How do you get ranked 106.5?  Was there a tie?

Ian, you are correct about what the Putnam fellows are.  There are a lot of ties.  If there are a number of tied people, they rank them all with the same rank, equal to the median rank that they would have.  (In other words, the people who got 46 were either the 106th and 107th, or 105th, 106th, 107th and 108th, or so on.)

As for knowing each other, I don't think I've personally ever met any of those three women, but there is a very good math-contest network.  Names get around, people get to know each other from contests and the web.  Just as an example, Yan Zhang, a member of the US IMO team last year, sat down at a table across from me at the beginning of the year.  We chatted briefly and then he said, "You're Joel Lewis, right?"  And I said "Yes, and aren't you Yan Zhang?"  So, without ever having actually met each other, we had managed to pick up (probably from ARML) who we actually were.

hmm...rumor has it that Ana is the top scorer with the 110, which if true is pretty unbelievable that Reid and Gabriel were beaten.

It's not so unbelievable that [i]someone[/i] scored higher than they did; what's most impressive is that she's only a first year University student.

[quote="ComplexZeta"]Where did you hear that rumor? I thought that the actual scores of the top 5 were not given out.[/quote]

There's a thread on mathlinks about that. Valentin was supposedly told by Ana that she thought she got 11 questions when she finished taking the exam.

They're not. Someone who knows her said that she said she got 11 right.


limit
logarithms
inequalities
probability
integration
binomial coefficients
real analysis

It's crazy! I'm trying to prove that

$\lim_{k\to \infty}\sum_{2^{x}< 3^{y}}\frac{\binom{x+k}{y}}{2^{k}3^{y}}> 0$

Also, I have a gut feeling that it's pretty close to $2/3$.

Edit: yes, "it" refers to the limit.

it's your limit?
If I understand everything right then your answer is wrong. Answer is $+oo$
because
your limit we can rewrite us $\sum_{x=0}^{+oo}\sum_{y>x\frac{\ln{2}}{\ln{3}}}\frac{(x+k)(x+k-1)...(k-y+1)}{2^{k}3^{y}y!}=\frac{D}{2^{k}}\sum_{y=0}^{+oo}\frac{1}{y!3^{y}}+A$ where $A$=$\sum_{x=1}^{+oo}\sum_{y>x\frac{\ln{2}}{\ln{3}}}\frac{(x+k)(x+k-1)...(k+1)}{2^{k}3^{y}y!}>0$ and $D$ we get when we consider this expression $\frac{k(k-1)...(k-y+1)}{y!}$easy seen that when $y\to+oo$ then expression $k(k-1)...(k-y+1)$ is like $k!$ therefore $\lim_{k\to+oo}\frac{k!}{2^{k}}=+oo$ and this mean that $D\to+oo$ when $k\to+oo$
therefore we get that your sum is more than $Dconst$  where $const=\sum_{y=0}^{+oo}\frac{1}{y!3^{y}}$. after is simpl.

I think this is doable. Here is one idea. Note that $\binom nk$ is about $2^{n}/\sqrt n$ if $\left|k-\frac{n}{2}\right|<\sqrt n$. This tells you that you have about $\sqrt k$ pairs of indices $(x,y)$ where the binomial coefficient is comparable to $2^{x+k}/\sqrt k$ and $2^{x}$ is comparable to $3^{y}$ (just solve the corresponding system of linear inequalities to see it) . They give you a contribution bounded from below by some number independent of $k$. If you want to find the limit or just show that it exists, you have to do much more work though. I'll leave the topic at this stage now because it'll take me quite a lot of time to run an actual computation of such a limit and right now I don't have this time. Hope, this helps a bit, nevertheless. :)

What are you going to do with your limit anyway and how is it relater to Collatz conjecture?

It should be around $2/3$ of the density of the inverse iterates of $4$ under the Collatz map (making lots of approximations)

Maybe it should be $8/3$ or $1/6$ or something, I didn't think about the exact value too much, I'm just trying to prove it's nonzero. It's also related to the density of the inverse iterates of ANY number, and that's where being positive is important.

Well, if that is divergent, then either I made a mistake or something REALLY pathological happens to my approximation, because "on average" the density of the inverse images of a number $k$ ought to be $c/k$ times that sum, where $c$ doesn't vary so much. And obviously, densities shouldn't be... well... more than 1, let alone infinite.

My approximation was that the number of inverses of "type" $(a,b)$ (for $3$ not dividing $k$) is on average $\frac{\binom{a+b}{a}}{2*3^{b-1}}$, where $a$ is the number of moves from $n$ to $n/2$ and $b$ is the number of moves from $n$ to $\frac{3n+1}{2}$.

Two news for zeb: one good, one bad. The good one is that the limit exists and is finite. The bad one is that it is well above $1$. Here is a "back of envelope computation" I tried on a long car drive. I believe it is correct but justification of every single detail will require a much longer post. First of all, note that $\binom{x+k}y\le 2^{x+k}$ and $2^{x}\le 3^{y}$ so the terms in the series never exceed $1$ and decay fairly fast if $y$ deviates much from $\frac{x+k}2$ and from $y(x)$ where $y(x)$ is the least integer for which $3^{y}>2^{x}$. Thus, we may consider only pairs $x$ and $y$, for which the equalities $y(x)\approx \frac{x+k}2$, $y\approx y(x)=[\eta x]+1$ hold with reasonable precision (here $\eta=\frac{\log 2}{\log 3}$). In this range both $x$ and $y$ are comparable to $k$ . Second, let us recall that, by the Central Limit Theorem $\binom nm \approx 2^{n}\sqrt{\frac 2{\pi n}}e^{-2(m-\frac{n}{2})^{2}/n}$ or something like that when $m\approx\frac{n}{2}$ (if I misplaced $2$ or $\pi$ somewhere, it won't matter in the slightest for the computation). Now, $n=x+k$ stays essentially fixed in our range and the binomial coefficients $\binom{x+k}y$ change fairly slowly when we change $y$ within our range. It means that the sum over $y$ for fixed $x$ is about $2^{-k}\binom{x+k}{y(x)}\sum_{y\ge y(x)}3^{-y}=\frac{3}{2}2^{-k}3^{-y(x)}\binom{x+k}{y(x)}$. Using the above formula for the binomial coefficient, we see that it is $2^{x+k}$ times the "normal distribution term", in which we essentially have the fixed $n$ in it and the difference $y(x)-\frac{x+k}2$, roughly speaking, increases with step $\eta-\frac{1}{2}$ ($y(x)\approx\eta x$, remember?). Since $2^{x+k}2^{-k}3^{-y(x)}=2^{x}3^{-y(x)}$, and the sum with step $1$ would give us $1$ (total probability), the sum with step $\eta-\frac{1}{2}$ would give $\frac 2{2\eta-1}$ if not for the factor $2^{x}3^{-y(x)}$. But this factor jumps up and down fairly fast and its average value is $\int_{0}^{1}3^{-t}\,dt=\frac 2{3\log 3}$ (here we use the fact that integer multiples of $\log 2$ are uniformly distributed $\mod \log 3$). Bringing all the factors together, we get $\frac{3}{2}\frac 2{3\log 3}\frac 2{2\eta-1}=\frac 2{\log 4/3}\approx \frac 2{\frac{1}{3}-\frac 1{18}}=\frac{36}5=7.2$ for the limit. I wonder how this prediction of mine agrees with numerical computations.

[quote="fedja"][quote="Extremal"]easy seen that when $y\to+oo$ then expression $k(k-1)...(k-y+1)$ is like $k!$ [/quote]
Extremal, think before you post! :) First of all, you obviously meant something else to say and, second, don't forget that when you take large $x$ you are forced to take large $y$.[/quote]
It' was mechanicaly mistake.  :oops:

Alright, looks like I ought to explain where I got the sum from, so we can figure out which constant that was supposed to be :)

Assume $3$ doesn't go into $k$. Let $f(k)$ be the density of the inverse iterates of $k$ in the natural numbers, and let $p_{k}(a, b)$ be the number of inverses of $k$ which use exactly $a$ steps of the form $x \to x/2$ and $b$ steps of the form $x \to \frac{3x+1}{2}$ before reaching $k$. As I stated before, if we keep $a, b$ constant and vary $k$, the average value of $p_{k}(a, b)$ is $\frac{\binom{a+b}{a}}{2*3^{b-1}}$. Thus,

$f(k) \approx \lim_{n\to \infty}\sum_{k \frac{2^{a+b}}{3^{b}}\le n}\frac{p_{k}(a,b)}{n}= \frac{1}{k}\lim_{n\to \infty}\sum_{\frac{2^{a+b}}{3^{b}}\le n}\frac{p_{k}(a,b)}{n}\approx \frac{1}{k}\lim_{n\to \infty}\sum_{\frac{2^{a+b}}{3^{b}}\le n}\frac{\binom{a+b}{a}}{2n3^{b-1}}= \frac{3}{2k}\lim_{n\to \infty}\sum_{\frac{2^{a+b}}{3^{b}}\le n}\frac{\binom{a+b}{a}}{n3^{b}}= \frac{3}{2k}\lim_{n\to \infty}\sum_{\frac{2^{a+b}}{3^{b}}\le 2^{n}}\frac{\binom{a+b}{a}}{2^{n}3^{b}}= \frac{3}{2k}\lim_{n\to \infty}\sum_{2^{a+b-n}\le 3^{b}}\frac{\binom{a+b}{b}}{2^{n}3^{b}}\approx \frac{3}{2k}\lim_{n\to \infty}\sum_{2^{x}\le 3^{b}}\frac{\binom{x+n}{b}}{2^{n}3^{b}}$.

