number theory
greatest common divisor
calculus
integration
algebra
function
domain

Hi, some time ago I studied some well known theorems about greatest common divisors (gcd), and from my own proof trials I have been led to the following question:

[b]Is there a commutative unitary ring R s.th. every two elements of R have a gcd, but there is an infinite subset S of R s.th. S does not have a gcd?[/b]

A gcd of a subset S of R is defined to be an element d, s.th. d divides every element of S (d is a common divisor) and every element c which divides every element of S must also divide d (among the common divisors d is the greatest).

[b]What is the answer if we restrict R to being an integral domain, a factorial ring or an euclidean ring?[/b]

LiMa

It can be done with modules; $\mathbb{Q}$ has this property as a $\mathbb{Z}$-module.

We need a more complicated construction for rings. Consider the ring of polynomials with rational coefficients on infinitely many  commuting variables $x_{1},x_{2},\dots,x_{n},\dots$, with relations $x_{1}^{2}=0$ and $x_{n+1}^{2}=x_{n}$. Every element with no constant term is nilpotent, so every element with a constant term is invertible. With some work, every element is the product of an invertible element and a monomial. The monomials are linearly ordered by divisibility, and thus, given any two elements, one divides the other and is a gcd for the pair.

On the other hand, the ideal of all nilpotent elements is not principal, and the set of nilpotent elements has no gcd.

Now that I think about it, there's a reasonably well-known integral domain with these properties: the ring of entire functions.

We can construct an entire function given any set of zeros (with multiplicity) that has no accumulation point. That allows us to construct the gcd of two functions by finding a function with zeros at the intersection of their zero sets.

On the other hand, there are nontrivial ideals that are not finitely generated, and their generating sets have no gcd.

It should be simple enough to see that every set has a gcd in factorial rings, and since a Euclidean ring is factorial, this takes care of the Euclidean case as well.

In general (even if we require the ring to be an integral domain) this is no longer the case. What follows is an example of this kind.

Let $K$ be a field, and let $\nu: K^{*}\to\Gamma$ be a [url=http://en.wikipedia.org/wiki/Valuation_ring]valuation[/url] whose image is the totally ordered abelian group $\Gamma$. We'll take our ring $A$ to be $\{a\in K\ |\ \nu(a)\ge 0\}$, as usual. For any totally ordered abelian group $\Gamma$, there is such a field $K$ (see, for example, Atiyah and Macdonald’s “Introduction to commutative algebra”, chapter $5$, where valuations are treated), so the construction is possible with any $\Gamma$ we choose.

If we choose $\Gamma$ such that its order is not complete, i.e. there is a set $S\subset\Gamma$, which we may as well assume consists of  non-negative elements, and which has no infimum (we can take $\Gamma$ to be the additive group of rational numbers with the usual order), then the ring $A$ has the required properties: every two elements have a gcd, because among every two elements one divides the other, but the set $\{a\in A\ |\ \nu(a)\in S\}$ has no gcd because of the choice of $S$.

Another example I had in mind, but which has zero divisors and so is superseded by the above, is an incomplete Boolean algebra (or Boolean ring; the two notions are equivalent). The phenomenon is basically the same: an element $p$ divides an element $q$ iff $q\le p$, so if the algebra is incomplete, any set with no supremum will have no gcd.

P.S.

Jmerry, I don't think the ring of entire functions works. In that ring $f|g$ iff the zero set of $f$ is contained in that of $g$ (here, "set" means "multiset with each point having finite multiplicity"). Moreover, every set with no limit point in the complex plane is the zero set of an entire function. Now, given a set of entire functions, any entire function having the intersection of their zero sets as its zero set will be a gcd (the gcd is determined up to multiplication by invertible elements, i.e. entire functions with no zeros).

Thank you both for your thoughts.

Jmerry, I thought about your first example (the polynomial ring), and came to the conclusion: Every two elements have a gcd, and there are nonprincipal ideals. But that alone is not enough for not having a gcd. I see that no monomial except 1 divides all nilpotents, and since up to units there are only monomials in the ring, my conclusion is that the gcd of the nilpotents is 1.
That the gcd cannot be presented as a linear combination is a different problem. Compare the situation of the ideal $(2, X)$ in $\mathbb Z[X]$.

grobber, I learned the algebraic closure of $\mathbb Q_{p}$ as an example of such a field (am I remembering right that its valuation group is $\mathbb Q$?). In its valuation ring A we have $a | b$ iff $v(a) \leq v(b)$.
Then by taking as S the subset of rationals greater than some positive irrational number, the set $\{a\in A\ |\ \nu(a)\in S\}$ has no gcd, because for any common divisor $d$ there is an element $c$ which has a greater valuation (so $c \nmid d$) but is still a common divisor.

And yes, grobber, it is indeed easy to see that in factorial rings every subset has a gcd. :)

LiMa

OK, I see the point. My examples work well from the ideal point of view (every finitely generated ideal is principal) but not from the strict elementwise point of view.
grobber's examples are still related; there are (partial) orders with infinite chains in every example listed here.

Incidentally, there is a way to salvage my first example. Every monomial $p$ can be associated to a dyadic rational $f(p)$ so that $p|q$ if and only if $f(p)\le f(q)$. The monomials $p$ with $f(p)\ge\frac13$ (or any other number that isn't a dyadic rational) have no gcd; the common divisors $q$ satisfy $f(q)\le\frac13$ (with the natural extension of $f$), and there's no maximum there. This is similar to grobber's valuation idea.

I stumbled over that "valuation" f in my calculations: The monomial $x_{m}^{n}$ can be associated to the fraction $n / 2^{m}$.

The ideal of nilpotent elements consists of all elements having a valuation greater than 0, and the valuations' infimum clearly is 0, with the element 1 having that valuation.
The set of elements with valuation greater than 1/3 does not have a gcd.
And similarly the set of elements with valuation smaller than 1/3 does not have a lcm.
Wow. :)

LiMa

A variation on my example which is an integral domain:
Drop the relation that $x_{1}^{2}=0$, and switch to formal power series rather than polynomials. We keep one restriction on these series: only finitely many of the variables are used in any one element.
This is effectively the ring of polynomial combinations of formal power series in $x$, $\sqrt{x}$, $\sqrt[4]{x}$, and so on. We use the same valuation to construct a set with no gcd or no lcm.


How many even four-digit numbers greater than $ 7000$ can be formed using the digits $ 3,4,5,7$, and $ 9$ when no repetition of digits is allowed in a number?

Since the number must be even, $ 4$ is fixed on the units digit.  So we have only one choice for the units digit.  For the thousands digit, it must either be $ 7$ or $ 9$, since the number must also be larger than $ 7000$.  Thus, there are $ 2$ choices for the thousands digit.  For the hundreds digit, we have either $ 3$, $ 5$, $ 7$ or $ 9$ (one of these will not be here, depending on which number was placed in the thousands digit).  This is $ 3$ choices.  Finally, the tens digit.  It has either $ 3$, $ 5$, $ 7$, or $ 9$ (two of these would not be here, since one of them is in the thousands digit, and the other in the hundreds).  This is a total of $ 2$ choices.  So our final answer is simply $ 2\cdot 3\cdot 2\cdot 1 \equal{} \boxed{12}$.  Sorry if this is a bit confusing...


conics
ellipse
function
real analysis
real analysis theorems

HELLO, (:-) October 23, 2007 6:43 p.m.
(Radius of the series full term $ z ^ n / sin (pi n y) which is$
Real non-rational)

In their 1946 article "Notes on the theory of series: a curious
Power series "procedings of the Cambridge Philosophical society vol 42 p 85-90
 Hardy and Littlewood, saying the earliest that the
lines Ray of the whole series of general term $ z ^ n / sin (pi n y)$
May take any value between 0 and 1, depending on the choice of non-y
Rational.

This is demonstrated by using fractions continued in the
Hervé remarkable article Queffélec p 127-138, no less
Impressive feat: numbers former and current problems
(Ellipses 2000)

If the authors believe that the radius of this series is well known,
Is that his study was made in a magazine, a book, an article:
I recheche any evidence to find the anticipation of this
Series (and indeed the author)
This is to complete an article that I wrote for the site culture
math,
To demonstrate the level of students in mathematics and special.

It's obvious that English isn't your first language, and I can't really understand what you're trying to say.

Related topics have come up several times here- here [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113870]are[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=99515]two[/url].

This is from MathReviews:
[quote="R. Salem"]the authors prove first that, $ \theta$ being any irrational number, if the radius of convergence of the series (S) $ \sum x^n/\sin n\pi\theta$ is $ \rho$, then the radius of convergence of the series (T) $ \sum x^n/(\sin\pi\theta\sin 2\pi\theta\cdots\sin n\pi\theta)$ is $ \rho/2$. [...] Since for almost all $ \theta$ the radius $ \rho$ is equal to 1, it follows that, for almost all $ \theta$, the radius of convergence of (T) is $ \frac12$.[/quote]
I think the original poster wants to know who was the first to study the radius of convergence of (S) as a function of $ \theta$.


calculus
derivative
abstract algebra
probability
algorithm
ratio

What is the difference between a hedge fund and an investment bank?
I should probably know this but i don't really see what the major difference is, sorry.

if you're in high school, you probably don't need to know this. if you colelge and not majoring in finance, you probably don't need to know it either hah. I'm in college majoring in applied computational math and economics with emphasis in finance. Let's see, Hedge Funds are nothing related to an ibank.

1. Hedge Funds are like your average mutual funds but they have less to no restriction by the SEC (Securities and Exchange Committee) and from this, they are able to pursue very aggressive strategies like futures, options, forwards, interest derivative trading and a lot more. They also strive to "hedge" risk which is to decrease it as much as posisble in order to maximize total profit. A hedge fund you usually need at least a million dollars to buy in that's why there's so few of them (only about 7000) and you don't hear about them that much because people like to investors like to keep it as private as possible. Many people base their careers off of hedge fund investing because a simple 1% profit return on a 5 billion dollar fund means 50 million dollars which is the yearly income of hundreds of investors.

2. Investment Banking is an ENTIRELY different ball game. Investment banking is geared entirely to big cooporations and handling a lot of the financial components of the companies including sales and trading of bonds, and stocks. They assist in raising funds in the capital markets (equity and debt). Two big things that I've been looking into as a career choice is M & A and IPOs. M & A is mergers and acquisitions that deals with companies merging together and when bigger comopanies take over smaller companies. A LOT of stuff goes into this process. IPOs are initial public offerings which is when a company first goes public (onto the stock market). Believe it or not, there is A LOT of work that needs to be filed in order to do this.

As you can see, there is little relation between the two fields. Hope that helps you out. There are whole courses dedicated to parts of Ibanking so it's hard to suimmarize it in such a small space.

According to Economist the separation decreases:

[quote="Economist"] Buttonwood
Identity crisis

Jun 28th 2007, From The Economist print edition, As the line blurs between hedge funds and banks, a bit of mystique goes missing

PEOPLE have trouble defining the term hedge fund. For some it simply conveys an aura of big money tinged with a dashing hint of menace. But within a few years the term may be even more meaningless than it is now, because hedge funds are rapidly becoming indistinguishable from the rest of the financial-services industry.

D.E. Shaw, an American group, is a case in point. It started as a “quantitative” manager, using sophisticated computer models to pick stocks and, with $ \$$26 billion under management at the end of 2006, was ranked as one of the four largest fund groups in the world.

But hedge funds were only the beginning; there is barely a financial activity in which D.E. Shaw is not now involved. In early June it announced a bid for James River, an insurance firm. The group already has an arm, Laminar Direct Capital, that makes direct loans to firms. It has considered moving into private equity and owns FAO Schwarz, a big toy store. As well as running hedge funds, it operates a “long-only” business, which buys assets in the hope they rise in price.

As hedge funds like D.E. Shaw move in one direction, investment banks and conventional fund managers are shifting in the other. Many have bought hedge-fund groups outright (such as JPMorgan Chase's purchase of Highbridge Capital Management) or have taken minority stakes in them (Lehman Brothers bought 20% of D.E. Shaw in March). Others either operate funds-of-hedge-funds (Goldman Sachs) or have set up separate hedge-fund arms (Gartmore and—less successfully of late—Bear Stearns).

There also seems to be a growing belief that there is more to investment than long-only management. The latest fashion is 130/30 funds, which use borrowed money to combine 130% long positions with 30% short (betting on falling prices). According to this philosophy, stopping fund managers from shorting stocks is like preventing Tiger Woods from using all the clubs in his bag; smart investors should be able to spot overpriced stocks as well as underpriced stocks. Such products, which have been dubbed “hedge funds lite”, allow investors such as pension funds to take their first steps into the world of “absolute return” investing.

It is not too difficult to work out why banks and traditional fund-management firms should want to be more like hedge funds. For a start, the annual management fees are a lot higher. Second, as the flotations of Fortress and Blackstone, two large and varied alternative-investment firms, have shown, the stockmarket is willing to pay a very high multiple for companies that earn performance fees.

