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relatively prime

Prove that, if $(n-1)|k$, then $(n-1)^{2}|(n^{k}-1)$. $n,k$ are natural numbers.

To show that $n^{k-1}+n^{k-2}+\cdots+1$ is divisible by $n-1$, we could also say that these numbers represent the cycle generated by multiplying repeatedly by n (which is relatively prime to $n-1$, our modulus, so this number should have some order $\pmod{n-1}$ dividing n-1). Since n-1 divides k, this order divides k, so in the k terms given we cycle through all the terms $\frac{k}{n-1}$ times, and we know that the sum of all things $\pmod{n-1}$ is congruent to 0.

Wow, I just noticed the simplest thing...

$n^{k-1}+n^{k-2}+\cdots+1 \equiv 1+1+\cdots+1 \equiv k \equiv 0 \pmod{(n-1)}$.

Hahaha. That's what I was getting at...

Only I didn't realize it was so obvious, overcomplicated it  :roll:

Hey I posted in your blog
I think i'm the first user lol

replace n by x: these are polynomials. so just divide the first polynomial $(P(x)=\sum_{i=0}^{k-1}x^{i})$ by $x-1$: the remainder theorem for polynomials says that $\frac{P(x)}{x-1}=P(1)=k$.

The remainder theorem for polynomials: if $A(x)=B(x)\cdot (x-\alpha)+R$, for $x=\alpha, A(\alpha)=0+R=R$

[quote="Tiago Guerreiro"]hi :)

I'm sorry to bore you but I have some questions about this:

[quote="paladin8"]
$n^{k}-1 = (n-1)(n^{k-1}+n^{k-2}+\cdots+1)$,

so we just have to show that $n^{k-1}+n^{k-2}+\cdots+1$ is divisible by $n-1$.

$n^{k-1}+n^{k-2}+\cdots+1 = (n-1)\left[n^{k-2}+2n^{k-3}+3n^{k-4}+\cdots+(k-1)\right]+k$

but $k = a(n-1)$, so

$n^{k-1}+n^{k-2}+\cdots+1 = (n-1)\left[n^{k-2}+2n^{k-3}+3n^{k-4}+\cdots+(k-1+a)\right]$,

as desired.[/quote]

How did you get:

$(n-1)\left[n^{k-2}+2n^{k-3}+3n^{k-4}+\cdots+(k-1)\right]+k$ from $n^{k-1}+n^{k-2}+\cdots+1$ ?

And how did you get:

$(n-1)\left[n^{k-2}+2n^{k-3}+3n^{k-4}+\cdots+(k-1+a)\right]$ from $n^{k-1}+n^{k-2}+\cdots+1$ ?

Thanks in advance  :lol:[/quote]

Ok for your first question,

(-1)(k-1) = -k +1.

The last term would be 1, but since we added -k, we must also add +k.

For your second question,

We know that k = a(n-1) where a is an integer, not 0 from the first part of the problem.

Hence, the +k that was on the outside of the first step get put inside since k=(n-1)a. So we get k-1+a.


The sides of a regular polygon of $ n$ sides, $ n > 4$, are extended to form a star. The number of degrees at each point of the star is:

$ \textbf{(A)}\ \frac {360}{n} \qquad\textbf{(B)}\ \frac {(n \minus{} 4)180}{n} \qquad\textbf{(C)}\ \frac {(n \minus{} 2)180}{n}$
$ \textbf{(D)}\ 180 \minus{} \frac {90}{n} \qquad\textbf{(E)}\ \frac {180}{n}$

[hide="Click for solution"]
Each such angle is the vertex angle of an isosceles triangle whose base angles are each, in degrees, $ 180\minus{}\frac{(n\minus{}2)(180)}{n}\equal{}\frac{360}{n}$.  Thus, $ \text{angle }\equal{}180\minus{}\frac{720}{n}\equal{}\frac{180n\minus{}720}{n}\equal{}\frac{180(n\minus{}4)}{n}$, or $ \boxed{\textbf{(B)}}$.
[/hide]


Find all primes P for which there exist integers x an y such that

p+1=2x 2
p 2+1=2y 2

Here is the main part of my solution, I think Thanliem solution is wrong, because there is solution only for p=7

first: p is odd, now locking modulo p, we find x 2 -y 2  is divisible by p, so either x+y or x-y is divisible by p, but from the first 2 equations one has that 0<x<y<p, (one can assume they are positive) so the only possibility is x+y=p, replacing this into the equations there are just a few calculations left.

Pascual


Let $ a,b,c$ be positive integers. Prove the following inequality:

$ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2)\plus{}4(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1) \geq 6(a\plus{}b\plus{}c)^2$

[quote="Mashimaru"]Let $ a,b,c$ be positive integers. Prove the following inequality:

$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \plus{} 4(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \geq 6(a \plus{} b \plus{} c)^2$[/quote]

It's equivolent to
$ 5a^2b^2c^2\plus{}6a^2b^2\plus{}6a^2c^2\plus{}6b^2c^2\plus{}8a^2\plus{}8b^2\plus{}8c^2\plus{}12 \ge 6a^2\plus{}6b^2\plus{}6c^2\plus{}12ab\plus{}12ac\plus{}12bc$
$ \leftrightarrow$
$ 5a^2b^2c^2\plus{}6a^2b^2\plus{}6a^2c^2\plus{}6b^2c^2\plus{}2a^2\plus{}2b^2\plus{}2c^2\plus{}12 \ge 12ab\plus{}12ac\plus{}12bc$
which is right becouse:
$ 2a^2b^2\plus{}2a^2c^2\plus{}2b^2c^2 \ge 2ab\plus{}2ac\plus{}2bc$
$ 4a^2b^2\plus{}4a^2c^2\plus{}4b^2c^2\plus{}12 \ge 8ab\plus{}8ac\plus{}8bc$
$ 2a^2\plus{}2b^2\plus{}2c^2 \ge 2ab\plus{}2ac\plus{}2bc$

[quote="quykhtn-qa1"][quote="Mashimaru"][quote="SaYaT"]Let $ a,b,c$ be positive integers. Prove the following inequality:

$ 2a^2b^2 \plus{} 2a^2c^2 \plus{} 2b^2c^2 \ge 2ab \plus{} 2ac \plus{} 2bc$
[/quote]

Why? :lol:[/quote]
Because $ a,b,c$be positive integers and not equality[/quote]

Consider $ a\equal{}0.1;b\equal{}0.2;c\equal{}0.3$ :roll:

[quote="Mashimaru"][/quote][quote="quykhtn-qa1"][quote="Mashimaru"][quote="SaYaT"]Let $ a,b,c$ be positive integers. Prove the following inequality:

$ 2a^2b^2 + 2a^2c^2 + 2b^2c^2 \ge 2ab + 2ac + 2bc$
[/quote]

Why? :lol:[/quote]
Because $ a,b,c$be positive integers and not equality[/quote][quote="Mashimaru"]

Consider $ a = 0.1;b = 0.2;c = 0.3$ :roll:[/quote]
$ a,b,c$ are integers :wink:

Oh no, I'm extremely sorry. The first problem was mis-posted. It should have been $ a,b,c$ are positive real numbers :blush:

I did not even find $ (a,b,c)$ for the equality :(


The function $ f(n)$ is defined for all positive integers $ n$ and takes on non-negative integer values values.Also , for all $ m,n$
$ f(m\plus{}n)\minus{}f(m)\minus{}f(n)\equal{}$$ 0$ or $ 1$, f(2)=0,f(3)>0 , and f(9999)=3333.
Find $ f(1982)$.

1982 IMO, #1
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60805[/url]


Hello,

Can someone take a look at this and explain how you get that derivation on page 27? Try to make as simple as possible please.

http://physics.gmu.edu/~ngoldman/ch%2021%20present%202%20p260%20sum.pdf

Actually the problem you had a doubt with "point dipole approximation" is on page 17 not 27.
So I guess you understand the things on page 16, that is why the electric field of the dipole is:

\[ E_x \equal{} \dfrac{q}{4\pi\epsilon_0}\left[\dfrac{x}{(x^2\plus{}(y\minus{}a)^2)^{3/2}}\minus{}\dfrac{x}{(x^2\plus{}(y\plus{}a)^2)^{3/2}}\right]\]

now we know that $ y>>a$ so we can say that: $ (y\pm a)^2 \approx y^2 \pm 2ay$. We also see that: $ x^2\plus{}y^2\equal{}r^2$. So the electric field is:

\[ E_x \equal{} \dfrac{qx}{4\pi\epsilon_0}\left[\dfrac{1}{(r^2\minus{}2ay)^{3/2}}\minus{}\dfrac{1}{(r^2\plus{}2ay)^{3/2}}\right]\]

Now lets take out $ r^2$ from the denominator, then we get $ (r^2(1\minus{}...))^{3/2}\equal{}r^3(1\minus{}...)^{3/2}$. So:
\[ E_x \equal{} \dfrac{qx}{4\pi\epsilon_0 r^3}\left[\dfrac{1}{(1\minus{}2ay/r^2)^{3/2}}\minus{}\dfrac{1}{(1\plus{}2ay/r^2)^{3/2}}\right]\]

Now lets use Taylor approximation: $ (1\plus{} x)^{\minus{}n} \approx 1\minus{}nx$ and $ (1\minus{}x)^{\minus{}n}\equal{} 1\plus{}nx$, hence:

\[ E_x \equal{} \dfrac{qx}{4\pi\epsilon_0 r^3}\left[1\plus{}\dfrac{3ay}{r^2}\minus{}1\plus{}\dfrac{3ay}{r^2}\right]\equal{}  \dfrac{qx}{4\pi\epsilon_0 r^5}6ay\]

But we know that by definition the $ p$ dipole moment is $ p\equal{}2a\cdot q$. Hence we get:
\[ E_x\equal{}\dfrac{3p}{4\pi\epsilon_0}\dfrac{xy}{r^5}\equal{}\dfrac{3p}{4\pi\epsilon_0}\dfrac{1}{r^3}\dfrac{x}{r}\dfrac{y}{r}\]

From the geometry we can see that:
\[ \sin \theta \equal{} \dfrac{x}{r} \;\;\;\;\;\;\; \cos\theta \equal{}\dfrac{y}{r}\]

Hence we get:
\[ E_x\equal{}\dfrac{3p}{4\pi\epsilon_0 r^3}\sin\theta \cos\theta\]

The other $ E_y$ component can be done similarly!

nice, I can actually undertand what you wrote. Thanks. How would I show what they got for E on page 33?
I'm sorry i'm just horrible at physics

come on guys. You guys are the smart ones


function
algebra
functional equation
algebra unsolved

Find all function $ f: \mathbb R \to [0,\plus{}\infty)$ such that \[ f(x^2\plus{}y^2)\equal{}f(x^2\minus{}y^2)\plus{}f(2xy), \quad \forall x,y \in \mathbb R.\]

This has definitely been posted before. I can't find a link so I'll write up a solution.

$ y\equal{}0\implies f(0)\equal{}0$ and $ x\equal{}0\implies f(a)\equal{}f(\minus{}a)\forall a\in \mathbb{R}$.
Let's denote $ g(x)\equal{}f(\sqrt{x})$ where $ g: [0,\infty)\to[0,\infty)$
For any $ a,b$ that are non negative the system
\[ a\equal{}x^2\minus{}y^2,b\equal{}2xy\]
always has a solution in $ (x,y)\in \mathbb{R}^2$
And therefore $ g(a^2\plus{}b^2)\equal{}g(a^2)\plus{}g(b^2)$ or
\[ g(x\plus{}y)\equal{}g(x)\plus{}g(y)\]
which is a Cauchy functional equation and therefore we get that
\[ g(x)\equal{}ax,\forall x\in [0,\infty)\cup \mathbb{Q}\]
translating back to our equation this simply means that
\[ f(x)\equal{}ax^2\forall x\in \mathbb{Q}\]
And the final observation is that $ f$ is monotone so therefore the only solutions to $ f$ are functions of the form $ ax^2$ where $ a\in [0,\infty)$

[quote="Albanian Eagle"]This has definitely been posted before. I can't find a link so I'll write up a solution.

$ y \equal{} 0\implies f(0) \equal{} 0$ and $ x \equal{} 0\implies f(a) \equal{} f( \minus{} a)\forall a\in \mathbb{R}$.
Let's denote $ g(x) \equal{} f(\sqrt {x})$ where $ g: [0,\infty)\to[0,\infty)$
For any $ a,b$ that are non negative the system
\[ a \equal{} x^2 \minus{} y^2,b \equal{} 2xy\]
always has a solution in $ (x,y)\in \mathbb{R}^2$
And therefore $ g(a^2 \plus{} b^2) \equal{} g(a^2) \plus{} g(b^2)$ or
\[ g(x \plus{} y) \equal{} g(x) \plus{} g(y)\]
which is a Cauchy functional equation and therefore we get that
\[ g(x) \equal{} ax,\forall x\in [0,\infty)\cup \mathbb{Q}\]
translating back to our equation this simply means that
\[ f(x) \equal{} ax^2\forall x\in \mathbb{Q}\]
And the final observation is that $ f$ is monotone so therefore the only solutions to $ f$ are functions of the form $ ax^2$ where $ a\in [0,\infty)$[/quote]

Nothing allow you to conclude that $ f(x)$ is monotone (and your conclusion $ f(x)\equal{}ax^2$ shows the contrary).

