Find the total volume of the steel rods and use that to find the weight. Find the total volume of the aluminum rods and use that to find the weight. Add up all of the weights. Remember weight = volume x weight density, and that by rods they mean very thin cylinders. Also remember to have consistent units.

[quote="guess"]15 pieces of 2" diameter solid steel rod 12' long and 8 pieces of 3" diameter solid aluminum rod 10' long are loaded on a skid that weighs 76 pounds. What is the total weight of the load,including the skid?
steel weighs 490 pds per cu' or 0.28pds per cu"
aluminum weighs 167pds per cu' or 0.097 pds per cu"[/quote]
[hide]1 rod of steel = $4\pi*12(12)=576\pi\approx1809.557\text{in.}^{3}\approx506.68\text{pounds}$
Total steel = $506.68*15=7600.2\text{pounds}$
1 rod of aluminum = $9\pi*10(12)=1080\pi\approx3392.9201\text{in.}^{3}\approx329.11\text{pounds}$
Total aluminum = $329.11*8=2632.88\text{pounds}$
Overall total = $7600.2+2632.88+76=10309.08$ (10309.05 if you don't round until the end).[/hide]

Quevvy,
  That is not one of the possible answers. I have 4 answers to choose from. I have come up with a couple of answers but none of mine match one of the possibilities.
Here is what I tried-Area=D2x 3.1416divided by 4
steel rod--4 x .785=3.14 sq."
15 rods 12' long=2160"
  2160 x 3.14=6782.4
6782.4 x.28=1899.07

aluminum rod--7.065 sq"
8 rods 10' long=960"
960 x 7.065 =6782.4

1899.07 + 6782.4 + 76 (skid weight)= 8757.47 This is not one of the possible answers either. I need help. I am stumped, actually it doesn't take much to do that,lol.


Let S be a finite set of positive integers such that none of them has a prime factor greater than three. Show that the sum of the reciprocals of the elements in S is smaller than three.

[hide="maybe?"]S consists of numbers of the form $ 2^x3^y$. Now, if there were an infinite amount of numbers in S, there would be a total sum of $ (1\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}...)(1\plus{}\frac{1}{3}\plus{}\frac{1}{9}\plus{}...)$. Summing both geometric sequences, we get $ (2)(\frac{3}{2})\equal{}3$ as the sum. However, we're given that S only has a finite set of members, so one number or more is missing and thus, the sum must be less than 3.[/hide]

I think I have somewhat the right idea, but I can't write proofs very well.

[quote="xpmath"]I think I have somewhat the right idea, but I can't write proofs very well.[/quote]

QFT, yeah.


A "robotic" number is a positive, base 10 integer that starts and ends with a 1, and every number in between is either a 1 or a 0. The set of robotic numbers is listed from lowest to highest. 1, 11, 101, 111, 1001, 1011, 1101, 1111, 10001, 10011, etc.

1) What is the lowest robotic number whose prime factorization contains the digit 8?

2) Are there any robotic numbers (other than 1) that are also perfect squares/cubes/4ths/5ths... ? If so find the lowest one.

3) How many robotic numbers are there less than or equal to a googol + 1? (A googol = 10^100)

4) How many of THOSE numbers are multiples of 27?

could you be moe specific for 1)?
Do you mean that there is a prime dividing the robotic number that contains the digit 8? or that p^8 exactly divides the number for some prime p?

I think he means it contains a factor like 83 or 89.

Yes, it contains a prime with the digit 8, but not the factor 8. In fact, it's easy to see that's impossible for robotic numbers.


For what integer a does x^2 − x + a divide x^13 + x + 90?

(i thought about this for a while and didn't get anywhere)

There is a hard way and an easy way to do this.  One of the easier techniques is to note the following:  if they divide each other, then their quotient is an integer polynomial.  In other words, if $ f(x)$ and $ g(x)$ are monic integer polynomials such that $ f(x) | g(x)$ then $ f(n) | g(n)$ for any integer $ n$.  How can you use this fact to narrow down the possible values of $ a$?  (And can you now test them?)

[hide="Just plug in numbers"]$  \; \begin{array}{ccc} x = 1 & \implies & a|92 & x = 0 & \implies & a|90 \end{array}  \} \implies a | (92 - 90) \iff a|2 \iff a \in \{ - 2, - 1,1,2 \}$

$ x^2 - x + a$ has a positive root for all $ a \leq 1/4$, but $ x^{13} + x + 90$ has no positive roots. So, $ a > 1/4$ , and thus $ a \in \{1,2 \}$

$ x = - 1 \implies a + 2 | 88$, which rules out $ a = 1$

Verify that $ \boxed{a = 2}$ actually works.[/hide]

[i]PS: He was 101 posts away from 9000[/i]


function
analytic geometry
integration
geometry
logarithms
real analysis
real analysis unsolved

Hey guys - if I take the function f(x) = |log|x||^p for |x|< 1, 0 elsewhere, is this function integrable over the real line for p<1. More generally, if we look at this function defined identically over R^n, is it integrable for p<1/n? I'm pretty that the first part is true, not so sure about the second.

Thanks!

It's $p>-1$ regardless of dimension. For functions of the form $f(x)=g(|x|)$ (i.e. radially symmetric $f$), you can use the "generalized polar/spherical coordinates" like this: $\int_{\mathbb R^{n}}f(x)\,dL^{n}(x)=\omega_{n-1}\int_{r=0}^{\infty}g(r)r^{n-1}dr$. Here $\omega_{n-1}$ is the area of $S^{n-1}$.
For $p<0$ the singularity of $g(r)=|\log r|^{p}$ is at $r=1$, where the factor $r^{n-1}$ does not affect the convergence.

For $p>0$ the singularity of $g(r)=|\log r|^{p}$ is at $r=0$, but it's logarithmic and therefore integrable regardless of $p$.


I don't think this should be hard but my brain is dysfunctional today.  Given constants $c_i$ and $a_i$, find $C$ and $A$ such that

$c_1\cos(\theta-a_1) + c_2\cos(\theta-a_2)=C\cos(\theta-A)$.

Is it supposed to come out nice?  I got
[hide]$C = \sqrt{c_1^2 + c_2^2 + 2c_1c_2\cos{(a_1-a_2)}}$

And $A$ is easy to get into any inverse trig function, but I don't know if I could simply it from there.

Work: expand the cosine differences, the match up $\cos{\theta}$ and $\sin{\theta}$ terms to get two equations.  To solve for $C$ I squared both so I could get rid of $A$, since $\sin^2{A} + \cos^2{A} = 1$.[/hide]


Ugly proof.
By schur's inequality we have:
$\sum_{sym}x^{6}+\sum_{sym}x^{4}yz \ge 2\sum_{sym}x^{5}y$
So we only need to prove:
$\sum_{sym}x^{5}y+\sum_{sym}x^{2}y^{2}z^{2}\ge \sum_{sym}x^{4}yz+\sum_{sym}x^{3}y^{2}z$
Which is equal to:
$\sum(x(y+z)(y^{2}+z^{2})-x^{2}yz)(y-z)^{2}\ge 0$
It can be proved by SOS.

A solution without SOS (my 'parallelogram trick' :D ): we have
$[600]+[222]\ge2\cdot[420]$ (Schur)
$[600]+[420]\ge2\cdot[510]$ (AM-GM)
$[222]+[420]\ge2\cdot[321]$ (AM-GM)
sum it all up. Done. :P

Nice! My solution used only SOS.

I've been trying to come up with a general rule for inequalities of the form:

$[a,b,c]+[d,e,f]\ge[g,h,i]+[j,k,l]$

Any conjectures? My most promising conjecture so far is this one:

Does $[a,b,c,d,e,f]\ge[g,h,i,j,k,l]$ imply $[a,b,c]+[d,e,f]\ge[g,h,i]+[j,k,l]$?

[quote="zeb"]I've been trying to come up with a general rule for inequalities of the form:

$[a,b,c]+[d,e,f]\ge[g,h,i]+[j,k,l]$

Any conjectures?[/quote]
This is being discussed in two threads:
http://www.mathlinks.ro/Forum/viewtopic.php?t=56369 and
http://www.mathlinks.ro/Forum/viewtopic.php?t=104731
 [quote="zeb"] My most promising conjecture so far is this one:

Does $[a,b,c,d,e,f]\ge[g,h,i,j,k,l]$ imply $[a,b,c]+[d,e,f]\ge[g,h,i]+[j,k,l]$?[/quote]
This looks like a very nice and [b]original [/b]idea!! For all cases I've checked, it works. :)  Note there are even cases like
$[15,9,0]+[8,8,8]\ge[12,6,6]+[11,11,2]$ which holds in spite of
$[15,9,8,8,8,0]\ge[12,11,11,6,6,2]$ being false. So we may hope that this condition is :showoff: "strong enough". :showoff: Indeed, a promising conjecture!

After retrying with some partitions one would not really suspect to work, I found that there must be something more required :maybe: . For example, we have
$[655220]\ge[654311]$, but $[622]+[550]\ge[631]+[541]$ is false (e.g. $(x,y,z)=(y^{3},y,1)\Rightarrow$ LHS has smaller degree than RHS, as $20<21$) :(
I think if we suppose wlog $a\ge d$ and $a\ge j$, where the 4 partitions $[a,b,c],...$ are of course each given in decreasing order, we must require at least $[a,b,c]\ge [g,h,i]$.

So the conditions would sensibly approach Soarer's, and I wonder if with the condition $[a,b,c,d,e,f]\ge[g,h,i,j,k,l]$ your conjecture may be altogether almost weaker (because of the more restrictive set of conditions) than Soarer's one.  :roll:
I mean this in the sense: if both conjecutres are true, [i]one doesn't contain the other[/i], but I think there are "more" quadruples of partitions satisfying Soarer's conditions.  :)


ratio LaTeX quadratics MATHCOUNTS calculus

golden ratio

too lazy to write it using latex

As if you didn't give away the answer.

Anyway, the equation $ x^2\equal{}x\plus{}1$ was already part of the square root problem. Just saying.

[quote="math154"]did you know LaTeX is actually pronounced as if the x were a k instead??? wow i'm so bored.[/quote]

Please try to keep your posts related to the topic.

And $ \sqrt {0 \plus{} \sqrt {0 \plus{} \sqrt {0 \plus{} ...}}} \equal{} 0,1$

EDIT: Didn't know if you wanted non-obvious answer.

:lol:

Not $ \minus{}1$...if you square it, you have the quadratic $ x^2\minus{}x\equal{}0$.

[quote="math154"]or 1 or -1[/quote]

the square root sign implies non-negative, and last time I checked, -1 is negative

-1 is negative??? :blush: I never knew that!

[quote="math154"]Evaluate $ \sqrt {0 \plus{} \sqrt {0 \plus{} \sqrt {0 \plus{} \cdots}}}$ and $ \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} \cdots}}}.$

Hint: One of them is my favorite number!!![/quote]

for first one, x = sqrt(0+x)

x=sqrtx
x^2=x
x^2-x=0
x(x-1)=0
x=0,1

for 2nd one

x=sqrt(1+x)
1+x=x^2
x^2-x-1=0
(1+/- sqrt(1--4))/2
(1+/- sqrt (5)/2

find solutions to that

It's not $ \frac {1\pm\sqrt {3}}{2}$. It's $ \frac {1\pm\sqrt {5}}{2}$--$ \phi$!!! You forgot your --4. I think you misread it as -4.

