modular arithmetic
Combinatorial Number Theory
number theory
prime numbers
IMO
IMO 1994
combinatorics

Show that there exists a set $ A$ of positive integers with the following property: for any infinite set $ S$ of primes, there exist [i]two[/i] positive integers $ m$ in $ A$ and $ n$ not in $ A$, each of which is a product of $ k$ distinct elements of $ S$ for some $ k \geq 2$.

Beautiful problem. Let's look at what it says for a moment. A version of Ramsey's Theorem says this: if we color the $k$-element subsets of $\mathbb N$ with two colors (it works for a fintie number of colors, of course, but we're interested in this case here), then there is an infinite $k$-homogemous set, i.e. an infinite set all of whose $k$-element subsets bear the same color. This problem says that this sort of result is not necesarily true any more if we try to do this for all $k\ge 2$ at the same time: if we color all the subsets of $\mathbb N$ which have at least two elements with two colors, then there need not exist an infintie set which is $k$-homogenous for all $k$. Here, of course, the set of primes represents $\mathbb N$, and the product $p_{i_1}\cdot\ldots\cdot p_{i_k}$ represents the set $\{\i_1,\ldots,i_k\}$. If a product of distinct primes (we don't work with non-square-free numbers) belongs to the set we want to construct, we'll say the corresponding set is red, while if it does not we'll say the corresponding set is blue. That's why from now on we'll work in this setting, using subsets of $\mathbb N$ instead of square-free positive integers.

We now prove that given $k\ge 2$ and some $x\in\mathbb N$, we can find a two-coloration of the $k$-element subsets of $\mathbb N$ such that $x$ belongs to no infinite $k$-homogenous set. This is almost obvious. We can simply color all the $k$-element sets which contain $x$ in red, and the rest of them in blue. We'll call this procedure $P(x,k)$. Now, in order to obtain what we want, we can apply the procedures $P(k,k)$ for all $k\ge 2$. In other words, in the two-coloring of the subsets of $\mathbb N$ which have at least $2$ elements that we obtain, $2$ cannot belong to any infinite $2$-homogenous set, $3$ cannot belong to any infinite $3$-homogenous set, and, in general, $n$ cannot belong to any infinite $n$-homogenous set. In other words, there are no infinite subsets of $\mathbb N$ which are $k$-homogenous for all $k\ge 2$.

Maybe another solution to this beautiful problem:

The set $ A$ of the square-free positive integers is defined as follows: A product $ n$ of $ k$ distinct primes $ p_{1},\ldots,p_{k}$ is a member of $ A$ if and only if $ n\equiv 1\pmod{k+1}$, i.e. $ p_{1}\cdot\ldots\cdot p_{k}\in A\Leftrightarrow p_{1}\cdot\ldots\cdot p_{k}\equiv 1\pmod{k+1}$ for all $ k$ and all primes $ p_{i}$.

Now, let $ S$ be an infinite set of primes and let $ p_{1}$ be the smallest and $ p_{2}$ be the second-smallest prime in $ S$. Let furthermore $ p$ be any prime with $ p>p_{2}$.

Lets now consinder $ S$ modulo $ p$. By the pigeonhole principle, some remainder modulo $ p$ must appear infinitely many times in $ S$, i.e. there exists an integer $ a$, so that $ q\equiv a\pmod p$ for infinitely many $ q\in S$. Obviously, $ p\nmid a$.

Take now $ p-1$ distinct primes $ q_{1},\ldots,q_{p-1}\in S$ and let $ m=q_{1}\cdot\ldots\cdot q_{p-1}$. Then $ m\in A$ because by Fermat's little theorem, $ m = q_{1}\cdot\ldots\cdot q_{p-1}\equiv a^{p-1}\equiv 1\pmod p$.

On the other hand, we have $ p>p_{2}>p_{1}$ by definition, so $ p_{2}\not\equiv p_{1}\pmod p$. Hence, at least one of the numbers $ p_{1}\cdot a^{p-2}$ and $ p_{2}\cdot a^{p-2}$ is not congruent to $ 1$ modulo $ p$. Thus, at least one of the numbers $ n_{1}= p_{1}\cdot q_{1}\cdot \ldots \cdot q_{p-2}$ and $ n_{2}= p_{2}\cdot q_{1}\cdot \ldots \cdot q_{p-2}$ is not congruent to $ 1$ modulo $ p$ and hence ist not a member of $ A$. $ \Box$

Another solution:
Interpretating A as a set of finite subsets of N, we say that $ \{n_{1},n_{2},\ldots,n_{k}\}\in A$, where $ n_{1}< n_{2}< \ldots n_{k}$, if and only if $ n_{2}\le n_{1}+k$.
Given an infinite set of naturals,  just take the first $ d$ elements, where d is the difference between the two least elements: this set belongs to A.
Now take the first element, not the second, an then all the next d-1 elements: this set does not belong to A.

And another one :D (but I'm not really sure if it is correct...):
We will construct two disjoint sets $ A, B$ consisting of only squarefree numbers such that for every infinite set of primes $ S$ there exist $ m \in A$ and $ n \in B$ such that $ m$ and $ n$ are product of the same number ($ \geq 2$) of elements of $ S$. We define:
$ p_{1}p_{2}...p_{n}\in A$ iff $ n$ has exactly two prime divisors from $ p_{1}, p_{2}, ..., p_{n}$ and $ p_{1}p_{2}...p_{n}\in B$ iff $ n$ has exactly one prime divisor from the same set (we don't care about the other numbers). Now let $ S=\{q_{1}, q_{2}, ...\}$. Consider the number $ m=q_{1}q_{2}q_{i_{1}}q_{i_{2}}...q_{i_{s}}$, $ n=q_{1}q_{i_{1}}...q_{i_{s}}q_{i_{s+1}}$, where $ s=q_{1}q_{2}-2$ and $ q_{i}$ are different elements of $ S$. From the definition we immidiately get that $ m \in A$ and $ n \in B$.

I think that there can be very many significantly distinct solutions to this problem.

I think I have another construction:
If $ p_{i}$ is the $ i$-th prime, and if $ n \equal{} p_{t_{1}}*p_{t_{2}}*...*p_{t_{m}}$, for some primes $ p_{t_{i}}$, assign to the number $ n$ the set $ (t_{1}, t_{2} ... t_{m})$ (clearly any squarefree number $ n$ can be written uniquely in such a form).  Define the set $ A$ as follows: a number $ n$ is in $ A$ iff the assigned set $ (t_{1}, t_{2}, ... t_{m})$ is such that $ m | t_{1} \plus{} t_{2} \plus{} ... \plus{} t_{m}$. Now suppose to the contrary this set does not satisfy the condition: then say an infinite set $ S$ of primes, $ {s_{1}, s_{2} ... }$ (where $ s_{i}$ refers to the prime $ p_{s_{i}}$, is such, so that for any $ k$, either [1] all $ k$-element subsets of $ S$ are in $ A$ OR [2] all $ k$ element subsets of $ S$ are not in $ A$.
If [2] occurs, then for any $ k$-element subset of $ S$, the sum of the elements in this $ k$-element subset is not divisible by $ k$. But this cannot happen, since using well-known Erdos result, in any $ 2k \minus{} 1$ numbers, some $ k$ have sum divisible by $ k$, and $ S$ is an infinite set.
Since [2] never occurs, that means [1] must occur for every $ k$. This means for any fixed $ k$, any $ k$-element subset $ S$ has sum divisible by $ k$. Choose one such set, $ (t_{1}, t_{2} ... t_{k})$. Now replace $ t_{1}$ by another element from $ S$ - since the subset still has sum $ 0$ mod $ k$, this means that all elements apart from $ (t_{2} .. t_{k})$ are the same as $ t_{1}$ in $ Z_{k}$. Doing the same thing, but with $ t_{i}$ for other values of $ i$ (ie, replacing $ t_{i}$ by some other element in infinite subset $ S$) we find that elements of $ S$ have the same value in $ Z_{k}$. Now pick any two elements $ s_{1}, s_{2}$ from $ S$. This means $ s_{2} \minus{} s_{1}$ is divisible by $ k$, for any $ k$. But this is clearly a contradiction, since we can choose $ k$ to be arbitrarily large.

Let $p_1 < p_2 < \cdots$ be the ordered sequence of positive prime numbers. Now let $A$ be the set of all positive integers which are the product of $k$ distinct prime numbers one of which is $p_k$ for each $k \ge 2$. Consider an arbitrary set $S$ of primes and let $p_i$ be the minimal prime number in $S$. Now consider a sequence of $i$ distinct prime numbers $a_1, a_2, \dots, a_i$ such that $a_j \in S$ and $a_j > p_i$ for each $1 \le j \le i$. By the definition of $A$, it follows that $p_i \cdot a_1 \cdot a_2 \cdots a_{i-1} \in A$ and $a_1 \cdot a_2 \cdots a_i \not \in A$ and $A$ satisfies the conditions required by the problem.

I don't understand the problem. Is k fixed or does it vary depending on what S is?

@viperstrike $k$ varies depending on $S$.

Sorry to resurrect; I was wondering about the validity of a possible construction.

Denote the set of all primes as $\{p_1,p_2,p_3,\ldots\}$. For integral $i \ge 2$, let $B_i = \{ \prod_{j_k \in D} p_{j_k} \mid |D| = i \text{ and } j_k \equiv j_{m} \pmod {2^{i-2}} \text{ for } 1 \le k < m \le i\}$. Similarly, let $C_i = \{\prod_{j_k \in E} p_{j_k} \mid |E| = i \text { and } \exists k,m \text{ such that } 1 \le k < m \le i \text{ and } j_k \not \equiv j_m \pmod {2^{i-1}} \}$.

Let $A_i = B_i \cap C_i$. Take $A = \bigcup_{i \ge 2}  A_i$.

The thought behind this was that if there exists at least one even-indexed prime and one odd-indexed prime in $S$, then since there are an infinite number of at least one of these classes represented in $S$, WLOG even-indexed primes, we can take a product of an even-indexed prime with an odd-indexed prime, which is in $A$, and an even-indexed prime with an even-indexed prime, which is not in $A$. This would correspond to the addition of the set $A_1$ to $A$. Motivation for larger $i$ works similarly. Notice also that the only set contributing $i$-factor products to $A$ is $A_i$.

For arbitrary $S$, let the two smallest elements be $p_a < p_b$ and let $n$ be the smallest natural number such that $b < 2^n$. Then $a$ and $b$ are different $\pmod {2^n}$. We then proceed through the following $n$-step iteration:

Step $1$: If not all indices in $S$ are odd and not all indices are even, then there exists at least one even-indexed $p_{2c}$ and one odd-indexed $p_{2d-1}$, the product of which is in $A_1$ and therefore in $A$. WLOG there are infinitely many odd-indexed primes in $S$. Then there exist $p_{2d-1}$ and $p_{2e-1}$ such that $p_{2d-1}p_{2e-1} \not \in A$. Done.

Step $2$: Assume the assumption of step 1 is not true. If not all indices in $S$ are $a \pmod 4$ and not all indices in $S$ are $a+2 \pmod 4$, then there exists at least one index which is $a \pmod 4$ and one which is $a+2 \pmod 4$, and one of these classes contributes an infinite number of indices to $S$. Assume the first. Then we can take $a \equiv c \equiv d-2 \pmod 4$ so that the product $p_a p_c p_d$ is in $A_2$ and therefore in $A$. Take $a \equiv c \equiv e \pmod 4$, so that $p_a p_d p_e \not \in A$. Done.

Step $3$: Assume the assumptions of steps $1$ and $2$ are not true. Proceed similarly to step 2, but work $\pmod 8$.

$\vdots$

Step $n$: Assume the assumptions of all previous steps are not true. Then all indices are the same $\pmod {2^{n-1}}$. Since $a,b < 2^n$, it automatically follows that there exist two indices which are not the same $\pmod {2^n}$, namely $a$ and $b$. There must exist an infinite number of indices in $S$ corresponding to $a \pmod {2^{n}}$ or to $a+2^{n-1} \pmod{2^{n}}$. WLOG assume the former. Let $b < i_1 < i_2 < \cdots < i_{n-1} < i_n$ be such that $a \equiv i_1 \equiv i_2 \equiv \cdots \equiv i_{n-1} \equiv i_n \pmod {2^{n}}$. Then $p_{a}p_{b}p_{i_1} \cdots p_{i_{n-1}} \in A_{n}$ and therefore is in $A$ but $p_{a}p_{i_1}p_{i_2}\cdots p_{i_{n-1}} p_{i_n} \not \in A$. Done; this iteration handled the final case.

I had the same idea as Mc Teague but the resulting solution worded in a quite different way.
let $\{p_i\}$ the set of all prime numbers, where $p_i < p_{i+1}$.
Let $A_k = \{p_{i_1} p_{i_2}.....p_{i_{k}}, k >=2   $where all $i_j$ are not all congruent to the same value $mod (k)\}$
At last, let $A =\cup_{k>=2} A_k$.

By contradiction : We suppose that for any $m \in A_k$ for any $k >=2$, there exists at least one prime divisor not in S.
Therefore, all prime divisor in S have an indice congruent to the same value $mod(k)$. From here, it's easy to get a contradiction : take $p_i$ and $p_j$ two prime divisors of S and let k big enough, you'll have $k>i$ and $k>j$ and  $i \equiv j [k]$, therefore $i=j$ : S has only one element, absurd.

