conics
ellipse
exterior angle
geometry
angle bisector
geometry proposed

[quote="thaithuan_GC"]Given the circle (O) and two fixed point A,B lie outside (O). Find the position of the point M on (O) such that the sum MA+MB attains its min value.[/quote]

Proposed & Own Problems is for questions you have a solution to. As for this one, I doubt it, since it is unsolvable with compass and ruler. It is equivalent to http://www.mathlinks.ro/viewtopic.php?t=143900 .

  darij

Sorry, I don't have a solution. Please help me to solve it. Who can? :help:

[quote="darij grinberg"]As I said it is unsolvable.

  darij[/quote]
Can you explain with more details? Why is this problem unsolvable? In addtion, please show the relation between my problem and problem in the link you mentioned. Thanks.

The point $ M$ on $ \left(O\right)$ that minimizes the sum $ MA\plus{}MB$ is one of the points $ P$ for which the tangent at $ P$ to $ \left(O\right)$ bisects the angle $ APB$ (because the ellipse with foci $ A$ and $ B$ passing through $ M$ must touch the circle $ \left(O\right)$ at $ M$ (since otherwise there would be some points $ Q$ on the circle $ \left(O\right)$ lying inside this ellipse and thus satisfying $ QA\plus{}QB<MA\plus{}MB$, contradicting the minimality of $ MA\plus{}MB$), so the tangent to the circle $ \left(O\right)$ at $ M$ must be the tangent to this ellipse at $ M$, i. e. the exterior angle bisector of the angle $ AMB$).

And as for the points $ P$ for which the tangent at $ P$ to $ \left(O\right)$ bisects the angle $ APB$, they all are (in general) not constructible with compass and ruler, as I showed at http://www.mathlinks.ro/viewtopic.php?t=143900 .

  darij


AMC AIME MIT college Yale Stanford USAMTS

What school would a good AIME score help?

Any school that knows what it is, so schools with a strong math / science program.  It won't help you get into a liberal arts school at all.  I will mention that MIT goes out of its way to specifically ask for AMC / AIME scores.

Stanford does not, unless they added something for it when they changed over to common app.  But they definitely know what it is, and will like to see high scores.

Are you sure it won't help at a liberal arts college?  They like people who are good at math, too.

I think Stanford has a section that asks for some number of additional standardized test scores of your choice, and they list the AMC as an example of a standardized test. I'm sure, by that broad definition, the AIME could be used as a standardized test, as well.

If you're ever worried that someone won't understand what the AIME is, you can always write it out, and all of a sudden it sounds more impressive. If this is on your resume and you did particularly well on it, you can add statistics / a [u]brief [/u]explanation of what it is.

That's pretty much what I did with the essays- I used the "Tell us about yourself" prompt to explain what my superlatives meant.
In practice, it didn't matter for me. I was in before they even looked at that part of the application.

At the time, UC Berkeley admitted half its class based purely on grades and test scores.

o haha i wish it was still like that. didn't you like make IMO though?

Yeah, I was hoping something to do with math, someplace where being successful in math contests (AMC) would help.

What about USAMTS?

I know Mathcounts doesn't really help: :lol: its Jr. HIgh

MIT, Caltech, and Brown were the schools I applied to that specifically asked for AMC scores.


Which cell organelles are found only in eukaryotes, which are found only in prokaryotes, and which are found in both?

prokaryotes have no nucleus but have other little helpers, while eukaryotes have cell-membrane bound nucleuses, at least that's all  i can remember at this moment. Prokaryotes depend on many cells working together to actually work....

I don't think that helped, u probably knew that but oh well

Is that the only difference though? I somehow know that there are a couple more I think.

basically eukaryotes are more "independent" while prokaryotes need to work with others because they don't have nucleus

I am gonna research on that later today. because i dont know much about this either

Prokaryotes have no membrane-bound organelles. They may have a cell wall, plasma membrane, thylakoids to generate energy, cilia/flagella, freely floating ribosomes.

Instead of a nucleus, they possess a nucleoid region. They also have a cytoplasm.

cool. thank you everyone!

i have ap biology test tomorow.

Prokaryotes also usually reproduce by binary fission, while eukaryotes use mitosis (or meiosis).  This is because of the different DNA structure.

[quote="ashpotter"]The DNA of Prokaryotes are in a string like form. Prokaryotes ofter are Archeon or Bacteria[/quote]
I believe you mean that Prokaryotes are either Archaea (or Archaebacteria or Eubacteria).  Note that some teachers prefer grouping bacteria into the Monera kingdom.


[color=blue]
Let $x_1 ,x_2 ,x_3$ be the roots of equation $x^3 - px^2 + qx - r = 0$
using Viete's relations, from:
$\left( {a + b + c} \right)\left[ {\left( {a + b + c} \right)^2 - 3\left( {ab + bc + ca} \right)} \right] = a^3 + b^3 + c^3 - 3abc \ \ \ \forall a,b,c\in\mathbb{R}$
for:
$A = \sqrt[3]{{x_1 }} + \sqrt[3]{{x_2 }} + \sqrt[3]{{x_3 }},B = \sqrt[3]{{x_1 x_2 }} + \sqrt[3]{{x_2 x_3 }} + \sqrt[3]{{x_3 x_1 }}$
we have:
$\left\{ \begin{array}{l} A\left( {A^2 - 3B} \right) = p - 3\sqrt[3]{r} \\ B\left( {B^2 - 3\sqrt[3]{r}A} \right) = q - 3\sqrt[3]{{r^2 }} \\ \end{array} \right.$
[/color]

For $x^3 - 4x^2 - 11x + 1 = 0$
 we have:
$\left\{ \begin{array}{l} A\left( {A^2 - 3B} \right) = 7 \\ B\left( {B^2 + 3A} \right) = - 14 \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} A^3 - 3AB = 7 \\ B^3 + 3AB = - 14 \\ \end{array} \right. \Leftrightarrow$

$\Leftrightarrow \left\{ \begin{array}{l} A^3 - 3AB = 7 \\ A^3 + B^3 = - 7 \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} A^3 - 3AB = 7 \\ B = \sqrt[3]{{ - A^3 - 7}} \\ \end{array} \right. \Leftrightarrow$

$\Leftrightarrow \left\{ \begin{array}{l} A^3 - 3A \cdot \sqrt[3]{{ - A^3 - 7}} = 7 \\ B = \sqrt[3]{{ - A^3 - 7}} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} A^3 + 3 \cdot \sqrt[3]{{\left( {A^3 } \right)^2 + 7A^3 }} = 7 \\ B = \sqrt[3]{{ - A^3 - 7}} \\ \end{array} \right.$

For $u = A^3$ we obtain equation: $u + 3\sqrt[3]{{u^2 + 7u}} = 7$

It is easy to prove that:
$\exists !u \in \mathbb{R}: u + 3\sqrt[3]{{u^2 + 7u}} = 7$
and
$u + 3\sqrt[3]{{u^2 + 7u}} = 7 \Leftrightarrow u = 1$

then: $A = \sqrt[3]{{x_1 }} + \sqrt[3]{{x_2 }} + \sqrt[3]{{x_3 }} = 1$
and $B = \sqrt[3]{{x_1 x_2 }} + \sqrt[3]{{x_2 x_3 }} + \sqrt[3]{{x_3 x_1 }} = - 2$

[quote="metru"]
For $u = A^3$ we obtain equation: $u + 3\sqrt[3]{{u^2 + 7u}} = 7$
[/quote]
Why? Explain please!  ;) Thanks.

Nice solution, Metru.
I meant to the next:
[quote="metru"]
For $x^3 - 4x^2 - 11x + 1 = 0$
 we have: $\left\{ \begin{array}{l} A^3 - 3AB = 7 \\ B^3 + 3AB = - 14 \\ \end{array} \right.$
[/quote]
Hence, $A^3=3AB+7$ and $B^3=-3AB-14.$
Hence, $A^3B^3=(3AB+7)(-3AB-14)\Leftrightarrow(AB+2)(A^2B^2+7AB+49)=0.$
Hence, $AB=-2$ and $A=1.$
Try the next:
$x_1,x_2,x_3$ are roots of equation $x^3-16x^2-57x+1=0.$
Prove that $\sqrt[5]{x_1}+\sqrt[5]{x_2}+\sqrt[5]{x_3}=1.$  ;)
Thank you for the interest.

For the fifth degree it would be too hard if not your suggestion, arqady.
$x^3-16x^2-57x+1$ has roots in the intervals $(-4,-2),(0,1),(18,20)$, so $x_1,x_2,x_3$ are all real.
$e_1=\sqrt[5]{x_1}+\sqrt[5]{x_2}+\sqrt[5]{x_3}\wedge e_2=\sqrt[5]{x_1x_2}+\sqrt[5]{x_2x_3}+\sqrt[5]{x_3x_1}\wedge e_3=\sqrt[5]{x_1x_2x_3}$
Using Newton identities and Viete's formulas
$e_3=-1$
$16=e_1^5-5e_1^3e_2-5e_2e_3+5e_1e_2^2+5e_1^2e_3$
$-57=e^5_2-5e_1e_2^3e_3+5e_2^2e_3^2+5e_1^2e_2e_3^2-5e_1e_3^3$
It suffices to check that $e_2=-2\wedge e_1=1$ satisfies both equations. QED

Cleaner formulation of metru's work.

$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$

So if $a^3,b^3,c^3$ are the three roots then:

$(a+b+c)^3-3(a+b+c)(ab+ac+bc)=0\implies(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$

$\implies(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]=0\implies a+b+c=0\vee a=b=c$

But since $x^3+3x^2-24x+1$ is note a perfect cube polynomial, so therefore we conclude $a+b+c=0$.

One interesting note:

$y=x+1\implies (y-1)^3+3(y-1)^2-24(y-1)+1=0\implies y^3=27y-27$

$3st=27$ and $s^3+t^3=-27$ means $s^3,t^3$ are roots of $u^2+27u+729=0$. This solves to $u=\frac{-27\pm 27i\sqrt{3}}{2}=27e^{\pm2\pi i/3}$.

So $x\in\left\{-1+6\cos\frac{2\pi}{9},-1+6\cos\frac{4\pi}{9},-1+6\cos\frac{8\pi}{9}\right\}$


If xy=c, 9xy=a, and 2a-5c=kxy, then k= ?

[hide]We see that 9c = a, so substituting,

18c - 5c = kxy.
13c = kxy.
since xy = c, k must be 13.[/hide]

[hide]i got k=13 too. i'm finally gettin the hang of this stuff :lol: [/hide]


calculus
derivative
integration
function
inequalities
calculus computations

x/ square root of (x^2 + 1) dx = square root of (x^2+ 1)

Just... do it. This is a routine exercise.

Trivial.
But I'm posting because I was thinking of something related the other day, just a diversion.

We want to solve $g(x)g'(x)=x$. Now integrate both sides, so $\int gg' dx=\frac{x^{2}}{2}+C$.
Now \[\int gg'=g^{2}-\int gg'\implies \int gg'=\frac{g^{2}}{2}\] (which could have been noted in reverse from chain rule on squares :roll: )
Thus \[\frac{g^{2}}{2}=\frac{x^{2}}{2}+C\implies g(x)=\sqrt{x^{2}+a}\] yields all function with a funny derivative like this.

Actually I was thinking about it after partial differentiating $\sqrt{x^{2}+y^{2}+z^{2}}$ in an inequality :)


polynomial
function
algebra
Galois Theory
superior algebra
superior algebra open

I've been struggling with this one for a while, as none of the "usual" methods for showing irreducibility seem to be applicable.

I'm fairly sure that you can't weasel the expression enough to apply Eisenstein, and my attempts to find an order n element of the Galois group have been fruitless.  Are there general techniques for this sort of question?

for x=0 it is equal to 1
and for x=1 it is equal to -(n-1)
so it has a zero since its continous.

Whether or not the function has a real zero is unrelated to the question of whether it is irreducible, that is, not the product of nontrivial polynomials.  (to be clear, I mean reducibility over the rational numbers.)

For example, the polynomial x^2 - 2 has two real zeros, but it is not the product of two nontrivial polynomials with rational coefficients.

[quote="-oo-"]I think the ground field is assumed to be $ \mathbb{Q}$.[/quote]

Yes; it's false over $ \mathbb{F}_{p}$, as $ {f}_{n}(x)$ has a root at $ 1$ for $ {p}|{n \minus{} 1}$

Here's an entertaining proof for when $ n\equal{}2^k\minus{}1$., $ k>\equal{}2$.

Consider $ f_{n}(x\plus{}1) \equal{} 1 \minus{} (x\plus{}1) \minus{} (x\plus{}1)^2 ... \minus{} (x\plus{}1)^n$.  We will apply Eisenstein with $ p\equal{}2$.  First observe that the constant term is $ 1\minus{}n\equiv 2\mod 4$.  Now, it's a fact that $ {2^k\minus{}1}\choose j$ is odd for all j. (this is a nice exercise in itself)  So we observe that $ [x^j]f_{n}(x\plus{}1)\equiv n\plus{}1\equiv 0\mod 2$, and we are done.

It's also possible to show that if we can apply Eisenstein to $ f_{n}(x\plus{}a)$ for some $ a$ and prime $ p$, then we must have $ n\equal{}2$ or $ n\equal{}2^k\minus{}1$.  Which is interesting, if not particularly useful.

If 2 is a primitive root mod some prime $ p$, then $ f_{p \minus{} 1}$ is irreducible, being irreducible mod $ p$: If $ \alpha$ is a root of $ f_{p \minus{} 1}$ over $ \mathbb{F}_p$, then $ \alpha^{p^k} \equal{} 2^k\alpha \minus{} (2^k \minus{} 1)$ is equal to $ \alpha$ iff $ p \minus{} 1|k$, so $ \alpha$ has degree $ p \minus{} 1$.  So this is a conjecturally infinite set of $ n$, at least.