[quote="wiseidiot"]another golden ratio question: what number is exactly 1 more than its reciprocal?[/quote]

*cough* My lovely favorite number, Phi? :P *luffs Phi*

Read the book "The Golden Ratio"? One of the most interesting non-fiction books I've ever read.

Lol, what number is exactly one less than it's square?

mathking123, try to use \sqrt instead of sqrt. And use \frac instead of /.

The obvious answer to the first problem is $ 0$. The non-obvious answer is $ 1$. Intuitively, that doesn't make sense. (The square root of $ 0$ is $ 0$, right? Then starting from inside out, $ \sqrt{0} \equal{} 0$, $ 0 \plus{} \sqrt{0} \equal{} 0$, $ 0 \plus{} \sqrt{0} \equal{} 0$, etc...) Is $ 1$ a valid answer?

Think of it in terms of the limit. It's kind of like 0+0+0+...+0...(infinity 0's)=1 (even though it is undefined.)

Actually, $ \sum_{i \equal{} 0}^\infty{0} \equal{} 0$ (take the limit of partial sums).

Unless you're directly talking about $ \infty \cdot 0$, which is indeed undefined.

Meh, if you think about it, it makes sense that $ \sqrt{0\plus{}\sqrt{0\plus{}...}}\equal{}1$. Kind of.

math154, this is not some kind of live chat, please don't post spam like "I'm tired" or "I'm bored."

No, actually, it's an undefined chat, as the "I'm tired" post was clearly removed (or turned invisible).

But we're getting off topic... the real question is whether $ \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} \sqrt{0 \plus{} ...}}}}}}}}}}}}}}}}}}}$ is really $ 1$.

It probably does. I was just wondering what other people would think.

[quote="alanchou"]the thing that seems weird to me is that if you try with any other nonnegative real number, you only get one solution, I think.  There's probably a counterexample though.  It makes sense that the continued root is 0, but it kinda also makes sense that it's 1...

it probably takes more knowledge of math than used in mathcounts to determine this[/quote]

are you sure?
for example,

sqrt6+sqrt6.....

(x-3)(x+2) -2,3

sqrt12+sqrt12.....

(x-4)(x+3) -3,4. etc.

yeah. Square root always implies positive.

$ \sqrt{0\plus{}\sqrt{0...}} \equal{} \sqrt{\sqrt{\sqrt...}}.$If we let this $ \equal{} x$, then $ \sqrt {\sqrt {\sqrt {...{\sqrt {x}}}}}$ = the infinity square root of $ x$. So $ x \equal{} x^{1/\infty}$. I really don't know whether i'm using infinity right, but... if you take both sides to the infinity power  :P it becomes: $ x \equal{} x^{\infty}$. I think $ 1 \equal{} 1^{\infty}$, but does $ 0^{\infty} \equal{} 0$? I don't know :P

infinity isn't a number, it's a concept.  And besides, that's not really a proper operation.  $ 1^{\infty}$ and $ 0^{\infty}$ are also indeterminate forms, as I believe they're called

EDIT: So is $ 0\cdot (\pm \infty)$ apparently

Actually, $ 0^\infty$ is [b]not[/b] an indeterminate form, only $ 1^\infty$ is. My calculus textbook says that $ 0^\infty$ evaluates to $ 0$.

hehe, oops, mixed it up with something else

but $ 0\cdot\infty$ still is one

Err... $ 0 \cdot \infty$ is another indeterminate form...

You should probably stop worrying about this. Whenever you have infinite towers of powers, infinite radicals, whatever, this is a valid method when you can be sure the value actually converges. If it doesn't converge, some strange stuff happens. The thing with MathCounts is, whenever you are given this type of problem, it's okay to assume it converges. But if you want to be a real baller (as in, be rigorous), you have to deal with convergence.

On another note, as far as I know, the symbol "$ \infty$" isn't well-defined, so stuff like $ 0\cdot\infty$ doesn't make much sense.

[quote="Yongyi781"]No, actually, it's an undefined chat, as the "I'm tired" post was clearly removed (or turned invisible).

But we're getting off topic... the real question is whether ${ \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + \sqrt {0 + ...}}}}}}}}}}}}}}}}}}}}$ is really $ 1$.[/quote]
Why aren't you guys solving this using the normal way to solve these?
$ x = \sqrt{0+\sqrt{0+\sqrt{0+...}}}$
$ x^{2} = 0 + \sqrt{0+\sqrt{0+\sqrt{0+...}}}$
$ x^{2} = 0 + x$
$ x^{2} = x$
$ x^{2} - x = 0$
$ x(x - 1) = 0$
$ x = 0 \mbox{ or } x = 1$
Oh I guess you're debating whether or not that's true.

Calling the number inside the radicals y, and x being the solution (I probably messed up somewhere)

x^2=x+y
x^2-x-y=0
(x-.5)^2=y+.25
x=+/- sqrt(y+.25)+.5
For there to be 2 possible vaues of x, sqrt(y+.25)<=.5
-.25<y<1/4

Sorry about messiness, I'm too lazy to learn Latex. Maybe someone could clean this up?

So while it doesn't solve the original question, for there to be more than one positive value for x, the number within the radicals must be in between -1/4 and 1/4

edit: While having a negative value of y may cause one to believe that you take the square root of a negative number, there is no term sqrt(y), only infinite terms which would cause a positive number.

Hey dude. Try to use more specific terms...that was kind of hard to read...

[quote="Yongyi781"]
Actually, 0^\infty is not an indeterminate form, only 1^\infty is. My calculus textbook says that 0^\infty evaluates to 0.[/quote]

Actually $ 1^{\infty}$ is 1.  However, when you take the limit of x as x approaches one of $ x^{\infty}$ you have something indeterminate.

Wikipedia says indeterminate forms include 0^0, 0/0, 1^∞, ∞ - ∞, ∞/∞, 0×∞, and ∞^0.  However, immediately before it, it mentions how indeterminate forms are involved with [b]limits[/b]

We shouldn't be talking about this anymore...

I think you misread the thing a bit :wink:

I think that we should not talk about this <edit>i.e. limits and indeterminate forms</edit> anymore...at least not in here<edit>I severely doubt, from experience, that mathcounts will test you on what 0^0 is</edit>.  Perhaps in High School stuffs?


[color=blue][b]Problem.[/b] Let $ x,y,z$ be three positive numbers such that $ x\plus{}y\plus{}z\equal{}1$. Find the maximum value of $ A\equal{}xyz(x^3y\plus{}y^3z\plus{}z^3x)$.[/color]

(Notice that the equality doesn't hold when $ x\equal{}y\equal{}z$).

I wait...

[hide="guess"]
$ x\equal{}\frac{1}{2}$
$ y\equal{}\frac{\sqrt{5}\minus{}1}{4}$
$ y\equal{}\frac{3\minus{}\sqrt{5}}{4}$
Lagrange multiplier should work
[/hide]

the maximum value of A is $ \frac{5\sqrt{5}\minus{}11}{128}$, apprioximately ,0.00140890535

Thank you modeler and yongyao. I will try an elementary solution to it.


prove that :in for real x $ \sum_{k\equal{}1,n}\alpha_{k}cos(i_{k}x)$=0 then $ \alpha_1\equal{}.. \equal{}\alpha_n\equal{}0$ ,where $ i_1...,i_n$ are n fixed real numbers

You have to add the hypothesis $ i_k^2 \neq i_j^2$ for all $ j \neq k$ otherwise it is clearly untrue.

Differenciate wo times the given equation. Thus, for all real $ x$, we have $ \sum_k \alpha_k i_k^2cos(i_kx) \equal{} 0$.
Then, do it again and again until you get $ n$ equations.
Then, write these equations for $ x\equal{}0$.
It leads to a linear homogeneous system where the unkowns are the $ \alpha_k$'s, and the determinant of the system is the Vandermonde of the $ i_k^2$'s. Since they are pairwise distincts, this determinant is nonzero, and the system has a unique solution, hence result.

Pierre.

i didn't learn matrice yet ,but i think it's not the vandermonde determinant,it's his transposite , i think $ M$ and $ Tr(M)$ have the same determinant?


calculus integration quadratics ARML AMC

For how many positive integral values of n is $n^3 -8n^2+20n -13$ a prime number?

       any hint would be nice thanks :lol:

EDIT: Texed by mod.

Hint: Factor it, examine what the factors can be

answer:

[hide]

[/hide]

hmmm can you show me the way to do it please :lol:

factor in terms of $n$ and it can be solved easily

Answer:[hide]First of all, we notice that $n^3-8n^2+20n-13=(n-1)(n^2-7n+13)$ This expansion is only prime if one of the terms is $=1$ and the other is prime. When $n=2$, the $n-1$ term is $=1$ and the expansion is $1*3$ which is prime. This is the only prime where $n-1=1$. Because $n^2-7n+13$ is a quadratic, there is a maximum of two possible points where $n^2-7n+13=1$ Solving, this occurs when $n=4$ or $n=3$. Plugging back in, we find that when $n=3$, $n^3-8n^2+20n-13=2$. When $n=4$, $n^3-8n^2+20n-13=3$. Therefore, the only positive primes for which this expression is prime are $n=2,3,4$. The answer is $3$.[/hide]

But be careful: The answer is not $n=2,3,4$, the answer is 3, as you're asked how many values there are.  ARML problems are notorious for doing things like this.

Don't forget, in general, each term except the prime must be 1 or -1, since you could end up with something like $-1\cdot-13$ or something. In this problem, it doesn't matter since $n-1$ is nonnegative.


function
algebra
functional equation
algebra unsolved

find all function f:R\to R that for every x,y\in Z this is true:
i. f(x)<=x
ii. f(x+y)<=f(x)+f(y)

This is true of every pathological solution satisfying the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url] with $ f(q) \equal{} q \forall q \in \mathbb{Q}$, so I think the condition should be strengthened a little.

I'm going to guess that the condition applies for all $ x,y \in \mathbb{R}$. Otherwise it doesn't even need to satisfy the Cauchy equation; nonintegers can be anything.

Condition 1 has $ f(0) \leq 0$, while condition 2 has $ 0 \leq f(0)$ (when $ x \equal{} 0$ is put in). Observe that $ 0 \equal{} f(0) \leq f(x) \plus{} f(\minus{}x) \leq x \plus{} \minus{}x \equal{} 0$. So $ f(x) \equal{} \minus{}f(\minus{}x)$. But now $ f(x) \leq x$ and $ f(\minus{}x) \leq \minus{}x \Rightarrow \minus{}f(x) \leq \minus{}x \Rightarrow f(x) \geq x$. So $ f(x) \equal{} x$ for all $ x$.


modular arithmetic
Additive combinatorics
counting
combinatorics
Additive Number Theory
IMO Shortlist

Let $A$ be a set of $N$ residues $\pmod{N^{2}}$. Prove that there exists a set $B$ of of $N$ residues $\pmod{N^{2}}$ such that $A + B = \{a+b|a \in A, b \in B\}$ contains at least half of all the residues $\pmod{N^{2}}$.