But $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ with $ g(x)$ : $ \mathbb R_0^\plus{}\to\mathbb R_0^\plus{}$ is enough to conclude $ g(x)\equal{}ax$ ($ g(x)$ is monotonous).

$ f(x^2\plus{}y^2)\geq f(x^2\minus{}y^2)$ so $ f$ is monotone in $ \mathbb{R}_{\plus{}}$ is what I meant  :)


NOOOOOOOOOOOOO!
so sam is leaving next year and hopefully some parents will be able to keep it running
lets make a topic to discuss how to keep it going, etc.

:(

Who do you think might be able to take his place?

jim li!!!!

or possibly we can make it so that it's just a problem session kind thingy. Like the ARML meets; we get a set of problems to work on, check answers, discuss the interesting ones. THEN I might actually go.

Huh, how good is Jim Li?

Sadly, I have Chinese School coinciding with Stanford Math Circles, otherwise I would go :( .


I have a Dell Inspiron 6000 with a pentium M cpu and 6 cell lithium ion battery.  I bought it last year for quite a high price (2.0 ghz pentium M, 1 GB DDR2, 128 mb video, 15.4 in wide screen ultrasharp lcd) and want to make sure that it can last all the way through college (i start next fall).

What are some things that you do to preserve your laptop?  I use mine as my primary computer - my desktop is incredibly weak - and plan to only use a laptop while in college.  I want to get an external 17 inch LCD to hook up to my laptop for better viewing.  I use my computer for things like MS word, LaTeX, internet, computer graphing, c++, and various games, so it's constantly on and running.  How can I make sure that things like my battery last as long as possible?

Thanks for the advice

[quote="Phoenix257"]I use my computer for things like MS word, LaTeX, internet, computer graphing, c++, and various games, so it's constantly on and running.  How can I make sure that things like my battery last as long as possible?[/quote]

Don't drop it.  Don't spill things on it.  Also, turn it off when you aren't using it (or change its power settings so it hibernates when you haven't been using it for a while).  Don't spill things on it  :wink:.

[url=http://www.laptopbattery.net/notebookbatteries_life.html]Notebook Battery Maintenance Tips[/url]. I never followed any of that advice myself.  :|  The battery of my laptop now lasts about 20 min.

So I should physically remove my battery when I'm plugged into the wall?  That seems kinda odd, will my laptop still operate properly with the battery physically missing?

Whether or not this is practical depends on your usage pattern. If you go weeks without using the battery, why not take it out? If you need the battery every other day, just leave it in place.

[quote="Phoenix257"]will my laptop still operate properly with the battery physically missing?[/quote]

Yes -- as long as it's plugged in, it gets the power it needs.

Put lots and lots of really brightly-colored duct tape on the cover and make it as ugly as you can, so nobody will want to steal it.

[quote="mlok"]Also, get the laptop engraved with your name, and don't leave it in a backpack outside of a weight room.[/quote]

Is this a reference to run-away RV?  :D

I always leave my laptop plugged into the AC while the battery is still connected, is this bad  :maybe:

You can always buy a new battery after a while.  $\$$ 100-150 isn't too bad in comparison with a new laptop.  However, by the time your battery dies you might be wanting to get a new laptop anyway.

I don't [i]think[/i] there's anything bad about leaving the battery in with the AC connected.  I usually do, but that's because I use my laptop on the run a lot so it needs to charge.  However, I know it's recommended to take it out.  There's really no need for it to be in there and it does reduce the risk of it exploding :D

I think some systems already implement those battery-saving tips. I've used my (Mac) laptop mostly plugged in with battery for a year and a half, and the battery life hasn't changed significantly.
Over long periods, the charge decreases slightly- I often see the machine display about 95 percent charge while plugged in, and it doesn't charge back to 100 percent until I use it unplugged for a while.

One advantage of the laptop is that it's easy to haul it in for service if necessary. I've had two major hardware problems, and being able to simply carry the computer to the Apple store to be looked at and fixed was a godsend.

I'm pretty sure my computer switches to AC once it's plugged in and leaves the battery alone once it fully charges back on.  I know on Windows you can go to Power Management and set it so that the screen turns/hard drives turn off after so long depending on whether or not the laptop is plugged in or not.  There are some presets here so just look for the right one for you.


A $ 16$-quart radiator is filled with water. Four quarts are removed and replaced with pure antifreeze liquid. Then four quarts of the mixture are removed and replaced with pure antifreeze. This is done a third and a fourth time. The fractional part of the final mixture that is water is:

$ \textbf{(A)}\ \frac{1}{4}\qquad
\textbf{(B)}\ \frac{81}{256}\qquad
\textbf{(C)}\ \frac{27}{64}\qquad
\textbf{(D)}\ \frac{37}{64}\qquad
\textbf{(E)}\ \frac{175}{256}$

After the first time: $ \frac{3}{4}$ water

Every time, this is multiplied by $ \frac{3}{4}$. So it is $ 1\times(\frac{3}{4})^4\equal{}\frac{81}{256}$. This is choice B.

$ Answer\equal{}\mathbf{(B)}$


Does anyone know where i can get semifinal exams from past years

What about this year?How can i find problems of this year? :maybe:


Hey, I was wondering if the other AOPS (mainly the introduction books) were worth getting seeing as I have both AOPS 1 & 2. I was thinking about the intermediate level books but I was wondering if I could get some advice on what you lot think? Is there anything that the introduction books cover that the classic series doesn't?
What about the intermediate books?

The Intro and Intermediate series basically go into everything way more in-depth, and have many more practice problems.


geometry
inradius
trigonometry
ratio
geometry unsolved

Let $ k\in \mathbb R$, $ k > 2$ and the triangle $ ABC$ with $ AC > AB$. Let $ AB \equal{} c$, $ BC \equal{} a$ and $ CA \equal{} b$. It is known that:
$ a \plus{} b \plus{} c \equal{} ka$. The median drawn from the vertex $ A$ separates the diameter of the incircle, which is perpendicular to the
side $ BC$, into $ 2$ segments. Find the report between these two segments.

$ a, b, c$ are triangle sides, $ r$ the inradius, $ s$ the semiperimeter, $ S$ the area. Incircle $ (I)$ is tangent to $ BC, CA, AB$ at $ D, E, F.$ $ JD \perp BC$ is a diameter of $ (I).$ $ M$ is midpoint of $ BC.$ Lines $ AM, IJ, EF$ are concurrent at $ X$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=147114[/url]). Perpendicular to $ EF$ from $ D$ cuts $ EF$ at $ Y$ and the incircle $ (I)$ again at $ K.$ $ DXJ, DYK$ are isogonals WRT the angle $ \angle FDE,$

$ \frac{DX}{XJ} \equal{} \frac{DY}{YK} \equal{} \tan{\widehat{DEF}} \tan{\widehat{EFD}} \equal{} \tan \frac{C \plus{} A}{2} \tan \frac{A \plus{} B}{2} \equal{} \frac{1}{\tan \frac{B}{2} \tan \frac{C}{2}} \equal{}$

$ \equal{} \frac{(s \minus{} b)(s \minus{} c)}{r^2} \equal{} \frac{s^2(s \minus{} b)(s \minus{} c)}{S^2} \equal{} \frac{s}{s \minus{} a} \equal{} \frac{k}{k \minus{} 2}$

[quote="Inequalities Master"] [color=darkred]Let $ ABC$ be a triangle with $ b > c$ . The $ A$ - median separates the diameter of the incircle

which is perpendicular to $ BC$ into two segments. Find the ratio between these segments.[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the points $ D\in BC$ , $ E\in CA$ , $ F\in AB$ where the incircle $ w$ touches the sides of $ \triangle ABC$ , the midpoint $ M$ of the side $ [BC]$ and the point $ T$ where the $ A$ - excircle touches the side $ [BC]$ . Thus, $ MD \equal{} MT \equal{} \frac {b \minus{} c}{2}$ and $ DB \equal{} TC \equal{} p \minus{} b$ . Consider the diameter $ [DK]$ of $ w$ and denote $ L\in AM\cap DK$ . Must ascertain the ratio $ \frac {LK}{LD}$ . I"ll use two remarkable properties $ \boxed {\ K\in AT\ \ \wedge\ \ L\in EF\ }$ (you can prove easily) . Denote $ k \equal{} \frac {a \plus{} b \plus{} c}{a} > 2$ , i.e. $ \frac {b \plus{} c}{a} \equal{} k \minus{} 1 > 1$ . Apply the [b]Menelaus' theorem[/b] to $ \overline {ALM}/\triangle DKT$ , i.e. $ \frac {AK}{AT}\cdot \frac {MT}{MD}\cdot\frac {LD}{LK} \equal{} 1$ $ \implies$ $ \frac {LK}{LD} \equal{} \frac {AK}{AT} \equal{}$ $ \frac {h_a \minus{} 2r}{h_a} \equal{}$ $ \frac {ah_a \minus{} 2ar}{ah_a} \equal{}$ $ \frac {2pr \minus{} 2ar}{2pr} \equal{}$ $ \frac {p \minus{} a}{p}$ $ \implies$ $ \boxed {\ \frac {LK}{LD} \equal{} \frac {p \minus{} a}{p}\ }$ , i.e. $ \frac {LK}{LD} \equal{} \frac {k \minus{} 2}{k}$ .

Show easily that $ \frac {LF}{LE} \equal{} \frac cb$ and  [b]Menelaus' theorem[/b] to $ \overline {DLK}/\triangle AMT$ $ \implies$ $ \boxed {\ \frac {LA}{LM} \equal{} \frac {2(p \minus{} a)}{a}\ }$ , i.e. $ \frac {LA}{LM} \equal{} k \minus{} 2$ .[/color]


Set up an equation for the tangent at the point M1(X1,Y1) to the circle given by X^2 + Y^2 = R^2

Please help

This is implicitly answered in my answer to your first circle tangent problem (if you still can't get it, try using similar right triangles).

sorry i thought it meant in this one that the point isnt on the circle

Well if $M_{1}$ isn't on the circle, it'll be a bit different, but you can still use similar right triangles

So lets say that M is (5/3, -5/3) and the circle is X^2 + Y^2 = 5 and M doesnot lie on the circle how would i find this.

let the line that is tangent to the circle be L, and suppose that it is tangent at a point T=(a,b), what can you say geometrically about O=(0,0), T, and M?

what can you say about a and b?

basically, you approach these problems by saying: how can i find a system of equations to solve for the line? well, i can take two points on the line, and that determines the line...so how do i find the second point? well what equations can i get relating a,b to other values, to get an equation to solve for them, etc


by fermat's theorem,

$ 2^{198}\equiv 1 \pmod{199}$

$ 2^{1980}\equiv 1 \pmod{199}$

$ 2^{1990}\equiv 2^{10}\equal{}1024\equiv 29 \pmod{199}$

also,

$ ({2^4})^{497}\equal{}2^{1988}\equiv 1 \pmod{5}$

$ 2^{1990}\equiv 4 \pmod{5}$

$ 2^{1990}\equiv 0 \pmod{2}$

$ 2^{1990}$ leaves a remainder  4 when divided by 5, 29 when divided by 199, 0 when divided by 2

$ 2^{1990}$ leaves a remainder x when divided by 1990 such that

$ x\equiv 4\pmod{5}$

$ x\equiv 29\pmod{199}$

$ x\equiv 0\pmod{2}$

$ x\equal{}1024$

[quote="nishantkumar"]what would be the remainder when 2^1990 is divided by 1990 ?????[/quote]
This is Indian RMO 1996 Problem.
I proved this by noting that
$ 2^{1990}\equiv 2^{10} (\mod 199)$
And $ 2^{1990}\equiv 2^{10}(\mod 10)$.
You can try to prove both the results :)
[hide="Hint for #1"] Use fermat, and make a substitution.[/hide]
[hide="Hint for #2"] What is the last digit of $ 2^{1990}$? [/hide]


number theory
relatively prime
number theory unsolved

Let n be an odd integer > 3. Let m be the number of integers that are relatively prime to n (Euler's totient function). Prove that there is a prime p such that p divides 2^m - 1, but p doesn't divide m.

$ n = q $ is an odd prime, then $ \phi (q) = q-1 $ is an even number.
Fermat's little gives that $ 2^{q-1}-1 \equiv 0 (mod \; q) $
And since $ q-1 $ is even....

Yeah, but if I have n = (p_1)^2 * (p_2)^2, for example, then m = p_1*p_2*(p_1 - 1)*(p_2 - 1) and we have that 2^m - 1 is divisible by p_1 and p_2 by Fermat. But how to show that there is some other prime in 2^m - 1 that is not in the even factors of m?