So there must exist $k$ and $m \in A$ where $m$ is a product of $k$ disctinct elements of S.

let $k$ such a number. As $S$ as infinitely many primes, we can find infinitely many primes for which all indices are congruent to the same value $mod (k)$. Take $n$ the product of $k$ such numbers of S. $n \notin A$  but n is a product of k disctinct element of S : done.

It boils down to prove that there exists a countable collection $\mathcal{F}$ of subsets of $\mathbb{N}$ with the following property. For every infinite subset $D$ of $\mathbb{N}$ there exists $k\ge 2,$ such that $D$ contains a $k$-element subset of $D$ which is a member of $\mathcal{F}$ as well as a $k$-element subset of $D$ which is not a member of $\mathcal{F}.$

[i]Proof.[/i] Arrange all 2-element subsets of $\mathbb{N}$ as $A_1,A_2,\dots.$ Map to each $A_i$ a family $\mathcal{F}_i$ of $i+1$-element subsets of $\mathbb{N}$ as follows
$$\mathcal{F}_i := \{A_i\cup X : X\subset \mathbb{N}\setminus A_i\,,\, |X|=i-1 \}$$
So, each set in $\mathcal{F}_i$ contains exactly $i+1$ elements. Let $\mathcal{F}:=\bigcup_{i=1}^{\infty} \mathcal{F}_i.$ This is the wanted collection of subsets. Indeed, let $D$ be an infinite set of numbers. Take any $x,y\in D.$ Let $A_n=\{x,y\}.$ Take any $A,B\subset D\setminus\{x,y\}$ such that $|A|=n-1, |B|=n.$ Now $A_n\cup A \in \mathcal{F}_n$ and $\{x\}\cup B\notin \mathcal{F}_n,$ which means $A_n\cup A\in \mathcal{F}$ and $\{x\}\cup B\notin \mathcal{F},$ since the only subsets in $\mathcal{F}$ with $n+1$ elements are those in $\mathcal{F}_n$

[b][u]Comment[/u][/b]. It was ISL 1994, N3, but it's a pure combinatorial one, just dressed as NT. I saw it given at [url=https://artofproblemsolving.com/community/c6h2978219]Germany 2010 - Problem 3.[/url] Something about motivation. One may try first to find a collection of, say, 2-element sets, because it is the simplest way. It means you have to connect some of numbers in $\mathbb{N}$ with edges, that is, the edges will be the required sets. The imposed requirement just says that there should be no infinite set of vertices that form a clique or an anti-clique. Unfortunately, the infinite version of Ramsey's theorem shows, that it's impossible. The same happens if we try to choose subsets with same size $k>2,$ because of a version of Ramsey's theorem for $k$-uniform hypergraphs. This means that we are doomed by trying to create a collection of same size sets. We should vary $k$ by choosing bigger and bigger sets.


geometry
topology
Princeton
college
Putnam
Columbia
number theory

im only in high school, so i don't know much about what you're asking, but on Jane Street Capital's website (the company Sandor Lehoczky works for) has a listing of colleges they recruit from on their website, and they also mention that you can still get an interview if you go to a different college, you just have to e-mail them with your resume.  http://www.janestcapital.com/careers.html at the bottom of the page is the list.

[quote="1234567890"]I was wondering from what colleges top quantitative companies recruit from. I know several of the administrators here worked at DE Shaw, so it would be nice if they could recall the list of colleges that were recruited. Specific colleges, please. Do they usually recruit physics/ math majors, or business majors?

How hard is it to get interviews in general for top companies (not only quants, but scientific companies, other financial groups, etc) from an above-average college (like University of Michigan)?  Or do companies snub all non-Ivies / non-elite tech schools grads?

How does recruiting work anyways?

Thanks for answering the numerous questions.[/quote]

Univ of Michigan is certainly on the 'will interview' list in some disciplines, though I admit it would be harder to get an interview from there than it would be from MIT.  As for what schools are on the definite list, it's pretty much the usual suspects.  Look at lists of top math/CS schools, and that will give you the list of schools Shaw/Jane Street would definitely pay attention to.

If you want to be a quant, don't major in business.  Physics, CS, Math.  And learn how to code.

The typical path to getting hired is:

1) Send in a resume, or have a headhunter send in a resume.
2) Have a phone interview.
3) Have an in-house interview.
4) Get hired.

Your college/awards/test scores/etc are probably most important in clearing those first couple hurdles.  Once you are in house, your performance there can overcome other areas that might not be as good as other applicants.

[quote="1234567890"]I actually want to attend University of Michigan in math. I think their math program is decent. So is it very hard to even get an interview?

Do firms look for coursework/GPA heavily?[/quote]

They look at coursework and GPA - GPA is an indicator of willingness to work.

It is tough to get an interview regardless of where you go.  Make sure you excel at your coursework, and learn how to program.

[quote="ffdbzathf"]math, cs, and physics were the exact subjects i plan on majoring when i go to college!  i didn't realize physics was helpful for being a trader or quant, but i guess i'm in all the better position already.[/quote]

It's not so much that physics is helpful in trading, but that people who master physics have shown an ability to model phenomena very well.  It's this ability that financial firms are interested in, not the actual physics.

[quote="ffdbzathf"]math, cs, and physics were the exact subjects i plan on majoring when i go to college!  i didn't realize physics was helpful for being a trader or quant, but i guess i'm in all the better position already.[/quote]

Majoring in three subjects like that inevitably comes at the expense of breadth in any one subject.

[quote="asjsd"]sorry, i may just ask a silly question.. :?
but i what to know is that  quants = quantitative finance??
if yes, then why dont just study a major called quantitative finance?
( actually i am not sure if this major exists or not...... :P )[/quote]

Some places do have a major similar to that, but most top-notch hedge funds will place their bets on the top physics/CS/math graduates over 'quantitative finance'.  The former is all but guaranteed to be much more challenging.  The hedge funds aren't hiring for financial knowledge - they're hiring for modeling/programming ability.  The finance can be learned relatively quickly.

blahblahblah, i disagree.  it solely depends on the work you're willing to put in.  i know people who have had triple majors and learned each topic in a great amount of depth.

At the end of next summer, I'll basically have finished a degree in math (that is, I won't have enough credits, but I'll have taken a majority of the math courses offered at my university)

That will include:

5 courses in analysis
4 courses in algebra
2 courses in number theory
2 courses on differential equations
1 course on algebraic topology
1 course on numerical analysis
1 course on problem solving
2 courses in statistics
2 courses in computer science
3 lower division math courses

This will have taken me 2 full years, including summers (I'm allowed to take a couple of courses while I hold NSERCs, which are the equivalents of REUs over here)

If I wanted to take similarly many courses in CS (my university is very strong in CS), I could easily take another 3 semesters doing so, and I would still be only two thirds of the way towards your hypothetical triple major.

blah^3, what field of math are you interested in.... and how did you take that many courses in just 2 years? How much math per semester? Just curious.

at certain colleges, they have programs where you can skip some mandated requirements and go straight to working towards your major, such as Harvard's advanced standing program where you use AP grades of 5 to replace many topics not necessary for your major.  Also, math, physics, and CS have some overlap in their courseload, mainly math and physics.

I'm most interested in analysis, but I like algebra and number theory as well.

Well, I got a full year's worth of credits by completing the IB Diploma, so I don't have to take any arts courses or experimental sciences or whatever.  I had 12 math credits coming in to university, so I basically got to skip the whole mind numbing calculus/linalg/discrete math introductory sequence.  So I've basically been taking 4 math courses a semester and 1 'non-math' course.  I also took a lot of courses without prerequisites, and found that it was a good thing, because it requires you to really focus (my GPA in courses I took without prereqs is actually higher than in courses I took with prereqs).  I'll also be taking two grad courses in each of the next two semesters (analysis, algebra, algebraic number theory, functional analysis)

I chose universities based on money.  I turned down Princeton for a considerably less prestigious school with an excellent math department, but certainly not one that compares with Princeton's.  So I figured that to make up for that, I would have to take as many math courses as I could and challenge myself as much as I can, and that's what I've done so far.  In future semesters, I plan on taking a few more economics courses and CS courses, but I'll still take at least 3 math courses every semester.

[quote="blahblahblah"]I chose universities based on money.  I turned down Princeton for a considerably less prestigious school with an excellent math department, but certainly not one that compares with Princeton's.[/quote]

I've been meaning to ask you about that. I thought Princeton's financial aid is about as good as it comes in the Ivy League. So did you get an "offer you couldn't refuse" from your current university, in the form of REALLY complete coverage of all your out-of-pocket costs? Do Canadian students have more actual expense if admitted to Princeton than do American students?

Maybe I should start a separate thread about this in the College Forum, in the interest of staying on topic.

I'll reply here, and it can be split later.

Anyways, as I remember, Princeton offered me about 33000 (USD) a year in financial aid, so it would have cost me about 11000-12000 a year out of my own pocket (my parents weren't willing to help out much).  After exchange rates and everything (they've gotten significantly better in my favor since then), it would have been about 15000-17000 per year, and that, of course, would rise a little each year.  So I was looking at paying roughly 70000 for a degree, minus whatever earnings I could have made in the summer.  I absolutely loved Princeton when I visited, and that's a bargain for the quality of university we're talking about here, but...

I was the top applicant at my local university (Simon Fraser University).  Tuition there is about 2500 (CDN) per semester - my scholarship is worth 4250 per semester over 8 semesters.  The rest goes back to me.  The math department there has achieved some fame for some of the computational work they've done (record calculations of pi, among other things) as well as their affiliation with the Maple CAS.  I also won a 5000 scholarship as one of the 13 top all-around students in my province and a 4000 scholarship from the federal government (a Millenium Excellence Award, for those who know what that means).  All told, I think I took in about 55000 in scholarships.  I also am pretty much guaranteed to have an NSERC research grant every summer (I just completed my first one) which is worth 6000-8000 over 4 months, depending on your supervisor and the year you're in.  So over the course of my degree (which I will probably complete in three years), I will make roughly 80000 and only have to pay about 20000 in tuition, which is a rather nice sum of money for someone just completing their undergrad.  I also do not plan to work immediately after - I will probably just go to grad school, so assuming that I get into the grad school of my choice, I will have sacrificed very little and gained very much, tangibly speaking.

All finances aside, I wouldnt've gone to SFU if I felt that I was seriously compromising the quality of my education.  Now, I don't think very much of it as a liberal arts university, but it is quite strong in mathematics and computer science (it has been top 10 on the Putnam before, and does very well in the ACM on a year to year basis, both events which I have (Putnam) or will (ACM) be involved in).  My professors have been excellent, and because of my IB Diploma, I have been free to take only the courses I like (whereas at Princeton, I would have had to take a very large number of non-mathematics courses).  I also have access to all of the grad courses at the University of British Columbia, which has a better grad program than SFU (and some intriguing grad courses on mathematical finance).  I do regret that I haven't been able to pursue interests in English lit and philosophy and whatnot, but I think that's a small price to pay for a lot of the benefits that I've gotten.

I would have enjoyed the experience of going to Princeton and interacting with a much brighter student body, as well as moving away from home and taking a more balanced set of courses, but I feel that for me, it's an indulgence that wasn't worth the price.

I'll also note that my family is fairly well off.  The other Canadians I visited Princeton with had financial aid awards as high as 40000 a year, which really makes it an easier choice (they also didn't apply to many local universities and hence didn't have any big canadian scholarships to turn down).

Blahblahblah,

I just wonder

1. if your parents were willing to help with that 11-12g per year, would you make the same choice? or

2. if your parents were quite wealthy that you are not qualified for any need based financial aid from Princeton. Your parents would provide full amount (44-45g per year for 4 years) to you regardless of Princeton or Simon Fraser, you could either pocket their money (total about 200g) and go to Simon Fraser, or you could spend all the money to go to Princeton, what would you do?

How about other bright students on this forum? How would you all make your choice(s)?

If there was no debt (or a very small amount) associated with going to Princeton, I would have gone there in a second.  I thought that it was an incredible opportunity, but more than that, it would be a once in a lifetime experience.  If I ever have kids, I will definitely encourage them to go to the best school they can get into.

I am very sympathetic to the situation of students whose parents don't pay out the full calculated "family contribution" for their children's education, because I was in that situation in my day. My conclusion then was to attend my state's flagship university, which had a low "list price." My conclusion now is that I will pay out every penny, and then some, up to the maximum limit of my creditworthiness, for my children's education, and certainly fully pay out what each desirable college calculates that I can afford to pay. I admire blahblahblah's resourcefulness in coming up with a good plan B--obviously one has to pay a certain price in terms of effort and achievement in the high school years to have multiple channels for covering college costs.

BTW, I would split this thread myself, except I haven't got moderation powers in this forum. If the Careers in Mathematics moderator chooses to split off these last few posts to a new topic "Paying for College" in the College Forum, I'll happily moderate it there from then on.

[quote="1234567890"]I actually want to attend University of Michigan in math. I think their math program is decent. So is it very hard to even get an interview?

Do firms look for coursework/GPA heavily?[/quote]

Hey, so are you in college yet? are you going to mich now?