Let $\delta = \lim_{n\to \infty}\sum_{2^{x}\le 3^{b}}\frac{\binom{x+n}{b}}{2^{n}3^{b}}$, then we expect $f(4) = f(8) = 1 \approx \frac{3}{2*4}\delta \approx \frac{3}{2*8}\delta$, so $\delta \approx \frac{8}{3}\approx \frac{16}{3}$. Hm, writing this up made me realize that $x$ could be slightly negative, which makes the approximation even worse...


algebra
system of equations
real analysis
real analysis unsolved

Show that the initial-value problem
$\frac{dx}{dt}=-x(x^{2}+y^{2})$; $\frac{dy}{dt}=-y(x^{2}+y^{2})$; $x(0)=x_{0}$ and $y(0)=y_{0}$ has a solution for all $t\geq 0$, but not for all $t<0$, unless $(x_{0},y_{0})=(0;0)$.

put $r(t)=\sqrt{x(t)^{2}+y(t)^{2}}$ and $\theta(t) = \arctan(\frac{y}{x})$. You will obtain a very simple system of equations, if my computations were rights ..

$z=x+iy$

then (1) $z'=-|z|^{2}.z$

Consider $u=|z|^{2}$ then find a differential equation of order one satisfy by $u$ solve it . Plug the expression of $u$ in (1),  it is easy to solve
It will give you the expression $x(t)$ and $y(t)$

In fact two solutions are the same. Thank you :)


search
combinatorics unsolved
combinatorics

1.into how many parts at most is a plane cut by n lines?
2.into how many parts is space divided by n planes in general position?

no reply?please tell me if it has been posted before.

Yes, it has (both parts), but I don't know what the best thing to search for is.

[quote]Yes, it has (both parts), but I don't know what the best thing to search for is[/quote]please give me the link then.

1.$ n \plus{} 2$ parts
2.$ n(n\plus{}2)$ parts

In order to find a link, I would need to know what to search for.  Here are a couple of old threads with discussions, but I'm sure there have been better and more recent ones, if you're willing to spend some time trying well-chosen searches.
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=514477548&t=151093
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=514477548&t=164902

It is hopefully unnecessary to point out that the answers suggested by the preceding poster are just completely wrong.  I don't even think it's possible to divide a plane into $ n \plus{} 2$ pieces with $ n$ lines for any $ n \neq 2$.


[url=http://www.image-share.com/image.php?img=77/299.jpg][img]http://www.image-share.com/upload/77/299m.jpg[/img][/url]

I'm having hard time solving this. Can someone give me some tips for getting started?

Hi huismjar.

$ (1\plus{}x)^{2n} \equal{} \binom{2n}{0} \plus{} \binom{2n}{1}x \plus{} ... \binom{2n}{n}x^{n} \plus{} ... \plus{} \binom{2n}{2n}x^{2n}$.

$ (1\plus{}x)^{n}\cdot (1\plus{}x)^{n} \equal{} (\binom{n}{0} \plus{} \binom{n}{1}x \plus{} ... \plus{} \binom{n}{n\minus{}1}x^{n\minus{}1} \plus{} \binom{n}{n}x^{n})^{2}$

$ \equal{}\binom{n}{0}\binom{n}{0} \plus{}[\binom{n}{0}\binom{n}{1} \plus{} \binom{n}{1}\binom{n}{0}]x \plus{}... \plus{}[\binom{n}{0}\binom{n}{n} \plus{} \binom{n}{1}\binom{n}{n\minus{}1} \plus{} ... \plus{} \binom{n}{n}\binom{n}{0}]x^{n} \plus{} ... \plus{}\binom{n}{n}\binom{n}{n}\cdot x^{2n}$.

Therefore, $ \binom{2n}{n} \equal{} \binom{n}{0}\binom{n}{n} \plus{} \binom{n}{1}\binom{n}{n\minus{}1} \plus{} ... \plus{} \binom{n}{n\minus{}1}\binom{n}{1} \plus{} \binom{n}{n}\binom{n}{0}$

$ \equal{}\binom{n}{0} ^{2} \plus{} \binom{n}{1}^{2} \plus{} ...\plus{} \binom{n}{n\minus{}1}^{2} \plus{} \binom{n}{n}^{2}$.

It's way WAY cleaner to prove it combinatorially.

[hide="Hint"]
What does the right hand side, $ \binom{2n}{n}$, mean combinatorially? Think of it in terms of choosing some objects out of a set of objects.

How can we count the number of ways to do that if the set of objects gets split into two equal sized sets of objects?
[/hide]

[hide="Answer; try it yourself before reading this"]
Let's take a pile of $ 2n$ objects and split it up into two piles of $ n$ objects. We now would like to choose $ n$ objects out of these two piles.

Suppose we take $ k$ objects out of the first pile. We'd obviously have to take $ n \minus{} k$ out of the second pile to get $ n$ objects. This can be done in $ \binom{n}{k}\binom{n}{n \minus{} k}$ ways. A well known (and easily proven) identity states that $ \binom{n}{k} \equal{} \binom{n}{n \minus{} k}$, so this value is $ \binom{n}{k}^2$.

If we sum over all the values of $ k$ possible (which is clearly from $ 0$ to $ n$), we get the total number of ways to choose $ n$ objects out of $ 2n$ objects.

However, the number of ways to do that is obviously $ \binom{2n}{n}$, and the conclusion follows.
[/hide]

A more general solution, but somewhat long
[hide]

You should note that $ \binom{n}{m} \plus{} \binom{n}{m\plus{}1} \equal{} \binom{n\plus{}1}{m\plus{}1}$

Expanding on this property, we can try to simplify:
$ \binom{n}{m}\binom{2}{0} \plus{} \binom{n}{m\plus{}1}\binom{2}{1} \plus{} \binom{n}{m\plus{}2}\binom{2}{2}$
And we get
$ \binom{n}{m} \plus{} 2\binom{n}{m\plus{}1} \plus{} \binom{n}{m\plus{}2}$
And we can break it up like so:
$ (\binom{n}{m} \plus{} \binom{n}{m\plus{}1}) \plus{} (\binom{n}{m\plus{}1} \plus{} \binom{n}{m\plus{}2})$
$ \binom{n\plus{}1}{m\plus{}1} \plus{} \binom{n\plus{}1}{m\plus{}2}$
$ \binom{n\plus{}2}{m\plus{}2}$

Using a little recursion we can do the same for:
$ \binom{n}{m}\binom{3}{0} \plus{} \binom{n}{m}\binom{3}{1}\binom{n} \plus{}{m\plus{}2}\binom{3}{2} \plus{} \binom{n}{m\plus{}3}\binom{3}{3}$
From what we know about the $ \binom{3}{x}$'s, we can expand these and collect two sets of $ \binom{2}{a}$'s.
$ (\binom{n}{m}\binom{2}{0} \plus{} \binom{n}{m\plus{}1}\binom{2}{1}\plus{} \binom{n}{m\plus{}2}\binom{2}{2}) \plus{} (\binom{n}{m\plus{}1}\binom{2}{0} \plus{} \binom{n}{m\plus{}2}\binom{2}{1} \plus{} \binom{n}{m\plus{}3}\binom{2}{2})$
And the left and right we use our formula, and we get:
$ \binom{n\plus{}2}{m\plus{}2} \plus{} \binom{n\plus{}2}{m\plus{}3}$
And we bring this to:
$ \binom{n\plus{}3}{m\plus{}3}$

I don't want to include a full proof here, but basically you use induction.
Just prove the first equation, and then show the expansion step (where we went from $ \binom{3}{a}$ to two $ \binom{2}{a}$ and just show how $ \binom{n}{a}$'s go to $ \binom{n\minus{}1}{b}$'s.

Now because of what was proven above, you can show the sum of squares is a case of this general form, and you get your identity.

[/hide]


[b]Prove that the sum of the reciprocals of [u][i]every[/i][/u] factor of a perfect number sum to 2[/b]

[hide="Proof"]For some perfect number $ P$, we have $ P = f_1\cdot f_{n} = f_2\cdot f_{n - 1} = \ldots$

Note that $ f_n = P$.

By definition, $ \sum_{k = 1}^{n}f_k = 2P$.

\begin{align*}S & = \frac {1}{f_1} + \frac {1}{f_2} + \ldots + \frac {1}{f_{n - 1}} + \frac {1}{f_n} \\
 \\
& = \frac {f_n}{P} + \frac {f_{n - 1}}{P} + \ldots + \frac {f_2}{P} + \frac {f_1}{P} \\
 \\
& = \frac {2P}{P} \\
 \\
& = 2\end{align*}[/hide]

Why do you post all of your problems in bold?

[quote="t0rajir0u"]See http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=132982 .[/quote]
darn it torajiru.  :lol:  i thought i had made a creative problem for once. oh well, back to the drawing board.
@carl, it's a signature thing. i don't see why that annoys you so much.  :)


limit
integration
trigonometry
calculus
calculus computations

Compute \[\lim_{n \to \infty}\int_{n}^{n+1}\left\{ x^{2}\cdot \sin \frac1{x}\right\}\, dx .\]

The limit is $\frac12.$

Let $f(x)=x^{2}\sin\left(\frac1x\right).$

We can write $f(x)=x-g(x)$ where $g(x)>0$ and $\lim_{x\to\infty}g(x)=0.$ We also claim that $f$ is eventually monotone increasing.

The is mostly based on this: for $\theta>0,\ \theta-\frac{\theta^{3}}6<\sin\theta<\theta.$

For sufficiently large $n,$ $f(n)=n-g(n)<n<f(n+1)=n+1-g(n+1).$

This shows that $f(x)=n$ has a solution in $(n,n+1).$ By the monotonicity of $f,$ this is unique. Write this solution as $x=n+\alpha_{n}.$

Note that $f(n+\alpha_{n})=n+\alpha_{n}-g(n+\alpha_{n})$ so that

$\alpha_{n}=g(n+\alpha_{n}).$ We conclude that $\lim_{n\to\infty}\alpha_{n}=0.$

Now we can write the integral as

$\int_{n}^{n+1}\{f(x)\}\,dx=\int_{n}^{n+\alpha}f(x)-(n-1)\,dx+\int_{n+\alpha}^{n+1}f(x)-n\,dx$

$=\int_{n}^{n+1}f(x)-n\,dx+\int_{n}^{n+\alpha}1\,dx$

$=\int_{n}^{n+1}(x-n)\,dx-\int_{n}^{n+1}g(x)\,dx+\alpha_{n}.$

But the first integral is equal to $\frac12,$ while

$\int_{n}^{n+1}g(x)\,dx\to0$ and $\alpha_{n}\to 0.$


conics
hyperbola
geometry
geometric transformation
reflection
circumcircle
perpendicular bisector

Circles $C(O_1)$ and $C(O_2)$ intersect at points $A$, $B$. $CD$ passing through point $O_1$ intersects $C(O_1)$ at point $D$ and tangents $C(O_2)$ at point $C$. $AC$ tangents $C(O_1)$ at $A$. Draw $AE \bot CD$, and $AE$ intersects $C(O_1)$ at $E$. Draw $AF \bot DE$, and $AF$ intersects $DE$ at $F$. Prove that $BD$ bisects $AF$.

