1. Evaluate $\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}$.

2. If $x=2-\sqrt{3}$, evaluate $\frac{\sqrt{3-x}-\sqrt{1-x}}{\sqrt{3-x}+\sqrt{1-x}}$.

3. Find the minimum value of $x^2+2y^2-6x+4y-2$, where $x$ and $y$ are real.

4. Find the minimum value of $x^2-4xy+5y^2-6x+6y-10$, where $x$ and $y$ are real.

5. If $a-b=3+\sqrt{5}$ and $b-c=3-\sqrt{5}$, compute $a^2+b^2+c^2-ab-bc-ca$.

[hide="Hint for 1"]There's more than one way to do this.  Try squaring it and then taking the square root back again. [/hide]
[hide="Hint for 2"] Rationalize the denominator. [/hide]
[hide="Hint for 3 and 4"] Complete the square. [/hide]
[hide="Hint for 5"] Can you think of a way to factor $T = a^2 + b^2 + c^2 - ab - bc - ca$?  How about $2T$?  :) [/hide]

[hide="Hint for #1"]
Or u can simplify each sqrts into $\sqrt{a}+\sqrt{b}$.
[hide="Simplified Form"]
$\sqrt5-2$.

[/hide][/hide]

[hide="1"] Set $\sqrt{9-4\sqrt5}+\sqrt{9+4\sqrt5}=x$. Square both sides: $9-4\sqrt5+9+4\sqrt5+\sqrt{(9-4\sqrt5)(9+4\sqrt5)}=x^2$. Simplify: $18+\sqrt{81-80}=x^2 \longrightarrow 19=x^2 \longrightarrow x=\boxed{\sqrt{19}}$. [/hide]
[hide="2"] First multiply $\frac{\sqrt{3-x}-\sqrt{1-x}}{\sqrt{3-x}+\sqrt{1-x}}$ by $\frac{\sqrt{3-x}-\sqrt{1-x}}{\sqrt{3-x}-\sqrt{1-x}}$. We get $\frac{4-2x-2\sqrt{(1-x)(3-x)}}{2} = 2-x-\sqrt{x^2-4x+3}$. Plugging in $x=2-\sqrt{3}$, we get $\boxed{\sqrt3-\sqrt2}$??  :? I don't know if this is right. [/hide]

CT_14, the next time you need to make posts in quick succession, just edit your previous post instead.

[quote="tz"][quote="lotrgreengrapes7926"][hide="1"] Set $\sqrt{9-4\sqrt5}+\sqrt{9+4\sqrt5}=x$. Square both sides: $9-4\sqrt5+9+4\sqrt5=TWO\sqrt{(9-4\sqrt5)(9+4\sqrt5)}=x^2$. Simplify: $18+\sqrt{81-80}=x^2 \longrightarrow 19=x^2 \longrightarrow x=\boxed{\sqrt{19}}$. [/hide]
[hide="2"] First multiply $\frac{\sqrt{3-x}-\sqrt{1-x}}{\sqrt{3-x}+\sqrt{1-x}}$ by $\frac{\sqrt{3-x}-\sqrt{1-x}}{\sqrt{3-x}-\sqrt{1-x}}$. We get $\frac{4-2x-2\sqrt{(1-x)(3-x)}}{2} = 2-x-\sqrt{x^2-4x+3}$. Plugging in $x=2-\sqrt{3}$, we get $\boxed{\sqrt3-\sqrt2}$??  :? I don't know if this is right. [/hide][/quote]

see problem 1. Since $(a+b)^2=a^2+TWOab+b^2$, shouldn't it be (see above)? so the answer should be $\sqrt{20}=2\sqrt{5}$?[/quote]
Oh yeah.  :blush:  :oops:

Yeah, most people got the right answers :)

Does anyone want to try #4?

[quote="Hamster1800"][quote="frt"]
4. Find the minimum value of $x^2-4xy+5y^2-6x+6y-10$, where $x$ and $y$ are real.
.[/quote]

[hide="4"]
$x^2 - 4xy+5y^2-6x+6y-10$

$x^2 - 2xy + y^2 + y^2 - 2xy + 3y^2 - 6x + 6y - 10$

$(x-y)^2 + y^2 - 2xy + x^2 - x^2 - 6x + 3y^2 + 6y - 10$

$(x-y)^2 + (x-y)^2 - (x^2 + 6x + 9) + 9 + 3(y^2 + 2y + 1) - 3 - 10$

$2(x-y)^2 - (x+3)^2 + 3(y+1)^2 - 4$

letting $y=x=-1$ we obtain $-8$.

Is that right? I don't know how to prove that it's the minimum.

EDIT: no it's not right. The minimum appears to occur at (5,1) with a value of -24. Any proof for this?
[/hide][/quote]
I think you're on the right track, but remember that we want to rewrite it as $(\text{something})^2+(\text{something})^2+...+(\text{constant})$.

[hide="hint"]
$x^2-4xy+5y^2-6x+6y-10=x^2-2(2y+3)x+5y^2+6y-10$

Then complete the square...
[/hide]


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I need to know How can I prove that $ \mathbb{Q}$ is not a proyective module.

A module $ P$ over a ring $ R$ is said  to be proyective if given $ g: A\rightarrow B$ an epimorfismo and $ f: P\rightarrow B$ a homomorphism, there exist an $ R$- module homomorphism $ h: P\rightarrow A$ sach that \[ g\circ h =f.\]

I apreciate any help.
  Adriana

projective modules may be characterized by direct summands of free modules. thus, over an PID (in our case the integers), they are even free. but obviously no two rational numbers are linearly independant, and the rationals are not cyclic.


superior algebra
superior algebra unsolved

Z[ sqrt(d)]=  { a+ b*sqrt(d)| a,b E Z}

for any square free integer d. Let

N(a + b*sqrt(d) )= (a+b*sqrt(d) ) * (a - b* sqrt(d) ) = a^2 - (b^2) * d

*--multiplacation
N- norm
sqrt- square root
a^2 means a squared

show that the norm can detect when an element of Z[sqrt(d) ] is a unit.

It's a unit iff its norm is $ 1$. Proof: If the norm isn't $ 1$, then since norm is multiplicative you can't multiply it by anything to get $ 1$. On the other hand $ N(a)$ is a multiple of $ a$, so if $ N(a) \equal{} 1$, then $ a$ is a unit.

The correct version is: it is a unit iff the norm is $ \pm 1$.


AMC
USA(J)MO
USAMO
combinatorics unsolved
combinatorics

There are 23 players and every time they decide to have a football match they choose one judge and then they make two teams of 11 players each. We know that for any judge we choose from the 23 players, the remaining 22 players can be divided in 2 teams so that the sum of the players' weights in one team is equal to the sum of the players' weights in the other team. Prove that all the 23 players have the same weight.

I guess we probably have to suppose otherwise and then choose the player with the minimal weight, but I can't find the contradiction.

Start with http://www.artofproblemsolving.com/Forum/viewtopic.php?t=312178 and follow links as necessary.  (Searching for "football weight" or other similar things works well, too.)


A group of 12 girl scouts had enough food to last for 8 days when they arrived in camp. However, 4 more scouts joined them without the amount of food being increased. How long will the food last if each scout is given the same daily ration as originally planned?

they had $ 12\cdot8\equal{}96$ sections of food. now, they have $ 12\plus{}4\equal{}16$ people, and $ \frac{96}{16}\equal{}6$, so $ \boxed{6}$ days.


calculus
integration
logarithms
calculus computations

Substitute $ u\equal{}xy\qquad I\equal{}\int_{0}^{1}\int_{0}^{x}\frac{\frac{u}{y}\minus{}1}{(1\plus{}u)\ln u}\frac{du}{y}dy$ and reverse integration order

$ I\equal{}\int_{0}^{1}\int_{u}^{1}\frac{\frac{u}{y^{2}}\minus{}\frac{1}{y}}{(1\plus{}u)\ln u}dy\, du\equal{}\int_{0}^{1}\frac{\ln u\plus{}1\minus{}u}{(1\plus{}u)\ln u}du \equal{}\underbrace{\int_{0}^{1}\frac{du}{1\plus{}u}}_{\equal{}\ln 2}\plus{}\underbrace{\int_{0}^{1}\frac{1\minus{}u}{(1\plus{}u)\ln u}du}_{\equal{}\ln\frac{2}{\pi}}$

For the last integral see that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=161896]link[/url].


I did some googling but came up empty.

Say I want to align a bunch of equations, but I want some text between the equations.  Can I do this without two separate align statements.  In other words, can you re-create this code without two align sections?

[code]\documentclass[12pt, bold]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[dvips]{graphics}

\newcommand{\solution}[2]{\bigskip \bigskip \noindent {#1}. \vtop{\noindent #2}}

\begin{document}
\solution{2}{
The variance is
\begin{align*}
\textrm{Var}(Z)  &= 10,\!000^2 \left(^2_{5|}\bar{A}_{30} - \left(_{5|}\bar{A}_{30}\right)^2 \right) \\
\end{align*}
Using CF
\begin{align*}
_{5|}\bar{A}_{30} &= _5E_{30} \cdot \bar{A}_{35} \\
 &= e^{-5(.05+.1)} \cdot \tfrac{5}{15} = .1575 \\
^2_{5|}\bar{A}_{30} &=  e^{-5(.05+.2)} \cdot \tfrac{5}{25}
\end{align*}
Finally
\[ \sigma_Z = 1,802.60 \]}
\end{document}[/code]

BTW, I know the first eqn is only one eqn and I could have used \[ \], but I just wanted to illustrate my question here.

\intetext will do what you want. Replace [code]\end{align*}
Using CF
\begin{align*} [/code]by [code]\intertext{Using CF}[/code]


Ring Theory
superior algebra
superior algebra unsolved

Let R be a commutative ring with unity.  Prove that the only nilpotent element in R is zero if zero is the only nilpotent element of the local ring Rp  for all p primes in R.

[quote="MrGaussian"]
Let R be a commutative ring.  Prove that the only nilpotent element in R is zero if zero is the only nilpotent element of the local ring Rp  for all p primes in R.
[/quote]

I guess we also have that the ring is unitary. Suppose $ 0 \neq x \in R$ is nilpotent. Let $ I \equal{} \text{ann}_R(x).$ Obviously $ I \neq R.$ So there exists a maximal ideal $ \mathfrak{m} \supseteq I.$ Now $ \frac {x}{1}$ is a nilpotent element of $ R_{\mathfrak{m}}$

and so it has to be $ 0$, by our assumption. Thus $ I \cap (R \minus{} \mathfrak{m}) \neq \emptyset.$ Contradiction!


geometry
trigonometry
projective geometry
geometry proposed

Let ABC be a triangle and $\omega$ its incircle with I incenter. Points K and L are the orthocenters of the triangles BIC and AIC respectively. Let T be the point where the excircle of ABC with respect to C touches the side AB. Prove that KL and CT intersect on $\omega$.

First, I restate the problem as below by slightly changing the notations:

Let ABC be a triangle and $\omega$ its incircle with I incenter. Points K and L are the orthocenters of the triangles BIA and CIA respectively. Let T be the point where the excircle of ABC with respect to A touches the side BC. Prove that KL and AT intersect on $\omega$.

It can be solved it by the following 3 results:
(1)Let $\omega$ touch BC at D. Then D lies on KL.
(2)Let $AT\cap\omega=Q$, then $DQ\perp AQ$.
(3)$\frac{\cos (\angle BAT-\angle BAK)}{\cos (\angle CAT-\angle CAL)}=\frac{\cos (\angle BAT+(B-A)/2)}{\cos (\angle CAT+(C-A)/2)}=\frac{AL}{AK}=\frac{\sin C/2}{\sin B/2}$.
The first two are easy to prove. The third one can be proved by some trigo computations.

[quote="Kuba"]I made 1) and 2) also
but 3? :-/ dunno
sorry if it is a stupid question.. what are you proving with 3)
how do you prove that $AT\perp KL$ (with your notations).
thx[/quote]
(3) is actually equivalent to $AT\perp KL$.
Now I'm thinking of a synthetic proof using the fact that triangle KLM and ABC are orthologic(M is the orthocenter of BIC), but so far I have no success... :(

Well....I think it can solved by Pascal Theorem. But then I have to manage 9 points on a circle, in which every 6 of them generate 60 Pascal lines and totally generate 9C6*60=5040 Pascal Lines!

edit: I think I have solved it by "brute force" using Desargue's Theorem....

After nearly 3 months, I realised my proof using Desargue is flawed :mad:
Anyway, I found a pretty simple proof today :)
[i]Problem
Let P and Q be the orthocenters of triangles AIB and AIC. Then $PQ\perp AT$.[/i]
Using the usual notations, let PQ intersect the incircle at X other than D. Let AT intersect the incircle at Y. Then it is well known that $DY\perp BC$. Z is the midpoint of EF. Invert line AT wrt the incircle. Then we have to show that I,X,Y,Z are concyclic. This is equivalent to $\triangle CDQ\sim\triangle IZY$, which follows from $\angle DCQ=\angle ZIY$ and $\frac{CQ}{CD}=\frac{AI\cot C/2}{s-c}=\frac{AI}{r}=\frac{r}{IZ}$.

Edit: the first problem which forced me to use inversion on mathlniks :(


What is the dimerization product of cyclohexene?


function
LaTeX
ratio
algebra
polynomial
geometric sequence

Find the sequence:
$a_1=2$
$a_{n+1}=\frac{1}{3}a_n+\frac{1}{3}$
This staff is quite new to me so I would be grateful if you explain to me how we do such things.