But why do hedge-fund groups want to move the other way? Part of the reason is the Darwinian environment in which they operate. They are constantly on the lookout for markets that are inefficient or areas that offer excess returns. In banking and insurance, for example, hedge funds may benefit because they lack either the costly infrastructure or regulatory burdens that impede the traditional operators; borrowers say hedge funds are much quicker than banks at deciding whether to make a loan.

For the individual hedge-fund manager, diversifying makes sense. Some strategies may be profitable for a while, but then have bad years, as convertible-arbitrage managers found out in 2005. If returns are bad enough, the business can disappear overnight. But that is far less likely to happen with a range of strategies.

There is a further level of protection if the manager raises “permanent capital” by issuing shares. Hedge-fund investors have the right to withdraw their capital, subject to lengthy notice periods. But if the manager is running a listed fund, investors can redeem their holdings only by selling them on the open market; the annual management fee is unaffected.

Permanent capital can also be raised in a different way if the hedge fund issues bonds (as Citadel did last year) or floats shares of the management company (this week London-based GLG Partners became the latest to aim for a New York listing). Such capital-raising exercises allow founders to cash in their holdings and also give the hedge funds some independence from their prime brokers, on whom they depend heavily when borrowing money.

But flotations also force hedge-fund managers to be more transparent, diluting the mystique on which their high fees partly depend. And they accelerate the process by which boutiques turn into broadly based financial groups, with all the bureaucracy that implies (bureaucracy that many managers went into the business to escape). Hedge funds may be gaining fame and fortune as they expand, but they may be losing part of their soul.[/quote]

I was wondering about the applied methods in the fields of investment banking and hedge funds. From my experience which is based on reading web sites, books and atttending career fairs is the following approaches from applied math are applied:

hedge fund: more statistical techniques, machine learning as (non-)parametric inference, Bayesian techniques in general, kernel methods, time series

investment banks: probability theory, stochastic calculus, partial differential equations

Is that roughly correct? What is your opinion?

I'm sorry for bumping this old post, but I am also interested in seeing if orl's list is right.

I think the list is true, for a particular function.
A higher % of HF jobs ask for those skills than ones at investment banks.
This applies particularly to machine learning, genetic algorithms, AI in general.

However there are so many jobs at IBs that you will find lots of roles that ask for "HF" skills.
Stats, Bayesian and time series methods are to be found in large quantities on banks.

The difference is that banks still do many things that are quite rare in banks so the ratio of skills is different, but there is conversion.

I'm not sure exactly how helpful I'm being here :(

If you're making decisions on choice of courses, I'd normally tell you to lean towards those ones you feel you will excel at rather than aiming at IBs or HFs.


Có ai chỉ cho mình phương pháp dồn biến mạnh :D
 :blush:

Thanks :D

vào trang này xem :) http://diendantoanhoc.net/forum/index.php?s=b7619e2a2ae399f424d1cd0927e56aa3&showtopic=22514&pid=128204&st=0&#entry128204

Nếu là SMV thì bạn vào đây:
http://reflections.awesomemath.org/2006_6/2006_6_mixing.pdf
hoặc: http://www.mathlinks.ro/viewtopic.php?t=113248


inequalities
triangle inequality
IMO 2004
IMO
IMO Shortlist
Hi
n-variable inequality

Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.

solution by induction: the hardest part is to show that it works for n=3. so assume $t_1\geq t_2\geq t_3$. it is easy to see that the right is strictly increasing in $t_3$. so if we show that the reverse inequality holds for $t_1=t_2+t_3$, we are done. a quick calculation is enough to show that(it is too boring to post). but if we know is true for n-1, then let's do the following computation:
$((t_1+...+t_{n-1})+t_n)(\frac{1}{t_1}+...+\frac{1}{t_{n-1}}+\frac{1}{t_n})\geq (t_1+...+t_{n-1})(\frac{1}{t_1}+...+\frac{1}{t_{n-1}})+2n-1$,
by AM-GM and AM-HM. so we get the inequality reduces to that one for n-1 variables.

Peter

It reminds me the following problem from China 1987/88 :

Let $n \geq 3$ be an integer. Suppose the inequality
$(a_1^2 + a_2^2 + ... + a_n^2)^2 > (n-1)(a_1^4 + ... a_n^4)$
holds for some positive real numbers $a_1,...a_n$. Prove that $a_i,a_j,a_k$ are the three sides of some triangle for all $i<j<k$.

Even there is an elegant direct solution, the most common way is induction.

Pierre.

When I saw this, my immediate thought was- an elegant direct solution.

The inequality
$n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + ... + \frac{1}{t_n} \right)$
reduces to $\displaystyle{n^2+1>\sum_{j=1}^n \sum_{k=1}^n{\frac{t_j}{t_k}}}$
$\displaystyle{n^2+1>n+\sum_{1 \le j<k \le n}{\frac{t_j}{t_k}+\frac{t_k}{t_j}}}$
Each of the terms in the last sum is at least 2, so if the sum of some $m$ terms is at least $2m+1$, the whole sum will be at least $n^2-n+1$ and the inequality will be violated.

Suppose $t_i+t_j \le t_k$. Then $\frac{t_i}{t_k}+\frac{t_j}{t_k} \le 1$
If we can show that $a+b+\frac{1}{a}+\frac{1}{b} \ge 5$ when $a+b \le 1$, we will be done by the argument above.
First, we have $a+\frac{1}{a}=\frac{a^2+1}{a} \ge \frac{(a-\frac{1}{2})^2+a+\frac{3}{4}}{a} \ge 1+\frac{3}{4a}$
and $b+\frac{1}{b} \ge 1+\frac{3}{4b}$
so $a+b+\frac{1}{a}+\frac{1}{b} \ge 2+\frac{3}{4}(\frac{1}{a}+\frac{1}{b})$
By AM-HM, $\frac{2}{\frac{1}{a}+\frac{1}{b}} \le \frac{a+b}{2} \le \frac{1}{2}$, so $\frac{1}{a}+\frac{1}{b} \ge 4$ and $a+b+\frac{1}{a}+\frac{1}{b} \ge 5$

In fact the case n=3 is sufficient to prove the desired fact for all n - we can simply pair off the other terms and only worry about the ones in our non-triangle.

it is, indeed, a very elegant solution jmerry!

Kantor

Well, I just adore this problem. So easy, beautiful and elegant. But try to think about the following problem, which I think it's much more difficult and interesting:
    For iven n>2, find the greatest constant k (which may depend on n) such that if positive reals $ x_1,...,x_n$ verify $ k>(x_1+...+x_n)(\frac{1}{x_1}+...+\frac{1}{x_n}) $ then any three of them are sides of a triangle. I think I've managed to solve this problem and if my solution is correct, then it's a really hard and nice problem.

If my computation is not wrong then it should be [tex] n^2+(2\sqrt10-6)n+19-6\sqrt{10} [/tex]

Yes, it is $ (n+\sqrt{10}-3)^2  $. I think this one is much more interesting.

btw, the creator of this problem is Hojoo Lee :)

Another solution:

Note that

\[ (t_1+ \cdots + t_n) ( \frac{1}{t_1} + \cdots  + \frac{1}{t_n}) = n^2 + \sum_{i<j} (\sqrt{\frac{t_i}{t_j}} - \sqrt{\frac{t_j}{t_i}} )^2 \]

And so if three of the terms satisfy $a \geq b + c$ then

\[{{ (\sqrt{a/b} - \sqrt{b/a})^2 + (\sqrt{a/c} - \sqrt{c/a})^2 =
\frac{ (a-b)^2}{ab} } + \frac{ (a-c)^2}{ac} }
\geq \frac{(a-b)c}{ab} + \frac{(a-c)b}{ac}
= c/b + b/c -(b+c)/a \geq 2 - 1 = 1 \]

And the result follows immediately.

(During the IMO, I accidentally forgot the square root, costing me a huge chunk of marks).

just wanted to say that this is *really* nice ...
btw, anh cuong/harazi could you please post the solution to the harder version (where n^2+1 is replaced by  ( n + :sqrt: 10 - 3 ) ^ 2 )? thanks.

Of course, Hojoo Lee is the official proposer. But I knew the case n=3 some four months ago, when I tried to create a problem for the junior TST in Romania.

[quote="harazi"]...find the greatest constant k (which may depend on n) such that if positive reals [tex]x_1,x_2,...,x_n[/tex]  verify  [tex]k(n)>(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)[/tex] then any three of them are sides of a triangle. [/quote]

[quote="harazi"]Yes, it is [tex](n+\sqrt{10}-3)^2[/tex] . I think this one is much more interesting.[/quote]
Consider [tex]f(x_1,x_2,...,x_n)=(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)[/tex] where [tex]x_1<=x_2<=...<=x_n[/tex].

Greatest constant [tex]k(n)=(n+\sqrt{10}-3)^2[/tex] might be found elementary. From the solution of Sung-yoon Kim:

[quote="Sung-yoon Kim"]...By Cauchy-Schwartz inequality,
..........
So [tex]n^2+1>(\sqrt((x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3))+(n-3)^2[/tex]

=> [tex](x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3)<(\sqrt(n^2+1) -(n-3))^2[/tex]

[/quote]

instead of [tex](n^2+1)[/tex] , use that [tex]k(n)[/tex] and instead of  [tex] x_3[/tex] , use  [tex] x_n[/tex]:
[tex](x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2[/tex]
Assume [tex]x_1+x_2<=x_n[/tex] then:
[tex](x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)>=10[/tex] , so we must have [tex]10>=(\sqrt(k(n)) -(n-3))^2[/tex],
which leads to the wanted answer. Indeed if:
[tex](x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2[/tex], then:
[tex]x_1+x_2>x_n[/tex] , so for any [tex]1<= i, j, k <= n[/tex] :
[tex]x_i+x_j>x_k[/tex] .

Of course, there is a quite simple (but not elementary) "forged" solution, using optimization theory:
[tex]k(n)=min f(x_1,x_2,...,x_n)[/tex] ,
when [tex]x_1<=x_2<=...<=x_n[/tex] and [tex]x_1+x_2<=x_n[/tex] , we may get:
[tex]k(n)=f(1,1,2\sqrt(2/5),...,2\sqrt(2/5),2)[/tex], where [tex] 2\sqrt(2/5)[/tex] appears [tex](n-3)[/tex] times.

I think it's pretty interesting.

Problem may be generalised as it follows:
Instead of a triangle, use a polygon of "given" [tex]M[/tex] edges , [tex]M>=3[/tex] so if the same inequality holds for any [tex]n>=M[/tex] positive reals (of course with something like [tex]k(n,M)[/tex] in the RHS) then any [tex]M[/tex] from our  [tex]n[/tex] real numbers can be the side lengths of a polygon with [tex]M[/tex] edges. Solution is basically the same.

Assume for the sake of contradiction the condition is false. WLOG, $t_1\geq t_2+ t_3$. By AM-GM, $\frac{t_i}{t_j} + \frac{t_j}{t_i} \geq 2$, so the sum of the $n^2-4$ fractions that don't have $t_1$ in the numerator or denominator (however, we do count $\frac{t_1}{t_1}$) is at least $n^2-4$. We claim that
\[
\frac{t_1}{t_2} + \frac{t_1}{t_3} + \frac{t_2}{t_1} + \frac{t_3}{t_1} = t_1 \left( \frac{t_2+t_3}{t_2 t_3}\right) + \frac{t_2+t_3}{t_1} \geq 5.
\]
It suffices to minimize the expression. Let $t_2+t_3 = s, t_1=s+x$, then $t_2t_3$ is maximized when $t_2=t_3=s/2$, and it remains to show $f(x) = \frac{4x}{s} + \frac{s}{s+x}\geq 1$. Taking the derivative with respect to $x$, we get $\frac{4}{s} - \frac{s}{(x+s)^2}$. It is easy to see this is positive for $x\geq 0$, so $f(x)$ is minimized when $x=0$ (over nonnegative $x$). Then the right hand side is at least $(n^2-4) + 5 \geq n^2+1$, contradiction.

Due to Cauchy Schwarz,
\[ -(n-1)^2 \geq -(t_1 + t_2 + \dots + t_{n-1}) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_{n-1}} \right). \]
Adding this to both sides of the equation and manipulating gives us the restriction
\[ \left( \frac{t_n}{t_1} + \frac{t_1}{t_n} \right) + \left( \frac{t_n}{t_2} + \frac{t_2}{t_n} \right) + \dots + \left( \frac{t_n}{t_{n-1}} + \frac{t_{n-1}}{t_n} \right) < 2n-1.\]

It suffices to prove that $t_n < t_1 + t_2$, because all the other inequalities we're asked to prove will follow by symmetry. Due to AM-GM, each of these $n-1$ terms is at least $2$; thus, subtracting, we have the restriction
\[ \left( \frac{t_n}{t_1} + \frac{t_1}{t_n} \right) + \left( \frac{t_n}{t_2} + \frac{t_2}{t_n} \right) < 5. \]
If we plug in $t_n = t_1 + t_2$, we have
\begin{align*}
& \left( \frac{t_1 + t_2}{t_1} + \frac{t_1}{t_1+t_2} \right) + \left( \frac{t_1+t_2}{t_2} + \frac{t_2}{t_1+t_2} \right)\\
= ~ & 1 + \left( t_1 + t_2 \right) \left( \frac{1}{t_1} + \frac{1}{t_2} \right) \\
\geq ~ & 1 + 2^2 = 5,
\end{align*}
and if we increase $t_n$ greater than $t_1 + t_2$, analyzing the previous expression we clearly see that both terms only increase.