Actually, we have the following for a prime $ p$: $ f_{p\minus{}1}$ is irreducible mod $ p$ iff 2 has order $ p\minus{}1$ mod $ p$ iff $ f_{p\minus{}1}$ is irreducible mod 2.  This is because mod 2, $ f_{p\minus{}1}$ is the $ p$-th cyclotomic polynomial $ \varphi_p$, and mod $ q$, $ \varphi_p$ splits into factors whose degree is the order of $ q$ mod $ p$.

Actually we can show $ f_p$ is irreducible for all primes $ p$.  Let $ \alpha$ be a root of $ f_p$ over $ \mathbb{F}_p$.  Then $ \alpha^{p\plus{}1}\equal{}2\alpha\minus{}1$, so $ \sigma(\alpha)\equal{}\alpha^p\equal{}\frac{2\alpha\minus{}1}\alpha$.  We find $ \sigma^k(\alpha) \equal{} \frac{(k\plus{}1)\alpha\minus{}k}{k\alpha\minus{}(k\minus{}1)}$.  Then $ \sigma^k(\alpha)\equal{}\alpha$ iff $ k(\alpha\minus{}1)^2\equal{}0$, which happens iff $ p|k$ (since $ \alpha\neq1$).  So again $ \alpha$ has degree $ p$ over $ \mathbb{F}_p$, hence $ f_p$ is irreducible mod $ p$.

Can we show that all of the roots are within some region in the complex plane, then get a contradiction if we can factor it?

Indeed!  Let $ h(z) \equal{} (1\minus{}z)f_n(z) \equal{} z^{n\plus{}1} \minus{} 2z\plus{}1$, and let $ g(z)\equal{}2z$.  Assuming $ n\geq 2$, have $ r^{n\plus{}1}\plus{}1 < 2r$ for $ r<1$ sufficiently close to 1.  Now on the circle of radius $ r$, $ |h(z)\plus{}g(z)|\equal{}|z^{n\plus{}1}\plus{}1| \leq r^{n\plus{}1}\plus{}1 < 2r \equal{} |g(z)|$, so $ h$ and $ g$ have the same number of roots inside this circle, namely 1.  Letting $ r\to 1^\minus{}$, we see that $ h$ (and therefore $ f_n$) has exactly 1 root $ \alpha$ inside the unit circle.  But if $ m_\alpha|f_n$ is the minimal polynomial of $ \alpha$, then the product of the roots of $ m_\alpha$ must be $ \pm1$, and the only way this can happen is if $ m_\alpha\equal{}f_n$ is irreducible.

(An algebraic integer such as $ \alpha^{\minus{}1}$ which lies outside the unit circle, but all its other conjugates lie inside, is known as a Pisot number.)

Very nice.  And it's convinced me to work on my complex analysis, as I did need to look up Rouché's theorem...

[quote="Eigenray"]Indeed!  Let $ h(z) \equal{} (1 \minus{} z)f_n(z) \equal{} z^{n \plus{} 1} \minus{} 2z \plus{} 1$, and let $ g(z) \equal{} 2z$.  Assuming $ n\geq 2$, have $ r^{n \plus{} 1} \plus{} 1 < 2r$ for $ r < 1$ sufficiently close to 1.  Now on the circle of radius $ r$, $ |h(z) \plus{} g(z)| \equal{} |z^{n \plus{} 1} \plus{} 1| \leq r^{n \plus{} 1} \plus{} 1 < 2r \equal{} |g(z)|$, so $ h$ and $ g$ have the same number of roots inside this circle, namely 1.  Letting $ r\to 1^ \minus{}$, we see that $ h$ (and therefore $ f_n$) has exactly 1 root $ \alpha$ inside the unit circle.  But if $ m_\alpha|f_n$ is the minimal polynomial of $ \alpha$, then the product of the roots of $ m_\alpha$ must be $ \pm1$, and the only way this can happen is if $ m_\alpha \equal{} f_n$ is irreducible.

(An algebraic integer such as $ \alpha^{ \minus{} 1}$ which lies outside the unit circle, but all its other conjugates lie inside, is known as a Pisot number.)[/quote]

What stops $ f_n$ from being the product of some $ m_\alpha$ (for some Salem number $ \alpha^{\minus{}1}$) with some cyclotomic polynomial? You could still get the product of the roots being $ \pm 1$, while there is exactly one root of $ f_n$ inside the unit circle (and exactly one outside).

[quote="Nukular"]What stops $ f_n$ from being the product of some $ m_\alpha$ (for some Salem number $ \alpha^{ \minus{} 1}$) with some cyclotomic polynomial? You could still get the product of the roots being $ \pm 1$, while there is exactly one root of $ f_n$ inside the unit circle (and exactly one outside).[/quote]
If $ |z|\equal{}1$ with $ f_n(z)\equal{}0$, then $ z\minus{}{1\over 2}\equal{}{z^{n\plus{}1}\over 2}$ and $ \left| z\minus{}{1\over 2}\right|\equal{}{1\over 2}$.
So $ z$ lies on the unit circle and the circle around $ {1\over 2}$ with radius $ {1\over 2}$.
And their intersection is $ z\equal{}1$, which contradicts $ f_n(1)\equal{}1\minus{}n\not\equal{}0$.
So every irreducible factor of $ f_n$ must have some root $ \alpha\in\mathbb{C}$ with $ |\alpha|<1$.


The first step is the generation of a benzyne.

OK.Thanks! Then diels-alder right?  :D

But how did we get benzyne?  :maybe:

I am not sure about the last step, because the double bond can also react.

yes.....even i am also thinking abt Y to Z  :maybe:  :|


function
number theory open
number theory

Please help me find the solutions to this equation. I've tried brute force on a computer, but there must be a better way. Thank you.

I suppose your definition of natural numbers includes zero, otherwise we have $ LHS\geq 6$ and the equation will not have a single solution. Consider the function $ F(a,b,c)\equal{}ab\plus{}bc\plus{}ac$. If $ abc\not\equal{}0$, then $ F(a,b,c)\geq 3$ with equality, iff $ a\equal{}b\equal{}c\equal{}1$. If $ abc\not\equal{}0$ and $ F(a,b,c)>3$, then $ F(a,b,c)>4$ holds. Since your equation is symmetric in each sub-tuple $ (x,y,z)$ and $ (u,v,t)$ and between sub-tuple sets, we will only analyse ordered solutions $ (x,y,z,u,v,t)$ in the sense of $ x\leq y\leq z$ and $ u\leq v\leq t$. Since a zero must occur in one of the sub-tuples, we distinguish two cases:

[b]Case 1: Only one sub-tuple has a zero entry[/b]. Without restriction set $ x\equal{}0$ and $ uvt\not\equal{}0$, then $ 4\equal{}yz\plus{}F(u,v,t)$ implies $ u\equal{}v\equal{}t\equal{}1$, which then yields $ y\equal{}z\equal{}1$ as well. Case 1 provides ordered solutions $ (0,1,1,1,1,1),(1,1,1,0,1,1)$

[b]Case 2: Both sub-tuples have a zero entry[/b]. Then we have $ x\equal{}u\equal{}0$ and $ 4\equal{}yz\plus{}vt$.

[b]Case 2a: For $ yzvt\equal{}0$[/b] assume without restriction $ y\equal{}0$, then $ v\equal{}t\equal{}2$ or $ v\equal{}1,t\equal{}4$ and $ z$ arbitrary. Case 2a provides ordered solutions $ (0,0,z,0,2,2),(0,0,z,0,1,4),(0,2,2,0,0,z),(0,1,4,0,0,z)$ with arbitrary $ z$

[b]Case 2b: For $ yzvt\not\equal{} 0$ [/b] we have $ 0<y,z,v,t<4$ and the highest value must be greater than $ 1$, which we assume to be $ z$ without restriction. $ z\equal{}3$ implies $ y\equal{}v\equal{}t\equal{}1$. $ z\equal{}2$ implies $ y\equal{}1,v\equal{}1,t\equal{}2$. Therefore, case 2b provides ordered solutions $ (0,1,3,0,1,1),(0,1,1,0,1,3),$ $ (0,1,2,0,1,2)$

Thank you very very much for your solution. It helped me a lot.


Este problema me parecio bien sencillo, pero bueno, esta chevere.

Sea $a_n$ una sucesion de enteros postivos tal que $(a_m, a_n)=(m,n)$. Demuestre que $a_n=n$

Edit: I thought this problem was from 93, but is actually from Russian 95

Indeed, it is to easy  :? Are you sure it is from Russia?

1. If we allowed to put $m=n$, then $a_n=(a_n,a_n)=(n,n)=n$ :)

2. If we are not allowed to put $m=n$ (I think it was a condition in the initial problem).
Then we have $(a_n,a_{n^2})=n$, so $n\mid a_n$. So $a_n=(a_n,a_{a_n})=(n,a_n)=n$ :)

Well, I'm sorry, i just forgot to put the condition that $m\not= n$, but still is very easy. This problem appears as problem 15 of 1995 Russian olimpiad in http://www.kalva.demon.co.uk/russian/rus95.html. :D


The circumference of a circle is $ 50\pi$ units. In terms of $ \pi$, how many square units are in its area?

[quote="Ebob"]50pi = 2pi*r, making r 25. $ 25^2 \cdot \pi \equal{} \boxed{625}$[/quote]

actually, it's $ 625\pi$


Let $ n\geq 2$ be an integer. Find the sum of the integers $ k$, $ 0<k<n$ and $ (k,n)\equal{}1$.

Alexandros

[hide="Hint"]If $ (k,n) \equal{} 1$ then $ (n \minus{} k,n) \equal{} 1$ then your sum is $ \frac {n\phi(n)}{2}$,($ \phi(n)$ is Euler funtion).[/hide]

ANOTHER APROACH:
Let $ a_1,a_2,....a_k$ all te natural so $ (a_i,n) \equal{} 1$ and let $ 1 \equal{} a_1 < a_2 < .... < a_k \equal{} n \minus{} 1$
Because $ a_i \plus{} a_{k \minus{} i \plus{} 1} \equal{} n$ the sum is $ nf(n)/2$
($ f(n)$ is Euler's function).

Well Done tdl and Αρχιμήδης 6!! My solution to this problem is the same as tdl's and I discovered it while I was solving some problems in Fields and Galois Theory and specially something about Norm and Traces. It's very interesting and cute result!

Alexandros

Thanks Alexandros.... :) You read Fields And Galois Theory....These are very interesting.... :)


geometry
geometric transformation
reflection
power of a point
radical axis
perpendicular bisector
geometry unsolved

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14572[/img]

 :(

To construct the radical axis of the non-intersecting circles $ (O_1), (O_2),$ draw an arbitrary circle $ (K)$ intersecting both $ (O_1), (O_2)$ (and possibly $ (O_3)$ as well). Radical axes of $ (O_1), (K)$ and $ (O_2), (K)$ meet at ther radical center $ R.$ Radical axis $ r_{12}$ of $ (O_1), (O_2)$ is the perpendicular to the center line $ O_1O_2$ through $ R.$ Radical axes $ r_{23}, r_{31}$ are constructed similarly (possibly using the same auxiliary circle intersecting $ (O_1), (O_2), (O_3)$).

The locus of centers of circles $ \left\{(C)\right\}$ bisecting 2 given circles $ (O_1), (O_2)$ in diametrically opposite points is the reflection $ q_{12}$ of their radical axis $ r_{12}$ in the perpendicular bisector of the segment $ O_1O_2.$ Furthermore, the circles $ \left\{(C)\right\}$ are coaxal, intersecting the center line $ O_1O_2$ at 2 fixed points $ U_1, U_2.$ All this was proved at [url]http://www.mathlinks.ro/viewtopic.php?t=57715[/url]. I might just add that the points $ U_1$, $ U_2$ do not have to be calculated. If 2 circles $ (C), (C')$ cut both $ (O_1), (O_2)$ with radii $ \varrho_1, \varrho_2$ in diametrically opposite points, then powers of $ O_1$ to $ (C), (C')$ are both equal to $ \minus{} \varrho_1^2$ and powers of $ O_2$ to $ (C), (C')$ are both equal to $ \minus{} \varrho_2^2,$ hence $ O_1O_2$ is the radical axis of $ (C), (C')$ and they obviously intersect the line $ O_1O_2.$ To find the points $ U_1, U_2,$ draw an arbitrary diameter $ P_1Q_1$ of $ (O_1).$ Perpendicular bisector of $ P_1Q_1$ cuts $ q_{12}$ at $ C$ and the circle $ (C)$ with center $ C$ and radius $ CP_1 \equal{} CQ_1$ cuts $ (O_2)$ also at diametrically opposite points $ P_2, Q_2.$ In addition, it cuts the center line $ O_1O_2$ at the fixed points $ U_1, U_2.$

Let $ q_{23}$ be the reflection of the radical axis $ r_{23}$ of the circles $ (O_2), (O_3)$ in the perpendicular bisector of the segment $ O_2O_3.$ Let $ Q$ be the intersection of $ q_{12}, q_{23}.$ Circle $ (Q)$ with radius $ QU_1 \equal{} QU_2$ cuts both $ (O_1), (O_2)$ at diametrically opposite points. Since it is centered on $ q_{23}$ and cuts $ (O_2)$ in diametrically opposite points, it cuts $ (O_3)$ in diametrically opposite points as well. But now, $ (Q)$ cuts both $ (O_3), (O_1)$ at diametrically opposite points, hence it is centered on the reflection $ q_{31}$ of the radical axis $ r_{31}$ in the perpendicular bisector of the segment $ O_3O_1$ $ \Longrightarrow$ the lines $ q_{12}, q_{23}, q_{31}$ concur at $ Q.$


Let $ a;b;c;d$ be positive real numbers .Prove that:
$ a^{5}\plus{}b^{5}\plus{}c^{5}\plus{}d^{5}\plus{}4abcd\geq 2(a\sqrt[3]{bcd}\plus{}b\sqrt[3]{acd}\plus{}c\sqrt[3]{abd}\plus{}d\sqrt[3]{abc})$

It is wrong.
Try $ a \equal{} b \equal{} c \equal{} d\equal{}0.5$.


real analysis
function
inequalities
real analysis unsolved

If $f\in L^{p}(R^{n})$ and $g\in L^{p\prime}(R^{n})$,where $1<p<\infty$ and $\frac{1}{p}+\frac{1}{p^\prime}=1$, then convolution $f\star g$ is a continuous function in $R^{n}$.