Never mind. I think I'm on the right path now. This was actually my initial idea that I threw away (for some strange reason). I write m = product(j = 1 to k)( p^(a_j - 1)*(p_j - 1) ) = 2^r*s, where s is odd. Now we factor 2^m - 1 in relatively prime integers, most of the form 2^((2^c)*s) + 1, where we have 2^s - 1 as the last product...

The problem is very simple if I use my idea, but change the problem to: prove that there is a prime p that divides 2^m - 1 but p doesn't divide n (instead of m as before). But if I have m instead of n, it becomes much more complicated...is that case true too?


[quote="4everwise"]If I randomly choose a 4-digit number not beginning with zero, what is the probabilty all 4 digits are distinct?[/quote]
[hide]
The total number of 4 digit numbers (not beginning with 0) possible is $9*10*10*10=9000$

The total number of 4 digit numbers (not beginning with 0) that has distinct digits is $9*9*8*7=4536$

So the probability is $\frac{4536}{9000}=\frac{63}{125}$[/hide]

[quote="nvidura"]considering the permutations say, _ _ _ _ is the number. To fill the thousandth you have 9 digits (excluding 0), hundredth also you have 9 digits (excluding the 1000th digit), tenth you have 8 digits and to fill the unit digit you have 7 numbers… if you multiply all

(9*9*8*7) = [b]4536 [/b][/quote]

Okay... what's the probability?

(welcome to AoPS  :) )


Let $f(x)=ax^2+bx+c$ with $a>0$. Equation $f(x)=x$ has two real roots $x_1$, $x_2$, which satisfy $x_1>0$, $x_2-x_1> \frac{1}{a}$.
If $0<t<x_1$, then determine the sign of $f(t)-x_1$.

i think it should be negative.
$f(x)-x$ has two  distinct real roots means that the discriminant $\Delta=(b-1)^2-4ac>0$. (i)

$x_1=\frac{-b+1-\sqrt{\Delta}}{2a}$
$x_2=\frac{-b+1+\sqrt{\Delta}}{2a}$
$x_2-x_1>1/a$ implies $\Delta>1$.

$x_1>0$ gives $-b+1-\sqrt{\Delta}>0$ and so $b<0$

Since from (i) we have a quadratic in $b$ that is always greater than zero, we have the discriminant is less than zero, or $4-4(1-4ac)<0$ from which we conclude $c<0$.

$f(0)=c <0$, and $f$ cannot change signs for $0\leq t <x_1$ because we have $x_2>x_1>0$.  (The function can only change signs once it passes through the zeros of the function).  Since $f(t)<0$ for all $t$ in that boundary, and $x_1>0$, we have $f(t)-x_1<0$.

I got positive, which means: $f(t)>x_1$.

$f(x)-x=ax^2+(b-1)x+c=0$, thus $x_1+x_2=-\frac{b-1}{a}=\frac{1-b}{a}$. Since we already have $x_2-x_1 \geq \frac{1}{a}$, so:
\[ x_1+x_1-x_2+x_1 \leq \frac{1-b}{a}-\frac{1}{a} \]
\[ x_1 \leq -\frac{b}{2a} \]
Thus $t<x_1< -\frac{b}{2a}$.

Since $f(x)$ is monotonically decreasing on $(-\infty, -\frac{b}{2a})$, so $f(t) > f(x_1)=x_1$.

OK that looks good to me.  another way you could arrive at the conclusion $x_1<-b/(2a)$ is to say $x_1<x_2$ and the vertex will have x-coordinate $-b/(2a)$, which will be in between $x_1$ and $x_2$, so $x_1<-b/(2a)$.

my previous post was wrong because the 'quadratic' in b was constant.


geometry
incenter
trapezoid
inequalities
circumcircle
arithmetic sequence
national olympiad

I congratulate my students, who took I and III diploma in 11-th grade, I, II and III diploma in 10-th grade!!!  :first:
I am so proud and happy! :) :D

These are selected problems.

9.2 http://www.mathlinks.ro/Forum/viewtopic.php?t=32160
Given 19 cards. Is it possible to write a nonzero digit on each card in such a way that you can compose from these cards an unique 19-digits number, which is divisible by 11?
([i]R. Zhenodarov, I. Bogdanov[/i])

9.4, 10.3 http://www.mathlinks.ro/Forum/viewtopic.php?t=32163
Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$.
([i]A. Badzyan[/i])

9.6, 10.6 http://www.mathlinks.ro/Forum/viewtopic.php?t=32164
Construct for each vertex of the trapezium a symmetric point wrt to the diagonal, which doesn't contain this vertex. Prove that if four new points form a quadrilateral then it is a trapezium.
([i]L. Emel'yanov[/i])

9.7 http://www.mathlinks.ro/Forum/viewtopic.php?t=32169
Is there an infinite arithmetic sequence $\{a_n\}\subset \mathbb N$ s.t. $a_n+...+a_{n+9}\mid a_n...a_{n+9}$ for all $n$?
([i]V. Senderov[/i])

8.8, 9.8, 11.8 http://www.mathlinks.ro/Forum/viewtopic.php?t=32161
a) 99 boxes contain apples and oranges. Prove that we can choose 50 boxes in such a way that they contain at least half of all apples and half of all oranges.
b) 100 boxes contain apples and oranges. Prove that we can choose 34 boxes in such a way that they contain at least a third of all apples and a third of all oranges.
c) 100 boxes contain apples, oranges and bananas. Prove that we can choose 51 boxes in such a way that they contain at least half of all apples, and half of all oranges and half of all bananas.
([i]I. Bogdanov, G. Chelnokov, E. Kulikov[/i])

10.2 http://www.mathlinks.ro/Forum/viewtopic.php?t=32170
Prove for all $x>0$ and $n\in\mathbb{N}$ the following inequality \[ 1+x^{n+1}\geq \frac{(2x)^n}{(1+x)^{n-1}}. \] ([i]A. Khrabrov[/i])

10.4, 11.3 http://www.mathlinks.ro/Forum/viewtopic.php?t=32162
Given $N\geq 3$ points enumerated with 1, 2, ..., $N$. Each two numbers are connected by mean of arrow from a lesser number to a greater one. A coloring of all arrows into red and blue is called [i]monochromatic[/i] iff for any numbers $A$ and $B$ there are no two monochromatic paths from $A$ to $B$ of different colors. Find the number of monochromatic colorings.
([i]I. Bogdanov, G. Chelnokov[/i])

10.7 http://www.mathlinks.ro/Forum/viewtopic.php?t=32168
Find all pairs $(a,b)$ of natural numbers s.t. $a^n+b^n$ is a perfect $n+1$th power for all $n\in\mathbb{N}$.
([i]V. Senderov[/i])

11.4 http://www.mathlinks.ro/Forum/viewtopic.php?t=32165
Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$. ([i]L. Emel'yanov[/i])

11.6 http://www.mathlinks.ro/Forum/viewtopic.php?t=32166
Construct for each vertex of the quadrilateral of area $S$ a symmetric point wrt to the diagonal, which doesn't contain this vertex. Let $S'$ be an area of the obtained quadrilateral. Prove that $\frac{S'}{S}<3$.
([i]L. Emel'yanov[/i])

11.7 http://www.mathlinks.ro/Forum/viewtopic.php?t=32167
Let $N$ be a number of perfect squares from $\{1,2,...,10^{20}\}$, which 17-th digit from the end is 7, and $M$ be a number of perfect squares from $\{1,2,...,10^{20}\}$, which 17-th digit from the end is 8. Compare $M$ and $N$.
([i]A. Golovanov[/i])

[quote="Myth"]I congratulate my students, who took I and III diploma in 11-th grade, I, II and III diploma in 10-th grade!!!  :first:
I am so proud and happy! :) :D
[/quote]I am also :D :first: May we know their names? :P

Note: it is not a final round.

So I wish good luck on the final!

11 grade
Belousov Kirill - I diploma
Morozkov Mikhail - III diploma

10 grade
Chernov Vadim - I diploma
Kuprin Sergei - II diploma
Smotrov Dmitrii - III diploma

Few questions that interest me: why does the numbering for each grade starts *.2? How much time are student given during the contest? Is there a link, where a *.pdf (no matter in Russian or English) containing these problems, can be found?

These are only selected problems, not the whole problem set.

I think contestants have 4-4.5 hours (I know it is 5 hours for final round) for 4 problems in each of two days.

You forgot to add information about who translated and typed it ;)

[quote="Myth"]These are only selected problems, not the whole problem set.[/quote]

Why don't you give the whole problem set ?

[quote="orl"][quote="Myth"]I don't want to translate all 32 problems. I have selected the most interesting problems, in my opinion.[/quote]

In the case you have the problems electonically available in Russian than post them in the Russian National Forum. Otherwise please scan them and provide them there as well. Thanks.  :)[/quote]

Please, if you will get all the 32 problems, although only in russian version, put them
in this forum. I will thank you! ;)

[quote="Myth"]I don't want to translate all 32 problems. I have selected the most interesting problems, in my opinion.[/quote]

   can you post all problems in Russian ? thank you .


If $a,b,c>0$ and $a+b+c=1$
prove that $\frac{a}{\sqrt{1-c}}+\frac{b}{\sqrt{1-a}}+ \frac{c}{\sqrt{1-b}}\leq\frac{5}{4}$.

it is equal to $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+ \frac{c}{\sqrt{c+a}}\leq\frac{5}{4} \sqrt{a+b+c}$
 it have been posted.and I remeber harizi give a nice solution.you can search.

Try to solve it in the form I gave it. I think we will have a better and nice solution.
BTW, I know harazi's solution


A certain kind of microbe dies or subdivides himself in two microbes of the same kind. If p=0.6 is the probability that the microbe subdivides himself, what is the probability that the descent he is generating  doesn't ever extinguish?

[quote="towbomb"]could you clarify what the probability is?  Other than that, I understand the problem.[/quote]

If you mean what was meant by 0,6 - that means 0.6, many countries use , and . in the opposite way from the rest.

p=0.6 means that if we have 10 microbes only 6 microbes divide themself.

Hm, nobody wants to try this? Ok, I give a hint at one possible solution:

[hide]Let x = P(starting with 1 microbe, everything eventually dies out) and y = P(starting with 2 microbes, everything eventually dies out). Find 2 equations relating y and x.[/hide]

solution in spoiler

[hide]

let x be the probability that a given microbe gives birth to an everlasting colony.  Then we can relate x to itself by noting that the probability that a microbe gives an everlasting colony, is the probability that it divides, and then at least one of it's children produces an everlasting colony( or 1 - the probability that no children produce an everlasting colony) so this is:

x = .6*(1-(1-x)^2).  Rewriting this and solving for x, we see that x can be either 0 or 1/3, so we are left with the task of determining which is correct.

in order to determine which is correct, we use a sequence that will have limit x as follows.  Let P_n be the probability that a given microbe colony lasts at least n generations.  Then by the same reasoning as above,

P_(n+1) = .6*(1-(1-P_n)^2), with the limit of P_n -> x.  By the definition of the P_n, it is evident that they are not increasing, so we must show that they are bounded below by 1/3.  But this can be proved using a trivial induction argument (based on the idea that if P_n > 1/3 => 1-P_n < 2/3 => (1-P_n)^2 < 4/9 => 1-(1-P_n)^2 > 5/9 => .6*(1-(1-P_n)^2) > 1/3 => P_(n+1) > 1/3)

so in fact the limit of P_n must exist, and since it is bounded below by 1/3, the probability must be 1/3 (since it was either 0 or 1/3).

[/hide]


algebra
polynomial
modular arithmetic
quadratics
number theory open
number theory

For all positive integers n, does there exist a prime p greater than the nth prime such that the first n primes are all not primitive roots mod p?

I think it should be true, but have no idea on how to prove it.

Let $ p_1, ... p_n$ be the first $ n$ primes and pick another prime $ q$.  Now $ p_i^{q\minus{}1} \equiv 1 \bmod q$, so no prime congruent to $ 1 \bmod (q \minus{} 1)^2$ can have $ p_1, ... p_n$ as primitive roots, and one can prove using cyclotomic polynomials (that is, without Dirichlet's theorem) that there are infinitely many such primes.

Sorry, I can't quite see how $ p_i^{q \minus{} 1}\equiv 1\pmod{q}$ implies that a prime congruent to $ 1 \pmod{(q \minus{} 1)^2}$ does not have $ p_i$ as a primitive root.
Like, for example $ 3^4\equiv 1\pmod{5}$ (obviously), but 3 is a primitive root mod 17. Or am I misunderstanding what you said?

Whoops.  Brain misfired.

Anyway, by quadratic reciprocity and Dirichlet's theorem we can find a prime $ q$ such that $ p_1, p_2, ... p_n$ are all quadratic residues $ \bmod q$, which implies they can't be primitive roots.


How could I show algebraically that a certain function has an inverse? Would it involve showing (algebraically) it's one-to-one? How would I start that?