Comment:

   It comes as a pleasant surprise to me that in Western part of China they do

   know about  Apollonious theorem  concerned with a  tangent to hyperbola!

   Thank you (Chairman Mao?).

   M.T.

I solved it,but I use at least 4 hours!!!
Oh,my God.

What's that?? I mean, stealing problems from other olympiads is OK if one has no phantasy; proposing ugly but original problems is also halfways OK, as a matter of honor, but stealing ugly problems from other sources sux, particularly if the other source is the ISL 1998.

To make it short: The problem is equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=18215 . Finding out why is not that easy, since the proposers at the CWMO apparently tried everything to make the situation as confusing as possible, so here are the details:

First let me rewrite the problem in a constructive way:

[color=blue][b]Problem.[/b] Let $k_1$ be a circle with center $O_1$, and let D and A be two points on this circle.

The tangent to the circle $k_1$ at the point A intersects the line $O_1D$ at a point C.

The perpendicular to the line $O_1D$ through the point A meets the circle $k_1$ at a point E (apart from A).

Let F be the feet of the perpendicular from the point A to the line DE.

Let $k_2$ be the circle through the points A and C which touches the line $O_1D$ at the point C. (You can easily see that there is only one such circle: The center of this circle must lie on the perpendicular bisector of the segment AC - since the segment AC is a chord of this circle - and on the perpendicular to the line $O_1D$ at the point C - since the circle touches the line $O_1D$ at the point C; hence, the center of this circle is uniquely determined as the point of intersection of these two lines; moreover, since we know a point on the circle - namely, the point A -, the radius is uniquely determined, too, so there is only one such circle.)

Let B be the point of intersection of the circles $k_1$ and $k_2$ (different from A).

Prove that the line BD bisects the segment AF.[/color]

[i]Solution.[/i] Let Y be the midpoint of the segment AF. Then, we want to show that the line BD passes through this point Y.

Let G be the point diametrically opposite to the point D on the circle $k_1$, i. e. the reflection of the point D in the center $O_1$ of this circle. Then, the segment DG is a diameter of the circle $k_1$, and thus < DAG = 90°.

Since the segment AE is a chord of the circle $k_1$, the perpendicular bisector of this segment AE must pass through the center $O_1$ of this circle $k_1$. In other words, the perpendicular bisector of the segment AE is the perpendicular to the line AE through the point $O_1$. But the perpendicular to the line AE through the point $O_1$ is the line DG (since the line DG coincides with the line $O_1D$, and $AE\perp O_1D$). Hence, the perpendicular bisector of the segment AE is the line DG. In other words, the point E is the reflection of the point A in the line DG.

Now, let the line DY intersect the circle $k_1$ at a point B' (apart from D). In order to solve the problem, it will be enough to prove that B' = B. In fact, the point B' is known to lie on the line DY; in other words, the line B'D passes through the point Y; hence, once it will be shown that B' = B, we will conclude that the line BD passes through the point Y, and the problem will be solved.

Now, consider our situation from another viewpoint: We have a triangle ADG with < DAG = 90°. Its circumcircle is $k_1$, and the tangent at the point A to this circumcircle $k_1$ intersects the line DG at the point C. The point E is the reflection of the point A in the line DG. The point F is the foot of the perpendicular from the point A to the line DE. The point Y is the midpoint of the segment AF. The line DY intersects the circle $k_1$ at a point B' (apart from D).

Hence, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=18215 , the line DC (i. e., the line $O_1D$) is tangent to the circumcircle of triangle ACB'. Thus, the circumcircle of triangle ACB' is the circle through the points A and C which touches the line $O_1D$ at the point C. Hence, the circumcircle of triangle ACB' is simply the circle $k_2$. Consequently, the point B' lies on the circle $k_2$. Thus, the point B' is the point of intersection of the circles $k_1$ and $k_2$ (different from A), and this yields B' = B, so the problem is solved.

  Darij

My solution:

Let $ M=AE \cap CD , G=AF \cap BD $ .

Easy to see $ M $ is the midpoint of $ AE $ .

Since $ \angle BCM=\angle BAC=\angle BEM $ ,
so we get $ B, C, E, M $ are concyclic ,
hence $ \angle BMA=90-\angle CMB=90-\angle CEB=90-\angle EDB=\angle BGA $ .
ie. $ A, B, G, M $ are concyclic
Since $ \angle AMG=\angle ABD=\angle AED $ ,
so we get $ MG \parallel DE $ . ie. $ AG=GF $

Q.E.D


Gauss
polynomial
group theory
abstract algebra
algebra
function
domain

[hide="Hint"]Gaussian Integers in the ring of omega, where omega is the third root of unity.[/hide]

Alternately, it can be written as an elliptic curve in Weierstrass form, and then SAGE or similar software can tell you that it's a curve of rank zero with a trivial torsion subgroup :D

i don't understand how gaussian integers can be in the ring of omega. Isn't omega an element of $ Q[\sqrt{\minus{}3}]$ while the gaussian integers are in $ Q[i]$?

[quote="Riders on the Storm"]prove it for exponent three:

x^3+y^3=z^3 has no nonzero solutions.[/quote]
I have found it: $ x\equal{}y\equal{}1$ and $ z\equal{}\sqrt[3]2.$ :maybe:  :D  :)

I think a relevant condition that is missing from the problem statement is $ x, y, z\in\mathbb{Z}$...

[quote="arqady"][quote="Riders on the Storm"]prove it for exponent three:

x^3+y^3=z^3 has no nonzero solutions.[/quote]
I have found it: $ x \equal{} y \equal{} 1$ and $ z \equal{}\sqrt [3]2.$ :maybe:  :D  :)[/quote]

Well, sorry for the misunderstanding, but I thought the title "fermat's last theorem" would suffice to show that we are talking about diophantine equations.

By the way, silow theory- did you mean lemmas?

[quote="Riders on the Storm"]
By the way, silow theory- did you mean lemmas?
[/quote]
Yes, I mean [b]lemmas[/b] not [i]lemons[/i].

My point, is that using $ \mathbb{Z}[\omega]$ is the way to go, as mentioned, but it is still very hard. You need to knew several properties about this domain.

[b]Lemma 1:[/b] The ring $ \mathbb{Z}[\omega]$ is an [i]Euclidean Domain[/i].

[b]Lemma 2:[/b] For $ p\in\mathbb{Z}$ a prime the following are equivalent: (a)$ p$ is irreducible (prime) in $ \mathbb{Z}[\omega]$, (b)$ x^{2}\plus{}x^{2}\plus{}1$ is an irreducible polynomial in $ \mathbb{Z}_{p}[x]$, (c)$ p\equiv 2(\bmod 3)$.

[b]Lemma 3:[/b] $ \mathbb{Z}[\omega]/\left< 1\minus{}\omega\right>\simeq\mathbb{Z}_{3}$.

[b]Lemma 4:[/b] Any solution $ x,y,z\in\mathbb{Z}[\omega]$ so that $ x^{3}\plus{}y^{3}\equal{}z^{3}$ then one of $ x^{3},y^{3},z^{3}$ is divisible by $ (1\minus{}\omega)^{4}$.

Now you factor,
$ (x^{3}\plus{}y^{3})\equal{}(x\plus{}y)(x\plus{}\omega y)(x\plus{}\omega^{2}y) \equal{} z^{3}$.
And apply unique factorization arguments....
(Easier said than done).

Sorry, but can anyone elaborate on their ideas? So far only silow theory has given more than a couple of ideas.

Hi Everyone,

Thanks very much for referencing my site.

In addition to the elementary proof for n=3:

[url]http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html[/url],

I also include a proof based on Eisenstein integers:

[url]http://fermatslasttheorem.blogspot.com/2005/07/proof-for-n3-using-eisenstein-integers.html[/url]

Since I am a mathematical amateur, all proofs are taken from books written by experts.  The proofs are valid but it is very possible to find mistakes in my understanding of the proof.

Even though I am math amateur, I am a professional in computer science so perhaps I am not a complete amateur.  I leave it to your judgment if you look at my site.

I applaud this web site and your curiosity toward Fermat's Last Theorem.

Feel free to post comments any time on my blog if you find any mistakes, inaccuracies, or claims that you believe require additional justification.

I am very glad to encourage math students to pursue the subject more deeply.

Cheers,

-Larry Freeman
[url]http://fermatslasttheorem.blogspot.com[/url]

The proof using the eisenstein integers isn't that bad at all.
[hide]
$ x^{3}\equal{}z^{3}\minus{}y^{3}$
$ x^{3}\equal{}(z\minus{}y)(z\minus{}y\omega)(z\minus{}y\omega^{2})$
By the Euclidean algorithm all the terms on the RHS are relatively prime. They also are all perfect cubes.
$ z\minus{}y\omega \equal{} (a\plus{}b\omega)^{3}$
$ z\equal{}a^{3}\plus{}b^{3}$
$ y\equal{}\minus{}3a^{2}b\omega\minus{}3ab^{2}\omega$
Clearly, $ 3|y$. By symmetry, 3|x also. Mod 3, 3|z. By infinite descent, no nontrivial solutions.
[/hide]

[quote="sylow_theory"]The proof using $ \mathbb{Z}[\omega]$ has no nontrivial solutions for $ x^{3}\plus{}y^{3}\equal{} z^{3}$ is [i]really reallly[/i] nasty. Something one is not expeted to come up with in a course of a few days. It has a tonful of lemmons.