I am lost trying to prove this! Any help please>?

Use that "nice" functions are dense in $L^{p}$ for $p<\infty.$

Find sequences $f_{n}$ and $g_{n}$ of continuous, compactly supported functions such that $f_{n}\to f$ in $L^{p}$ and $g_{n}\to g$ in $L^{p'}.$ You can show that $f_{n}*g_{n}$ is continuous and compactly supported. Your remaining task is to show that $f_{n}*g_{n}$ converges uniformly to $f*g.$ Hölder's inequality should be useful.


geometry
circumcircle
incenter
analytic geometry

Say we have a triangle with vertices at: A(a,b) B(c,d) and C(e,f).  The centroid of the triangle can be expressed as: $\displaystyle \left(\frac{a+c+e}{3},\frac{b+d+f}{3}\right)$

How would the orthocenter, circumcenter, and incenter be expressed in terms of the coordinates of the 3 vertices?

[quote="Hokkage"]Idea for the circumcenter: use the standard equation of a circle and plug in values for the vertices to make a system of equations. (Haven't tried, so I'm not sure how well it will work.)[/quote]

That's basically what im trying to do for the circumcenter: i'm putting the distances from (x,y) to the three vertices equal (equalling the radius). Only two equations are necessary because we're only solving for two variables.

It's not looking too pretty at the moment...
[hide]
We have that
$(x-a)^2+(y-b)^2 = (x-c)^2 + (y-d)^2$ so $(2x-a-c)(c-a)=(2y-b-d)(b-d)$
Similarly, $(x-c)^2+(y-d)^2 = (x-e)^2 + (y-f)^2$ so $(2x-c-e)(e-c)=(2y-d-f)(d-f)$

I really don't feel like doing all of that expanding, but basically divide by $c-a$ then add $a+c$ and divide by $2$ in the first equation, substitute, solve. Wash. Lather. Repeat.  :P[/hide]

An alternative is to use the fact that the circumcenter is the intersection of the perpendicular bisectors of the triangle sides.

Obviously generalization will look quite ugly, but if we have specific numbers for the coordinates, then it will work nicely.

Cool, thanks.  I want to know this kind of stuff, becuase I usually transpose regular geo problems into the coordinate plane.

How about the incenter (intersection of angle bisectors) and orthocenter (intersection of altitudes)?

There are closed-form expressions for all the four points. See here: [url]http://mathforum.org/dr.math/faq/formulas/faq.ag2.html#twotriangles[/url] (some involve short notation using determinants)

yikes! those would be difficult to solve for  :) i wonder if there is a conversion from trilinear cooridnates to cartesian, because all of those points have simple trilinear coordinates...search before asking...i suppose those just come from the system that you solve with the perpendicular bisectors for instance and using cramer's rule

Actually, they are not so difficult to solve for, since they all have a unit column, which is very suitable for determinant expansion. (Moreover, exactly because of those unit columns, I suspect the expressions are not Cramer-originated, but just a shorthand for expressions of a Shoelace Theorem type.)


limit
vector
Functional Analysis
real analysis
real analysis unsolved

$X$ and $Y$ are Banach spaces. $A_n$ are bounded linear operators from $X$ to $Y$. $\{A_n x\}$ converges in $Y$ for all $x\in X$. So a linear operator $A$ is defined as $Ax = \lim A_nx$.
Show that \[\|A\|\leq \varliminf_{n\to\infty} \|A_n\|.\]

I have to say, I don't get this.. :?

Take some $x\in X$ with $\|x\|=1$. There is a sequence $k_n\to\infty$ s.t. $\|A_{k_n}\|\to\varliminf_{n\to\infty}\|A_n\|$ as $n\to\infty$ (even if that inferior limit is $\infty$). We have $\|Ax\|=\lim_{n\to\infty}\|A_{k_n}x\|\le\lim_{n\to\infty}\|A_{k_n}\|=\varliminf_{n\to\infty}\|A_n\|$. Since this holds for all $x$ with $\|x\|=1$, it means that $\sup_{\|x\|=1}\|Ax\|\le\varliminf_{n\to\infty}\|A_n\|\ (*)$.

We might eventually need $X$ to be Banach in order to have the uniform boundedness principle, so that the sequence $\|A_n\|$ is bounded, but I really think we do not need $Y$ to be Banach, just a normed vector space. $(*)$ holds whenever $X,Y$ are normed vector spaces, unless I'm overlooking something. :?.


trigonometry
LaTeX
geometry unsolved
geometry

Show that the triangle whose angles satisfy the equality \[ \frac{sin^2(A) + sin^2(B) + sin^2(C)}{cos^2(A) + cos^2(B) + cos^2(C)} = 2 \] is a rectangular triangle.

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
- second link
etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

Multiplying out the fraction and adding $\cos^2A+\cos^2B+\cos^2C$ to both sides
gives
$3=3(\cos^2A+\cos^2B+\cos^2C) \Rightarrow 1=\cos^2A+\cos^2B+\cos^2C$.
Finally, using the trigonometry identity
$1=\cos^2A+\cos^2B+\cos^2C+2\cos{A}\cos{B}\cos{C}$ we have
$\cos{A}\cos{B}\cos{C}=0$, so it must be a rectangular triangle.


function
vector
topology
real analysis
real analysis theorems

1) I want to prove that if (x(n)) is a sequence of positive numbers in R with lim [x(n+1)/x(n)] = 1, then lim [x(n)] diverges if and only if (x(n)) is unbounded (there exists no K>0 such that for every natural number n, x(n)<=K).

If (x(n)) is unbounded then (x(n)) diverges. To prove the theorem when (x(n)) diverges is where I'm having to trouble.

I wanted to go by contradiction and show that (at least for large enough n), x(n+1) = x(n)(1+a(n)), a(n)>=0, where a is a monotonic decreasing non-negative function on R. EDIT: I just realized a needn't be monotonic decreasing but must satisfy lim [a(n)] = 0. Not sure how to show such a function exists though. :(

But I don't know how to show this from the given: lim [x(n+1)/x(n)] = 1

What if I take any a>0 and an arbitrarily large open interval containing a (namely, (0, b) ) and show that the consequence of divergence implies there is always at least one x(n) greater than b? I think that would be sufficient to prove the theorem but my intuition tells me the procedure I proposed earlier would work too so I'd appreciate any help on that.

2) I want to show that a set in R(p) is connected if and only if it is convex. I know there's some theorem dealing with polygonal curves and connected sets but I'd much rather deal with this theorem on its own terms. I'm not sure how to go about it though.

3) How can I show rigorously that the sequence x(n) = sin (n) is divergent?

Oh, and can anyone show me how to prove that the unit interval [0,1] is compact without using the Heine-Borel theorem? I have seen the proof in Bartle's Elements of Analysis but the entire proof is unclear.

Thank you for your time

1) is false. For example, set $x_{1}=3$ and $x_{n+1}=x_{n}e^{\pm 1/n}$ for $n\ge 1$, where the sign is chosen as follows: it is $+$ if $x_{n}<1/2$, it is $-$ if $x_{n}>2$, and otherwise it is the same as for the preceding value of $n$.

mlok can you teach me how your counter example is divergent?

oh, and how did you come up with it?

Since the harmonic series is divergent, the signs $\pm$ will never stabilize. Indeed, if they are $+$ for all $n\ge N$, then $x_{n}/x_{N}=\exp(\sum_{k=N}^{n-1}1/k)$ grows without bound, a contradiction. Same for $-$. Therefore, the sequence will switch from $+$ to $-$ infinitely many times, which means that it contains infinitely many terms that are smaller than $1/2$ and infinitely many terms that are greater than $2$. Its limit does not exist. However, the sequence is bounded from above by $2e$ and from below by $1/(2e)$.

Concerning 2): what is R(p)?

R(p) denotes the vector space of p-tuples.

Then (2) is false for any $p>1$: there are lots of connected subsets of $\mathbb R^{p}$ that are not convex. For instance, the set of all $p$-tuples $(x_{1},\dots,x_{p})$ except those for which $x_{1}>0$ and $x_{2}=\dots=x_{p}=0$. Or just draw any non-convex connected figure in the plane.

(2) is true when $p=1$, because then the only connected subsets are intervals.

Hello mlok,

I haven't had the time to think deeply about your reply but just so you can guide my thinking, but do you think (2) would be correct if I added that the set must be open? I am wondering because I have seen a theorem that says that an open set G is connected if and only if any x,y in G can be joined by a polygonal curve lying entirely in G. This is from Bartle's elements in analysis. It makes me think that a similar statement may be made with the language of  convex sets.

Thanks anyway for pointing out the errors in my brainstorming. I have heard that errors are even more common than breakthroughs for a mathematician. I will try to come back and answer better after I am done with finals..

A polygonal curve is defined to be 'the union of a finite number of line segments' connected 'end to end'. a line segment might joining x and y might be {x+t(y-x) | t is in the unit interval}. I think you can see where I got the idea that convex sets might be related from.

[quote="§outh§tar"]I haven't had the time to think deeply about your reply but just so you can guide my thinking, but do you think (2) would be correct if I added that the set must be open? I am wondering because I have seen a theorem that says that an open set G is connected if and only if any x,y in G can be joined by a polygonal curve lying entirely in G. [/quote]
See http://www.mathlinks.ro/Forum/viewtopic.php?t=121766

Thanks Kalle. I am familiar with the general proof for open sets in R(n). But my question is whether the terminology of polygonal curves used in the proof is equivalent to a theorem that requires the open set be convex. It appears to me to be true.

A connected open set needs not be convex. An example: the set bounded by two concentric circles of different size.

It may be advisable to spend a few minutes drawing several examples of two-dimensional connected open sets.


For a given positive integer $n$, find all pairs of integers $(a,b)$ such that

${a-n\over b-n}={a^n\over b^n}$


number theory
complex analysis
complex analysis unsolved

1.  Show that no proper divisor of a perfect number can be perfect.

2.  Verify that no power of prime can be a perfect number

Why posting it here a second time¿
http://www.mathlinks.ro/Forum/viewtopic.php?t=85282 (I moved it to Number Theory since it obviously belongs there and changed the name because the old one was redundant and said nothing to the reader).


Hello.

I know how to solve second grade equations ([tex]ax^2 + bx + c = 0[/tex]) but how
do you solve third grades?
[tex]ax^3 + bx^2 + cx + d[/tex]

And maybe also fourth grade?

you would solve equations higher than degree 2 by factorizing the equation with something linear like x-1, or x-2 until you are eventually left with a quadratic equation.

There is a formula for the every third equation, but it is very complex and people hardly use it......

I think there was a discussion [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41829[/url]

Use synthetic division, which factors the equation, and also gives you the roots.

Okay, thank you all for the replyes!

I've been reading a little about factorization (I guess Iversonfan2005 and h78reg was talking about the same thing) now. I have a question:

But to do this, will I not have to find a root first?
If 5 is a root then I'll use (ax^3 + bx^2 + cx + d)(x - 5). How do I find a root without a calculator when doing higher degree equations?

you can use Decartes rule of signs and eventually make an educated guess regarding possible roots in a higher degree equation. for example, try x=1 or x=2 in the equation until you get a zero-value.  the number that you used as x in the equation would then be your root and you would use synthetic division to factorize.

You can also try substituting:you can substitute something like $x^2$ with $y$, and after solving for $y$ you can set $x^2$ equal to that and solve for $x$.

You could also use Newton-Raphson's method to find the roots.

use rational roots theorem as long as a and d don't have a million factors each. If it has a rational roots, it is some form of x/y or y/x where x|a and y|b

[quote="1+1"]Iversonfan2005:

Let's say we have

[tex]2x^3 - 4x^2 - 22x + 24 = 0[/tex]
t = x^2

[tex]2t^2 - 4t - 22\sqrt{t} + 24 = 0[/tex]

But isn't this harder to solve?

h78reg and kahsi: I shall take a look at them.

hmoonster: I didn't find "rational roots theorem". But i dound something called "rational zero theorem", is it this one?[/quote]

No, it's much easier say if you use it in situations like:

$x^4+x^2=10$, something like that, and then substitute $x^2$ with something like $y$.

rational root theroem:

possible rational roots are the plus or minus of the last term divided by first terms coefficient

[quote="Qingchenl"]There are a couple of things to keep in mind while solving for the roots:
1) There must be at least one rational root if all of the co-efficients are rational and the degree is odd. This means that if you have a cubic equation with rational coefficients, you would be able to find a rational factor by rational roots test and then apply the quadratic formula.

2) There is a zero between an adjacent local minimum and local maximum (and vice versa) if and only if there is a sign change. This allows you to limit the possible locations of roots and makes Newton's method a lot easier.[/quote]
No, you are wrong. Number 1 is wrong. $x^3+3=0$ no rational roots but coefficients are rational and degree is odd.