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

In fact, we can even get $N + (N-1) + \cdots + 1$ residues. We do this by adding elements to $B$ one-by-one. We want to make sure that when we add the $i$th element to $B$, this expands $A+B$ by at least $n+1-i$ elements. But if, for the $i$th choice, we simply add a random element to $B$, the expected value of the increase in $|A+B|$ is at least $n+1-i$, so this is always possible.

Let $S=\{B\}$. Summing up the total number of distinct values taken by $A+B$ over all possible $B$ by complementary counting,
\begin{align*}
\sum_{B\in S}|A+B| = \sum_{B\in S}(N^2-|\overline{A+B}|)
&=\binom{N^2}{N}N^2-\sum_{i=1}^{N^2}\sum_{B\in S}(i\not\in A+B)\\
&=\binom{N^2}{N}N^2-\sum_{i=1}^{N^2}\binom{N^2-N}{N}\\
&=\binom{N^2}{N}N^2-N^2\binom{N^2-N}{N}.
\end{align*}Thus
\[E(|A+B|)=N^2-N^2\frac{\binom{N^2-N}{N}}{\binom{N^2}{N}}>\frac{N^2}{2}\]since
\begin{align*}
\frac{\binom{N^2}{N}}{\binom{N^2-N}{N}}
&=\left(\frac{N^2}{N^2-N}\right)\cdots\left(\frac{N^2-N+1}{N^2-2N+1}\right)\\
&=\left(1+\frac{N}{N^2-N}\right)\cdots\left(1+\frac{N}{N^2-2N+1}\right)\\
&>1+N\left(\frac{1}{N^2-N}+\cdots+\frac{1}{N^2-2N+1}\right)\\
&>1+N\frac{N}{N^2-N}>2.
\end{align*}

Choose a random set $B$ of $n$ residues modulo $n^2$.  Let $x$ be the number of residues that can be written as $a+b$.  The number of ways you can choose a $b$ such that $i$ can be written as $a+b$ where $a\in A$ is $n$ since $\#(A)=n$.  Therefore the probability that $i$ cannot be written in the form $a+b$ is $\left(1-\frac{n}{n^2}\right)^n$ and so the probability that $i$ can be written in the form $a+b$ is
\[1-\left(1-\frac{n}{n^2}\right)^n.\]
Thus $\mathbb{E}(X)=n^2\left(1-\left(1-\frac{n}{n^2}\right)^n\right)$ and so what is left to prove is that
\[1-\left(1-\frac{n}{n^2}\right)^n\geq \frac12.\]
This is true because $1-\left(1-\frac{n}{n^2}\right)^n=1-\left(1-\frac{1}{n}\right)^n\leq 1-\frac{1}{e}\le \frac12$ and we are done.

We can even show that the expected value $\mathbb{E}(X)=n^2\left(1-\left(1-\frac{n}{n^2}\right)^n\right)$ is greater than or equal to $\dfrac{n^2+n}{2}$

$\Longleftrightarrow 1-\left(1-\frac{n}{n^2}\right)^n \ge \dfrac{n+1}{2n}$

$\Longleftrightarrow \dfrac{n-1}{2n} \ge \left(\frac{n-1}{n}\right)^n$

$\Longleftrightarrow \dfrac{1}{2} \ge \left(\frac{n-1}{n}\right)^{n-1}$

$\Longleftrightarrow 2 \le \left(1+\frac{1}{n-1}\right)^{n-1}$

Which is true by Bernoulli's Inequality since $\left(1+\frac{1}{n-1}\right)^{n-1} \ge 1+\frac{1}{n-1}\cdot(n-1) = 2$

The last part could be done by other ways, too.

For a fixed $x\in\mathbb{Z}/n^2\mathbb{Z}=:Z$, and a random $b\in Z$, the probability that $x\in b+A$ is clearly $n/n^2=1/n$. Thus, if we choose each $n$ elements of $B$ randomly one at a time (this is different than choosing $B$ randomly), the size of the $B$ we get is at most $n$, and the probability that $x\in B+A$ is $1-(1-1/n)^n\ge 1/2$. Thus, if the probability that any element $x$ is in $B+A$ is at least $1/2$, then the expected size of $B+A$ is at least $n^2/2$, so we're done.

Consider a randomly chosen set $B$ with $|B| = n$ and $B \subseteq Z$. We will show that
\[\sum_{k \in A + B} 1 \ge \frac{1}{2}n^2.\]
We consider the sets $A' = \{a_1 - a_2 \mid a_1, a_2 \in A\}$ and $B' = \{b_1 - b_2 \mid b_1, b_2 \in B\}$. Clearly, these sets each contain at most $n^2$ elements, meaning that $|A' \cap B'| \le n^2$. This means that we have at most $n^2$ quadruples $(a_1, a_2, b_1, b_2)$ such that $a_1 - a_2 = b_1 - b_2$, or equivalently, $a_1 + b_2 = a_2 + b_1$. We call quadruples satisfying this property [i]reinforcing[/i]. Since there are $n^4$ possible pairs $(a_1, a_2, b_1, b_2)$ with $a_1, a_2 \in A$ and $b_1, b_2 \in B$, the probability that any given one of these quadruples is reinforcing is less than or equal to $\frac{1}{n^2}$. Since the pair $(a_1, a_2)$ is included in $n^2$ of these quadruples, the expected number of quadruples that $(a_1,a_2)$ is included in that are reinforcing is less than or equal to $\frac{n^2}{n^2} = 1$. Thus, for any $c \in A + B$, the expected value of the number of times it can be expressed as $a+b$ for $a \in A$ and $b \in B$ is less than or equal to $2$. Now, we observe that
\[2\mathbb{E}\left[\sum_{k \in A + B} 1\right] = \mathbb{E}\left[\sum_{k \in A + B} 2\right] \ge \sum_{a \in A, b \in B} 1 = n^2,\]
and we are done by the probabilistic method.

Let $A=(a_1, .., a_n)$. Randomly choose $n$ distinct residues for $B$. If $k$ is not in $A+B$ that means $k-a_1, k-a_2, ..., k-a_n$ (all taken mod $n^2$) were all not chosen for $B$, which has a probability of $\frac{{n^2-n\choose n}}{{n^2\choose n}}$ of occurring. Thus the probability that an individual residue is in $A+B$ is $1-\frac{{n^2-n\choose n}}{{n^2\choose n}}>\frac{1}{2}$ (since $\frac{{n^2\choose n}}{{n^2-n\choose n}}>(1+\frac{1}{n})^n>2)$, so by linearity of expectation we're done.

Let $A=\{a_1,a_2,\ldots,a_n\}$. We will randomly pick a set of residues $B=\{b_1,b_2,\ldots,b_n\}$, and we will show that $E[|A+B|] \geq \frac{n^2}{2}$. Take a residue $c$ mod $n^2$. The chance that $c-a_i$ is not equal to one of the $b_i$'s is $1-\frac{1}{n}$. This means that the chance $c$ is not in $A+B$ is $(1-\frac{1}{n})^n$, so the chance $c$ is in $A+B$ is $1-(1-\frac{1}{n})^n$. Multiplying by all $n^2$ residues, we get that $E[|A+B|] = n^2-n^2(1-\frac{1}{n})^n$. It remains to show this is at least $\frac{n^2}{2}$. \\

This is equivalent to showing that $(1-\frac{1}{n})^n \leq \frac{1}{2}$ for all positive integers $n$. This is obvious for $n=1$, so assume $n>1$. Taking reciprocals gives $(1+\frac{1}{n-1})^n \geq 2$. In fact, we can show that $(1+\frac{1}{n-1})^{n-1} \geq 2$, since $1+\frac{1}{n-1}$ is increasing for $n \geq 2$ and $(1+\frac{1}{2-1})^{2-1}=2$. This proves the desired inequality.

Fix $A$ and consider a random set for $B$. For some value $x$, there are $n$ elements that could be in $B$ in order for $x$ to be in $A+B$. The probability that none of these values is in $B$, which is also the probability $x$ is not in $A+B$, is
\[ \left( 1 - \frac{n}{n^2} \right)^n = \left( 1 - \frac{1}{n} \right)^n < \frac{1}{e} < \frac12 .\]
Therefore, $x$ is in $A+B$ with more than $\frac12$ probability, so there must be a $B$ such that $A+B$ has $\frac{n^2}{2}$ values.

Fix a residue $i$. Note that for every single element in $A$, there is exactly one corresponding residue such that $a + b \equiv i$. Hence, there are actually $n$ elements that we can put in $B$ to get equivalence. We will use this fact to calculate the sum. \newline
\newline
From this, the probability that a particular residue $\pmod n^2$ isn't covered by some $a + b$ is thus $\left(1 - \frac{1}{n} \right)^n$. Hence the probability it is covered is $1 - \left(1 - \frac{1}{n} \right)^n$.

Then the expected number of elements that are covered is $$n^2 \left(1 - \left(1 - \frac{1}{n} \right)^n \right), $$ which we can easily check satisfies the desired amount. In fact, we get the better bound of $\frac{e - 1}{e}$.

 This bound is ridiculously weak.
Fix $A$ and consider a random set $B$. We will compute the expected value of the number of residues which are not in $A+B$. By Linearity of Expectation, this is equal to $N^2$ times the probability that an arbitrary residue mod $N^2$ is not in $A+B$—WLOG let it be $1$. Observe that there are $\binom{N^2}{N}$ choices of $B$ unconditionally. There are also $N$ residues $r$ such that there exists an element $a \in A$ with $a+r \equiv 1 \pmod{N^2}$, so there are $\binom{N^2-N}{N}$ choices of $B$ such that $r \not \in A+B$. We now compute:
\begin{align*}
\frac{\binom{N^2-N}{N}}{\binom{N^2}{N}}&=\frac{N^2-N}{N^2}\cdot \frac{N^2-N-1}{N^2-1}\cdots \frac{N^2-2N}{N^2-N}\\
&\leq \left(\frac{N^2-N}{N^2}\right)^{N+1}\\
&=\left(1-\frac{1}{N}\right)^{N+1}\leq \frac{1}{e}.
\end{align*}
Therefore, for a random set $B$ with $A$ fixed, the expected number of residues mod $N^2$ which are not in $A+B$ is at most $\tfrac{N^2}{e}$. This implies that there exists some set $B$ of $N$ residues such that $A+B$ contains at least $N^2\tfrac{e-1}{e}>N^2/2$ residues mod $N^2$. $\blacksquare$

Take some random set $B$. Then, the probability that any residue is in $A+B$ is
\[1-\left(1-\frac{1}{N}\right)^N > 1-\frac{1}{e} > \frac{1}{2}\]
thus, by linearity of expectation the number of valid residues is $>\frac{1}{2}$, so by pigeonhole there exists some set $B$ which covers at least half the residues.

Nothing new here, but since I liked the problem, I will post my solution.
Choose $B$ at random. Then a number $t$ is not in $A+B$ if all numbers $t-a_i$ are not in $B$, so the probability this happens in $\frac {\binom {n^2-n}{n}}{\binom{n^2}{n}}<\frac{1}{2}$ (calculations are omitted) . Hence, the expected number of the missing residues is less than $\frac{n^2}{2}$, so we are done.