EDIT: and err while we're on the topic, how would I algebraically show that this limit doesn't exist:

$ \lim_{x\rightarrow7}{\frac {|x \minus{} 7|}{x \minus{} 7}}$

Thanks.

For the first question, you could try swapping x and y and see if you can solve for y.

For the second question,
As $ x$ approaches 7 from the left, we have

$ \lim_{x\rightarrow7}{\frac {7 \minus{} x}{x \minus{} 7}} \equal{} \minus{}1$

As $ x$ approaches 7 from the right, we have

$ \lim_{x\rightarrow7}{\frac {x \minus{} 7}{x \minus{} 7}} \equal{} 1$

Since there are two possible limits, the limit DNE.

ahh.. so I just need to separate the absolute value like that.. Thanks a bunch!

hmm .. f(x) has the limitation if and only if $ lim_{x\minus{}>a\minus{}}f(x) \equal{} lim_{x\minus{}>a\plus{}}f(x)$
in this problem , we can show that $ lim_{x\minus{}>7\plus{}}f(x)$ is different from $ lim_{x\minus{}>7\minus{}}f(x)$ then the lim doenst exist

yea.
i think the arrow is \rightarrow
$ \rightarrow$

The absolute value is a piecewise function and it's almost always easier to treat it as such, splitting into the two cases.

\lim makes your limits look nice (and puts the _{stuff} underneath it), and \to also makes a respectable right arrow.


inequalities
trigonometry
inequalities proposed

Let $ \alpha$, $ \beta$ be acute angles. Find the maximum value of
\[ \frac{\left(1-\sqrt{\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha+\cot\beta}\]

[quote="April"]Let $ \alpha$, $ \beta$ be acute angles. Find the maximum value of
\[ \frac{\left(1-\sqrt{\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha\cot\beta}. \]
[/quote]

Something is wrong here. If we take $ x^{2}=\tan\alpha\tan\beta$ then $ \frac{\left(1-\sqrt{\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha\cot\beta}=(x(x-1))^{2}$ and since $ \tan\alpha \in (0;\infty)$ with $ \alpha \in \left(0;\frac{\pi}{2}\right)$ so given expression doesn't have maximum value.

In pdf file here: http://www.mathlinks.ro/Forum/viewtopic.php?t=161544 it is given different form of the expression: $ \frac{\left(1-\sqrt{\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha+\cot\beta}$. However this also doesn't have maximum value with given conditions.

hello, i think it is
$ \frac{\left(1-\sqrt{\tan(\alpha)\tan(\beta)}\right)^{2}}{\cot(\alpha)+\cot(\beta)}\le \frac{2}{27}$.
Sonnhard.

[quote="April"]Let $ \alpha$, $ \beta$ be acute angles. Find the maximum value of
\[ \frac {\left(1 \minus{} \sqrt {\tan\alpha\tan\beta}\right)^{2}}{\cot\alpha \plus{} \cot\beta}
\]
[/quote]
Let $ x\equal{}\tan \alpha$
      $ y\equal{}\tan \beta$
Then $ P\equal{}\frac{xy (\sqrt{xy}\minus{}1)^2)}{x\plus{}y}$
Let $ x\equal{}y$
$ P\equal{}\frac{x(x\minus{}1)^2}{2}$
If $ x\to\infty$ then $ p\to \infty$
Imply that $ P$ has not max value.

Maybe $ \alpha \plus{}\beta$is constant ?!

[quote="Extremal"]maybe we need minimal value?
 :maybe:[/quote]
I don't think that because $ \tex{min} P = 0$ when $ xy = 1\Longleftrightarrow \alpha + \beta = \frac {\pi}{2}$

click on the first pdf:
https://www.google.com/search?q=Chinese+Northern+Mathematical+Olympiad+2007&rlz=1C1HIJA_enUS786US786&oq=Chine&aqs=chrome.0.69i59j0l2j69i57j0l2.844j0j7&sourceid=chrome&ie=UTF-8

the problem is on the second page

Wait so does anyone know what the problem should be, cause I also got that it can approach infinity

It should be determine  the maximum value of $\[ \frac{\left(1-\sqrt{\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}\right)^{2}}{\cot\alpha+\cot\beta}\]$.


Two small spheres P and Q of equal radius have masses m and 5m respectively. They lie
on a smooth horizontal table. Sphere P is moving with speed u when it collides directly
with sphere Q which is at rest. The coefficient of restitution between the spheres is e,
where e > .
(a) (i) Show that the speed of P immediately after the collision is (u/6)(5e-1).

I got the answer v= (u/6)(1-5e), which it says is correct, then the markscheme says thus speed=(u/6)(5e-1).
Why have they negated it?

Minus only points out direction so it doesn't matter if you need it's absolute value.

[quote="JRav"]I haven't worked out the problem but it seems that your sign conventions may be different.[/quote]

I don't know what you mean by this.

Flame- that's what I thought but it seems to matter.

Exam on this tomorrow. :S

Then recheck your work. It's pretty common "to misplace" a minus in physics... at least to me :)

[quote="JasonSullivan"]I don't know what you mean by this.[/quote]
Sign convention is how you choose to orient your axes when doing calculations.  Since velocity is a vector and vectors have direction, you need to choose which direction is positive and which is negative.

[quote="JasonSullivan"]I got the answer v= (u/6)(1-5e), which it says is correct, then the markscheme says thus speed=(u/6)(5e-1).
Why have they negated it?[/quote]

You got the correct answer, but remember that the question asks for the speed, not the velocity.

When you got $ V_{P_f} \equal{}\frac{u}{6}(1 \minus{} 5\mathrm{e})$, that's the velocity, which is in the opposite direction wrt the initial velocity $ u$. Therefore, if when you set up the problem you chose the direction of $ u$ to be positive, the natural thing to do since in the beginning $ P$ is the only thing in motion, then $ V_{P_f} < 0$.  The speed, the solution to the problem, is $ |V_{P_f}|$.

If $ V_{P_f} < 0$, then $ |V_{P_f}| \equal{} \minus{}V_{P_f}$.

Hence, $ |V_{P_f}| \equal{} \frac{u}{6}(5\mathrm{e} \minus{} 1)$.


[quote="Treething"]Bob the Builder had a bag of dozen bananas on sale for 3 dollars. He adds some bananas to the bag, but does not change the price tag. Now how much does a dozen bananas cost?

http://media.putfile.com/banana39[/quote]

??

[hide]
n = number of banannas added.
$\frac{ 36 }{ 12 + n }$
[/hide]

Is this a joke?
What's the audio for anyway? :huh:

edit: huh? does this render incorrectly for anyone else?

The left side of the page is definitely not loading correctly for me, but I don't know exactly how to describe the problem.


Let $ x_n$ be a sequence with $ x_0\equal{}x_1\equal{}1$ and $ x_{n\plus{}1}\equal{}\frac{x_n^2\plus{}1}{x_{n\minus{}1}}$,$ n\geq 1$
Prove that $ x_n$ is a positive integer.

The fairly common invariant (such as in [[2007 AIME I Problems/Problem 14]]) can be used: subtracting \begin{align*}
x_nx_{n+2} &= x_{n+1}^2 + 1\\
x_{n-1}x_{n+1} &= x_n^2 + 1
\end{align*}
and rearranging yields $ \frac{x_{n-1} + x_{n+1}}{x_{n}} = \frac{x_n + x_{n+2}}{x_{n+1}}$. Then $ \frac{x_{n+1}+x_{n-1}}{x_n} = \frac{x_2 + x_0}{x_1} = 3 \Longrightarrow x_{n+1} = 3x_n - x_{n-1}$, so $ x_n \in \mathbb{Z}^{+}$.

This is the converse of a result like the type given [url=https://www.artofproblemsolving.com/Forum/weblog_entry.php?t=175257]here[/url].  See also, if you're interested, the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url].


induction
combinatorics unsolved
combinatorics

$n$ towns are connected with each others by $2n-1$ roads with a single direction. It is possible to go from any towns to any towns. Show we can remove a road such that the precedent property is still true.

The problem is quite easy:

In fact, if we have an oriented graph $G$ on $n$ vertices such that there is an oriented path from every vertex to every other vertex, then there is a subgraph $H$ of $G$ with at most $2n-2$ edges having the same property.

We will prove that statement by induction $n$. For $n=1$ it is trivial.

Now assume the statement is true for any integer $n'<n$. We will show that it is true for $n$. So, let $G$ be any graph on $n$ vertices with the above property.
Choose any two different vertices $A,B$ of the $G$. Then there is an oriented path from $A$ to $B$ and an oriented path from $B$ to $A$. Walking along these paths from $A$ to $B$ and back to $A$, we get an oriented cycle $C$ from $A$ to $A$ with length at least 2.

Now delete all the edges of the cycle $C$. Let $G_1,...,G_m$ be the connected components of the remaining graph. Then we now that the statement is true for all the graphs $G_1,...,G_m$, so we get subgraphs $H_1,...,H_m$. It is now easy to see that the subgraph $H$ of $G$ which is the union of $H_1,...,H_m$ and $C$ has the claimed property again and has at most $2n-2$ vertices.

This proves the statement. But, of course, the problem is just a special case of that problem.


How do i make link hidden so i just have to clik something like "this" and the link will open?

[code]
[url=INSERT WEB URL HERE] this [/url]
[/code]

For example,

[code]
[url=www.artofproblemsolving.com]this[/url]
[/code]

will produce [url=http://www.artofproblemsolving.com]this[/url]

When ever you want to find out how to use something like that, if you let your mouse thing hover over the button thing, the blue border thing above the text thing shows how to use it. For example, if you hover over "URL", it will say:

[code]Insert URL: [url]http://url[/url] or [url=http://url]URL text[/url] (alt + w)[/code]


Solve in R the following equation

$x^8y^4+2y^8+4x^4-6x^4y^4=0$

a really nice problem i think  :D  i liked it alot.

[hide]good appliance of AM-GM:
let $a=x^4$ and $b=y^4$ that $a,b\in\mathbb R_0^+$
then we know:
$6ab=a^2b+2b^2+4a\ge 3\cdot \sqrt[3]{ab^2\cdot 2b^2\cdot 4a}=6ab$
so $6ab\ge 6ab$ with equality only if $a^2b=2b^2=4a$
one trival solution is $(0,0)$ and we see if $a$ or $b$ is zero the othe is zero as well.
otherwise we can conclude:
$a^2b=2b^2 \Leftrightarrow a^2=2b$
$2b^2=4a \Leftrightarrow b^2=2a$
$a^2b=4a \Leftrightarrow ab=4$
and $a^2\cdot ab = 2b \cdot 4 \Leftrightarrow a^3=8$
so $a=2$ and $b=2$

the solutions are $\{(x,y)|(0,0),(\sqrt[4]2,\sqrt[4]2),(-\sqrt[4]2,\sqrt[4]2),(\sqrt[4]2,\sqrt[4]2),(-\sqrt[4]2,-\sqrt[4]2)\}$[/hide]

edit: only hid the solution

I don't suppose there is straight method of doing this. Can we make some perfect squares etc........ :(


function
trigonometry
AMC
USA(J)MO
USAMO
real analysis
real analysis unsolved

Let $ q\in\mathbb{Q}_+$
prove that there exists $ n\in\mathbb{N}$  and functions \[ f_1,\dots,f_n\in\{\sin,\cos,\tan,\arcsin,\arccos,\arc\tan\}\] such that \[ q=(f_n\circ\dots\circ f_1)(0)\]


This problem in a magazine :
    Solve the equation : (R)
   x^5 +10x^3 +20x - 18 =0

This can be since it is a polynomial with odd degree. It must have a real root.
Indeed, 0,7111 is an approximation of its unique real root (Maple).

Pierre.

Yes 0,7111 is a real root . Indeed , there is not another root . However , the way to find it is really difficult ....

Take $x=u-\frac2u$ then $0=x^5+10x^3+20x-18=u^5-\frac{32}{u^5}-18$.
From here $u^5=9\pm\sqrt{113}$. Then find a very nice ;) formula of x.


Does a "convex, equilateral hexagon" mean a regular hexagon?

No, try varying the lengths of opposite sides while keeping the length the same.

Equilateral only means the sides are congruent.  A regular polygon also has congruent angles.
A rhombus is an equilateral quadrilateral, but only squares are regular quadrilaterals.


A theater group has eight members, of which four are females. How many ways are there to assign the roles of a play that involve one female lead, one male lead, and three different inanimate objects that can be played by either gender?

There are $ 4$ choices for the female lead, and $ 4$ for the male lead. The 3 inanimate objects can be played by any gender. Since two people already have roles(the female lead and the male lead), there are $ 6$ choices for the first object, $ 5$ for the next, and $ 4$ for the third.

Now we just multiply.

$ 4 \times 4 \times 6 \times 5 \times 4 \implies \boxed{1920}$


ratio
geometry
power of a point
radical axis
perpendicular bisector
geometry unsolved

A circle center $O$ meets a circle center $O'$ at $A$ and $B.$ The line $TT'$ touches the first circle at $T$ and the second at $T'$. The perpendiculars from $T$ and $T'$ meet the line $OO'$ at $S$ and $S'$. The ray $AS$ meets the first circle again at $R$, and the ray $AS'$ meets the second circle again at $R'$. Show that $R, B$ and $R'$ are collinear.