CIA has decided to create a special telephone directory for its agents. The first 2 pages of the directory contain the name of the directory and instructions for agents, telephone number records begin on the third page. Each record takes exactly one line and consists of 2 parts: the phone number and the location of the phone. The phone number is 4 digits long. Phone numbers cannot start with digits 0 and 8. Each page of the telephone directory can contain not more then K lines. Phone numbers should be sorted in increasing order. For the first phone number with a new first digit, the corresponding record should be on a new page of the phone directory. You are to write a program, that calculates the minimal number P pages in the directory. For this purpose, CIA gives you the list of numbers containing N records, but since the information is confidential, without the phones locations.

Is this solution correct:
 We sort all phone numbers. Then we go through the numbers and fill them into the directory. Every time time when the page is fool or the number had new first digit we take the new page.

may b N will b too much big for NlgN. think of a linear algo.

Mahbub


function
algebra
polynomial
algorithm
combinatorics open
combinatorics

Hi, I need to solve a problem that is specified as follows: we have some group of nodes with various associated variables.  For every possible group of three nodes, there is an associated weight which is given by a function of the variables associated with three nodes.  The goal is to choose a grouping of all nodes into groups of three such that the sum of the weights is maximized.  Is this solvable? Solvable under certain conditions, maybe with only a few discrete values allowed for the node variables?  I've been reading a lot and unable to find any concrete examples of a solution or a conclusion that there is none.  Thanks in advance for any help.

-mgibson

The problem is too generalized.

Whether it is solvable depends on your weighting function.

The problem would be [i]NP-complete[/i] and cannot be solved in polynomial time if your weighting function were too complicated.

If the function is easy such as $\max(a, b, c)$ or $(a+b+c)^2$, then just run the greedy algorithm :) .

Ah, sorry.

It's just a linear programming problem(but not [i]NP-complete[/i]).

The Karmarkar's algorithm proves the polynomial time upper-bound of this problem.


Find all P(x) be polynomial such that
$ P(x^2)\equal{}P(x)P(x\minus{}1)$ for all $ x\in\mathbb{C}$

Seems to me that $ (x^2\minus{}x\plus{}1)^n$ covers it. But you could wait for other answers  :)

If $ r$ is root to $ P(x)\equal{}0$. So are $ r^2$ and $ (1\plus{}r)^2$.
Hence, $ |r|\equal{}|1\plus{}r|\equal{}1$, otherwise, $ P$ will have infinitely many roots.
For $ |r|\equal{}|1\plus{}r|\equal{}1$, $ r$ can only be $ w$ or $ w^2$, where $ w$ solves $ w^2\minus{}w\minus{}1\equal{}0$.
Now $ P(x)\equal{}(x\minus{}w)^k(w\minus{}w^2)^m$.
Plug back to the original equation. You can verify hsbhatt is correct.

A general result:
If $ P(h(x))\equal{}P(g(x))\plus{}P(f(x))$ an $ deg(h)\equal{}deg(f)\plus{}deg(g)$ then there is a unique solution P(x).


logarithms
inequalities unsolved
inequalities

[quote="Inequalities Master"]Let $ a,b,c > 0$ s.t. $ abc\geq 1$.Prove that:

$ (a)$ $ a^{\frac {a}{b}}b^{\frac {b}{c}}c^{\frac {c}{a}}\geq 1$.[/quote]
Min will be then $ abc\equal{}1$;
we must prove :$ \frac{a}{b}\ln(a)\plus{}\frac{b}{c}\ln(b)\plus{}\frac{c}{a}\ln(a) \ge 0$;
$ f(x)\equal{} x\ln(x)$ - concave . So we have : $ \frac{a}{b}\ln(a)\plus{}\frac{b}{c}\ln(b)\plus{}\frac{c}{a}\ln(c) \ge (\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\ln(\frac{a/b\plus{}b/c\plus{}c/a}{1/a\plus{}1/b\plus{}1/c})$ ;
so we must prove : $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$  then $ abc\equal{}1$;
let $ a\equal{}x/y$ , $ b\equal{}y/z$ , $ c\equal{}z/x$; $ <\equal{}> (xy)^3\plus{}(yz)^3\plus{}(zx)^3 \ge (x^2y\plus{}y^2z\plus{}z^2x)xyz$
$ xy\equal{}a_1,yz\equal{}b_1, zx\equal{}c_1<\equal{}> a_1^3\plus{}b_1^3\plus{}c_1^3 \ge c_1^2b_1\plus{}a_1^2c_1\plus{}b_1^2a_1$ - (Which is easy) :)


floor function
number theory
real analysis
real analysis unsolved

Let $ p\geq 2$ be a prime number and let

$ \mu_{p}(n!)=\sum_{k=1}^{\infty}\left\lfloor{\frac{n}{p^{k}\right\rfloor}}$, be the exponent of $ p$ in $ n!$.

Study the convergence of the following series:

$ \sum_{n=1}^{\infty}\frac{\mu_{p}(n!)}{n^{\alpha}}$,

where $ \alpha>0$ is a positive real number.

This may be more of a number theory question.  Anyways,
$ \nu_{p}(n!)=\sum_{k=1}^{\infty}\left\lfloor{\frac{n}{p^{k}\right\rfloor}\approx \frac{n}{p}}$.
where we use $ f\approx g$ to mean that there exists constants $ c,C$ such that $ cf\le g \le Cf$.
  So,  $ \sum_{n=1}^{\infty}\frac{\nu_{p}(n!)}{n^{\alpha}}\approx \sum_{n=1}^{\infty}\frac{1}{pn^{\alpha-1}}$.  As $ p$ is fixed, your series of interest is
$ \approx \sum_{n=1}^{\infty}\frac{1}{n^{\alpha-1}}$
which converges iff $ \alpha>1$.


[quote="April"][color=darkred]Let be given a quadrilateral $ABCD.$ Let $I,\,J,\,H,\,K$ be the midpoints of $AC,\,BD,\,AB,\,CD,$ respectively.
Define the points $E=AJ\cap BI,\,F=CJ\cap DI$ Show that $EF\parallel HK$   :rotfl: [/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] I"ll use the vectorial method. I choose the origin $O\in AC\cap BD$ and denote $\frac{OC}{OA}=m>0$, $\frac{OD}{OB}=n>0$, i.e. $C=-mA$ and $D=-nB$. Thus,

$2\cdot \overline{HK}=2\cdot (K-H)=2K-2H=(C+D)-(A+B)=-mA-nB-A-B$ $\Longrightarrow$ $2\cdot \overline{HK}=-(1+m)A-(1+n)B$

and $\{\begin{array}{c}2\cdot I=A+C=(1-m)\cdot A\\\ 2\cdot J=B+D=(1-n)\cdot B\end{array}$. Therefore, $\{\begin{array}{ccc}E\in AJ\cap BI & \Longleftrightarrow & (\exists )\ \{u,v\}\subset\mathrm R\ , E=(1-u)\cdot B+u\cdot I=(1-v)\cdot A+v\cdot J\\\\ F\in CJ\cap DI & \Longleftrightarrow & (\exists )\ \{x,y\}\subset\mathrm R\ ,\ F=(1-x)\cdot D+x\cdot I=(1-y)\cdot C+y\cdot J\end{array}\|$ $\Longrightarrow$

$\{\begin{array}{ccc}E=(1-u)\cod B+\frac{u(1-m)}{2}\cdot A & = & (1-v)\cdot A+\frac{v(1-n)}{2}\cdot B\\\\ F=-n(1-x)\cdot B+\frac{x(1-m)}{2}\cdot A & = &-m(1-y)\cdot A+\frac{y(1-n)}{2}\cdot B\end{array}$. After some calculus obtain :

$\{\begin{array}{c}[4-(1-m)(1-n)]\cdot E=(1-m)(1+n)\cdot A+(1+m)(1-n)\cdot B\\\ [4mn-(1-m)(1-n)]\cdot F=m(1-m)(1+n)\cdot A+n(1-n)(1+m)\cdot B\end{array}\|$ $\Longrightarrow$

$[4-(1-m)(1-n)][4mn-(1-m)(1-n)]\cdot \overline{EF}=-2(1-m^{2})(1-n^{2})\cdot \overline{HK}$ $\Longrightarrow$ $EF\parallel HK$.[/color]

[quote="April"][color=darkred]Let be given a quadrilateral $ABCD.$ Let $I,\,J,\,H,\,K$ be the midpoints of $AC,\,BD,\,AB,\,CD,$ respectively. Define the points $E=AJ\cap BI,\,F=CJ\cap DI$
Show that $EF\parallel HK$  :D [/color][/quote]

[size=167][color=darkred][b][u]An useful idea :[/u][/b] $\boxed{\ S\in AD\cap BC\Longrightarrow S\in EF\ }$ ![/color][/size]

[color=darkblue][b][u]Proof.[/u][/b] Indeed, I"ll apply the [b]Pappus' theorem[/b] for the lines $\overline{AIC}$ and $\overline{BJD}$ : $\left\{\begin{array}{c}E\in AJ\cap BI\\\ S\in AD\cap BC\\\ F\in DI\cap CJ\end{array}\right\|$ $\Longrightarrow$ $S\in EF$.

[b][u]Remark.[/u][/b] Thank you, [b]April[/b], for your very nice proposed problem ! Until now, it is the best !

Otherwise, $\left\{\begin{array}{cccc}\left\{\begin{array}{ccc}E\in AJ & \Longrightarrow & [EAB]=[EAD]\\\ E\in BI & \Longrightarrow & [EAB]=[EBC]\end{array}\right\| & \Longrightarrow & [EAD]=[EBC]\\\\ \left\{\begin{array}{ccc}F\in DI & \Longrightarrow & [FCD]=[FAD]\\\ F\in CJ & \Longrightarrow & [FCD]=[FBC]\end{array}\right\| & \Longrightarrow & [FAD]=[FBC]\end{array}\right\|$ $\Longrightarrow$ $S\in EF$.

I denoted $[XYZ]$ -  the area of $\triangle XYZ$ and I used the following remarkable geometrical locus :
[b][u]the set of the points $L$ for which $[LAD]=[LBC]$ is the reunion of two lines which pass by the intersection $S\in AD\cap BC$.[/u][/b][/color]

[size=134][color=darkblue]Thank you, [b]April[/b], for your very nice proposed problem. Until now, it is the best ! From where have you this problem ?![/color][/size]

[color=darkred][size=167][b][u]An useful idea.[/u][/b][/size][/color] [color=darkblue] Define $S\in AD\cap BC$ and I"ll apply the [b]Pappus' theorem[/b] for the lines $\overline{AIC}$ and $\overline{BJD}$ : $\{\begin{array}{c}E\in AJ\cap BI\\\ S\in AD\cap BC\\\ F\in DI\cap CJ\end{array}\|$ $\Longrightarrow$ $\boxed{S\in EF}$.
[size=134][b][u]Proof of the proposed problem.[/u][/b][/size] Define the area $[XYZ]$ of $\triangle XYZ$, the points $\{\begin{array}{c}M\in [SA\ ,\ SM=AD\\\ N\in [SB\ ,\ SN=BC\end{array}$ and the middlepoint $P$ of $[MN]$.

Thus,$\{\begin{array}{ccccc}\{\begin{array}{ccc}E\in AJ & \Longrightarrow & [EAB]=[EAD]\\\ E\in BI & \Longrightarrow & [EAB]=[EBC]\end{array}\| & \Longrightarrow & [EAD]=[EBC] & \Longrightarrow & [ESM]=[ESN]\\\\ \{\begin{array}{ccc}F\in DI & \Longrightarrow & [FCD]=[FAD]\\\ F\in CJ & \Longrightarrow & [FCD]=[FBC]\end{array}\| & \Longrightarrow & [FAD]=[FBC] & \Longrightarrow & [FSM]=[FSN]\end{array}\|$. From the topic

 http://www.mathlinks.ro/Forum/viewtopic.php?t=145368 obtain that " the set of the points $L$ which belong to the interior of the angle $\widehat{MSN}$

 and for which $[LSM]=[LSN]$ is the ray $[SP$", i.e. $\{E,F\}\subset [SP\Longrightarrow S\in EF$.

Denote the intersection $R\in AC\cap BD$. Prove easily that $[HBC]-[HAD]=[KBC]-[KAD]=k$, where $k=\frac{1}{2}\cdot ([RBC]-[RAD])$, i.e.

$[HSN]-[HSM]=[KSN]-[KSM]=k$. From the same topic obtain that " the set of the points $L$ which belong to the interior of the angle $\widehat{MSN}$

 and for which $[LSN]-[LSM]=k$ (constant) is included in a parallel line $d$ to the line $SP$". Therefore, $\{H,K\}\subset d$, $d\parallel SP$ $\Longrightarrow$ $HK\parallel EF$.

[b]Remark.[/b] $E\in HK\Longleftrightarrow F\in HK\Longleftrightarrow S\in HK\Longleftrightarrow AB\parallel CD$.[/color]

[quote="April"][color=darkred]Let be given a quadrilateral $ABCD.$ Let $I,\,J,\,H,\,K$ be the midpoints of $AC,\,BD,\,AB,\,CD,$ respectively. Define the points $E=AJ\cap BI,\,F=CJ\cap DI$
Show that $EF\parallel HK$ [/color][/quote]

Thank you very much for your interest and for your nice solution, Mr. Virgil Nicula!
And here is the solution:
[color=indigo]
[b]Solution. (Use the notations as Mr. Virgil)[/b]
First, we have: $S\in EF$
Now, we'll prove that: $P\in EF,$ too.
Indeed, we have: $S(ABE)=S(ADE)=S(BCE)\Longrightarrow S(SME)=S(SNE)$ $\Longrightarrow E\in SP\Longrightarrow P\in\overline{SEF}\quad (1)$

And now, we'll prove: $SP\parallel HK$

Contruct the parallelograms $AHXD,\,BHYC$
We have:
$\begin{cases}HX%Error. "parralel" is a bad command.
\ SM\;\textrm{and}\;HX=AD=SM\\ HY%Error. "parralel" is a bad command.
\ SN\;\textrm{and}\;HY=BC=SN\end{cases}\Longrightarrow\begin{cases}HX=SM, \ HY=SN\\ \widehat{XHY}=\widehat{MSN}\end{cases}\\ \Longrightarrow\triangle HXY=\triangle SMN \Longrightarrow\triangle HXK=\triangle SMP\Longrightarrow\angle XHK=\angle MSP\\ \Longrightarrow HK\parallel SP\quad (2)$

From $(1)$ and $(2)$ we have: $HK\parallel EF$[/color]


function
real analysis
real analysis unsolved

Let (f_n) be a sequence of increasing functions from the set of real numbers into [-1,1]. Prove that we can find a subsequence of this sequence that converges simply to a real function with values in [-1,1]. Is it true that we must find a subsequence that converges uniformly at least on one compact of the real numbers?