Therefore, $t_n \leq t_1 + t_2$, and all likewise inequalities follow by symmetry.

Can someone check this? It seems much shorter than the ones in this thread..

Assume that without loss of generality we have $t_3 \geq t_1 + t_2$. Note then that it suffices to show,
\[ (t_1 + t_2 + t_3)\left( \frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \right) + \sum_{i = 4}^n \sum_{j = 1}^3 \left( \frac{t_j}{t_i} + \frac{t_i}{t_j} \right) + \left(\sum_{i = 4}^n t_i \right) \left( \sum_{i = 4}^n \frac{1}{t_i} \right) \geq n^2 + 1.\]
Now using cauchy on the last term, and AM-GM on the middle, it suffices to show,
\[ (t_1 + t_2 + t_3)\left( \frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \right) \geq 10.\]
However we can reduce this to showing,
\[
\left( \frac{t_1}{t_2} + \frac{t_1}{t_3} + \frac{t_2}{t_1} + \frac{t_2}{t_3} + \frac{t_3}{t_1} + \frac{t_3}{t_2} \right) \geq 7
\]
Letting $t_3 = t_2 + t_1$ it is easy to see this holds by AM-GM. However it is also easy to see that as $t_3$ grows larger the terms $\left( \frac{t_3}{t_1} + \frac{t_1}{t_3} \right)$ and  $\left( \frac{t_3}{t_2} + \frac{t_2}{t_3} \right)$ are strictly increasing, so we are done.

WLOG assume $t_i \le t_j$ iff $i \le j$. It suffices to show $t_1+t_2>t_n$.

For the sake of contradiction, assume that $t_1+t_2 \le t_n$. Notice that

\[\left(\sum_{i=1}^n t_i \right)\left(\sum_{i=1}^n \frac{1}{t_i} \right) = \left(\sum_{i=1}^{n-1} t_i \right)\left(\sum_{i=1}^{n-1} \frac{1}{t_i} \right)+\frac{t_n}{t_1}+\frac{t_n}{t_2}+\frac{t_1}{t_n}+\frac{t_2}{t_n}+\sum_{i=3}^{n-1} \left(\frac{t_n}{t_i}+\frac{t_i}{t_n} \right)+1\]

Then, we have

\[\left(\sum_{i=1}^{n-1} t_i \right)\left(\sum_{i=1}^{n-1} \frac{1}{t_i} \right) \ge (n-1)^2\]

by Cauchy. Additionally,

\[\sum_{i=3}^{n-1} \left(\frac{t_n}{t_i}+\frac{t_i}{t_n} \right) \ge 2n-6\]

by AM-GM. Finally,

\[\frac{t_n}{t_1}+\frac{t_1}{t_n}+\frac{t_n}{t_2}+\frac{t_2}{t_n}\ge \frac{t_1+t_2}{t_1}+\frac{t_1}{t_1+t_2}+\frac{t_1+t_2}{t_2}+\frac{t_2}{t_1+t_2}=3+\frac{t_2}{t_1}+\frac{t_1}{t_2}\ge 5\]

and substituting all of this into our original expression yields

\[\left(\sum_{i=1}^n t_i \right)\left(\sum_{i=1}^n \frac{1}{t_i} \right) \ge n^2+1,\]

a contradiction. $\square$

[hide=Cute]
If FTSOC $t_1\ge t_2+t_3$:
\begin{align*}
    \sum t_i\cdot\sum\frac{1}{t_i}&=\frac{t_1}{t_2}+\frac{t_1}{t_3}+\frac{t_2}{t_1}+\frac{t_3}{t_1}+n+\sum\left(\frac{t_i}{t_j}+\frac{t_j}{t_i}\right)\\
    &\ge\frac{t_1}{t_2}+\frac{t_1}{t_3}+\frac{t_2}{t_1}+\frac{t_3}{t_1}+n^2-4,
\end{align*}
so it suffices to show $\tfrac{t_1}{t_2}+\tfrac{t_1}{t_3}+\tfrac{t_2}{t_1}+\tfrac{t_3}{t_1}\ge5$.
If WLOG $t_1=1$, the statement reduces to
\[x+y\le1\stackrel{?}{\implies}x+y+\frac{1}{x}+\frac{1}{y}\ge5.\]
But clearly $x+y+\tfrac{1}{x}+\tfrac{1}{y}\ge x+y+\tfrac{4}{x+y}=S+\tfrac{4}{S}$ where $S:=x+y$, and it is easy to see that the minimmum of $f(S):=S+\tfrac{4}{S}$ on the interval $(0,1]$ is $5$ at $S=1$, as desired.

Sketch:

Without loss of generality, let the side lengths be $t_1 \leq t_2 \leq t_3 \dots \leq t_n$. If all triplets form a triangle, then the condition allowing this is $t_1 + t_2 \geq t_n$. We can contradict this by saying that $t_1 + t_2 < t_n$. If this occurs, then $\frac{t_1}{t_n} + \frac{t_2}{t_n} < 1$. Now with either algebraic identities/HM-AM/calculus, you can readily derive that $\frac{t_1}{t_n} + \frac{t_2}{t_n}  + \frac{t_n}{t_1} + \frac{t_n}{t_2} > 5$. However, this makes the value of $(\sum_{i = 1}^{n}t_i)(\sum_{i = 1}^{n}\frac{1}{t_i}) = n + \sum_{1 \leq i < j \leq n}(\frac{t_i}{t_j} +  \frac{t_j}{t_i})$ to be larger than $n^2 + 1$. as each pair has a sum of at least $2$, however two of the pairs have a sum of $5$.

we prove the contrapositive.for $n=3$ set $t_1,t_2,t_3=x,y,1$ and we need to show $\frac xy+\frac yx+x+y+\frac1x+\frac1y\ge 7$ for$x+y\le 1.$ This is because $\frac xy+\frac yx\ge 2,\frac1x+\frac1y\ge\frac4{x+y}$ and $(x+y)+\frac4{x+y}\ge5$ for $x+y\le 1.$ Now for $n\ge 4$ we have $(t_1+t_2+\cdots+t_n)\left(\frac1{t_1}+\frac1{t_2}+\cdots+\frac1{t_n}\right)=n+\sum_{1\le i<j\le n}\frac{t_i}{t_j}+\frac{t_j}{t_i}$ and wlog $t_1+t_2\le t_3.$ Then from earlier we have $\sum_{1\le i<j\le 3}\frac{t_i}{t_j}+\frac{t_j}{t_i}\ge10$ and for all other $i,j$ we have $\frac{t_i}{t_j}+\frac{t_j}{t_i}\ge 2$ so summing finish

to generalize. to find the strong bound we show the contrapositive again. wlog $i<j\implies t_i\le t_j$ and assume $t_1+t_2\le t_n.$ we can notice that replacing $t_i,t_j$ with $\frac{t_i+t_j}2,\frac{t_i+t_j}2$ decreases the expression by amgm when $3\le i,j\le n-1$ and replacing $t_1,t_2$ with $\frac{t_1+t_2}2,\frac{t_1+t_2}2$ also decreases the expression. thus we may assume $t_1=t_2,t_3=\cdots=t_{n-1}.$ now notice that upon expanding we get the expression is $n+\sum_{i\le i<j\le n}\frac{t_i}{t_j}+\frac{t_j}{t_i}$ and when $j=n$ then decreasing $t_n$ increases all pairs $\frac{t_i}{t_n}+\frac{t_n}{t_i}$ by convexity. thus we may assume $t_n=2t_1.$ now since it is homogeneous we assume $t_1=t_2=1,t_n=2.$ thus if we take $t_3=\cdots=t_{n-1}=x$ then we want to minimize $(4+(n-3)x)\left(\frac52+\frac{n-3}x\right).$ upon expanding this equals $n^2-6n+19+\frac{5n-15}2x+\frac{4n-12}x\ge n^2-6n+19+2(n-3)\sqrt{10}$ by amgm. this is sharp at $x=\sqrt{1.6}$ so equality does hold.

I don't remember my solution being this messy :P

We proceed via induction.

[b]Base case:[/b] set $n = 3$. Suppose without loss of generality that $t_1 + t_2 \leq t_3$. Now we let \[a = \frac{t_1}{t_3}, b = \frac{t_2}{t_3}\] for convenience so that $a + b \leq 1$. We have (where $s = a + b \leq 1$)
\begin{align*}
    (t_1 + t_2 + t_3)\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3}\right) &= 3 + \frac{a}{b} + \frac{b}{a} + a + \frac{1}{a} + b + \frac{1}{b} \\
    &= 3 + \frac{a}{s - a} + \frac{s - a}{a} + a + \frac{1}{a} + s - a + \frac{1}{s - a} \\
    &= 2 + s + \frac{a + 1}{s - a} + \frac{s + 1}{a} \\
    &= 2 + \frac{a^2 (1 - s) + a(s^2 - s) + (s^2 + s)}{a(s - a)}\text{.}
\end{align*}
Now consider the inequality
\begin{align*}
    a^2 (1 - s) + a(s^2 - s) + (s^2 + s) &\geq 8a(s - a) \\
    a^2 (9 - s) + a(s^2 - 9s) + (s^2 + s) &\geq 0\text{.}
\end{align*}
We will show that this always holds. Here the discriminant of the quadratic in $a$ is \[\Delta = (s^2 - 9s)^2 - 4(9 - s)(s^2 + s) = s(s - 1)(s - 4)(s - 9) \leq 0\text{.}\] Furthermore, the leading coefficient is $9 - s$ which is always positive so this inequality always holds, as desired.

Now note that since $0 < a < s$, $a(s - a)$ is positive so we have
\begin{align*}
    2 + \frac{a^2 (1 - s) + a(s^2 - s) + (s^2 + s)}{a(s - a)} &\geq 2 + 8 \\
    &= 10 \\
    &= n^2 + 1\text{, a contradiction.}
\end{align*}

[b]Inductive step:[/b] suppose the proposition holds for some fixed $n \geq 3$. Let \[S_n = (t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\text{.}\]
Assume without loss of generality that $t_1 \leq t_2 \leq t_3 \leq \dots \leq t_{n + 1}$.

Hence
\begin{align*}
    (n + 1)^2 + 1 &= n^2 + 2n + 2 \\
    &> S_{n + 1} \\
    &= S_n + 1 + \sum_{i = 1}^n \left(\frac{t_i}{t_{n + 1}} + \frac{t_{n + 1}}{t_i}\right) \\
    &\geq n^2 + 1 + \sum_{i = 1}^n \left(\frac{t_i}{t_{n + 1}} + \frac{t_{n + 1}}{t_i}\right) \\
    2n + 1 &\geq \sum_{i = 1}^n \left(\frac{t_i}{t_{n + 1}} + \frac{t_{n + 1}}{t_i}\right)\text{.}
\end{align*}

Once again let $a_i = \frac{t_i}{t_{n + 1}}$ for all $1 \leq i \leq n$. Suppose that $t_i + t_j \leq t_{n + 1}$ or equivalently $a_i + a_j \leq 1$ for some $i < j$. We have (where $s = a_i + a_j \leq 1$)
\begin{align*}
    a_i + \frac{1}{a_i} + a_j + \frac{1}{a_j} &= s + \frac{1}{a_i} + \frac{1}{s - a_i} \\
    &= -\frac{a_i^2 s - a_is^2 - s}{a_i (s - a_i)}\text{.}
\end{align*}
Now consider the inequality
\begin{align*}
    -a_i^2 s + a_is^2 + s &\geq 5a_i (s - a_i) \\
    a_i^2 (5 - s) + a_i (s^2 - 5s) + s &\geq 0
\end{align*}
We will show that this always holds. Here the discriminant of the quadratic in $a$ is \[\Delta = (s^2 - 5s)^2 - 4(5 - s)s = s(s - 1)(s - 4)(s - 5) \leq 0\text{.}\] Furthermore, the leading coefficient is $5 - s$ which is always positive so this inequality always holds, as desired.

Now note that since $0 < a_i < s$, $a_i(s - a_i)$ is positive so we have \[-\frac{a_i^2 s - a_is^2 - s}{a_i (s - a_i)} \geq 5\text{.}\]

Thus
\begin{align*}
    \sum_{i = 1}^n \left(\frac{t_i}{t_{n + 1}} + \frac{t_{n + 1}}{t_i}\right) &= \sum_{i = 1}^n \left(a_i + \frac{1}{a_i}\right) \\
    &= 5 + \sum_{k \in \{1, 2, \dots, n\} \setminus \{i, j\}} \left(a_k + \frac{1}{a_k}\right) \\
    &\geq 5 + 2(n - 2)\text{ (by AM-GM)} \\
    &= 2n + 1\text{.}
\end{align*}

Equality holds in the AM-GM part if and only if $a_k = 1$ for all $k \in \{1, 2, \dots, n\} \setminus \{i, j\}$ or in other words $t_k = t_{n + 1}$. However, $t_i, t_j \neq 0$ so it follows that $t_i, t_j < t_{n + 1}$, and that with the assumption that $t_1 \leq t_2 \leq \dots \leq t_{n + 1}$, yields that we must have $i = 1, j = 2$. But then by the inductive hypothesis $t_i + t_j > t_3 = t_4$, a contradiction.