The base length of an isosceles triangle is 24 feet, and the side length is 20 feet. What is the number of feet in the altitude to the base?


What is the maximal number of knights that can be placed on the usual 8x8 chessboard so that each of then threatens at most 7 others?

[hide]
60
first fill the boards with knights
in order to avoid the situation that the center knight threatens 8 other knights, we just need to take the 4 center knights away
[/hide]


parameterization
calculus
integration
function
algebra
linear equation
calculus computations

Can anyone help me with this problem?

Consider the following method of solving the general linear equation of 1st order:

y' + p(t) y = g(t)

If g(t) is not everywhere zero, assume that the solution is of the form y = A(t) exp [- integral of p(t) dt]

where A is now a function of t. By substituting for y in the given differential equation, [b]show [/b]that A(t) must satisfy the condition

A'(t) = g(t) exp [integral p(t) dt]

How can I show this?

Thanks.

I think it will help to you:

$I(t)=e^{\int p(t)dt}$ is integrating factor and if we product the both side of the equation then it changs in the form $\frac{d(yI)}{dt}=Ig(t)$ and integrate both side of this last equation with respect to $t$ , and then solve the resulting equation for the $y$. The general solution is $y=\frac{\int I(t)g(t)+c}{I(t)}$ where $c$ is the constant of integration.


Find the (x,y,z) that satisfy
(x+y)(x+y+z)=72
(x+z)(x+y+z)=96
(y+z)(x+y+z)=120

[hide]Add all three equations:
$2(x+y+z)^{2}= 288$
Hence, $x+y+z = \pm12$. Thus, plugging in,
$x+y = \pm 6$
$x+z = \pm 8$
$y+z = \pm 10$

Using $x+y+z = \pm 12$, subtract all three equations from this one to get:
$z = \pm 6$
$y = \pm 4$
$x = \pm 2$
$(x,y,z) = (\pm 2,\pm 4,\pm 6)$[/hide]

EDIT: Wasn't this in the AOPS II Book?

[hide="i didnt see vishal s solution"] let $k=x+y+z$. summing the three equations we get $2k(x+y+z)=288=2k^{2}$ thus $k=\pm12$.
so $(k-z)k=72\Rightarrow z=\frac{k^{2}-72}{k}=\frac{144-72}{\pm 12}=\pm 6$
similarly $y=\frac{144-96}{\pm 12}=\pm 4$
and $x=\frac{144-120}{\pm 12}=\pm 2$. [/hide]


Construct the semicircle $ h$ with the diameter $ AB$ and the midpoint $ M$. Now construct the semicircle $ k$ with the diameter $ MB$ on the same side as $ h$. Let $ X$ and $ Y$ be points on $ k$, such that the arc $ BX$ is $ \frac{3}{2}$ times the arc $ BY$. The line $ MY$ intersects the line $ BX$ in $ D$ and the semicircle $ h$ in $ C$.
Show that $ Y$ ist he midpoint of  $ CD$.

let $ O$ be center of $ k$ and $ a \equal{} < YOB$,

we have $ < XMY \equal{} < XMB \minus{} < YMB \equal{} 1/2( < XOB \minus{} < YOB) \equal{} a/4$,(because $ BX$ is $ 3/2$ times

$ BY$). therfore $ < CDB \equal{} 90 \minus{} a/4$ (since $ < DXM \equal{} < BXM \equal{} 90$).

and since $ MC \equal{} MB$ then $ 2 < DCB \plus{} < CMB \equal{} 90$ .

so $ < DCB \equal{} 90 \minus{} 1/2( < CMB) \equal{} 90 \minus{} a/4$(because $ < CMB \equal{} < XMB \equal{} a/2$).

so $ < CDB \equal{} < DCB$ and since $ BY$ is orthogonal to $ CD$ then $ YC \equal{} YD$.


[quote="bjh7790"]Think of a triangle ABC with $ \angle$B=30.

The orthocenter of ABC is H.

The center of gravity(?) of ABH is G. (EDIT : G=the intersection point of the three medians)

CG meets with AB at M and meets with AH at N.

Show that AMN a equilateral triangle

Thanks.[/quote]
 Let Q be the midpoint of $ AB, AH\cap BC \equal{} \{J\}$.Draw a line d which pass through C and parallel to $ QJ. d\cap AH \equal{} \{N'\}$
We have $ \angle N'CJ \equal{} \angle BJQ \equal{} 30^o \equal{} 1/2\angle HCJ$ then $ \frac {HN'}{JN'} \equal{} \frac {HC}{JC} \equal{} 2 \equal{} \frac {HG}{GQ}$
So $ GN'//QJ$ we get $ G,N',C$ are collinear.
$ \Rightarrow N\equiv N'$ and $ MC//QJ$
Since AQJ is the equilateral triangle we have q.e.d


i'm waiting...
if nobody has a solution then its ok but i will never post mine (until i change my mind) :P .

[quote="nayel"]let $a_{i}>0$ such that $a_{1}+a_{2}+...+a_{n}=n$. Prove that for each $r\in N$,${{a_{1}^{a_{1}}a_{2}}^{a_{2}}...a_{n}}^{a_{n}}\ge a_{1}^{r}a_{2}^{r}...a_{n}^{r}$[/quote]

I don't understand why $r$ has to be a natural number. The inequality is true for any $r \in \mathbb{R}^{+}$.

[quote="nayel"]please dont post your solution. how was the problem?[/quote]

Don't post my solution? Sure. The problem was easy to me though.

[quote="nayel"][quote="weiquan"][quote="nayel"]which formula did you use?[/quote]

If you mean inequality, I used just Jensen's Inequality.[/quote]

can you state the inequality?[/quote]

$\frac{1}{n}\sum_{i=1}^{n}{f\left(a_{i}\right)}\geq f\left(\frac{1}{n}\sum_{i=1}^{n}{a_{i}}\right)$ if $f(x)$ is convex over the range of $\{a_{i}\}$ and the inequality is reversed when $f(x)$ is concave. Now why [i]can't[/i] I go offline whenever I like? I can't possibly stay online [i]forever[/i]. And besides I did not go offline yet contrary to what you said.


Let

$f\left( a;\;b;\;c\right) =\left| \frac{\left| b-a\right| }{\left| ab\right| }+\frac{b+a}{ab}-\frac{2}{c}\right| +\frac{\left| b-a\right| }{\left| ab\right| }+\frac{b+a}{ab}+\frac{2}{c}$

Prove that
$$
f\left( a,b,c\right) =4\max \left( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right) .
$$

For any $n$ numbers $a_{1},$ $a_{2},$ $...,$ $a_{n},$ we use the notation $ \max \left( a_{1}, a_{2}, \ldots, a_{n}\right)$ for the greatest of the numbers $a_{1}, a_{2},\ldots a_{n}.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

LaTeX back then was different. I think this is the updated latex:
$ f(a,b,c)=4\max(\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$
[color=#f00]Amir: I just fixed it, thanks.[/color]

Use $|x-y|+x+y = 2\max\left(x,y\right)$ two times then

$$f(a,b,c) = 2\max \left( \frac{\left| b-a\right| }{\left| ab\right| }+\frac{b+a}{ab}, \frac{2}{c}\right) = 2\max \left( \left| \frac{1}{a}-\frac{1}{b} \right| + \frac{1}{a}+\frac{1}{b}, \frac{2}{c}\right) = 2\max \left( 2\max \left( \frac{1}{a}, \frac{1}{b} \right), \frac{2}{c}\right) = 4\max \left( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right)$$


In triangle $ABC$, let $K$ be midpoint of $AB$ and $L$ be on $AC$ such that $AL=LC+CB$.

Prove: $\angle KLB$ is right if and only if $AC=3CB$

[hide]Let $E$ be a point in the extension of $AC$ such that $CE=CB$

$AL = LC+CB \Rightarrow  AL=LC+CE \Rightarrow  AL=LE$

So, $L$ is the midpoint of $AE$
Also, $K$ is the midpoint of $AB$. So $KL \parallel BE$

Let $F$ be the symmetric point of $E$ w.r.t. $C$
The circle with diameter $FE$ passes through $B \Rightarrow  BF \perp BE$

Let $k$ be the positive real number such that $AC = k\cdot BC$

$AE = AC+BC \Rightarrow  AE = (k+1)BC$
$AF = AC-BC \Rightarrow  AF = (k-1)BC$
$AL=\frac{AE}{2}\Rightarrow  AL = \frac{k+1}{2}BC$

$KL \perp BL \Leftrightarrow BE \perp BL \Leftrightarrow F \equiv L \Leftrightarrow k-1=\frac{k+1}{2}\Leftrightarrow k=3$

[color=brown]Note
We have $BE \perp BL \Rightarrow  F \equiv L$, because only one perpendicular line is drawn to $AC$ through the point $B$.
And the points $F,L$ are on $AC$[/color][/hide]

[hide="Part 1"]If: $\measuredangle KLB=90^\circ$

It is enough to prove $BC=CL$

Draw a parallel to $KL$ through $B$, it intersects $AC$ at $X$.

Then we get $BC=CX$ and $\measuredangle XBC= \measuredangle CXB= \measuredangle KLA$

Also: $\measuredangle BCL= 2 \measuredangle XBC$.

We can see that $BCL$ is isosceles then: $\boxed{BC=CL}$  $\mathbb{QED}$[/hide]

[hide="Part2"]If: $AC=3CB$

$X$ is defined the same way.

We have: $AL=2LC=2BC=2CX$, then $\measuredangle CBL= 90^\circ$

$BC//KL$: $\measuredangle CBL= \boxed{\measuredangle KLB= 90^\circ}$  $\mathbb{QED}$[/hide]

[b][u]Very nice and ... easy problem ![/u][/b]

[hide="Here are two proofs and an easy extension."][color=darkblue][b]Proof 1 (metric).[/b]

$\{\begin{array}{c}LA=\frac{b+a}{2}\\\\ LC=\frac{b-a}{2}\end{array}$ $\Longrightarrow$ $\{\begin{array}{c}4b\cdot BL^{2}=[(2a-b)(a+b)^{2}+2c^{2}(b-a)]\\\\ 4b\cdot KL^{2}=[a(a+b)^{2}-ac^{2}]\end{array}\ .$

$\boxed{\ LK\perp LB\ }$ $\Longleftrightarrow$ $LK^{2}+LB^{2}=KB^{2}$ $\Longleftrightarrow$ $4b\cdot LK^{2}+4b\cdot LB^{2}=4b\cdot (\frac{c}{2})^{2}$ $\Longleftrightarrow$

$a(a+b)^{2}-ac^{2}+(2a-b)(a+b)^{2}+2c^{2}(b-a)=bc^{2}$ $\Longleftrightarrow$

$(3a-b)[(a+b)^{2}-c^{2}]=0$ $\Longleftrightarrow$ $\boxed{\ b=3a\ }\ .$

[b]Proof 2 (synthetic).[/b] Denote the point $S\in (BK)$ for which $\overline{BS}=2\cdot\overline{SK}\ .$ Prove easily that $\overline{AB}=3\cdot \overline{SB}$ and $\frac{\overline{AB}}{\overline{AK}}=2=-\frac{\overline{SB}}{\overline{SK}}\ ,$ i.e. the points $A,S$ are (harmonical) conjugate w.r.t. the pair of the points $\{B,K\}\ .$ Therefore, $b=3a$ $\Longleftrightarrow$ $LA=2a$ and $LC=a$ $\Longleftrightarrow$ $LS\parallel BC$ and the ray $[LB$ is the bisector of the angle $\widehat{CLS}$ $\Longleftrightarrow$ $LK\perp LB\ .$

[b]An easy extension.[/b] Let $ABC$ be a triangle for which exists the point $L\in (AC)$ so that $2\cdot LA=(k-1)\cdot (LC+CB)\ ,$ where $k>1\ .$ Denote the point $K\in (AB)$ for which $\frac{KA}{KB}=\frac{k-1}{2}\ .$ Prove that $LK\perp LB$ $\Longleftrightarrow$ $b=ka\ .$

[b]Remark.[/b] For $k: =3$ obtain the proposed problem.[/color][/hide]


geometry
real analysis
real analysis unsolved

Prove that the intersection of 2 neighborhoods $N_1$ and $N_2$ of a point x is as well a neighborhood of point x

Would "neighborhood" be a better word here than "area"?


algebra
system of equations
algebra unsolved

Find all values of $ a$ and $ b$ for which the system of equations
 $ xyz \plus{} z \equal{} a$
$ xyz^2 \plus{} z \equal{} b$
$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 4$
has only one real solution.

Let $ a$ and $ b$ be real numbers satisfying the condition.
Observe that if$ x$ or $ y$ is not zero, and $ (x,y,z)$ is a solution, then $ ( \minus{} x, \minus{} y,z)$ will be another solution.
Hence, we have $ x \equal{} y \equal{} 0$.
Then it follows that $ a \equal{} b \in \{ \minus{} 2,2\}$.
And now we have to check which value of $ a,b$ that satisfies the condition.

(i) $ a \equal{} b \equal{} 2$ doesn't work since we have $ (x,y,z) \equal{} (\frac {\sqrt {5} \plus{} 1}{2},\frac {\sqrt {5} \minus{} 1}{2},1)$ be a solution.
(ii) $ a \equal{} b \equal{} \minus{} 2$ doesn't have real solutions.