For first question.
Let's enumerate all rational numbers $q_1$, $q_2$, ...
Define $f_n^{0}$ as $f_n$. Now $f_n^{k}$ is a subsequence of $f_n^{k-1}$ s.t. $f_n^{k}(q_k)$ converges. Finally, take $f^{k}=f_k^k$.
We see that $f^k$ converges in all rational points. It easily follows that $f^k$ converges in all real points.
Right?

I will have to study the solution, since their solution is at least complicated and quite difficult.

Oh, sorry, I mean closed bounded intervals.

Myth's first argument is 'just' a diagonal one.

Pierre.

[quote="pbornsztein"]Myth's first argument is 'just' a diagonal one.[/quote]
Just!!!!  :what?:  :mad:  :mad:  :mad:  :mad:

 :lol:  :lol:

[quote="Myth"]Of course, we can't find uniformly converging subsequence in general.
Just consider saltus functions $f_n(x)=\sum_{q_k<x} h_k^n$, where $h^n=1/2^n,1/2^{n-1},...,1/2,1/2^{n+1},1/2^{n+2},...$.[/quote]
I just wonder is it true or not?  :blush:  :blush:  :blush:


calculus derivative LaTeX AMC AIME function videos

This is stolen from an ancient thread, that I appear to be unable to find. Basically, you complete the blank. Some examples:

- Your friend says, "I have a problem!" and you whip out your calculator.

- You call little children "epsilons".

- You try to write QED at the end of your English paper.

This was stolen from the "You know your an AoPSer if..." shirt

[quote="r15s11z55y89w21"][quote="ytrewq"]- You [b]try[/b] to write QED at the end of your English paper.

This was stolen from the "You know your an AoPSer if..." shirt[/quote]

Try!?!?!?[/quote]

As in, you "try hard not to because you can't resist the temptation". XD

1. You cringe at the word "combination" because it's so close to "combinatorics".
2. @ the above: You have to consciously stop yourself from writing QED at the end of an English paper.

- When you start to call a kid at your school, named Minmin, as $ \minus{}m^2n^2$

When you try to impress hot chicks by saying..."I wish I were your derivative so that I could lie tangent to your curve!"

Stolen from FB

Or else when you think $ \pi$ is the best thing that happened to you.

1) When you stop going to math club for the ice cream and go instead for the math.
2) When you try hard not to cry if you get a less than perfect score on your trivial school math test.
3) When you spend more than 5 hours a day on AoPS.

[quote="ak909090"]1) When you stop going to math club for the ice cream and go instead for the math.
2) When you try hard not to cry if you get a less than perfect score on your trivial school math test.
3) When you spend more than [b]24[/b] hours a day on AoPS.[/quote]

@ak909090 I corrected it for you.

When you start writing iff instead of if.

[quote="ak909090"]1) When you stop going to math club for the ice cream and go instead for the math.
2) When you [b]cry[/b] if you get a less than perfect score on your trivial school math test, [b]no matter how hard you try not to[/b].
3) When you spend more than 5 hours a day on AoPS.[/quote]

@1: in my case, it's pastries.

1) When you start doing a math problem on your sofa(not actually drawing...just outlining with my finger)
2) When you start trying to write a proof during a civics lecture
3) When you want to go to different math clubs to see at what level you are and to add to your array of math problems instead of going for the pizza {and possibly showing off to a couple people(mostly girls)}

what's Q.E.D.?

QED is written at the end of proofs. Google/Wikipedia it for more info.

you start making jokes about math that no one gets.

When you start to be called by your AoPS username by your co-AoPS users that you know in real life...

you find out that you haven't been playing enough Starcraft.
you consider anything less than or equal to a perfect score, an EPIC FAIL

[quote="Poincare"][quote="isabella2296"][quote="dragon96"]

When someone uses :D in a chat, you think of guassy, lotsofmath, aleph0, ...[/quote]

Me. :D

...When your idea of a break is to study some chemistry instead.[/quote]

Wait wait wait wait...
Weren't you like a complete razz person like 2 months ago (i've been inactive as you can....)?[/quote]

:D and :P made a truce. And I meant that the statement completely described me.

@Ducky: Same here. :P

...When while answering this thread, you started typing a new problem, like in a marathon. (Literally me, just now.)

When you choose your girlfriend according to her mathematical capability.

[quote="Poincare"]Hmmm... I wonder why pythag is inactive these days? Is he at MOP?[/quote]

He started coming on starting last night.

When you call people even more trivial than TST.

[quote="PowerOfPi"]When you perfectly fit the description in the song "White and Nerdy". (My friend sings that with me as the person. "________ is white and nerdy. ... He is on the chess team. ...")[/quote]

This is kind of reviving, but I just noticed I was born on the same day as Weird Al Yankovic...

You posted a malformed binary!  (I looked it up.)

You know you've been studying too much math when your favorite activity is memorizing pi.

...When you see the number 179, and think of that FTW problem with the Roman numerals (me, this morning.)

when you're in a vocab competition that's coming up in 2 weeks, but study more math than vocab...

When you spend 3*the amount of time you watch TV, on AoPS... :rotfl:  :rotfl:  :rotfl:

...When your idol is someone math-related, e.g. Melanie Wood, Zuming Feng, etc.

[hide="lol..I just posted the lyrics to white and nerdy"]They see me mowin’
My front lawn
I know they’re all thinking I’m so white and nerdy
Think I’m just too white and nerdy
Think I’m just too white and nerdy
Can’t you see I’m white and nerdy?
Look at me, I’m white and nerdy
I wanna roll with
The gangstas
But so far they all think I’m too white and nerdy
Think I’m just too white and nerdy
Think I’m just too white and nerdy
I’m just too white and nerdy.
Really really white and nerdy.

First in my class here at MIT
Got skills, I’m a champion at D&D
MC Escher - that’s my favorite MC
Keep your 40, I’ll just have an Earl Grey tea
My rims never spin, to the contrary
You’ll find that they’re quite stationary
All of my action figures are cherry
Steven Hawking’s in my library
My MySpace page is all totally pimped out
Got people beggin’ for my top eight spaces
Yo, I know pi to a thousand places
Ain’t got no grills but I still wear braces
I order all of my sandwiches with mayonnaise
I’m a whiz at Minesweeper - I could play for days
Once you see my sweet moves you’re gonna stay amazed
My fingers’ movin’ so fast I’ll set the place ablaze
There’s no killer app I haven’t run
At Pascal, well I’m number one
Do vector calculus just for fun
I ain’t got a gat but I got a soldering gun
Happy Days is my favorite theme song
I could sure kick your butt in a game of ping pong
I’ll ace any trivia quiz you bring on
I’m fluent in JavaScript as well as Klingon

They see me roll on
My Segway
I know in my heart they think I’m white and nerdy
Think I’m just too white and nerdy
Think I’m just too white and nerdy
Can’t you see I’m white and nerdy
Look at me, I’m white and nerdy
I’d like to roll with
The gangstas
Although it’s apparent I’m too white and nerdy
Think I’m just too white and nerdy
Think I’m just too white and nerdy
I’m just too white and nerdy
How’d I get so white and nerdy

I’ve been browsin’, inspectin’
X-Men comics, you know I collect ‘em
The pens in my pocket, I must protect ‘em
My ergonomic keyboard never leaves me bored
Shopping online for deals on some writable media
I edit Wikipedia
I memorized Holy Grail really well
I can recite it right now and have you ROTFLOL
I got a business doing websites
When my friends need some code, who do they call?
I do HTML for ‘em all
Even made a homepage for my dog
Yo, I got myself a fanny pack
They were havin’ a sale down at The Gap
Spend my nights with a roll of bubble wrap
Pop, pop, hope no one sees me
Gettin’ freaky
I’m nerdy in the extreme
And whiter than sour cream
I was in AV Club and Glee Club and even the Chess Team
Only question I
Ever thought was hard
Was do I like Kirk
Or do I like Picard
Spend every weekend at the Renaissance Fair
Got my name on my underwear

They see me strollin’
They laughin’
And rollin’ their eyes ‘cause I’m so white and nerdy
Just because I’m white and nerdy
Just because I’m white and nerdy
All because I’m white and nerdy
Holy cow, I’m white and nerdy
I wanna bowl with
The gangstas
But, oh well, it’s obvious I’m white and nerdy
Think I’m just too white and nerdy
Think I’m just too white and nerdy
I’m just too white and nerdy
Look at me, I’m white and nerdy
[/hide]

I'm planning on making a remix.......anyone want to help? Maybe an AoPs remix :lol:  :rotfl:

When you can trivially prove the Riemann-Zeta hypothesis, and explain it in a way that five-year-olds can understand.

When you realize that I mispunctuated "Riemann zeta hypothesis".

2nd one was me. :D

- When this happens:

dragon96: happy are you?
me: happ = ou
hap^2 = ou
dragon96: therefore h, a, p, or o must be 0


Let $ a,b,c$ be positive real numbers. Prove that:
\[ a^{3}+b^{3}+c^{3}+ab^{2}+bc^{2}+ca^{2}\ge 2(a^{2}b+b^{2}c+c^{2}a)\]
An easy one. I want to a  non-SID-CID solution :P

$ a^{3}+ab^{2}\ge 2a^{2}b$, by AM-GM, and we're done.

thank you for your problem. I will try, it is really nice and of course the inequality is stronger than the first inequality  :blush:

a related one: http://www.mathlinks.ro/Forum/viewtopic.php?t=148414

okie, in this topic have a nice and easy problem of 10math_tp06_09
Let $ a,b,c$ be positive real numbers. Prove that:
\[ a^{3}+b^{3}+c^{3}+2(a^{2}b+b^{2}c+c^{2}a)\ge 3(ab^{2}+bc^{2}+ca^{2})\]
I solved it in some minutes  :P It is really easy one. Try it  :wink:

[quote="zaizai-hoang"]
Let $ a,b,c$ be positive real numbers. Prove that:
\[ a^{3}+b^{3}+c^{3}+2(a^{2}b+b^{2}c+c^{2}a)\ge 3(ab^{2}+bc^{2}+ca^{2}) \]
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157639

yes, it is also my solution  :blush: I used substitution as same as your. I am really amazing because it was posted in Mathlinks Forum. I didn't see all topic in Mathlinks Forum :P


A right rectangular prism has a base area one-quarter the base area of a larger right rectangular prism. The height of the smaller prism is half the height of the larger one. What is the ratio of the volume of the smaller prism to the volume of the larger prism? Express your answer as a common fraction.

The area of a rectangular prism is the area of the base times the height.
The larger is 4*2=8 times more volume, making the ratio $ \boxed{\frac{1}{8}}$


trigonometry
geometry
trig identities
Law of Sines

[quote="Attila the Hun"]If a, b, c are the sides of a triangle and A, B, C are its angles, prove that

a^3 cos(B-C) + b^3 cos(C-A) + c^3 cos(A-B) = 3abc[/quote]

$ S\equal{}\frac{1}{2}R^{2}(sin2A\plus{}sin2B\plus{}sin2C)$
$ a^{3}cos(B\minus{}C)\equal{}8R^{3}siin^{3}Acos(B\minus{}C)\equal{}8R^{3}siin^{2}AsinAcos(B\minus{}C)\equal{}8R^{3}*\frac{1\minus{}cos2A}{2}*\frac{1}{2}(sin(A\minus{}B\plus{}c)\plus{}sin(A\plus{}B\minus{}C))\equal{}2R^{3}(1\minus{}cos2A)(sin(\pi\minus{}2B)\plus{}sin(\pi\minus{}2C))\equal{}2R^{3}(1\minus{}cos2A)(sin2B\plus{}sin2C)$-->

$ a^3 cos(B\minus{}C) \plus{} b^3 cos(C\minus{}A) \plus{} c^3 cos(A\minus{}B)\equal{}2R^{3}(1\minus{}cos2A)(sin2B\plus{}sin2C)\plus{}2R^{3}(1\minus{}cos2B)(sin2A\plus{}sin2C)\plus{}2R^{3}(1\minus{}cos2C)(sin2B\plus{}sin2A)\equal{}2R^{3}(2(sin2A\plus{}sin2B\plus{}sin2C)\minus{}(cos2Asiin2B\plus{}cos2Asin2C\plus{}cos2Bsin2A\plus{}cos2Asin2C\plus{}cos2Csin2B\plus{}cos2Csin2A))\equal{}2R^{3}(2(sin2A\plus{}sin2B\plus{}sin2C)\minus{}(sin(2A\plus{}2B)\plus{}sin2A\plus{}2C)\plus{}sin(2B\plus{}2C))\equal{}2R^{3}(2(sin2A\plus{}sin2B\plus{}sin2C)\minus{}(\minus{}sin2C\minus{}sin2B\minus{}sin2A))\equal{}6R^{3}(sin2A\plus{}sin2B\plus{}sin2C)\equal{}12RS\equal{}12R*\frac{abc}{4R}\equal{}3abc$

the solution is quite simple.

a^3 cos(B-C) + b^3cos(C-A) + c^3 cos(A-B) = 3abc

<=> sum [a{a^2cos(B-C) - bc}] = 0.....................(1)

now, we use the law of sines to find that

sin^2(A)/a^2 = sinBsinC/bc

                     = [cos(B-C) - cos(B+C)]/2bc

                     = [ cos(B-C) + cosA]/2bc.

from which,

a^2 cos(B-C) = 2bc*sin^2(A) - a^2 cosA.

we substitiute this into (1) and simplifying, we find that we have to prove

sum [a {bc*cos(2A) + a^2 cosA}] = 0

<=> sum[a{2bc*cos^2(A) + a^2cosA - bc}] = 0

<=> sum [a{ cosA {2bc*cosA + a^2} - bc}] = 0

<=> sum [a {cosA(b^2 + c^2) - bc} ] = 0

<=> sum [a/bc {b^4 + c^4 - a^2(b^2 + c^2)}] = 0

<=> sum [a^2 {b^4 + c^4 - a^2(b^2 + c^2)} ] = 0

<=> sum [(a^2b^2)(b^2 - a^2) + a^2c^2(c^2 - a^2)] = 0.