[url=http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html]Here's a more elementary approach[/url]. (To be honest I didnt have time to check and read it, so I dont know if this is useful).[/quote]
Uhhh, at Awesomemath we started learning the most basic algebraic number theory (no one had ever seen it) and an hour and a half later we were given this problem. It was solved by one of the students in under 10 minutes.

Larry, this is awesome. Thanks a lot.

This could be the same, or an alternate approach, I don't know. But would it make a difference if instead of taking step 3 (in Larry Freeman's post) we instead factored $ x^{3}\plus{}y^{3}$ into $ (x\plus{}y)(x\plus{}\omega y)(x\plus{}\omega^{2}y)$?

Can we use infinite descent on this? If so, please explain how.

Thanks.

By the Euclidean algorithm, $ \gcd(z\minus{}y,z\minus{}y\omega) \equal{}\gcd(z\minus{}y,y(1\minus{}\omega))$ Since $ \gcd(z,y) \equal{} 1$, $ \gcd(z\minus{}y,y(1\minus{}\omega)) \equal{}\gcd(z\minus{}y,1\minus{}\omega)$. But $ 1\minus{}\omega$ is a Eisenstein prime because it's norm is 3, a prime. and since the only real multiple of $ 1\minus{}\omega$ is 0, $ \gcd(z\minus{}y,1\minus{}\omega) \equal{} 1$ I think you can do something similar for the others.

[quote="diophantient"]
$ x^{3}\equal{} (z\minus{}y)(z\minus{}y\omega)(z\minus{}y\omega^{2})$
By the Euclidean algorithm all the terms on the RHS are relatively prime. They also are all perfect cubes.
[/quote]

could you explain how the Euclidean algorithm is used to explain this?

Hi SplashD,

Thanks very much for your comments.  There is one mistake in your reasoning.

You write:

[quote]But 1-w is a Eisenstein prime because it's norm is 3, a prime. and since the only real multiple of 1-w is 0, gcd(z-y,1-w) = 1[/quote]

But this is not correct since (1 - w) divides 3.

This can be shown since:

(1 - w)(1 - w^2) = 1 - w - w^2 + 1 = 2 + -w + -w^2

Now, we know that since w is a primitive root of unity:

1 + w + w^2 = 0

[In my blog, I show the proof for this at:  see Lemma 2 at [url]http://fermatslasttheorem.blogspot.com/2006/05/basic-properties-of-cyclotomic.html[/url]

This means that:

1 = -w -w^2

So that we have (1 - w)(1 - w^2) = 2 + 1 = 3

-Larry

Also notice that the norm of $ a\plus{}b\omega$ is $ (a\plus{}b\omega)(a\plus{}b\omega^{2})$, so the number always divides its norm.

We have to show that it's impossible for $ 3$ to divide $ z\minus{}y$. This is equivalent to $ z\equiv y\pmod{3}$. Well let $ z \equal{} 3k\plus{}a$, $ y\equal{}3j\plus{}a$. $ (3k\plus{}a)^{3}\equal{} 27k^{3}\plus{}27k^{2}a\plus{}9ka^{2}\plus{}a^{3}$. Thus, $ z^{3}\equiv y^{3}\pmod{9}$. Thus, $ x^{3}\equiv 0\pmod{3}\implies x^{3}\equiv 0\pmod{27}\implies z^{3}\equiv y^{3}\pmod{27}$, or $ 9ka^{2}\minus{}9ja^{2}\plus{}a^{3}\minus{}a^{3}\equiv 0\pmod{27}\implies ka^{2}\minus{}ja^{2}\equiv 0\pmod{3}$. If $ a\equiv{0}\pmod{3}$, we're done, so we have $ k\minus{}j\equiv 0\pmod{3}$, or $ z\equiv y\pmod{9}$. Now, $ z \equal{} 9k\plus{}a$ (different $ k$, $ a$) and $ y\equal{} 9j\plus{}a$. $ (9k\plus{}a)^{3}\minus{}(9j\plus{}a)^{3}\equal{} 729(k^{3}\minus{}j^{3})\plus{}243a(k^{2}\minus{}j^{2})\plus{}27a^{2}(k\minus{}j)$.

There should be some way to get a contradiction...but I'm not seeing this.

In case it helps anyone, here's a similar proof for all pythagorean triples (very short)
$ x^{2}\plus{}y^{2}\equal{} z^{2}$
$ (x\plus{}iy)(x\minus{}iy)\equal{}z^{2}$
$ (x\plus{}iy, x\minus{}iy) \equal{} (x\plus{}iy, 2iy)$ which divides $ 2 \equal{} (1\plus{}i)(1\minus{}i)$. Notice that $ (1\plus{}i)(a\plus{}bi) \equal{} (a\minus{}b)\plus{}(a\plus{}b)i$ and $ (1\minus{}i)(a\plus{}bi) \equal{} (a\plus{}b)\plus{}(b\minus{}a)i$. So, if the gcd is not $ 1$, then $ x\equiv y\pmod2$, but this requires that they are both $ 0\mod 2$ or else $ z^{2}\equiv 2\pmod{4}$, a contradiction. But this leads to $ (x,y) \equal{} 2$, which is also a contradiction. Thus, $ (x\plus{}iy, x\minus{}iy) \equal{} 1$, so they are both squares of gaussian integers. Thus, $ x\plus{}iy \equal{} k^{2}\equal{} (p\plus{}qi)^{2}\equal{} p^{2}\minus{}q^{2}\plus{}2pqi$, or in other words, $ x \equal{} p^{2}\minus{}q^{2}$ and $ y \equal{} 2pq$. And checking this indeed gives pythagorean triples.

[quote="SplashD"][quote="Hamster1800"]We have to show that it's impossible for $ 3$ to divide $ z\minus{}y$. This is equivalent to $ z\equiv y\pmod{3}$. [/quote]

Doesn't this imply $ x\equiv 0\pmod{3}$ and then by symmetry $ y\equiv 0\pmod{3}$[/quote]

No. We are assuming that $ 3|z\minus{}y$, which implies $ x\equiv 0\pmod{3}$, but it doesn't imply that $ y\equiv 0\pmod{3}$. Our additional condition is asymmetric.

[quote="LarryFreeman"]Hi Everyone,

Fascinated by the proof just offered.  I hope that you don't mind that I add an additional posting to your forum.

If I understand the proof just stated, here is the argument:

(1)  Assume that there is a solution in terms of Eisenstein Integers for x^3 + y^3 = z^3

(2)  We can assume that x,y,z are relatively prime.

[This is easily shown since if there is a common factor, say u, then there exists x',y',z' such that (x=x'u, y=y'u, z=z'u) and it can be shown that (x')^3 + (y')^3 = (z')^3.  We can use this same method to eliminate any common factor]

(3)  From step #1, we have x^3 = z^3 - y^3

(4)  Using Eisenstein integer w, x^3 = z^3 - y^3 = (z - y)(z - yw)(z - yw^2)

[This factorization is not so obvious, but it is correct since w is a 3rd root of unity; that is w^3 = 1; I show a similar proof here: [url]http://fermatslasttheorem.blogspot.com/2006/05/cyclotomic-integers-factoring-fermats.html[/url]]

(5)  z-y, z-yw, and z-yw^2 are relatively prime.

[I'm not sure that this step is easily proven.  I would be glad to see more details.  In my blog, I show that either this is true or (w - 1) is a factor; see step #6 in [url]http://fermatslasttheorem.blogspot.com/2006/06/fermats-last-theorem-kummers-proof-for.html[/url]]

(6)  Z[w] is Euclidean

[In my blog, I show this in the second lemma at [url]http://fermatslasttheorem.blogspot.com/2005/07/euclidean-integers.html[/url]]

(7)  Therefore z-y, z-yw, and z-yw^2 are perfect cubes.

[This follows from step #4, step #5, and step #6 and the additional reasoning which I provide at [url]http://fermatslasttheorem.blogspot.com/2005/05/infinite-descent.html[/url]]

(8)  Since z-yw is a perfect cube of an Eistenstein integer, we know that there exists a,b such that:

z -yw = (a + bw)^3

(9)  z = a^3 + b^3

(a + bw)^3 = a^3 + 3a^2bw + 3ab^2w^2 + b^3w^3 = a^3 + b^3 + 3a^2bw + 3ab^2w^2

So that:

z = a^3 + b^3

And:

-yw = 3a^2bw + 3ab^2w^2

So:

y = -3a^2b - 3ab^2w

[NOTE:  You incorrectly include a w in the proof.  It doesn't change the proof, but should be pointed out]

(10)  3 divides y.

(11)  Since for step #3, we can just as easily say y^3 = z^3 - x^3, we can also conclude using the same logic that 3 divides x.

(10)  Since z^3 = x^3 + y^3, then 3 must necessarily divide z.

(11)  Now, this is impossible since by step #2, we assumed that x,y,z are relatively prime.  Therefore, we have a contradiction and we can conclude that there is no such solution.

Did I understand it correctly?  If so, could you please provide additional details on my step #5.

Thanks very much,

-Larry[/quote]
4) The factorization comes somewhat simply if you consider difference of cubes and the Norm function.
5) It's relatively simple to prove that they are relatively prime. For example,
$ (z, y)\equal{}1\Rightarrow (z, z\minus{}y\omega) \equal{}1$. We can prove the other things similarly.
6) I thought this was a relatively well known fact, at least well known enough to cite on IMO  :maybe:
7) This isn't that hard of a step, consider $ abc\equal{}y^{3}$. Every factor of $ y^{3}$ appears 3 times but each of a, b, and c can only contribute to distinct factors (since they're pairwise relatively prime). So each of the factors that a, b, and c contribute are given three times so they're each cubes. Again, I thought that this is well known enough to cite on IMO  :maybe:
9) :oops: thanks for pointing that out.

I think that covers all points :maybe:

For step #5, you write:

[quote]5) It's relatively simple to prove that they are relatively prime. For example,
(z, y)=1 --> (z, z-yw) =1. We can prove the other things similarly.
[/quote]

But consider w divides 3.  Let z=9 and y=2.