Thus equality does not hold, and so \[\sum_{i = 1}^n \left(\frac{t_i}{t_{n + 1}} + \frac{t_{n + 1}}{t_i}\right) > 2n + 1\] which is a contradiction.

Huh? This felt rather difficult for an IMO Problem 4.

Assume for the sake of contradiction that $t_1, t_2, t_3$ don't form the sides of a triangle; that is, assume WLOG that $t_1 + t_2 \le t_3.$ By simply expanding the product, we get
\[
n^2 + 1 > n + \sum_{1 \le i < j \le n} \left(\frac{t_i}{t_j} + \frac{t_j}{t_i}\right).
\]
We can take out $\sum_{1 \le i < j \le 3} \left(\frac{t_i}{t_j} + \frac{t_j}{t_i}\right)$ from the sum to get
\[
n^2 + 1 > n + \sum_{1 \le i < j \le 3} \left(\frac{t_i}{t_j} + \frac{t_j}{t_i}\right) + \sum_{\text{rest of terms}} \left(\frac{t_i}{t_j} + \frac{t_j}{t_i}\right),
\]
where the number of terms in the last summation is $\frac{n^2 - n - 6}{2}.$ By AM-GM, each term in the last summation is greater than or equal to $2,$ hence that entire summation is greater than or equal to $n^2 - n - 6.$ Therefore,
\[
n^2 + 1 > n + \sum_{1 \le i < j \le 3} \left(\frac{t_i}{t_j} + \frac{t_j}{t_i}\right) + n^2 - n - 6 = n^2 - 6 + \sum_{1 \le i < j \le 3} \left(\frac{t_i}{t_j} + \frac{t_j}{t_i}\right).
\]
Redefining $a = t_1, b = t_2,$ and $c = t_3,$ we get
\[
7 > \frac{a}{b} + \frac{c}{b} + \frac{b}{a} + \frac{c}{a} + \frac{a}{c} + \frac{b}{c}.
\]
By AM-GM, we get $\frac{a}{b} + \frac{b}{a} \ge 2,$ so plugging that in and rearranging terms gives
\[
7 > 2 + c \left(\frac{1}{a} + \frac{1}{b}\right) + \frac{a+b}{c}.
\]
Now let $x = \frac{a+b}{c} \le 1.$ By AM-HM, we get $\frac{1}{a} + \frac{1}{b} \le \frac{4}{a+b},$ so plugging that in gives us
\[
5 > \frac{4c}{a+b} + \frac{a+b}{c} = \frac{4}{x} + x.
\]
Multiplying both sides by $x$ and rearranging gives
\[
x^2 - 5x + 4 = (x-1)(x-4) < 0,
\]
which is a contradiction since $x \le 1.$

Suppose not, then we have $a,b,c\in t_i$ such that $a+b\le c$. But note that the terms in the RHS besides
\[S=\frac ac+\frac bc+\frac ca+\frac cb\]
will add up to at least $n^2-4$ by AM-GM, so we have $S<5$. WLOG $c=1$, so
\[S=\left(1+\frac{1}{ab}\right)(a+b).\]
For a fixed $s=a+b$, $1+\frac{1}{ab}$ is minimized when $a=b$. Therefore,
\[S\ge 2a+\frac{2}{a}\ge 2\cdot\frac12+\frac{2}{\frac12}=5,\]
contradiction. $\blacksquare$

10 mohs alg is harder than 20 mohs geo :stretcher:
Suppose for the sake of contradiction there exists $a,b,c \in t_i$, such that $c= a+ b+ \epsilon$ for $\epsilon \ge 0$
then by cauchy
$$\left(t_1 + t_2 + \dots + t_n\right)
  \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) \ge \left(\sqrt{(a+b+a+b+\epsilon)\left(\frac{1}a+ \frac{1}b+ \frac{1}{a+b+\epsilon} \right)} + (n-3) \right)^2$$
  We claim
  $$(a+b+a+b+\epsilon)\left(\frac{1}a+ \frac{1}b+ \frac{1}{a+b+\epsilon} \right) > 10$$

This  is true because it equals
$$3+\frac{a}{b}+\frac{b}{a}+\frac{\epsilon}{a}+\frac{\epsilon+a+b}{a}+ \frac{\epsilon +a+b}{b}+ \frac{a+b}{a+b+\epsilon}$$
$$\ge 9+ \frac{\epsilon}{a}+ \frac{\epsilon}{b}+1 -\frac{\epsilon}{a+b+\epsilon}$$
$$\ge 10+ \epsilon(\tfrac{1}{a}+\tfrac{1}{b}- \tfrac{1}{a+b+\epsilon})$$
$$>10$$
Now
$$\left(t_1 + t_2 + \dots + t_n\right)
  \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) > (\sqrt{10}+n-3)^2$$
  Which is greater than $n^2+1$ for $n\ge 3$ contradiction $\square$

Assume FTSOC that $t_n \geq t_1 + t_2$. We claim that $\frac{t_1}{t_n} + \frac{t_1}{t_2} + \frac{t_n}{t_1} + \frac{t_n}{t_2} + \frac{t_2}{t_n} \geq 7$ Notice that as $t_n$ increases the expression decreases. Thus, we just need to prove that when $t_n = t_1+t_2$ we have the equality case.

This is easy since we have \[\frac{t_1}{t_n} + \frac{t_1}{t_2} + \frac{t_n}{t_1} + \frac{t_n}{t_2} + \frac{t_2}{t_n} = 3 + \frac{2t_1}{t_2} + \frac{2t_2}{t_1} \geq 7\] Now we can use AM-GM on all of the other terms in the expansion of our product to see the result. $\blacksquare$.

Very similar to @asdf334


conics ellipse geometry perimeter articles calculus integration

Okay, i know that i have previously posted something on this forum about how to find the circumference of an ellipse. Everyone said that it couldn't be found with a nifty formula, but.....  :idea:

After working through a million tests, I have looked at all of the perimeter of an ellipse questions, and seen that all of the "correct" answers follow the pattern:
\[ 2\pi \sqrt\frac {a^{2} \plus{} b^{2}}{2}
\]
Are there any objections/ideas/corrections/better formulas that you guys know/can think of?

Please test this out and give me some feedback :P

That formula is a worthwhile [i]approximation[/i] to remember, but you need to keep in mind that it is nothing but an approximation. It does not give the correct value (except for rare combinations, such as a=b, ie, a circle).

Are these local contests? Do the questions ask, "which is closest to the actual value?" If I were to write a contest, I would never make a problem that requires using this kind of memorized approximation; to me, that's bad problem writing.

You can read more about the "circumference" of an ellipse [url=http://en.wikipedia.org/wiki/Ellipse#Circumference]in this article[/url]. It introduces the elliptic integral, and also mentions several approximations.


How hard are the MathCounts Nationals?  I've never been to a national competition (lost at state from stupid mistakes), and you can't get national competitions tests online.

Look through the national mathcounts math jam transcripts, there's a good variety of problems there. The problems aren't hard, but you have do them quickly and accurately.

The sprint wasn't hard if you know how to do the problems and can do them fast.(which I can do, I just made like 8 extremely careless errors) The target wasn't that hard either.(That is if you do the problem right one of the first five times you do it.)

Are the MathCounts problems at the end of the handbooks National Competition level?  My teacher gave me the handbooks from the past 3 years.

Hmmm... the problems at the back are about nationals sprint round I think, but they're all beginning to look the same for me, so I'm not sure.  Probably you should just get past nationals contests.  I think there are quite a few people who have like all of them, and they can illegally copy -- *ahem* -- give you some problems.  I think this website by "Elias Saab" has good problems too.  The only problem with those is that they are worded like "If n is integer between 1 and 5?  What is the probablity is prime."  If you can read it, though...

i thought the problems in the handout would be more of a practice for chapter, or at most, state

Actually, I heard [u]most[/u] of those problems are at the "school" competition level.

Hmm... that's odd.. I found the questions useful though, since they used the basic skills recurring in nats problems that you had to be fast at.  There are also an occasional few that are kind of challenging.  Of course, these are only the workouts, and they map to only the easy to medium nats sprint round problems.  Then again, I can't really tell the difference anymore.

If you want a few samples, the MATHCOUNTS Web site typically lists a few questions in their "Problem of the Week" after each competition window.  The newsletters also sometimes contain sample questions.

[quote="nr1337"]Are the MathCounts problems at the end of the handbooks National Competition level?[/quote]

This is a difficult comparison.  Your best bet, really, is to order the Competition Sets through Sports Awards.  If you and your school are serious about Mathcounts, you should order as much competition prep material from Sports Awards as possible.

[quote="rcv"]If you and your school are serious about Mathcounts, you should order as much competition prep material from Sports Awards as possible.[/quote]

My school is planning to cut MathCounts since a levy failed, so they aren't serious about it!  I'm going to find a way to participate in MathCounts anyway.  I'll take a look at the competitions that they are selling.


If $ m \neq 0$, what is the arithmetic mean of $ \frac{m\plus{}3}{m}$ and $ \frac{m\minus{}3}{m}$?

Adding vies us $ \frac{2m}{m}\equal{}2$ so the average is $ \frac{2}{2}\equal{}\boxed{1}$.


calculus
derivative
function
analytic geometry
graphing lines
slope
calculus computations

Consider the function $ f(x)=xe^{\frac{-x^{2}}{2}}$

(i)Find the coordinates of the stationery points A and B

(ii)Find the gradient of the tangent to the curve at the origin O. Hence find the set of values of the real number k such that the eqn $ f(x)=kx$ has 3 real roots

Some vocabulary:

In (i), a typical American textbook would say "critical point" instead of "stationary point."

In (ii), a typical American textbook would say "slope" instead of "gradient."

But it's a pretty straightforward problem. What have you done so far? In particular, what is $ f'(x)?$

(i) $ f'(x)$ = $ e^{\frac{-x^{2}}{2}}-x^{2}e^{\frac{-x^{2}}{2}}$

at critical point $ \frac{dy}{dx}$=0 therefore critical points at $ (1,e^\frac{-1}{2})$ and $ (-1,e^\frac{-1}{2})$

(ii) at origin (0,0) $ =\Rightarrow$ $ y-0=1(x-0)$
y=x

but I dont udnerstand the final part (fx)kx has 3 real roots

---------------------------------
In Australia critical point can mean stationery points/intercepts/max. pt/min. pt etc etc

thanks bye

You should sketch what the graph looks like.

You'll notice that the tangent line through the origin intersects the graph exactly once. You can see that algebraically by setting $ x=xe^{-\frac{x^{2}}2}.$

Now, with that picture in mind, consider drawing other lines through the origin. What happens if the slope of the line is less than $ 1$ (that is, less than the slope of the tangent line)? What happens if the slope of the line is greater than $ 1?$


I have a few questions about emailing solutions.

What should the subject of the email be?
Is there anything specific we need to write in our email (besides the attachments)?
what should be the name of the tex and pdf file that contains our solutions?
also, if the green check doesn't appear when we log on to USAMTS after a while after we send in our solutions, should we send it again just in case?

[quote="USAMTS Website"] In the subject line, include your name, your user ID, and the round number, like this: 'Richard Rusczyk, 4, Round 3.' [/quote]
The next 2 probably don't matter.

[quote="matt276eagles"]Don't expect to see the green check until about a week after you submit your solutions, especially if you submit them near the deadline.  I'm pretty sure they don't want you submitting your solutions twice, that would just create a mess.[/quote]

Everything in this post is correct.

Don't worry about the name of the file.  Do include some text in the body of the email; some spam blockers block no-text messages.


A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 100 seniors, 150 juniors, and 200 sophomores who applied. Each senior's name is placed in the lottery 3 times; each junior's name, 2 times; and each sophomore's name, 1 time. What is the probability that a senior's name will be chosen?

There are $3 \times 100=300$ senior tickets, $2 \times 150=300$ junior tickets, and $1 \times 200=200$ sophomore tickets.

So, the probability is $P=\frac{\text{\# senior tickets}}{\text{\# total tickets}}=\frac{300}{300+300+200}=\frac{300}{800}=\frac{3}{8}$

So, $\boxed{P=\frac{3}{8}}$

Hope that helps.


function
algebra
polynomial
algebra unsolved

can you solve with many solution ?
                          solve equation :
$x+ \frac{\sqrt{2}x}{\sqrt{x^2+1}} =1$

[quote="tranthanhnam"]can you solve with many solution ?
                          solve equation :
$x+ \frac{2 \sqrt{2}x}{\sqrt{x^2+1}} =1$[/quote]

let$x = tant , t \in$( $-\pi/2 ;\pi/2$)
              i hope this solution is nice and good .
 but i don't like it .