For any value $x$, $x\notin A+B$ if and only if $x-B$ is disjoint from $A$. The probability that they are disjoint for a random choice of $B$ is \[\tfrac{\tbinom{n^2-n}{n}}{\tbinom{n^2}{n}}=\frac{(n^2-n)(n^2-n-1)\dots (n^2-2n+1)}{n^2(n^2-1)\dots (n^2-n+1)}.\] It suffices to show that is less than one-half. Indeed, \[\frac{(n^2-n)(n^2-n-1)\dots (n^2-2n+1)}{n^2(n^2-1)\dots (n^2-n+1)}<\left(\frac{n-1}{n}\right)^n<\frac{1}{e}<0.5\]

Fix $A$ and randomly choose set $B$ of $N$ residues. Set $A+B$ doesn't include number $x$ iff $A$ doesn't include any of $N$ numbers $x-b$ for $b\in B,$ which has probability $$\frac{\binom{N^2-N}{N}}{\binom{N^2}{N}}=\prod_{i=0}^{N-1} \frac{N^2-N-i}{N^2-i}\leq \left( \frac{N-1}{N}\right) ^N<\frac 1e<\frac 12.$$ In particular, the expecting size of $A+B$ is at least $\frac{N^2}{2},$ so the conclusion follows.

$n=1$ and $n=2$ are trivial. Assume $n\geq3$ from now on.\\

Main Claim: For all positive integers $n\geq3$, we have $$2{n^2-n\choose n}\leq {n^2\choose n}$$ (I promise this will be relevant later). Multiplying this by $n!(n^2-n)!(n^2-2n)!,$ it becomes $$2((n^2-n)!)^2\leq (n^2)!(n^2-2n)!$$ Cancelling as much as possible, this is $$2(n^2-2n+1)(n^2-2n+2)\cdots (n^2-n)\leq (n^2-n+1)(n^2-n+2)\cdots (n^2).$$ Rewrite this as $$\frac{n^2}{n^2-n}\cdot \frac{n^2-1}{n^2-n-1}\cdots \frac{n^2-n+1}{n^2-2n+1}\geq 2.$$ Note that $x/(x-n)\geq 1+\frac{1}{n}$ for all $x\leq n^2$, so $$\frac{n^2}{n^2-n}\cdot \frac{n^2-1}{n^2-n-1}\cdots \frac{n^2-n+1}{n^2-2n+1}\geq (1+\frac{1}{n})^n\geq 2,$$ which shows the claim.\\

Consider choosing the set $B$ randomly. Now, consider the probability that a residue $r$ with $0\leq r\leq n^2-1$ appears in the set $A+B\pmod{n^2}$. Since there are $n$ elements in $A$, if any of $n$ corresponding elements appear in $B$, then $r$ appears in $A+B\pmod{n^2}$. Hence, the probability that $r$ appears is $$1-\frac{{n^2-n\choose n}}{{n^2\choose n}},$$ so the expected number of distinct residues is $$n^2(1-\frac{{n^2-n\choose n}}{{n^2\choose n}}).$$ However, due to our main claim, $$n^2(1-\frac{{n^2-n\choose n}}{{n^2\choose n}})\geq \frac{1}{2}n^2,$$ so we are done.

[hide=Solution]
We show there exists some set $B$ of $n$ residues such that
\[|{A+B}| \geq n + n + (n-1) + (n-2) + \cdots + 2 = \frac{n^2+3n-2}{2}.\]
First observe that it is safe to ignore numbers in $|{A+B}|$; this will make calculations easier.

Start by adding any number, say $0$, to $B$. Then, suppose we have added $i \geq 1$ numbers to $B$ and
\[
    |{A+B}| = M \geq n + n + (n-1) + \cdots + (n-i+2)
    = in - \tfrac{(i-1)(i-2)}{2}.
\]
Ignore some numbers in $|{A+B}|$ so that we have equality ($M = \cdots$) above.

Summing over all possible new $b$ we could add to $B$, on average we have
\begin{align*}
    \phantom={}\frac{(n^2-M)n}{n^2-i}
    ={}&\frac{n^3 - in^2 + \tfrac{(i-1)(i-2)}{2} n}{n^2-i} > n-i \\
    \iff& n^3 - in^2 + \tfrac{(i-1)(i-2)}{2} \cdot n > n^3-in^2-in+i^2 \\
    \iff& \tfrac{(i-1)(i-2)}{2} \cdot n \geq 0 > i(i-n)
\end{align*}
Hence there must exist some new $b$ that hits at least $n-i+1$ new residues; inducting this gives the bound above. \qed
[/hide]

Fix $A$ and choose $B$ randomly with uniform probability. The probability that a residue $x$ is not in $A+B$ is \[ \frac{\binom {n^2-n}{n}}{\binom{n^2}{n}} < \left(1-\frac{1}{n}\right)^n. \] Furthermore, we bound the probability that $x$ is in $A+B$ as below: \[ 1- \left(1-\frac{1}{n}\right)^n > 1-\frac{1}{e} > \frac{1}{2}. \] Thus, by Linearity of Expectation, the expected number of residues contained in $A+B$ (mod $N^2$) is greater than $\frac{N^2}{2}$, so there exists $B$ such that number of residues contained in $A+B$ (mod $N^2$) is greater than $\frac{N^2}{2}$, and we are done.

For any fixed $A$, the probability some residue $r$ is not in $A + B$ is \[ p = \frac{\dbinom{n^2 - n}{n}}{\dbinom{n^2}{n} } = \frac{(n^2  - n)(n^2 - n - 1) \cdots (n^2 - 2n + 1)}{n^2(n^2 - 1)\cdots (n^2 - n + 1)} \] Now, notice the fractions \[ \frac{n^2 - n}{n^2}, \frac{n^2 - n - 1}{n^2  - 1}, \ldots, \frac{n^2 - 2n + 1}{n^2 - n + 1} \] are all less than or equal to $\frac{n-1}{n}$, so \[ p \le \left(\frac{n-1}{n} \right)^n = \left( 1 - \frac{1}{n} \right)^n < \frac{1}{e} < \frac{1}{2}\]

By LOE, the expected number of residues in $A+B$ is equal to $n^2(1 - p)$, which is greater than $\frac{n^2}{2}$, so there must exist some set of $B$ for which $A + B$ contains at least $\frac{n^2}{2}$ residues.

One of the peaks of a GLOBAL problem, though incredibly easy :) the problem literally screams expected value

    Select B randomly with replacement; we have $$\forall c\in Z,c\notin A+B\Leftrightarrow\forall b\in B,c-b\not\in A;\quad P(c-b \not\in A)= 1 - \frac1n\Rightarrow P(\text{omit element})=\left( 1 - \frac1n \right)^n < \frac 1e < \frac12,$$ meaning there's some B where $<\frac{n^2}2$ elements are omitted, as desired. $\blacksquare$

also i added those hints on arch so dont think that i literally copied the exact same thing as arch, i infact made the hint from this

Suppose not. Let's choose one element of $B$ at a time. On each move, since there are at least $\frac12 N^2$ residues left, by LoE there is some element $b$ that can get $A+b$ to have at least $\frac12 N$ new elements. Obviously this $b$ won't be already in $B$, so we get $\frac12 N\cdot N=\frac12 N^2$, contradiction.

Let us fix $A$, and pick $B$ randomly. We can prove that the expected value of $|A + B|$ is $\geq \frac{1}{2} n^2$.
Note that for some element in $|A + B| x$, and some element $a$ in $A$, there is only one element $b$ in $B$ that satisfies $a + b \equiv x$.
$\newline$

Then the probability that some element in $Z$ is not covered by $|A + B|$ is $(1 - \frac{1}{n})^n$, because for every $n$ elements in $A$, there is a $1 - \frac{n}{n^2}$ chance that there isn't an element in $B$ that adds up to the element in $Z$.
$\newline$

Then $(1 - \frac{1}{n})^n \ge \frac{1}{e}$ (apparently), so taking the complement, we get $1 - \frac{1}{e} \ge \frac{1}{2}$, so we are done.

Clearly the probability that a residue is  covered is $1-\left(1-\frac{1}{n} \right)^{n}$, so the $\mathbb{E[X]} = n^2\left(1- \left(1-\frac{1}{n}\right)^n \right) > n^2\left(\frac{e-1}{e}\right)>\frac{n^2}{2}$.

It suffices to prove that $E[|A + B|] \ge \frac{1}{2} n^2.$ Fixing $A$, an element $i \in Z$ is in $A + B$ if $\exists a \in A$ s.t $i - a \in B.$ Hence the probability is,
\begin{align*}
	1 - \dfrac{\binom{n^2 - n}{n}}{\binom{n^2}{n}} &= 1 - \prod_{i = 0}^{n-1} \frac{n^2 - n - i}{n^2 - i} \\
	&\ge 1 - (\frac{n^2 - n}{n^2})^n \\
	&= 1 - (1 - \frac{1}{n})^n.
\end{align*}
Observe that $(1 - \frac{1}{n})^n$ is a monotonic increasing function and let $y = \lim_{x \to \infty } (1 - \frac{1}{x})^x.$ Then,
\begin{align*}
	\ln(y) &= \lim_{x \to \infty} \ln (1 - \frac{1}{x})^x \\
	&= \lim_{x \to \infty} \frac{\ln(1 - \frac{1}{x})}{1/x} \\
	&= \lim_{x \to \infty} \frac{\dfrac{1/x^2}{1 - \frac{1}{x}}}{-1/x^2} \\
	&= \lim_{x \to \infty} -\frac{1}{1 - \frac{1}{x}} = -1 \\
	y &= \frac{1}{e}.
\end{align*}
Hence,
\begin{align*}
	E[|A + B|] &= n^2 (1 - (1 - \frac{1}{n})^n) \\
	&\ge n^2(1 - \frac{1}{e}) \ge \frac{n^2}{2}.
\end{align*}
So we are done by PhP.

We want to prove that $\Pr(x\not\in (A+B))<\frac12$. We see that $\Pr(x\not\in (A+B))=\frac{\binom{n^2-n}{n}}{\binom{n^2}{n}}=\prod_{k=0}^{n-1}\frac{n^2-n-k}{n^2-k}<(1-\frac{1}{n})^n<\frac1e<\frac12$, as desired.


modular arithmetic
number theory unsolved
number theory

Hi :D .Show that for $ P(x)$ $ \epsilon$ $ Z\left[ x\right]$ we can find $ k$ consecutive numbers like $ x_1 , x_2 , ... , x_k$ such that all of $ P(x_1)$ , $ P(x_2)$ , .... , $ P(x_k)$ are composite .
(Sorry for my poor English :blush: )

thank you !  :D . i hope that i can solve it .....

by using t0rajir0u's hint,
just see that if $ S\equal{}\{p_1,p_2,....\}$ is the set of primes dividing $ P(x)$( we can easily prove the infinity of $ S$)  then there exist $ x_1,...x_k$ such that $ p_i|P(x_i),p_i \neq |P(x_i)|$ then choose $ X$ such that $ X \equiv x_1 \pmod {p_1},X \equiv x_2\minus{}1 \pmod {p_2},...,X \equiv x_k\minus{}k\plus{}1 \pmod {p_k}$.(note that $ X$ can be chosen in such a way that $ |P(X\plus{}i\minus{}1)| \neq p_i$)  and we are done!


The premise of this discussion is so incredibly ridiculous I don't even know what to say.  http://realclimate.org/ is a website run by a group of climate scientists.  http://www.ipcc.ch/ is the home of the Intergovernmental Panel on Climate Change.  Go.  Read.  Learn.