I'll not prove this,since it will be long (like the chain of theorem). you'll find a lot of informaton about this on internet or discrete mathbooks.
as we know the recurrence relations Ord 1 (i.e $a_n = 3n$) have the form $a_n =r^n$, so if we subs this to the equation $3a_{n+1}-a_n =1$ we will get
$3r^{n+1} - r^n =1$. to solve this, we consider the case when it is homogenous.
$3r^{n+1} -r^n = 0$ and we have $r = \frac{1}{3}$ so $a{}_n{}_{h} = \frac{1}{3^n}$
now look at the constant that have just ignored that is $1$.it has contribute in determining the particular solution. that is $a{}_n{}_{p} = C$. the genenral solution is
$a_n = A (a{}_n{}_{h}) + C_2 (a{}_n{}_{p}) =A \frac{1}{3^n} + B$
let $n=1$ and $n=2$ respectivly we have $a_1 = 2 = \frac{A}{3}+B$ and $a_2 = 1 = \frac{A}{9} + B$, solve this system,  $A=\frac{9}{2}$ and $B=\frac{1}{2}$.so
$a_n = \frac{1}{2} \left ( 9 \frac{1}{3^n} +1 \right) = \frac{1}{2} \left ( \frac{1}{3^{n-2}} +1 \right)$

I know this is far from clear, but I don't know there are simpler way to expain this. I'm just learning it from my old brother's book.  :P

Well, there are a couple of ways to tackle homogeneous recurrence relations. My personal favorite is actually to use generating functions but that may be a little too difficult for this level of the forums, and actually in this case I would use the technique known as unwinding.

Since I'm not sure how to post with LaTeX I won't be able to use it :\

 a(n) = 1/3*a(n-1) + 1/3 = 1/3*(1/3*a(n-2) + 1/3) + 1/3 = 1/3*(1/3*(1/3*a(n-3) +1/3) + 1/3) + 1/3 ...
         = 1/3 *(1/3 * (1/3 * (1/3 ...* (1/3 * a(1) + 1/3) + ...) + 1/3) + 1/3) + 1/3

 We try to expand a few terms and notice that this ends up being
  a(1)*(1/3)^(n-1) + (The sum from k = 1 to n-1 of (1/3)^(k)) .

Using the formula to sum up geometric progressions and recalling that a(1) = 2, we get

2*(1/3)^(n-1) + ((1 - (1/3)^n)/(2/3) - 1) = 2*(1/3)^(n-1) + 3/2 - 3/2*(1/3)^(n) -1

This is equal to 1/2 + (1/3)^(n-1) * (2 - (3/2)*(1/3)) = 1/2 + (3/2)*(1/3)^(n-1)

So, a(n) = 1/2 + (3/2)*(1/3)^(n-1)

Sorry that this might be hard to follow without it being in LaTeX, but this method is pretty standard for simple homogeneous recurrence relations.

Consider $\alpha$ such that $\alpha=\frac{1}{3}\alpha+\frac{1}{3}$, from this equation and the given recurrsive formula we realize that two points $(a_n, a_{n+1}), (\alpha, \alpha)$ lie in the line $y=\frac{1}{3}x+\frac{1}{3}.$Now consider the gradient of the line, we have $\frac{a_{n+1}-\alpha}{a_n-\alpha}=\frac{1}{3}$ which implies the sequence $\{a_n-\alpha\}$ is a geometric sequence with ratio $\frac{1}{3}.$

Could someone link me to a page that details the art of creating a closed from expression for some recursive sequence?
Basically any help on this topic would be fantastic, especially links.

There are many ways to solve linear recurrence relations. Generating functions is a particularly good method. Here's an explaination of another method which is not so different.

[quote="blahblahblah"]google 'Generatingfunctionology'.[/quote]

As for me, I don't think generating function is necessary unless hard reccurence. ;)

I agree - I use characteristic polynomials whenever possible, as I tend to view recursions from a linear algebraic perspective.  But the method of generating functions is fantastic, and I think is quite instructive when looking at nonhomogeneous recursions.  Plus the resource is free :)

[quote]Except that your terminology is wrong, because that is anything but homogeneous...[/quote]

Yes, you are correct, sorry for the mixup.


Let $ AM$,$ BN$ and $ CP$ be three concurrent cevians in the triangle $ ABC$.The parallel from $ N$ to $ AB$ meets $ PM$ in $ E$, and the parallel from $ M$ to $ AB$ meets $ PN$ in $ F$. Prove that the straight lines $ MN$,$ EF$, and $ PC$ are concurrent.

Let $ I\equal{}MF\cap PC,K\equal{}MF\cap AC,D\equal{}NE\cap BC,J\equal{}NE\cap PC$,we only need to prove that $ IF\equal{}IM$
in fact we have $ \frac{IM}{IK}\equal{}\frac{PB}{PA}(1)$

$ \frac{IF}{IK}\equal{}\frac{IF}{NJ}*\frac{NJ}{IK}\equal{}\frac{BM}{BD}*\frac{CN}{CK}\equal{}\frac{BM.CN}{MC.NA} (2)$

and apply Ceva's theorem we have $ \frac{BM.CN.PA}{MC.NA.PB}\equal{}1 (3)$
from $ (1),(2),(3)$ obtain the result


Consider the following sequence: 1, 5, 9, 13, 17, 21, …. Which term of this sequence is 2009?
a. the 501st
b. the 502nd
c. the 503rd
d. the 504th
e. None of these

Because the sequence increases by $ 4$ and starts at $ 1$, we know that the $ n^{\text{th}}$ number of the sequence can be expressed as $ 1\plus{}4(n\minus{}1)$  Setting this equal to $ 2009$, we get
\[ 1\plus{}4n\minus{}4\equal{}2009\implies 4n\equal{}2012\implies n\equal{}503\]
Thus, our answer is $ \boxed{\textbf{c.}}$

$ \frac{2009\minus{}1}{4} \plus{} 1 \equal{} 503$ so the answer is 503, or C.


Choosing any two different integers at random, will their sum more likely be an even integer or an odd integer?

My original proposition was that

The reason for this is because given a set, let's say {1, 2, 3,... , 100}
There are 50 odds and 50 evens.
In order to make an odd sum you must choose an even number and an odd number.
In order to make an even sum you must either choose two even numbers or two odd numbers.
There  1,225 ways to choose two even numbers and 1,225 ways to choose two odd numbers. So added together there are 2,450 ways to choose an even sum.
Furthermore, there are 4,950 ways to choose 2 different numbers out of  100.
Thus, there are 2,500 ways to choose an odd sum.

but is there a fallacy to this?

Why do you think there's a fallacy?

Suppose we didn't have the different numbers restriction. Essentially, each of 100*100 = 10000 ordered pairs would be equally possible. Now, exclude the 100 ordered pairs with the same number twice. Note that all of these pairs add to an even number.

Without the restriction of numbers being different, it is fairly easy to show that 5000 pairs add to an even number and 5000 to an odd number. So now, you're left with 4900 that add to an even number.

You can divide 4900 and 5000 by 2 if you don't want to count permutations separately, so you get 2450 and 2500, respectively, which are the numbers from your post.

This doesn't answer the question, however: "choosing any two integers at random" is an imprecise term (there is no uniform distribution on $ \bf Z$), but what it typically means is that you are to make a density calculation.  That is, let $ P_n$ be the probability of some event you want when you choose integers uniformly at random from $ [ \minus{} n, n]$ (or $ [0, n]$ if you want nonnegative integers, or $ [1, n]$ if you want positive integers); then the density of this event is $ \lim_{n \to \infty} P_n$.  Properties like "is divisible by 2" have obvious densities (in this case, $ \frac {1}{2}$).

If you're going to ignore what I've just said to do it as you've done, you should at least be sensitive to possible dependence on the parity of the endpoints.  For example, if you use {1, ..., 101}, it's no longer true that you have the same number of even and odd numbers in your set.


A triangle with sides of lengths a, b, and c is partitioned into two smaller triangles by the line which is perpendicular to the side of length c and passes through the vertex opposite that side. Find integers a < b < c such that each
of the two smaller triangles is similar to the original triangle and has sides of integer lengths.

It seems that the original triangle should be a right triangle, so $ c^{2}=b^{2}+a^{2}$.
$ \{ \begin{array}{l}a^{2}=m^{2}-n^{2}\\ b^{2}=2mn\\ c^{2}=m^{2}+n^{2}\\ \end{array}(m,n\in N)$

You didn't answer the question, though.

[hide="A derivation"] The perpendicular clearly creates right triangles.  It follows that the original triangle is right.

It's not hard to write out an infinite family of solutions, but for now I will construct only the smallest example: let the smallest triangle in the partition have the side lengths $ (9, 12, 15)$.  The next smallest has side lengths $ (12, 16, 20)$ and the largest has side lengths $ \boxed{ (15, 20, 25) }$. [/hide]

[quote="t0rajir0u"]You didn't answer the question, though.[/quote]
I'm sorry that I had a mistake when understanding the problem... :(


If $ \boxed{abc\equal{}1}$ and $ \boxed{\frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\equal{}\frac{3}{2}}$, then at least one of the numbers $ a,b,c$ is equals $ 1$.

[quote="moldovan & Arrange your tan"]If ${ abc = 1}$ and ${ \frac {1}{1 + a} + \frac {1}{1 + b} + \frac {1}{1 + c} = \frac {3}{2}}$, then at least one of the numbers $ a,b,c$ equals $ 1$.[/quote]

This problem has been on one of these threads before.
Here is my attempt:

[hide]
$ abc = 1$

$ a = \frac {1}{bc}$

Substitute this into the first fraction and simplify the complex fraction (and the
whole equation) into:

$ \frac {bc}{bc + 1} + \frac {1}{1 + b} + \frac {1}{1 + c} = \frac32$

$ 2bc(1 + b)(1 + c) + 2(1 + c)(bc + 1) + 2(bc + 1)(1 + b) = 3(bc + 1)(1 + b)(1 + c)$

$ 2b^2c^2 + 4b^2c + 4bc^2 + 6bc + 2b + 2c + 4 = 3b^2c^2 + 3b^2c + 3bc^2 + 6bc + 3b + 3c + 3$

$ b^2c^2 - b^2c - bc^2 + b + c - 1 = 0$

$ b^2c(c - 1) - b(c^2 - 1) + 1(c - 1) = 0$

$ (c - 1)[b^2c - b(c + 1) + 1] = 0$

$ c - 1 = 0$

$ c = 1$

[/hide]

[quote="JBL"][quote="Arrange your tan"]$ (c \minus{} 1)[b^2c \minus{} b(c \plus{} 1) \plus{} 1] \equal{} 0$

$ c \minus{} 1 \equal{} 0$[/quote]  This does not follow.  (E.g., the first is satisfied when $ c \equal{} \pi$ and $ b \equal{} \frac {1}{\pi}$.)  You're most of the way there, though -- just factor a bit more.[/quote]

I am wrong above.
JBL showed I am wrong.

$ (c \minus{} 1)[b^2c \minus{} b(c \plus{} 1) \plus{} 1] \equal{} 0$

$ (c \minus{} 1)(b^2c \minus{} bc \minus{} b \plus{} 1) \equal{} 0$

$ (c \minus{} 1)(b \minus{} 1)(bc \minus{} 1) \equal{} 0$

$ c \minus{} 1 \equal{} 0$ or $ b \minus{} 1 \equal{} 0$ or $ bc \minus{} 1 \equal{} 0$

$ c \equal{} 1$ or $ b \equal{} 1$ or $ bc \equal{} 1$

kunny deleted his post, so I'm agreeing with him.
[hide="I agree."]Finding a common denominator in the second sum and expanding both the numerator and the denominator gives

$ \frac {ab \plus{} ac \plus{} bc \plus{} 2(a \plus{} b \plus{} c) \plus{} 3}{abc \plus{} ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c \plus{} 1} \equal{} \frac {3}{2}$

Cross multiplying and moving things around yields $ ab \plus{} ac \plus{} bc \equal{} a \plus{} b \plus{} c$. Let $ a \plus{} b \plus{} c \equal{} y$. Therefore $ a$, $ b$, and $ c$ are roots of the polynomial $ p(x) \equal{} x^3 \minus{} yx^2 \plus{} yx \minus{} 1$. Note that $ p(1) \equal{} 0$. Therefore one of $ a$, $ b$, or $ c$ must be 1. [hide="What I agreed with:"](He said that Vieta's kills it)[/hide][/hide]

[quote="moldovan"]If $ \boxed{abc \equal{} 1}$ and $ \boxed{\frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \equal{} \frac {3}{2}}$, then at least one of the numbers $ a,b,c$ is equals $ 1$.[/quote]

Let $ a\plus{}b\plus{}c\equal{}p,\ ab\plus{}bc\plus{}ca\equal{}q$, $ a,\ b,\ c\in{x^3\minus{}px^2\plus{}qx\minus{}1\equal{}0}$, so we can write $ x^3\minus{}px^2\plus{}qx\minus{}1\equal{}(x\minus{}a)(x\minus{}b)(x\minus{}c)\ \cdots [*]$

$ \therefore \ln |(x\minus{}a)(x\minus{}b)(x\minus{}c)|\equal{}\ln |x^3\minus{}px^2\plus{}qx\minus{}1|.$

Differentiating both sides with respect to $ x$,

$ \frac{1}{x\minus{}a}\plus{}\frac{1}{x\minus{}b}\plus{}\frac{1}{x\minus{}c}\equal{}\frac{3x^2\minus{}2px\plus{}q}{x^3\minus{}px^2\plus{}qx\minus{}1}$

Plugg $ x\equal{}\minus{}1$ into the both sides,

$ \minus{}\left(\frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\right)\equal{}\minus{}\frac{3\plus{}2p\plus{}q}{2\plus{}p\plus{}q}$

$ \Longleftrightarrow \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\equal{}\frac{3\plus{}2p\plus{}q}{2\plus{}p\plus{}q}\equal{}\frac 32$, yielding $ p\equal{}q.$

Thus from $ [*]$, we get $ x^3\minus{}px^2\plus{}px\minus{}1\equal{}0\Longleftrightarrow (x\minus{}1)\{x^2\minus{}(p\minus{}1)x\plus{}1\}\equal{}0$, which completes the proof.