Let $M$ be, the intersection point of the common tangent $TT',$ from the segment line $AB,$ as the radical axis of $(O),$ $(O').$

So, we have that $MT = MT'$ $($ $(MT)^{2}= (MA)\cdot (MB) = (MT')^{2}$ $)$ and hence $KS = KS',$ where $K\equiv OO'\cap AB.$

We denote as $A',$ $A'',$ the antidiametric ( = diametric opposite ) points of $A,$ with respect to $(O),$ $(O'),$ respectively.

It is easy to show that the points $A',$ $B,$ $A'',$ are collinear $($ $\angle ABA'+\angle ABA'' = 180^{o}$ $).$

$\bullet$  From similar right triangles $\bigtriangleup STO,$ $\bigtriangleup S'T'O',$ we have that $\frac{OS}{O'S'}= \frac{OT}{O'T'}= \frac{OA}{O'A}$ $,(1)$

Because now, the triangles $\bigtriangleup AOS,$ $\bigtriangleup AO'S',$ have $\angle ASO+\angle AS'O' = 180^{o}$ $($ from the isosceles triangle $\bigtriangleup AS S'$ $)$ and two homologous sidelines with the same ratio, we conclude that $\angle OAS = \angle O'AS'$ $,(2)$

From cyclic quadrilaterals $ABA'R$ and $ABR'A''$ $\Longrightarrow$ $\angle RBA' = \angle RAA' = \angle R'AA'' = \angle R'BA''$ $,(3)$

From $(3),$ we conclude that the points $R,$ $B,$ $R'$ are collinear and the proof is completed.

Kostas Vittas.

Nice proof vittasko! I have an different idea.

I will use vittasko's notation. Name the circle with center $O$ as $o$ and the circle with center $O'$ as $o'$. Since $SS'$ is a perpendicular bisector of $AB$ triangle $ASB$ is isosceles and we have: $\angle BSR = 180^{\circ}-\angle BSA = 180^{\circ}-(180^{\circ}-2\angle BAS) = 2\angle BAS$. We of course have $\angle BOR = 2 \angle BAR = 2\angle BAS$ so points $S, O, B, R$ lie on one circle, call it $\omega$. We analogously prove that points $S', O' B, R'$ lie on a circle $\omega'$. Now, consider the intersection $P$ of lines $OO'$ and $TT'$ (I don't consider the case if they are parallel, since the problem becomes very easy). Right-angled triangles $PST$ and $PTO$ are similar, so $|PT|^{2}= |PS|\cdot |PO|$. It means that the powers of point $P$ with respect to the circles $o$ and $w$ are the same, so $P$ lies on their radical axis, which is the line $BR$. In the same manner we prove that $P$ lies on a radical axis of circles $o'$ and $w'$, namely the line $BR'$. It follows that points $R, B, R'$ are collinear and the proof is finished.

Thank you dear [b][size=109]''Tomcio''[/size][/b], for a clever solution.

I like your proof very much and I will preserve it in the future. So, is it posibble to know your name and where do you come from?

Best regards, Kostas vittas.

It's nice to hear that you've liked my proof! My name is Tomasz Kobos and I am from Poland (Cracow) :).

If the two circles have the same radii then the problem holds trivially; suppose not.
Let $E = OO' \cap TT'$ be the exsimilicenter of the two circles. We use directed angles.

Note that $E$ and $S$ are conjugate pairs under inversion about $\omega_1$. Hence $A,O,R,E$ are concyclic. Thus
\[ \measuredangle{ERB} = \measuredangle{ERA} + \measuredangle{ARB} = \measuredangle{EOA} + \measuredangle{AOO'} = 0^\circ \]
So $E,R,B$ are collinear. Similarly, $E,R',B$ are collinear and the problem is finished.


Prove that $ \left\|\begin{array}{c}
\{x,y,z\}\subset (0,\infty )\\\\
xy\plus{}yz\plus{}zx\equal{}1\end{array}\right\|$ $ \Longrightarrow$ $ \boxed {\sum x(y^2\plus{}z^2)(x^2\minus{}yz)\ge 0}\ .$

[quote="Virgil Nicula"]Prove that $ \left\|\begin{array}{c} \{x,y,z\}\subset (0,\infty ) \\
 \\
xy \plus{} yz \plus{} zx \equal{} 1\end{array}\right\|$ $ \Longrightarrow$ $ \boxed {\sum x(y^2 \plus{} z^2)(x^2 \minus{} yz)\ge 0}\ .$[/quote]
Maybe without $ xy \plus{} yz \plus{} zx \equal{} 1$ $ ?$ :wink:

[quote="arqady"]$ \left\|\begin{array}{c} \{x,y,z\}\subset (0,\infty ) \\
 \\
xy \plus{} yz \plus{} zx \equal{} 1\end{array}\right\|\Longrightarrow\boxed {\sum x(y^2 \plus{} z^2)(x^2 \minus{} yz)\ge 0}\ .$

Maybe without $ xy \plus{} yz \plus{} zx \equal{} 1$ $ ?$ :wink:[/quote]
[color=darkblue][size=117][b]You are right, Arqady ![/b][/size] Observe easily that

$ \sum x(y^2 \plus{} z^2)(x^2 \minus{} yz)\ge 0$ $ \Longleftrightarrow$ $ \sum x^3(y^2 \plus{} z^2)\ge 2xyz\sum x^2\ .$

From $ y^2 \plus{} z^2\ge 2yz$ a.s.o. obtain $ x^3(y^2 \plus{} z^2)\ge 2x^3yz$ a.s.o. $ \Longrightarrow$

$ \sum x^3(y^2 \plus{} z^2)\ge 2xyz\sum x^2\ .$ [size=117][b]Is it O.K. ? Thank you ![/b][/size]  [/color] :bye:


AMC
USA(J)MO
USAMO
AMC 12
AMC 10
AIME
ratio

If more than 120 people in 10th grade or below score the floor value, then the AMC people are going to decide based on USAMO index, now how are they going to convert AMC12 scores to AMC10 scores?

Don't worry, you'll make it. The cutoff obviously won't be a 9.  ;)

From this one post I read before, I think that they take half from AMC 10 and half from AMC 12, or something like that.

I don't know, as this procedure has never been tested.  It's pretty unfair to use some people's AMC10 indices to compare against AMC12 indices.  I suspect if this becomes a problem, there will be some new system created to remedy the unfairness.  If a new system becomes necessary, I think you can count on the AMC folks to create one that's fair.  After all, they did a great job of doing that in 2002.  Yes, there might be a little chaos along the way (like roughly 75 people trying to book flights to Boston a week in advance), but it'll work out.

I looked at the AMC 10 and 12 of 2003.
13 questions from the AMC 10 were used in the AMC 12.
Answering 11 of these questons correctly from the AMC 12 would qualify someone to take the AIME. Wheras, If you took the AMC 10 and answered the same 11 questions correctly, you still would need to answer 6 more to qualify. This is why you should take the AMC 12.

The ratio of AMC12 students to AMC10 students that take the USAMO is around 2:1.

[quote="randomdragoon"]I'm pretty sure they want to encourage more people to take the AMC 12.

[quote]In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, it will be to their advantage to take the AMC 12 level contests.[/quote]
--from the official USAMO page[/quote]

oh w0rd, i bubbled my 137 AMC 12 instead of my 150 AMC 10

[quote="beta"]Well I'm not worrying but I'm just wondering since they specifically said about using index for tie-breaker on the USAMO manual

[quote="tcs09"][quote="MithsApprentice"] To simplify things, it was basically assumed that any qualifying AMC 12 score would be equated as a 150 on the AMC 10.[/quote]

That would be completely unfair. If you look at an AMC 10 and an AMC 12 in the same year you would find that there are enough questions on the 12 to qualify by [i]only[/i] answering [i]10[/i] questions ;) .

EDIT: That's AMC 10 questions, not 10 questions :D .[/quote]

Well AMC did warn that you will have more advantage taking the 12, plus for most people scoring above the AIME floor value, getting a 150 on AMC10 pretty much only takes a good day. So Daniel's suggesting is definitely possible.[/quote]

Well, would that method say that anything less than 150 on AMC 10 would be non qualifying?

A friend of mine last year had 1 small mistake on 10, but made 137 on the 12...

[quote="MithsApprentice"]Not that you have to worry Feiqi, but...

A discussion about how AMC 10 scores would be converted to AMC 12 scores came up on this board about 2 years ago when we were debating the floor for sophomores and frosh in 04. To simplify things, it was basically assumed that any qualifying AMC 12 score would be equated as a 150 on the AMC 10. I certainly don't know that this is the case and would not assume it to be so, but it seems reasonable that AMC 12 takers would have a decent advantage no matter how the scores were converted.

Moral of the story: You're going to make USAMO so stop worrying. But for the underclassmen that might actually be affected by a AMC + AIME selection of some sort, I'm willing to bet that the AMC 12 indices will get priority somehow.[/quote]

I know that people who take the AMC 12 will have at least some kind of advantage, but could a 112 on the AMC 12 make it?  Because that's kind of low compared to most people who are trying for USAMO.

[quote]    *  Selection for the USAMO will be made according to the follwing rules:

         1. The goal is to select about 250 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
         2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.
         3. The first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
         4. The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.
         5. The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
         6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.
         7. The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.
[/quote]

So I assumed that your higher score, AMC 10 or 12, will be used in the second selection. The reason that taking the 12 is more beneficial to qualify is because that the first 160 people selected are based on AMC12 + AIME indices only (although I'm sure the numbers will be tweaked a little bit to accomodate the increase in number of participants this year). [/quote]

[quote="Phelpedo"]
So I assumed that your higher score, AMC 10 or 12, will be used in the second selection. The reason that taking the 12 is more beneficial to qualify is because that the first 160 people selected are based on AMC12 + AIME indices only (although I'm sure the numbers will be tweaked a little bit to accomodate the increase in number of participants this year). [/quote]

Hmm wow people who took 12 are at a great disadvantage then, because for most people their 10 score will be higher than their 12 score. Being able to qualify from AMC12+AIME index is not beneficial at all since at best you set the floor value, you will qualify from the floor anyway. Yeah that method will be extremely unfair to people who take 12. It seems there is no fair way to do this, maybe AMC should just make AMC10 not part of AIME qualification, like AMC 8.

I'm strongly inclined to believe that AMC 12 and AMC 10 scores would not be treated equally if it came down to an index selection among underclassmen. The AMC website and teacher's manual clearly state that there would be an advantage associated with the 12 on terms of USAMO qualification. But beyond that, I'm not sure exactly not relevant this question is. A vast majority of USAMO potentials would have taken the 12 voluntarily anyway, and those that did not scored extremely well on the AMC 10. (I mean, personally I can't envision someone with a 120 AMC10 and, say, 7 AIME.)

[quote]It seems there is no fair way to do this, maybe AMC should just make AMC10 not part of AIME qualification, like AMC 8.[/quote]

There are seveal arguments against this, the first of which is the fact that the AMC12 is much less accessible for many students than the 10. A vast majority of current AMC10 takers would probably simply stop taking the AMCs if AIME qualification (which is quite an honor for those who don't consistently compete) was removed as an incentive.

[quote]That would be completely unfair. [/quote]

I fail to see how this is unfair. The AMC 12 content is a much better indicator of USAMO-level material than the AMC 10. Feiqi already pointed out that there was an explicit warning to those USAMO hopefuls who had the option of taking either. "Rewarding" (although, I believe that may even be an overstatement) underclassmen who chose the 12 would not be irrational.

While I didn't have this problem, I'd be willing to bet that some high schools didn't allow their <=10th graders to take AMC-12. There are always annoying things like that out there. I think it'd make more sense to take a proportionate amount from the 12 and the 10, depending on the number of AIME qualifiers from each.

[quote="beta"]Hmm wow people who took 12 are at a great disadvantage then, because for most people their 10 score will be higher than their 12 score. Being able to qualify from AMC12+AIME index is not beneficial at all since at best you set the floor value, you will qualify from the floor anyway. Yeah that method will be extremely unfair to people who take 12. It seems there is no fair way to do this, maybe AMC should just make AMC10 not part of AIME qualification, like AMC 8.[/quote]
Ya but what if you got a 150 on AMC 12 and a 7 on AIME? Then yes, you would probably set the floor value, but if you had taken the 10, then the floor value may have been set by someone else. So under extreme circumstances, taking is AMC 12 is a bit beneficial.

[quote="inyoung"]it's totally possible that a person with amc 10 with 120 gets an aime 7... i got a 122.5 on amc 10 and 8 on aime... so it is possible..
I substantially failed the 10's and 12's
but i got 114.5 on the 12 and 122.5 on the 10
which score do you think i should have used?
i used the 114.5[/quote]

OKay, I stand corrected. But you are certainly one in less than a handful.

hehe, in 8th grade (2 years ago) i had a 12b score of 112.5 after finding out about the test 2 days before and a 8 on the AIME

Yes they can.
The 120 floor qualifiers will probably be half and half or 2:1 between AMC 12 and AMC 10 takers.