Finally, there doesn't exist any value of $ a,b$ which satisfies the condition.
 :)


In the next three days, 2000 people will move to Florida. How
many people, on average, will move to Florida each hour?
Express your answer to the nearest whole number.

There are 72 hours in 3 days. On average, $ \frac {2000}{72}$, approximately equal to 28, migrate every hour.

Edited: Added a 3.


calculus
integration
logarithms
calculus computations

Evaluate $ \int_1^e \frac{1\minus{}x(e^x\minus{}1)}{x(1\plus{}xe^x\ln x)}\ dx$.


probability
geometry
AMC
AIME
MATHCOUNTS
calculus
number theory
\/closed

The Art of Problem Solving Spring Schedule is available [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].  The classes we offer this fall and winter are listed below.
[b]Note[/b]: We have added a second session of the Algebra 1 class, so you can enroll in either the Wednesday night class or the Thursday night class.  The two offerings are the same.

[b]Introductory level: Grades 6-10[/b]

*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=41]Algebra 1[/url].  Session one: Wednesdays starting Feb 24.   Session 2: Thursdays starting Feb 25.
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=1]Introduction to Counting & Probability[/url].  Mondays starting March 1.
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=2]Introduction to Number Theory[/url].  Fridays starting March 5.
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=42]Algebra 2[/url].  Tuesdays starting Feb 23.
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=10]Introduction to Geometry[/url].  Wednesdays starting March 3.
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=5]Advanced MATHCOUNTS/AMC 8[/url].  Fridays starting Feb 12.

[b]Intermediate level: Grades 8-12[/b]

*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=4]Algebra 3[/url].  Thursdays starting March 4.
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=31]Intermediate Number Theory[/url].  Tuesdays starting March 2.
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=11]Precalculus[/url].  Mondays starting March 8.
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=8]Special AIME Problem Seminar[/url].  Saturday-Sunday Mar 6-7.

Classes marked with a (*) have an accompanying textbook.  If you plan to enroll in one of these courses, you should do so well before the start of the course, so that you receive the text before the class starts.

All of our courses offer full transcripts for each class, so if you're going to miss a class or two, you'll still have access to everything that happened in class.

If you have any questions about a specific class, please contact us at classes@artofproblemsolving.com.

What will the Intermediate Number Theory seminar help with (as in level of contests)?

[quote]Classes marked with a (*) have an accompanying textbook. If you plan to enroll in one of these courses, you should do so well before the start of the course, so that you receive the text before the class starts.[/quote]

Uhh, I don't think Intermediate Number Theory has a book (unless you were planning to surprise us? :O).

when do the signups need to be in for intro to counting and number theory?

We recommend signing up for classes with textbooks at least 2 weeks before they start, so you can make sure to receive the text and have time to work a little on the first couple chapters.

Im enrolled in Intro 2 Number Theroy... :)  :)  :)

[quote]Im enrolled in Intro 2 Number Theroy...

[/quote]
That class is fun!!! I took it in the fall.

Another question: If i'm going to enroll in advanced mathcounts, should I do Intro to geometry, beause I have took Number theory, and Intro to counting and probabilty and I took geometry in school??? :?:

Could someone please help me?
Thanks in advance :)

Take a look at the diagnostic test for the geometry course [url=http://www.artofproblemsolving.com/Classes/IntroGeom/PostTest.pdf]here[/url].  If you can handle most of the problems on this, then you probably don't need the Intro Geometry course.

Umm... if you click the "Online Classes" button then it gives an in-depth explanation on all of the classes. You can find length of Intro to C&P there.

[quote="Letti"]how many classes are there in intro to counting and probability???[/quote]12

May I just ask... what's the difference between Algebra 3 and Precalculus?

Take a look at the course descriptions [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=4]here[/url] and [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=11]here[/url] to see the difference between the two courses.

Also for calculus, look [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=40]here[/url]. Seems like a long description.

By the way, is the Calculus textbook close to finishing? I remember it is supposed to come out in late January, and I'm really curious what it's going to be like.

We're hoping to get it to the printer in the next month.

are you gonna offer a class for calculus though?

I was looking at the online classes and came upon Olympiad Geometry and I am interested in that. However it has no future date. Are you planning to do it or just abandon it? If you are doing it, can you do it in the summer? Thank you!

We haven't decided when we will next offer the Olympiad Geometry.  It's unlikely to be summer, since many of the students who might be interested in that class will likely be in some summer program.

[quote="JITTT"]oh dang! well what would the cost be of that class?[/quote]

We haven't decided.  I think last time we offered it, the cost was around $ \$250$.

Since the Special AIME Seminar is sort of a wrap-up class, how many questions is it going to have? Does it also discuss the test-taking strategies and message board problems?

It doesn't have message board problems, but there will be some discussion of test-taking strategies.  The number of problems depends on how well the students are doing in the class.  We do as many as we can get through in the 5 hours of class.

We haven't decided yet.  Most likely, it will be in spring, 2011.

That is a long time away. Would WOOT be too for me? I have not taken a formal AIME yet, but looking at the old AIMES, I can do almost all geometry problems. I also did 5/6 on the Are You Ready? for Olympiad Geometry. Now besides the geometry, everything else is average. Would you recommend Precalculus or AIME Problem Series?

The Precalculus is much more thorough than the AIME Problem Series; in general, if you have the time, the subject classes will give you a much broader and deeper exposure than the Problem Series classes.

If you keep working at math between now and the fall, it's likely you'll be ready for WOOT, though I'd want to hear more about your math background before I'd say that for sure.  That said, you can learn plenty between now and September, so if you're anywhere close to ready for WOOT now, you'll likely be ready by September.

Thank you sir, I think I will take Precalculus.


What were your favorite problems and what were the hardest problems from this year's contest?

Not many of the problems were very original, but it was a good written test none the less. Some of the problems had too much computation, such as the one asking for the number of nines in $ 999999899999^2$ and the one asking for the number of partitions of $ 9$. The former could be simplified by rewriting the given number as $ (10^{12} \minus{} 10^5 \minus{} 1)^2$, but it was still fairly annoying from there. I don't think there was any non-brute-force way to do the latter.
There was not much of a difficulty gap in my opinion. I think that most of the problems were at essentially the same difficulty level, resulting in a number of very low scores and a number of very high scores (three-way tie for first place). Maybe the tie-breakers should have been a little more difficult because both the first and second place finishers got both of them correct, so the first place plaque was given at the judges' discretion.

I thought it was a nicely written test. Do you guys know when the results will be posted online?


function
vector
superior algebra
superior algebra solved

Let $A$  be a ring and $a\in A$ a non-invertible element. Denote by $X_a=\{x\in A: ax=1\}$. Prove that $X_a=\Phi$ or $X_a$ is infinite.

Simple algebraic manipulation shows that if $ax=1$ but $a$ is not left invertible then the elements of the fom $\alpha_n=xa^{n+1}-a^n,\ n\ge 0$ are all distinct (we take $a^0=1$), and satisfy $a\alpha_n=0,\ \forall n$. This means that we can take our infinite set of right inverses to be $x+\alpha_n$.

I foud a proof for it in a book. Here it is:

Assume that $X_a\neq \Phi$and $X_a$  is finite, and choose $b$ one of it's elements. Define $f: X_a\to X_a$, $f(x)=xa-1+b$ this function well defined: $a(xa-1+b)=ab=1$, is injective:  $f(x)=f(y)\Rightarrow ax=ay\Rightarrow a(x-y)=0\Rightarrow ba(x-y)=0\Rightarrow x=y$.

Since $X_a$ is finite it is surjective, hence $\exists c$ s.t. $b=f(c)=ca-1+b\Rightarrow ca=1$ hence $a$ has leftinverse, hence it is invertible.

Can anyone give me an example in which that set is infinite?

Try taking $A$ to be the ring of endomorphisms of an infinite-dimensional vector space (over any field), and $a$ a surjective but not injective endomorphism.


conics
ellipse
search
geometry
ratio
analytic geometry
trigonometry

Two different ellipses are given. One focus of the first ellipse coincides with one focus of the second ellipse. Prove that the ellipses have at most two points in common.

There's a [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=646639421&t=29201]very old thread[/url] in the Geometry section where this is proved. I think posts #7 and 8 in that topic add up to a proof.

The idea is that given two ellipses $ \mathcal E_i,\ i\equal{}1,2$ with common focus $ F$, (a) all their points of intersection lie in one of the four regions into which the directrices of $ \mathcal E_i$ corresponding to $ F$ cut the plane, and (b) for all the intersection points $ A$, the ratio between the distances from $ A$ to those two directrices is the same. It follows that all the intersection points are collinear, so, since they all belong to an ellipse, there can't be $ 3$ (or more) of them.

I have a very interesting solution, different and shorter than the official one. It came to me on the train, on my way home. :)

Given 3 points $ A,B,C$ and a focus $ F$, we want to prove that there is only one ellipse containing $ A,B,C$ and having one focus $ F$.
Consider the circles $ \mathcal{A},\mathcal{B},\mathcal{C}$ of centers $ A,B,C$ and rays $ FA,FB,FC$ respectively. There exists at most one circle $ \mathcal{D}$ which contains all other three circles and is tangent to them. If one such circle exists, then its center is the second focus of the ellipse, so the ellipse is uniquely determined. If such circle does not exist, then there is no ellipse with the required proprieties.

So in our problem, if the two ellipses have 3 points in common and one focus in common, then they are not different anymore. :)

Here's a more calculative approach.
The equation in polar coordinates $ (r,\theta)$ of an ellipse with a focus at the origin and the big axis aligned with $ \theta\equal{}0$ is $ r\equal{}\frac{b^2}{a\minus{}c\cos\theta}$
Thus, the equations for the 2 ellipses can be written as $ r\equal{}\frac{b_1^2}{a_1\minus{}c_1\cos(\theta\minus{}\theta_1)}$ and $ r\equal{}\frac{b_2^2}{a_2\minus{}c_2\cos(\theta\minus{}\theta_2)}$.
Therefore, the a point of intersection $ (r,\theta)$ satisfies $ \frac1{b_1^2}(a_1\minus{}c_1\cos(\theta\minus{}\theta_1))\equal{}\frac1{b_2^2}(a_2\minus{}c_2\cos(\theta\minus{}\theta_2))$, which is equivalent to $ \cos(\theta\minus{}\theta_0)\equal{}\alpha$ for some constants $ \theta_0$ and $ \alpha$. The last equation has at most 2 solutions in $ \theta$.

Another way is to apply the transformation which maps one of the ellipses onto the circle: the common focus is mapped to the center. From this the question becomes obvious.

Why is it *obvious*?

Ellipses and circles can have four points of intersection in principle.

Just adding the geogebra picture :D [geogebra]29df07b3e987dbe200c2eb9b47d17de706915fcd[/geogebra]

Let $\mathcal E_1$ be an ellipse with foci $F$ and $F_1$ and $\mathcal E_2$ be an ellipse with foci $F$ and $F_2$.  Suppose for the sake of contradiction that three intersection points $P_1$, $P_2$, $P_3$ exist.  Then \[P_iF_1 - P_iF_2 = (P_iF_1 + P_iF) - (P_iF_2 + P_iF)\quad\text{for }i=1,2,3,\] so $P_1$, $P_2$, and $P_3$ all lie on the same branch of a hyperbola with foci $F_1$ and $F_2$.  But such a branch can only intersect an ellipse in two places, contradiction.


Find all triples $ (a,b,c)$ of positive integers such that $ a^2\plus{}2^{b\plus{}1}\equal{}3^c$.

I think it is an old problem .
$ 3^{c}\equiv (\minus{}1)^{c}(\mod 4)$
Therefore $ 4|a^2\minus{}(\minus{}1)^{c}$
It gives $ c\equiv 0(\mod 2)$ ,then exist an integer m such that $ c\equal{}2m$
$ (3^m\minus{}a)(3^m\plus{}a)\equal{}2^{b\plus{}1}$
Exist $ (x,y)\in N_0$ such that :
$ 3^m\minus{}a\equal{}2^x,3^m\plus{}a\equal{}2^y,x\plus{}y\equal{}b\plus{}1$
$ \Rightarrow 2.3^m\equal{}2^x\plus{}2^y$
It follows that $ x\equal{}1$ ,then we have :
$ 3^m\equal{}1\plus{}2^{y\minus{}1}$
This is catalant equation ,it have only solutions :
$ (m,y)\equal{}(1,2),(2,4)$

[quote="TTsphn"]I think it is an old problem .
$ 3^{c}\equiv ( \minus{} 1)^{c}(\mod 4)$
Therefore $ 4|a^2 \minus{} ( \minus{} 1)^{c}$
 [/quote]
Sorry , but just currious why could you find out $ 4|a^2\minus{}(\minus{}1)^c$ in paragraph I quoted.I think your result only happens  if  b+1 is even

[quote="nkht-tk14"][quote="TTsphn"]I think it is an old problem .
$ 3^{c}\equiv ( - 1)^{c}(\mod 4)$
Therefore $ 4|a^2 - ( - 1)^{c}$
 [/quote]
Sorry , but just currious why could you find out $ 4|a^2 - ( - 1)^c$ in paragraph I quoted.I think your result only happens  if  b+1 is even[/quote]
It is clearly$ a\equiv b (\mod c)$ so $ c|a - b$
But his proof is not suitable for a contes (catalan equation is very hard)
Who can solve by an other way

Catalant equation is not very hard in the case :
\[ |2^a\minus{}3^b|\equal{}1
\]
This is one of problems in my school contest .
You can find it in our forum  :)

[quote="TTsphn"]Catalant equation is not very hard in the case :
\[ |2^a \minus{} 3^b| \equal{} 1
\]
This is one of problems in my school contest .
You can find it in our forum  :)[/quote]
Yes,thanks ,of course in that case,it is not really hard,I remember that case was found first by a France

Good solutions.