<=> sum [ (a^2b^2)(b^2 - a^2)] + sum[a^2b^2(a^2 - b^2)] = 0

which is true. :clap2:

$ \sum a^3 \cos (B\minus{}C) \equal{} R \sum a^2 (2 \sin A \cos (B\minus{}C)$

$ \equal{} R \sum a^2 [2 \sin (B\plus{}C) \cos (B\minus{}C)] \equal{} R \sum a^2 (\sin 2B \plus{} \sin 2C)$

= $ 4R^3 \sum \sin^2 A (\sin 2B \plus{} \sin 2C) \equal{} 2R^3 \sum (1\minus{} \cos 2A)(\sin 2B \plus{} \sin 2C)$

$ \equal{} 2R^3 \sum (\sin 2B \plus{} \sin 2C) \minus{} 2R^3 \sum (\cos 2A \sin 2B \plus{} \cos 2A \sin 2C)$

Now, by expanding and regrouping $ 2R^3 \sum (\cos 2A \sin 2B \plus{} \cos 2A \sin 2C) \equal{} \minus{}2R^3 \sum \sin 2A$

Also, $ 2R^3 \sum (\sin 2B \plus{} \sin 2C) \equal{} 4R^3 \sum \sin 2A$

Hence the summation simplifies to $ 6R^3 \sum \sin 2A$

But $ \sin 2A \plus{} \sin 2B \plus{} \sin 2C \equal{} 4 \sin A \sin B \sin C$

So the expression equals $ 24R^3 \sin A \sin B \sin C \equal{} 3abc$


inequalities
algebra
polynomial
superior algebra
superior algebra unsolved

Prove that \[ \left[ k(\alpha,\beta): k\right]\leq\left[K(\alpha): k\right]\cdot\left[K(\beta): K\right].  \]

$[K(a,b): K] = [K(a,b): K(a)][K(a): K]$
If $f$ is a minimal polynomial in $K[x]$ which has b as a root, it is also a polynomial in $K(a)[x]$, whereas the polynomial may not be minimal. So we see that $[K(a,b): K(a)] \leq [K(b): K]$. QED.

Another way: use http://www.mathlinks.ro/Forum/viewtopic.php?t=71108 (the proof gives the needed bound) for the special case of fields.


Given $ B\hat{A}D\equal{}D\hat{A}C\equal{}45^{\circ}$, prove that $ \alpha>45^{\circ}$.

Notice that quadrilateral $ ABDE$ is cylic since $ \angle{BAE}\plus{}\angle{BDE}\equal{}180$.
Now $ \angle{EBD}$ and $ \angle{EAD}$ subtend to the same arc. Therefore I get that $ \alpha\equal{}45$ not so sure where I went wrong.

You must be right. I got same result. Mistake must be in answer of the book where i got this problem.

I did a few changes in original problem, but idea is the same.


ABCDEFGH is a polyhedron. Face ABCD is a square side 12. Face ABFG is a trapezoid with GF parallel to AB and GF = 6, AG = BF = 8. Face CDE is an isosceles triangle with ED = EC = 14. E is a distance 12 from the plane ABCD. The other faces are EFG, ADEG and BCEF. Find EG^2.

[color=cyan]You 3D geometry wimps . . .

Just kidding, don't worry.  Give this one a try -- it's a really sweet problem.  [/color]


inequalities
topology
real analysis
real analysis unsolved

Let $(X,d)$ be a metric space, and $\mu^{*}$ an outer measure on $X$. One says that $\mu^{*}$ is a metric outer measure if for every subsets $A,B \subset X$ satisfying $d(A,B)>0$ the following inequality holds:
$\mu^{*}(A \cup B) = \mu^{*}(A)+\mu^{*}(B)$

Prove that for every closed subset $F$ and every subset $A$ of $X$ :
$\mu^{*}(A) = \mu^{*}(A \cap F)+\mu^{*}( A \setminus F)$

(hence the measure $\mu$ induced by $\mu^{*}$ is Borel)

We can forgert about $A$ and simply show that for all closed subsets $F$ of $X,\ \mu^{*}(X)=\mu^{*}(F)+\mu^{*}(X\setminus F)$ (since we can work on $A$ instead of $X$). This is clear when $\mu^{*}(X)=\infty$, so we assume that $\mu^{*}(X)$ is finite.

Let $A_{1}$ be the set of points at distance $\ge 1$ from $F$, and, for all $n\ge 2$, let $A_{n}$ be the set of points whose distance from $F$ belongs to the interval $\left[\frac{1}{n},\frac 1{n-1}\right)$. Notice that $d(A_{m},A_{n})>0$ whenever $|m-n|\ge 2$. Finally, let $S_{n}=\bigcup_{i=1}^{n}A_{i}$.

Suppose $\sum\mu^{*}(A_{n})=\infty$. We can then find sets $B_{n},\ n\ge 1$ such that (a) each $B_{n}$ is a finite union of $A_{k}$'s, (b) each $A_{k}$ appears in at most one $B_{n}$, (c) if $A_{k}, A_\ell$ are subsets of some $B_{n},B_{m}$ respectively (allowing for the possibility $m=n$), then $|k-\ell|\ge 2$, and (d) $\mu^{*}(B_{n})\ge t>0$ for all $n$ and some fixed $t$. All of this, however, implies that $\forall n,\ \mu^{*}(X)\ge \mu^{*}\left(\bigcup_{i=1}^{n}B_{i}\right)=\sum_{i=1}^{n}\mu^{*}(B_{i})\ge nt$, contradicting $\mu^{*}(X)<\infty$. This means that we must have $\sum\mu^{*}(A_{n})<\infty$, so, in particular, $\mu^{*}((X\setminus F)\setminus S_{n})\le\sum_{k\ge n+1}\mu^{*}(A_{k})\to 0$. We have $\mu^{*}(X\setminus F)\le\sup_{n}\mu^{*}(S_{n})+\sup_{n}\mu^{*}((X\setminus F)\setminus S_{n})$, which, from the preceding remarks, equals $\sup\mu^{*}(S_{n})$.

We have $\mu^{*}(X)\ge\sup_{n}\mu^{*}(F\cup S_{n})$. Because $\mu^{*}$ is metric and $d(F,S_{n})>0$, the RHS equals $\mu^{*}(F)+\sup_{n}\mu^{*}(S_{n})=\mu^{*}(F)+\mu^{*}(X\setminus F)$. On the other hand, $\mu^{*}(X)\le\mu^{*}(F)+\mu^{*}(X\setminus F)$ by subadditivity. Together, these two inequalities give the equality we're after: $\mu^{*}(X)=\mu^{*}(F)+\mu^{*}(X\setminus F)$.


calculus
integration
algebra
function
domain
abstract algebra
topology

let $ X$ be a $ k$-scheme, $ L/k$ a field extension. consider the natural projection $ p : X \times_k L \to X$. is $ p$ an open mapping of the underlying topological spaces?

I'll (try to :)) prove it, but first, here's a proof of the result you said you were initially interested in, -oo- (I read your second message before you deleted it). If I recall correctly, you wanted to show that For any $ k$-scheme $ X$, the dimension of $ X$ does not exceed that of $ X' = X\times_k\bar k$, where $ \bar k$ is the algebraic closure of $ k$.

Let $ Z_0\subset Z_1\subset\ldots\subset Z_n$ be any strictly ascending chain of closed irreducible subsets of $ X$. We want to find one such chain on $ X'$ as well. We can find an affine open subset $ U\subset X$ around the generic point of $ Z_0$, and replace $ X,X'$ with $ U,U' = U\times_k\bar k = p^{ - 1}(U)$ respectively, $ p$ being the projection $ X'\to X$; nothing changes if we do this: the intersections of the $ Z_i$ with $ U$ still form a strictly ascending chain of irreducible closed sets, and if we find another chain like this on $ U'$, then the closures of those sets in $ X'$ will form the desired chain on $ X'$. The problem has now been reduced to the following:

Let $ A$ be a commutative $ k$-algebra. Then, the Krull dimension of $ A$ does not exceed that of $ A' = A\otimes_k\bar k$. Just notice that $ A'$ is an integral extension of $ A$, hence the map induced on spectra by the inclusion $ A\to A'$ is surjective, and we can apply the [url=http://en.wikipedia.org/wiki/Going_up_and_going_down]Going Up Theorem[/url] to find a strictly ascending chain of $ n + 1$ prime ideals of $ A'$ lying over any chain of $ n + 1$ prime ideals of $ A$.

Now for the problem at hand:

Again, notice that we can assume $ X$ to be affine. The problem is now showing that for any $ k$-algebra $ A$ and any extension $ L/k$, the map $ i^*$ induced on spectra by the inclusion $ i: A\to A' = A\otimes_kL$ is open. A [i]basic open set[/i] of the spectrum of a commutative ring $ B$ is one of the form $ \mbox{Spec}(B)_f = \{\frak p\in\mbox{Spec}(B)\ |\ f\not\in\frak p\}$ for some element $ f\in B$. These form a basis for the spectrum of $ B$, and in our case it is enough to show that $ i^*$ maps every basic open subset of the spectrum of $ A'$ onto an open subset of the spectrum of $ A$.

Every field extension breaks up as a separably generated one (i.e. separable algebraic over a purely transcendental extension) followed by an algebraic, purely inseparable one. Hence it suffices to consider the two cases:

(1) [i]L/k is separably generated[/i]

Let $ f\in A'$ be an arbitrary element. We seek to characterize those prime ideals $ \frak p$ of $ A$ which are not contractions of primes in $ A'$ not containing $ f$. If we manage to prove that the set of such primes $ \frak p$ is closed, we'll be done, since this set is precisely the complement of $ i^*(\mbox{Spec(A')_f})$.

Denote by $ k(\frak p)$ the residue field of the prime ideal $ \frak p$ (i.e. the quotient field of the domain $ A/\frak p$). The condition that $ f$ be contained in all primes of $ A'$ which lie over $ \frak p$ is that it (or rather its image through the canonical map) be contained in all primes of $ k(\frak p)\otimes_AA'\cong k(\frak p)\otimes_kL\ (*)$, i.e. that it be nilpotent in $ (*)$. Now, because $ L$ is separably generated over $ k$, the ring $ (*)$ is reduced, and the conclusion is that $ f$ must be zero in $ (*)$, or, equivalently, in $ A/\frak p\otimes_kL$; yet another way of putting it is $ f\in \frak p\otimes_kL\ (\#)$. If $ e_j,\ j\in J$ form a basis of $ L/k,\ f$ can be written uniquely as $ \sum a_j\otimes e_j,\ a_j\in A$, and $ (\#)$ is equivalent to $ a_j\in\frak p,\ \forall j\in J$. Hence the set of primes we're after is the set of primes containing the ideal generated by $ a_j,\ j\in J$. It is indeed closed in $ \mbox{Spec}(A)$, so that's that.

(2) [i]L/k is purely inseparable[/i]

Let $ f$ be an element of $ A'$. Then, because $ L/k$ is purely inseparable, some power $ f^n$ of $ f$ lies in $ A$ (we can take $ n$ to be a large enough power of the characteristic of $ k$). The prime ideals of $ A'$ which do not contain $ f$ are precisely those which do not contain $ f^n$, of course, so in order to avoid carrying that exponent $ n$ around, we may as well assume that $ f\in A$ to begin with. The map $ i^*$ from the spectrum of $ A'$ to that of $ A$ is surjective because $ A'$ is integral over $ A$. From this and $ f\in A$ it follows that the primes of $ A'$ which do not contain $ f$ map precisely onto the primes of $ A$ which do not contain $ f$, which takes care of case (2).

impressing :-)

@owk: I'll try to read it :)
@grobber: I deleted the second post because I came up with the same proof as you, but I don't need going up or integral extensions. if $ p$ is prime in $ A$, let $ p'$ be the kernel of the natural homomorphism $ A \otimes_k \overline{k} \to \overline{k(p)}$. then $ p'$ is a prime ideal and $ p \subseteq q$ implies $ p' \subseteq q'$.

the question if $ p : \overline{X} \to X$ is always open arose as I was trying to prove directly $ dim(X) \leq dim(\overline{X})$. I thought that open, surjective mappings have the property, that preimages of irreducible, closed subsets are also irreducible, closed (but I cannot reconstruct the proof, maybe it's wrong ...).