How is the premise of this discussion wrong?  :maybe:

Yeah, I know the UN and climate scientists and everyone is all "WOOT Global Warming". And I believe them. But some people from the other side(s) are saying things that make sense and this confuses me.

what is the other side saying that makes sense? :maybe:

The basic stock argument is :
[quote]
"The essence of the issue is this. Climate changes naturally all the time, partly in predictable cycles, and partly in unpredictable shorter rhythms and rapid episodic shifts, some of the causes of which remain unknown. We are fortunate that our modern societies have developed during the last 10,000 years of benignly warm, interglacial climate. But for more than 90 per cent of the last two million years, the climate has been colder, and generally much colder, than today. The reality of the climate record is that a sudden natural cooling is far more to be feared, and will do infinitely more social and economic damage, than the late 20th century phase of gentle warming."[/quote]

This recent climate change makes everything that has ever happened even "unexpected" changes PALE by comparison, it's like an exponential function. I'm pretty sure this is beyond the realm of what is natural.

Yeah whatever. Stupid crazy people feeding me bad ideas.

Go ahead and delete this thread.

In fact, Ignite isn't quite correct about that.  The Earth has (somewhat unsurprisingly) undergone far greater shifts in climate over the past 4.5 billion years than it has over the past 100 years.  Of course, most such massive changes happen very slowly relative to human timescales, and none of them have happened within the past 10,000 years (i.e., within the lifespan of the human species).

I have to run to dinner now, but I'll be back to explain what I consider to be the most serious flaws in the paragraph presented.

[url=http://www.ucsusa.org/assets/documents/global_warming/exxon_report.pdf]Here[/url] is an article that gives an overview of where most of the anti-global warming ideas are coming from.

Warning: It is a large pdf file, so if you are using dial-up :o , then it might take a while to load.

[quote="PenguinIntegral"]So I really don't know about this global warming things. It's like:

Hippies : We're all gonna die and we have scientists supporting us.

Libertarians : We're not all gonna die. We have scientists supporting us.

I want to know the truth. And I am too lazy to do my own research.

So please cite thingies and convince me of your view: are humans making the planet hot (and not just a little, like uberly so)?

And because I know this board is full of hippies, can someone take the side of the libertarians just for fun and give them a run for their money? Good debate is always fun to watch.[/quote]

Nobody said things about dying/not dying, as far as I nkow (not a significant number of people).  But seriously, if you know a single scientist who disagrees with the existence of global warming, please tell me.

So, as promised, I'm going to spend some time taking apart the paragraph P.I. provided us at length.

The most egregious absurdity in that paragraph, in my view, is the following: the argument it offers, at the broadest level, is that climate changes all the time, our understanding of it is imperfect, and so we shouldn't be worried about climate changing.  (Incidentally, this closely mirrors a lot of crank science writing, and in particular a wide array of creationist literature in opposition to evolution.)  This argument is grossly, horribly misleading, in a variety of different (and quite subtle) ways:
The one that first jumps out at me: the climate scientists who have given us our understanding of global warming [i]already know[/i] about natural changes in climate.  They understand them better than anyone else on earth.  And, given that knowledge, they have come to consensus opinions like those embodied in the IPCC reports on the effects of human carbon dioxide outgassing on climate.  The notion that, somehow, climatologists overlooked the fact that climate changes all the time is just absurd.  The thing that is so terrifying about global warming is that we're already seeing effects that are very difficult to attribute to natural cycles, and when we turn our understanding (often in the form of climate models) of past climate to predict the future, they tell us that a lot of really massively bad things are possible.

The second major dishonesty that jumps out at me is this one: "the late 20th century phase of gentle warming."  Adding carbon dioxide to the atmosphere results in a global warming, but that happens with a certain amount of lag time.  Thus, if humans simply stopped producing carbon dioxide right now it would take some time (okay, I admit, I don't know the exact timescale) for the atmosphere and global average climate to equilibrate, and the result will be an earth slightly warmer on average than it is at present.  The "gentle warming" of the late 20th century has given us, in the past decade, 8 (I think ... that's by memory) of the warmest years in the past 100.  Thus, stopping all carbon production now would leave us with a world (subject to other future climate motions) already warmer than what industrialized humanity has experienced so far.  Of course, this involves cutting off carbon production entirely -- we're talking about messing directly with $f$ here, not tinkering around the edges of $f''$.  This involves no discussion of what happens if Americans continue chugging along with our ever-sprawling cities and ever-lengthening commute times, if China continues building a massive collection of coal-fired power plants.  The implication that "doing nothing" will result in some sort of "gentle warming" that we're equipped to handle, no problem, is badly, badly wrong.

Finally, I'm going to pick out two last misleading aspects of one part of the paragraph: the comparison to glacial ages over the past 2 million years.  The first one is one that every numerically-capable and politically-aware individual should constantly be on the lookout for: quantitative claims with no obvious support behind them.  In this case, we have a comparison between two numbers: the amount of economic and social damage caused by an ice age and the amount of economic and social damage caused by global warming.*  Where are the calculations of that?  Where's the evidence to support that view?  Right now, it's a quantitative assertation with no definitions and no calculations and no evidence.  The very worst kind of argument, all wrapped up to sound reasonable.
So, the second nasty little bit of misleadingness wrapped in that paragraph: Why make this comparison?  After all, ice ages just don't happen quickly, and there's no particular evidence that another one is iminent.  The only real property of an ice age that seems relevant for the author's argument is that they are unavoidable, i.e. humans don't have the knowledge or capacity to prevent an ice age on earth.  The author associates GW with ice ages in order to fix in our minds some notion about the inevitability of climate change and how we cannot (and therefore should not) do anything to prevent it.  But, of course, humans have both the knowledge and capacity to prevent global warming -- the only thing we seem to be lacking at present is the political will.**
(As an aside, I suspect that the factoid about how often it's been colder than now is just wrong, but I don't actually have evidence for that belief at present.  In geologic timescales, i.e. millions or billions of years, it's quite cold on earth right now: after all, we have ice at both poles, which is fairly atypical in earth's history.)

* In fact, there is a third dishonesty wrapped up here: the sentence actually compares the damage of an ice age to the damage of the warming [i]that's already happened[/i], as if this somehow is supposed to reassure you about damage that hasn't happened yet.  I don't think I have to say any more about this.

** I might as well add, for those few of you who will indulge my ego by reading this far, that I am a global warming pessimist: I don't think we're going to be able to deal with it effectively.  I think the American southwest is going to be more-or-less abandoned in my lifetime, and I don't even want to think about the effects on places like Africa (especially near but not in the Sahara), Bangladesh and the various Pacific island nations.  On the other hand, I turn off my computer or computer screen when I'm not using them, use compact fluorescent bulbs, live in a city where I don't need a car to get around and turn the heat down in the winter and the AC down in the summer.  (Consuming locally-produced or organic food is good, too, but difficult in a dining hall.)  So, I guess you could say that I'm hedging my bets.

What's the question? Whether or not "global warming" is actually caused by humans?

[url=http://www.crichton-official.com/speeches/speeches_quote04.html]Aliens Cause Global Warming.[/url]

Personally, I'm skeptical about the whole "humans cause global warming" thing.

[quote="mathnerd314"]Personally, I'm skeptical about the whole "humans cause global warming" thing.[/quote] I commend you on your capacity to share your beliefs without feeling compelled to provide any evidence for them whatsoever.  It helps prevent threads from being cluttered with "facts" or "information."

Actually, scientists have measured the amount of oxygen/CO2 IN layers of ice which date back far beyond 10,000 years. The change is not even close to anything that's ever been seen before, not even close.

Look through Al Gore's powerpoint, if you want to say it doesn't prove humans cause global warming, then the part I can even see you debating is that the information he cites is wrong to begin with and considering not a single scientific article in the past 10 years has disputed global warming, you're gonig to have a hard time.

Was that addressed to me?  The Vostok ice core goes back approximately 650,000 years; Dome C (I hope that's the right name ... it's "Dome something") may go back even further.  (Those are two sites in Antarctica, for those who are interested.  There are also cores from Greenland and high altitude glaciers in warmer parts of the world (ice cores from Peru, for example).)  650,000 is a very, very small percentage of 4.5 billion.  Over 4.5 billion years, the Earth has experienced far greater climate shifts than humans are ever going to cause.  (Try searching, for example, for "snowball earth.")  These changes were entirely natural -- it also happens that nothing like them is going to happen any time soon, so it's not the sort of thing one needs to design a climate policy around.

Also, (and it makes me sad to admit this), that wonderful (I mean this non-sarcastically) graph of Gore's is slightly misleading for the following reason: it's true that CO2 plays an important role in forcing temperature, but there are other things (methane, for example) that play a role forcing both CO2 and temperature, and temperature plays a role forcing CO2.  So the causation is a bit more complicated than he lets on.  (Of course, because CO2 forces temperature, filling our atmostphere with CO2 is still going to (indeed, already has) raise the temperature.)

Addition/clarification/correction: the [i]magnitude[/i] of the change is not unprecedented among natural events.  The [i]rate[/i] of the change very well may be unique in Earth's history.

Fair enough :).  That actually also brings to mind the example of denialists (AGW, creationist, and doubtless others) who complain about the "orthodoxy" of the scientific view (the existence of AGW, or evolution, or whatever) that won't accept alternate explanations.  It's so silly because it entirely misses the fact that the predominant view ("AGW is happening & accelerating") didn't pop, fully formed, out of the head of some priest.  Rather, someone had to have the first idea that this might be happening, and had to publish a paper (supported with some damn good evidence) that it was, and had to convince other scientists that it was, and more research had to have been done to verify the observed effects, and as it seemed to be holding up to inspection even more researchers were drawn in to studying the phenomenon, and so on.  When scientific ideas are born, they don't automatically become the consensus view and get labelled "good science;" rather, good science makes its mark and becomes a consensus view.*  There is a certain parallel here with Gore: his investment of time, effort and money followed the establishment of global warming as credible science, rather than helped cause it.

* There're reasons denialists don't like to bring this up, of course: it destroys their absurd notion of a scientific cabal, but it also suggests that the reason they're laughed out of scientific circles is because they [i]don't have any good science[/i] to support their view.

Ok, I'm not denying nor endorsing the GW theory, I'm not qualified to do so, I'm simply tired of being manipulated by some obscure people in power...In the 70ies they promised a "New Ice-Age", and they got it wrong , in the 80ies a nuke war and worlwide famine, 90ies ebola, AIDS, Y2K, Hanta virus, killer bees,mad-cow..., 00s SARS, avian flu...
When they got it wrong on again these new green-fascists (the catastrophists like Hulot in France) needed sth to beat us back to the stone-age and the GW theory sprung-up, first its only a theory and new serious research suggests that climate patterns aren't that simple to understand.When some brave scientists try to question the GW religion they are labeled as "bp,exxon-sponsored scientists", the consensus you are talking about isn't actually one, many very serious scientists who have resigned from the IPCC board actually refused to have their name added in the latest IPCC doc, but still they were added against their will, if we are so certain about AGW why can't anyone question it without being immediately attacked and subjected to abuse ? This whole GW thing has become the new Trojan horse of the greenist movement.Your claim that does who question (not deny) the GW theory are like creationists is simply untrue, actually those who question it are the new Galileos and Brunos, those who condemn them are the new inquisitors.
GW is more and more becoming a religion and from merely a theory it has grown into a dogma that nobody dares question sometimes at the peril of their own peril.Also claiming that our ridiculous output of CO2(which is not the most potent greenhouse gas)in comparison to natural emissions,essentially from decaying micro-organisms, is simply ludicrous.It's extraordinarily arrogant to assume that we humans have the power to change our climate in one century, this sort of thinking is quite similar to that of the heliocentrists who claim that WE are at the center of the Universe.