(9,2)=1 But (9, 9-2*w)=w

-Larry

Since $ \omega$ is a root of unity it's a unit in $ \mathbb{Q}[\sqrt{\minus{}3}]$.

So when we talk about the $ GCD$ its preferable to consider this as the common divisor with greatest norm.
Both $ N(\omega)$ and $ N(1)$ equal $ 1$, so we can say $ (z,z\minus{}y\omega) \equal{} 1$.

You keep saying that it's simple to show that they are relatively prime. Sure, $ (z, z\minus{}y\omega)\equal{}1$, but we want $ (z\minus{}y, z\minus{}y\omega)$, not $ (z, z\minus{}y\omega)$. Would you care to show your proof of $ (z\minus{}y, z\minus{}y\omega)\equal{}1$, rather than just sketching it, since none of us are seeing it.

[quote="Hamster1800"]You keep saying that it's simple to show that they are relatively prime. Sure, $ (z, z\minus{}y\omega) \equal{} 1$, but we want $ (z\minus{}y, z\minus{}y\omega)$, not $ (z, z\minus{}y\omega)$. Would you care to show your proof of $ (z\minus{}y, z\minus{}y\omega) \equal{} 1$, rather than just sketching it, since none of us are seeing it.[/quote]

I think this is the argument diophantient used...
[hide]
First, we define norm in the ring $ \mathbb{Z}[\omega]$.  The norm of $ a\plus{}b\omega$ will be $ a^{2}\minus{}ab\plus{}b^{2}$ (the same as absolute value for the complex numbers).  Check, and you will see this is multiplicative.

By WOP, we can select the smallest solution if there exist solutions.  In the smallest solution, we have $ (x, y, z)\equal{}1$.

Use the factoring mentioned before, $ x^{3}\equal{}(z\minus{}y)(z\minus{}y\omega)(z\minus{}y\omega^{2})$

$ (z\minus{}y, z\minus{}y\omega)\equal{}(z\minus{}y, y(1\minus{}\omega))$
But $ (z\minus{}y, y)\equal{}(z, y)\equal{}1$ (keep in mind that this is in the ring $ \mathbb{Z}[\omega]$.  We know $ (z, y)\equal{}1$ because $ (N(z), N(y))\equal{}1$.)

Therefore, we must calculate $ (z\minus{}y, 1\minus{}\omega)$.  Now calculating norms, we have that this is $ 1$ unless $ z\equiv y\mod{3}$.  But if $ z\equiv y\mod{3}$, then $ 3|x$.  If we have $ z$ and $ x$ are not equivalent modulo $ 3$, then we have $ (z\minus{}x, 1\minus{}\omega)\equal{}(z\minus{}x, 1\minus{}\omega^{2})\equal{}1$, and so we use the argument with $ z$ and $ x$.  If $ z\equiv x\mod{3}$, then we have $ 3|x, y z$, contradiction.

In any case, we get down to the conclusion that $ z\minus{}y$, $ z\minus{}y\omega$, and $ z\minus{}y\omega^{2}$ are perfect cubes.  Now the rest of diophantient's argument takes over.[/hide]

Of course, we do need to show that $ \mathbb{Z}[\omega]$ is a Euclidean domain, as sylow_theory said.

[quote="Hamster1800"]So you're saying that if $ z\not\equiv x\pmod{3}$, then we switch to use $ y^{3}\equal{} z^{3}\minus{}x^{3}\equal{} (z\minus{}x)(z\minus{}x\omega)(z\minus{}x\omega^{2})$, and $ 3\not|z\minus{}x$, so these are all relatively prime, yes? That's a good argument.[/quote]

Yes: $ x$ and $ y$ are interchangeable here, so it suffices to show that one of them satisfies the conditions you're looking for.  And indeed, it suffices to prove this weaker statement here.  Thank you.

I've asked this before on this thread, sorry, but I want to bring it up again.

Can anyone offer a solution if we factored $ x^{3}\plus{}y^{3}$ into $ (x\plus{}y)(x\plus{}\omega y)(x\plus{}\omega^{2}y)$ and used the equation $ (x\plus{}y)(x\plus{}\omega y)(x\plus{}\omega^{2}y) \equal{} z^{3}$?

Say we've proved that $ (x\plus{}y) \equal{} p$, $ (x\plus{}\omega y) \equal{} q$ and $ (x\plus{}\omega^{2}y) \equal{} r$ are pairwise relatively prime except for factors of $ \lambda \equal{}\frac{3\plus{}\sqrt{\minus{}3}}{2}$, and that the three are perfect cubes besides these factors of $ \lambda$.

Then how can we use infinite descent to complete the proof?

Yes, I'm answering my own question.

We must prove that if $ x^{3}\plus{}y^{3}\equal{}\epsilon z^{3}$, where $ x$, $ y$, and $ z$ are integers in $ \mathbb{Q}[\sqrt{\minus{}3}]$ and $ \epsilon$ is a unit, then $ x$, $ y$, or $ z$ equals $ 0$.

Assume $ x$, $ y$, and $ z$ are relatively prime, because otherwise we could simply reduce to find a solution where $ (x,y,z) \equal{} 1$.

Note that $ x^{3}\plus{}y^{3}$ factors into $ (x\plus{}y)(x\plus{}\omega y)(x\plus{}\omega^{2}y)$, where $ \omega \equal{}\frac{\minus{}1\plus{}\sqrt{\minus{}3}}{2}$. $ \omega$ is also a unit in $ \mathbb{Q}[\sqrt{\minus{}3}]$.

So we have $ (x\plus{}y)(x\plus{}\omega y)(x\plus{}\omega^{2}y) \equal{}\epsilon z^{3}$. Let $ x\plus{}y \equal{} p$, $ x\plus{}wy \equal{} q$, and $ x\plus{}w^{2}y \equal{} r$. So $ pqr \equal{}\epsilon z^{3}$.

We will show that $ p$, $ q$, and $ r$ are relatively prime, or have $ \omega\minus{}1$, or $ \frac{\minus{}3\plus{}\sqrt{\minus{}3}}{2}$, (which is $ \lambda\omega$, where $ \lambda \equal{}\frac{3\plus{}\sqrt{\minus{}3}}{2}$, and thus an associate of $ \lambda$) as a common factor.

[b]Proof that p, q, and r are relatively prime:[/b] Assume that a prime $ P$ divides $ p$ and $ q$. Then it must also divide $ q\minus{}p$, which is $ (\omega\minus{}1)y$, and $ q\minus{}\omega p$, which is $ \minus{}(\omega\minus{}1)x$. We stated that $ x$ and $ y$ are relatively prime, so $ P$ must divide $ \omega\minus{}1$.

Assume that a prime $ P$ divides $ p$ and $ r$. Then it must divide $ r\minus{}p$, which is $ (\omega^{2}\minus{}1)y$, and $ q\minus{}\omega^{2}p$, which is $ \minus{}(\omega^{2}\minus{}1)x$. So $ P$ divides $ (\omega^{2}\minus{}1) \equal{} (\omega\plus{}1)(\omega\minus{}1)$. $ \omega\plus{}1$ is a unit, so $ P$ must divide $ w\minus{}1$.

Assume that a prime $ P$ divides $ q$ and $ r$. Then it must divide $ r\minus{}q$, which is $ (\omega^{2}\minus{}\omega)y$, and $ r\minus{}\omega q$, which is $ \minus{}(\omega\minus{}1)x$. So $ P$ divides $ w\minus{}1$. (Note: $ \omega^{2}\minus{}\omega \equal{}\omega(\omega\minus{}1)$ and $ \omega$ is a unit.)

This means that $ p$,  $ q$, and $ r$ are relatively prime or have $ \lambda$ as a common factor.

If $ \omega\minus{}1$ divides $ a$ and $ b$ in the set $ [a,b,c] \equal{} [p,q,r]$ then it divides the last term $ c$ because $ \omega\minus{}1$ divides both $ b$ and $ b\minus{}c$.

If the element of $ {p,q,r}$ with the least factors of $ \lambda$ has $ n$ factors of $ \lambda$, then reducing $ pqr$ by $ \lambda^{3n}$ gives us new relatively prime factors $ p'$, $ q'$, and $ r'$.

Because $ \mathbb{Q}[\sqrt{\minus{}3}]$ is a $ UFD$ $ p'$, $ q'$, and $ r'$ are all relatively prime perfect cubes.

So we have $ p'q'r' \equal{}\epsilon (\frac{z}{\lambda^{n}})^{3}$.

Note that $ x \equal{}\frac{p\plus{}q\plus{}r}{3}$ and that $ y \equal{}\frac{\minus{}(\minus{}2p\plus{}q\plus{}r)}{3}$. Let $ x' \equal{}\frac{p'\plus{}q'\plus{}r'}{3}$, and $ y' \equal{}\frac{\minus{}(\minus{}2p'\plus{}q'\plus{}r')}{3}$. Then $ x'\plus{}y' \equal{} p'$.

Also note that $ y \equal{}\frac{\minus{}(\minus{}2q\plus{}p\plus{}r)}{3\omega}$. Let $ x' \equal{}\frac{p'\plus{}q'\plus{}r'}{3}$ and $ y' \equal{}\frac{\minus{}(\minus{}2q'\plus{}p'\plus{}r')}{3\omega}$. Then $ x'\plus{}wy' \equal{} q'$.

Also note that $ y \equal{}\frac{\minus{}(\minus{}2r\plus{}p\plus{}q)}{3\omega^{2}}$. Let $ x' \equal{}\frac{p'\plus{}q'\plus{}r'}{3}$ and $ y' \equal{}\frac{\minus{}(\minus{}2r'\plus{}p'\plus{}q')}{3\omega^{2}}$. Then $ x'\plus{}w^{2}y' \equal{} r'$.

Let $ z' \equal{}\frac{z}{\lambda^{n}}$. $ \epsilon z'^{3}\equal{} p'q'r' \equal{} (x'\plus{}y')(x'\plus{}\omega y')(x'\plus{}\omega^{2}y') \equal{} x'^{3}\plus{}y'^{3}$.