I took the AMC12B at my house unoficially yesterday and scored 136.0.
I'm so taking the AMC 12 next year.

[quote="Rep123max"]Sophmores who only took the 12 can still qualify via the floor, right?[/quote]
Yes.


Obvious solutions are $ (x, y) \equal{} (0, 5)$ and $ (x,y) \equal{} (1,6)$ and are the only solutions for $ x < 2$
Since $ 33^x > 2^{5x}$ we have $ 2^{5x}|33^x \plus{} 31 \equal{} (32 \plus{} 1)^x \plus{} 32 \minus{} 1 \equal{} (2^5 \plus{} 1)^x \plus{} 2^5 \minus{} 1$
But since $ (2^5 \plus{} 1)^x \equiv 1 \bmod 2^{5x}$ We have $ 2^{5x} |32$ and then no solutions for $ x \ge 2$

[quote="Mathias_DK"]But since $ (2^5 \plus{} 1)^x \equiv 1 \bmod 2^{5x}$ We have $ 2^{5x} |32$ and then no solutions for $ x \ge 2$[/quote]
You are wrong. $ (2^5\plus{}1)^x\equal{}1\mod 2^{5\plus{}v_p(x)}\not \equal{} 2^{5x}$.
For example $ 2^8|(2^5\plus{}1)^7\plus{}32\minus{}1$.

[quote="Rust"][quote="Mathias_DK"]But since $ (2^5 \plus{} 1)^x \equiv 1 \bmod 2^{5x}$ We have $ 2^{5x} |32$ and then no solutions for $ x \ge 2$[/quote]
You are wrong. $ (2^5 \plus{} 1)^x \equal{} 1\mod 2^{5 \plus{} v_p(x)}\not \equal{} 2^{5x}$.
For example $ 2^8|(2^5 \plus{} 1)^7 \plus{} 32 \minus{} 1$.[/quote]
I see :(

obviusly $ (x,y) \equal{} (1,6)$ is only solution with $ x\leq1$ and $ y\leq6$
now suppose that $ x \equal{} a \plus{} 1$ and $ y \equal{} b \plus{} 6$ with $ a,b\in\mathbb{N}$
$ \Longrightarrow 33(33^a \minus{} 1) \equal{} 2^6(2^b \minus{} 1)$
since $ 2^6|33^a \minus{} 1$ we get that $ 2|a$ therefore $ 33^2 \minus{} 1|33^a \minus{} 1$ and $ 17|33^2 \minus{} 1$
hence $ 17|2^b \minus{} 1$ since $ 8|b$ and $ 5|2^8 \minus{} 1|2^b \minus{} 1$ so $ 5|33^a \minus{} 1$ therefore $ 4|a$ and therefore $ 2^7|33^a \minus{} 1$ and so we must have $ 2^7|2^6(2^b \minus{} 1)$ therefore $ b \equal{} 0$ and $ a \equal{} 0$ so the only solution is  $ (x,y) \equal{} (1,6)$

Don't know what i was thinking...

Start is the same as before:
Obvious solutions are $ (x, y) \equal{} (0, 5)$ and $ (x,y) \equal{} (1,6)$ and are the only solutions for $ x < 2$
Since $ 33^x > 2^{5x}$ we have $ 2^{5x}|33^x \plus{} 31 \equal{} (32 \plus{} 1)^x \plus{} 32 \minus{} 1 \equal{} (2^5 \plus{} 1)^x \plus{} 2^5 \minus{} 1$

But the end is slightly different(almost the same):
$ (2^5 \plus{} 1)^x \plus{} 2^5 \minus{} 1 \equal{} \sum_{j \equal{} 1}^{x} \binom {x} {j} 2^{5j} \quad \plus{} 2^5 \equal{}$
$ \sum_{j \equal{} 2}^{x} \binom {x} {j} 2^{5j} \quad \plus{} (x \plus{} 1) \cdot 2^5$
Then we have $ x \plus{} 1 \equal{} 2^k \cdot p$
$ 2^{5x} | \left(\sum_{j \equal{} 2}^{x} \binom {x} {j} 2^{5j \minus{} k \minus{} 5} \quad \plus{} p\right) \cdot 2^{k \plus{} 5} \iff 5(x \minus{} 1) \le k$
Which obviosly only holds for x = 1.

Just a little comment: If $ k \ge 6$ it isn't a problem either, since $ \binom{2^k \cdot p \minus{} 1}{2} \equiv 1 \bmod 4$...

[b]PS:[/b] What does $ v_p(x)$ mean?

[quote="Mathias_DK"]
$ 2^{5x} | \left(\sum_{j \equal{} 2}^{x} \binom {x} {j} 2^{5j \minus{} k \minus{} 5} \quad \plus{} p\right) \cdot 2^{k \plus{} 5} \iff 5(x \minus{} 1) \le k$
[/quote]
I think this step is wrong(note that $ \sum_{j \equal{} 2}^{x} \binom {x} {j} 2^{5j \minus{} k \minus{} 5}$ can be a odd number...)
[quote]PS: What does$ v_p(x)$ mean?[/quote]
$ v_p(x)\equal{}e_p(x)$

[quote="Mathias_DK"][quote="ali666"][quote="Mathias_DK"]
$ 2^{5x} | \left(\sum_{j \equal{} 2}^{x} \binom {x} {j} 2^{5j \minus{} k \minus{} 5} \quad \plus{} p\right) \cdot 2^{k \plus{} 5} \iff 5(x \minus{} 1) \le k$
[/quote]
I think this step is wrong(note that $ \sum_{j \equal{} 2}^{x} \binom {x} {j} 2^{5j \minus{} k \minus{} 5}$ can be a odd number...)[/quote]
It's only odd if $ k \equal{} 5$..  It's easy to see that $ 2^{5x}$ doesn't divide it anyway. For $ x \ge 6$ see my comment. So this time it is good enough :)[/quote]
can you tell me,why for $ k \equal{} 5$ , $ 2^{5x}$ doesn't divide it?I don't see why...

[quote="Sylwek"]Solve an equation in $ x,y \in \mathbb{N}_0$:
$ 33^x\plus{}31\equal{}2^y$[/quote]
Considering this equation modulo 315, we can easily derive that $(x,y)=(0,5)$ and $(1,6)$ are the only solutions.


geometry
Euler
circumcircle
incenter
angle bisector
geometry solved

Show that the line connecting feet of the external angle bisectors of ABC
    is orthogonal to line connecting centers of incircle and circumcircle.

    Col. Pestich.

Just a sketch:

It's been proven before on the forum that the line $OI$ is, in fact, the Euler line of the triangle formed by the external bisectors (i.e. it also contains the orthocenter of the intouch triangle). This means that the statement is equivalent to the fact that in a triangle, the trilinear polar of the orthocenter is perpendicular to the Euler line.

[quote="pestich"]Show that the line connecting feet of the external angle bisectors of ABC
    is orthogonal to line connecting centers of incircle and circumcircle.[/quote]

Is this really unsolved for you? I mean, there is a well-known theorem (see e. g. [url=http://de.geocities.com/darij_grinberg/Schroeder/Schroeder.html#6]Lemma 1 in the [Schroeder6] posting in my Schröder points database[/url]) stating that:

[b]Lemma 1.[/b] If X, Y, Z are the feet of the altitudes of a triangle ABC issuing from the vertices A, B, C, and if the lines YZ, ZX, XY intersect the lines BC, CA, AB at the points X', Y', Z', respectively, then the points X', Y', Z' lie on one line perpendicular to the Euler line of triangle ABC.

Now, we want to prove that, for every triangle ABC, the points where the external angle bisectors of the angles CAB, ABC and BCA meet the lines BC, CA and AB, respectively, lie on one line, and that this line is perpendicular to the line OI, where O is the circumcenter and I is the incenter of triangle ABC.

In fact, if we introduce the excenters $I_a$, $I_b$, $I_c$ of triangle ABC, then the external angle bisectors of the angles CAB, ABC and BCA are the lines $I_bI_c$, $I_cI_a$ and $I_aI_b$, and it remains to prove that the points where the lines $I_bI_c$, $I_cI_a$ and $I_aI_b$ meet the lines BC, CA and AB, respectively, lie on one line, and that this line is perpendicular to the line OI.

Now, since the internal and the external angle bisector of an angle are perpendicular to each other, we have $II_a\perp I_bI_c$, $II_b\perp I_cI_a$, $II_c\perp I_aI_b$; thus, the lines $II_a$, $II_b$, $II_c$ are the three altitudes of triangle $I_aI_bI_c$, the common point I of these lines is the orthocenter of triangle $I_aI_bI_c$, and the points A, B, C are the feet of the altitudes of triangle $I_aI_bI_c$. Hence, the points where the lines $I_bI_c$, $I_cI_a$ and $I_aI_b$ meet the lines BC, CA and AB, respectively, play a role in the triangle $I_aI_bI_c$ which is equal to the role of the points X', Y', Z' of Theorem 1 with respect to the triangle ABC. Hence, Theorem 1, applied to the triangle $I_aI_bI_c$, yields that the points where the lines $I_bI_c$, $I_cI_a$ and $I_aI_b$ meet the lines BC, CA and AB, respectively, lie on one line perpendicular to the Euler line of triangle $I_aI_bI_c$. Now, of course, everything will be clear when we show that the Euler line of triangle $I_aI_bI_c$ is the line OI.

But this is very easy to show: Since the points A, B, C are the feet of the altitudes of triangle $I_aI_bI_c$, they lie on the nine-point circle of the triangle $I_aI_bI_c$, so that the circumcircle of triangle ABC is the nine-point circle of the triangle $I_aI_bI_c$, and the circumcenter O of triangle ABC is the nine-point center of the triangle $I_aI_bI_c$. Also, we know that the point I is the orthocenter of the triangle $I_aI_bI_c$. Thus, the points O and I both lie on the Euler line of triangle $I_aI_bI_c$; in other words, the line OI is the Euler line of triangle $I_aI_bI_c$. Proof complete.

  Darij


algebra
polynomial
quadratics
absolute value
number theory unsolved
number theory

We are given two polynomials $f,g \in \mathbb{Z}[X]$ irreducible over $\mathbb{Z}[X]$. Prove (disprove?) that there exist infinitely many positive integers $m$ with the following property: $\forall_{t \in \mathbb{Z}_{m}}f(t) \equiv 0 \mod m \iff g(t) \equiv 0 \mod m$.

It followed from: there are infinetely many primes p, s.t. f(t) and g(t) irreducible mod p.

That is not true, Rust. Examples of irreducible integer polynomials which are reducible modulo [i]every[/i] prime have been given on the forum before.

Edit:

There seems to be something missing here, Megus. $f(x)=x$ and $g(x)=x+1$ offer an obvious counterexample.

Just assume that degrees of polynomials are strictly higher than $1$ - I couldn't find any counterexample for higher degrees. Also when $deg(f)=deg(g)=2$ then the theorem follows from quadratic reciprocity law (as having root $\mod m$ is equivalent to $\sqrt{t}$ for some $t$ belongs to $\mathbb{Z}_{m}$) - for absolute value of determinants sufficiently large (say more than 10  :) , but I haven't checked it exactly)

Interesting question, it seems that it would follow from QR for degree 2 as you said, does this generalize to all the other degree polynomials that have "explicit formulas" for their roots?  That is to say, the existence of roots reduces to a congruence in some mod?  Also, a random observation: if there are no roots in mod m there aren't any roots in any multiple of mod m which may make it easier to get infinitely many since the most obvious way to satisfy your condition seems to be to make sure our two polynomials have no roots period.

I wonder if anyone can come up with a solution or a counterexample for this?  :)


integration
function
calculus
trigonometry
limit
complex analysis

Find $ \int_0^1\frac {dx}{\sqrt [3]{1 \minus{} x^3}}$.
Use the contour http://www.worldofstock.com/closeups/TRC1668.php

Hi Spasty.  This is what I did:

1.  Define the three sheets:

$ s_1[z] = \frac {1}{\sqrt [3]{1 - z^3}}$

$ s_2[z] = \frac {Exp[ - 2\pi i/3]}{\sqrt [3]{1 - z^3}}$

$ s_3[z] = \frac {Exp[ - 4\pi i/3]}{\sqrt [3]{1 - z^3}}$

2. Get a good understanding on the sheet interleaving.  Best way for me to do that is to draw all sheets via the code.  That's the upper left plot below.