PS:Even the boss of NASA recently questionned the GW theory.
Regards,

[quote="Appolonius"]Ok, I'm not denying nor endorsing the GW theory, I'm not qualified to do so[/quote]You should have just stopped here.  It's obvious that you have no relevant qualifications, and you certainly aren't approaching this conversation with a respect for facts, knowledge or truth.
[quote]I'm simply tired of being manipulated by some obscure people in power[/quote]
You mean who, exactly?  The large, multinational corporations that are forcing you to buy fuel efficient vehicles?  Or the US government, whose agents sneak into your home and turn off the lights in empty rooms or the television when you're not watching it?  Give me a break.
[quote]...In the 70ies they promised a "New Ice-Age"[/quote]Really, if you're going to repeat lies, you should at least choose lies that haven't been dealt with [i]on this very thread[/i].  It creates pointless extra work for me.
[quote] AIDS[/quote]Um, what?  There are 40 million people currently infected with HIV and 25 million have already died from it.  I mean, the rest of your post is crazy, but this may take the cake.  In fact, what you've written is such a bizarre collection of conspiracy theories, lies, and mistaken impressions (all provided without any evidence other than your assertion, of course), but your including AIDS on this list of non-disasters should, I think, be enough to impeach your credibility sufficiently that I don't actually have to slog through it all for the benefit of others.

[quote="JBL"]When scientific ideas are born, they don't automatically become the consensus view and get labelled "good science;" rather, good science makes its mark and becomes a consensus view.[/quote]
As an interesting exercise in this, try reading some 18th and 19th century (European and American) geology. Those guys did some fantastic work: mapping the strata in England as the first model; developing the idea that higher was younger and lower was older; beginning to utilize fossils to give relative dates to strata; identifying "erratic" boulders that were in places the didn't belong; finding fossils of dinosaurs and other extinct creatures. And, at least at first, they always wrote it up within the Judeo-Christian biblical narrative. The erratic boulders had been moved around by Noah's flood; the dinosaurs had missed the boat. For a while they stuck with this narrative even as they accumulated more and more evidence that undermined it at every turn. How could there be [i]so many[/i] strata and so many different groupings of fossils of apparently different ages? How could there be seashells on mountaintops? How could Mt. Etna have erupted so many times (as seen in the overlapping lava flows) and built up such a pile in just six thousand years or so? Those guys did not set out to overturn Genesis; in fact, intellectually, many of them fought as hard as they could to preserve that narrative. They didn't really want to consider the unthinkable idea of "deep time," of a very old Earth - but their own work kept piling up evidence that kept pulling them in that direction whether they wanted to go there or not.

[quote="JBL"]6) Get China to use anything other than coal for energy.  Seriously. [/quote]
Have you ever been to China? China is currently too dependent on coal power to switch to any other source; at least not any time soon.

And, in one of my classes at school my group is doing a news report on global warming; and research across the internet we've done has turned up some interesting facts. For example, the current trend of our Earth as a WHOLE is that our climate's temperature is going down, not up. (Although this variation is >1 degree Celsius.) Another is that most research done on the trends of our planet's climate is that almost all research has been done over land and completely ignored water. This means 70% of the Earth's surface is being ignored in research on global warming.

While some parts of the Earth are most certainly warming up, other parts are cooling down. Overall, the variation of the climate as a whole hasn't changed very much in the past 100 or so years. And some research suggests that the Earth was much warmer in the 7th and 8th centuries than it currently is now.

Sorry about a lack of any links; I'm typing this from memory off the printouts that some of my other group members brought in. I might look up sources later; Snopes was one of them.

Hello Angela, welcome to AoPS :)

[quote="angelafan11"]Have you ever been to China? China is currently too dependent on coal power to switch to any other source; at least not any time soon.[/quote]

That's also why parts of China have so much air pollution, including Beijing (where I'm living right now, as you know :P).

[quote]And, in one of my classes at school my group is doing a news report on global warming; and research across the internet we've done has turned up some interesting facts. For example, the current trend of our Earth as a WHOLE is that our climate's temperature is going down, not up. (Although this variation is >1 degree Celsius.) Another is that most research done on the trends of our planet's climate is that almost all research has been done over land and completely ignored water. This means 70% of the Earth's surface is being ignored in research on global warming.

While some parts of the Earth are most certainly warming up, other parts are cooling down. Overall, the variation of the climate as a whole hasn't changed very much in the past 100 or so years. And some research suggests that the Earth was much warmer in the 7th and 8th centuries than it currently is now.

Sorry about a lack of any links; I'm typing this from memory off the printouts that some of my other group members brought in. I might look up sources later; Snopes was one of them.[/quote]

Make sure that you're getting your information from reputable sources. Many of the people "on the Internet" don't know what they're talking about, as people on this thread have already pointed out. I'd suggest you read the IPCC report that JBL linked to.

And which Snopes article are you talking about? I did a search and found nothing terribly interesting.

Since this thread was made before I was born, I think the whole forum (or most of it at least) has come to a consensus that global warming is real.

The conditions 13 years ago are very different from the conditions now.

[quote=justJen]Global warming is really bad for our earth. So is plastic. Seriously, America. Why can't we just stop? No, the way to do it is NOT through useless and inconsequential memes; it's through actual changes.[/quote]

Internet culture is actually a great way to spread information (so long as that information is accurate).

I know that but admit it, sksksksks save the turtles was NOT for the benefit [It may have been once upon a time], it was fake.

A movement to reduce the use of plastic in an (admittedly somewhat misplaced) attempt to improve the environment? It seems good.

How so? They use metal straws instead of plastic ones, metal water bottles instead of plastic ones, and reusable hairbands instead of single-use, plastic ones.

Granted, it was partially a parody, yet many people did do the above actions to help the environment.

Yeah, that's what I said- it's partially helpful but not real... parody.


There is a step in the proof for the classification of pythagorean triples I'm having difficulty with.

b is the length of a leg of a right triangle and c is the length of the hypotenuse where gcd(b,c,)=1. Why must the greatest common divisor of $ c\plus{}b$ and $ c\minus{}b$ be less than or equal to two?

Hi.
If $ gcd(c\plus{}b,c\minus{}b)\equal{}d \Rightarrow (c\plus{}b)\vdots d$ and $ (c\minus{}b)\vdots d\Rightarrow ((c\plus{}b)\plus{}(c\minus{}b))\equal{}2c\vdots d$ and $ ((c\plus{}b)\minus{}(c\minus{}b))\equal{}2b\vdots d$ . So if $ d\neq 2, then (b,c)\neq 1$.

I think the $ \vdots$ just means $ |$, as in $ a|b$.

Wait, shouldn't d be smaller than $ c\plus{}b$? How can $ c\plus{}b$ be a factor of $ d$?
Does the colon mean "has the factor?"

Also, why can't $ (b,c)$ be 1 in Armenia post?

Just guessing about the colon, but he might have meant it the other way around.

Like a:b would mean b is a factor of a, rather than a is a factor of b.

(b,c)=1 'cause otherwise xb=sqrt(b^2+y^2){x E N &=/1,y is the other leg}
=>b^2*x^2=b^2+y^2.
=>(x^2-1)b^2=y^2.
=>y=b*sqrt(x^2-1)
sqrt(x^2-1) E/ N.
As pythagorian triplets are about integers ,(b,c)=1


how would you calculate the pH when there are 2 acids?

you are given the number of moles of each and the amount of water they are diluted in

Thanks
jli

If both acids are strong, then simply sum the number of moles from each and divide by the new volume. For example, when you mix 50 mL of HNO3 0.10 M with 150 mL of HCl 0.03 M, for example, then, as both acids are very strong, the total number of moles of hydronium ion will be

$n(H_{3}O^{+}) = 0.10 \times 0.050+0.03 \times 0.150 = 9.5 \times 10^{-3}mol$

and so $[H_{3}O^{+}] = \frac{9.5 \times 10^{-3}mol}{0.2 L}= 0.0475 M$, and the pH would be 1.32.
If one or both acids are weak, then a possible strategy would be calculate the "instantaneous" concentration of hydronium from each acid alone, and then consider the shift of each equilibrium to the new H+ concentration. Perhaps if you can provide some example it would be more simple to explain.


superior algebra
superior algebra unsolved

[img]http://epsilon.uscoders.net/ML_Problem3.png[/img]

Thank you.

There really is kind of an obvious thing to do.  If $ \sigma \in S_{n \minus{} 2}$, where might you send
it to in $ A_n$?  Well, if it is already even, we should send it to itself and if it is odd, we might
as well send it $ \sigma (n \minus{} 1 \hspace{.1 cm} n)$.    Now, check this a homomorphism and use an Isomorphism theorem.

Hopefully, that $ n\minus{}1$ $ n$ part comes out ok [it is the transposition that sends $ n\minus{}1$ to $ n$ and
vice versa]


function algebra polynomial Vieta calculus integration

er.... I am always having problems with problems which ask you to prove there are infinitely many integers or somethin like that ,
Can anyone give some appropriate sums with solutions explaining some brief idea ?

Sorry if this is dumb

A problem that might help is proving that there are an infinite amount of primes. You can achieve this by contradiction. Suppose there is a set of primes $ p_{1},p_{2},p_{3}...p_{n}$ and $ p_{n}$ is the largest prime. But look at this $ (p_{1})(p_{2})(p_{3})...(p_{n}) \plus{} 1 \equal{} p_{k}$. If $ p_{k}$ is prime than thats ok. Suppose it isn't, than there must be a prime number that can divide $ p_{k}$. That prime number can't be part of $ p_{1},p_{2},p_{3}...p_{n}$ because than it would have to divide the difference of $ p_{k} \minus{} p_{1})(p_{2})(p_{3})...(p_{n}) \equal{} 1$ which isn't possible. Contradiciton.

Possibly, he means proving that there are infinite integer solutions to some problem, or something like that.

Most of those problems can be solved by finding one solution $ (m_1, m_2, ... m_n)$ and showing that all solutions $ (k_1, k_2, ... k_n)$ work, where $ k_i$ is some function in terms of $ m_i$. Some times you can do this by Vieta Jumping, or just guessing what form to use.

For example, prove that there are infinite many integer solutions to $ a^2\plus{}b^2\equal{}c^2$ (yes I know this is a very basic example)

We see that if $ (a_1, a_2, a_3)$ is a solution, then so is $ (ka_1, ka_2, ka_3$). One solution is $ (3, 4, 5)$, so all solutions of the form $ (3k, 4k, 5k)$ for integral $ k$ work.