Setting $ a \equal{} \frac {x}{y},b \equal{} \frac {y}{z},c \equal{} \frac {z}{x}$ we have after some computations:
$ x^2z \plus{} y^2x \plus{} z^2y \equal{} x^2y \plus{} y^2z \plus{} z^2x \Longleftrightarrow (x \minus{} y)(y \minus{} z)(z \minus{} x) \equal{} 0$.
So $ x \equal{} y$ or $ y \equal{} z$ or $ z \equal{} x$.Then at least one of the $ a,b,c$ is equal to $ 1$.


In a party of 10 people, you are told that among any 3 people, there are atleast 2 who don't know each other. Prove that the party contains a set of four people, none of whom know the other three.

[hide]
We will assume, for contradiction, that no such set of 4 people exists.
First, we prove that there exists a set of 3 people who do not know eachother.

Notice that if a single person knows at least 4 other people, then 4 of those people must be strangers to the other 3. This contradicts our hypothesis that there exists no such set, so we will assume that no person knows 4 other people.

Take any 3 people and focus on one pair of 2 people in those 3 who do not know eachother. Denote these "powerful strangers."

Now, consider the relationships between the powerful strangers and the 7 other people not in the group.
Among the 7 relationshops between one of the powerful strangers and the 7 other people, there must be at least 4 non-acquaintances. If there are not, then this person knows at least 4 people, and this is not possible.
The same is true for the other person of the powerful strangers.
By the pigeonhole principle, there must be at least one person of the 7 who is strangers to both of the powerful strangers. These 3 people constitute a set of 3 people who are all strangers to the other 2.

[that was step 1, I suppose]

Now, consider this set of 3 people of mutual strangers and denote 2 of these people "ultimate strangers."

Again, both of these ultimate strangers must not know at least 4 of the other 7.
Neither of the ultimate strangers can know 3 out of the 7 people all of which the other does not know because we can create a set of 4 people from the 3 out of the  people and the latter ultimate stranger.
Therefore, these ultimate strangers must both be strangers to at least 2 of the other 7.
Proof: If among the 14 relationships between on of the 7 people and one of the ultimate strangers are 5 or less acquaintances, we are done.
If there are only 6, each ultimate stranger has 3 acquaintances. These must be distributed among 5 people, since we have seen that they cannot be distributed among 6.
So, these ultimate strangers are both strangers to at least 3 of the 8 non-ultimate strangers. Denote 3 of these "unfortunate strangers"
Now in order for a set of 4 to not appear, we must have acquaintances between each of the unfortunate strangers. (statement 1)
If any 2 of them are strangers, then we can take those 2 and the ultimate strangers, and they are all strangers to the other 3.
But statement 1 is impossible due to a condition in the problem. Therefore, we are done.

[/hide]

[hide="Solution"]Interpret this as a graph of ten vertices where an edge represents two people knowing each other. The condition that there's no group of three people who all know each other is equivalent to there being no triangles in the graph.

*To make this argument easier to visualize, draw $ 10$ points in a circle and number them from $ 1$ to $ 10$.

For the sake of contradiction, assume that all groups of four people contain one person who knows the other three people. Start with an arbitrary group containing an arbitrary point $ 1$, and pick three vertices to connect to. Now choose a group not containing $ 1$, and number one point in the group $ 2$; pick three vertices to connect to. Then choose a group not containing $ 1$ or $ 2$, and number one point in the group $ 3$; pick three vertices to connect to, etc. This works until there are only three unnumbered points left, and we have numbered points $ 1$ through $ 7$. Note that point $ 7$ must connect to the three remaining points (which will now also be denoted by $ 8,9,10$). If point $ 6$ connects to $ 7$, then it can't connect to any of the remaining points, or a triangle will be formed, so it must connect to the three remaining points as well. Now, if $ 5$ connects to $ 6$ or $ 7$, then it can't connect to any of the three remaining points or a triangle will be formed, and since at least one of the three emanating from $ 5$ must be in the remaining three points by the Pigeonhole Principle, we again have that all three must connect to $ \{8,9,10\}$. Similar analysis yields that $ 4$ either connects to $ \{5,6,7\}$ or $ \{8,9,10\}$. Note that no edges of $ \{5,6,7\}$ or $ \{8,9,10\}$ can be filled, or else a triangle will be created. Thus for the groups $ \{4,5,6,7\}$ and $ \{4,8,9,10\}$, $ 4$ must be the one that has $ 3$ edges emanating from it, so there are edges from $ 4$ to $ \{5,6,7,8,9,10\}$. But this creates a triangle since $ \{7,8\}$ is connected, and we have the desired contradiction. Hence some group of four where nobody knows the other three must exist.[/hide]

[b]Edit:[/b] [hide="Faster Solution"]As [b]stevenmeow[/b] noted, the degree of each vertex is at most $ 3$, so there are at most $ \frac12\cdot3\cdot10 \equal{} 15$ edges. But there are at least $ \frac {3\binom{10}4}{\binom82} \equal{} 22.5$ edges from the assumption, since there are at least $ 3$ edges for every group of four, and each edge is counted at most $ \binom82$ times. Thus we have a contradiction.[/hide]

[b]Note:[/b] The number of people can be reduced to $ 7$, and this can be shown easily by both of these solutions (the first by the fact that $ 1,2,3$ were essentially not used, and the second by simple calculations). However, I don't believe [b]stevenmeow[/b]'s solution generalizes, but it probably can with some tinkering.

I think you were misled into thinking that we only need to prove the existence of 3 people who are mutual strangers. Or, you made the assumption that we need to disprove that in every group of 4, one person knows the other 3.

This does not work for 7 people, as the graph could look like a regular heptagon.
(There is no triangle, and among 4 people, there must be 2 'adjacent' people.)
For 8 people, we could have a regular octagon with its 'major diagonals' (connecting opposite vertices)
It does, however, work for 9 people.
I'm not sure of this proof, but this is just one of the Ramsey numbers
(3,4)
[url]http://mathworld.wolfram.com/RamseyNumber.html[/url]

On your faster solution, i'm not too sure if it's valid.
I don't understand why there must be at least 3 edges for every group of 4.
there could be 2 or 1 edge per group.

I misread the problem. :(
[b]
Edit:[/b] Unless I'm being stupid again, though, here is hopefully a different proof.
[hide="Solution"]Consider any vertex and its $ 9$ edges (let solid and dotted lines represent two people knowing and not knowing each other, respectively). At most $ 3$ of these are solid, or else there is a dotted $ K_4$ (since we are given that no solid $ K_3$ exists). Thus there are $ 6$ dotted lines. Label the original point $ P$, and the non-$ P$ vertices $ A,B,C,D,E,F$. Obviously, there is at least one dotted line in $ \{A,B,C,D,E,F\}$, or else solid $ K_3$ exist; WLOG, let it be $ AB$. Considering each of the triangles among $ B,C,D,E$, at least one of the dotted lines of one of the triangles (each triangle must have at least one dotted line) must come from $ B$, or else $ P,C,D,E$ is a dotted $ K_4$. WLOG, let that dotted edge be $ BC$. Now, $ AC$ must be solid or we have $ P,A,B,C$ a dotted $ K_4$. Similarly, consider $ A,C,D,E$; WLOG, dot $ AD$ so that $ BD$ is solid. Now, consider $ B,D,E,F$; WLOG, dot $ BE$ so that $ CE$ is solid, $ AE$ is dotted, and we have a dotted $ K_4$: $ P,A,B,E$; contradiction.[/hide]

[b]Edit:[/b] Gahhh.....this can be shortened.

[hide="Much Quicker"]As before, create $ P,A,B,C,D,E,F$ where $ PA,PB,\ldots$ are all dotted. Note that if a solid or dotted triangle occurs in the remaining $ 6$, then we have a solid $ K_3$ or a dotted $ K_4$, respectively. It is well known that $ R(3,3)\equal{}6$, so we're done.[/hide]


linear algebra
matrix
probability
function
linear algebra theorems

If the sample covariance matrix $ \widehat\Sigma$ has the form $ \widehat\Sigma = \frac{1}{n-1}\sum_{i=1}^{n}(\vec{x_{i}}-\vec{X})(\vec{x_{i}}-\vec{X})^\top$, where $ \vec{X}$ is the sample mean, does that not mean that the determinant of the sample covariance matrix will be zero? If it is zero, then how can the normal probability density function $ \frac{\det(\widehat\Sigma)^{-1/2}}{(2\pi)^{N/2}}\exp\left(-{1 \over 2}(\vec{x}-\vec{\mu})^\top \widehat\Sigma^{-1}(\vec{x}-\vec{\mu})\right)$ make sense?

The sum of $ n$ rank 1 matrices can have rank $ n$; this happens with probability 1, unless the random variable $ x$ has a singular covariance matrix.


linear algebra
matrix
linear algebra unsolved

What is the algebra generated by the nilpotent matrices in $M_n(C)$?

Isn't it the entire $\mathcal M_n(\mathbb C)$?

First of all, all the matrices having $1$ in the position $(i,j)$ for some $i\ne j$ and $0$ everywhere else are nilpotent, so they belong to the algebra. Next, given $1$, we can consider the matrix faving $1$ in positions $(i,i)$ and $(i,j)$ and $-1$ in positions $(j,i),(j,j)$. It's nilpotent, and from the first observation it follows that the matrix having $1$ on $(i,i)$ and $-1$ on $(j,j)$ belongs to the algebra. This shows that it's square, which has $1$ on $(i,i),(j,j)$ also belongs to the algebra, so their sum divided by $2$, which has $1$ on $(i,i)$ and $0$ everywhere else, is in the algebra.

We have just shown that for all $i,j$, the matrix having $1$ on $(i,j)$ and $0$ everywhere else belongs to the algebra, so the algebra must be the entire $\mathcal M_n(\mathbb C)$.

Maybe I misunderstood something? :?

and what are those matrices that are a product of nilpotent matrices ?

I think we have already discussed that, alekk... but I'm sure you are not satisfied with the hints of killer... ;)
  To grobber: no, I did not say it is hard problem, just nice. My idea was to use the fact that any matrix of trace 0 is linear combination of nilpotent matrices and thus the dimension is at least $n^2-1$. But I've shown that a hyperplan cannot be an algebra.


Find all pairs of positive integer a and b such that : $ q^{2}\minus{}r\equal{}2007$ , where q and r are respectively the quotient and the remainder obtained when dividing $ a^{2}\plus{}b^{2}$ by $ a\plus{}b$


linear algebra
matrix
vector
abstract algebra
linear algebra unsolved

Let $A_n\ (n=1,2,3,\cdots)$ be  a column vector of $2\times 1$ matrix.
Find $A_n$ which satisfies the following conditions.

\[\ A_1=\left(\begin{array}{c} 2 & 1\\ \end{array}\right), \ A_2=\left(\begin{array}{c} 3 & 1\\ \end{array}\right), \]

\[\ A_{n+2}=\left(\begin{array}{cc} 1 & 1\\1 & -1\\ \end{array}\right)A_{n+1}+\left(\begin{array}{cc} 0 & -1\\1 & 0\\ \end{array}\right)A_n\ \ \ (n=1,2,3,\cdots)\]

One could consider this to be a one-step recursion with the following matrix:

$M=\begin{bmatrix} 0&0&1&0 \\ 0&0&0&1 \\ 0&-1&1&1 \\ 1&0&1&-1 \end{bmatrix}.$

This matrix has eigenvalues $1,1,-1,-1$ and is not diagonalizable. Hence the dominant long-term shape should be linear growth, governed by vectors in the kernels of $(M+I)^2$ and $(M-I)^2$. I don't have time right now to chase the details of that.


So what do all think of telekinesis? By telekinesis, I mean both the power to move objects with your mind and the power to tap into someone else's mind. Do you know that it is humanly achievable by some?

I don't know how humans come with all these crazy, stupid, retarded ideas. But I do have the ability to predict what Bush is going to do every day. (Like today he's going to be in a tantrum and then he's going to ask rummy and cheney for ways to diss Spain.

I'm a pretty skeptical person (as mith can tell you..)  I have trouble believing things like this without witnessing it first-hand or something.  However, I can't entirely doubt this "force's" existence, either, as I have no proof of that.. I'll just say that if it does exist, that's pretty cool, but could sort of collapse our current way of life ;)

When I saw the title of this thread, I thought we were talking about Star Wars.  I like Star Wars, and was ready to declare that the Force is really out there.

But then I realized that we're talking about ordinary telekinesis.  What a letdown!  No, I don't believe in telekinesis.

Fierytycoon wrote:Do you know that it is humanly achievable by some?

What are the persons doing with it?

(See spoiler, if you like, for a little more.)

[hide]

It would be a very good idea to cast a skeptic's eye on any such claim as ability to move objects through nonphysical means merely by power of human thought. The history of the power of human thought has often been the history of human beings deluding themselves into believing what they would like to be true rather than what is demonstrably true to neutral observers. [/hide]

That's one of my favorite questions to ask, whenever I hear about claims of extraordinary powers. What is the person doing with the extraordinary power, mere parlor tricks, or solving some important problem and making the world better?