Left expression is increase continiuosly function f(x), therefore only one solution, because $f(-\infty)<1<f(\infty).$

$\frac{1}{x^2}=\frac{8}{x^2+1}$ which is simple by taking the reciproces...but x>0 otherwise the LHS<0<1, so only 2-1=1 solution

oh , i can't understand thic solution .
 Rust said :  Left expression is increase continiuosly function f(x), therefore only one solution is very true but what  root of equation is ?
   and JohnConstantine's solution  doesn't make me understand it .
 can you explain ?

dividing by x both the sides and squaring, we can invert numerator and denominator for each side, and getting a second degree expression without the first degree term, so the roots of the equation will have opposite signs: if roots are real (and they are, since the function is continouos) , taking the negative one, both the elements of the LSH will be negative, so LHS<0<1, wrong; solution is only 2 (degree of the polynomial)-1 give us only the positive root as solution :)

hello, after squaring and simplifying i got
$-x^4+2x^3+2x-1=0$
only $x=\frac{1}{2}+\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}\cdot 3^{1/4}$ fulfills our original equation.
Sonnhard.


Prove that there aren't positive integers x and y such that:
 $ x^{3}\plus{}y^{3}\equal{}2^{2009}$

To elaborate on greenhand's solution: [hide="My Solution"]$ x^3 \plus{} y^3\equiv2^{2009}\pmod{7}$
$ x^3 \plus{} y^3\equiv2^{2007}*2^2\pmod{7}\Rightarrow x^3 \plus{} y^3\equiv 8^{669}*4\pmod{7}$
$ x^3 \plus{} y^3\equiv( \minus{} 1)^{669}*4\pmod{7}$
$ x^3 \plus{} y^3\equiv \minus{} 1*4\pmod{7}\Rightarrow x^3 \plus{} y^3\equiv3\pmod{7}$
As $ x^3$ and $ y^3$ can either be $ 1$ or $ \minus{} 1$ (all cubes $ \mod{7}$ are either $ 1$ or $ \minus{} 1$), there cannot be a solution to $ x^3 \plus{} y^3\equiv3\pmod{7}$
[/hide]

Very nice, but there are other method without using Mod ?
I don't understand very much of Mod, if there are others methods, please , explain me.

I would recommend learning at least a little about mods because they are extremely useful, especially in proofs.
[hide="Another possibility"]$ x^3 \plus{} y^3 \equal{} 2^{2009} \Rightarrow (x \plus{} y)(x^2 \minus{} xy \plus{} y^2) \equal{} 2^{2009}$
As the only factors of $ 2^{2009}$ are $ 2$'s $ x \plus{} y \equal{} 2^m$ and $ x^2 \minus{} xy \plus{} y^2 \equal{} 2^n$ for some m and n where $ m \plus{} n \equal{} 2009$
As $ x \plus{} y \equal{} 2^m$ x and y must be of the same parody (they must either both be odd or both be even) As $ x^2 \minus{} xy \minus{} y^2 \equal{} 2^n$ They cannot both be odd for otherwise $ 2^n$ would have to be odd (odd-odd+odd=odd).
So x and y are even. We can safely assume that $ x \plus{} y\not \equal{} 2^{2009}$ for $ x^2 \minus{} xy \plus{} y^2$ cannot equal 1. Therefore we can divide both sides of $ (x \plus{} y)(x^2 \minus{} xy \plus{} y^2) \equal{} 2^{2009}$ by $ 2^r$ such that $ \frac {x \plus{} y}{2^r}$ is odd (dividing out all of the prime factors of 2 in $ x \plus{} y$. Then we have $ \frac {x \plus{} y}{2^r}(x^2 \minus{} xy \plus{} y^2) \equal{} 2^{2009 \minus{} r}$.
$ 2^{2009 \minus{} r}$ is obviously even (as $ r\not \equal{} 2009$), but the other side is odd (odd*even = odd), a contradiction. Therefore$ x^3 \plus{} y^3 \equal{} 2^{2009}$ has no solution.[/hide]


function
calculus
derivative
calculus computations

a.) For the function f(x, y) = x^4 - xy + y^4, find all the critical points (the points where the gradient vanishes). Determine for each critical point whether
it is a local minimum, a local maximum or a saddle point.

b.)Find the absolute maximum and minimum values of f in (a) if there are any. Justify your answer.

c.) What numbers x, y come closest to satisfying the 3 equations x - y = 1, 2x + y = -1, x+ 2y = 1? Square and add the errors, (x -y - 1)^2 + A + B. Then
minimize. (Find appropriate A and B, and then finish the minimization.)

$ \frac {\partial f} {\partial x} \equal{} 4x^3 \minus{} y$ and $ \frac {\partial f} {\partial y} \equal{} \minus{} x \plus{} 4y^3\;\;\;$ (1)
$ \frac {\partial^2 f} {\partial x^2} \equal{} 12x^2$ and $ \frac {\partial^2 f} {\partial y^2} \equal{} 12y^2\;\;\;$ (2)
$ \frac {\partial^2 f} {\partial xy} \equal{} \minus{} 1\;\;\;$ (3)
$ \nabla f$ vanishes when $ \frac {\partial f} {\partial x} \equal{} \frac {\partial f} {\partial y} \equal{} 0\;\;\;$ (4)

So the partials exist.
If the partials exist and equal zero then $ f$ has relative extremum at $ (x_0,y_0)$

So, do (4) and arrive at $ (x_0,y_0) \equal{} (1/2,1/2)$ and $ (x_1,y_1) \equal{} ( \minus{} 1/2, 1/2)$ and $ (x_2,y_2) \equal{} ( \minus{} 1/2, \minus{} 1/2)$

Use these points and apply the second derivative test.  The second derivative test will show whether we have max or mins at these points.  Draw conclusions using the theorem. That should take care of (a)

(b)
The extreme value theorem guarantees the absolute extreme values occur at points we just found or on a boundary. No boundary has been given so we assume the whole of $ R^3$.  Assuming the math is correct then $ (x_0,y_0)$ and $ (x_1,y_1)$  and $ (x_2,y_2)$ are extreme values.

(c)
Hmmm...


Fabio has to write a series of natural numbers.
The 1° number is chosen by Fabio, between $1$ and $2004$.
Then, $a_{k+1}=a_{k}+1$, if $a_{k}$ is odd, and $a_{k+1}=\frac{a_{k}}{2}$, if $a_{k}$ is even.

For example, if the first number is $10$, the series is $10;5;6;3;4;2;1$, which has $7$ numbers (the series stops when you arrive to the number $1$).

Which is the largest series (the series with most quantity of numbers), and, in that series, which is the 1° term (the one chosen by Fabio)?

[hide] Interesting problem... just some ideas. We want to maximize the number of odd terms because then we don't divide by $2$ as much. If we hit a power of $2$, it's over and we go straight to $1$. So trying $2^{k}+1$ should give good results. In fact, this will allow us to alternate between even and odd until we reach $4$. [/hide]

If $a_{1}=2^{k}+1$, then $a_{2}=2^{k}+2$ and $a_{3}=2^{k-1}+1$  Obviously, it will take 2k steps from $a_{1}$ to get to $2^{0}+1=2$, so there will be 2k+2 numbers in the sequence.  The greatest number $2^{k}+1\le 2004$ is $1025=2^{1}0+1$, so there will be $2\times 10+2=22$ numbers in the sequence.

the easiest way is to begin from the end. obviously we want odd even...so, 1*2*2=4-3-6-5-10...which is $2(2(2\cdot3-1)-1)-1...$which equal $2k+1$.
[quote="José"][quote="t0rajir0u"][hide="The answer"] $10000000001_{2}$

Doing the calculations in base $2$ should make the reasoning clear. ;) [/hide][/quote]

Sorry, the result in the normal base is?

EDIT: Can you explain your solution too?

What about another solution without using other base?[/quote]
we divide in 2 so it's a good idea to use base 2!
just look at it in base 2 $5=101_{2},1010,1001...$


I [i]would[/i] play, but I'm trying to do more math in the upcoming months.
Good luck though.

[quote="perfect628"]We will be starting shortly after the USAMO.[/quote]
Good idea.

I would like to say, it was my idea.

Also it wouldn't have been used had I not won (or at least, been the winning player) in the other game. I believe that was the deal I made wiht perfect628.

I submitted my application. Hope I get in! :D (Doubt it...  :mad:)

[quote="perfect628"]Expect around a 2 month commitment.[/quote]
...unless, of course, you get eliminated really early...

Of course. Oh and Carl the deal was if I made USAMO, it stated afterwards. Otherwise, I was starting earlier to spite you.

I'm getting a good amount of apps. Keep them coming! :D

I would play, but I'm planning on making at least one of USAICO and MOP, both of which fall within two months of the start date. Oh well. I'll definitely play again if I'm around and there's a Mole 3 or 4.

[quote="randomdragoon"]I would play, but I have some very good reasons not to.

Some of you know what those reasons are.[/quote]

I think I know what that reason is.

Wait, do we have to answer the application questions truthfully?

Yeah... pretty much...

See the last game, there were tasks based on the answers to

the questions, so unless you can remember what you put,

answer truthfully........................

well, you don't even need to remember if you don't answer truthfully, since it's all in your pm sentbox.

UPDATE: I'm accepting applications until April 26, 4:00 PM EDT. Then, I will pm The Mole, and all accepted players, informing them of their status. I expect a reply by May 1, the day we are beginning, or else you will be removed from the game, and replaced. No exceptions.

So, keep the apps coming, nothing is decided yet!

Um... I am going to be away from home for a school thingie from the 26th of April-1st of May. If I am accepted, can I have mine on the 24, as to respond?

[b]SLIGHTLY CHANGED UPDATE[/b]
OK, here's some info. There will be 14 beinning at the start, and the pace will be faster than the first Mole. I'm accepting 10 "early applicants," whom I am currently pming with their status. I need responses by April 26, or your early applicant status will be revoked. Then, on the 26th, I will accept 4+x more, where x is the number of early ones who didn't respond/opted out. These can consist of people who already applied, or those who apply in the next few days. I will pm them, and then the process continues with the others on the 30th. We will begin on May 1, the final date for responses for replacements for my original 14.

So, the first 10 acceptances are going out now: keep applying!

Um, so do i need 2 apply again? I already applied...um. I'm confuzzledz.  :o

if you didn't get a pm, then you're not in the "early" list..
you don't have to re-apply.

Yeah basically. Don't tell if you've already been accepted please.

C'est le 26 avril! (so sad I'm dropping french next year... :( ) Anyway, 9/10 early acceptances have confirmed. They are...

CatalystOfNostalgia
BOGTRO
distracted523
ProtestanT
hm29168
zapi2007
Klebian
pianoforte
Nerd_of_the_Ages

Yay. Now, 5 others are receiving acceptance pms. One of them MAY be the one who never responded before. Now, they have until USAMO day #1(the 29th) to confirm, or their spots will be revoked.

I can't wait to start up again!

[quote="perfect628"]Based on the success(we even had spectators!  :) ), I'm running a second AoPS edition of The Mole. If we're going to start up before everyone starts leaving, we'd better start pretty soon. So, here's the application. See the other Mole thread for details and game play. It will be essentially the same idea. We will be starting shortly after the USAMO. (April 30)

[b][u]Application:[/u]
What is your birthday? MM/DD/YY please.

What is your favorite food? Be as specific as possible, as applicable.

How many countries have you visited, besides the one in which you live?

Can you juggle?

On a scale of 1-10, how good are you at crossword puzzles?

Do you have access to Photoshop, or an equivalent?

List any foreign languages you know or take in school.

Have you seen the Mole? If so, did you like it?

What is your strategy to win?

Do you want to be The Mole? Elaborate.

Will I have to replace you? Explain.[/b]

Send your application to me via pm. Pose any questions here. This game will be slightly larger than the original, but will run at a faster pace. Expect around a 2 month commitment. Good luck on being accepted![/quote]

MEEEEEEEE

lolz

I love to

 :spam:

Sorry, this was unclear. APPLICATIONS ARE CLOSED.

PI-Dimension's post didn't look like much of an application...

[quote="perfect628"]so sad I'm dropping french next year...[/quote]

indeed for us too. It means that on top of contrived clues based on latin, we now will have contrived clues based on greek too!


geometry
rectangle
combinatorics proposed
combinatorics

Today was realized the National Olimpiad in Cuba, this is the 3rd problem of the second day:

Prof that we can color the lattice points in the plane with two color so that every rectangle with vertices in the lattice points and edges parallels to the co-ordinate axis that have area 2^n is not monocromatic [/img]

Then color each point with the same colour... :wink:

I think there is something missing...

Pierre.

could you please post the complete test? if possible also in spanish just in case there some doubts about it.

Pierre: If we color each point with the same color then all the rectangles will be monocromatic.

Please ask me if you have any doubt about the statement

mszew: Do you speak spanish?

The problem in spanish:

Probar que es posible colorear los puntos latice del plano tal que no exista ningun rectangulo (con lados paralelos a los ejes coordenados y vertices en 4 puntos latice) con area igual a una potencia de 2 que tenga sus 4 vertices del mismo color

yes, I do!  :D

Could you be so kind to post the other problems?...