The latest equations can solve also with graph, $ y \equal{} 2^{x\minus{}1} \plus{} 1$ and $ y \equal{} 3^x$  :lol:

Seeing modulo $4$ we have $a^2\equiv_4(-1)^c$, therefore $a$ is odd and $c$ is even. So put $c=2d$. Then we have $(3^d-a)(3^d+a)=2^{b+1}$, which means $3^d+a$ and $3^d-a$ are powers of $2$ whose difference is $2a$, and since $2\parallel2a$, we have $3^d-a=2$, therefore we have $3^d=2^{b-1}+1$. Now consider the following cases:
Case 1: if $b=1$, then we have $3^d=2$, which is impossible.
Case 2: if $b=2$, then we have $d=1$, $c=2$ and $a=1$, from where we obtain $(a,b,c)=(1,2,2)$, which is a solution.
Case 3: if $b\ge3$, seeing modulo $4$ we have $(-1)^d\equiv_41$, therefore $d$ is even. So put $d=2e$. Then we have $(3^e-1)(3^e+1)=2^{b-1}$, which means $3^e+1$ and $3^e-1$ are powers of $2$ whose difference is $2$, and since $2\parallel 2$, we have $3^e-1=2$, which means $e=1$, $d=2$, $c=4$, $b=4$ and $a=7$, from where we obtain $(a,b,c)=(7,4,4)$, which is a solution.

Considering $\mod{4}$,$a$ is odd and $c$ is even.Let $c=2c'(c'\in \mathbb N)$,then $(3^{c'}+a)(3^{c'}-a)=2^{b+1}$.For $m>n\ge 0$,
\[\left\{\begin{array}{cc}
3^{c'}+a=2^m \\
3^{c'}-a=2^n \\
\end{array}\right.\]
$\rightarrow 2a=2^n(2^{m-n}-1)$.Since $a$ is odd,$n=1$ and $2\cdot 3^{c'}=2^m+2\Leftrightarrow 3^{c'}=2^{m-1}+1$.

If $m=1$,then $3^{c'}=2$ which is absurd.Thus $m\ge 2$.Considering $\mod{3}$,$m$ is even.

[b]Case1:[/b]$m=2$
$c'=1\rightarrow c=2,a=1,b=2$ which satisfies the condition.

[b]Case2:[/b]$m=4$
$c'=2\rightarrow c=4,a=7,b=4$ which satisfies the condition.

[b]Case3:[/b]$m>4$
By [b]Zsigmondy's theorem[/b],∃$p$(prime) s.t. $p|2^{m-1}+1,p\nmid 2+1=3$ which is absurd.

From case$1,2,3$,the answer is
$\boxed{(a,b,c)=(1,2,2),(7,4,4)}\blacksquare$ :coool:

good problem I solved it with a great fun hard but nice

@fighter
Why don't you use $\LaTeX$? $\LaTeX$ make mathematical expression be easy to see. :D

please give me a tutorial also turn my answer into latex

easy just find that $c$ is even and factor

@fighter
I don't know tutorial of $\LaTeX$.I've studied $\LaTeX$ by seeing other people's code.It  is easy.You just enclose mathematical expression with '$' mark.Try by all means!!! :D

@fighter
You can click other person's mathematical formula and see the code.

Or see here.
http://www.artofproblemsolving.com/wiki/index.php/LaTeX

LaTeX makes your solution beautiful. :-D

First of all notice that $a^2+2^{b+1}+1\equiv0\text{ or }1\pmod 4$ and $3^c\equiv(-1)^c\pmod 4$ thus $c$ is even and $a\equiv1\text{ or }3\pmod 4$
So now let $c=2k$ for some $k\in\mathbb{Z}^+$
If $b=1$ we have that $a^2+4=3^c$ however $a^2+4\equiv1\text{ or }2\pmod 3$ while $3^c\equiv0\pmod 3$ which is a contradiction. From now on assume that $b\ge2$
Using $c=2k$ we can transform in the original equation into $a^2+2^{b+1}=(3^k)^2\Longrightarrow2^{b+1}=(3^k-a)(3^k+a)$
Therefore $3^k-a=2^x\text{ and }3^k+a=2^y\text{ such that }x<y\text{ and }x+y=b+1$
Furthermore notice that summing the equations yields $2\cdot3^k=2^x+2^y\Longrightarrow3^k=2^{x-1}+2^{y-1}$
However $\text{LHS}\equiv1\pmod 2$ thus $\text{RHS}\equiv 1\pmod 2$ which is only the case if $y-1=0\Longleftrightarrow y=1$ this implies that $x=b$
Plugging this in yields $3^k-a=2\text{ and }3^k+a=2^b$ with a bit of algebraic manipulation we obtain $3^k-2=a\text{ plugging this into the other equation yields }3^k-2^{b-1}=1$

[b][color=#f00]Case 1:[/color][/b] $k=1\text{ and }b-1=1\Longleftrightarrow b=2$
This implies $c=2$, plugging this in yields $a^2+2^3=9\Longrightarrow a=1$ thus $(a,b,c)=(1,2,2)$ is a solution.

[b][color=#134F5C]Case 2:[/color][/b] $k,b-1>1$
Now notice that by the Catalan conjecture the only solution for $3^k-2^{b-1}=1$ for $k,b-1>1$ is $k=2\text{ and }b-1=3\Longrightarrow c=4\text{ and }b=4$
Plugging this in yields $a^2+2^5=3^4\Longrightarrow a^2=49\Longleftrightarrow a=7$ thus $(a,b,c)=(7,4,4)$ is the only other solution.

So, to sum up $\boxed{(a,b,c)=(1,2,2)\text{ and }(7,4,4)\text{ are the only solutions for }a,b,c\in\mathbb{Z}^+}$ $\blacksquare$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

My solution for the easy part (a permutation must lie in A_15)

We label the blank as 16. Then the required permutation is an element in $ S_{16}$ with 16 fixed.
There are two types of moves:
Type A: Horizontal, in which the two numbers in the corresponding transposition differ by 1.
Type B: Vertical, in which the two numbers in the corresponding transposition differ by 4.
We want to show that there are even number of both types of transpositions.

For type A, consider this:
Any required permutation is a product $ (16 ,a_1)(a_1,a_2),...,(a_n,16)$
meaning the blank moves to $ a_n$, then $ a_{n \minus{} 1}$,...,then $ a_1$, then move back the the original square. Note that the sum $ 16 \minus{} a_n \plus{} ... \plus{} a_2 \minus{} a_1 \plus{} a_1 \minus{} 16 \equal{} 0$. In this sum, a type A move accounts for 1 or -1, and a type B move accounts for 4 or -4. Since the total sum is even, there are even number of type A moves.

For type B, consider this:
A type B move shifts a "level" to another. (There are 4 levels, Level 1 = {1,2,3,4}, Level 2 = {5,6,7,8}, Level 3 = {9,10,11,12} and Level 4 = {13,14,15,16}).
We start at the fourth level. Again write a required permutation as a product $ (16 ,a_1)(a_1,a_2),...,(a_n,16)$. In which the total sum of "level changes" = 0 (since we go back to the original block eventually). Since a type A move doesn't affect the level, and a type B move changes the level by 1 or -1, there must be even number of type B moves.

There is a brute force way for the other half, that is to show that $ (123)$,...,$ (13,14,15)$ all can be done. Since it lies in $ A_{15}$ and it is obviously a subgroup, all 3-cycles can then be generated and thus $ A_{15}$.

[quote="ehsan2004"]The well-known $ 15$-puzzle consists of a shallow box filled with $ 16$ small squares in $ 4 \times 4$ array. The bottom right corner square is removed, and the other squares are labeled as ;(all in this order) $ 1,2,3,4$  in first row ,$ 5,6,7,8$ in second row, $ 9,10,11,12$ in third row and $ 13,14,15$ in the end row.By sliding the squares around (without lifting them up),Show that the set of possible permutations that can be obtained with the bottom right square blank is precisely $ A_{15}$. (There is no known easy proof that all elements in $ A_{15}$ must occur.)[/quote]

Here's a proof that any element $ s \in A_{15}$ must occur.

Given $ s$, you can move 2 to the top left corner, then 3 to the right of 2, then 4 to the right of 3. Finally move 1 to the square below 2:

[code]2 3 4 *
1 * * *
* * * *
* * * *
[/code]

Now move the empty square to the top right position. Move 4 right, 3 right, 2 right and 1 up. This completes the first row. In this way, you can complete the second row as well:

[code]1 2 3 4
5 6 7 8
* * * *
* * * *
[/code]

Now move 13 to the square below 5, and hope that 9 does not lie below 13:

[code]1 2 3 4
5 6 7 8
13* * *
* * * *
[/code]

In that case, move 9 to the right of 13, move the empty square to below 13, then move 13 down, 9 left. This completes the column for 9 and 13. If 9 were below 13, we're stuck temporarily - but fortunately not for long. Move the empty square to the right of 13, then perform: RULLDRRULDRULDLURDRUL, and now 9 and 13 are in place.

Similarly, move 10 and 14 in place. This leaves us with:

[code]1  2  3  4
5  6  7  8
9  10 *  *
13 14 *  *
[/code]

Move 11 to its rightful position and the empty square to the bottom right. Now the remaining two squares must be in their correct position - otherwise, it'll be an odd permutation. :)


geometry
circumcircle
geometric transformation
reflection
symmetry
homothety
perpendicular bisector

Consider $\triangle{ABC}$ with circumcircle $\omega$, and let $K$ be an interior point (to triangle). Let $AK$, $BK$ and $CK$ meet $\omega$ at $A_1$, $B_1$ and $C_1$, respectively.
Now, denote by $A_2$, $B_2$ and $C_2$ the reflection of $A_1$, $B_1$ and $C_1$, respectively, in the midpoints of $BC$, $CA$ and $AB$.
Prove that the circumcircle of $\triangle{A_2B_2C_2}$ passes through a fixed point (when $K$ varies).

Note: It’s quite obvious that such a fixed point is $H$ - the orthocenter of $\triangle{ABC}$, considering the trivial case: $K=O$.

I think this is from 2003 year. It is not easy(for me).

Note that the perpendicular bisector of $A_2H$ is the image of line $AK$ through a symmetry centered at $F$, the midpoint of $OH$, also the center of the nine point circle (see figure, here $AA'$ is a diameter). Thus, the perpendicular bisectors of $A_2H,B_2H,C_2H$ are concurrent, hence the conclusion.
I used well known facts, like $H$ is symmetric to $A'$ with respect to the midpoint of $BC$...

1. Thanks to Enescu for a cute solution!

2. Thanks to moderator for notification in the source :D , it's much better now :P !

Great solution, [b]enescu[/b]. I have another solution, not so ellegant. First note that $HA_2BC$ can be inscribed in a circle. The same goes for $HB_2AC$ and $HC_2AB$. Take an inversion with pole $H$ and power that invariates circle $\omega$. Let $X'$ be the image of $X$ in this inversion. Clearly we have $B', C'$ and $A_2'$ on a line. The same goes for $A',C',B_2'$ and $A',B',C_2'$. Now we have to show that $A_2',B_2' and C_2'$ are on a line. Some anglechasing shows that $\angle A_2'HB' = \angle KAC$ and $\angle A_2'HC' = 180-\angle KAB$. Thus $\frac{A_2'B'}{A_2'C'} = \frac {HB'sin(KAC)}{HC'sin(KAB)}$. (And the same for $B_2'$ and $C_2'$). Multiply these 3 identities and use [b]Menealaos Theorem[/b] for triangle $A'B'C'$ and the desired result follows easily.

enescu,i didn't understood what the quefitient of your homotety is.


Ok, basically this is a game where I start by listing a food.

The next person must list a new food that starts with the last letter of my food.

You 1 point for each new food that you list.

If someone catches a repeated food, that person receives 10 points!

READY?? I will list the first food in bold.

[b]Oatmeal[/b]

It is like the Geography game.

[b]Lasagna[/b] (if that is how you spell it)

PS Do you lose points for mispelled words? If so, don't make me lose points because I didn't know.

EDIT: I misspelled misspell.  :rotfl:  :rotfl: (not on purpose, though)

no lost points for mispellings  :lol:  don't worry I am generous!

[b]Apple[/b] unfortunately, this is the only one I could think of.

By the way, if the food is a compound word, go with the last lettter.

EGG

My message is too small, so I am typing this to make it longer.

[b]tuna[/b]

will this include drinks and brand-name snacks?