The classic example is bending a spoon just by looking at it, as Uri Geller was (is?) famous for at least claiming to do.

hmm.  No I really don't believe in it, but stranger things have happened.  My sister belives in it though she freaks when the door closes and no one shut it (makes for a very easy way to scare her).  Most of the time reprots like this come from air pressure doing things, right?

well, if yoda can do it, it must be real!
i actually think it does exist (not because of yoda), but i'm still waiting for some proof.

To those who can't wield it, they are oblivious to the existance. To those who can wield it, they have suppresed their desire to show off in the process of aquiring that skill.

I do believe in the force, however I don't recommend using it...well to try and learn it is the difficult part. Most likely you'll die trying.

when i saw star wars for the first time, i thought about how cool it would be to be able to do that and to use a lightsaber.  ive been hoping that one day a jedi knight would come and take me to the star wars place and teach me.  o well :lol:


[quote="maokid7"]question: are the numbers in the numerator the fibbanaci numbers? or is it some other pattern.[/quote]

yes, it's fibonacci


Again, do not form a new thread and please post in the right forum.

Where is the thread to post answers in??

miyomiyo,
[hide]didn't you mean 17 dots, not 17 days?[/hide]

[quote="miyomiyo"][hide="Answer"]17 days.

If only one samurai were dotted, he would go to lunch, see nobody with a dot, and kill himself.  >1 DAY<

If two samurai were dotted, one would go to lunch and see one with a dot.  He knows that if the one he sees doesn't kill himself the next night that somebody else has a dot.  He doesn't see anybody else with a dot, so he kills himself.  The same can be said with the other dotted samurai.  >2 DAYS<

And so on until 17 days.[/hide][/quote]


Color the plane in $2$ colors. Prove that one of these colors contains pairs of points at every mutual distance.

I am not sure how to interpret this problem; I assume that every mutual distance means every distance.

I guess it is right interpretation.

[hide="easy solution"]Just take triangle wich all sides equal to $d.$[/hide]

[quote="Phelpedo"]I think the problem is trickier than that; it asks you to prove that a single color has pairs of points at every distance. Your construction only shows that, at every distance, there is a pair of points that is the same color.[/quote]

With this interpretation:

[hide]
Let's say that the two colours are red and blue. Assume to the contrary. So there are no two red points $m$ apart, and no two blue dots $n$ apart. WLOG, assume that $m$<$n$. We can assume that there is at least one red dot (as if there are none, the problem is trivial). We can now draw a blue circle of radius $m$ with this point at its center. We can the draw a red circle of radius $m+n$ about the red dot (as every point on this circle is $n$ away from a point on the blue circle). But there will be two points on this circle that are $m$ apart. Contradiction.
[/hide]


Under the Community tab, there's always a Forum and Math Jams sub-tab thingy, but sometimes there's also a My Classes sub-tab and sometimes there isnt...is that supposed to happen?

If you're logged in & enrolled in a class, that tab will appear on all non-Forum pages (long story why it doesn't appear in Forum pages, has to do with difficulty of embedding PHP in Forum template files).

If you're not logged in, that tab won't appear anywhere.  If you're logged in but not enrolled in a class, it won't appear, either.


Given n distinct numbers x_1, ..., x_n

What is the value of a that will minimize

|x_1-a| + ... + |x_n-a|

Hint, the answer is not the avg of the x's. Consider, e.g.,
0 0 0 0 0 0 0 0 0 1. [/code]

This has shown up many times; the answer is $a$ is the median of the $x_{i}$'s.

You can find discussion in any of these threads:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=median+minimum&t=50407

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=median+minimum&t=46127

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=median+minimum&t=19601

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=median+minimum&t=16630

We can imagine that the $x_{i}$ are points on a number line and that $|x_{i}-a|$ is the distance between a and the point $x_{i}$. WLOG, assume that they are arranged in increasing order $x_{1}, x_{2}, ...x_{n}$. Note that the distance from a to $x_{i}$ plus the distance from a to $x_{n-i}$ is greater than equal to the distance between $x_{i}$ and $x_{n-i}$ (easily verified by the triangle inequality). Additionally, equally occurs when a is between the two points. Continuing in this way, we find that the minimum occurs when a is the median point if n is odd and if a is anywhere in between the two middle points if n is even.


Hi all

Here's my problem:

I don't know how to cut ( \not ) I'll be more specific

When I type: (with de dollar symbols) \not tanx or something like that,
this is the result: $\not tanx$ instead of cuting it all.

How can I correct it ?

Thanks

Use the [url=http://texcatalogue.sarovar.org/entries/cancel.html]cancel[/url] package with \cancel{\tan x}. It also provides effects like this
[img]http://sixthform.info/steve/wordpress/latexrender/pictures/a95924092167431f92ff877604ae3c62.gif[/img]


Let $a \in R$ number so that $ |a| \leq4 $.  If $x,y \in R$  then $x^{4}+y^{4}+axy+2 \geq 0$

cheers! :D :D :cool:


This is from: [url=http://www.icoool.com/index.asp]http://www.icoool.com/index.asp[/url]

一、选择题:

1.“仙剑”的全名为：
A.仙剑奇侠传 B.仙剑奇侠 C.仙剑传 D.金庸群侠传

2.仙剑的女主角不包括：
A.赵灵儿 B.林月如 C.阿奴 D.香兰

3.李逍遥自己炼出的第一个招数为：
A.凝神归元 B.天罡战气 C.天师符法 D.飞龙探云手

4.四种基本毒性不包括：
A.赤毒 B.瘴毒 C.毒丝 D.虫毒

5.仙剑中最差的剑是：
A.竹剑 B.木剑 C.铁剑 D.长剑

6.游戏开始时李逍遥的体力与真气分别是：
A.150、100 B.100、150 C.200、150 D.150、200

7.仙剑中补充体力的最佳药品是：
A.紫菁玉蓉膏 B.天仙玉露 C.赤血蚕 D.A、C都对

8.初期李逍遥的小村庄里的大夫姓：
A.李 B.丁 C.洪 D.王

9.以下哪颗灵珠的取得与战斗无关：
A.土 B.雷 C.风 D.水

10.鬼将军身上可偷到：
A.铁鳞甲 B.青铜甲 C.精铁战甲 D.藤甲

11.下列字注音全对的一组是：
A.瘴(zhàng)毒 赎(shú)魂灯 水(shuǐ)果 B.紫菁(qīng)玉蓉膏 五气朝(cháo)元 舍利(lì)子 C.醍(shì)醐香 金罡(gāng)珠 鬼降(jiàng) D.龙涎(yán)草 金蚕蛊(gǔ) 菩提加裟(shā)

12.下列字形全部正确的一组是：
A.炼狱真火 万剑决 芦苇漂 B.万蚁食象 磐龙剑 鹿茸 C.紫罂粟 玉菩提 霓虹羽衣 D.九节鞭 鲤鱼 金蚕王

13.下列诗句与原文有误的一句是：
A.伴月愿做一颗星 B.结伴江湖侠侣行 C.盼君拾得惜相连 D.仗剑江湖为红颜

14.下列哪个物品在隐藏情节中才会出现：
A.钓竿 B.情书 C.天书 D.竹笛

15.消耗赵灵儿真气最多的法术为：
A.武神 B.梦蛇 C.火神 D.狂雷

16.下列哪种情况下灵力加的最高：
A.装备圣灵珠 B.服用42个试炼果 C.装备玉佛珠 D.装备圣灵披风

17.下列哪种情况能最快走出有敌人的迷宫（不看地图）：
A.键盘微操 B.使用十里香 C.使用驱魔香 D.碰见敌人就用最厉害的法术

18.下列哪项法术为“李赵林奴”四人中两个人共有的：
A.元灵归心术 B.观音咒 C.天雷破 D.还魂咒

19.下列哪项真气可能加的最多：
A.一个天仙玉露 B.一个雪莲子和一个灵山仙芝 C.六个龙涎草 D.一个灵蛊

20.七大毒蛊不包括：
A.血海棠 B.孔雀胆 C.无影毒 D.毒龙胆

21.合招一共能合出___种招数:
A.5 B.8 C.11 D.12

22.阿奴的体重为___Kg?
A.40 B.42 C.44 D.48

23.鬼头杖头颅中囚禁__了条生魂?
A.48 B.49 C.50 D.51

24.鬼牙刀的别名叫___?
A.鬼头刀 B.勾魂刀 C.鬼魂刀 D.天邪刀

25.金罡珠是___的物品?
A.大罗金仙 B.金天罡仙 C.金地煞仙 D.小神仙

二、多选题(每题3分):

1.仙剑整个游戏中不能单独操作的人物有：
A.李逍遥 B.赵灵儿 C.林月如 D.阿奴

2.下列能分身的敌人有：
A.地上的一团烟雾 B.血池中的红色妖怪 C.锁妖塔中的蓝色跳跳娃 D.骷髅头

3.下列能在战斗中“招兵”的敌人有：
A.赤鬼王 B.女飞贼 C.镇狱明王 D.拜月教主

4.下列合招不能成功的是：
A.三人都是健康状态 B.两人健康一人虚弱且由健康者发出合招命令 C.两人健康一人虚弱且由虚弱者发出合招命令 D.一人健康两人虚弱且由健康者发出合招命令

5.雪蛤蟆的作用是：
A.武术+3 B.灵力+3 C.防御+3 D.身法+3

6.下列兵器装备后（普通攻击时）能自动攻击两次的有：
A.巫月神刀 B.玄冥宝刀 C.仙女剑 D.芙蓉刀

7.雄黄的作用是：
A.解赤毒 B.解瘴毒 C.解尸毒 D.解毒丝

8.下列哪些是体力真气皆补的药品：
A.蜂王蜜 B.灵葫仙丹 C.神仙茶 D.糖葫芦

9.当正装备着土灵珠时，要从迷宫中逃脱，这时选择使用土灵珠时，土灵珠三个字是什么颜色：
A.红色 B.白色 C.黄色 D.绿色

10.当李第一次上仙灵岛时，如果先去水月宫，李会被一股力量推回来，这时李会说什么：
A.！！！ B.？？？ C.怎么回事？ D.奇怪！

三、填空题(每题4分):

1.（语文题）仙剑中有一物品在《神雕侠侣》中出现过，请问它是：

2.（英语题）仙剑中HP代表什么：

3.（物理题）水中用什么法术威力大增：

4.（化学题）树林中用什么法术威力大增：

5.（音乐题）游戏开始时客栈音乐与后期哪个城市的音乐相同：

太难了......连我这个仙剑狂都感到力不从心了......
大家一起努力,争取做出一份正确答案!

等着吧，我再把仙剑玩一遍（能做出来吗）

[quote="wawl"]等着吧，我再把仙剑玩一遍（能做出来吗）[/quote]
游戏里面没提过阿奴的体重吧? 如果玩一遍的话,就要把所有的合体招数都打出来, 而且还要下到试炼窟的最底层......不知道需要多少时间...记得我第一次打出所有支线剧情是上个世纪,那次干了整整4天!


rotation
geometry
geometric transformation

Here a problem for you:  I just figured it out and am so proud.  Can you figure it out?

STEPS:
1) Rotate an object 90deg around the X-axis
2) Rotate the object 90deg around the Y-axis
3) Find a single rotation for this object.  Include the axis and the angle.

I will post the answer in two days if no one gets it...

[quote="lotrgreengrapes7926"]Rotations are always countercloskwise, right? :blush:[/quote]
I think so. :?:


inequalities
induction
inequalities open

Suppose [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ae33363502b6ec738774cf22bb596029.gif[/img] are positive reals such that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b0098ad3b8a176642c814b6328d5a55d.gif[/img].
Prove that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/0660a40b71ae56d8353312b43ef9dfb1.gif[/img]
Instruction:This is a very very hard ineq(at least to me),anyone who solves it must be a great genius. :D  :D

If $n=3$, it is just Schur.  Unfortunately, I don't know how to proceed for larger $n$, and the method of Schur does not work then.

Here is the homogenized form (indices are cyclic mod $n$)

$\sum_i x_i^n+\sum_i x_i^{n-1}x_{i+1}+n(n^2-3n+1)\prod_i x_i\geq (n-1)\sum_i x_i\prod_{j\not=i} x_j$

[quote="suweijie"]The ineq is so hard that I even have no courage to do it. :blush:[/quote]
don't be that ;)
and i guess it can be proved by induction
use the skill for suranyi,i'm not sure

I know Suranyi' inequality,but how to deal with it?

I can prove a weaker inequality only,
${\sum a_1^{n-1} + n(n-2) \prod a_i^{\frac{n-1}{n}} \ge (n-1) \sum (\prod_{i \neq j} a_i})$

[quote="suweijie"]How to prove it?Could you post your solution?[/quote]
It is a direct consequence of popoviciu's inequality. I'm still thinking if we can use the karamata inequality to prove the initial one.

well, what is the source of this problem? sss, do you mean prof 冷崗松? I'm not sure because I saw the same inequality in mr. 劉保乾's "用代數法證明代數不等式" p. 11 question 1. Maybe that suggests us to use hte method of generation to prove it.