Thus we have found another solution for $ x$, $ y$, and $ z$ such that $ x^{3}\plus{}y^{3}\equal{}\epsilon z^{3}$. All of the characteristics of our original $ x$, $ y$, and $ z$ apply to $ x'$, $ y'$, and $ z'$.

So we should be able to find another triple in the same way: $ x'',y'',z''$, where $ z''$ has less factors of $ \lambda$ than $ z'$, just as $ z'$ had less factors of $ \lambda$ than $ z$.

This means that we should be able to continue in this manner of reducing $ z$ by factors of $ \lambda^{k}$, where $ k$ is a rational integer, indefinitely. However, $ z$ only has finitely many factors of $ \lambda$, so by infinite descent, we cannot have an infinite series.

This is a contradiction, so we have proved that $ x^{3}\plus{}y^{3}\equal{}\epsilon z^{3}$ does not have any solutions unless $ x$, $ y$, or $ z$ is $ 0$, which provides trivial identity solutions of the form $ 0 \equal{} 0$ or $ n^{3}\equal{} n^{3}$.

$ \mathbb{Q}[\sqrt{\minus{}3}] \equal{}\mathbb{Q}(\sqrt{\minus{}3})$ is a field, so everything is a unit and unique factorization has no meaning. You want $ \mathbb{Z}[\sqrt{\minus{}3}]$. Also, just to be picky, you actually show that $ p,q,r$ are associates of cubes, since $ (\omega a^{3})(\omega b^{3})(\omega c^{3})\equal{}a^{3}b^{3}c^{3}$, for example.

Other than that, looks good.

Can I get this cleared up? I thought we were using $ \omega$ as a third roots of unity, $ \omega\equal{}e^{\frac{i2\pi}{3}}\equal{}\minus{}\frac{1}{2}\plus{}\frac{i\sqrt{3}}{2}\not \equal{}\sqrt{\minus{}3}$ So do people talk about $ \mathbb{Z}[\sqrt{\minus{}3}]$ when $ \omega\not\in\mathbb{Z}[\sqrt{\minus{}3}]$ since $ \mathbb{Z}[\sqrt{\minus{}3}]\equal{}\{ a\plus{}b\sqrt{\minus{}3}\ |\ a,b\in\mathbb{Z}\}$

My bad. $ \mathbb{Z}[\sqrt{\minus{}3}]$ isn't integrally closed...not even close to a UFD. We REALLY want $ \mathbb{Z}[\omega]$ (but note $ \mathbb{Q}[\sqrt{\minus{}3}] \equal{}\mathbb{Q}[\omega]$)

Sorry; when I say Q[sqrt(-d)] I usually mean the integers of that field. It looks like I'm taking about rational numbers from the Q, but I use the Q to indicate "quadratic."

The book I took this from isn't up to date, or something, I guess.


trigonometry
calculus
integration
real analysis
real analysis unsolved

can you sum this series... :D

$ \frac {1}{2}\tan\,2x\;\tan^2\,x \;\plus{}\; \frac {1}{2^2}\tan\,2^2x\;\tan^2\,2x \;\plus{}\; \frac {1}{2^3}\tan\,2^3x\;\tan^2\,2^2x \;\plus{}\; \frac {1}{2^4}\tan\,2^4x\;\tan^2\,2^3x \;\plus{}\; \frac {1}{2^5}\tan\,2^5x\;\tan^2\,2^4x \;\plus{}\;\ldots\;\;\;\ldots$

$ \text{i.e. }\;\;\;\quad\sum_{n\equal{}1}^\infty\;\frac {1}{2^n}\tan\,2^nx\;\tan^2\,2^{n\minus{}1}x$

(thought i should post something that doesn't involve solving an integral, i think i will leave that to borislav m  :rotfl: )

[hide="Hint"]Use telescopic sum.  :wink: [/hide]


Find all $n\in \mathbb Z^+$, so that \[ z_n = \underbrace{ 101\dots101}_{2n+1 \text{ digits} } \] is prime.

$z_n=\frac{100^{n+1}-1}{100-1}$ may be prime if n+1=p prime. Primes $\frac{a^p-1}{a-1}$ are generalized Mersene primes (Mersene consider case a=2).

First of all you have to find all solutions:
[quote="Rust"]$z_n=\frac{100^{n+1}-1}{100-1}$ may be prime [b]if[/b] n+1=p prime. Primes $\frac{a^p-1}{a-1}$ are generalized Mersene primes (Mersene consider case a=2).[/quote]
(it should be [b]iff[/b] if they are all)

second of all you are wrong! ($100$ is no simple $a$  ;) )

[quote="Rust"]Yes, only $z_1=101$ is prime.[/quote]

how to prove it is the only one ?

The answer is $101$ only.

	Consider the polynomial $x^{2n}+x^{2n-2}+\ldots+x^2+1=a^n+a^{n-1}+\ldots+a+1=\frac{a^{n+1}-1}{a-1}=\frac{x^{2n+2}-1}{x^2-1}.$ For $x=10$, this is
	\[\frac{(10^{n+1}-1)(10^{n+1}+1)}{99}.\]
	If $n>1$ is odd, $\frac{10^{n+1}-1}{99}$ is an integer, and thus the number clearly isn't prime. Otherwise, $10^{n+1}-1$ is divisible by $9$ and $10^{n+1}+1$ is divisible by 11.


[quote="Round1 Problem2"]$A'\in(VA), B'\in (VB), C'\in(VC), D'\in(VD)$[/quote]
Does (XY) mean the segment XY without the points X and Y?

[url=http://www.mathlinks.ro/Forum/contest/6th_Edition/6th_edition_round_1.pdf]problems of round 1[/url]


What is the conjugate of $ \frac {3 \minus{} 2i}{12 \plus{} 5i}$?

I'm not sure what it means by the conjugate.

[hide="I think it means..."]
... well, the conjugate of $ a \plus{} bi$ is $ a \minus{} bi$, so:
$ \dfrac{3 \minus{} 2i}{12 \plus{} 5i} \equal{} \dfrac{(3 \minus{} 2i)(12 \minus{} 5i)}{(12 \plus{} 5i)(12 \minus{} 5i)} \equal{} \dfrac {36 \minus{} 15i \minus{} 24i \plus{} 10i^2}{144 \minus{} 25i^2} \equal{} \dfrac {26 \minus{} 39i}{169} \equal{} \dfrac {2 \minus{} 3i}{13}$, which has conjugate $ \boxed{\dfrac{2 \plus{} 3i}{13}}$.

Or... you start with conjugate $ \dfrac {3 \plus{} 2i}{12 \minus{} 5i}$, which gives you the same result.
[/hide]

Yes, the conjugate of a number in the form $ a\plus{}bi$ will be $ a\minus{}bi$.  And like Brut3Forc3 did, will allow you to "get" the complex number out of the denominator.

[quote="Bicameral"]I see what you all are saying about this problem, but wouldn't it be $ \frac {1}{2 \minus{} 3i}$[/quote]

no it's not reciprocal

No.  You have to multiply it evenly, so in using the conjugate you would multiply both the top and bottom by a fraction.  In this case it would be $ \frac{12\minus{}5i}{12\minus{}5i}$ multiplied by the original.

Multiplying the complex numbers is like multiplying binomials.  This is like the difference of two squares and will have the middle terms cancel out, allowing you to simplify the fraction better.

$ \frac{1}{2\minus{}3i} \equal{} \frac{2\plus{}3i}{13}$

Generally you want an integer in the denominator of a fraction, which is why the RHS is preferred, at least here. However this is not always the case; you might see why later.


How can I make something like this:

[img]http://img518.imageshack.us/img518/8382/asdasdar5.png[/img]

The color part I can't help you with.  The rest of it isn't too bad.  The bold text can be done with the command "\textbf".  The italicized from "\textit".  The math is all basic syntax.  To get rid of the indent, use the command "\noindent".

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=ntheorem]ntheorem package[/url] does exactly as you want.

Thanks a lot stevem, but I'm a bit lost, how'd you produce the above result? (Of course, not the whole thing, just a little example.) What do I have to put in the preamble?

You must have the latest version of ntheorem. The one that comes with MiKTeX is rather old.
You can get it from [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/ntheorem.html]here[/url], extract and then run latex ntheorem.ins.

You will need to read the documentation carefully to get what you want but one of the examples they give is:
[code]\documentclass[12pt]{article}
\usepackage{framed}
\usepackage{pstricks}
\usepackage[framed]{ntheorem}
\begin{document}
\theoremclass{Theorem}
\shadecolor{cyan}
\newshadedtheorem{moreImportantTheorem}{Theorem}
\begin{moreImportantTheorem}[More Important Theorem]
This is a more important theorem.
\end{moreImportantTheorem}
\end{document}[/code]This gives the attached pdf. Since it uses pstricks you must use LaTeX => PS => PDF when compiling.

Thanks again, it's working now, but, I want to make some changes:

1) "Theorem 1 (More Important Theorem)." If I want to edit this as I want, how do I make it?

2) "[i]This is a more important theorem[/i]." How do I change this so that the text appears normal?

[quote="Healker"]1) "Theorem 1 (More Important Theorem)." If I want to edit this as I want, how do I make it?[/quote]Not sure what you mean? ntheorem is very much like the theorem environment so the text is defined in the \begin statement eg \begin{moreImportantTheorem}[Put whatever you like here]
[quote="Healker"]2) "[i]This is a more important theorem[/i]." How do I change this so that the text appears normal?[/quote]But you asked for it in italics in your original post.
\theorembodyfont{\upshape} changes to upright font.

Perfect!

Yeah, I know, I asked for italics in my original post. Just one more thing: I want to change the word "Theorem 1" to any word I want, how can I make it?

(Forget it, I figured out by myself. Thanks a lot for your help.)

Any ideas in how to produce the following?

[img]http://img523.imageshack.us/img523/2999/dfgdgtj5.png[/img]

I just made what stevem did and it worked fine, but I'd like to know if the above result can be produced.