Orange

This message is most certainly NOT too small

@abcak, I would [i]drink it[/i], if it was some smart brain liquid... :rotfl:  :rotfl:
[b]
An impaled scorpion dipped in garlic and tomato sauce, freshly caught and poisoned from the desert[/b]

I WILL [u]NEVER [/u] EAT THAT^^^^^^^

NEVER

[quote="abcak"]liquified human brains served fear factor style in a real skull.[/quote]

There was a case where some workers inhaled liquified pig brains and got a neurotic disorder. Read about it [url=http://www.startribune.com/lifestyle/health/15089906.html]here[/url].

How did you know that :huh:  :huuh:  :what?:

wow, and in my state too.

Our Life Science teacher told us. It took him 30 mins to comment about it and it wasn't even on topic (I think we were studying parts of a cell or something).


Dera Mathlinkers,
let ABC be a triangle, (1) the Euler's circle of ABC,
X a point, PQR the X-pedal triangle of ABC, (2) the circumcircle of PQR, X* the isogonal of X wrt ABC
and (3) the Euler's circle of X*BC.
Prouve that (1), (2) and (3) are concurrent.
Sincerely
Jean-Louis

[b]This problem is equivalent to the following:[/b]
$ \mbox{Let triangle ABC and point D. The four Euler's circles of triangles BCD, ACD, ABD, ABC and} \\
\mbox{the circumcircle of D \minus{} pedal triangle of ABC are concurent (have a common point E).} \\
\mbox{The point E trrough and the circle circumscribed of A \minus{} pedal triangle of } \triangle{BCD},\dots$


Let $ X$ be a connected, normal topological space with more than a single point, prove that $ X$ is uncountable . ?

Any idea is highly appreciated . Thanks

What is "normal" for you? Does "normal" imply $ T_2$ ? If not then every indiscrete space is normal and connected, hence your claim would be false.
If "normal" implies hausdorff then every normal space is in particular a tychonoff space and has therefore for every $ x\in X$ and every closed subset $ \emptyset\neq A\subseteq X$ not containing $ x$ a continouos mapping $ f: X\to[0,1]$ with $ f(x)\equal{}0$ and $ f(A)\equal{}\{1\}$. Since $ X$ is connected, $ f(X)$ must be connected too. Thus $ f(X)\equal{}[0,1]$ and so $ |X|\geq |[0,1]|$. (the required $ x$ and $ A$ exist becauce $ X$ has more than one point and is hausdorff)


Find the maximum value of A = (x+y+z)*(1/x+1/y+1/z) with 2<= x,y,z <= 4 .
Help me with this problem , please !

Does this help you? ;)

http://www.mathlinks.ro/Forum/viewtopic.php?t=19244

  Darij


[hide]Plugging in $-1$ in  the function we have $3+B+4=12$. Solve for B and get $\boxed{5}$.[/hide]

[quote="Interval"]If f(x) = 3x^2 - Bx + 4 and f(-1) = 12, what is the value of B?[/quote]

ugh no offense but i think u have been here at AoPS long enough to learn and use latex. im getting sick and tired of looking at "^" symbols.

[quote="ckck"]anirudh, try doing the first one again. You subtracted wrong.[/quote]

It was a typo :blush:


Find all polinomials $ P(x)$ with real coefficients, such that

$ P(\sqrt {3}(a \minus{} b)) \plus{} P(\sqrt {3}(b \minus{} c)) \plus{} P(\sqrt {3}(c \minus{} a)) \equal{} P(2a \minus{} b \minus{} c) \plus{} P( \minus{} a \plus{} 2b \minus{} c) \plus{} P( \minus{} a \minus{} b \plus{} 2c)$
for any $ a$,$ b$ and $ c$ real numbers

Let $ Q(a,b,c)\equal{}P(\sqrt{3}(a\minus{}b))\plus{}P(\sqrt{3}(b\minus{}c))\plus{}P(\sqrt{3}(c\minus{}a))$, $ Q'(a,b,c)\equal{}P(2a\minus{}b\minus{}c)\plus{}P(\minus{}a\plus{}2b\minus{}c)\plus{}P(\minus{}a\minus{}b\plus{}2c)$

Let $ P(x)\equal{}a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_1x\plus{}a_0$.

So the terms of degree $ k$ of $ Q(a,b,c)\equal{}a_k(\sqrt{3})^k[(a\minus{}b)^k\plus{}(b\minus{}c)^k\plus{}(c\minus{}a)^k]$
that of $ Q'(a,b,c)\equal{}a_k[(2a\minus{}b\minus{}c)^k\plus{}(\minus{}a\plus{}2b\minus{}c)^k\plus{}(\minus{}a\minus{}b\plus{}2c)^k]$

So if $ a_k \neq 0$, then

$ (\sqrt{3})^k[(a\minus{}b)^k\plus{}(b\minus{}c)^k\plus{}(c\minus{}a)^k]\equal{}(2a\minus{}b\minus{}c)^k\plus{}(\minus{}a\plus{}2b\minus{}c)^k\plus{}(\minus{}a\minus{}b\plus{}2c)^k$.
Clearly it's possible for $ k\equal{}0$ it's true. Put $ a\equal{}b\equal{}\minus{}c$ we know it's not possible for $ k \ge 1$.

So $ P(x)\equal{}C$, where $ C$ is any real constant , are the only possible polynomials.

[quote="stephencheng"] ...
$ (\sqrt {3})^k[(a \minus{} b)^k \plus{} (b \minus{} c)^k \plus{} (c \minus{} a)^k] \equal{} (2a \minus{} b \minus{} c)^k \plus{} ( \minus{} a \plus{} 2b \minus{} c)^k \plus{} ( \minus{} a \minus{} b \plus{} 2c)^k$.
Clearly it's possible for $ k \equal{} 0$ it's true. Put $ a \equal{} b \equal{} \minus{} c$ we know it's not possible for $ k \ge 1$.

...[/quote]

You are wrong , because the equality holds for $ k \equal{} 0,k \equal{} 1,k \equal{} 2$ . Let's suppose now $ k > 2$
For $ a \equal{} 1,b \equal{} 0,c \equal{} 0$ you obtain $ (\sqrt 3)^k(1 \plus{} ( \minus{} 1)^k) \equal{} 2^k \plus{} 2( \minus{} 1)^k$, and only $ k \equal{} 4$ may be possible . You have to check this case.
:cool:

$ P(\sqrt {3}s) \plus{} P(\sqrt {3}t) \plus{} P(\sqrt {3}u) \equal{} P(s \minus{} u) \plus{} P(t \minus{} s) \plus{} P(u \minus{} t)$ for $ s,\ t,\ u\in{\mathbb{R}}.$

[color=red]Last edit[/color]

$ s\plus{}t\plus{}u\equal{}0.$

[quote="kunny"]$ P(\sqrt {3}s) \plus{} P(\sqrt {3}t) \plus{} P(\sqrt {3}u) \equal{} P(s \minus{} u) \plus{} P(t \minus{} s) \plus{} P(u \minus{} t)$ for $ s,\ t,\ u\in{\mathbb{R}}.$[/quote]

There is a mistake here. This equation is only true when $ s \plus{} t \plus{} u \equal{} 0$
Here is my solution:

Denote $ x \equal{} a \minus{} c, y \equal{} b \minus{} a, z \equal{} c \minus{} b$, which implies $ x \plus{} y \plus{} z \equal{} 0$
The first equation is equivalent to
$ P(\sqrt{3}x) \plus{} P(\sqrt{3}y) \plus{} P(\sqrt{3}z) \equal{} P(x \minus{} y) \plus{} P(y \minus{} z) \plus{} P(z \minus{} x)$

Assume that $ deg P \equal{} n$. Compare the coefficients of degree n follows
$ (\sqrt{3})^n(x^n \plus{} y^n \plus{} z^n) \equal{} (x \minus{} y)^n \plus{} (y \minus{} z)^n \plus{} (z \minus{} x)^n (*)$.

Let $ x \equal{} \minus{}1, y \equal{} 0, z \equal{} 1$ which takes
$ (\sqrt{3})^n[(\minus{}1)^n \plus{} 1^n] \equal{} 2.(\minus{}1)^n \plus{} 2^n (1)$

Let $ x \equal{} 1, y \equal{} 0, z \equal{} \minus{}1$ which takes
$ (\sqrt{3})^n[(\minus{}1)^n \plus{} 1^n] \equal{} 2 \minus{} 2^n (2)$

Combine (1) and (2) gives us
$ 2^n \plus{} 2.(\minus{}1)^n \equal{} 2 \minus{} 2^n,$
$ 2^n \equal{} 1 \plus{} (\minus{}1)^n (3)$

It's easy to verify that $ n \equal{} 0$ and $ n \equal{} 2$ are only values satisfying (3).
- If $ n \equal{} 0$, it follows that $ P(x) \equal{} C$. Indeed this polynomial satisfies the first equation
- If $ n \equal{} 2$, which follows that $ P(x) \equal{} t_2x^2 \plus{} t_1x \plus{} t_0$. This polynomial satisfies the first equation

And the solution is $ P(x) \equal{} C$ and $ P(x) \equal{} ax^2 \plus{} bx \plus{} c$

P/S: My English is very poor  :(

[quote="ll931110"]There is a mistake here. This equation is only true when $ s \plus{} t \plus{} u \equal{} 0$[/quote]

You are right, I forgot to type the condition.
My solution is almost same as yours.

[quote="ll931110"] P/S: My English is very poor  :([/quote]

We can understand your English well. :)

The answer is $ P(x) \equal{} a_4x^{4} \plus{} a_2x^{2} \plus{} a_1x \plus{} a_0$.

$ p(x)\equal{}ax^4\plus{}bx^2\plus{}cx\plus{}d$ is the correct answer...

ll931110: check at $ (2)$, you must have $ 2\plus{}2^n$ and not $ 2\plus{}(\minus{}2)^n$...

[quote="campos"]$ p(x) \equal{} ax^4 \plus{} bx^2 \plus{} cx \plus{} d$ is the correct answer...

ll931110: check at $ (2)$, you must have $ 2 \plus{} 2^n$ and not $ 2 \plus{} ( \minus{} 2)^n$...[/quote]

Sorry. I didn't check when I posted it.

It's easy to verify that n is even. Set $ n \equal{} 2k$, where k is an integer
Compare the coefficient of degree n follows
$ 3^k(x^{2k} \plus{} y^{2k} \plus{} z^{2k}) \equal{} (x \minus{} y)^{2k} \plus{} (y \minus{} z)^{2k} \plus{} (z \minus{} x)^{2k}$ for all x, y, z satisfying $ x \plus{} y \plus{} z \equal{} 0$

Put $ x \equal{} 2, y \equal{} z \equal{} \minus{}1$ gives
$ 4^k \plus{} 2 \equal{} 2.3^k$, and we could verify the roots of this equation is $ k \equal{} 0,1,2$. So we have the answer is
$ P(x) \equal{} ax^4 \plus{} bx^2 \plus{} cx \plus{} d$


For how many positive integers $n$ is  $\left(2003+\frac{1}{2}\right)^n +\left(2004+\frac{1}{2}\right)^n$ an integer?

I only found 3 that work ($0,1,3$), but I wasn't using a formula; there might be more.

binomial expansion is $(a\pm b)^n=\sum_{k=0}^{n}(\pm 1)^k\binom{n}{k}a^{n-k}b^{k}$

$\left(2003+\frac12\right)^n+\left(2004+\frac12\right)^n=\left(2004-\frac12\right)^n+\left(2004+\frac12\right)^n$

if $n$ is even, last term of sum of both expansion is $2\cdot\left(\frac12\right)^n$ which is not integer for $n\geq 2$, so $n$ cannot be even (except for zero, of course).

assume $n$ is odd. then last term of sum of both binomials is zero so the penultimate term is $2\cdot(\pm 1)^{n-1}\binom{n}{n-1}2004^1\left(\frac12\right)^{n-1}$. because $2004=2^2\cdot 501$, $n-1\leq 3$. only odds less than 4 are $n=1,3$

answer $n=0,1,3$

Well, say $t_n = (2003 + \frac{1}{2})^n + (2004 + \frac{1}{2})^n$. For now, say $2003 + \frac{1}{2} = 2004 - \frac{1}{2} = a$ and $2004 + \frac{1}{2} = b$.