Wow,you are right,just from Leng Gangsong

This inequality can be proven by (n-1)EV Principle.
Assume that $x_1 \le x_2 \le ... \le x_n$.
For $x_1=0$ it is true, and for $x_1>0$ it reduces to the case $0<x_1 \le 1$ and $x_2=...=x_n=1$.
To prove the last part, first write the inequality in the form
$S\sum{x_1^{n-1}+n^2(n-2)P - (n-1)PS\sum{\frac{1}{x_1}} \ge 0}$,
where $S=x_1+...+x_n$ and $P=x_1...x_n$. For $S=constant$ and $P=constant$, the expressions $\sum{x_1^{n-1)}}$ and $(-\sum{\frac{1}{x_1}})$
are minimal for $x_1 \le x_2=x_3=...=x_n$ (see (n-1)EV Principle).

[quote="suweijie"]WHAT IS [color=blue]see (n-1)EV Principle[/color]?Could you tell me,Vasc? :)[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?p=227570#p227570


Let $ (x_n )$ be a sequence of positive real numbers such that  $ \sum_{n\equal{}1}^{\infty} x_n$ is convergent.

Show that there exists a sequence $ (y_n )$ such that $ y_n /x_n \to \infty$ as $ n \to \infty$, but $ \sum_{n\equal{}1}^{\infty} y_n$ is still convergent.

define the increasing sequence $ a_k$ such that $ \sum_{a_k}^{\infty} x_n \leq 3^{\minus{}n}$ and for $ a_k \leq n \leq a_k\minus{}1$ define $ y_n\equal{}2^k x_n$.


Let $ M$ be the set of all natural non-zero numbers that doesn`t contain the digit $ 9$. Prove that for any $ n\in N^{*}$ and any $ r_{1}<r_{2}<...<r_{n}\in M$, we have $ \frac{1}{r_{1}}+\frac{1}{r_{2}}+...+\frac{1}{r_{n}}<80$.

I am not sure how many times I have seen this problem before, but it used to be one of my favorites in high school.

We begin by observing that the number of numbers in $ M$ that are less than $ 10^{n}$ is $ 9^{n}-1$.

So:  $ \sum_{r_{i}\in M}\frac{1}{r_{i}}= \sum_{i \geq 1}\sum_{10^{i-1}\leq r_{i}\leq 10^{i}}\frac{1}{r_{i}}\leq \sum_{i \geq 1}\frac{9^{i}-1}{10^{i-1}}=\frac{800}{9}<90$.

Hmm, that shows convergence but the bound is not as tight as you would like.

Equally:  For an $ i$ digit number we have $ 8$ ways to choose the first digit and $ 9$ ways to choose each additional digit.  So we could write that as:

$ \sum_{r_{i}\in M}\frac{1}{r_{i}}\leq \sum_{i}8(9)^{i-1}\frac{1}{10^{i-1}}= 80$.  Okay, that works.
 :)

(It has been awhile; thank you for reminding me of this problem)


I thank you for your help.  Is there a good math website that explains complex numbers in a non textbook manner?

This should definitely NOT be in intermediate. You should read the posting guidelines for the forums before you post. ;)

[quote="F u r u y á"]Other way is using this practice rule:

To solve $i^{763}$, for example, divide 763 by 4, then the rest will be an equivalent exponent:

$763 \ \underline{| 4 \ \ \ } \\36 \ \ \ 190\\ \mbox{ }\ 03$

So, $i^{763}=i^3=-i$   ;)[/quote]

That's not an "other way".  It's exactly the same as what blahblahblah did.

[quote="Altheman"]how is this not intermediate, this is really simple algebra 2[/quote]

[quote="Posting Guidelines"]The Intermediate Topics board is for problems and questions that are beyond the basics, but not quite olympiad level. These are problems that might be on the AMC, AIME, Mandelbrot, or ARML.[/quote]

I have a feeling this problem would never show up on any high school math competition. It should probably be in Basics or maybe Getting Started in my opinion. It's stuff you learn in school anyway. ;)

Okay...183 people viewed this question and 9 replies.
I had no idea that complex numbers were so popular here.

[quote="joml88"]That's not an "other way".  It's exactly the same as what blahblahblah did.[/quote]

I know it is strictly the same thing, but is faster I think, especially with greater exponents

You suit your response to the level of the individual that is asking the question.

I could have easily said that if $a\equiv b\bmod 4$, then $i^a=i^b$, but what would that have meant to the person who was asking the question?


Me - 14 minutes -- not great, but likely not the worst -- anyone do it in under 6? What prize did you get for that? I got a 10 cent crooked superball!

Hey, something else - I post an average amt, but have no rating! (or do I?) Ok, maybe I post less than average - but can someone be nice and offer a fair rating? Am I allowed to ask that? If no, please disregard - thanks.

David Field did it in 7 en got 5 the next day. I got an abysmal 22 with people telling me where to go.

ya I saw Carl in the maze and he's like..."well this looks familiar"...and Frost is yelling "Carl...now keep going...there!"

I didn't get to do the maze...
Was it hard?

It looks small but its harder than it looks.
I got about 9 minutes the first time and 15 minutes the second

i went through the maze and got a time of 6 minutes. they gave me this thing filled with water and it was strechy..it had a spider in of the "thing" and some confeity(sp error) in it. you could put your finger through the hold and it would strech and then come back to you.

I got 10 minutes or so, I received a bouncy ball.
Which I ended up giving to my team member who was directing me.  :D

EDIT: All my teammates got 5 minutes.  :rotfl:
I have no idea how.

I got it in 15 min. My teammate said there was a letter under the tower but it was near the tower so I messed up.

How was the maze like? In fact, what was it, how big...

It wasn't that big, it was just confusing :D
20 minutes here

Hmm, I really didn't like the rodeo, like that one girl who spent half an hour singing,  country music at the top of her lungs -_-.

Maybe luck is involved more than "skill"?  I thought with all of many of our "spatial relation" skills, we would have Owned that maze compared to the average tourist going through -- interesting, don't ya think?

15...

I'm making the message longer before submitting...

I could have done without that country singer too .. plus far too much time "setting up" between action, I felt.. but I guess the bull riding - even if they lasted only a few secs was pretty cool

I don't want to jinx it  :P  but I see my rating went up slightly - then again, though I read the site, I've NO Clue whatsoever what these ratings really mean, or how to rate someone else -- Is anyone else on the site in the dark about this as much as me?? Am I supposed to rate someone Every time I read a post? Oh, yeah, that'd be hard since I've no clue How!!!!!!!!

Oh, got off topic - the maze, the people on the lookout pt that I called to for help seemed more clueless than me in offering directions! Anyone else seek their help?

Just click one of the stars on other posts...

Were there questions or something in this maze?

the j person

rating went way down - whatever - I think it's sort of pointless - maybe I'm just not "getting the pt of it" - if someone posts a topic I find interesting, I'll answer it or read it - regardless if they have a 0 or a 80 rating .. so I don't get it -- ANYWAY - to answer the last poster, - nah, the maze didn't have questions - just had to figure out how to walk through it from start to end and get to 4 distinct areas where you punched a letter in a piece of paper - when you spelled out M A Z E  you could leave and you got paper time stamped at start and end so you could see how long you took! It was harder than it sounds, though

Ah, I now understand what you are talking about. There was a similar maze in Niagara, Canada.


Let $ABCD$ be an orthodiagonal trapezoid such that $\measuredangle A = 90^{\circ}$ and $AB$ is the larger base. The diagonals intersect at $O$, $\left( OE \right.$ is the bisector of $\measuredangle AOD$, $E \in \left( AD \right)$ and $EF \| AB$, $F \in \left( BC \right)$. Let $P,Q$ the intersections of the segment $EF$ with $AC,BD$. Prove that:

(a) $EP=QF$;

(b) $EF=AD$.

[i]Claudiu-Stefan Popa[/i]

[hide]a) $\Delta AEP \sim \Delta ACD \implies \frac{EP}{AE}=\frac{CD}{AD} \implies CD=EP.\frac{AD}{AE}$

$\Delta BFQ \sim \Delta BCD \implies \frac{FQ}{BF}=\frac{CD}{BC} \implies CD=FQ.\frac{BC}{BF}$

$\Longleftrightarrow EP.\frac{AD}{AE}=QF.\frac{BC}{BF}$ considering that $EF \| AB$  by Thale's theorem, $\frac{BC}{BF}=\frac{AD}{AE}$ consequently:

$EP=FQ$.

b) $\Delta DEQ \sim ABD \implies \frac{EQ}{DE}=\frac{AB}{AD} \implies EQ=\frac{DE.AB}{AD}$ and we know that $EP=FQ=\frac{AE.CD}{AD}$ so ,

$EF=\frac{DE.AB+AE.CD}{AD}$

Since the trapezoid is orthodiagonal we have that, $\Delta ADE \sim \Delta ACD \sim \Delta ABD$, simutaneously,with the  bisector theorem, we get:

$\frac{AE}{AO}=\frac{DE}{DO} \implies \frac{AE}{DE}=\frac{AO}{DO}=\frac{AD}{CD} \implies CD.AE=AD.AE$ analogously $DE.AB=AD.DE$ thus:

$EF=\frac{DE.AB+AE.CD}{AD}=AE+DE=AD$     [/hide]


S22

A bee can fly or walk in a straight line between any two corners on the inside of a cubic box of length 1. The bee managed to move so that it visited every corner of the box without passing through the same point twice in the air or on the wall of the box. What is the largest possible length of such a path?

[quote="Kyron"][hide="hint"]
The spacial diagonals (opposite vertices) are the longest ways to travel about the inside of a cube. Try to use as many of these as possible.
[/hide][/quote]
[hide="but..."]All the space diagonals intersect each other. Also, once you use a face diagonal, you won't be able to use the other face diagonal on the same face.[/hide]

[quote="aidan"]Yeah we have got to be careful here.. [b]without passing through the same point twice in the air or on the wall of the box[/b]. So we know that a length of $ \sqrt{3}$ can occur at most once because the space diagonal may be used just once.[/quote]
Oh  :blush:  dang, hadn't noticed that part. However, make sure to include that one, then.

[hide="Is this the answer?"]
Well, to start the problem, we need to identify the ways in which the bee can travel.
There are 3 ways: 1st way is the shortest way, from one vertex to another vertex with only 1 dimension changed and we shall try to avoid this way of travelling as much as possible to get the longest travelling distance.
2nd way is travelling diagonally in a 2 dimensional plane. lets call this method A
3rd way is from one end of the cube to the other end of the cube, space diagonally across. Lets call this method B

we start from one point and use method A 3 times, den method B once and method A 3 times again and by then we have covered 7 vertices, without crossing the same path and space. And this is all it needed to travel since it started from one of the vertices.

Hence answer = 6 * sqroot2 + sqrt3
sry i dunno how to type sqroot into the presentable form
[/hide]

[hide]
The path can consist of at most 7 segments, since there are 8 corners to be visited and no corner is visited twice.

There can be at most one space diagonal in the path, as all three space diagonals meet at the center of the cube.

Therefore, the maximum possible length of a path consists of 6 face diagonals and 1 space diagonal, which has length $ 6\sqrt{2}\plus{}\sqrt{3}$. It is pretty easy to construct such a path.

(Labeling the cube with face ABCD, E opposite A, F opposite B, G opposite C, ad H opposite D)
AFCHDBGE is an example a path.
[/hide]


use a combination of the following to simplify the problem.

$\log_{a}b=\frac{1}{\log_{b}a}$

$\log_{a^{x}}b=\frac{1}{x}\log_{a}b$

$\log_{a}b^{x}=x\log_{a}b$

Well, I used something like that, turned it all into base x, got to a quadratic which didn't have any solutions in $R$ so basically I got that there aren't any solutions...

No idea if that's correct however.


A solution containing 296.6 g of Mg(NO3)2 per liter has a density of 1.114 g/mL.  What is the molarity of the solution?

I've converted the mass of the solute to moles:

296.6 g Mg(NO3)2/148.33 g Mg(NO3)2 = 1.9996 moles Mg(NO3)2.

Now, I know I need to convert the 1.114 g/mL of solution to liters.  That's where I get confused.  Please help.

[quote]Now, I know I need to convert the 1.114 g/mL of solution to liters. [/quote]

Actually, you don't.  You already know that the solution has 296.6 g of Mg(NO3)2 per liter, which according to your calculations is the same as 2.000 moles Mg(NO3)2 (note that there there should only be 4 sig figs) per liter of solution....

So what we really have is mols of solute per liter of solution... sound familiar?  That's right, it's just molarity.  Hence, the molarity of the solution is 2.000 M.  Notice that the 1.114 g/mL was not necessary at all; don't let an excess of information get to you.[/quote]


First, I took the number of cans of soda in my pantry (eighteen), divided it by the number of cans of soda in the trash can in the kitchen (three), and added six to get [b]12[/b].

[quote="Ignite168"]Can you please hide your answers? That problem was like, team round #10 at nationals and I had trouble doing it but then I look and then I see the answer and the problem is ruined :([/quote]

Yes please, hide your solutions.... well, a good explanation of the problem ??? anyone??

[quote="Treething"][quote="José"]Yesterday, I had $84$ photos on album $a$, and $20$ photos on album $b$. Today, I put $x$ photos on each album. Now, the number of photos in album $a$ is three times the number of photos of album $b$. Find $x$.[/quote]

[hide="Alternate method"]
84 + x = 3(20 + x)
84 + x = 60 + 3x
24 = 2x
$\boxed{12}$ = x
[/hide][/quote]

That's exactly what I wanted

[quote="fireemblem13"][hide]or.... if you don't have paper and pen handy,

$\frac{84-20}{2}-20=12$[/hide][/quote]

why? Can you explain it with more detail?

[quote="kevyy"]same thing as treething[/hide][/quote]

OK

[hide]The difference 64 constitutes 2/3 of the new a, or twice the new b. So the new b is 32, thus we added 12.[/hide]

[quote="mathgeniuse^ln(x)"][quote="kevyy"]same thing as treething[/hide][/quote]

Same thing as kevyy.[/quote]

OK

[quote="mathgeniuse^ln(x)"][quote="kevyy"]same thing as treething[/hide][/quote]

Same thing as kevyy.[/quote]
Same thing as mathgeniuse^ln(x). (just kidding)

[hide="my solution"]
I would set up an algebraic equation.
Since he added x photos to each album we have
$84+x$ and $20+x$
But since the number of photos in album $a$ is three times the number of photos of album $b$, we mulitply $20+x$ by $3$.
Here's our equation
$84+x=3 \cdot (20+x)$
Simplifying we get:
$x=12$[/hide]


trigonometry
function
calculus
derivative
geometry
3D geometry
algebra

Prove that the system of equations:
20x+3y+5z=4+cos(x)
25y+3x+4z=5+2cos(y)
30z+4y+5x=6+3cos(z)

admits a solution (x,y,z) in the set (0,1)*(0,1)*(0,1)

Consider map A
\[
(x,y,z)\to (\frac{\cos x}{20}-\frac{3y}{20}-\frac{5z}{20}+\frac{1}{5}, ...)
\]
It is easy to show A:[1/100,1/3]3->[1/100,1/3]3 and moreover A is a contracting mapping. Hence A has unique fixed point in [1/100,1/3]3.

I like your solution! But how do you prove a maps (1/100,1/3)^3 into itself?

Actually this topic is dead but Myth has suggested me to post my solution:
Consider the function
$f(x,y,z)=10x^2+12.5y^2+15z^2+3xy+4yz+5xz-4x-5y-6z-\sin x-2 \sin y- 3 \sin z$.
The conditions are equivalent the to $f_x'(x,y,z)=f_y'(x,y,z)=f_z'(x,y,z)=0$.
Now take $x_0,y_0,z_0$ be point of minimum on $[0;1]$.
It's obvious that $x,y,z<\frac 12$ otherwise the derivative in corresponding variable is  positive and we could decrease it decreasing the expression.
Now we can prove that $x,y,z>0$ otherwise the derivative in corresponding variable is  negative and we could increase it decreasing the expression.
Hence the minimum occurs in the interior of the cube an thus the partial derivatives are zero
QED


[img]http://bbs.pep.com.cn/attachments/20071128_8fadcfa14f3975159142eoNsh4kMmgtq.jpg[/img]

I wonder how to represent this thing by using $ \prod_{}{}$ or sth .

I've tried , but not work out . Can you ? guys

After working by hand, and fixing my errors (computation and LaTeX) over and over...
[hide="This better render correctly..."]
$ \displaystyle e \equal{} 2\sqrt2 \prod^{\infty}_{n \equal{} 2} {\sqrt [2^n]{\dfrac {\displaystyle \prod^{2^{n \minus{} 1} \minus{} 1}_{m \equal{} {2^{n \minus{} 2}}} {(2m)(2m \plus{} 2)}}{\displaystyle \prod^{\dfrac {n^2 \minus{} n}2} _{m \equal{} {\dfrac{n^2 \minus{} 3n \plus{} 4}2}} {(2m \plus{} 1)^2}}}}$
[/hide]
There has got to be a better way... like $ \sum_{n\equal{}0}^{\infty}{\dfrac1{n!}}$ xD.

It's wrong because the poster inexplicably used exponential bounds on the product in the numerator and polynomial bounds in the product in the denominator.  Change them both to exponential and it should work out fine.

Yeah, I noticed that myself after a while... too lazy to fix it...


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

I thing these two integrals will be interesting for you. They are proposed in Bulgarian Math Magazine. I think they are not too easy or too difficult. But they seems to be good. Proposer is Zdravko Startz - Serbia and Montenegro.

Prove that:
$ \int^1_0 \frac {2x \plus{} (1 \plus{} x^{2})^{2}arctgx}{(1 \plus{} x^{2})(1 \plus{} (1 \plus{} x^{2})arctgx)} \,dx \equal{} 1 \minus{} ln\frac {2 \plus{} \pi}{4}$
$ \int^1_0 \frac {2x \minus{} (1 \plus{} x^{2})^{2}arccotgx}{(1 \plus{} x^{2})(1 \minus{} (1 \plus{} x^{2})arccotgx)} \,dx \equal{} 1 \plus{} ln2$

P.S. :-) Do not use integration software.

Let $ \tan ^{ \minus{} 1}x \equal{} \theta \Longrightarrow x \equal{} \tan \theta$,

$ \int_0^1 \frac {2x \plus{} (1 \plus{} x^2)^2\tan ^{ \minus{} 1}x}{(1 \plus{} x^2)\{1 \plus{} (1 \plus{} x^2)\tan ^{ \minus{} 1}x}\ dx$

$ \equal{} \int_0^{\frac {\pi}{4}} \left\{\frac {1}{\cos ^2 \theta} \minus{} \frac {(\theta \plus{} \cos ^2 \theta )'}{\theta \plus{} \cos ^2 \theta }\right\}\ d\theta$

$ \equal{} \left[\tan \theta \minus{} \ln (\theta \plus{} \cos ^2 \theta )\right]_0^{\frac {\pi}{4}} \equal{} 1 \minus{} \ln \left (\frac {\pi}{4} \plus{} \frac {1}{2}\right)$.

How did you come to this idea?! Have you ever seen these integrals or similar before? Was it difficult to you?

Thank you! :)

This is the first time to me and it is easy to me. Sorry I have omitted 4 lines. :lol:

Idea: For $ m \equal{} \tan \frac {\theta }{2}$, we have $ \sin \theta \equal{} \frac {2m}{1 \plus{} m^2},\ \cos \theta \equal{} \frac {1 \minus{} m^2}{1 \plus{} m^2}$.

As I think - people are different ... and I believe you have more experience than me with integrals.

Let $ \cot ^{ \minus{} 1} x \equal{} y\Longrightarrow x \equal{} \cot y$, Use $ \tan ^{ \minus{} 1} x \plus{} \cot ^{ \minus{} 1} x \equal{} \frac {\pi}{2}$,

$ \int_0^1 \frac {2x \minus{} (1 \plus{} x^2)^2\cot ^{ \minus{} 1} x}{(1 \plus{} x^2)\{1 \minus{} (1 \plus{} x^2)\cot^{ \minus{} 1} x}\ dx$

$ \equal{} \int_{\frac {\pi}{4}}^{\frac {\pi}{2}} \left\{\frac {1}{\sin ^ 2 y} \plus{} \frac {(\sin ^ 2 y \minus{} y)'}{\sin ^ 2 y \minus{} y}\right\}\ dy$

$ \equal{} \left[ \minus{} \frac {1}{\tan y} \plus{} \ln |\sin ^ 2 y \minus{} y|\right]_{\frac {\pi}{4}}^{\frac {\pi}{2}}$

$ \equal{} 1 \plus{} \ln \frac {\frac {\pi \minus{} 2}{2}}{\frac {\pi \minus{} 2}{4}} \equal{} 1\plus{}\ln 2$.

Did you use integration solving software for hints?

Great, kunny! You are very smart then - it were not easy at all. I tried even with software ... but without results.


articles geometry algorithm algebra polynomial

What can you do with a degree in math? (preferably something that makes a lot of money)

Well, with a Ph.D in math you can do basically anything you want, from a chef to the president of your country (I am serius, José Joaquín Trejos Fernández studied mathematics and became the presiden of Costa Rica in 1966, and actually, is the second last president that people felt bad when his period ended, in Costa Rica a president cannot be reelected, what is an stupidicity if you ask me.)

But know, let's be serius, and talk a little more in real. If you have a Ph.D. in math, may be in the so called third world, there is not that much that you can really do, but if you are in the first world, that include actually a lot of places, you can do a lot of stuff. There are three main things you can do, if you want to stay somehow in the area:

1) You can be a university professor, that in the first world is a good paid profession, if you compare to the pressure that you have to deal with.
2) Be a researcher for the goverment (in the US, NSA, CIA, Department of Defense, etc) or for private center (Bell Labs, Microsoft, Hewlett Packard, just to mention a few)
3) Finance. (What I am saying about finance is not first hand experience, but things I have heard from some people around the department)

You can relatively easy jump from 1 to 2 back and forth, specially if you are a good researcher. The stippends, I would say, are in increasing order in the list I said. Unfortunatelly, once you jumped to Finance is a little hard to go back to any of the other two (hard does not mean impossible) mainly because you are not doing the same kind of research that is spected from you in the first two, and there is a gap in your publications, etc.

Nevertheless, I am not going to mention names, but a guy that I know resently stopped by, he told me that he has gotten a job in a Finance Firm, and that he was earning more than $\$300K a year, and besides, his chrismas bonus was A LOT bigger than his year salary as a Prinston Professor with a tenure track.
As far as I have heard, in Finance people is not always looking for people with a lot of knowledge, but very smart people, and if you have a Ph.D in mathematics, you are by definition a very smart person.
This may make sound like finance is a GREAT job, and well, as far as I have heard, it is, but it is also a very unstable one. Most of the people don't stay more than a few years in business, principally because if a firm trust you to manage accounts of around $\$1billion, well, they are probably going to chop your head if you do some mistake that means a 0,01% of an account, because only that is A LOT OF MONEY, that means, it is a job where you cannot screw up at all.

You can also go to some not totally related, but someshow math applied job, as some software development job, etc.

I hope this helps.

Best regards,

[quote="djimenez"]Well, with a Ph.D in math you can do basically anything you want, from a chef to the president of your country (I am serius, José Joaquín Trejos Fernández studied mathematics and became the presiden of Costa Rica in 1966[/quote]
Well, Ahmed Chalabi has a Ph.D. in Mathematics from the same university that I do: the University of Chicago. Given the news out of Iraq regarding Chalabi's activities over the last 10 or 20 years, I'm not so sure I want to claim him.

People in this forum often talk about people who like math/have degrees in math getting jobs in "finance".  Which jobs are being reffered to be "jobs in finance"?  Does it mean like mutual fund managers, or market analysts or executives?  it would seem to me that all three of these things would take quite a bit of experience and would take a while to get to.

figure out some of those million dollar questions and after you answer only about a thousandof them, you will be a billionnare :D

Yeah, but there's only seven of them, and I think two have already been solved.

all of them are the things under our sn and ive been lead to believe none of them are solved yet

Well, I know there's a solution of the Poincare Conjecture out there, but mathematicians who have looked it over are still not sure if the proof is valid.

[quote="LynnelleYe"]Which two?[/quote]

There are proofs that haven't been neither accepted or rejected of the Poincare Conjecture (by a russian, I believe) and the Riemann Hypothesis (by a French who is currently teaching in Purdue), or at least that is what I have heard!

Best,

[quote="h_s_potter2002"]I may be wrong, but I thought that P=NP was also solved.[/quote]

Well, usually I don't read too much about algorithm complexity, but I could almost swear that it hasn't. I do work  a little in Cryptograpy (not as main research but as a parallel area), and it would have a deep impact if someone proves either $P=NP$ or $P\neq NP$.

$P=NP$ would create a HUGE scandal in mathematics if it would be solved. See, nobody have ever came up with even an attempt of a solution, because it somehow require to proof that something cannot be done in polynomial time, and there are (as far as I know) no techniques for such problems. If someone comes up with a solution, even if it is wrong, it would be heard for thousands very fast.  Of course, nobody either propose something not  seius, because if it is not a serius attempt, probably that could destroy a mathematician's carrier.

The problem with, for example, Poincare Conjecture, but principally with Riemann's Hypothesis, is that some people have came up with solutions to these problems in the past, and someone has found mistakes in those problems, therefore experts are not that optimistic when someone else come with a "proof". Either way, the Clay Institute supposly would have to pay Grigori Perelman $1 million for his solution of the Poincare Conjecture, if no body is capable to find an error on it, because next April it would be 2 years since he submited his papers, and that is the period stipulated by the Clay Institute.

Best regards,

[quote="ComplexZeta"]Isn't it two years after being published? I don't think Perelman's paper has been published.[/quote]

It depends on your definition of pubilshing a paper. If your definition is that to publish means to prepare and issue (printed material) for public distribution, well, may be not, I am not sure, but if, on the other hand, you define to publish as to bring to the public attention, he certainly has done it. I believe that the problem is that, being a theorem that important, people would like to check before printing it.

Best,

[quote="Treething"][quote="miraculouspostmaster"]I'm not too familiar with the occupation, but Actuaries, who apparently evaluate risk indexes for clients of insurance companies, make a good deal of money.  I'm not sure how much more schooling (if any) would be required beyond a general mathematics degree.  Maybe someone else could elaborate?[/quote]

Lucky guessing is all you need for that job =X .

Remember CNN's presidential election projection blunder in the 1970s :P ? I bet the predictor got fired after that xD .[/quote]

Well, usually a Master is required to obtain the Actuarian License, and a bunch of exams, as far as I have heard, not easy ones, most of the people have to present them more than three times!

The problem, as far as I know, is that some of the papers that a company has to present to place some products in the market have to be done by a licensed actuarian.

Best,

[quote="manuel"]Hey if somebody is thinking to study mathematics and is asking what can you do with mathematics. DO NOT STUDY MATHEMATICS! you should study mathematics if you love it.. it does not matter if you win any money ... just do mathematics beacuse of curiosity and love. Same as a painter, musician or any artist..[/quote]

Yes manuel, the painter loves to paint, but usually sell his paintings, the musician loves his instrument and loves to play it, but he (indirectly) charges you to go to his concert or to have his/her CD. They need to make themselves a living (and some of them do it pretty well  :P )

I love to do math, but I don't live from air, I have to make a living. Probably many people is wondering about that (specially that people that comes from a working class family, I include myself) because we know that lives is usually not easy. It is a healthy couriosity to ask about how can you make a living with it.

I agree with manuel, in the sence that, if you don't love mathematics, you shouldn't study mathematics, because, it would be worthless. But you can be sure that there is plenty of work for mathematicians, and some specialities that we can choose  :first:

Best regards,

[quote="manuel"]That's true.. a mathematician needs food and a house.. but you can live "well" beinga profesor of any university...Or maybe you do not need a house.(like Erdos) and if you want to get married look for a woman that is a mathematician or an artist or that has the same ideas that you have and live happy with her in a small house[/quote]

Sorry, I didn't get your point. :?

On the other hand, in many countries, professors are the professionals more underpaid, specially considering their qualifications!  :(

And forgive my ignorance, but how is that you say about Erdos not needing a house?

Best regards,

Erdos did not have a house... he lived on the move with his fellow researchers ..
What I was trying to say is that every professor lives well..
I haven't heard of any profesor that doesn't have any money to eat.

Any way, eventually you would like to have more than just something to eat. Unfortunatelly, in some countries that is not always something easy to do with the salary of a professor.

For example, I was doing some research about possitions for mathematicians, and I found that most of the postdoc in the US pay around $\$35K to $\$45K, what, for a professional, is not that much (at least not inside the US), but the private investigation centers have salaries most of the time for around $90K (at least the few I saw), and if you see what mathematicians in finance are earning, well, figures go up, usually A LOT. Even if you love to do math, wouldn't you like to gain a lot better, be able to afford a house that you can pay in five years or so, than to be paying for it for the next thirty years? and well, a lot of things like that.

I know that money don't give happiness, but isn't a good salary something that most of the people wants!

Best regards,

Actuaries have to write 7 (or 9 if you want to go into Casualty exams), and according to a friend of mine who is in it, you need good math skills (and/or a good amount of hard work, depending on your cleverness) for the first 3 (or 4) exams (the math ones), and then a lot of hard work for the next few ones since it is more on non-mathematical theory, like finance/economics topics. You have to study quite hard (and regularly) to pass your exams. My friend studied for 4 months, 2 hours every day (and he wasn't exaggerating) to prepare for his 5th exam, and he's still not sure if he passed it or not.

You don't need all the exams to start working though, and many people finish their exams while they work, though all have said it is very time-consuming. Employers tend to want you to have at least an Undergraduate mathematics degree and a few exams (exams are looked at by employers with a lot more importance, I think).

In all, I've heard the whole process of [i]becoming an actuary[/i] is quite difficult but if you make it in the end, and if the job market is still good as you finish your exams (better in the US than Canada), you can be well-paid. Historically, in the past, actuaries have been very well-paid, but today there are many more people wanting to become actuaries. But if you are good at math (and have or develop other skills like communication and business), then the job is nice, not only in pay, but supposedly the work conditions are good as well, so I've heard. (My friend in actuarial science who is in co-op (you go on work terms with companies while you are at University) contests this fact, as he says it depends on the person, as some of his managers are not as happy as others, and that he himself has had to work late at times. However, he says it depends on which field in actsci -- some jobs in the field allow you to have nicer hours than others. However, third-party written sources state that the conditions, in general, in actsci are quite good compared to other jobs, in terms of stress.)

Of course, I know first-hand that there are a lot more people going to actuarial science at University now, especially since last/this year.

I think the SOA (Society of Actuaries -- they write and mark the exams) design the exams so 40-50% of the people who write each exam pass it, so you may have to rewrite an exam more than once before you pass it.

In University, there is a lot of politics that go on, and I do think there is quite a bit of pressure to publish results and fulfil administration responsibilities at your University, besides teaching. If you receive tenure, your job tends to be quite secure, however.

I think the more seniority you have (and/or the quantity and quality of results you publish) the more you make, naturally. I think Professors who have been in their field for many years get paid quite handsomely, and quite possibly the new people who are the head of the field publishing groundbreaking results also (though I think years of experience/amout of results count for a lot).

Of course, I am only inferring this from various pieces of (generally-valid) second-hand information, as I am not a professor myself.

These comments come from my 4-year experience as a postdoc at the University of Washington.

[quote="daleliop"]In University, there is a lot of politics that go on, and I do think there is quite a bit of pressure to publish results and fulfil administration responsibilities at your University, besides teaching. If you receive tenure, your job tends to be quite secure, however.[/quote]All true.
[quote]I think the more seniority you have (and/or the quantity and quality of results you publish) the more you make, naturally. I think Professors who have been in their field for many years get paid quite handsomely, and quite possibly the new people who are the head of the field publishing groundbreaking results also (though I think years of experience/amout of results count for a lot).[/quote]This is not necessarily true.  People who get groundbreaking results (and we are talking about a handful of people a year) can win big prizes, but these are generally from outside organizations (e.g. AMS), not part of their regular salary.  Also, seniority and/or voluminous research results do not necessarily mean higher pay, especially at public universities where the vagaries of the legislative agenda can affect university funding.  Needless to say, no one (in my experience) becomes a professor for the money; if it is primarily money that one cares about, there are so many other options that pay much better.  You become a professor becuase you love mathematics research.

If you want some idea of what it pays to be a mathematics professor, read this survey published by the AMS: [url]http://www.ams.org/notices/200502/04first-report.pdf[/url]

[hide="True AoPS answer to the questio is obviously .."] :) Same as $\frac {\pi}{180}$ radians :).[/hide]

[quote="tetrahedr0n"]Could someone post that article (or parts of it) here? An AMS subscription is required to access it.[/quote]
You need to register on the AMS web site, but you don't need to be an AMS member.

[quote="Gyan"][quote]Professors who have been in their field for many years get paid quite handsomely,[/quote]
To put the things in perspective when I taught  for a short time at the City University of New York, The starting salary of an  assistant professor was less than the  starting salary of  a trash collectors - Both of them NY City employees. :(

(The salary of a Nobel Laurreate at the state university was oh... about 1/3 of  a good physician..)[/quote]

I think trash collectors [i](also called sanitation engineers)[/i] do get paid a good sum, so I've heard.

[quote="djimenez"]There are proofs that haven't been neither accepted or rejected of the Poincare Conjecture (by a russian, I believe) [/quote]

There is a highly condensed but very credible (i.e. either a proof or a significant breakthrough) proof by Perelman, available on the web along with expansions and exegeses by various people (websearch "Kleiner Lott Perelman" for links).

[quote] and the Riemann Hypothesis (by a French who is currently teaching in Purdue), or at least that is what I have heard!  [/quote]

There is a serious but not at all credible attempted proof by Louis de Branges, who did solve the Bieberbach conjecture earlier.   His paper is on the web and has been scrutinized, if I remember both the paper and a rebuttal are at  arxiv.org.

Well, now in 2018, 13 years later, things are different... xd


Sa se determine functiile $f : \mathbb{Z}$ $\rightarrow$ $\mathbb{N}$ pentru care avem:

$6f(k+3)-3f(k+2)-2f(k+1)-f(k)=0$, pentru orice $k$ din $\mathbb{Z}$

La ora de istorie am facut ceva din ea. [hide]Am luat mai intai

\[6f(a+3)-3f(a+2)-2f(a+1)-f(a)=0\]

si apoi

\[6f(a+2)-3f(a+1)-2f(a)-f(a-1)=0\]

Calcule, substitutii, alea alea, nu am finalizat rezolvarea, m-am mai uitat o data pe 'subiect' si am observat ca $f(a)=x$, adica $f$ este o functie constanta.
[/hide]

Rezolvare completa, any?

asa "ochiometric" $f(x) =n$, $n \in \mathbb{N}$ , dar o rezolvare completa cu substitutii nu prea vad , deoarece cum iti faci o substitutie iti apare o alta necunoscuta in loc .

[hide="idee ... "]

Avem $6f(k+3) = 3f(k+2)+2f(k+1)+f(k)$

Daca f este crescatoare , atunci
\[f(k) \le f(k+3) \]
[b](1)[/b]
\[2f(k+1) \le 2f(k+3) \]
[b](2)[/b]
\[3f(k+2) \le 3f(k+3) \]
[b](3)[/b]

Insumand (1) , (2), (3) , obtinem
\[3f(k+2)+2f(k+1)+f(k) \le 6f(k+3) \]
Analog tratam pentru f descrescatoare , si atunci observam ca ... pentru egalitate avem :

$f(k) = f(k+1) = f(k+2) = f(k+3)$ - adica f -functie constanta

[/hide]

O altă abordare:

Asta e de fapt o recurenţă liniară de ordinul trei.
Ecuaţia asociată e $6 x^{3}-3 x^{2}-2 x-1 = 0$, i.e.
\[(x-1) \left( 6 x^{2}+3x+1 \right) = 0 \, \, \Longleftrightarrow \, \, (x-1) \left( x-u \right) \left( x-\overline u \right) = 0 . \]
Soluţiile sunt date de $f(n) = \alpha 1^{n}+\beta u^{n}+\gamma \overline u^{n}$, unde $\alpha, \beta, \gamma$ sunt determinate de "valorile iniţiale" $f(0),f(1),f(2)$.
Observăm că $|u| < 1$, deci $\lim_{n \to \infty}\left( \beta u^{n}+\gamma \overline u^{n}\right) = 0$.
Cum $f(n) \in \mathbb Z$, $\left\{ \beta u^{n}+\gamma \overline u^{n}\right\}$ trebuie să fie eventual $0$, de unde ne iese $\beta = \gamma = 0$, adică $f(n) = \alpha$.

Orice funcţie constantă e bună, aşa că astea-s toate.

[quote="pohoatza"]da, asta e ideea de baza, cu o mica rectificare, nu e nevoie sa presupunem ca e crescatoare. cum Im-u functiei e inclus in $\mathbb{N}$, atunci are un minim, si luam $a$-ul acela minim.[/quote]

Aha ... Atunci cred ca am putea "largi" codomeniul functiei , adica :
\[f : \mathbb{Z}\rightarrow \mathbb{Z}\]
:?:

mie solutia lui perfect_radio imi pare corecta, si intr`adevar, e suficient ca "a doua multime" (cum ii zice ? :blush: ) sa fie $\mathbb{Z}$.
pe de alta parte, o functie $f$ arbitrara nu este neaparat crescatoare sau descrescatoare. adica poate sa nu fie in niciun fel. (e foarte usor de vazut). partea cu elementul minim din imaginea lui $f$ e altceva.


function
inequalities proposed
inequalities

Let $ a,b,c\geq0$. Prove that

$ \sum\frac{a^2bc\plus{}a^2c^2\minus{}2b^2c\plus{}bc^2\plus{}b^3}{c(b\plus{}c)(b\minus{}c)^2}\geq\frac{5}{2}$

----------------------
 :blush:  Panagiote LIGOURAS

[quote="Ligouras"]Let $ a,b,c\geq0$. Prove that

$ \sum\frac {a^2bc \plus{} a^2c^2 \minus{} 2b^2c \plus{} bc^2 \plus{} b^3}{c(b \plus{} c)(b \minus{} c)^2}\geq\frac {5}{2}$

----------------------
 :blush:  Panagiote LIGOURAS[/quote]
Your function is not homogeneous, are you sure?


For how many single-digit positive integers $ d$ is there an integer $ n$ for which the units digit of $ d^n$ is 6?

First, note that $ d$ must be even by considering modulo $ 2$.  Then:

Does $ 0$ work?  No.

Does $ 2$ work? $ 2,4,8,6,\ldots$ Yes.

Does $ 4$ work? $ 4,6,\ldots$ Yes.

Does $ 6$ work? $ 6,\ldots$ Yes.

Does $ 8$ work? $ 8,4,2,6\ldots$ Yes.

This is a grand total of $ \boxed{4}$.


integration
calculus
calculus computations

the limit as a approaches infinity of arctan(a/5)

It's the same as the limit of the arctangent itself, namely $\frac{\pi}2.$ Note that $\frac{a}5$ tends to infinity as $a$ does.

Related: show that $\int_{0}^{\infty}\frac{b\,dx}{b^{2}+x^{2}}$ is the same for all positive values of $b.$

si is the answer phi/2 or is it infinity?? cause I think it's infinity

[quote="EquinoX"]... cause I think it's infinity[/quote]

Can you give your reasons for thinking this?


我出生在一個非常貧苦的家庭，記得小時候，爸爸的生活很無聊，整天只能數錢，

媽媽也是，整天都在掃錢；我家住在深山中，每次要出去買東西，都相當麻煩，開保時

捷要5小時，開賓士也要4小時；雖然家裡有直昇機，但是開到那邊很難找到停車位，相

當的不方便；我偶爾也會出去逛街，爸媽怕我被綁架，隨身有20多個保鑣保護著，大家 看到我都嚇跑了，所以我從小就開始被排斥........

因為家裡是在深山，每次到了冬天都非常冷，爸爸也說，出去買東西很麻煩，所以

冬天沒有暖爐...沒有棉被...每天只能陪著爸媽燒錢取暖...而睡覺的時候，也只能蓋

著錢睡覺(我覺得英鎊比較溫暖)。

記得小時候，有一次，因為房間太大，還來不急跑出房間，就尿褲子了，所以，爸

爸在我房間放了一台小綿羊機車，好讓我能在10分鐘以內，衝出1000坪的房間，穿越5

公里的走廊，到達800坪的廁所(我常常在走廊迷路)；爸爸又另外叫人加蓋了20間600坪

裡面有著250坪小廁所的破房間，說:以後如果尿濕了，直接換房間，如果房間不夠或是

太小的話，再告訴爸爸，爸爸再叫人來多蓋幾間，我們的生活很辛苦，你要忍著點!!現 在想起來，爸爸真是個善良的人!!