Yes, you can draw a box round the theorem. See the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/ntheorem/ntheorem.pdf]nthorem documentation[/url], section 4.2 for examples.

On post #5, on that PDF appears:

[b]Theorem 1 (More Important Theorem)[/b] [i]This is more important theorem.[/i]

Is there a way to remove [b]Theorem 1 (More Important Theorem)[/b] ? Because in \begin{moreImportantTheorem}[More Important Theorem] I eliminated More Important Theorem and leaving that as \begin{moreImportantTheorem}[], but the problem is that appears a 1, is it possible to eliminate it?

[] denotes optional parameters so if you don't want anything you can leave it out as in \begin{moreImportantTheorem}

There are two no number styles you can use:
\theoremstyle{nonumberplain} which gives
Theorem This is a more important theorem.
or
\theoremstyle{nonumberbreak} which uses a new line as in:
Theorem
This is a more important theorem.

See the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=ntheorem]ntheorem documentation[/url]


function
floor function
algebra
number theory open
number theory

Calculate $\sum_{i=1}^{1000} <\sqrt{2} i>$ , where $<x>$ means integer part of $x$

In standard notations, the integer part is known as the greatest integer function denoted as $\lfloor x\rfloor$.
Now, for the solution:
[hide="Solution"]
$\lfloor \sqrt{2}\rfloor=1$
$\lfloor 2\sqrt{2}\rfloor=2$
$\lfloor 3\sqrt{2}\rfloor=4$
...
$\lfloor 999\sqrt{2}\rfloor=1412$
$\lfloor 1000\sqrt{2}\rfloor=1414$.
The difference of the sums of every 2 consecutive terms is 6.
There are $\frac{1000}{2}=500$ terms (i.e. $\lfloor \sqrt{2}\rfloor+\lfloor 2\sqrt{2}\rfloor=1+2=3$ would be the first term).
The first term has a value of $1+2=3$.  The last term has a value of $1412+1414=2826$.
As a result,
$\sum_{i=1}^{1000} \lfloor i\sqrt{2}\rfloor=\frac{2826+3}{2}\cdot 500=\boxed{707250}$
[/hide]

Masoud Zargar

[quote="boxedexe"]
The difference of the sums of every 2 consecutive terms is 6.
[/quote]

Are you sure?  $\lfloor 5 \sqrt{2} \rfloor + \lfloor 6 \sqrt{2} \rfloor = 15$ and $\lfloor 7 \sqrt{2} \rfloor + \lfloor 8 \sqrt{2} \rfloor = 20$... it does not seem like the sequence would be that regular.

:(, However, I learnt a lesson.  The lesson is to never rely on the first 10 numbers and the last few numbers. :)  I'll think more.

by using excel, I obtained the answer is 707314.

I was pretty close :(

Do you have a proper proof?

I think we can give a recursive formula this one:

Let $f(n) = \sum_{k=1}^{n}[\sqrt{2}k]$

where [$x$] is integer part of $x$.

Define the sequence

$a_n = [\sqrt{2}n]$

Now there is a theorem that states that if $\alpha$ and $\beta$ are positive irrational numbers such that $\frac{1}{\alpha} + \frac{1}{\beta} = 1$ then the two sequences $[\alpha n]$ and $[\beta n]$ are disjoint and contain the positive integers (I think these sequences are called Beatty sequences).

So consider the sequence
$b_n = [(2 + \sqrt2)n] = 2n + a_n$

$a_n$ and $b_n$ are complimentary and fill up the positive integers.

Suppose $L(n)$ is the greatest integer such that $b_{L(n)} < a_n$

Then, we must have that
$f(n) = \sum_{k=1}^{a_n}k - \sum_{k=1}^{L(n)}b_k$

Since $b_n = 2n + a_n$

We have that

$f(n) = \sum_{k=1}^{a_n}k - \sum_{k=1}^{L(n)}2k - f(L(n))$

where $L(n) = [\frac{a_n} {2+\sqrt 2}]$

This should make it easy to calculate $f(1000)$


Prove that

               $ s\geq\frac{ab}{a\plus{}b}\plus{}\frac{bc}{b\plus{}c}\plus{}\frac{ca}{c\plus{}a}>\frac{s}{4}$

It is well-known that $ \frac {2ab}{a \plus{} b}\le\frac {a \plus{} b}{2}$ , so $ \frac {ab}{a \plus{} b}\le\frac {a \plus{} b}{4}$ . Therefore $ \sum\frac {ab}{a \plus{} b}\le\frac {2(a \plus{} b \plus{} c)}{4} \equal{} s$ .

For the right side of the inequality note that $ \frac {ab}{a \plus{} b} > \frac {ab}{a \plus{} b \plus{} c} \equal{} \frac {ab}{2s}$ .

Summing the similar inequalities $ \implies\ \sum\frac {ab}{a \plus{} b}>\frac {ab \plus{} bc \plus{} ca}{2s} \equal{} \frac {s^2 \plus{} r^2 \plus{} 4Rr}{2s}$ .

Now it's obvious that $ \frac {s^2 \plus{} r^2 \plus{} 4Rr}{2s} > \frac s4$ .

[quote="Ligouras"]Prove that

               $ s\geq\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a} > \frac {s}{4}$[/quote]

Is also true:

$ s\geq\frac {ab}{a \plus{} b} \plus{} \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a} \geq\frac {2[ABC]}{R} > \frac {s}{4}$


Given triangle ABC and $ M\in BC, N\in CA, P\in AB$ such that AM=BN=CP. It appears that MN=NP=PM.
Can we say that AB=BC=CA, too?

As I can see by the drawing, when it is true the only one condition of $ AM \equal{} BN \equal{} CP,$ then the triangle $ \bigtriangleup ABC,$ is not necessary equilateral.

But, if it is true simultaneously, the condition also of $ MN \equal{} NP \equal{} PM,$ then I conjecture ( I have not in mind any proof ) that the triangle $ \bigtriangleup ABC,$ is only equilateral.

Sorry if I am mistaken and I am looking forward in to any approach, show us that the triangle $ \bigtriangleup ABC,$ is not necessary equilateral.

Kostas Vittas.

I skipped the condition MN=NP=PM, now the answer is "Yes" - it is my arguments:

1. It is easy to see that equaliteral triangle have desired properties.
2. It is easy to see that if two of the sides of the triangle are equal - the triangle with these properties is equaliteral.
3. We will prove that the case all sides of ABC are different is impossible.
4. Without loss of generality (checked) we may assume that a>b>c.
5. Let Angles AMP, MNB, NPC are M, N, P. AM=BN=CP=x. Side of MNP is y.
6. From 4 it is easy to see A>B>C, sinA>sinB>sinC, P>M>N.
7. sin(120-P)=(x/y)sinA, sin(120-M)=(x/y)sinB, sin(120-N)=(x/y)sinC (standard notations - all angles are in degrees).
8. It is easy to see that M, N, P are all less than 120 degrees.
9. It is easy to see that at least two of M, N, P are greater than 30 degrees (proved by contradiction). Without loss of generality (checked) we may assume M, N > 30 degrees.
10. From 6, 7, 8, 9 follows that N>M.
11. From 6 and 10 we get a contradiction and it follows the only triangle with required properties is the equaliteral triangle.
Q.E.D

What about my open problem? (http://www.mathlinks.ro/Forum/viewtopic.php?t=171622) Is ABC equaliteral?


inequalities
group theory
abstract algebra
function
number theory proposed
number theory

Let $ S_{7,3}(n)$ is the number of divisor of the form $ 10k\plus{}7,10k\plus{}3$ of n.
      $ S_{1,9} (n)$ is the number of divisor of the form $ 10k\plus{}1,10k\plus{}9$ of n.
Prove that :
$ S_{1,9} (n)\geq S_{7,3} (n)$
Note : I have post a problem with the same method but also have some different.

Let $ E_m$ is Euler's group ($ E_m\equal{}(Z/mZ)^*$). Let $ G$ is subgroup, suth that $ E_m: G\equal{}2$, then $ E_m\equal{}G  \cup gG$.
Let $ D_0(n)$ is set of divisors n, suth that $ k\mod m \in G$ and $ D_1(n)$ is set of divisors n, suth that $ k\mod m \not \in G$.
Let $ S_0(n)\equal{}|D_0(n)|,S_1(n)\equal{}|D_1(n)|, S(n)\equal{}S_0(n)\minus{}S_1(n)$.
If $ n\mod m \in gG$, then $ S_0(n)\equal{}S_1(n)$, because $ f: D_1(n)\to D_0(n)$, suth that $ f(k)\equal{} \frac{n}{k}$ bijective.
If $ n\mod m \in G$ and for all prime divisor p $ p\in D_0(n)$, then $ |D_1(n)|\equal{}S_1(n)\equal{}0<S_0(n)$.
Else exist prime divisor $ p\in D_1(n)$. Let $ f: D(\frac np )\to D(n): k\to kp$.
It give $ S(n)\equal{}S_0(n)\minus{}S_1(n)\equal{}S_0(\frac{n}{p^{ord_p(n)}})\minus{}S_1(\frac{n}{p^{ord_p(n)}})$.
Let $ n\equal{}n_0n_1$, were all prime divisor od $ n_0$ belong $ D_0(n)$ and all prime divisor $ n_1$ belong $ D_1(n)$.
Then $ S(n)\equal{}0$ if and only if  $ n_1$ is not square. $ S(n)\equal{}|D(n_0)|\ge 1$ if $ n_1$ is perfect square.

Are there any proof use arithmetic function?


Suppose $ a, b, c, d, e, f$ are all rational numbers. If

$ a\plus{}b\sqrt{2} \plus{} c\sqrt{3}\equal{} d\plus{} e\sqrt{2} \plus{}f\sqrt{3}$

does it follow that $ a\equal{}d, b\equal{}e$ and $ c\equal{}f$?