[code]
s1[z_] := 1/(1 - z^3)^(1/3);
s2[z_] := 1/(E^((2*Pi*I)/3)*(1 - z^3)^(1/3));
s3[z_] := 1/(E^((4*Pi*I)/3)*(1 - z^3)^(1/3));
reim = Im;
p1 = Plot3D[Evaluate[reim[f[z] /. z -> x + I*y]], {x, -2, 2},
     {y, -2, 2}, PlotStyle -> Red ]
p2 = Plot3D[Evaluate[reim[f2[z] /. z -> x + I*y]],
     {x, -2, 2}, {y, -2, 2}, PlotStyle -> Blue
     ]
p3 = Plot3D[Evaluate[reim[f3[z] /. z -> x + I*y]],
     {x, -2, 2}, {y, -2, 2}, PlotStyle -> Green
     ]
Show[{p1, p2, p3}, BoxRatios -> {1, 1, 2}]
[/code]

2.  Now, anneal continuations of the principal sheet $ s_1(z)$ into one analytic sheet by connecting the sheets between the principal branch-cuts:

$ A(z) = \{\begin{array}{rcl} s_1(z) & \mbox{for} & 0 < \theta \leq 2\pi/3 \\
s_2(z) & \mbox{for} & 2\pi/3 < \theta \leq 4\pi/3 \\
s_3(z) & \mbox{for} & 4\pi/3 < \theta \leq 2\pi \end{array}$

That's the top right plot.  The bottom right plot is looking directly down onto the top right plot.  You can see the Mercedes-Benz cut in the unit circle  although the plotting code shows false discontinuity outside the unit circle.

$ A(z)$ is now analytic outside the cut. This is the code to draw the imaginary sheet of $ A(z)$:

[code]
reim = Re;
slope1 = Sin[2 \[Pi]/3]/Cos[2 \[Pi]/3];
slope2 = Sin[4 \[Pi]/3]/Cos[4 \[Pi]/3];
p1 = Plot3D[Evaluate[reim[f[z] /. z -> x + I*y]],
     {x, -2, 2}, {y, 0, 2}, PlotStyle -> Red,
     PlotRange -> {{-2, 2}, {-2, 2}, {-2, 2}},
     RegionFunction -> Function[{x, y}, y >= slope1*x]]
p2 = Plot3D[Evaluate[reim[f2[z] /. z -> x + I*y]],
     {x, -2, 0}, {y, -2, 2}, PlotRange ->
       {{-2, 2}, {-2, 2}, {-2, 2}}, PlotStyle -> Blue,
     RegionFunction -> Function[{x, y}, y <= slope1*x &&
           y >= slope2*x]]
p3 = Plot3D[Evaluate[reim[f3[z] /. z -> x + I*y]],
     {x, -2, 2}, {y, -2, 0}, PlotRange ->
       {{-2, 2}, {-2, 2}, {-2, 2}}, PlotStyle -> Green,
     RegionFunction -> Function[{x, y}, y <= slope2*x]]
Show[{p1, p2, p3}, BoxRatios -> {1, 1, 2}]
[/code]

3.  Now, since $ A(z)$ is analytic except at the Benz cut, the following is true:

$ \mathop\int\limits_{Green} A(z)dz \equal{} \mathop\int\limits_{{\substack{Blue \\
and Red}}} A(z)dz$

(see lower left plot)

4.  The green contour can be evaluated directly by a suitable change of variable. The blue and red contours have to be done in 9 separate integrals but they collapse to a relatively simple expression.

I get for the final expression:

$ \left\{(e^{ \minus{} 4\pi i/3} \minus{} 1) \plus{} (e^{2\pi i/3} \minus{} 1) \plus{} (e^{2\pi i/3} \minus{} 1)\right\}\int_0^{1}\frac {1}{\sqrt [3]{1 \minus{} x^3}}dx \equal{} \frac {2\pi i}{\sqrt [3]{ \minus{} 1}}$

which yields the desired answer although I left some interpretation out.
[edit] made a few corrections

The result is $ \frac{2\pi}{3\sqrt{3}}$.
In general we have $ \int_0^1\frac{dx}{\sqrt[n]{1\minus{}x^n}}\equal{}\frac{\pi}{n\sin{\frac{\pi}{n}}}$.

The figure below I think helps to better understand this problem although I can't say I fully understand it myself.  If we let the principal value of $ \frac {1}{\sqrt [3]{1 - z^3}}$ be $ P(z)$ then we can represent $ A(z)$ in terms of $ P(z)$ as the three functional regions shown in the figure.  The light-red background represents the function $ A(z)$ which is now seen as ``continuous and analytic'' except for the white branch-cuts lines.  The dashed lines represent the demarcation lines for each functional region controlled by the argument of $ (1 - z^3)$. We then take the limit of the integrals over the Mercedes-Benz contour $ M$, as this contour approaches a contour $ C$ exactly ON the branch cuts and around the branch points in the limit.  It's now a simple matter to represent the integral over each leg of the contour in terms of the functional region the contour resides in.  For example, take the green leg $ G_1$ extending from the branch point $ e^{4\pi i/3}$ to the origin.  In the limit, we let $ z = re^{4\pi i/3}$ and integrate the function $ P(z)e^{ - 4\pi i/3}$:
\[ \mathop\lim\limits_{M\to C}\mathop\int\limits_{G_1} A(z)dz = \int_{1}^0 \frac {Exp[ - 4\pi i/3]}{\sqrt [3]{1 - r^3}}(e^{4\pi i})dr = - \int_0^1 \frac {1}{\sqrt {1 - r^3}}dr
\]
Likewise, for the leg $ B_2$ shown in the figure, let $ z = re^{4\pi i/3}$:
\[ \mathop\lim\limits_{M\to C}\mathop\int\limits_{B_2} A(z)dz = \int_{0}^1 \frac {Exp[ - 2\pi i/3]}{\sqrt [3]{1 - r^3}}(e^{4\pi i})dr = e^{2\pi i/3}\int_0^1 \frac {1}{\sqrt {1 - r^3}}dr
\]
The same is done for the other straight legs.  In the limit, the contours around the branch points tend to zero because the integrands can be shown to tend likewise to zero although in the case of the branch point at $ z = 1$ a little more work is involved I think.


linear algebra
matrix
limit
linear algebra unsolved

Let $|||\cdot|||$-be usual operator norm on $M_n(\mathbb{C})$. Prove that $|||A|||^2=max({|\lambda|, \lambda \in Spec(A*A)})$.
I can prove it using polar decomposition and than the fact that every matrix $A$ can be represented in the form $A=UDV$, where $U,V$-unitary, $D$-diagonal.
I'm interested whether there is easier approach non-using for example polar decomposition.

Prove first that for a symmetric matrix $T$ we have $max_{N(x)=1}{<x,T(x)>}$ is $(max_{a in Sp T}{abs(a)})^2$ (this is very easy, just diagonalize it in orthonormal basis) and then observe that $N^2(AX)=<X,A^{*}AX>$.

Diagonalizing means using the spectral decomposition of a Hermitian matrix, and this is precisely what eugene doesn't want to do :). I don't know a [b]completely[/b] elementary proof, based only on simple algebraic manipulation. Am I allowed to use the formula for the spectral radius of a linear operator $T$ for example? (i.e. the fact that it's $\lim_{n\to\infty}\|T^n\|^{\frac 1n}$) Just how "easy" do you want the proof to be, eugene? These types of questions are always a bit murky.

No, he said only that he does not want to use polar decomposition, not spectral matrix theorem.

Thanks you both for you replies, i think the spectral decomposition of a Hermitian matrix will be ok.


all i can say at present is that this result is stronger than schur's inequality :D  :P

Very nice and very difficult inequality.
It suffices to show that
$\sum{a(a-b)(a-c)}\geq \sum\frac{bc(b-c)^2}{2(b+c)}$,
that is
$\sum({b+c-a-\frac{bc}{b+c})(b-c)^2}\geq0$.
Suppose that  $a\geq b\geq c$ and have
$(a+b-c-\frac{ab}{a+b})\geq0$
and
$(b+c-a-\frac{bc}{b+c})+(c+a-b-\frac{ca}{c+a})>0$ .

Yes,that is what I have done too. :) And
[tex] (b+c-a-\frac{bc}{b+c})+(c+a-b-\frac{ac}{a+c}) \geq 0 [/tex]

While trying to solve this problem, i attempted to prove the stronger inequality[tex]\sum a(a-b)(a-c)\geq \sum bc(\sqrt{b}-\sqrt{c})^2[/tex], but failed. Is it possible for this inequality to be true?

I found that it is not true in the case $a=b=1$ and $c=0.1$.

Just one more thing, the general form for three numbers is:
[tex]\sum a^k(a-b)(a-c)\geq \sum \frac{b^{k}c^{k}(b-c)^2}{2(b^{k}+c^{k})}[/tex] where $k \geq1$:)

[quote="Anh Cuong"]Give $ a,b,c\geq0$. Prove that

$ a^3\plus{}b^3\plus{}c^3\plus{}3abc\geq bc\sqrt{2b^2\plus{}2c^2}\plus{}ca\sqrt{2c^2\plus{}2a^2}\plus{}ab\sqrt{2a^2\plus{}2b^2}.$[/quote]$ \left(3b^2\plus{}2bc\plus{}3c^2\right)^2\minus{}8(b\plus{}c)^2\left(b^2\plus{}c^2\right)\equal{}(b\minus{}c)^4\geq0\Longrightarrow$

$ \sqrt{2b^2\plus{}2c^2}\leq \frac{3b^2\plus{}2bc\plus{}3c^2}{2(b\plus{}c)}.$[quote="Vasc"]It suffices to show that[/quote]$ 3abc\plus{}\sum a^3\minus{}\sum \frac{bc\left(3b^2\plus{}2bc\plus{}3c^2\right)}{2(b\plus{}c)}$

$ \equal{}\sum{\frac{(b\minus{}c)^2(b\plus{}c\minus{}a)^2\left[3\left(a^2\plus{}bc\right)\plus{}5a(b\plus{}c)\right]}{8(b\plus{}c)(c\plus{}a)(a\plus{}b)}}\geq0.$

By the way, Anh Cuong's inequality holds for [b]real[/b] $ a,b,c$ with $ a\plus{}b\plus{}c\geq 0\leq abc.$

Given $ a, b, c \geq\ 0.$ Prove that:
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc \geq\ ab\sqrt[3]{4(a^3\plus{}b^3)} \plus{} bc\sqrt[3]{4(b^3\plus{}c^3)} \plus{} ca\sqrt[3]{4(c^3\plus{}a^3)}$
Is it trues ?  :wink:

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq\ ab\sqrt [3]{4(a^3 \plus{} b^3)} \plus{} bc\sqrt [3]{4(b^3 \plus{} c^3)} \plus{} ca\sqrt [3]{4(c^3 \plus{} a^3)}$
Is it trues ?  :wink:[/quote]

It's wrong for $ a\equal{}b\equal{}1,1$ and $ c\equal{}0,8$.

[quote="Ji Chen"][quote="Anh Cuong"]Give $ a,b,c\geq0$. Prove that

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\geq bc\sqrt {2b^2 \plus{} 2c^2} \plus{} ca\sqrt {2c^2 \plus{} 2a^2} \plus{} ab\sqrt {2a^2 \plus{} 2b^2}.$[/quote]$ \left(3b^2 \plus{} 2bc \plus{} 3c^2\right)^2 \minus{} 8(b \plus{} c)^2\left(b^2 \plus{} c^2\right) \equal{} (b \minus{} c)^4\geq0\Longrightarrow$

$ \sqrt {2b^2 \plus{} 2c^2}\leq \frac {3b^2 \plus{} 2bc \plus{} 3c^2}{2(b \plus{} c)}.$[quote="Vasc"]It suffices to show that[/quote]$ 3abc \plus{} \sum a^3 \minus{} \sum \frac {bc\left(3b^2 \plus{} 2bc \plus{} 3c^2\right)}{2(b \plus{} c)}$

$ \equal{} \sum{\frac {(b \minus{} c)^2(b \plus{} c \minus{} a)^2\left[3\left(a^2 \plus{} bc\right) \plus{} 5a(b \plus{} c)\right]}{8(b \plus{} c)(c \plus{} a)(a \plus{} b)}}\geq0.$

By the way, Anh Cuong's inequality holds for [b]real[/b] $ a,b,c$ with $ a \plus{} b \plus{} c\geq 0\leq abc.$[/quote]

Congratulations! Incredible Solution!
but how did you know at first
$ \left(3b^2 \plus{} 2bc \plus{} 3c^2\right)^2 \minus{} 8(b \plus{} c)^2\left(b^2 \plus{} c^2\right) \equal{} (b \minus{} c)^4\geq0$

It is unbelievable identity.. maybe it is famous.. sorry for my ignorance..