So, $t_n(a+b) = (a^n + b^n)(a+b) = a^{n+1} + ab^n + a^n b + b^{n+1}$

$= a^{n+1} + b^{n+1} + ab(a^{n-1} + b^{n-1}) = t_{n+1} + ab t_{n-1}$

Since we have numerical values for $a$ and $b$, we know $a+b = 2004 - 1/2 + 2004 + 1/2 = 4008 = 501(2^3)$ and also $ab = (2004 - \frac{1}{2})(2004 + \frac{1}{2}) = 2004^2 - \frac{1}{4} = (501^2)(2^4) - (2^{-2})$

Using these, we get:
$t_{n+1} = t_n(a+b) - ab t_{n-1} = t_n(501(2^3)) - t_{n-1}((501^2)(2^4) - (2^{-2}))$

Since the $t_{n-1}$ term is being divided by 4 everytime we generate a new term $t_{n+1}$, we must be aware of the greatest power of 2 that divides a given $t_i$. So, $t_0 = a^0 + b^0 = 2$ and $t_1 = a^1 + b^1 = 501(2^3)$. So, we know the greatest powers of 2 that divide $t_0$ and $t_1$ are 1 and 3, respectively. So, lets define $[x]$ to be the greatest power of 2 that divides $x$. So,
$[t_{n+1}] =[t_n(501(2^3)) - t_{n-1}((501^2)(2^4) - (2^{-2}))]$

$= \min(3 + [t_n], 4 + [t_{n-1}], -2 + [t_{n-1}] ) = \min(3 + [t_n], -2 + [t_{n-1}] )$

Now we are more concerned with the new sequence $f_n = [t_n]$ where $f_0 = 1$ and $f_1 = 3$ and $f_{n+1} = \min(3 + f_n, -2 + f_{n-1} )$. When $f_n$ is always negative, we can have no more integers in our sequence. So, say for some $j$, $f_j$ and $f_{j-1}$ are negative, then $f_{j+1} = \min(3 + f_j, -2 + f_{j-1} )$. Since $-2 + f_{j-1}$ is always negative due to the fact that $ f_{j-1}$ is and the minimum must be less than or equal to $-2 + f_{j-1}$, $f_{j+1}$ must always be negative. So, if we have 2 consecutive $f_i$'s negative, then all remaining are negative due to line of infinite descent. So all we need do is find the first consecutive pair of negatives.
$f_0 = 1$

$f_1 = 3$

${f_2 = \min(3 + 3, -2 + 1} ) = -1$

${f_3 = \min(3 + -1, -2 + 3} ) = 1$

${f_4 = \min(3 + 1, -2 + -1} ) = -3$

${f_5 = \min(3 + -3, -2 + 1} ) = -1$

So, we have 2 consecutive and we are done with only $0,1$ and $3$ producing positive powers of 2 and thus being integers.

Heh, I got it right by guess-and-test. :P

Nice formulae, though.

hehe, guess-and-test this one :D

For how many positive integers $n$ is $\left(2005!-\frac{1}{2}\right)^n +\left(2005!+\frac{1}{2}\right)^n$ an integer? ;)

[quote="236factorial"]Well, you know there are immediately two, 0 and 1. The other one can be found out.[/quote]

hmm yea, that was a bit too obvious,  :spam:  :?:   :rotfl:

Refer to my old post for bits and pieces of this if need be.
Say $a = 2005! - \frac{1}{2}$ and $b = 2005! + \frac{1}{2}$. So $a+b = 2(2005!)$ and $ab = (2005!)^2 - \frac{1/4}$. So, we need to know how many powers of 2 are in 2005. We can easily find this by finding the sum $\displaystyle\sum_{i=1}^{\infty}\left\lfloor\frac{2005}{2^i}\right\rfloor$. This sum is just $1002 + 501 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 1997$. So, we get a recursion and other cool stuff but it comes down to $f_{n+1} = \min(1998 + f_n,-2 + f_{n-1})$ With $f_0 = 1$ and $f_1 = [a+b] = 1998$. Sorry I gotta go for now.


since $ a,b,c,d$ are distinct, the slope (m) of the line passing through these four points is not equal zero.
Let the line passing through these four points be$ L: y \equal{} mx \plus{} c$
let $ a \equal{} x_1$, $ b \equal{} x_2$, $ c \equal{} x_3$, $ d \equal{} x_4$,
then$ \frac {{x_i}^3 \minus{} 5}{x_i \minus{} 2} \equal{} \frac {{x_i}^4}{x_i \minus{} 2}m \plus{} c$
we have $ 0 \equal{} m{x_i}^4 \minus{} {x_i}^3 \plus{} cx_i \minus{} 2c \plus{} 5$........(*)$ i \equal{} {1,2,3,4}$
so $ a,b,c,d$ are the roots of (*).
then $ a \plus{} b \plus{} c \plus{} d \equal{} \frac {1}{c}$, $ abc \plus{} bcd \plus{} acd \plus{} abd \equal{} \minus{} \frac {c}{m}$,
$ abcd \equal{} \frac {5 \minus{} 2c}{m} \equal{} 5(\frac {1}{c}) \plus{} 2( \minus{} \frac {c}{m}) \equal{} 5(a \plus{} b \plus{} c \plus{} d) \plus{} 2(abc \plus{} bcd \plus{} cda \plus{} dab)$


A sequence is defined by $ u_{1}\equal{}8$ and $ u_{n\plus{}1}\equal{}u_{n} \plus{} 3.$

Find the value of $ N$ such that $ \sum_{n\equal{}1}^{2N} u_{n} \minus{} \sum_{n\equal{}1}^{N} u_{n} \equal{} 1256$.

[hide="Solution"]From the recursive equation $ u_n \equal{} u_{n \minus{} 1} \plus{} 3$, we can find the closed equation $ u_n \equal{} 8 \plus{} 3(n \minus{} 1)$. So,

$ \sum_{n \equal{} 1}^{2N} u_n \minus{} \sum_{n \equal{} 1}^{N} u_n \equal{} (8 \plus{} 11 \plus{} ... \plus{} u_{2N}) \minus{} (8 \plus{} 11 \plus{} ... \plus{} u_N)$

$ \equal{} u_{N \plus{} 1} \plus{} u_{N \plus{} 2} \plus{} ... \plus{} u_{2N}$

$ \equal{} (8 \plus{} 3N) \plus{} (8 \plus{} 3(N \plus{} 1)) \plus{} ... \plus{} (8 \plus{} 3(2N \minus{} 1))$

$ \equal{} 8N \plus{} 3\frac {N(3N \minus{} 1)}{2} \equal{} 1256$

$ 16N \plus{} 9N^2 \minus{} 3N \equal{} 2512$

$ 9N^2 \plus{} 13N \minus{} 2512 \equal{} 0$

$ (N \minus{} 16)(9N \plus{} 157) \equal{} 0$

$ \therefore \boxed{N \equal{} 16}$[/hide]


Princeton
college
geometry
search
LaTeX
Science Olympiad

Not anything specific unless you want.  :P

Northern Jersey for me.

north jersey!!!!!!!!!!!! :w00t:  :winner_first:  :winner_first:

Smack dab in the middle...
Phew, it's chilly...

Northern New Jersey, Morris County more specifically.

northern new jersey for me :) i live RIGHT next to the george washington bridge its like a 5 minute walk away from the house haha

Princeton. Like right next to the University.

[quote="Chuthecommie"]Woot me too!!!

Me lives near Princeton because my mom works at the university :O[/quote]
That's so cool! I live near the Princeton area too!  :w00t:

[quote="JSGandora"][quote="dingdong162"]Go nyc jersey-ians!!!![/quote]
What's NYC?  I doubt that it stands for New York City...[/quote]

Some of the New York metropolitan area is in New Jersey.

Yeah, like the part where filetmignon821, BOGTRO, and I live.

A very important part that also includes the great AAST.

Also, BOO THOMAS GROVER AND COMMUNITY FOR WINNING SCIENCE OLYMPIAD AND MATHCOUNTS.  BOOOOOOOOO!!!!!!!!!!!!

[quote="JSGandora"][quote="Chuthecommie"]
Since you are in Grover, I am obliged to say this out of loyalty to my science olympiad team: GROVER S.O. IS WORSE THAN MONTY S.O.[/quote]
Which is why we placed 2nd and you guys placed (checking...) 4th.  :)[/quote]

Yea, we were the school that sat next to you guys and I was one of the Pink People...yeah

It's funny that we did not receive a top-6 finish and we were sitting between Thomas Grover and Community.  Ironic, no?

central, more specifically  mercer county

@middlesex, WW-P, Community Middle School

bergen county
were all the asians live LOL

[quote=ironpanther29]middlesex county[/quote]

SAME!

Do you guys all go to the Princeton Campus?

[quote=ironpanther29]middlesex county[/quote]

same

reeee i was born in Princeton and nj is my favorite state

NJ even looks cool
is fat in some parts and skinny in others


A house worth $ \$9000$ is sold by Mr. A to Mr. B at a $ 10\%$ loss. Mr. B sells the house back to Mr. A at a $ 10\%$ gain. The result of the two transactions is:
$ \textbf{(A)}\ \text{Mr. A breaks even} \qquad\textbf{(B)}\ \text{Mr. B gains }\$900 \qquad\textbf{(C)}\ \text{Mr. A loses }\$900\\
\textbf{(D)}\ \text{Mr. A loses }\$810 \qquad\textbf{(E)}\ \text{Mr. B gains }\$1710$

[hide="Hehe, nice title."]
The cost that Mr. A sells the house for is
$ 9000(.9)\equal{} 8100$
The cost that Mr. B sells the house back at is
$ 9000*(.9)*(1.1) \equal{} 8910$
Thus, the answer is

$ \boxed{\text{D}}$[/hide]


Let $r$ be the inradius of an acute triangle. Prove that the sum of the distances from the orthocenter to the sides of the triangle does not exceed $3r$.

Ehasan2004, i think you posted this problem once before. Anyway, it is easy to see that $HA' =2R cosB cosC,$ etc... So the inequality becomes \[ cosA cosB +cosB cosC +cosC cosA\leq\frac{3r}{2R}. \] but since
 $cosA cosB +cosB cosC +cosC cosA=\frac{p^{2}+r^{2}-4R^{2}}{R^{2}}$ the inequality reduces to a well-known one. ;)

[quote]Ehasan2004, i think you posted this problem once before[/quote]

Well ,because I post it long time ago and I want to see the Iranians solution  ;)


find  $y(x)$ from this differential equation
\[ (2+2x \sqrt{y})x dx + (2+x^2 \sqrt{y})y dy =0 \]
I'm just learning differential eqution about three weeks and just know separable, finding integrate factor, excact diff, bernoulli, and rucatti. Is any of this method works?

sorry for this confused. I miss typo it, I've edit it. Just give me a hint, if it can be solved with the method that I've metioned above.

anybody solve this please? I'm stuck. I've working on it 5 days ago.Still, I don't get the answer.come on I'm askking for hint please...
Or this problem use advance tehnique  :?


What is the formal defenition of this.. I was told they couldn't be ordered but I could think of a few ways to do it so I think I may have misinterpreted what "ordered" meant..

thanks for any help

The actual statement is that the complex numbers cannot be given an order that makes them an ordered field. Of course you can invent a total order for the set of complex numbers - lexicographical order, for example - but you can't make that order cooperate with the arithmetic.

A field $F$ is an [b]ordered field[/b] iff it is a field an satisfies the following properties:

There exists a set $P\subset F,$ informally known as the positive elements of $F.$

For all $x\in F,$ exactly one of the following is true: $x\in P,\,-x\in P,$ or $x=0.$

If $x,y\in P,$ then $x+y\in P.$

If $x,y\in P,$ then $xy\in P.$

We can prove the following theorem: in any ordered field, if $x\ne 0,$ then $x^2\in P.$ This is simple: either $x$ or $-x$ is in $P$, and either way, $x^2=(-x)^2\in P.$

This proves that $1\in P,$ since $1=1^2.$

But in the complex numbers, $-1$ is also a square, being $i^2,$ hence positive. That violates the trichotomy axiom.

One can also prove that no field of finite characteristic can be an ordered field.


invariant
combinatorics proposed
combinatorics

Is it possible to cover a $ 6\times 6$ square perfectly, only using long (4x1) and T-shaped tetromino's?

I can't find an example nor a good invariant.

This solution works for every (4k+2)*(4k+2) square!!!

1. EVEN:
we paint the square like a chess. And then the number of black and white 1*1 squares are equal, so we have that the number of T-shaped tetromino's is even.

2. ODD:
we paint 1*1 squares with 4 colors.\[ color(i,j)\equiv (i\plus{}j)(mod 4 )\]
the number of the 1*1 squares with the same color is are 8,9,10,9.

i) if we cover the 6*6 square with only long 4*1, we have 9, 1*1 squares from each color.
ii) if we replace a T-shaped tetromino instead of a long 4*1, one of the colors have 1 more and the opposite color have one less, because each T-shaped tetromino covers 2 of a color and none of the opposite color.

from i and ii we have that the number of the T-shaped tetromino's is odd.

But the number of T-shaped tetromino's can't be even and odd at the same time, contradiction!!! :lol:

[quote="sinaj"]ii) if we replace a T-shaped tetromino instead of a long 4*1, one of the colors have 1 more and the opposite color have one less, because each T-shaped tetromino covers 2 of a color and none of the opposite color.[/quote]Can you please clarify this? What is "the opposite color" when you have 4 colors?

is the meaning of tetromino a T-shape domino with 4 squares?

A tetromino is any domino with 4 squares. A T-shaped tetromino is the T-shaped domino with 4 squares.

To be letter-perfect,[color=red] a domino has $ 2$ squares[/color] (the initial "d" from "deuce"), so Solomon Golomb introduced the term [color=blue]polyomino[/color].

Therefore a tetromino is a polyomino with $ 4$ squares, a pentomino one with $ 5$ squares, etc.

We don't need an invariant for this one. First we check that a I-piece cannot cover a corner, for, if say, we place a horizontal I-piece on the top left square then clockwise we need I-pieces to cover the rest of the corner squares. It's easy to see that you end up trying to fill a 2x2 square, which is impossible.

Then we need T-pieces to cover each corner. But placing one of them completely determines how to place the other ones. After placing the four of them, we have 2 forced I-pieces,  and we end up with a 4x4 square with corners removed, which is clearly impossible to cover.


probability
combinatorics proposed
combinatorics

In a class of $n$ students, what is the probability of existence of two students, whose birth dates differ by $k$ days?