Here's another example:

Prove that there are infinitely many integral solutions to $ x^4\plus{}y^4\equal{}z^5$.


algebra
polynomial
induction
inequalities unsolved
inequalities

let $p = min(x_i)$ and $q = max(x_i)$.
let the polinomial $f(x) = x^2 - (p+q)x +pq$.
see that the roots of this polinomial are $p$ and $q$.
easily we see that for $p<k<q$, $f(k) <0$.

so,

${x_i}^2 -(p+q){x_i} +pq \leq 0$ for all $i$

$\Longrightarrow \sum{{x_i}^2 -(p+q){x_i} +pq} \leq 0$

$\Longrightarrow 1+npq \leq 0 \Longrightarrow pq \leq -1/n$.  :)

[quote="zhaobin"][quote="hardsoul"][quote="Mathx"]Given $n$ real numbers $x_{1}$, $x_{2}$, ..., $x_{n}$ which satisfy $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x_{i}^{2}=1$. Prove that there exist two numbers $i$ and $j$ from the set $\left\{1,2,...,n\right\}$ such that $x_{i}x_{j}\leq \frac{-1}{n}$.[/quote]

Induction
We prove more general theorem:
let  $\sum_{i=1}^{n}x_{i}=0\ ,\ \sum_{i=1}^{n}x_{i}^{2}=a$ prove that there are $i,j$ such that:
$x_{i}x_{j}\le \frac{-a}{2},a>0$
 Proof:
Suppose the assertion holds for $n$ , then consider the following expressions
Let WLOG $x_{n+1}>0$ and $x_{n}<0.$ $\sum_{i=1}^{n+1}x_{i}=0\ ,\ \sum_{i=1}^{n+1}x_{i}^{2}=a$
Now $\sum_{i=1}^{n-1}x_{i}+(x_{n}+x_{n+1})=0\ ,\ \sum_{i=1}^{n-1}x_{i}^{2}+(x_{n}+x_{n+1})^{2}=a-2x_{n}x_{n+1}$ so there exist $x_{i}$ and $x_{j}$ such that $x_{i}x_{j}\le \frac{-(a-2x_{n}x_{n+1})}{2}\leq \frac{-a}{2}$ the proof is complete...
I'm not sure about this proof... :?[/quote]
if one of $x_{i}$ or $x_{j}$ in $x_{i}x_{j}\le \frac{-(a-2x_{n}x_{n+1})}{2}\leq \frac{-a}{2}$  is $x_{n}+x_{n}+1$ can it be true?
if n is even.we can let half of $x_{i}$ is $x_{i}=\sqrt{\frac{1}{n}}$ the rest is $-\sqrt{\frac{1}{n}}$ then your conclution will be wrong. ;)[/quote]

You are right...I must be sleeping...  :sleeping:

[quote="Mathx"]Given $n$ real numbers $x_{1}$, $x_{2}$, ..., $x_{n}$ which satisfy $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x_{i}^{2}=1$. Prove that there exist two numbers $i$ and $j$ from the set $\left\{1,2,...,n\right\}$ such that $x_{i}x_{j}\leq \frac{-1}{n}$.[/quote]

We can WLOG assume that $x_{1}\leq x_{2}\leq ...\leq x_{n}$. Then, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=57588 , we have

$x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}+nx_{1}x_{n}\leq 0$,

so that $nx_{1}x_{n}\leq-\left(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}\right)=-1$, and thus $x_{1}x_{n}\leq\frac{-1}{n}$. And it's done.

  Darij


Three congruent squares are placed side-by-side to form a
rectangle with a perimeter of 104 inches. What is the number
of square inches in the area of one of the original squares?

if the side length of an original square is x, the perimeter is 2*(3x+x)=8x.

8x=104, x=13

13*13=169

answer : 169

This rectangle will be $ x*3x$ where $ x$ is the side length of the square. So the perimeter is $ 2(x \plus{} 3x) \equal{} 8x$.

$ 8x \equal{} 104$
$ x \equal{} 13$

So $ x^2 \equal{} 169$.

Edit: Sorry, Fantasy. I didn't see your post.  :blush:


How many roots are there $ x^{3}\plus{}(x\plus{}1)^{3}\plus{}(x\plus{}2)^{3}\equal{}(x\plus{}3)^{3}$ in positive numbers?

hello, after expanding and sorting your equation i have got $ 2x^3\minus{}12x\minus{}18\equal{}0$ this can be factorized in $ (x\minus{}3)(x^2\plus{}3x\plus{}3)\equal{}0$, so the only positive root is $ x\equal{}3$.
Sonnhard.


Hey, I'm studying for Science Olympiad, and I need a good genetics book.

What's the best AP Bio book, then a more detailed book focused exclusively on genetics/molecular biology/biotechnology?

AP Bio Book: [u]Biology[/u] by Campbell.  No exceptions.
Genetics Book: I like [u]Principles of Genetics[/u] by Snustad and Simmons, but it may be a little dense.


inequalities
Euler
inequalities proposed

Prove that

$ \frac{1}{r}\leq\frac{r_a}{r_br_c}\plus{}\frac{r_b}{r_cr_a}\plus{}\frac{r_c}{r_ar_b}\leq\frac{4R\minus{}5r}{3r^2}$

[quote="Ligouras"]Prove that

$ \frac {1}{r}\leq\frac {r_a}{r_br_c} \plus{} \frac {r_b}{r_cr_a} \plus{} \frac {r_c}{r_ar_b}\leq\frac {4R \minus{} 5r}{3r^2}$[/quote]

we have
\[ \sum {\frac {{r_a }}{{r_b r_c }}} \equal{} \frac {{(4R \plus{} r)^2 \minus{} 2s^2 }}{{s^2 r}}\]
and use Gerretsen Inequality $ 4R^2 \plus{} 4Rr \plus{} 3r^2 \ge s^2 \ge 16Rr \minus{} 5r^2$ and Euler Inequality $ R \ge 2r$ can prove it.


Prove that there are no positive integers $ n$ satisfying $ 2n^2 \plus{} 1$,$ 3n^2 \plus{} 1$,$ 6n^2 \plus{} 1$ are perfect squares.

[hide]
let x = 4p +/- 1 for any non-negative p
2x^2 + 1 = 3 (mod 4)
3x^2 + 1 = 0 (mod 4)
6x^2 + 1 = 3 (mod 4)
Since there is no square with 3(mod 4) so rejected.
when x = 4p, case obviously can be rejected
when x = 4p + 2
2x^2 + 1 = 2 (mod 7)
3x^2 + 1 = 6 (mod 7) [contradiction as no squares with 5 when mod 7]
6x^2 + 1 = 4 (mod 7)
So proves.
[/hide]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=23649 and http://www.mathlinks.ro/viewtopic.php?t=156645 for a more general version.
[quote="xienohp"]
when x = 4p, case obviously can be rejected[/quote]
Because¿
[quote="xienohp"]
when x = 4p + 2
2x^2 + 1 = 2 (mod 7)
3x^2 + 1 = 6 (mod 7) [contradiction as no squares with 5 when mod 7]
6x^2 + 1 = 4 (mod 7)
So proves.
[/quote]
You seem to have made some mistakes concerning modulo. $ 4p \plus{} 2$ can have any value $ \mod 7$.


You visit a remote desert island inhabited by one hundred very friendly dragons, all of whom have green eyes. They havent seen a human for many centuries and are very excited about your visit. They show you around their island and tell you all about their dragon way of life (dragons can talk, of course).

They seem to be quite normal, as far as dragons go, but then you find out something rather odd. They have a rule on the island which states that if a dragon ever finds out that he/she has green eyes, then at precisely midnight on the day of this discovery, he/she must relinquish all dragon powers and transform into a long-tailed sparrow. However, there are no mirrors on the island, and they never talk about eye color, so the dragons have been living in blissful ignorance throughout the ages.

Upon your departure, all the dragons get together to see you off, and in a tearful farewell you thank them for being such hospitable dragons. Then you decide to tell them something that they all already know (for each can see the colors of the eyes of the other dragons). You tell them all that at least one of them has green eyes. Then you leave, not thinking of the consequences (if any). Assuming that the dragons are (of course) infallibly logical, what happens?If something interesting does happen, what exactly is the new information that you gave the dragons?

Anyone get it?

Yes... a very interesting problem indeed.

Although a variation has been posted before at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14830, and my solution is on page 2.

So does it mean...

[hide]they all transform into long-tailed sparrows?[/hide]

I don't think there is  a concequence becuase everyone know that eyeone else's eyes are green.  The guy went to the island says that at least one of them have green eyes, so the alligators might think it's just one, two, or mabye even all but one alligotar that have green eyes..

What they learn from the statement is about time. They'll never lose their dragon powers without this statement because they won't know when to begin counting, more or less.

[quote="RC-7th"]I'm not sure, but I'm guessing that one the 100th day after your departure, all the dragons commit suicide at midnight. Am I off by a bit?[/quote]

Eum...That would be correct if the problem stated there were 100 dragons :D

So we just say the $n$-th day, where $n$ is the amount of dragons :P

[quote="Ars"]It is 100 there were 100 dragons. :)  :)  :)  :D  :D  :D  ;)  ;)  :cool:  :cool:[/quote]

What the ... ?

I don't get it.  :?

See this on word says it  over here to Peter VDD's answer   :downto:
[quote="lunmu"]You visit a remote desert island inhabited by one hundred very friendly dragons, all of whom have green eyes.
[quote]

[quote="Peter VDD"]Eum...That would be correct if the problem stated there were 100 dragons :D

So we just say the $n$-th day, where $n$ is the amount of dragons :P[/quote]

Did you read the problem?

yea yea... if 1 person points out my mistake that's ok :D

[quote="Jophus"]Can someone please explain why the dragons would turn into sparrows?[/quote]

Because as of this moment, there are no dragons, which means the story had to have taken place in the past, meaning the race of dragons are now extinct, and replaced by long tailed sparrows. Since the dragons must relinquish their powers when turning into long-taileld sparrows, they will no longer be immortal. But, they could mate with other long-tailed sparrows and reproduce. So the next time you pas a long tailed sparrow, you can think to yourself, "I may have seen the offspring of a dragon."

Seriously, I don't understand the logic.

Ok so suppose there are only two dragons then. Both have green eyes, but each knows only that the other has green eyes. The midnight after the man leaves, neither dragon turns into a sparrow since neither can figure out whether or not he has green eyes. But suppose one of them had non-green eyes. Then the green-eyed one would have been able to figure out that he had green eyes, and he would have turned into a sparrow. So when each one sees the other, he knows that he must have green eyes, as the other one is still a dragon. So at midnight of the second day, both of them turn into sparrows. The case for $n$ dragons is the same.

Ok, thanks. I understand 2 dragons but I'm not sure about more than 2.
There are 3 dragons now. And in the eyes of one green-eyed dragon, he waits for the other 2 to figure out that they are both green-eyed. When they don't transform on the midnight of the 2nd night, he realizes that they are waiting for him. So they all transform into sparrows on the third night.
   So is my logic correct?


inequalities
function
inequalities proposed

If $ a,b,c >0$ and $ a < b\plus{}c$ prove that :

$ \frac{a}{1\plus{}a} < \frac{b}{1\plus{}b}\plus{}\frac{c}{1\plus{}c}$

[quote="arqady"][quote="Tiger100"]If $ a,b,c > 0$ and $ a < b \plus{} c$ prove that :

$ \frac {a}{1 \plus{} a} < \frac {b}{1 \plus{} b} \plus{} \frac {c}{1 \plus{} c}$[/quote]
Also for all $ k\geq0$ and with the same conditions we obtain:
\[ \frac {a}{k \plus{} a} < \frac {b}{k \plus{} b} \plus{} \frac {c}{k \plus{} c}\]
[/quote]
It obviously true after expanding.....

[quote="Tiger100"]If $ a,b,c > 0$ and $ a < b \plus{} c$ prove that :

$ \frac {a}{1 \plus{} a} < \frac {b}{1 \plus{} b} \plus{} \frac {c}{1 \plus{} c}$[/quote]
The function $ f(x) \equal{} \frac{x}{x\plus{}1}$ is increasing and concave. So $ RHS \equal{} f(b)\plus{}f(c) \ge f(b\plus{}c)\plus{}f(0) \equal{} f(b\plus{}c) > f(a) \equal{} LHS$


In a triangle $ABC$ it is $\angle B= 75$ and $BC = 2AD$ , where $AD$ is the altitude from $A$. Prove tht $\angle C= 30$

 [u]Babis[/u]

  [b]NOTE[/b]

  [color=blue]I have seen the problem in the booklet '' South Africa and the IMO , 1995''. I'd like a straight solution with pure synthetic geometry , if possible.[/color]

[color=darkblue][b]Remark.[/b] It is well-known that $bc=2Rh_{a}$ and $\frac{1}{\tan B}+\frac{1}{\tan C}=\frac{a}{h_{a}}\ .$ Therefore, $a=\lambda\cdot h_{a}\Longleftrightarrow$ $\frac{1}{\tan B}+\frac{1}{\tan C}=\lambda\ .$

[b]A metrical proof.[/b] It is well-known that $\tan 75^{\circ}=2+\sqrt 3\ .$ For $\lambda =2$ obtain :
$\frac{1}{\tan B}+\frac{1}{\tan C}=2$ $\Longleftrightarrow$ $2-\sqrt 3+\frac{1}{\tan C}=2$ $\Longleftrightarrow$ $\tan C=\frac{1}{\sqrt 3}$ $\Longleftrightarrow$ $C=30^{\circ}\ .$

[b]A synthetical proof.[/b] Denote the middlepoint $M$ of the side $[BC]$ and the interior point $E$ for which the triangle $ABE$ is equilateral.
$AB=BE$, $AD=BM$, $\widehat{BAD}\equiv\widehat{EBM}$ $\Longrightarrow$ (s.a.s.) $\triangle ABD\equiv\triangle BEM$ $\Longrightarrow$ $EM\perp BC$ $\Longrightarrow$
The point $E$ is the circumcenter of the triangle $ABC$ $\Longrightarrow$ $AB=R$- the circumradius $\Longrightarrow$ $C=30^{\circ}\ .$  [/color]

[quote="silouan"]Quite same solution . $bc=2Rh_{a}$  so $c=4R$ but
$\frac{c}{sinC}=2R$  so $sinC=\frac{1}{2}$ , $C=30$ .

***
I think that a synthetic solution will come up from the fact .If we extend the height $AD$  so $AD=DE$  then the triangle $ACM$ is equilateral and also the circle with center $C$ and radius $CA$ passes through $A,M,B$[/quote]

You don't know that $ACM$ is equilateral - in fact, that's equivalent to the original statement, since that gives $AC=BC\implies\angle C=30^\circ$

[quote="silouan"]Quite same solution . $bc=2Rh_{a}$  so $c=4R$ but
$\frac{c}{sinC}=2R$  so $sinC=\frac{1}{2}$ , $C=30$ .

[/quote]

My solution is the above . The text that foolows the **** it is only a hint for a synthetical solution
 :wink:

I present in the my last message a [b]synthetical proof[/b] of this nice problem.

VERY VERY Nice solution !!!!!!!!!!!!!

    I'm happy to  read it :lol:

Thank you Virgil.

 [b]Babis[/b]

[quote="stergiu"][quote="silouan"]I think that a synthetic solution will come up from the fact .If we extend the height $AD$  so $AD=DE$  then the triangle $ACM$ is equilateral and also the circle with center $C$ and radius $CA$ passes through $A,M,B$[/quote]

 Do you mean $\triangle ACE$  (and not $\triangle ACM$)?

[u]Babis[/u][/quote]

Yes it was a typo .
 :blush: BTW Very nice solution Virgil  :D


group theory
abstract algebra
advanced fields
advanced fields unsolved

How do you show that symetric, connected nbhs of e form a fundamental nbh system for e?
(symetric fundamental nbh system for topological group is clear. But what if this group is locally connected, like Lie group, can we also asks for connectness?  I asked one proffessor, he said need to use Lie Algebra..., is there an elementry proof?)

$Ui$ is connected local base then $Ui \bullet Ui^{-1}$ is what is needed


Remember even roots always come with 2 solutions $ \pm x$

@sniped: That came out wrong.

What he's referring to is the conjugate root theorem:
if $ a\plus{}bi$ is a root of $ f(x)$, then $ a\minus{}bi$ is a root of $ f(x)$.

There are a total of four roots to the expression.
[hide="HINT"]
DeMoivre.
[/hide]

[quote="Christine"]Is x=(-16)^(1/4) => x = 2*sqrt.(i) ?[/quote]
[hide]
DeMoivre's isn't necessary.
\[ x\equal{}(\minus{}16)^{\frac{1}{4}}\] \[ x^4\equal{}\minus{}16\] \[ x^4\plus{}16\equal{}0\] \[ (x^4\plus{}8x^2\plus{}16)\minus{}8x^2\equal{}0\] \[ (x^2\plus{}4)^2\minus{}(2x\sqrt{2})^2\equal{}0\] \[ (x^2\minus{}2x\sqrt{2}\plus{}4)(x^2\plus{}2x\sqrt{2}\plus{}4)\equal{}0\] Plugging each into the quadratic equation, we obtain that $ x\equal{}\frac{2\sqrt{2}\pm\sqrt{8\minus{}16}}{2}$ and $ x\equal{}\frac{\minus{}2\sqrt{2}\pm\sqrt{8\minus{}16}}{2}$.  Simplifying, we obtain that $ x\equal{}\boxed{\frac{\pm\sqrt{2}\pm i\sqrt{2}}{2}}$
[/hide]

$ x=-16^{\frac{1}{4}}$ =>

let $ x$ = $ cos\theta + isin\theta$

$ cos\theta + isin\theta$=$ -16^{\frac{1}{4}}$

$ \cos4\theta + isin4\theta = -16$

$ \cos4\theta + isin4\theta$ = $ 2^4[cos(\pi+2\pi{k}) + isin(\pi+2\pi{k})]$

how do i proceed?


geometry
circumcircle
Euler
power of a point
radical axis
geometry proposed

[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given and let $ D,\ E,\ F$ be, the tangency points of its $ A$-excircle $ (I_{a}),$ to the sidelines $ BC,\ AC,\ AB,$ respectively. We denote the points $ M\equiv I_{a}B\cap DE,\ N\equiv I_{a}C\cap DF$ and let be the point $ S\equiv FM\cap EN.$ Prove that the line segment $ I_{a}S,$ passes through the circumcenter $ O,$ of the given triangle $ \bigtriangleup ABC.$[/color][/size][/b]

Kostas Vittas.

That's a very nice problem but I have a very ugly solution with calculation.

I use the fact that $ \frac {sin\angle SI_aF} {sin\angle SI_aE}\equal{}\frac {sin\angle OI_aF} {sin\angle OI_aE}$ and this is enough to prove that $ S\minus{}O\minus{}I_a$ lie on the same line.

After some calculation we obtain that $ \frac {sin\angle SI_aF} {sin\angle SI_aE}\equal{}\frac {sin\frac {C} {2}sin(B\plus{}\frac {C} {2}\minus{}90^{\circ})} {sin\frac {B} {2}sin(C\plus{}\frac {B} {2}\minus{}90^{\circ})}$ and $ \frac {sin\angle OI_aF} {sin\angle OI_aE}\equal{}\frac {a\plus{}b} {a\plus{}c}$

It remains to prove that $ \frac {sin\frac {C} {2}sin(B\plus{}\frac {C} {2}\minus{}90^{\circ})} {sin\frac {B} {2}sin(C\plus{}\frac {B} {2}\minus{}90^{\circ})}\equal{}\frac {a\plus{}b} {a\plus{}c}$ which can be easily deduced.

Inversion in the A-excircle $ (I_a)$ takes the circumcircle $ (O)$ of the $ \triangle ABC$ into the 9-point circle $ (N_a)$ of the A-extouch $ \triangle DEF$ with circumcircle $ (I_a)$ $ \Longrightarrow$ $ ON_aI_a$ is Euler line of this extouch triangle, its orthocenter $ H_a \in OI_a$ is on this Euler line. $ I_a$ is orthocenter of the $ \triangle DMN.$ The quadrilaterals $ BDMF, CDNE$ are kites $ \Longrightarrow$ the isosceles $ \triangle DMF \sim \triangle DNE$ are similar $ \Longrightarrow$ the quadrilateral $ EFMN$ is cyclic with a circumcircle $ (P).$ Let $ (K), (L)$ be circles with diameters $ FM, EN,$ respectively. From $ (P),$ $ SF \cdot SM \equal{} SE \cdot SN$ $ \Longrightarrow$ $ S$ is on the radical axis of $ (K), (L).$ Powers of the orthocenter $ H_a$ of the A-extouch $ \triangle DEF$ to $ (K), (L)$ are both equal to half the power of $ H_a$ to the circumcircle $ (I_a)$ of this triangle, because both  $ (K), (L)$ are centered on midlines of this triangle and passing through its opposite vertices. Similarly, powers of the orthocenter $ I_a$ of the $ \triangle DMN$ to $ (K), (L)$ are equal. Then $ SI_aH_a$ is radical axis of the circles $ (K), (L)$ $ \Longrightarrow$ the lines $ SI_aH_a \equiv I_aH_aO$ are identical and the points $ S, I_a, H_a, O$ collinear.

Hello!

Let $ G_{a}$ be the centroid of the triangle $ DEF$.

It's known that the points $ O$, $ I_{a}$, $ G_{a}$ are collinear     (*)

{$ X$}$ \equal{} BI_{a}\cap DF$ and {$ Y$} $ \equal{} CI_{a}\cap DE$.
$ E,M,Y\in DE$ and $ F,N,X\in DF$.
{$ S$} $ \equal{} EN\cap FM$, {$ I_{a}$} $ \equal{} MX\cap NY$ and {$ G_{a}$} $ \equal{} YF\cap XE$
The Pappus' theorem implies that the points $ S, I_{a}, G_{a}$ are collinear     (**)
(*), (**) $ \Rightarrow$ the points $ S,O,I_{a}$ are collinear. :)

Thank you very much dear [b][size=100]Petry[/size][/b], for a clever solution.

I would like to preserve in the future this nice proof with your name and if it is possible, tell me please about it.

Best regards, Kostas Vittas.

PS. I will post here next time, the solution I have in mind of the proposed peoblem.

Hello!

   I'm happy you like my solution and I agree that you preserve this proof with my name.
I'm also looking forward to seeing your solution.

  Best regards, Petrisor Neagoe