Yes. [hide="Solution"]
Let $ A\equal{}a\minus{}d, B\equal{}b\minus{}e,C\equal{}c\minus{}f$ and not all of them is equal to zero. Then:
\[ A\plus{}B\sqrt{2}\plus{}C\sqrt{3}\equal{}0\\
A\plus{}B\sqrt{2}\equal{}\minus{}C\sqrt{3}\\
A^2\plus{}2\sqrt{2}AB\plus{}2B^2\equal{}3C^2\\
2\sqrt{2}AB\equal{}3C^2\minus{}A^2\minus{}2B^2\]
RHS of last expression is rational, so LHS also must be rational. But it's possible if and only if $ AB\equal{}0$. So we have 2 variants:
1)$ A\equal{}0$:Then from initial condition we have:$ B\sqrt{2}\plus{}C\sqrt{3}\equal{}0$ - it's impossible, because $ \sqrt{\frac{2}{3}}$ is irrational
2)$ B\equal{}0$:Then:$ A\plus{}C\sqrt{3}\equal{}0$ - it's impossible, because $ \sqrt{3}$ is irrational
So must be $ A\equal{}B\equal{}C\equal{}0$
[/hide]


When I try to view the "Recent Changes" page, it will only let me see the last 10 or so changes, even when I set it to a larger number of changes and a longer window of time. I'm using Firefox version 3 on some sort of linux machine, and I use the default wiki skin (same style as the AoPS forum skin).

(Apologies: I first posted this in the AoPSWiki forum, but it probably should have been placed here.)

It appears that is all the data we have. Perhaps the Wiki software gobbled it up?


A confectioner sells two types of nut mixtures. The standard-mixture package contains 100g of cashews and 200g of peanuts and sells for 1.95 dollars. The deluxe-mixture package contains 150g of cashews and 50g of peanuts and sells for 2.25 dollars. The confectioner has 15kg of cashews and 20kg of peanuts available. Based on past sales, he needs to have at least as many standard as deluxe packages available. How many bags of each mixture should he package to maximize his revenue?

[hide="Hint"]Use linear programming. Let $ m$ be the number of standard packages, and $ n$ be the number of deluxe packages. He needs $ 100m \plus{} 150n$ cashews and $ 200m \plus{} 50n$ peanuts, which must be less than $ 15$kg and $ 20$kg, respectively. Also, $ m \ge n$.

We see that we need to maximize $ 1.95m \plus{} 2.25n$.

Graph the three given inequalities, and examine the boundary points to see which one gives the maximum. [/hide]

Can you explain it a little more for me? I'm still confused


Claire jogs 6 miles per hour and Megan jogs 5 miles per hour. They start together at their campsite and jog to an outpost 18 miles away. When Claire gets there, she immediately turns around and heads back towards Megan. How many miles will they be from the outpost when they meet? Express your answer as a mixed number.

It takes Claire 18/6= 3 hours to get there.

By then Megan is 3 miles away from the out post.

So, they need to travel 3 miles at a combined rate of 11 miles per hour.

Claire will travel 6/11 of this distance which is 18 /11 miles.

So when they meet, Claire will be 1 and 7/11 miles away from the outpost!


[quote="Megus"]There's nothing to solve - it's just reangarrement[/quote]
Hmm... Maybe I am missing something, but you are wrong... It is not a simple rearrangement.
Excuse, if I am wrong...  :blush:

Maybe I think wrong:

first take 2 sequences: $x_1,...,x_n$ and $x_2,...,x_n,x_1$ and "add" them. [so we have $\sum x_ix_{i+1}=\sum x_ix_{i+1}$]
now we are taking $\frac{1}{x_1},...,\frac{1}{x_n}$ to the right side [which is "the worst" option because we multiply e.g. $x_1 \frac{1}{x_1}$ the smallest and the biggest elements] and some permutation of it to the left. The left side is obviously bigger.

P.S. modified to make it clearer

It was posted here:
http://www.mathlinks.ro/Forum/topic-23743.html

[quote="Megus"]Maybe I think wrong:

first take 2 sequences: $x_1,...,x_n$ and $x_2,...,x_n,x_1$ and "add" them.
now we are taking $\frac{1}{x_1},...,\frac{1}{x_n}$ to the right side [which is "the worst" option because we multiply e.g. $x_1 \frac{1}{x_1}$ the smallest and the biggest elements] and some permutation of it to the left. The left side is obviously bigger.[/quote]
I understood nothing, but if harazi says it is very difficult...

Mhm... strange - so I'd like someone to point me out what is wrong in my thinkin  :? Why it can't be done by just reangarrement - I saw [b]blahblahblah[/b] say at first glimpse to use reangarrement as well, but then he cut off from this so maybe he can point me what's wrong.  :?

As I said, I understood nothing, so I can't point your mistake (if there is one).

[quote="Myth"]
I understood nothing, but if harazi says it is very difficult...[/quote]

I see you followed to redirection links in order to find harazi's post. As most people are not aware that some problems are on the forum already and don't look for it I wonder what the longest link/pointer chain is we can create on ML/AoPS.  :D


From now on, anyone who can answer the following problem will be considered "smart", "good", and "worthy of being moderator", while anyone who cannot will be considered "utterly devoid of intelligence", and "unworthy of washing a moderator's sandals".  Thus, anyone who can solve the following problem gets to be moderator:

[i]An elastic strip of length 1m is attached to a wall and the back of a car. A fly starts at the wall and crawls on the strip towards the car, which at the same time drives away from the wall at a speed of 1m/s.

Assuming that the strip can be stretched infinitely long, will the fly ever reach the back of the car?  If so, how long will it take if the fly walks at a speed of x meters per second?  If not, why not?[/i]

Stephen, you'd better solve this one.

Neat! There's a math problem here. This is a (slight) variant on a famous puzzle, but can it be solved without calculus? I think not...

This is ridiculously hard....not only will we require logarithm but we need a variation of the formula used by gause202 on my famous "Complicated?" problem in MATHCOUNTS...I give up.

Btw the answer: [hide]Yes, but it takes like a huge amount of time, astronomical I believe, and I can't do this by hand [/hide]

Of course its not possible without calculus

Hmm lets see.

[hide]

Lets let L(t) = the length of the elastic at time t, and let s(t) be the distance the fly has travelled. The fly walks at a speed of x m/s.

We have s + ds = ((L+dL)/L)s + x dt. Simplifying that, s + ds = s + s dL/L + x dt, or ds = s dL / L + x dt.

Now put y = s/L. Then s = yL, and ds = y dL + L dy. Equating these two, we get y dL + x dt = y dL + L dy, so x dt = L dy and dy = x dt / L.

But we know that dL/dt = 1, since the car is going 1 m/s. So dy = x dL / L.

Integrate both sides to get y = x ln(L) + c. At the start, y=0 as the fly hasn't moved, so c = - x ln(L(0)). So y = x ln (L/L(0)). L(0)=1m long, so y = x ln(L).

When/if the fly reaches the car, y=1. So L = e^(1/x). The time taken is thus L-1 (starts at 1m, increasing at 1m/s) which is t=e^(1/x) - 1.

So yes, it always reaches the car, after e^(1/x) - 1 seconds.

[/hide]

18, but whats that got to do with anything?

Actually forget all the above proof. It can be done in about two lines..

[hide]

Define the wall to be 0, the car to be 1, p to be his position. Then dp/dt = x/(t+1). Solves for t = e^(1/x) - 1.

[/hide]

...mathfanatic, defining those who can not solve a problem involving calculus as "utterly devoid of intelligence" is not a very good description, and comparing the mathematical level of those who are in middle school with those in college using the same scale is not a good idea either. Course this should be common sense, but in case you haven't realized that. I believe the above answers your question TripleM.

And may I also point out that mathematical skills have nothing to do with the ability to moderate Fun and Games whatsoever since Fun and Games does not have math in it...oh well.

P.S. Yay! I got it right! (well at least the first half :))

Say what is this problem doing here?   I have hardley any idea what calculus is!

It's mathfanatic trying to diss the mods again, just ignore him. I'll send this off to Intermediate. It'll be up to the mods there to change the title.

Can one of the mods please send this back to fun and games? BHorseMath, if you know what calculus is, then you don't have to try and solve it..

I think this is a harmonic series, which diverges.  Thus, the ant will eventually reach the car, but it will take a long time.  I forgot how to find the sum; a year without calculus really rots your brain.

It's a harmonic series if the car moves every second or something like that, but that's not the case. It moves continuously, so it's logarithmic/exponential instead.

Oops.  I think it's pretty similar to the harmonic series, though.  Stupid budget cuts.


I dont understand dividing decimals into decimals like 0.03/0.4 like how you set the problem up and how you solve it.

Please write with correct grammar and spelling (wow, I have no room to talk, but whatever). I cannot understand the whole second half of your question. Here's your example:

$ \dfrac{0.03}{0.4}\equal{}\dfrac{0.03}{0.4}\times\dfrac{100}{100}\equal{}\dfrac{3}{40}$

or:

$ \dfrac{0.03}{0.4}\equal{}\dfrac{\frac{3}{100}}{\frac{40}{100}}\equal{}\dfrac{3}{40}$

well if you don't divide correctly, you obviously won't be able to check it....
BTW HAPPY [size=200]13TH BIRTHDAY[/size](early by like idk 3 hours or something)
you use the identity property of multiplication...
so $ \frac{0.03}{0.4}*1\equal{}\frac{0.03}{0.4}$
now you know $ \frac{x}{x}\equal{}1$ for any number $ x$ other than 0
so you substitue $ \frac{100}{100}$ for $ 1$
and then you just multiply fractions.
when you multiply the answer you get by $ 0.4$ you should get $ 0.03$

The key is that the [b]denominator must be an integer.[/b] Fix this by rewriting it as a fraction and multiplying by the reciprocal or multiplying both numbers by a power of ten.

i have this pre-algebra book that really specifies on this.
i'll get the title then edit this post...  :)

Wow.  I'm getting rusty on dividing huge decimals.  I get to use a calc in =my Alg. 1 class and when we don't we use kid-friendly numbers.  Then a kid asked me for help with division homework...............and I know I'm a lot worse at it than I was before because I just don't practice it much.

Also, once you get done with Pre Algebra(or even Math 177 depending on your schooll) you don't really have to be too worried about big complicated division as long as you know HOW to do it.