Do you mean [i]exactly[/i] $k$ days, or $\leq  k$ days?


Prove that there exists a natural number n such that
sin n>2003/2004

More generally, you may prove that the sequence {sin(n)/ n is a positive integer} is dense in [0,1].

Pierre.

Yes, Pierre, how can you prove that the sequence sin n is dense in
0 \leq x \leq 1? (Ive asked a teacher who said that it might not be true)

Ok, I give you the steps, I left the details :

Let a>0 be an irrational number.
Let U(n) = {an} = an - [an], where {.} denotes the decimal part, and [.] denotes the integer part.
Let E = { U(n) / n > 0}.

1) Use that a is irrational to deduce that if n,m are two distinct integers then U(n) =/= U(m).

2) Use the pigeon-hole principle to deduce that, for each integer N > 0 there exists p,q in {1,...,N+1} such that p =/= q and 0 < U(p) - U(q) < 1/N.

3) Deduce from above that there exists a non-zero integer k = p-q (not necessarly positive) such that 0 < U(k) < 1/N.

Now let x be in [0,1[ and e > 0 be given. Let I = ]x-e,x+e[.
Let N > 0 be an integer such that 1/N < e. Use 3) to get k.

4) If k > 0.
Let F = { {ka}, 2*{ka}, ...., [1/{ka}]*{ka} }.
Prove that I \cap F contains at least one element, and deduce that I \cap E contains at least one element.

5) If k < 0.
Let r = -k. Then 1 - 1/n < U(r) < 1. Let U(r) = 1-d.
Let G = {1,2,..., [1/d]}. For i in G, we have {i*r*a} = 1 - i*d.
As in 4), deduce that I \cap E contains at least one element.

6) Deduce that E is dense in [0,1].

Let x be in [-1,1], and w in [0,2 \pi [ such that sin(w) = x. Let t = w/(2 \pi ), thus t is in [0,1]. Let e > 0 and I = ]x-e,x+e[.

7) Use 6) for a = 1/(2 \pi ). Deduce that there exists a positive integer n such that |2 \pi *{n/(2 \pi } - w| < e.

8) Note that, for every real u,v we have |sin(u) - sin(v)|  \leq  |u-v|.
Use it to deduce that | sin(2 \pi *{n/(2 \pi )}) - x| < e.

9) Note that 2 \pi *{n/(2 \pi )} = n - 2 \pi *[n/(2 \pi )], and deduce that |sin(n) - x| < e.
Thus { sin(n) / n > 0 is an integer} is dense in [-1;1] (and therefore in [0,1] ).

Pierre.

Let k \in R\Q, the sequence  (e^(i \pi nk))_{n \in Z} is dense in
S^1={ z \in C; |z|=1}

You are right Moubi, and in fact this latter result is easier to prove.
But, the original problem ask for a natural number, thus the work is a little bit difficult...

Pierre.


calculus
integration
real analysis
trigonometry
calculus computations

With that convolution in there, I would highly recommend a transform method.

[hide="Fixed below."]Laplace transforms, with $Y(s)$ being the Laplace transform of $y(t):$

$Y(s)+\frac1{s-1}Y(s)=\frac{s}{s^{2}+1}$

$sY=\frac{s(s-1)}{s^{2}+1}$

$Y=\frac{s-1}{s^{2}+1}$

$y(t)=\sin t-\cos t.$[/hide]

OR

$\int_{0}^{t}e^{-v}y(v)dv=e^{-t}(\sin t-y)$

Differentiate and you get that

$y'=\cos t-\sin t$. It follows that $y(t)=\sin t-\cos t$.

Your answers are wrong, two guys.
Isn't $y(0)=0?$

Now that kunny mentions it ... I had the Laplace transforms of sine and cosine reversed. Let's try this again.

Laplace transforms, with $Y(s)$ being the Laplace transform of $y(t):$

$Y(s)+\frac1{s-1}Y(s)=\frac{1}{s^{2}+1}$

$sY=\frac{s-1}{s^{2}+1}$

$Y=\frac{s-1}{s(s^{2}+1)}=\frac{A}{s}+\frac{Bs+C}{s^{2}+1}$

$Y=\frac{s+1}{s^{2}+1}-\frac1s$

$y(t)=\cos t+\sin t-1.$


$ \mathcal F$ is a family of 3-subsets of set $ X$. Every two distinct elements of $ X$ are exactly in $ k$ elements of $ \mathcal F$. It is known that there is a partition of $ \mathcal F$ to sets $ X_1,X_2$ such that each element of $ \mathcal F$ has non-empty intersection with both $ X_1,X_2$. Prove that $ |X|\leq4$.

Let $ |X_1| \equal{} m$, $ |X_2| \equal{} n$ and $ |X| \equal{} \ell$.  ($ \ell \equal{} m \plus{} n$)

For each pair $ \{a,b\}$ from $ X_1$ we have $ p$ different subsets $ A_i$

Then there are in total $ {m\choose 2}\cdot p$ different subsets $ A_i$ (beacuse each $ A_i$ cuts both $ X_1$ and $ X_2$)

Also there is $ {n\choose 2}\cdot p$ (beacuse of $ X_2$)

So we have at least $ {m\choose 2}\cdot p \plus{} {n\choose 2}\cdot p$ subsets $ A_i$:

$ k \geq {m\choose 2}\cdot p \plus{} {n\choose 2}\cdot p \geq {\ell ^2 \minus{} \ell\over 4}\cdot p$

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

On the other side each pair $ \{a,b\}$ from $ X$ is ''connected'' with $ p$ diffrent subsets $ A_i$
and each subset $ A_i$ is connected with exactly $ 3$ pairs $ \{a,b\}$, so we have
\[ {\ell \choose 2}\cdot p \equal{} 3k
\]
-------------------------------------------------------------------------------------------------

Joining both equations we get $ \ell\leq 4$.


analytic geometry
graphing lines
slope
function
rotation
conics
ellipse

Consider a graph of a line segment with slope zero from (0,0) to (1,0). This line segment contains all real x coordinates from 0 to 1. Now consider a sloped line from (0,0) to (1,1). This line segment is longer. On a graph of real numbers, if you move either way, inevitably you'll either increase or decrease your coordinate numbers since it's imposible to equally space coordinates you can't list. So if a line is longer, wouldn't it contain more points? Since the first line contains all the real x coordinates, doesn't this mean that there are more than one y coordinate for each x coordinate? That would mean that the graph of a sloped line isn't a function, right? I think my logic is wrong with the spacing, but i don't understand how a line can be longer without containing more points, especially when there are an infinite amount of points.

Infinity's a wierd concept, because some infinite quantities are greater than other infinite quantities, and some infinite quantities are the same as other infinite quantities. For example, there are the same number of even numbers as integers, because you can multiply every integer by two.

In your question, the infinite quantities are the same. If you graphed them side by side, you could move from the horizontal line a certain distance up such that you hit  the other line. Therefore every point on the bottom line corresponds to a point on the top line, and vice versa.

Hello, filletwho!

As RC-7th pointed out, Infinity is a mind-boggling concept.
We are simple carbon-based lifeforms and cannot completely understand it.

Consider the dilemma of a shepherd who has no number system.
How can tell if all his sheep have returned to the pen?

At the beginning of the day, he releases the sheep one at a time.
As each sheep leaves the pen, he places a pebble in his pocket.

And the end of the day, he removes a pebble for each sheep that returns.
If he has any pebbles left over, he knows there are some sheep missing.
[Note: He may not know that there are "three" sheep missing, but he knows
that there are as many sheep missing as there are pebbles in his pocket.]

This is a primitive method for determining which group is larger (sheep or pebbles).

We are equally baffled when comparing infinite sets.
We have no "number system" to handle them.

Which has more points:
the line segment from (0,0) to (0,1) or the line segment from (0,0) to (0,2) ?

Since we can't really count them, we will pair them up (like pebbles and sheep).

Every point on (0,1) has a coordinate, some real number assigned to it.
Every point on (0,2) has a coordinate, too.

We will pair them up like this:
the point $x$ on (0,1) will correspond to the point $2x$ on (0,2).

Imagine that we can (somehow!) pair up ALL of them . . .
Will there be any "left over"?  (Will one set be larger than the other?)

The answer is NO.
For every point $x$ on (0,1), there is a corresponding point $2x$ on (0,2).
For every point $y$ on (0,2), there is a corresponding point $\frac{y}{2}$ on (0,1).

There is a [i]one-to-one correspondence[/i] between the points in the two intervals.
Neither set will "run out of number" first.

We are forced to conclude that the two sets have [i]the same number of points[/i].

This is contrary to our Common Sense, of course.
But then, Common Sense does not apply to concepts like "Infinity" or "forever".

Cantor proved that number of decimal numbers that are bigger than zero but smaller than one is greater than the number of counting numbers. By doing so, he showed that there are different sizes of infinity.

Assume that the above-mentioned sets are the same size.

Let every counting number $a_n$ correspond to a number $0.d_{n1}d_{n2}...$. After $d_{n9}$, we go on to $d_{n10},d_{n11}...$ or something like that.

What about $0.x_{1}x_{2}x_{3}...$, where $x_{1}$ is something not equal to $d_{11}$, $x_{2}$ is not equal to $d_{22}$, and so on?

Clearly it doesn't correspond to any $a_n$ because it difffers from every $a_{n}$ by at least one digit.

We have a contradiction, proving that there are more decimal numbers between 0 and 1 than there are counting numbers.

With that logic, there are more points in the segment from (0,0) to (1,1) because there is a greater distance between each set of corresponding points.

That makes a lot more sense to me--one infinity can be greater than another.

For instance, the limit of x as x -> :inf: is greater than the limit of (x/2) as x -> :inf:

(well, technically, neither limit exists...)

No, that's not true at all. If you can find a bijection between two sets, they have to have the same size. (That's the definition.)

One infinity can only be larger when speaking with respect to the orders of sets.  X and X/2 are both unbounded as X increases, and the value of X is always greater than that of X/2 for every X>0, but that doesn't make the limit larger.

Infinity from different perspective.

Imagine a circle viewed from a point on the line normal to the center.
That is you are looking straight on at the circle.

Then draw east-west &  north-south diagonals.
Next count all of the points in each direction.
(It may require a more that a few seconds, but the result is infinity.)

Next rotate the plane of the circle until you have an ellipse, such that
the east-west diagonal is 10 times longer that the north-south diagonal.

If points cannot warp, stretch, elongate, contract, or change, then when
you recount the points on each diagonal you will have 10 times as many
in the east-west direction as in the north-south direction.

Actually, you haven't changed anything except your viewpoint.  No matter
how the circle in observed there is only an infinity of points.  Just as you
change you perspective it appears as if the values alter.

Ten infinities or 20 or 100 infinities do not change infinity.
It would seem likely that 20 times Zero is a lot bigger that 2 times Zero.
Thus, an infinite infinities are not larger that 1 infinity.

At infinity ALL LINE SEGMENTS are the same length.

You just just need to view your coordinate system from the necessary
point of view to get the correct count.

I still hold that one infinity can be larger than another infinity.

First of all, it can be shown that $\lim_{x\rightarrow c}\frac{a}{b} = \frac{\lim_{x\rightarrow c} a}{\lim_{x\rightarrow c} b}$

Therefore $\lim_{x\rightarrow + \infty}\frac{x^3}{x^2}  = \frac{\lim_{x\rightarrow + \infty}{x^3}}{\lim_{x\rightarrow + \infty}{x^2}}$

and we can simplify

$\lim_{x\rightarrow + \infty}\frac{x^3}{x^2} = \lim_{x\rightarrow + \infty}{x} $

and, obviously, $\lim_{x\rightarrow + \infty}{x}> 1$

So therefore :
$\lim_{x\rightarrow + \infty}{x^3} / \lim_{x\rightarrow + \infty}{x^2} > 1$

which leads to:
$\lim_{x\rightarrow + \infty}{x^3} > \lim_{x\rightarrow + \infty}{x^2}$

Even though ${\lim_{x\rightarrow + \infty}{x^3} = \infty_1}$ and ${\lim_{x\rightarrow + \infty}{x^2} = \infty_2}$

so $\infty_1>\infty_2$

[size=150]
Not all infinities are created equal  :P [/size]

With cantor's argument, you can show that there are indescribable real numbers. :D  :P

[quote="boy_21"]First of all, it can be shown that $\lim_{x\rightarrow c}\frac{a}{b} = \frac{\lim_{x\rightarrow c} a}{\lim_{x\rightarrow c} b}$    [/quote]

This isn't true at all, only under certain conditions.  One of those conditions is that both the numerator and denominatory [i]not[/i] approach infinity or 0.  What your argument amounts to saying is that "different functions approach infinity at different speeds."  This doesn't tell you anything about sizes of infinities.  There are in fact differently-sized infinities, but they mean something entirely different that how you're treating them.  It's possible to define systems in which there are infinities more like the one that you are trying to use, but such systems aren't the standard real numbers that we're used to working with.  For example, they admit of "infinitesimal numbers," positive numbers which are less than $\frac{1}{n}$ for all positive integers n -- something that we don't have in the "normal" reals.