[quote=can_hang2007][quote="Anh Cuong"]Give $ a,b,c\geq0$.Prove that:
$ a^3\plus{}b^3\plus{}c^3 \plus{} 3abc \geq ab\sqrt{2(a^2\plus{}b^2)}\plus{}bc\sqrt{2(b^2\plus{}c^2)}\plus{}ca\sqrt{2(c^2\plus{}a^2)}$ :)[/quote]

This inequality is [b]Problem 2.44[/b], page 98 of new book (where it is presented with three simple AM-GM proofs):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.[/quote]

<=>
$x,y,z>0$,prove that

\[\frac{1}{2}\,\sqrt {2}\geq {\it \sum} \left( xy\sqrt {{\frac {{x}^{2}+{y}^{2}}{
 \left( {x}^{3}+{y}^{3}+{z}^{3}+3\,xyz \right) ^{2}}}} \right) \]

$W=xy$

By Holder,need to prove
\[\frac{1}{2}\geq {\it \sum} \left( xy \right) {\it \sum} \left( {\frac {xy \left( {
x}^{2}+{y}^{2} \right) }{ \left( {x}^{3}+{y}^{3}+{z}^{3}+3\,xyz
 \right) ^{2}}} \right) \]
<=>
\[bds=\left( {x}^{3}+{y}^{3}+{z}^{3}+3\,xyz \right) ^{2}-2\, \left( xy+xz+y
z \right)  \left( {x}^{3}y+x{y}^{3}+{y}^{3}z+y{z}^{3}+{x}^{3}z+x{z}^{3
} \right) \geq 0\]
<=>
\[\frac{1}{2}\,{\it \sum} \left(  \left( z-x \right) ^{4} \left( -z-x+y \right)
^{2} \right) +2\,{\it \sum} \left( {x}^{3} \left( y+z \right)  \left(
x-y \right)  \left( x-z \right)  \right)\geq 0\]

can do....

Or this:

${\it \sum} \left(  \left( y-z \right) ^{2} \right) {\it bds}$
$={\it
\sum} \left(  \left( y-z \right) ^{4} \left( -y-z+x \right) ^{2}
 \left( {y}^{2}+{z}^{2} \right) +\frac{5}{4}\, \left( y-z \right) ^{2} \left(
{y}^{2}-2\,xy+yz+{z}^{2}-2\,xz+{x}^{2} \right) ^{2}yz+\frac{3}{4}\, \left( x-y
 \right) ^{2} \left( {z}^{2}-{x}^{2}-{y}^{2}+xy \right) ^{2}xy
 \right)$
$ \geq 0$

so
\[bds\geq 0\]
Done!

[quote=Mildorf]Does the solution not also use $c+a-b-\frac{ca}{a+c} > 0$ after cancelling $\left(b-c\right)^2 \cdot \left( \left(c+b-a-\frac{bc}{b+c}\right) + \left(c+a-b-\frac{ac}{a+c}\right)\right) \ge 0?$

My question: what inspired this solution? How did you come up with the [incredibly sharp!] substitution

$\frac{3(b+c)^2-4bc}{2(b+c)} \ge \sqrt{2b^2 + 2c^2}?$[/quote]

$\sqrt{2b^2+2c^2} = \sqrt{(b+c)^2+(b-c)^2} = (b+c)\sqrt{1 + \frac{(b-c)^2}{(b+c)^2}} \le (b+c) \left(1 + \frac{1}{2}\frac{(b-c)^2}{(b+c)^2}\right)
= \frac{3(b+c)^2-4bc}{2(b+c)}$
since $\sqrt{1+x} \le 1 + \frac{1}{2}x$  for $x\ge 0$.


I was wondering whether anyone in here is a libertarian and what you think of this ideology. I believe this political philosophy is more superior then the 2 main political parties in the US because it is in favor of pure free markets and believes in the full liberty of the individual, it is like economically conservative and politically liberal.

I'm not at all a pure free market libertarian but very much so in terms of person freedoms. I guess that makes me a leftist libertarian.

You're just a liberal then.

Liberal = socially liberal + economically liberal(big government, welfare, medicare, more controls in the economy)
Libertarian = socially liberal + economically conservative(small government, no taxes, etc etc)

I am personally a liberal; while I think to some level, the economic policies advocated by liberals(like rawls) are quite utopian, I find the minimalist state libertarianism(by Nozick, the libertarian party did not exist before his "Anarchy, State, and Utopia") much more problematic. To clarify, I am refering to underlying philosophy rather than party platform.

The problem with Nozick begins with his assumption that property transfers have absolute legitimacy, and private property is absolutely inviolable; However, there's several problems with this approach: 1. It assumes that there's an initial transaction of property which is absolutely legitimate; we can never verify this. But furthermore, historically, inequalities have existed between rich countries and poor countries and thus nozick's own theory would justify market intervention to recify those inequalities. In addition, the very assumption of the inviolability of property can be called into question: why is property inviolable? What if at minimal expense to you, the lives of others can be improved greatly? Also, what if you were simply born into opulence while others were born into poverty, what justifies your absolute right to this property?

At any rate, that's my short list of a few reason I disagree with libertarianism.

why is property inviolable? What if at minimal expense to you, the lives of others can be improved greatly? Also, what if you were simply born into opulence while others were born into poverty, what justifies your absolute right to this property?

To the first statment, a person acting in their own rational self interest does not want to have a minimum expense to benefir others greatly (Unless for some reason it gives them a marginal benefit which I doubt happens to most people).

To the second statement, in certain aspects you can say that those born into riches because of ancestry is unfair but free markets are the only normal and free way of doing it. Its unfair that some people get less chances from the beginning but, suppose you have an ancester who is rich and wants to give you money, it is obvious that he/she has the full right to do so which makes this transaction fair. This gives unfair opporunities to others but the actual events that make those opportunites arise are fair.

Antoni, why r u not for free markets? Are you a socialist?

Hello:
1. The first question is a question of ethics. Is it ethical that you should not care about others when helping them is minimal expense to you?

2. Well, no, clearly free market isn't the only way to lessen those inequalities. The solution is to have a welfare system(and I am not suggesting any implementation here) so that people who are born into poverty can have some chance of escaping from it. The question here is in association with free will:

Let us assume that people are morally responsible for choices that they themselves make(so free will, at some level, is a presumption), then people cannot be held morally responsible for choices they do not make, such as where they are born, who they are born to, and the conditions in which they are born. You see, no one chooses their ancestors(or I would've chosen Bill Gates). So why would it be that you deserve your great riches more than I do, when I had no choice in doing so? It is your good fortune you were born richly, but to say the way you come upon it is necessarily legitimate(morally) is a stretch.

Furthermore, keep in mind that the unbridaled free market capitalism is rife with inequalities,  that is, unjust property transactions(also, note that there is no criterion for just property transaction), but I think we can agree child labor and such are unjust.

I might miss some stuff that I want to write back but here is some of it:

No, its fair that if one person was born in a rich family and another in a poor family that it happends. I mean that it is not fair from the perspective that the one in the rich family might not be better (more productive because of greater human capital) then the poor one but it is the only natural way for a transaction to happen in the case that a relative has the full right to provide for their own son, daughter or whoever else it might be.

Yes, free markets contribute to inequality greatly but who says that it isnt fair? If someone is more demanded in the market then someone else, why should they not receive more? Well guess what, they should. The same argument can be made that the rich and poor have to pay an equal tax, and guess what, that is the only fair solution.

Cant remember what else I wanted to say, maybe I will post again.

In response to hello's last post:

1. Altruism isn't ethical?  In this case, altruism (allowing a loss of one's own resources in order to benefit others with fewer resources) is simply one way of using one's resources.  This is certainly ethical, especially in a free market where individual choice regarding resource use is valued.

2. No, welfare isn't strictly "fair" in a free-market sense.  But then again, the practice of inheriting money is only "half fair": allowing people to pass on their money and resources is fair, but allowing descendents to receive it is arguably unfair, since they haven't earned it.  Secondly, it could be said that the benefits of greater income equality produced by a welfare system outweigh the disadvantage of a less free market (although the other side of that argument has merit as well).  Thirdly, keep in mind that the teachings of some religions can be construed to mean that [i]not[/i] having a welfare system is neither fair nor ethical.  I personally subscribe to all three of these arguments.

1. I think altruism is fine if the person giving to another is doing it by free will, not by force. For example, taxation, progressive taxation, is not free will but by force.

2. I agree that inhertinance is partially fair, but it is given by free will and fair by free markets. As far as welfare vs markets it depends who u ask. The rich like markets& the poor dont.

[quote="hello"]1. The first question is a question of ethics. Is it ethical that you should not care about others when helping them is minimal expense to you?

Yes, it is ethical, at least by the free market system,  because a rational individual behaves in their own self-interest, you should read Rand who will tell you that oh boy its very ethical and that altruism isnt ethical

2. Well, no, clearly free market isn't the only way to lessen those inequalities. The solution is to have a welfare system(and I am not suggesting any implementation here) so that people who are born into poverty can have some chance of escaping from it. The question here is in association with free will:

Welfare will decrease inequalities but it will do so by being unfair to their rich by such methods as higher tax rates for the rich which is not a free market principle.[/quote]

1. I have indeed read Rand, and I don't agree with her(But I must admit I've only read the fountainhead, can you give a logical derivation of why selfishness is the only moral philosophy? If you do, I will try to refute it.). You should realize that there are different system of ethics. And one in which you are not in any way obligated to help others(ever) doesn't coincide with most people's intuition. (I am giving an intuitionist argument here)

2. Silas responds to this.

But I'd like to point out a flaw within your thinking: Essentially, all your arguments have been tautological, you have given justification for the free market, by using the free market.
As an example, here:
[quote]
 I agree that inhertinance is partially fair, but it is given by free will and fair by free markets.
[/quote]

to 1 of your second post: What you have highlighted is the clash between equality and freedom. Which one you value higher is of course dependent on your own philosophical convictions and I do not imagine I could change them. However, from a policy stand point, I believe equality makes a lot more sense. You may have heard Mill's famous saying "your right to swing your arm ends at another person's nose". I think most of us would agree with this, nothing but anarchy would result if some state(a monopoly on force) enforced this. Thus there are limits to freedom, reasonable limits at that. Certainly no one should have unfettered freedom to do whatever they want(not in a state anyways, and we are talking about states)

Progressive Taxation is critical towards providing social services and a government. If you are advocating a flat tax rate, that is essentially regressive because only rich people will be able to afford basic things. From a policy perspective already, that's pretty bad; when people are dying on the streets, the government will probably go down pretty soon. But from a moral perspective, I do not believe you can give a sufficient justification for why selfishness is the only moral philosophy.

and just as a side note, I think you misinterpret what I say about free will, I make a presumption of some level of free will in my arguments; I do not assume it should be unrestrained.

Anyways, you ignore the argument I make about people not choosing their circumstances and moral responsibility:

[quote]Let us assume that people are morally responsible for choices that they themselves make(so free will, at some level, is a presumption), then people cannot be held morally responsible for choices they do not make, such as where they are born, who they are born to, and the conditions in which they are born. You see, no one chooses their ancestors(or I would've chosen Bill Gates). So why would it be that you deserve your great riches more than I do, when I had no choice in doing so? It is your good fortune you were born richly, but to say the way you come upon it is necessarily legitimate(morally) is a stretch.

Furthermore, keep in mind that the unbridaled free market capitalism is rife with inequalities, that is, unjust property transactions(also, note that there is no criterion for just property transaction), but I think we can agree child labor and such are unjust.[/quote]

Its true that extreme free markets will have very little support by voters at the polls but it is the most fair way of doing things. Intervention using welfare and other sch things is unfair because its taxing the rich and giving to the poor. Like I said, is there a justification from taking from some people without their will and giving to others? Even if they r easily capable of doing it, it doesnt mean they have to. The ideal system is free markets because it guarantees freedom (which is more important then equality) and is purely fair as apposed to social welfare, I understand that it has contributed to inequality which might not be popular among the poor but why should the more productive  not receive more then the less productive? I believe that they should receive more, which justifies my argument fully. All exchanges should be voluntary.

[quote="hello"]1. I think altruism is fine if the person giving to another is doing it by free will, not by force. For example, taxation, progressive taxation, is not free will but by force.
[/quote]

I think I misunderstood you before; my mistake.  Yes, progressive taxation is essentially altruism by force (although that phrase puts a pretty negative spin on it).  I happen to think that the benefits outweigh the disadvantages.  This topic does raise some interesting questions about imposing one's value systems on others, but that's a discussion for another thread.

[quote="hello"]2. I agree that inhertinance is partially fair, but it is given by free will and fair by free markets. As far as welfare vs markets it depends who u ask. The rich like markets& the poor dont.
[/quote]

Exactly.  By the way, I should clarify my last post in that I would not agree with using religious teachings as a reason for a secular government (like ours) to have a welfare system.  That's just one of my reasons for supporting it.  And on the inheritance question: I agree with you here, too.  I would also say that the partial unfairness of inheritance is one reason to have the estate tax.

Well, economists very often disagree based on their values rather then the results of their economics. I believe that it is important to use proper philosophical foundations before working on the actual economics.


Prove that among any $ 2^{n\plus{}1}$ natural numbers, there are $ 2^{n}$ natural numbers whose sum is divisible by $ 2^{n}$.

Perhaps you start out by taking residues mod $ 2^n$ and then arranging them in increasing order. Then you subtract a constant amount from each so the smallest is 0. Then I am stuck.