I've a question about your case (2) (which is the most interesting, I think): you tacitly use the equivalence of the conditions

"there is some prime $ q \subseteq A'$ with $ f' \notin q$ and $ p \subseteq i^{ \minus{} 1}(q)$"
"there is some prime $ q \subseteq A'$ with $ f' \notin q$ and $ p \equal{} i^{ \minus{} 1}(q)$"

why are the equivalent?

//now it is case (1)

[quote="-oo-"]
I've a question about your case (2) (which is the most interesting, I think): you tacitly use the equivalence of the conditions

"there is some prime $ q \subseteq A'$ with $ f' \notin q$ and $ p \subseteq i^{ \minus{} 1}(q)$"
"there is some prime $ q \subseteq A'$ with $ f' \notin q$ and $ p \equal{} i^{ \minus{} 1}(q)$"
[/quote]

It doesn't look to me like I'm doing anything of the sort. What I'm showing there is this:

Let $ A\subseteq B$ be an integral extension, and let $ f\in A$. Then the contractions of the primes of $ B$ which do not contain $ f$ are the primes of $ A$ which do not contain $ f$.

One direction is clear: if a prime $ \frak q$ of $ B$ does not contain $ f$ then of course, $ \frak q\cap A$ does not contain $ f$ either. Conversely, because $ B$ is integral over $ A$, we already know that every prime $ \frak p$ of $ A$ is the contraction of a prime $ \frak q$ from $ B$. It's quite obvious that if $ f\not\in\frak p$ then $ f$ cannot belong to $ \frak q$ either.

At some point my proof was divided into three cases rather than two, and I only realized the extra case was unnecessary after posting. Then I edited my post, but what is now case 2 has always been case 2, so your error is inexcusable :).

[quote="-oo-"]ok ... my problem is the following equivalence in your proof:

$ f$ lies in all primes of $ A'$ which lie over $ p$ (i.e. whose intersection with $ A$ is $ p$, right?) if and only if $ f$ lies in all primes of $ k(p) \otimes_A A'$.[/quote]

Let $ \varphi: A\to B$ be any homomorphism rings (all rings unital and commutative, all homomorphisms unital), inducing the continuous map $ \varphi^*: Y\equal{}\mbox{Spec}(B)\to X\equal{}\mbox{Spec}(A)$. Let $ \frak p$ be a prime of $ A$, i.e. a point of $ X$, and, just as above, let $ k(\frak p)$ be the residue field of $ A$ at $ \frak p$. Finally, let $ f$ be an element of $ B$. The basic open subset of $ Y$ determined by $ f$ will be denoted by $ Y_f$, and I'll use similar notation for spectra and elements of other rings.

What I used there is this: the canonical map $ \psi: B\to B\otimes_Ak(\frak p)$ induces a bijection (homeomorphism, actually, but this is not important) $ \psi^*$ of the spectrum $ Z$ of $ B\otimes_Ak(\frak p)$ onto the subset $ \left(\varphi^*\right)^{\minus{}1}(\frak p)$ of $ Y$ with the subspace topology it inherits from $ Y$. That's the set of primes of $ B$ lying over $ \frak p$. Since $ \left(\psi^*\right)^{\minus{}1}(Y_f)\equal{}Z_{\psi(f)}$, you get the desired result: all primes in $ B$ lying over $ \frak p$ contain $ f$ iff all primes of $ B\otimes_Ak(\frak p)$ contain $ \psi(f)$.

well that's a reformuation of the problem ..


calculus
integration
trigonometry
complex analysis

[quote="nueng"]Proof that
\[ \Gamma \left( x \right)\Gamma \left( {1 \minus{} x} \right) \equal{} \pi \csc \pi x
\]
[/quote]

start with weierstrass.

$ \dfrac{1}{\Gamma(x)}\;\equal{}\;x\,e^{\gamma\,x}\prod_{k\equal{}1}^\infty\;e^{\minus{}\dfrac{x}{k}}\cdot \left(1\plus{}\dfrac{x}{k}\right)$

so,
$ \dfrac{1}{\Gamma(x)}\cdot\dfrac{1}{\Gamma(\minus{}x)}\;\equal{}\;x\,e^{\gamma\,x}\,(\minus{}x)\,e^{\minus{}\gamma\,x}\prod_{k\equal{}1}^\infty\;e^{\minus{}\dfrac{x}{k}}\cdot \left(1\plus{}\dfrac{x}{k}\right)\cdot\,e^{\dfrac{x}{k}}\cdot \left(1\minus{}\dfrac{x}{k}\right)$

put $ \Gamma(\minus{}x)\;\equal{}\;\dfrac{\Gamma(1\minus{}x)}{\minus{}x}$ to get

$ \dfrac{1}{\Gamma(x)}\cdot\dfrac{1}{\Gamma(1\minus{}x)}\;\equal{}\;x\prod_{k\equal{}1}^\infty\;\left(1\minus{}\dfrac{x^2}{k^2}\right)$

recall

$ \sin\,(\pi\,x)\;\equal{}\;(\pi\,x)\prod_{k\equal{}1}^\infty\;\left(1\minus{}\dfrac{x^2}{k^2}\right)$

$ \therefore \dfrac{1}{\Gamma(x)}\cdot\dfrac{1}{\Gamma(1\minus{}x)}\;\equal{}\;\dfrac{\sin\,(\pi\,x)}{\pi}$


1. If cos2x-2cosx=2, find cos2x+2cosx.

2. If 3sinA+4cosB=(sqrt5), what is the max value of 4sinA+3cosB?

3. For 0<x<180, find the sum of the solutions to tan^2 x - 2009tanx +1 =0.

hello, to your first problem. Note that $ \cos(2x)\equal{}\cos^2(x)\minus{}\sin^2(x)\equal{}2\cos^2(x)\minus{}1$, inserting this we have
$ 2\cos^2(x)\minus{}2\cos(x)\minus{}3\equal{}0$, solving the quadratic equation we got
$ x_1\equal{}\minus{}\arccos(\frac{1}{2}(1\minus{}\sqrt{7}))\plus{}2k_1\pi$
$ x_2\equal{} \arccos(\frac{1}{2}(1\minus{}\sqrt{7}))\plus{}2k_2\pi$
Sonnhard.

hello, inserting $ x_1$ or $ x_2$ in the term $ \cos(2x)\plus{}2\cos(x)$ we get $ 4\minus{}2\sqrt{7}$.
Sonnhard.


trigonometry
algebra
system of equations

I recently saw a problem asking for an infinite sum involving the term $\tan^{-1}\left(\frac{1}{2n^{2}}\right)$.
The solution involved rewriting this term as $\tan^{-1}(2n+1)-\tan^{-1}(2n-1)$.
How would one go about deducing this? I think I have seen other questions involving inverse tangent and cotangent with algebraic terms inside but didn't know how to solve them, so if anyone has any problems like this, please post them! Thanks.

The idea is the identity:

$\tan(a+b)=\frac{\tan a+\tan b }{1-\tan a \tan b}$

with a little creativity, many problems branch off.

This yields:
$\tan^{-1}a+\tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$
which is the basis for all identities.

This can be made to telescope like in the problems you have seen, I think it is more beneficial to try to write your own problems then offer them to people to see if your ideas hold water...

Well if I didn't see a solution at first, then I would write out the system of equations, then if I still didn't see a solution, I might try to solve it keeping in mind what the terms most likely look at... but by then, usually you have already found what it really is...

[quote="pkothari13"]Thanks me@home!  :)  I had already figured out the tan(a+b) identity, but what I was trying to ask was:

If I gave you, say, $\tan^{-1}\left(\frac{1}{2n^{2}}\right)$, how would you know what to rewrite it as?

Would you say $1=a+b$ and $2n^{2}= 1-ab$ and then solve for $a$ and $b$? That seems somewhat cumbersome.[/quote]

Sometimes it might not be nice...

[quote="Altheman"]you have to assume that the problem has a nice solution, and you try various techniques until one works, telescoping is one of the first ideas that i would try[/quote]

Yeah, but how would you go about knowing what to rewrite it as such that terms will cancel?

If you are certain it telescopes, then looking at powers and coeffs you can tell which terms must cancel which terms from then it is easy.
Worst case, algebra can be done by guessing a form putting in variable coeffs and solving until something works.
Don't stress this type of problem


well probably the best explaination for this is experimental observation.well i could have very welll  answered taht it has $ 8$ electrons in the outer shell and hence don't react....but these are all secondary reasons
but at the same time there do exist compounds like $ \text{XeF_{4}},\text{XeF_{6}}$

Welcome to this forum Math^_^.

The reactivity of atoms is experimentally measured by the ease with which they can form different compounds with other atoms, and can be theoretically explained on basis of their electronic configuration: atoms with incomplete valence shells "have a tendency" to complete them by either gain or loose electrons, because a closed shell configuration is very stable. Noble gases, as they have closed shell configurations, have therefore very low tendency to gain or loose electrons and so their chemical reactivity is relatively low. However, as we go down the 18th group, the ionization energy decreases, and for Xenon it equals the ionization energy of the oxygen molecule. So, there are several known compounds of Xenon with other elements, most of them with fluorine and oxygen (why?). Some examples are $ XeF_{2}$, $ XeF_{4}$, $ XeF_{6}$, $ XeF_{5}^{\minus{}}$, $ XeF_{7}^{\minus{}}$, $ XeF_{8}^{2\minus{}}$, $ XeO_{3}$, and the perxenate ion $ XeO_{6}^{4\minus{}}$. There are very few know compounds of Kripton  (for example, $ KrF_{2}$), because this element has a higher ionization energy compared with Xenon, and so they are much harder to prepare. On the other hand, Radon has an even lower ionization energy than Xenon and it would be expected to be more easy to prepare Radon compounds: however, its radioactivity makes working with this gas very dangerous.

The gases of the 18th group are sometimes known by three names: rare gases, inert gases, and noble gases. The first two terms are inappropiate (since some of them are relatively abundant, and Xenon has an inexpected rich chemistry), and so it would be more correct to refer to them as noble gases.

The unexpected chemistry of Xenon reminds me the common view in organic chemistry that alkanes are the inertest organic compounds. However, just like Xenon, under the appropriate conditions they can show quite an interesting chemistry.

[quote="Carcul"]
The gases of the 18th group are sometimes known by three names: rare gases, inert gases, and noble gases. The first two terms are inappropiate (since some of them are relatively abundant, and Xenon has an inexpected rich chemistry), and so it would be more correct to refer to them as noble gases.[/quote]

Thank you for your answer Carcul, I just wonder why do chemists refer them as noble gas?

We may see a noble person as someone who relates only with some special persons: you will not see a noble, an aristocrat relating socially with a "normal" person. Likewise, noble gases don't react with other atoms under normal conditions, but Xenon for example has as interesting chemistry under special conditions and only with certain reagents.


Prove that there exit three diffirent integers a,b,c in the set {${n^2 , n^2 +1 ,...., n^2,n +[3+\sqrt{n}] }$} so that c divides ab

I do not see the link between $a$ & $b$ with the question. What other conditions are there?

I don't understand your question  :blush:
But in my post , there is some thing wrong  :lol:

And I correct it:

$n \in N$ and a,b,c in the set : {$n^2, n^2+1,......,n^2 + [3+\sqrt{n} ]$}


a,b,c>o and

\[(a+b)(b+c)(c+a) = 1\]

prove that

\[ab+bc+ca \ge \frac{3}{4}\]

[quote="kimnimalar"]a,b,c>o and
\[(a+b)(b+c)(c+a) = 1 \]
prove that
\[ab+bc+ca \ge \frac{3}{4}\]
[/quote]
It's wrong. Try $a\rightarrow0^{+}$ and $b=c\rightarrow\frac{1}{\sqrt[3]{2}}.$ :wink:

I think he meant, the JBTST 2005, Romanian problem proposed by Cezar, with the sign reversed: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=232780#p232780[/url].


[quote="epatjn"]Again--what's the easiest approach to teaching how to approach this type of question?

The prime numbers are added in order starting with
2: 2, 2 + 3, 2 + 3 + 5, and so on.

How many of the first 12 such sums are also prime?[/quote]
[hide]well it's obvious that all other primes after 2 are odd so an even number of odds creates an even;
 i actually saw this one before where? in the handbook i think; of course my handbook has been taken away from me. i think i relied on that fact and brute force to do it.  :ewpu: it worked though.[/hide]

what are you talking about?

the ones with even amount of numbers are odd!

jorian

This problem is only the first 12 sums, so it's as fast to brute force it as to use casework.


*When a plague of locusts attack a wheat crop,the proportion of the crop destroyed after [b]t[/b] hours is denoted by [b]x[/b].In a model,it is assumed that at the rate at which the crop is destroyed is proportional to  [b]x(1-x)[/b] .A plague of locusts is discovered in a wheat crop when one quarter of the crop has been destroyed.Given that the rate of destruction at this instant is such that if remained constant, the crop would be completely destroyed in a further 6 hours.

a)Show that dx/dt= 2x/3(1-x)
b)find the percentage of the crop destroyed 3 hours after the plague of locusts is first discovered.

thanks :D


Basically post a random video that you would like to share. You can make them interrelated if you want.
[youtube]qRuNxHqwazs[/youtube]
400 BABIES!!!!

toldya it was random.

???????

There already is a topic like this...

[youtube]MTsfVbN7UOM[/youtube].

$ \color{blue} \text{VITA} \color{red} \text{MEATA} \color{green} \text{VEGA} \color{black} \text{MIN}$.

[quote="AstroPhys"][youtube]MTsfVbN7UOM[/youtube].[/quote]

I have that episode. It was from the first season, which I have. :)

[youtube]HbPDKHXWlLQ[/youtube]

I can't find the Gamma song...

WHY DO YOU HATE MY THREAD?