I swear I didn't see the 'not' before monochromatic :blush:
Okay, so I have to give an answer :

Let's color $ (x,y)$ red iff $ x\plus{}y \neq 1 \mod[3]$ and blue otherwise.
In the following $ [..]$ means area.

- Let $ B$ be a rectangle with all vertices blue, sides parallel to the grid lines.
Let $ (a,b)$ and $ (c,d)$ be two of its vertices.
Then $ a\plus{}b \equal{} a\plus{}d \equal{} 1 \mod[3]$ so that $ b\minus{}d \equal{} 0 \mod [3]$.
Thus $ [B]$, which is a multiple of $ b\minus{}d$, is divisible by $ 3$, and so cannot be a power of $ 2$.

- Let $ R$be a rectangle with all vertices red and sides parallel to the grid lines.
Let $ X(a,b),Y(a,d),Z(c,d),T(c,b)$ be its vertices, with $ c>a$ and $ d>b$.
Let $ x\equal{}a\plus{}b$.
Assume, for a contradiction, that $ [R]  \equal{} (c\minus{}a)(d\minus{}b)$ is a power of $ 2$.
It follows that $ c\equal{}a\plus{}2^p$ and $ d\equal{}b\plus{}2^q$ for some non-negative integers $ p,q$.
Moreover, since the four vertices are red, we have :
$ x \neq 1 \mod[3]$ and $ x\plus{}(\minus{}1)^p \neq 1 \mod[3]$ and $ x\plus{}(\minus{}1)^q \neq 1 \mod[3]$ and $ x\plus{}(\minus{}1)^p \plus{} (\minus{}1)^q \neq 1 \mod[3]$.
Thus, modulo $ 3$, we have
$ \{x,x\plus{}(\minus{}1)^p \} \equal{} \{0,2 \} \equal{} \{x,x\plus{}(\minus{}1)^q \}$. It follows that $ x \plus{} (\minus{}1)^p \equal{} x \plus{} (\minus{}1)^q \mod [3]$.
Therefore, $ p,q$ have the same parity. But, this forces $ x,x\plus{}(\minus{}1)^p,x\plus{}(\minus{}1)^p\plus{}(\minus{}1)^q$ to be pairwise distinct modulo $ 3$. In particular, one of them must be $ 1 \mod [3]$, a contradiction.

Hence result.

Pierre.

Oh my god!!
I can't believe that. Yesterday in the olympiad, I think the problem in this way too (construct a coloration) and in 1½ hours I tried with mod 2^k, I wrote the co-ordinates in binary and I taked their 2-adic value and more and every failed.
Thanks you very much Pierre.

mszew: the first five problems are easy and it's posible too that they aren't original. However I will to post them here.

First day:

1- We place the numbers from 1 to 81 in a 9x9 board. Prove that exist k in {1,2,...,9} so that the product of the numbers in the k-th column is diferent to the product of the numbers in the k-th row.

2- Let H a regular hexagon and let P a point in the plane of H. Let V(P) the sum of the distances from P to the vertices of H and let L(P) the sum of the distances from P to the edges of H.
    a)Find all points P so that L(P) is minimun
    b)Find all points P so that V(P) is minimun

3-A boy write three times the number n in a blackboard. He can make the followen move:
   He flock one of the numbers and write in its place the sum of the two others less 1. After several moves one of the three numbers in the blackboard is 900. Find all the posible values of n

Second day:

1- a^2+a+b^2≥a^4+a^3+b^4      a,b≥0
    Proof that (1-a^4)/a^2≥ (b^2-1)/b

2- Let ABC an acute-angle triangle. Let R a rectangle with vertices in the edges of ABC. Let O de center of R.
   a)Find the locus of all the points O
   b)Decide if there is a poit that is the center of three of this rectangles.

3- Prove that we can color the lattice points in the plane with two color so that every rectangle with vertices in the lattice points and edges parallels to the co-ordinate axis that have area 2^n is not monocromatic.

mszew: Where are you from?


Solve these differential equations:

A) $yy'=y+(y')^2$

B) $y'''(1+(y')^2)=2y'(y'')^2$

C) $xy''=y'+x^2$

Help plizzzz.  :?

[quote="AYMANE"][quote="WildChildX"]Solve these differential equations:

A) $yy'=y+(y')^2$

Help plizzzz.  :?[/quote]

Derivate once :

$yy''+(y')^2=y'+2y''y'$
$(y')^2=y'+y''y'$ then
$y'=1+y''$or $y'=0$
And its easy to continue...........[/quote]
Can you solve the exersice this way?
y'=p
Y''=dp/dx=(dy/dx)*(dp/dy)=p*(dp/dy)
 then the original ecuation goes like this:

 y*p*(dp/dy)=y+p^2

Can you continue the exercise with Bernoulli or similar method?


I have recently started dealing with more complex recursion, I'm wondering if there is any general was to solve a recursion like

f(n,k) = f(n-1,k) + f(n-2,k-1) + f(n-3, k-2) with appropriate initial conditions.

Is there a good free resource for this kind of problems?

(by the way, I just made up the above recursion, and cannot guarantee that it even has an explicit for mula..)

thanks in advance!

As given, there is really a constant and a variable - you can express k in terms of n and a constant - let c=n-k, then you should be able to solve this as a single variable problem with a constant term possibly.  For this problem you'd get that f(n)=A1 * (b1)^n + A2((b2)^n+(b3)^n) where A1, A2 are coefficients determined by an initial choice of f(n0,k0), f(n0+1,k0+1), f(n0+2,k0+2) and b1 is the real root of x^3-x^2-x-1=0, b2, b3 are the complex roots giving a sinusoidal element with their sums.  This may not help in the more general 3 level dependency of two variables, but I guess all I can say is that it is a class with many subclasses that may have well described solutions.

I don't know if this will help you, but it's worth a shot:

http://www.math.ucla.edu/~hbe/amil/rpf.pdf


absolute value
number theory unsolved
number theory

Are there exist four distinct positive integers that the absolute value of the difference between any pair of them is a perfect square.

Let $ x_1<x_2<x_3<x_4$ these numbers. Then $ x_2\minus{}x_1\equal{}a^2,x_3\minus{}x_2\equal{}b^2,x_4\minus{}x_3\equal{}c^2$, and your problem equavalent to problem about parallepepid with integer sides a,b,c, suth that $ a^2\plus{}b^2\equal{}d_1^2,b^2\plus{}c^2\equal{}d_2^2,a^2\plus{}b^2\plus{}c^2\equal{}d_3^2$.
These problem had solutions, but unknown general solution.

I have the idea of this one because of the other problem:

color every positive integers using one of the three colors , prove that there exists two distinct positive integers a,b that |a-b| is a perfect square.

I didn't get this one in the competition because I was all thinking about if there is four integers that the difference of any pair of them is perfect square.(then the result follows). But the fact is that we really don't need to use this to prove the color problem.

and why a^2+b^2=d^2, b^2+c^2=e^2 and a^2+b^2+c^2 have general solutions, how do you judge that

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} d_3^2$ has the general solution

$ a \equal{} m^2 \plus{} n^2 \minus{} p^2 \minus{} q^2$
$ b \equal{} 2(mp \plus{} nq)$
$ c \equal{} 2(np \minus{} mq)$
$ d_3 \equal{} m^2 \plus{} n^2 \plus{} p^2 \plus{} q^2$

(actually, the choice of $ b$ is not arbitrary).  It remains to find values of $ m, n, p, q$ such that two of the sums of pairs of distinct terms on the LHS are also themselves squares, which is rather tedious.  In particular, $ d_3^2$ is a sum of two squares.

Edit:  The extra condition that $ a^2 \plus{} c^2$ is a sum of squares gives a [url=http://mathworld.wolfram.com/PerfectCuboid.html]perfect cuboid[/url], none of which are known.  But MathWorld gives the solution $ a \equal{} 672, b \equal{} 153, c \equal{} 104$ for this problem.

Nice, it's better that I know this in the contest!


What conditions should 3 vectors [b][i]a[/i][/b],[b][i]b[/i][/b],[b][i]c[/i][/b] satisfy so that they can be sides of a triangle?

Here is a problem related to your question from Japanese university entrance exam.

1968　Gunma University entrance exam.

[color=blue]Given three non zero vectors [/color]$\overrightarrow{a},\ \over ri\overrightarrow{b},\ \overrightarrow{c}$ [color=blue]such that any two vectors don't have the same direction or the opposite one. Add to this, if the sum of any two vectors have the same direction or the opposite one with the rest vector among [/color]$\overrightarrow{a},\ \over ri\overrightarrow{b},\ \overrightarrow{c},$ [color=blue]then prove that these vectors can be the three sides of a triangle.[/color]

[quote="perfect_radio"]Their sum must be $0$. But what does this have to do anything with Linear Algebra?[/quote]
He may have thought that they were a part of Linear Algebra since in most introductory Linear Algebra books, they teach vectors.

Masoud Zargar

Actually, he posted in in Calculus-Analysis. I'm the one who moved it here. I suppose I should have moved it to a high school forum, but in my mind, vectors are a part of linear algebra, albeit a fairly low-level part of the subject.

OK, thanks for the clarification.

[quote="kunny"][color=blue]Given three non zero vectors [/color]$\overrightarrow{a},\ \over ri\overrightarrow{b},\ \overrightarrow{c}$ [color=blue]such that any two vectors don't have the same direction or the opposite one. Add to this, if the sum of any two vectors have the same direction or the opposite one with the rest vector among [/color]$\overrightarrow{a},\ \overrightarrow{b},\ \overrightarrow{c},$ [color=blue]then prove that these vectors can be the three sides of a triangle.[/color][/quote]

Indeed, I did not think that those vectors could be oriented both ways. So the correct answer should be: ... iff there are $\lambda_1,\lambda_2,\lambda_3 \in \left\{ -1,1 \right\}$ such that $\lambda_1 \overrightarrow a + \lambda_2 \overrightarrow b + \lambda_3 \overrightarrow c = \overrightarrow 0$.


A type of algae grows continuously so that its population doubles in 3 days. Given a beginning population of 100 algae cells per millileter of water, how many algae cells per millileter would you expect at the end of 10 days? Express your answer to the nearest whole number.

[hide="hint"]the answer's NOT 1067!![/hide]

[hide]
After 9 days, it has doubles 3 times.
100->200
200->400
400->800

For the last day, consider how by how many times the number of algae grow in a day.
You multiply the total number a fixed number of times 3 times to get 2.
Thus it is $\sqrt[3]{2}$
$800\cdot\sqrt[3]{2}\approx1008$[/hide]

[hide]for the first 9 days multiply it by 8 to get 800
then find the cube root of 2 and multiply it to 800 to get [b]1008[/b][/hide]

in $LaTeX$

[hide]$2*2*2*100 = 800$
$800*\sqrt[3]{2} = 1008$[/hide]

I think i kinda get it... Would it be like 800^2/3

isnt it just:
$2^{(10/3)}*100$ ???

[size=75][color=blue]Moderator kyyuanmathcount's note: Make sure you put { } around the exponent, or it does $n^21$ instead of $n^{21}$[/color][/size]


AMC AIME USA(J)MO USAMO function algebra polynomial

I'm going into ninth grade, and I just completed geometry last year. How much math exactly does the AMC cover? If say, I complete AoPS V1 this summer, would I at least have the skills to do all the problems in AMC, AIME,etc.? Skills as in like, algebra, trigonometry, not like problem solving skills.  And if not, should I still try the contests, or should I wait a couple of years?

i may not be the best one to answer this question, ive only taken practice tests and such. but ill try.

on the amc 10 it covers everything before calculas.
you most likely wont be able to solve all the problems with aops v1 alone.
you could pretty much easaly qualify for aime though with that. aime is another thing. read v2 for that one. but you can still do pretty well i guess. it really does matter on what else youve done.

you asked if you should wait a couple of years. theres no point in that. no harm in taking it. cause its free and such. i reccomend the amc10 if you just completed geometry. and ill let other people post their suggestions. make sure to do ALL of the questions in the aops v1 though. they have bunches of ahsme questions, which is amc12 level. thats what it used to be called.

please dont start insulting me or attacking me if i made some mistake or somat. just correct, no need for the flame war.

personally im going to most likely opt to take the amc10, instead of amc8 this year, unless something changes my mind, but that most likely wont happen. cause you need practice on it. and cause ill most likely make aime.

hope i helped
good luck

i suggest going to http://www.unl.edu/amc to get more info[/url]

Oh, definitely try the contests!  It can't hurt you, and you'll probably do great!

I started taking the AMC the year I took Geometry (7th grade) and there weren't any concepts that I hadn't covered, really.  I hadn't even read AoPS until after I took it.  Granted, I didn't know how to do all the problems yet(me being a silly little guy back then...), but I'm sure AoPS will help with that, considering both its title and its content.  Anyway, the AMC10 generally covers Algebra, Geometry, and maybe a bit of Algebra II (but you'll be taking that next year, I presume, so you'll know plenty of math by February).  The AMC12 covers about the same topics, just using slightly harder questions.

So again, definitely take the contests if you can.  It will never hurt you to do more math or more math contests.

As long as you won't feel discouraged if you don't do great, take the test by all means! The AMC10 doesn't cover topics that are taught after algebra 2 I believe, so you should be able to get all the math you need from AoPS 1. The problem solving is the more important issue. To get that you must work hard.

I agree with mathfanatic that there is no reason not to start taking the AMC exams.  If nothing else you will expose yourself to the kinds of problems that will need to learn how to solve in order to take the AIME or even the USAMO.  Most of the people I knew who took the USAMO took the AHSME (old version of the AMC) for at least a couple of years while working up to a higher level of math and problem solving.

The Art of Problem Solving books should help cover topics that you haven't been exposed to in Algebra I and Geometry.  The AMC has some number theory, combinatorics, functions, polynomials, etc. that you may not necessarily encounter in your school curriculum (or maybe you will if you have a good program).  Also, maybe you can get ahold of some of the AHSME, AIME, AMC problem books from your school or a library (or buying them) -- they are very instructive and the people I knew who scored well on those exams had almost invariably taken dozens of those past tests.

Syntax, you know you can take both the AMC8 and AMC10, since the former is in November and the latter is in February?  And you can take the '10 twice, or the '10 and the '12, which is what I did, since there's the A and the B days.

I'd highly recommend the Art of Problem Solving books - wish I'd had them when I started taking the AMC exams.  Huh, huh, now I'm starting to sound like The Great Rusczyk and The Great Crawford, two names I'm sure I've heard somewhere....

And yes, AMC past exams are cheap and easy to get - just look on [url]http://www.unl.edu/amc[/url] for ordering forms, etc.  They're something like $1 each, which is, as we math nerds say, [i]trivial[/i] compared to other resources you may find.  (For instance, I believe Zeitz' book is $70!).

Good luck and happy solving!

- mathfanatic

cool, i didnt know that i could take both amc8 and amc10. all the more awards. cool beans

[quote="mathfanatic"]I'd highly recommend the Art of Problem Solving books - wish I'd had them when I started taking the AMC exams.  Huh, huh, now I'm starting to sound like The Great Rusczyk and The Great Crawford, two names I'm sure I've heard somewhere....
[/quote]

I owe a great debt of gratitude to Rusczyk & Crawford for teaching me how to solve problems but I believe that there is a little too much over-idolizing of these peoples... It is excellent to express your gratitude but the fine line between expressing gratitude and over-idolizing, a line which is becoming finer and finer as I browse these boards is slowly diminishing.  This wasn't geared at you fanatic of math but just something which inspired me to write this.

I apologize if I have offended anyone, these are just my thoughts on this matter.

Personally however, I have a picture of 1*10^6 digits of pi and Lenny Ng in a little room in which I light incense and daily recite the first 500 digits. Hell, I should start a cult.

Peace,
IW.OAV

Instead of starting a cult, why don't you use that time to solve math problems? By the way, the over-idolizing was starting to bother me as well, but I wasn't going to say anything about it. I'm glad you did.

[quote="Idiot_without_a_village"]It is excellent to express your gratitude but the fine line between expressing gratitude and over-idolizing, a line which is becoming finer and finer as I browse these boards is slowly diminishing.  This wasn't geared at you fanatic of math but just something which inspired me to write this.

I apologize if I have offended anyone, these are just my thoughts on this matter.
[/quote]

I'm not over-idolizing anyone.  Let me explain my reasoning behind this:

I think Syntax and some others referred to Mr. Rusczyk as "The Great Rusczyk" several times before.  But notice: no one referred to Mr. Crawford as "The Great Crawford".  Now how is that for unfairness?  There are two guys running this site, two guys doing math jams, two guys teaching classes, and only one of them gets to be called "Great" while the other is just "hey you".  So I figured I would set things on an equal playing field.

Besides, they [i]are[/i] always referring to "back in the day" when we had to walk 5 miles uphill both ways, or "we wish we'd had books like the one you are holding in your hand when we were doing math competitions".  Although technically, that would be Misters Rusczyk and Lehoczky....

Anyway, I don't see how "over-idolizing" a person is any worse than over-idolizing a number like pi.  Sure, I use it, I push the button on my calculator, I use radians sometimes, but I don't memorize hundreds of digits.  That, in my mind, is equally over the limit, and it takes up far more time than it does for me to append a "The Great" in front of a person's name.  Besides, what's wrong with expressing your approval for a person's deeds?  Or making a reference to another person's comments - what if I were using "The Great" as a reference to Syntax Error's calling them that?

But I'm spending less time on the AoPS website each day.  More math, less internet, and I'm going to get better at math...

my apologies to anybody that i have offended durning my stay on this site. deepest apologies. everybody is considered "Great" in something. especially frosted flakes, but thats another story...

[quote="Idiot_without_a_village"]
Personally however, I have a picture of 1*10^6 digits of pi and Lenny Ng in a little room in which I light incense and daily recite the first 500 digits. Hell, I should start a cult.
[/quote]

Good God, I can't believe you took me seriously.. I was jokin' I mean I realize that you guys are enthusiastic about math and all, but I didn't think someone actually thought I would do that? Who ever recites the first 500 digits of pi while burning incense in front of Lenny Ng needs to be shot. Come on guys, we like solving problems, we're not a god damn cult.

My 2 rupee.
IW.OAV

How are we supposed to know whether to take you seriously or not? I have never met you, and I assume that most people here haven't either, so how can we tell if it seems out of character or not? I'm sure there are some people who do rituals such as the one you discussed.

[quote="Idiot_without_a_village"][quote="Idiot_without_a_village"]
Personally however, I have a picture of 1*10^6 digits of pi and Lenny Ng in a little room in which I light incense and daily recite the first 500 digits. Hell, I should start a cult.
[/quote]

Good God, I can't believe you took me seriously.. I was jokin' I mean I realize that you guys are enthusiastic about math and all, but I didn't think someone actually thought I would do that? Who ever recites the first 500 digits of pi while burning incense in front of Lenny Ng needs to be shot. Come on guys, we like solving problems, we're not a god damn cult.[/quote]

If you were trying to make a burlesque of Syntax and mine and whoever else's on-and-off usage of the word "Great", it was a pretty poor one, considering that no one who read it realized it.  Especially considering that you have posted before "admiring" the name Lenny.  Were you joking there, too?  Because in that case, you would be making fun of someone, which is not a nice thing to do.

Plus, if you are joking when you talk about burning incense and chanting in front of a picture, then you are making fun of rituals where, for instance, family members may mourn the death of a loved one.  In that case, you are making fun of a whole culture, which, aside from being wrong, mean, and ignorant of others' beliefs, seems perfectly fine according to you.

WOW, it's amazing how one topic can lead to another distant topic!  :wink:  :roll:

-interesting_move

The integrity of this thread has become corrupted... People take what I say and try to fine tune it so that it looks like I was intending to offend someone. Instead of making it look like I offend anyone intentionally so that YOU feel better, then I suggest that you ought to get a life.

Quit the flamin' and chill out.
:-x  'Nuff said.

By the way, about the whole incense and chanting or whatever. I'm a hindu, so I'm pretty damn aware what the hell I'm talkin' about. I was just joking but because you math peeps have the pencils so far stuck up your - apparently, you've forgotten the meaning of a joke.

[quote="Idiot_without_a_village"]The integrity of this thread has become corrupted... People take what I say and try to fine tune it so that it looks like I was intending to offend someone. Instead of making it look like I offend anyone intentionally so that YOU feel better, then I suggest that you ought to get a life.

Quit the flamin' and chill out.[/quote]

IW.OAV, it doesn't matter whether you were intending to offend someone.  If you DO offend someone, you should consider what you said to offend them and consider not saying it again.   Otherwise, you'll get in all sorts of flame wars, because people get offended by what you say, not what you mean to say.

[quote]I was just joking but because you math peeps have the pencils so far stuck up your - apparently, you've forgotten the meaning of a joke.[/quote]

Spare us the lecture on the sense of humor thing, as well as the rude expressions concerning the stick-up-my-butt that I don't have.  I'm pretty sure I have one [a sense of humor], and I'm pretty sure that making fun of people or their beliefs and practices does not appeal to that sense of humor.  As for starting flame wars, I don't believe you're helping things any by being rude to others - or that you don't share your own part of the blame by making unnecessary comments in the first place.

Anyway, I apologize for misinterpreting your comments, but kindly stop making fun of things I say or do, and don't tell me to get a life.  I already have one.  I already know what I like doing.  Math.  So why should I not have role models that I may call "great"?   I'm sure you've called Strongbad "great" sometime in your life, and has anyone complained about your doing that?  Respect others' opinions.

On a lighter note, yes, interesting_move, don't we have some interesting conversations?   :D

and please dont cuss on this forum. there are younger people viewing, that might learn bad bad things. i dont mind cussing, but just dont do it here.

[quote="s0mp"]we need jimmy carter![/quote]
I'm not sure how that will help...
Can we register for AMC 8 as an individual, not as a school?

[quote="Tare"][quote="s0mp"]we need jimmy carter![/quote]
I'm not sure how that will help...

This thread has gone way off. Somebody needs to put it back on the track.
If no one minds, I'll nudge it back a bit.

Can we register for AMC 8 as an individual, not as a school?[/quote]

s0mp is referring to the one and only Jimmy Carter, whose peacemaking efforts finally resulted in the Nobel Peace Prize earlier this year (or was it last?).  Anyway, here's a funny article about him.  Just remember that The Onion is a satirical piece.  http://www.theonion.com/onion3930/former_president.html

I see you're trying out the spoiler.  Hmph.  I feel saddened that my efforts at making a flame war were undermined.   :twisted:  </jk>

I'm sure you can register as an individual, but you'll be paying the fees, so watch out.  You would do well to find a nearby school and take it there.  You could give them five bucks or something.

[quote="Syntax Error"]gosh, whyd you have to put spoiler text.[/quote]
In case it offended you, I deleted it. I'm sorry if I offended you.

it didnt offend me at all, but i dont see the reason why you did. dont expect us to highlight every post to check for spoiler text. sheesh. but im not offended. just teaching a lesson

[quote="mathfanatic"]
s0mp is referring to the one and only Jimmy Carter, whose peacemaking efforts finally resulted in the Nobel Peace Prize earlier this year (or was it last?).  Anyway, here's a funny article about him.  Just remember that The Onion is a satirical piece.  http://www.theonion.com/onion3930/former_president.html
[/quote]
Yes, I know about The Onion. My former science teacher tricked me into believing that *erased*.
[quote="mathfanatic"]I see you're trying out the spoiler.[/quote]
Just mentally note that most of what I write in spoiler is a joke
[quote="mathfanatic"]I'm sure you can register as an individual, but you'll be paying the fees, so watch out.  You would do well to find a nearby school and take it there.  You could give them five bucks or something.[/quote]
Interesting. Though I doubt that 5 bucks part will work...

[quote="Tare"]Can we register for AMC 8 as an individual, not as a school?[/quote]

Yes, you can.  And I would encourage it.  You might also want to look into the AMC-10 and AMC-12.

What is the advantage of registering as an individual? Doesn't it cost more money? The only thing I can think of is that you might be able to take the AMC 10/12 on both test days if your school only offers one.

I'm not sure what he meant; and yes, you could take it twice, but you'd have to pay the registration fee twice.  Whereas you could take it with your school on the A day, and do the B by yourself if you're the only one...

I think Mr. Crawford was just saying "yes, shoot for it" and take the AMC 8 - not that he should take it by himself if his school sponsors it anyway.  That would be both silly and wasteful to do so.

[quote="mathfanatic"]I hope you're joking when you don't realize that I'm joking.  I put the </jk> thing to signal the end of the joking.  Oh boy, here we go again...[/quote]
So that's what </jk> was...
...
...[size=59]oops :( [/size]


induction
number theory unsolved
number theory

Let X be the set {1,2,...,2001}. Find the smallest n so that given any n-subset of X we can always find an element which is a power of 2 or two elements whose sum is a power of 2.

I think the smallest n is 1001.
the problem consists in finding the greatest n for which any n-subset doesn't contain 2 elements such that their sum is a power of 2, or 1 element that is a power of 2.i found this number this number by induction of m, where m is the exponent of 2:
Suppose the from 1 to 2^m we have a(m) such numbers(maximum).
From 2^m to 2^(m+1) we can add maximum 2^m-1-a(m).
thus
                        a(m+1) = a(m) + 2^m -1 -a(m)= 2^m - 1
indeed till 8 we have only three elements: 3,6,7;
till  16 we might have maximum 7 elements: 3,6,7, 11,12,14,15.
thus a(10) (numbers till 1024) is equal to 2^9 -1 that is 511.
Say we have t numbers from 1025 to 2001 which satisfy the condition of the problem. the sum of any of these t numbers is greater than 2048 and smaller than 4096 and none of them is a power of 2.
till 2048 we would have 2^10-1 = 1023 numbers.
thus we have to subtract from 1023 the number of numbers from 2002 to 2048(which apeear in the good subset).
Consider the numbers from a set satisfying the condition smaller or equal to 46.
they are 3,6,7,11,12,14,15,19,22,23,24,27,28,30,31,35,38,39,43,44,46. in total 21 numbers.
so we have exactly 46-21=25 numbers from 2002 to 2047 (those for which the sum with the first 21 numbers isn't 2048).
thus from 1 to 2001 we may have maximum 1023 - 25 = 998 such numbers (the sum of any two isn't a power of 2, and any number isn't a power of 2).
thus the minimum n required in the problem
is 2001 - 998 = 1001 numbers.
hope i didn't mistakes in mysolution. :)

My solution is nearly the same as yours, but i think you got a typo below?
[quote="freemind"]the problem consists in finding the greatest n for which [b]any[/b] n-subset doesn't contain 2 elements such that their sum is a power of 2, or 1 element that is a power of 2.[/quote]
Moreover shouldn't the answer be 998+1=999?

yes, you're right
we have to add 1 to 998.
In rest , i think everything is clear, i think


calculus
function
derivative
calculus computations

Prove that the mixed partials of a function of 2 or more variables are equal.

Your missing a hypothesis... (see [url=http://mathworld.wolfram.com/PartialDerivative.html]Partial Derivative[/url]


source: IRMO
how many many members doest the set $A$ has if
$A=\{(x,y)\in\mathbb{N}*\mathbb{N}|x<y, 2x^2-5xy+2y^2-x-y-1=11\}$

A)$0$                             B)$1$                C)$2$               D)$3$         E)$\infty$

Sorry for, are you sure $2x^2-5xy+2y^2-x-y-1=11$, amirhtlusa? :?

I think $2x^2-5xy+2y^2-x-y-1=0$.


Neil and Jean each choose a number at random from the set $ \{1,2,3,4,5,6,7,9,10\}$; their numbers are allowed to be the same. What is the probability that the product of those two numbers is even?

[hide]the probability that the product of those two numbers is even
= the probability that at least one number is even
= 1 - the probability that both of them are odd
= $ 1 \minus{} (5/9)^2\equal{}56/81$[/hide]

[quote="1077400"][hide]the probability that the product of those two numbers is even
= the probability that at least one number is even
= 1 - the probability that both of them are odd
= $ 1 \minus{} (5/9)^2 \equal{} 56/81$[/hide][/quote]

[quote="GameBot"]Neil and Jean each choose a number at random from the set $ \{1,2,3,4,5,6,7,9,10\}$; [b]their numbers are allowed to be the same.[/b] What is the probability that the product of those two numbers is even?[/quote]

The probability that the product is odd is $ \frac59\cdot\frac48 \equal{} \frac5{18}$, so the probability that it's even is

$ 1 \minus{} \frac5{18} \equal{} \boxed{\frac {13}{18}}$.

Edit: Sorry!

[quote="math154"]
[hide]The probability that the product is odd is $ \frac59\cdot\frac48 \equal{} \frac5{18}$, so the probability that it's even is

$ 1 \minus{} \frac5{18} \equal{} \boxed{\frac {13}{18}}$.[/hide][/quote]

"their numbers are allowed to be the same"

I lost Power XP because I didn't see the missing 8 at first. Rekt!

[quote=trinath]the answer is 13/78[/quote]

Hide solution and an explanation?

[hide = What?]Use complementary counting. $\text{odd} \cdot \text{odd}$ will be odd. There are $5$ odd numbers and therefore $5 \cdot 5 = 25$ ways to get an odd product. In total $9 \cdot 9 = 81$ ways, so $81 - 25 = 56$, out of $81$, so wouldn't the answer be $\boxed{\frac{56}{81}}$?[/hide]

[quote=AppleCEO]It's 56/81 and here is how. the only way it could be odd is 1,3,5,7,9 x 1,3,5,7,9 . That is 25 possibilities. 81-25=56. the answer should be 56/81.[/quote]

Hide sol

[quote=laffytaffy]Sorry for responding to an old thread, but why isn't 10 included? I got this problem wrong because of that as well.[/quote]

no it was that 8 wasn't included :P

Ohhh I can't believe I didn't look at all of the numbers included  :wallbash:  :rotfl:

[quote=awesomeguy856][quote=MathHayden]did anyone else get trolled at first thinking that there were 10 numbers from 1 to 10?[/quote]

yep, i thought it was 3/4[/quote]

yes me too :(

[quote=purplepenguin2]3/4 right?
oops I got trolled by the fact that there was no 8 lol[/quote]

same

I didn't get trolled!

[hide=Solution]Let's check which ones wouldn't make it even. Neil and Jean have to pick a odd number, so we get $\frac59\cdot\frac59 \rightarrow \frac{25}{81}.$
Then subtracting from 1, we get $1-\frac{25}{81} \rightarrow \frac{56}{81}$
Let's double check with cases.
Case 1: Jean picks an even number. Neil can pick anything. We get $\frac49\cdot\frac99 \rightarrow \frac49.$
Case 2: Neil picks an even number. Neil can pick anything but a even number due to duplication of cases. We get $\frac49\cdot\frac59 \rightarrow \frac{20}{81}.$
Adding them we get $\frac49+\frac{20}{81}\rightarrow\frac{56}{81}$[/hide]


Find all possible sets $A \subset \mathbb{Z}$ and $B \subset \mathbb{Z}$ that satisfy the following conditions:
1) $A \cup B = \mathbb{Z}, \ A \cap B = \emptyset, \ |A| = \infty, \ B \neq \emptyset,$ and
2) for every $0 \neq i \in \mathbb{Z},$ there exists $j \in B$ such that $i+j \in A.$
Have fun!

$we\ take\ for\ i\neq 0,\ f_{i}: P(\mathbb{Z})\to P(\mathbb{Z})\ such\ that\ f(A)=\{x-i|\ x\in A\}$
we have $\forall i\neq 0,\exists j\in B,\ such\ that\ i+j\in A,\ \Longrightarrow \ j\in f_{i}(A)$
so $\forall i\neq 0,\ f_{i}(A)\cap B\neq \varnothing\ \Longleftrightarrow \ \forall i\neq 0, f_{i}(A)\neq A$

so, if $\forall i\neq 0, f_{i}(A)\neq A$ and $|A|=+\infty$
we take $B=\mathbb{Z}-A$
and we have $|B|\neq 0$ be cause $A\neq \mathbb{Z}$

for example:
$A=\mathbb{Z}-\{5\}$, and $B=\{5\}$
because $\forall i\neq 0, f_{i}(\mathbb{Z}-\{5\})=\mathbb{Z}-\{5-i\}\neq \mathbb{Z}-\{5\}$

or $A=\{\pm p^{k}|\ k\in\mathbb{N}\},\ B=\mathbb{Z}-A\ for\ p\neq-1,0,1$
or $A=\{\pm primes \},\ B=\{\pm composite\ numbres\}$


Hi,

Let $ a > 1$ fixed and $ b \equal{} a^2  \plus{} a \plus{} 1$. The sequence $ \left( {x_n } \right)_{n \ge 1}$ is defined as: $ x_1  \equal{} \sqrt b$, $ x_2  \equal{} \sqrt {b \minus{} \sqrt b }$and $ x_{n \plus{} 2}  \equal{} \sqrt {b \minus{} \sqrt {b \plus{} x_n } }$. Prove that $ \left| {x_{n \plus{} 2}  \minus{} a} \right| \le {1 \over {a\left( {a \plus{} 1} \right)}}\left| {x_n  \minus{} a} \right|$ and study the convergence of the given sequence.

Thanks!

$ x_{n \plus{} 2} \minus{} a \equal{} \sqrt {a^2 \plus{} a \plus{} 1 \minus{} \sqrt {a^2 \plus{} a \plus{} 1 \plus{} x_n}} \minus{} a$
=$ \frac {a \plus{} 1 \minus{} \sqrt {a^2 \plus{} a \plus{} 1 \plus{} x_n}}{a^2 \plus{} a \plus{} 1 \plus{} \sqrt {a^2 \plus{} a \plus{} 1 \plus{} x_n}}$
$ |x_{n \plus{} 2} \minus{} a|\leq \frac {1}{a(a \plus{} 1)}|x_n \minus{} a|$
Imply that $ |x_{n \plus{} 2} \minus{} a|\to 0$
Imply that $ \lim_{n\to\infty}x_n \equal{} a$

$ x_{n\plus{}2}\minus{}a\equal{}\sqrt{b\minus{}\sqrt{b\plus{}x_n}}\minus{}a\equal{}\frac{b\minus{}\sqrt{b\plus{}x_n}\minus{}a^2}{\sqrt{b\minus{}\sqrt{b\plus{}x_n}}\plus{}a}$

$ \equal{}\frac{a\plus{}1\minus{}\sqrt{b\plus{}x_n}}{\sqrt{b\minus{}\sqrt{b\plus{}x_n}}\plus{}a}$

$ \equal{}\frac{(a\plus{}1)^2\minus{}(b\plus{}x_n)}{(a\plus{}\sqrt{b\minus{}\sqrt{b\plus{}x_n}})(a\plus{}1\plus{}\sqrt{b\plus{}x_n})}$

$ \therefore |x_{n\plus{}2}\minus{}a|\equal{}\left|\frac{a\minus{}x_n}{(a\plus{}\sqrt{b\minus{}\sqrt{b\plus{}x_n}})(a\plus{}1\plus{}\sqrt{b\plus{}x_n})}\right|\leq \frac{1}{a(a\plus{}1)}|x_n\minus{}a|$. Q.E.D.

Since for $ a>1$, we have $ 0<\frac{1}{a(a\plus{}1)}<\frac{1}{2}$, clasiffying $ n$ is even or odd, we will have $ \lim_{n\to\infty} x_n\equal{}a$.

What's the source of the problem?

The problem has been given recently at a math contest for high-school students (11th grade).

I see. This kind of problems are often posed in Japanese university entrance exam.


inequalities
trigonometry
induction
rearrangement inequality
algebra unsolved
algebra

Suppose $\alpha_i > 0, \beta_i > 0$ for $1 \leq i \leq n, n > 1$ and that \[ \sum^n_{i=1} \alpha_i = \sum^n_{i=1} \beta_i = \pi. \] Prove that \[ \sum^n_{i=1} \frac{\cos(\beta_i)}{\sin(\alpha_i)} \leq \sum^n_{i=1} \cot(\alpha_i). \]

Inductions on $n$;
For $n=3$ it's easy from a conclusion
$x^{2}+y^{2}+z^{2}\geqslant 2xy\cos C+2yz\cos A+2zx\cos B$

The result is immediate by Lagrange multipliers.

It's not hard with higher methods, but I think the primary thoughts are also beautiful :lol:

@shalex
Could you please be more specific? How could induction be applied here?

shalex, I would be interested in your solution as well.

EDIT: wrong solution below.
Meanwhile let me say that the problem follows from the rearrangement inequality. In fact, any two distinct indices $i$ and $j$ satisfying $\alpha_i \geq \alpha_j$ must satisfy $\cos\alpha_i \leq \cos\alpha_j$ (obviously, because $\alpha_i$ and $\alpha_j$ are between $0$ and $\pi$) and $\sin\alpha_i \geq \sin\alpha_j$ (this requires using $\alpha_i+\alpha_j\leq \sum_{k=1}^n\alpha_k = \pi$).

The first instinct when you see such a problem is to use the rearrangement inequality, and the second is "oh no, it won't work because $\sin$ is not monotonic on $\left[0,\pi\right]$; even if it works, I will have to distinguish between several cases". This is a good example of how one should not blindly follow instincts.

I don't think it is given that $(\beta_i)$ is a permutation of the $(\alpha_i)$, so how is the rearrangement inequality working?

Could some mathlinkers provide a solution by lagrange multipliers as mentioned above by probability1.01,thanks.


topology
real analysis
real analysis unsolved

I cannot prove the following problem:

Let $ U$ be an open set in $ \mathbb R^n$ and $ C \subset U$ is proper compact set of $ U$. Show that there exists a compact set $ D$ such that $ C$ is a subset of the interior of $ D$ and $ D \subset C$.

Any suggestion  :idea: ?

well, on one hand, $ C\subset \text{int}(D)\subset D\subset C$, and $ \text{int}(D)\not\equal{}D$ unless $ D$ is empty, so something seems wrong in the statement.

If $ D \subset U$ is required instead of $ D \subset C$, then this follows from that $ \mathbb{R}^n$ is normal.

But we can directly prove this. Each $ x \in C$ is an element of $ U$, so there is $ \delta_x > 0$ such that $ B(x, \delta_x) \subset U$, where $ B(x, r) \equal{} \{ y \in \mathbb{R}^n \, | \, |x \minus{} y| < r \}$ is an open ball of radius $ r$ centered at $ x$.

Then $ \{ B(x, \delta_x / 2) \}$ is an open cover of $ C$, so there is a finite subcover $ \{ B(x_1, \delta_{x_1} / 2), \cdots, B(x_n, \delta_{x_n} / 2) \}$ of $ C$.

Now define $ D$ as the closure of $ B(x_1, \delta_{x_1} / 2) \cup \cdots \cup B(x_n, \delta_{x_n} / 2)$. It is not hard to confirm that $ D$ satisfies $ C \subset \text{int} \, D \subset D \subset U$.