還記得有一次，家裡遭小偷，因為他用炸藥炸壞了我爸的保險箱，使得裡面的金幣

不斷的往外滾出來，結果竟把那個小偷壓死了......我覺得那個小偷好可憐......金幣

壓死人是很痛的!!他不像之前那個被鈔票悶死的小偷一樣幸運。

媽媽也常常告訴我，我們家生活非常困苦~~要我學著吃苦~~所以我從小養成了刻苦

耐勞精神，我將來的志願，是要找200隻會吃錢的怪物，好把家裡的錢都吃光，為那些 可憐的小偷報仇!!

老師評語: 去死啦!!

[quote="ifai"]
老師評語: 去死啦!![/quote]
Oh, man!
If that teacher were me, I'll just go tell the student, you've already got one of these monsters-- yup, that's me! :lol:
frank

[quote="ifai"]我出生在一個非常貧苦的家庭，記得小時候，爸爸的生活很無聊，整天只能數錢，

O, gosh!
What a student!


search
induction
number theory unsolved
number theory

$f(x)=x^{3}+17$prove that for each natural number n>=2, there is a natural number $x$ for which $f(x)$ is divisible by $3^{n}$but not $3^{n+1}$
and how can i search?

[quote="Seung-Hoo, Lee"][quote="fermat3"]your result flow derictly from hensel lemma :lol:[/quote]

Yes! I want to tell about Hensel Lemma!
I forgot the name  :P[/quote]

what is hensel lemma?

Look at http://www.jmilne.org/math/CourseNotes/math679.pdf page 10.

Can someone post a  solution  to this one' like the  one with induction please at  least a hint.  thanks

by assumption exist $y$satify $3^{n}\parallel y^{3}+17$then let $x=r*3^{n-1}+y$where$r$is an arbitrary integer.then we have$x^{3}+17\equiv r*3^{n}+y^{3}+17/mod{3^{n+1}}$.so we can chose $r$ satify $3^{n+1}\parallel x^{3}+17$


[quote="Gauss's multiplication formula"]\[ \prod_{i\equal{}1}^{m\minus{}1}\Gamma\left(\frac{i}{m}\right) \equal{} \frac{(2\pi)^{\frac{m\minus{}1}{2}}}{\sqrt{m}}\]
\[ \prod_{i\equal{}0}^{m\minus{}1}\Gamma\left(z\plus{}\frac{i}{m}\right) \equal{} m^{\frac{1}{2}\minus{}mz} (2\pi)^{\frac{m\minus{}1}{2}} \Gamma(mz)\]
[/quote]
So,
$ \Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right) \equal{} \frac{2\pi}{\sqrt{3}}$
$ \Gamma\left(\frac{3}{3}\right)\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{5}{3}\right) \equal{} 3^{\minus{}\frac{5}{2}} 2\pi \Gamma(3) \equal{} \frac{4\pi}{9\sqrt{3}}$
$ \Gamma\left(\frac{6}{3}\right)\Gamma\left(\frac{7}{3}\right)\Gamma\left(\frac{8}{3}\right) \equal{} 3^{\minus{}\frac{11}{2}} 2\pi \Gamma(6) \equal{} \frac{80\pi}{81\sqrt{3}}$
Thus,
\[ \prod_{r\equal{}1}^{8}\Gamma\left(\frac{r}{3}\right) \equal{} \frac{2\pi}{\sqrt{3}} \times \frac{4\pi}{9\sqrt{3}} \times \frac{80\pi}{81\sqrt{3}} \equal{} \boxed{\frac{640\pi^3}{2187\sqrt{3}}}\]


The distance from $ A$ to $ D$ is $ 3$ intervals, so each interval is of length $ \frac{2.83\minus{}1.63}3\equal{}\frac{1.2}3\equal{}0.4$, so $ E\equal{}D\plus{}0.4\equal{}\boxed{3.23}$.


Set systems
linear algebra
Coloring
combinatorics
IMO 1988
IMO

Let $ n$ be an even positive integer. Let $ A_1, A_2, \ldots, A_{n \plus{} 1}$ be sets having $ n$ elements each such that any two of them have exactly one element in common while every element of their union belongs to at least two of the given sets. For which $ n$ can one assign to every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly $ \frac {n}{2}$ zeros?

First of all, the incidence structure is determined uniquely by the given conditions: every element of $A=\bigcup A_i$ belongs to [i]precisely[/i] two $A_i$'s, and $|A|=\frac{n(n+1)}2$. We can see this as follows:

Suppose we can find $x\in A_1\cap A_2\cap A_3$. Then $A_1\cup A_2\cup A_3$ has, except for $x$, at least $3(n-1)$ elements. Every $A_i,\ i>3$ covers at most $3$ of these (one from each of $A_1,A_2,A_3$), so in order to cover these elements one more time we would need at least $n-1$ more $A_i$'s, but we don't have that many. From this, the fact that $A$ has $\binom{n+1}2$ elements follows by a simple counting argument.

Now let's observe that $n$ being a multiple of $4$ is a necessary condition for the existence of the $0-1$ labeling. This is beause if we count the $0$-labeled elements every time they appear in a set we get $\binom{n+1}2$ (since there are $n+1$ sets, each one containing $\frac n2$ of them), and an even number, since every $0$-labeled element appears in precisely two sets, so it's counted twice. This shows that $\binom{n+1}2$ must be even, which, in turn, means that $4|n$.

Conversely, there is such a $0-1$ labeling when $4|n$, and we can prove it inductively. First of all, let's construct an $(n+1)\times n$ table $T_n$ containing numbers between $1$ and $\binom{n+1}2$ as follows:

The rows of the table will be the sets $A_i$ in order, and the numbers in a row will be the elements in the respective set. The first row is just $1,2,\ldots,n$. In the $i$'th step (with $i\in\overline{2,n+1}$) we construct the $i$'th row by setting $a_{i,j}=a_{j,i-1}$ when $j\in\overline{1,i-1}$, and making $a_{i,i},a_{i,i+1},\ldots,a_{i,n}$ the smallest consecutive numbers larger than all the numbers which have appeared so far.

The subtable formed by the last $n+1-4$ rows and $n-4$ columns has the same structure as $T_{n-4}$ (i.e. we can construct an isomorphism from $T_{n-4}$ to it induced by a bijection between the numbers $1,2,\ldots,\binom{n-4+1}2$ and those which appear in the mentioned subtable), so, by the induction hypothesis, we can find the desired $0-1$ labeling of this subtable. We divide the rest in three other subtables: $X$, the subtable formed by the first $4$ lines and $3$ columns, which we color $\begin{pmatrix}1&1&0\\1&0&1\\1&0&0\\0&1&0\end{pmatrix}$, $Y$, the one formed by the first $4$ lines and $n-3$ columns, whose columns we color alternately $\begin{pmatrix}0\\0\\1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\1\\0\\0\end{pmatrix}$ (starting with the first one), and the rest of the initial table, which lies in the lower-left corner, and is just the transpose of $Y$ (so its coloration is determined).

Actually drawing these tables should help clarify the whole thing.

another way to see that there are $\frac{n(n+1)}{2}$ elements in $X=\bigcup_i A_i$ is to count the number of pairs $(x,i)$ where $x\in X,x\in A_i$; that counts that $n(n+1)\geq 2|X|$; on the other hand counting the number of triples
 $(i,j,x),x\in X,x\in A_i\cap A_j$ gives that $\sum_{x\in X} d_x^2=2n(n+1)$ where $d_x$ is the degree of each $x$(i.e. the number of sets that contains $x$). Use Cauchy and it follows that $2|X|\geq n(n+1)$.
in fact, if you observe, the set collection here is unique upto isomorphism(relabelling of the points).

Another way to look that 4/n is sufficient is to think the n+1 sets as the points of a graph, and their members as edges. So we need a n/2 regular graph on n+1 vertices which is easily constructed from the following general fact, Kn is the disjoint union of hamilton circuit iff n is odd.

 I didnt read Grobber post carefully but its probably constructing this.

Answer: All $n$ such that $4|n$

We first make the following $2$ claims:

Claim $1$: Each element of union belongs to exactly $2$ subsets.
Proof:
Consider a subset $A_i$. Assume that some element $x \in\ A_i$ also $\in A_k, A_l$. There are $n-1$ elements remaining in $A_i$ and there are $n-2$ subsets to choose from. By pigeon hole principle, at least $1$ of the remaining elements in $A_i$ must $\in A_k$ or $\in A_l$. This contradicts the assumption that any $2$ subsets have only $1$ element in common.

Claim $2$: $4|n$
Proof:
Now, since each element in $A_i \in$ exactly $1$ other subset, total number of elements present in the union = $n*(n+1)/2$.
If each subset must have $n/2$ elements assigned a value of $1$, the total number of elements assigned value of $1$ = $n/2*(n+1)/2 = n*(n+1)/4$.
Thus $4$ must divide $n$.

Now we make our final claim:

Claim $3$: $4|n$ is a sufficient condition to assign every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly $ \frac {n}{2}$ zeros.

Proof:
Consider a regular polygon consisting of $n+1$ vertices where each line joining two vertices $A_i, A_j$ represents the element which $\in A_i, A_j$. Clearly there are a total of $n*(n+1)/2$ such lines representing the total number of elements of the union where each vertex is connected to $n$ vertices, meaning each of the $n$ elements of $A_i$ is part of $1$ other subset.

Starting with $i = 1$, let us start coloring all lines joining vertices $A_i$, $A_{i+1}$ with color Red, all lines joining $A_i$, $A_{i+2}$ with color White, $A_i$, $A_{i+3}$ with color Red, $A_i$, $A_{i+4}$ with color White and so on ... $A_i$, $A_{i+n/2}$ with color White.

Clearly each line from vertex $A_i$ alternates Red, White for first $n/2$ lines and then alternates White, Red for remaining $n/2$ lines implying that we could have exactly $n/2$ red lines emanating from each vertex $A_i$. But these $n/2$ lines represent $n/2$ elements of each subset $A_i$ which could each be assigned a value of $0$.  This completes the proof.


calculus
integration
analytic geometry
calculus computations

I know the definition, but I don't know how to apply into this case, don't know where to start.

[img]http://farm4.static.flickr.com/3309/3254433977_6ec4e5b3d4_o.jpg[/img]

That's not a "hint". That's telling you exactly what to do. Do it.

anyone know?

I couldn't get zero

I choose n to be 2, so I have 4 squares. 1/2 as side

check coordinate, and the f(x,y)=....
some point have the same coordinate, some dont, so I ended up like this

m=lower   M=upper

m11=1         M11=1
m21=0         M21=0
m12=0         M12=0
m22=1         M22=1

its not zero

You have not calculated the lower sums correctly. Also, the zero appears as a limit- you're not going to get it just by looking at a special case.


algebra
polynomial
geometry
function
algebra proposed

Prove that for two non-zero polynomials $ f(x,y),g(x,y)$ with real coefficients the system:
\[ \left\{\begin{array}{c}f(x,y)\equal{}0\\ g(x,y)\equal{}0\end{array}\right.\]
has finitely many solutions in $ \mathbb C^{2}$ if and only if $ f(x,y)$ and $ g(x,y)$ are coprime.

when we can say that $ f(x,y)$ and $ g(x,y)$ coprime please?

We say $ f(x,y),g(x,y)$ are coprime, if there does not exist a polynomial $ d(x,y)$ of degree at least one that $ d(x,y)|f(x,y)$ and $ d(x,y)|g(x,y)$.

we take $ f(x,y) = d(x,y)F(x,y),g(x,y) = d(x,y)G(x,y)$ where $ F$ and $ G$ are coprime.
we suppsoe that $ \{\begin{array}{c}f(x,y) = d(x,y)F(x,y) = 0\\ g(x,y) = d(x,y)G(x,y) = 0\end{array}$ has finitely many solutions.
then $ d(x,y) = 0$ has finitely many solutions.
so $ \exists a\in C:\ \forall x\in C: P_{a}(x) = d(x,a)\neq 0$
so $ P_{a}$ is constant then $ \forall (x,y)\in C^{2}:\ d(x,y) = h(y)$ for some $ h\in C[X]$
and like the same we get $ \forall (x,y)\in C^{2}:\ d(x,y) = k(x)$ for some $ k\in C[X]$
so $ d(x,y) = constant$ and $ f$ and $ g$ are coprime

the prof is not complete

this is a basic fact in algebraic geometry, usually proven by considering the resultant of the two functions as polynomials in k[x] with coefficients in k[y] or vice versa, as in http://en.wikipedia.org/wiki/Resultant#Applications

(it would probably be a bit tricky to come up with this if you hadn't seen it before ?)

In particular, this is a trivial consequence of [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem]Bezout's theorem[/url].  As such, it doesn't make a great Olympiad problem.

eh, I don't actually know algebraic geometry (especially basic algebraic geometry), but the proof of that theorem usually follows in some way from the technique outlined above, I believe? (so it would be a bit silly to apply bezout).

My point is that Bezout's Theorem should be citable on an Olympiad, which makes this problem a pretty bad idea.

I think not -- my general experience with olympiads is that if you can cite some result that makes the problem utterly trivial, such a solution rarely receives credit and certainly not full credit.