[youtube]yY1JUiao9xs[/youtube]

[quote="AstroPhys"][youtube]MTsfVbN7UOM[/youtube].[/quote]

That actress really did an extraordinary job pretending to be drunk...

Best youtube video ever...
[youtube]nbZ9zJ22WfQ[/youtube]

some more funny videos
[youtube]vmVGQR3NNdg[/youtube]
reallly really funny its a parody of Yu-Gi-Oh :rotfl:
[youtube]VuBvZMhk_PI[/youtube]

nearly pointless, but I love this guy's screaming :D

[youtube]gW8gwg6pWoY[/youtube]


function
combinatorics proposed
combinatorics

Find the smallest number $n \geq 5$ for which there can exist a set of $n$ people, such that any two people who are acquainted have no common acquaintances, and any two people who are not acquainted have exactly two common acquaintances.

[i]Bulgaria[/i]

I've done it long time ago.And I remember the answer is 16.
The main idea is to prove the numbers of everyone's acquaintances are the same.
And suppose that's $m$,for every $m$ there is at most one graph.And maybe $m=4$.

[quote="Ahwingsecretagent"]Finally when $ k \equal{} 5,n \equal{} 16$ which should work.[/quote]  Specifically, the hypercube skeleton plus long diagonals works nicely.  (That is, your vertices are $ \{0, 1\}^4$ and two vertices are adjacent if you can get from one to the other by either changing a single entry or by changing all four entries.)

The beginning part of the argument is similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78677]this question from APMO 1990[/url] -- that question asked for all non-acquainted people to have exactly one common acquaintance, and the smallest such graph is the 5-cycle.

What's the smallest $ n$ such that there is a triangle-free graph on $ n$ vertices such that every two nonadjacent vertices have exactly $ k$ common neighbors?

[quote=Ahwingsecretagent]Since no one posted a solution and I recently solved this for practice I will post my solution. First we prove that if $ A$ and $ B$ are aquainted then they have the same number of friends. In fact we show there is an injection from the set of friends of $ A$ into the set of friends of $ B$. This is sufficient because of the symmetry.
Since $ A$ knows $ B$ none of $ A$'s friends are friends of $ B$. Let $ C$ be a friend of $ A$, as $ B,C$ are not aquainted there is a $ D$, a friend of both $ B$ amd $ C$ other than $ A$, call $ D  f(C)$. It is easy to check this function is an injection. If $ f(x)\equal{}f(y)$, then $ f(x)$ and $ A$ have at least 3 common aquaintances ($ x,y,B$), so $ x\equal{}y$. So we know that if two people know each other then the number of their friends is equal. Now if $ X$ and $ Y$ are any two people who don't know each other, then they have a common friend $ A$. Now $ X$ has the same number of friends as $ A$, who has the same number of friends as $ Y$, so $ X$ and $ Y$ have the same number of friends.

Therefore everyone has $ k$ friends, where $ k$ is a positive integer. Now pick a random person $ A$. $ A$ has $ k$ friends. We count all the friends of these $ k$ friends. By the problem hypothesis, everyone who is not a friend of $ A$ is counted twice, $ A$ is counted $ k$ times, and none of $ A$'s friends are counted at all. Therefore we have
$ k^2\minus{}k\plus{}1\minus{}(n\minus{}k\minus{}1)\equal{}n\minus{}k$
$ k^2\plus{}k\equal{}2n\minus{}2$.
Also if $ A$ knows $ B$ and $ 2k>n\minus{}2$ then $ A$ and $ B$ have a common aquaintance by the pigeon hole principle, which cannot happen according. Therefore $ 2k\le n\minus{}2$.
When $ k\equal{}1,2 n\equal{}2,4$, but since $ n\ge 5$ we can rule these out. When $ k\equal{}3,n\equal{}7$ which contradicts the inequality. When $ k\equal{}4,n\equal{}9$ it can be easily checked that this doesn't work. Finally when $ k\equal{}5,n\equal{}16$ which should work.[/quote]

I think when $ k=4,n=11 $


Let all the irreducible fractions whose denominators don't exceed 99 written in increasing order from left to right: 1/99, 1/98,..., a/b, 5/8, c/d,... What are the fractions a/b and c/d on either side of 5/8 ?

Yeah, I know Farrey numbers. It's one of the possible approaches in order to solve the problem.

Well, then:

5b-8a=1;8c-5d=1 because (a/b,5/8) and (5/8,c/d) are pairs of consecutive fractions in the Farey sequence F99. From the same thing (they are CONSECUTIVE numbers), we can conclude that (a+5)/(b+8) and (5+c)/(8+d) can't belong to F99, so both b+8 and 8+d are >99, so b,d>91, so b,d belong to {92, 93, 94, 95, 96, 97, 98, 99}.

We can see that b is 5 (mod 8) because 8|1-5b, so b=93, meaning that a=(5b-1)/8=58, so a/b=58/93.

For the second fraction we know that d=3 (mod 8), so d=99 and c=(1+5d)/8=62, so c/d=62/99.

Is it correct?


an irregular quadrangle ABCD,E,G and F,H trisect AD and BC,then prove
that the sguare measure of quadrangle EFGH equals 1/3 of the sguare measure of quadrangle ABCD.

[quote]Let ABCD be a quadrilateral.  Points E and H trisect AD and points F and G trisect BC.  Show that the area of quadrilateral EFGH is one third the area of ABCD.[/quote]

[hide="Steps to take"]First, you want some casework: if $AD \parallel BC$, you have trapezoids and the result follows easily from the area of a trapezoid.  Otherwise, extend $AD$ and $BC$ until they meet at a point $J$ and then use what you know about triangle areas (most importantly, that if you fix side $XY$ and angle $\angle X$ of $\triangle XYZ$, then the area of $\triangle XYZ$ is proportional to the length $XZ$.  After that, it should all fall out with a little algebra.[/hide]


donbaleye $\{a_n\}$ be soorat e zir tarif shode ast
$a_0=0, |a_{n+1}|=|a_n+3|$

minimum e ebareteh zir ra biyabid:

$|a_1+a_2+...+a_{2006}|$

2ros e :yup:

edit: albateh rah haye dige ham hast, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78383]here[/url]


Find the coefficient of $ x^{6}$ in the expansion of $ (2\minus{}\frac{1}{4}x^{2})^{7}$.

The $ x^6$ term occurs when $ -\frac{1}{4}x^2$ is put to a power of $ \frac{6}{2}=3$, meaning the $ 2$ is put to a power of $ 7-3=4$.  By the Binomial Theorem, the desired term is
\begin{align*}\binom{7}{3}(2)^4\Big(-\frac{1}{4}x^2\Big)^3&=\frac{7\cdot6\cdot5}{3\cdot2}\cdot16\cdot\Big(-\frac{1}{64}\Big)x^6\\
&=35\cdot-\frac{1}{4}x^6\\
&=-\frac{35}{4}x^6\end{align*}
Thus, the coeficient of $ x^6$ is $ \boxed{-\frac{35}{4}}$


inequalities
trigonometry
inequalities proposed

1/sqrt (1+a^2)  + 1/sqrt(1+b^2) + 1/ sqrt (1+ c^2) =<3/2  prove this inequality where a,b,c are positive reals and satisfy a+b+c=abc. Determine when equality holds?[/hide]

Set $ a\equal{}\tan\alpha, b\equal{}\tan\beta, c\equal{}\tan\gamma$ and the rest follows with Jensen.

hello, so it would be better:
Prove the following inequality
$ \frac{1}{\sqrt{1\plus{}a^2}}\plus{}\frac{1}{\sqrt{1\plus{}b^2}}\plus{}\frac{1}{\sqrt{1\plus{}c^2}}\le\frac{3}{2}$
where $ a,b,c$ positive reals and satisfy $ a\plus{}b\plus{}c\equal{}abc$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, so it would be better:
Prove the following inequality
$ \frac {1}{\sqrt {1 \plus{} a^2}} \plus{} \frac {1}{\sqrt {1 \plus{} b^2}} \plus{} \frac {1}{\sqrt {1 \plus{} c^2}}\le\frac {3}{2}$
where $ a,b,c$ positive reals and satisfy $ a \plus{} b \plus{} c \equal{} abc$
Sonnhard.[/quote]

We can write it into:
Given $ x, y, z \geq\ 0$ and $ x\plus{}y\plus{}z\equal{}1.$
Prove that: $ \sqrt{\frac{xy}{xy\plus{}z}} \leq\ \frac{3}{2}$
It's obviously trues because of Am-Gm and : $ \sum\ \frac {x}{x \plus{} y} \plus{} \sum\ \frac {y}{x \plus{} y} \equal{} 3$
 :)

But isn't trigonometric substitution very much easier? You just have to prove that $ cosA \plus{} cosB \plus{} cosC \le \frac {3}{2}$ which is obvious because $ cosx$ is concave down for all $ x$ in the interval $ 0$ to $ \frac{\pi}{2}$, not equal to $ \frac{\pi}{2}$. Anyway, nice normalization nguoivn.


algebra
polynomial
Vieta
search
number theory theorems
number theory

Vieta-jumping method is used to prove some class problem .

Ex.1,Let $ a$ and $ b$ be positive integers such that
$ ab\plus{}1$ divides $ a^{2}\plus{}b^{2}$. Show that
$ \frac{a^{2}\plus{}b^{2}}{ab\plus{}1}$ is the square of an integer.

2, If $ a, b, c$ are positive integers
such that $ 0 < a^{2}\plus{}b^{2}\minus{}abc\le c$,
show that $ a^{2}\plus{}b^{2}\minus{}abc$ is a perfect square

3,Let $ x$ and $ y$ be positive integers such that $ xy$ divides $ x^{2}\plus{}y^{2}\plus{}1$. Show that
$ \frac{x^{2}\plus{}y^{2}\plus{}1}{xy}\equal{} 3$.

4,Let $ x$ and $ y$ be positive integers such that $ xy\minus{}1$ divides $ x^{2}\plus{}y^{2}$. Show that
$ \frac{x^{2}\plus{}y^{2}}{xy\minus{}1}\equal{} 5$. :)

I don't usually do this but here are some links to problems solved by "Vieta-Jumping"

[hide="10 links"]
http://www.mathlinks.ro/viewtopic.php?search_id=788196387&t=26232
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=159987
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=150374
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=150373
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=150371
http://www.mathlinks.ro/viewtopic.php?t=150369
http://www.mathlinks.ro/viewtopic.php?t=150372
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=605111911&t=6233
http://www.mathlinks.ro/viewtopic.php?search_id=1682761494&t=94
http://www.mathlinks.ro/viewtopic.php?t=5935[/hide]

[quote]I don't usually do this but here are some links to problems solved by "Vieta-Jumping"

http://www.mathlinks.ro/viewtopic.php?search_id=788196387&t=26232
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=159987
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=150374
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=150373
http://www.mathlinks.ro/viewtopic.php?search_id=1597502048&t=150371
http://www.mathlinks.ro/viewtopic.php?t=150369
http://www.mathlinks.ro/viewtopic.php?t=150372
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=605111911&t=6233
http://www.mathlinks.ro/viewtopic.php?search_id=1682761494&t=94
http://www.mathlinks.ro/viewtopic.php?t=5935
 [/quote]
Thank you very much :lol:

Who can solve problem 5 ( i have posted above ). It is a very nice problem was solved by Vieta - Jumping  :D

Sorry which problem?
(You wrote only four problems above... :maybe: )


A positive integer $ n$ has 60 divisors and $ 7n$ has 80 divisors. What is the greatest integer $ k$ such that $ 7^k$ divides $ n$?

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ 2\qquad
\textbf{(D)}\ 3\qquad
\textbf{(E)}\ 4$

Firstly assume, that $7\not|n$. $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ None of primes $p_i$ is equal to $7$.
$n$ has $(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1)=60$ divisors.
$7n$ has $120$ divisors according to the formula for the number of divisors, which is contradictory with our task.
So $n=7^kp_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$    $7n=7^{k+1}p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$
$(k+1)(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1)=60$
$(k+2)(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1)=80$
$\frac{k+1}{k+2}=\frac{3}{4}$, so $k=2$ and this is the answer.

[quote="JesusFreak197"][hide="Answer"]$60=20\cdot 3$, $80=20\cdot 4$
Adding one factor of $7$ raises the number of $20$'s by $1$, so there were previously $2$ factors of $7$, so $k=2\Rightarrow \boxed{C}$.[/hide][/quote]

Can you please be (a lot) more detailed with that explanation? That's quite a leap of logic that I wasn't able to take.  :oops:

dondigo, that was a beautiful solution. Thank you! :)

The only leap in logic seems to be the formula for finding the number of factors.

[quote="JesusFreak197"][quote="Hikaru79"]Can you please be (a lot) more detailed with that explanation? That's quite a leap of logic that I wasn't able to take.  :oops:[/quote]

$n$ is the product of primes. The number of divisors is found by $(a+1)(b+1)...$, where $a,b,...$ are the powers which the primes are raised to. If $7$ was not a factor before, then when it is added, it would multiply the number of factors by at least $2$, since you'd add $(1+1)$ into the multiplication. So, the only factor must be $7$, and take it from there.[/quote]

Oh, I understand *that* version of it. But that's closer to what dondigo posted; I was referring to your solution.

That's what I did in my solution, I just didn't post it all.

All three of us have the same solution. ;)


Let $ g: R\rightarrow R$ be given by g(x)=3+4x.If  $ g^{n}(x)\equal{}gogo....og(x)$ ,show that $ g(x)\equal{}(4n\minus{}1)\plus{}4^nx$ .Also find $ g^{\minus{}n}(x)$ where $ g^{\minus{}n}$ denotes inverse of $ g^{n}$

There is an error in the statement . :D  The correct formula is $ g_n{(x)}\equal{}4^n\minus{}1\plus{}4^nx$ .. :cool: