For $ a,b,c$ positive numbers

$ \sum \sqrt {\frac {a^2(b^2\plus{}c^2)}{a^2\plus{}bc} } \leq \sum a$

[quote="manlio"]For $ a,b,c$ positive numbers

$ \sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} } \leq \sum a$[/quote]
It's easy by Am-Gm! :)

[quote="manlio"]For $ a,b,c$ positive numbers

$ \sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} } \leq \sum a$[/quote]
Wow! After AM-GM we need prove only $ \sum_{cyc}(a^6bc\minus{}a^3b^3c^2),$ which is true.

Please, how to use AM-GM?  :blush:

Thank you very much. :)

manlio, we can use AM-GM like this:
$ \sum_{cyc} \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} } \leq a \plus{} b \plus{} c\Leftrightarrow$
$ \Leftrightarrow(a \plus{} b \plus{} c)^2\prod_{cyc}(a^2 \plus{} bc)\geq\left(\sum_{cyc}\sqrt {a^2(b^2 \plus{} c^2)(b^2 \plus{} ac)(c^2 \plus{} ab)}\right)^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6bc \plus{} 2a^5b^2c \plus{} 2a^5c^2b \plus{} 2a^4b^4 \plus{} 2a^4b^2c^2 \plus{} 2a^4b^3c \plus{} 2a^4c^3b \plus{} 3a^3b^3c^2)\geq$
$ \geq2\sum_{cyc}ab(c^2 \plus{} ab)\sqrt {(a^2 \plus{} c^2)(b^2 \plus{} c^2)(a^2 \plus{} bc)(b^2 \plus{} ac)}.$
But by AM-GM we obtain:
$ 2\sqrt {(a^2 \plus{} c^2)(b^2 \plus{} c^2)(a^2 \plus{} bc)(b^2 \plus{} ac)}\leq(a^2 \plus{} c^2)(b^2 \plus{} ac) \plus{} (b^2 \plus{} c^2)(a^2 \plus{} bc).$
Id est, it remains to prove that
$ \sum_{cyc}(a^6bc \plus{} 2a^5b^2c \plus{} 2a^5c^2b \plus{} 2a^4b^4 \plus{} 2a^4b^2c^2 \plus{} 2a^4b^3c \plus{} 2a^4c^3b \plus{} 3a^3b^3c^2)\geq$
$ \geq\sum_{cyc}ab(c^2 \plus{} ab)\left((a^2 \plus{} c^2)(b^2 \plus{} ac) \plus{} (b^2 \plus{} c^2)(a^2 \plus{} bc)\right),$
which is equivalent to $ \sum_{cyc}(a^6bc \minus{} a^3b^3c^2)\geq0.$

The following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \sqrt {\frac {a^2(a^2 \plus{} bc)}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2(b^2 \plus{} ac)}{a^2 \plus{} c^2}} \plus{} \sqrt {\frac {c^2(c^2 \plus{} ab)}{a^2 \plus{} b^2}}\geq a \plus{} b \plus{} c\]

[quote="stevenwang1992"]can  we use SOS ?[/quote]
I have tried it but without success.

[quote="manlio"]For $ a,b,c$ positive numbers

$ \sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} } \leq \sum a$[/quote]
Applying the Cauchy Schwarz Inequality, we have
$ \left( \sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc}}\right)^2 \le \left(\sum {a(b^2 \plus{} c^2)}\right)\left( \sum\frac {a}{(a^2 \plus{} bc)}\right).$
Because $ \sum a(b^2 \plus{} c^2) \equal{} (b \plus{} c)(a^2 \plus{} bc) \plus{} a(b^2 \plus{} c^2),$ the above estimation is equivalent to
$ \left(\sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc}}\right)^2 \le \sum \frac {a[(b \plus{} c)(a^2 \plus{} bc) \plus{} a(b^2 \plus{} c^2)]}{a^2 \plus{} bc} \equal{} \sum a(b \plus{} c) \plus{} \sum \frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc},$
The inequality is therefore reduced to
$ a^2 \plus{} b^2 \plus{} c^2 \ge \frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} \plus{} \frac {b^2(c^2 \plus{} a^2)}{b^2 \plus{} ca} \plus{} \frac {c^2(a^2 \plus{} b^2)}{c^2 \plus{} ab}.$
To prove it, we can see that $ \sum \frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} \equal{} \sum \frac {a^2b^2}{a^2 \plus{} bc} \plus{} \sum \frac {c^2a^2}{a^2 \plus{} bc} \equal{} \sum \frac {a^2b^2}{a^2 \plus{} bc} \plus{} \sum \frac {b^2c^2}{c^2 \plus{} ab},$ so it is equivalent to
$ \sum b^2\left( 1 \minus{} \frac {a^2}{a^2 \plus{} bc} \minus{} \frac {c^2}{c^2 \plus{} ab}\right) \ge 0,$
or
$ \sum b^2\cdot \frac {ac(b^2 \minus{} ac)}{(a^2 \plus{} bc)(c^2 \plus{} ab)} \ge 0,$
that is
$ \sum b^2ac(b^2 \minus{} ac)(b^2 \plus{} ca) \ge 0.$
Since $ (b^2 \minus{} ac)(b^2 \plus{} ac) \equal{} b^4 \minus{} c^2a^2,$ this last inequality is equivalent to
$ abc\sum b(b^4 \minus{} a^2c^2) \ge 0,$
or
$ a^5 \plus{} b^5 \plus{} c^5 \ge abc(ab \plus{} bc \plus{} ca).$
Of course, the last one is obvious from the AM-GM Inequality, so we are done.

[quote="nguoivn"][quote="arqady"]
The following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \sqrt {\frac {a^2(a^2 \plus{} bc)}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2(b^2 \plus{} ac)}{a^2 \plus{} c^2}} \plus{} \sqrt {\frac {c^2(c^2 \plus{} ab)}{a^2 \plus{} b^2}}\geq a \plus{} b \plus{} c\]
[/quote]
It's stronger than ductrung's ineq (by CS). And I obtain the strongest:
$ \sum\ a^6 \plus{} 2\sum\ a^3b^3 \plus{} 15a^2b^2c^2 \plus{} 8abc\sum\ a^3 \geq\ \sum\ a^5(b \plus{} c) \plus{} \sum\ a^4(b^2 \plus{} c^2)$ $ \plus{} 6abc \sum\ ab(a \plus{} b)$
But SOS is so ugly for this last ineq  :wink:[/quote]
I's easy by Am-Gm,my friend!
We have $ \sum\sqrt{\frac{a^2(a^2\plus{}bc)}{b^2\plus{}c^2}}\geq\sum\frac{2a(a^2\plus{}bc)}{a^2\plus{}b^2\plus{}c^2\plus{}bc}$ (by Am-Gm)
And
$ \sum\frac {2a(a^2 \plus{} bc)}{S \plus{} bc} \minus{} RHS \equal{} \sum\frac {a(a \minus{} b)(a \minus{} c)}{S \plus{} bc} \plus{} \sum\frac {abc(a \minus{} b)^2}{(S \plus{} bc)(S \plus{} ca)}\geq0$ (with $ S \equal{} a^2 \plus{} b^2 \plus{} c^2$) :lol:

[quote="arqady"]nguoivn, 2424, there is solution of my inequality by one line.  :wink:[/quote]
Oh :o .Please post your solutions :)

I think arqady meant

$ \left(\sum \sqrt{\frac{a^2(a^2\plus{}bc)}{b^2\plus{}c^2}}\right)\left( \sum \sqrt{\frac{a^2(b^2\plus{}c^2)}{a^2\plus{}bc}}\right) \ge (a\plus{}b\plus{}c)^2,$
from the Cauchy Schwarz Inequality, and since $ \sum\sqrt{\frac{a^2(b^2\plus{}c^2)}{a^2\plus{}bc}} \le a\plus{}b\plus{}c$ (we have proved above), it is clear that
$ \sum \sqrt{\frac{a^2(a^2\plus{}bc)}{b^2\plus{}c^2}} \ge a\plus{}b\plus{}c.$

[quote="manlio"]For $ a,b,c$ positive numbers

$ \sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} } \leq \sum a$[/quote]$ \sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc} } \leq\sum{\frac {a[2a(b^2 \plus{} bc \plus{} c^2) \plus{} (b \plus{} c)(2b^2 \minus{} bc \plus{} 2c^2)]}{2[a^2(c \plus{} b) \plus{} b^2(a \plus{} c) \plus{} c^2(b \plus{} a)]}} \equal{} \sum a.$
[quote="can_hang2007"]Applying the Cauchy Schwarz Inequality, we have

$ \left( \sum \sqrt {\frac {a^2(b^2 \plus{} c^2)}{a^2 \plus{} bc}}\right)^2 \le \left(\sum {a(b^2 \plus{} c^2)}\right)\left( \sum\frac {a}{a^2 \plus{} bc}\right)$[/quote]$ \equal{}(a\plus{}b\plus{}c)^2\minus{}\frac{abc(a^5\plus{}b^5\plus{}c^5\minus{}ab^2c^2\minus{}bc^2a^2\minus{}ca^2b^2)}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}\leq(a\plus{}b\plus{}c)^2.$


Galois Theory
number theory solved
number theory

hey, I've looked at IMO2003 problem 6 on kalva webpage
and his solution seems rather magical? Where does this solution
come from?
Has anyone a more natural solution ?

There seems to be an other solution which uses Galois theory (something with Chebotarev density and Galois closure and ...) but that's really Chinese to me - if you're interested, though, you can find it at the official imo03 website ... but for me it's absolutely incomprehensible  :? but maybe this is "natural" for you ?
I didn't see any other solution, it always comes down to showing that p^p - 1 has a prime divisor q which is not equal to 1 mod p^2 and proving that this q indeed satisfies the desired condition.

However this question is way to difficult for me ... I'm happy with my honourable mention  :)


limit
superior algebra
superior algebra unsolved

for what real $x$ is $x-\frac{1}{x}$ integer , except for $-1$ and $1$ ?

$\lim_{x\to\infty}x-\frac{1}{x}=+\infty$ and $\lim_{x\to-\infty}x-\frac{1}{x}=-\infty$ and $f(x)=x-\frac{1}{x}$ is continous so $\text{N}\supset \text{Im}f$ . More than that $f(x)=n$ yields $x^{2}-nx-1=0$ which gives the asked numbers.


what did you get on the AMC 10B? So i decided to just improve my pathetic AMC 10A score of [hide="shut up"]132[/hide], so i just did 1-22 and 24, and barely read the other two problems and i got all 23 corect, so yeah 141 for me, do people think a 7 on AIME is safe for making USAMO...

Yes, 7 is almost definitely safe.

AMC10A -138
AMC10B -150 (unless I misbubbled something)

that is so dramatic that on your last chance to get a perfect on AMC 10, you get one... nice job fanatic, i am proud of you...  :)

Haha yeah, 5th time's the charm =P.

Thanks, and gl to you too next year. 141 > what I had last year on the 10 (132). So I expect a 150 out of you next year  :P

dude
I decided to improve my score of 115.5 on teh 10A lol
I fail a lot.
I should attempt to not make stupid mistakes which brought a 10 on teh AIME last year for me to a 5


$*)$If $x\ge\pi$ then the equation is equivalent to
$2\pi\cos x=x-x+\pi$
$\Leftrightarrow x=\pm\frac{\pi}{3}+2k\pi$ $k\in\mathbb{Z}$ and $k\ge1$
$*)$If $x\le 0$, by the same way we have
$\cos x=-\frac{1}{2}$
$x=\frac{2\pi}{3}+2k\pi$ $k\le-1$ or
$x=-\frac{2\pi}{3}+2k\pi$ $k\le0$
$*)$ If $0\le x\le \pi$
$2\pi\cosh=2x-\pi$
Let $f(x)=2x-\pi-2\pi\cosh$
$f'(x)=2+2\pi\sinh\ge0\forall x\in(0;\pi)$
And $f(\frac{\pi}{2})=0$ so $\frac{\pi}{2}$ is the only root of the equation $f(x)=0$ with $x\in (0;\pi)$


[quote="mathgeniuse^ln(x)"][quote="Agr_94_Math"]Find integer solutions suhc that :
$ 35x^3 \plus{} 66x^2 y \plus{} 42xy^2 \plus{} 9y^3 \equal{} 9$.[/quote]

[hide]Notice this is equivalent to $ (3x \plus{} 2y)^3 \plus{} (2x \plus{} y)^3 \equal{} 9$.  Now, we can find all pairs of integers such that $ a^3 \plus{} b^3 \equal{} 9$, which is easy to do.[/hide][/quote]

Nice transformation :)


assume that $G$ is barpartite graph with two part $X$ and $Y$ such that $|X|\geq |Y|$ .prove that there exists at least one maximal path with at least one end point in $X$.

Maybe, it is connected, or at least $X$ does not contain isolated vertices?

isolated points considered as maximal path :)

Do you mean maximal by inclusion, not maximal by the number of edges?

At first, we may suppose that there are no isolated vertices in $X$. Vertices from $X$ are denoted by $x$-s, and from $Y$ --- by $y$-s.

Lemma:  there exist some edges $x_1y_1,\,x_2y_2,\,\dots,x_sy_s$ such that no vertice $x_i\, (i=1..s)$ is joined with any vertex $y$, different from $y_1,\,y_2,\dots,y_s$.

In wedding language, if $X$ means boys and $Y$ means girls, it sounds as follows: it is possible to marry several boys (on familiar girls) so that no married boy does not know any unmaried girl.

Proof of the lemma: if any $k$ boys no at least $k$ girls, we may marry all boys by Hall lemma, and since $|Y|\le |X|$, all girls are married too. If not, find the least $k$ such that some $k$ boys no less than $k$ girls, say, they know only $l<k$ girls. Then choose arbitrary $l$ boys from these $k$. For them, Hall lemma conditions holds, so we may marry them. It is just what we needed. Note that $k>0$, since there are no isolated vertices in $X$.

It is easy to deduce a statement from a lemma. Consider a path $y_1-x_1$. If it may not be proceeded from vertice $x_1$, then it is contained in some maximal path with endpoint $x_1$. So, it may be proceeded, say, $x_1$ is joined with $y_2$. Then consider path $y_1-x_1-y_2-x_2$. If it may not proceeded from $x_2$, we again find a maximal path with endpoint $x_2$, and so on. But this process must be finite, so we find a maximal path with endpoint in $X$.


algebra
function
domain
vector
topology
integration
real analysis

show that the 1-form deifned on $ \mathbb{R}^2 \setminus
\{(0,0)\}$ by
$ w = \frac { - y}{r^2} + \frac {x}{r^2}dy$
is closed but not exact
also show the 2-form defined on the spherical shell by
$ w = \frac {x}{r^3}dy\wedge dz + \frac {y}{r^3}dz\wedge dz + \frac {z}{r^3}dx\wedge dy$
is closed but not exact.

These should follow immediately from definitions.

Take $ d\omega$ and you should get 0. This means closed. The first one is not exact since the partial of the $ dy$ coefficient is nonzero, so it can't match up with the partial of the $ dx$ term which is zero.

For the second one, try to construct a potential. You should get some sort of contradiction.

I assume that Mathx's original post contains a typo, and that first form should have been

$ w \equal{} \frac { \minus{} y}{r^2}\,dx \plus{} \frac {x}{r^2}\,dy,$ where $ r^2 \equal{} x^2 \plus{} y^2.$

It's the classic example, of course. And $ dw$ does equal zero, only its domain is not a simply connected region.

The simplest way to show that it's not exact would be to compute $ \oint_Cw$ where $ C$ is a curve that encloses the origin - the unit circle would be the easiest.

[u]The Second Form[/u]: The proof that $ \omega$ is closed is straightforward; you can write $ \omega \equal{} g\sum_{cyc}xdy\wedge dz$, where $ g \equal{} \frac {1}{r^3}$, and then use the product rule.  You can find that $ \omega|_{S^2} \equal{} \sum_{cyc}xdy\wedge dz$.  But notice that $ (x,y,z)$ is the unit normal vector of the manifold $ S^2$ at the point $ (x,y,z)$, so $ \omega|_{S^2} \equal{} dA$.  .  Considering $ \partial\bar{B^3} \equal{} S^2$, Stokes' Theorem implies that if $ d\eta \equal{} \omega$, then $ 0 \equal{} \int_{S^2}dA$, which is a contradiction.


function
symmetry
geometry
geometric transformation
reflection
3D geometry
sphere

Consider a function f(z), meromorphic on region G , which contains the disc |z| $ {\le 1}$.
Further f is to satisfy that |z|=1 implies |f(z)|=1.
Let F(z)= f(z)  if |z| $ {\le 1}$
           $ \equal{} f(\frac {z}{\overline{z}})$ if |z| >1

a) Deduced F(z)=f(z) on G
b) Show that if $ {\alpha}$ is a zero of F then 1/$ \overline{\alpha}$

There is probably some misprint in your definition of F for |z|>1. As you defined it, F is not analytic, not meromorphic, and has no zeros in |z|>1.

Pym.

Hi Pym

I am sorry for the mistake and the corrected one is given below

Consider a function f(z), meromorphic on region G , which contains the disc |z|$ {\le 1}$ .
Further f is to satisfy that |z|=1 implies |f(z)|=1.
Let F(z)= f(z) if |z| $ {\le 1}$
           = $ \overline{f(1/\overline{z})}$ if |z| >1

a) Deduced F(z)=f(z) on G
b) Show that if  is a zero of F then 1/$ \overline{\alpha}$

Achatter,
This is called the Symmetry Principle. All textbooks contain it with the symmetry with respect
to a line, but there is also symmetry with respect to a circle. If the circle is the unit circle,
the symmetry is given by the formula s(z)=1/overline{z}.
If a function f is meromorphic in the unit disc, continuous in the closed unit disc and maps
the unit circle to itself, it can be extended to the outside of the disc by the formula
F(z)=s(f(s(z)) (which is the same as your formula for the extension).
(It is analogous to the formula overline{f(overline{z})} in the usual reflection principle).
The extended function is  meromorphic: it has poles at those points outside
the unit circle which are symmetric to its zeros inside. And the other way around.
Because the point infinity is symmetric to zero
with respect to the unit circle.

The extended function is meromorphic on the whole Riemann sphere, thus it is rational.
More advanced textbooks contain this stuff.
Hope this helps.
Pym.


Given a right triangle ABC with $\angle A=\frac{\pi}{2}$. Prove the following inequality.

\[\sqrt{2}BC \geq CA+AB\]

I love this one!
Cauchy it: $(AB+AC)^2 \leq (1+1)(AB^2+AC^2) = 2BC^2$.
Hence: $AB+AC \leq \sqrt{2} BC$

Thank you for your answer,shobber.

By the way, how can we solve this problem in geometrically? :)

kunny

I realized it that you are asking for a geometrical solution after I posted....Because Cauchy makes it so easy.

So here is another proof, I don't know if it is ok.

\[(AC+AB)^2 \leq 2BC^2\]

\[\Longleftrightarrow AC^2+AB^2+2AB\cdot AC \leq 2BC^2\]

\[\Longleftrightarrow 2 AB \cdot AC \leq BC^2\]

\[\Longleftrightarrow 4 \triangle{ABC} \leq BC^2\]

\[\Longleftrightarrow \triangle{ABC} \leq \frac{1}{2} \cdot \frac{BC}{2} \cdot BC\]

Since $\angle{A}=90^{o}$, so $\frac{BC}{2}$ equals to the median of side BC, and the median of side BC is greated or equal than the height of side BC, which implies $\frac{1}{2}\cdot\frac{BC}{2}\cdot BC \geq \frac{1}{2}\cdot h_{BC} \cdot BC =\triangle{ABC}$.

Equality holds if and only if AB=AC. ($h_{BC}=m_{BC}$).


[quote="mitdac123"]Try $ x \equal{} y;z \equal{} 0$ ;)[/quote]
you are wrong mitdac123,$ x,y,z,$ are the side lengths of a triangle,i think it can be proved by mixing variables

:P . I don't think so . $ x\equal{}y,z \rightarrow 0$ ;)

Yes, mitdac is true.
@tranvanluan: You should check your ineq before post on any forum :D
@Allnames: if you can solve any problem, you should post the full solution. I think can_hang has talked to you about it :D

[quote="tranvanluan"]Let x;y;z are the side lengths of a triangle.
Prove that
$ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{x \plus{} z}} \plus{} \sqrt {\frac {z}{y \plus{} x}}\ge \frac {3}{\sqrt {2}}$[/quote]
It's wrong.
Try triangle with side lengths $ 14,$ $ 14$ and $ 1.$ :wink:

[quote="tranvanluan"]Let x;y;z are the side lengths of a triangle.
Prove that
$ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{x \plus{} z}} \plus{} \sqrt {\frac {z}{y \plus{} x}}\ge \frac {3}{\sqrt {2}}$[/quote]

  Inequality true is

 $ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{x \plus{} z}} \plus{} \sqrt {\frac {z}{y \plus{} x}}>2$

[quote="Marius Mainea"]

  Inequality true is

 $ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{x \plus{} z}} \plus{} \sqrt {\frac {z}{y \plus{} x}} > 2$[/quote]
The following inequality is true too:
Let $ x,$ $ y$ and $ z$ are non-negative numbers which not two of them equal to 0. Prove that:
\[ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{x \plus{} z}} \plus{} \sqrt {\frac {z}{y \plus{} x}} \geq 2\sqrt{1\plus{}\frac{xyz}{(x\plus{}y)(x\plus{}z)(y\plus{}z)}}\]

<=>$ \sum x(x\plus{}y)(x\plus{}z) \plus{} 2\sum (x\plus{}y)\sqrt{xy(x\plus{}z)(y\plus{}z)} \ge 4(\sum x)(\sum xy)$
By am-gm +schur, we hvae :
+$ LHS \ge \sum x(x\plus{}y)(x\plus{}z)\plus{}12xyz\plus{}2\sum xy(x\plus{}y)$
+$ x(x\plus{}y)(x\plus{}z)\plus{}12xyz\plus{}2\sum xy(x\plus{}y)\minus{}4(\sum x)(\sum xy) \equal{} \sum x^3\plus{}3xyz\minus{} \sum xy(x\plus{}y) \ge 0$

[quote="arqady"][quote="Marius Mainea"]

  Inequality true is

 $ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{x \plus{} z}} \plus{} \sqrt {\frac {z}{y \plus{} x}} > 2$[/quote]
The following inequality is true too:
Let $ x,$ $ y$ and $ z$ are non-negative numbers which not two of them equal to 0. Prove that:
\[ \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{x \plus{} z}} \plus{} \sqrt {\frac {z}{y \plus{} x}} \geq 2\sqrt {1 \plus{} \frac {xyz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)}}
\]
[/quote]
With this problem we have 2 way to solved it
The way 1:it is similar to mitdac123 `s proof
The way2 (me)
$ (\sum a^2(b \plus{} c)).LHS^2 \geq (a \plus{} b \plus{} c)^3$ *
let $ a \plus{} b \plus{} c \equal{} 1 \equal{} p,\sum ab \equal{} \frac {1 \minus{} q^2}{3},abc \equal{} r$
We will prove
$ < \equal{} > \frac {4(1 \minus{} q)^2 \minus{} 3(1 \minus{} q^2)}{36(1 \minus{} q^2) \minus{} 9}\leq r$
Use $ r\leq \frac {1 \plus{} q)^2(1 \minus{} 2q)}{27}$
we will prove this ineq
$ 12(1 \minus{} q)^2(1 \plus{} q) \minus{} 9(1 \minus{} q)\leq (1 \plus{} q)(1 \minus{} 2q)(4(1 \minus{} q^2) \minus{} 1)$
$ < \equal{} > q^2(4q \minus{} 1)^2\geq\ 0$
Am i wrong

I haven't check your solution :) .But this tech's too strong ;) .


assume that we have hexagon and we construct three rectangle on even sides inwardly. as figure below.prove that it is possible for odd sides.

I 'll explain my notion first
    Let ABCDEF is an hexagon and in AB, CD, EF, we can construct inwardly 3 rectangles like the figure. Then we can do that in BC, DE, FA.
      First we notice this two lemma
       1) Let ABCD be a rectangle. Then the equality holds for every M: $MA^2+MC^2=MB^2+MD^2$
       2) Let ABCD be a quadilateral. Then ABCD is a triangle if and only if the equality $MA^2+MC^2=MB^2+MD^2$holds for all M in the plane (ABCD).
    Proving these two lemma isn't very difficult.
    According to our problem and my first lemma, we obtain for every M we have $MA^2+MC^2+ME^2=MB^2+MD^2+MF^2$.
    Now, through B we draw a line which is perpendicular to BC. It cuts a line through A which is perpendicular to AF at K. Now construct the rectangle BKLC and AKNF. Then for every M, we have $ML^2+ME^2= MN^2+MD^2$. Thus DENL is a rectangle.
     I've done some research about this diagram before and found some beautiful properties. I call the two triangles ACE and BDF "twin triangles". But I lost my writng about it. But I still remember some of them
           1) the midperpendicular of the odd sides are concurent at X.
           2) the midperpendicular of the even sides are concurent at Y.
           3) The total area of the three rectangles contructed in the odd sides are equal to the even sides'.
           4) Let's call $A_1, C_1, E_1$ is the small triangle we obtain after constructing in AB, CD, EF. And $B_1, D_1, F_1$ is the small triangle we obtain after constructing in BC, DE, FA. Then $A_1E_1C_1$ and $B_1D_1F_1$ are orthologic.
        The rest I'll post in my next post.


My question is in two parts:

1- if I have a probability that an event would occur after one year, how can I compute this probability after 6 months and after 2 years?

2- Are the computations above going to change if I consider the probability of occurence discrete or continuous? if so how can I account for the fact that the event is discrete or continuous
Thx

Hmm...I'm not too good with probabilities, but maybe to find it after 6 months, you can raise the probaility to the $1/2$ power, and after two years, you can raise it to the $2nd$ power.


In October 1929, a middle aged mycologist died shortly after eating a mushroom stew he prepared from warty caps ([i]Amanita Rubescens[/i]) collected in the nearby wood. Cause of death: poisoning by Muscarine(figure given below), an alkaloid found in the fly agaric ([i]Amanita Muscaria[/i]).

You are Sir James Lubbock, Home Office Analyst, and you have been aasked to solve a problem crucial to the investigation: whether (a) a deadly [i]Amanita muscaria[/i] found its way into the mess of a closely similar, but harmless [i]Amanita rubescens[/i]; or (b) a lethal dose of synthetic muscarine (filched from a London laboratory) was deliberately added to the stew pot - perhaps by the lover of the beautiful Mrs. Harrison.

You have available a solution of Muscarine isolated from the leftover stew, a 1929 laboratory and 10 minutes. Can you provide a definite answer to the question: Was there a fly agaric in Mr. Harrison's soup- or did a second cook wilfully and with malice aforethought, spoil the broth?

[hide="Hint to the chemistry student"]Most of the alkaloids are produced by plants in onlyone of 2 possible enatiomeric forms and hence they are optically active.[/hide]

[hide="Suggestion to the biology student"]We expect a "chemical" answer. :rotfl:
No offence meant.  :wink:  :P  [/hide]

[quote="rituraj007"]i think it is a question from Morrison Boyd

and the ans is he was deliberately poisoned (that's clear from hint :rotfl: )[/quote]

How did you decide without experimenting?  :?:

well i just guessed
on basis of a bit of expts :P


Convert the following into $ \alpha$ napthol  :D

[hide]I am not actually worried abt the yield and other such things.......  :P  just propose a scheme  [/hide]

[hide="Method?"]

1) Put NBS to get Br at allylic (not at the protruding methyl gp)

2) Put excess SeO2 to allow OH to ccupy both available allylic positions.
3) aq. KOH to get the ketone.
4) conc. H2SO4 to expand ring and subsequent removal of proton to get double bond.

The above obtained product tautomerizes to the more stable alpha napthol.
 :) [/hide]


Here are my choices:

MIT
Duke
Harvard
Wake Tech... ROFL j/k

....

....

I hate that californian one >_>

Hmm, I think I want to go to Stanford :wink:.

I think the topic has been discussed here: http://www.mathlinks.ro/viewtopic.php?t=146351  :wink:
By the way, I'm going to MIT :)


As I see lots of Iranian people do not know LaTeX, here is a nice pdf file by Valentin Vornicu for learning LaTeX. I started learning LaTeX four years ago with this document.

The "Not So Short Guide to $ \text{\LaTeX}$" is also really good.  It covers a lot of stuff and does a good job with the basics.  Sometimes, though, if there is something that the guide doesn't have, the wiki is really, really good.  It has a lot of really nice tidbits and examples.

You can use [url=http://www.dessci.com/en/products/TeXaide]Texaide[/url]  if u r a lazy bone like me .


[hide]

x^2 = 1 mod 2

Thus x is odd, or x=2k+1.

2^n + 7 = 4k^2 + 4k + 1

2^n = 4k^2+4k-6

2^(n-2) = k^2 + k - 3/2

Since k^2 + k is an integer, the denominator of 2^(n-2) must be 2; thus 2^(n-2) = 1/2 and n=1.  So

2^1 + 7 = 9 = x^2

x = :pm: 3[/hide]

[quote="MCrawford"]Find all solutions of
\[ 2^n + 7 = x^2 \]
in which $n$ and $x$ are both integers.  Prove that there are no other solutions.[/quote]

I assume someone's already done this, but mod 4, for n >= 2, we have:
3 == x^2 mod 4, which is obviously impossible.

Hence, n<2.  It is impossible for n < 0, so we try n=0 and n=1.
8 is not a square, 9 is, hence (1,3) is the only solution.

[quote="nr1337"]
x^2 = 1 mod 2
Thus x is odd, or x=2k+1.. [/quote]

Just be a little careful..[b] that's only true for n>0[/b]. HTH

(and for n>1 we get x^2 = 3 mod 4 an imposibility  so we have to check only for n=0,1... as pointed out by others..)

We can be sure of it that there are only 2 roots of the equation.this is because the graphs of 2^n +7 and x 2 cut only twice.also from the graph it can be seen that one value is -ve and other is +ve.

Sriram

Pay attention to this graphical method of analyzing possible solutions.  It comes in handy over and over again when comparing functions whether or not the equation is Diophantine.


[img]http://i57.photobucket.com/albums/g206/despair89/orangeballs-1.jpg[/img]

[b]Calculate the exact number of balls that can be placed inside this container with the cover, using the given dimensions.[/b]

*Fitting the cover onto the container left me quite perplexed. Not sure hwo to do it though. This question is from a local contest.. no other additional info has been given so i just assumed that the container is a perfect cone..

Can anyone try this? get the exact answer?

hint[hide]See similar problems about stacking spheres. Then, create a general problem for n number of spheres and then solve[/hide]

What kind of shape is the cover? Looks like a part of a sphere?

yes it is is part of a sphere... i am just assuming. The contest form made no mention of that.


a and b are real solutions of $ x^4\equal{}4x\plus{}1$ find (2-a)(2-b).

[quote="bjh7790"]a and b are real solutions of $ x^4 \equal{} 4x \plus{} 1$ find (2-a)(2-b).[/quote]

Let $ x^4 \minus{} 4x \minus{} 1 \equal{} (x^2 \plus{} ux \plus{} v)(x^2 \minus{} ux \plus{} w)$, with $ u > 0$

We get :
$ v \plus{} w \equal{} u^2$
$ uw \minus{} uv \equal{} \minus{} 4$
$ vw \equal{} \minus{} 1$

So :
$ (E1)$ : $ w \equal{} \minus{} \frac {1}{v}$

$ (E2)$ : $ v \minus{} \frac {1}{v} \equal{} u^2$

$ (E3)$ : $ v \plus{} \frac {1}{v} \equal{} \frac {4}{u}$

$ (E2)^2 \minus{} (E3)^2$ $ \implies$ $ u^4 \minus{} \frac {16}{u^2} \equal{} \minus{} 4$ and so $ u^2 \equal{} z$ and $ z^3 \plus{} 4z \minus{} 16 \equal{} 0$, with a unique trivial real root $ z \equal{} 2$ and so $ u \equal{} \sqrt 2$ (we choosed $ u > 0$ in the first line).

$ (E2) \plus{} (E3)$ $ \implies$ $ 2v \equal{} u^2 \plus{} \frac {4}{u}$ and so $ v \equal{} 1 \plus{} \sqrt 2$ and so $ w \equal{} 1 \minus{} \sqrt 2$

So : $ \boxed{x^4 \minus{} 4x \minus{} 1 \equal{} (x^2 \plus{} x\sqrt 2 \plus{} 1 \plus{} \sqrt 2)(x^2 \minus{} x\sqrt 2 \plus{} 1 \minus{} \sqrt 2)}$

So the two real roots $ a$ and $ b$ are the roots of $ Q(x) \equal{} x^2 \minus{} x\sqrt 2 \plus{} 1 \minus{} \sqrt 2$ (the only quadratic with real roots)

And the required result is $ Q(2) \equal{} 5 \minus{} 3\sqrt 2$

Now we know the result, there is a quicker method :

$ x^4\minus{}4x\minus{}1\equal{}(x^2\plus{}1)^2\minus{}2(x\plus{}1)^2$

$ x^4\minus{}4x\minus{}1\equal{}(x^2\plus{}1\plus{}(x\plus{}1)\sqrt 2)(x^2\plus{}1\minus{}(x\plus{}1)\sqrt 2)$

$ x^4\minus{}4x\minus{}1\equal{}(x^2\plus{}x\sqrt 2\plus{}1\plus{}\sqrt 2)(x^2\minus{}x\sqrt 2\plus{}1\minus{}\sqrt 2)$

And the result.


HEY EVERYONE IF U UPGRADE MY CRACTER IN BATTLEON username:haider6
I WILL GIVE U A LEVEL 50 ACCOUNT IN BATTLEON username:deepak50

                 :lol:

Hey somebody is already playing as bubka!

Try turning the keyboard upside down and shaking it or tapping the corner on the table to get out whatever's causing your caps lock key to stick.

how are we supposed to upgrade your character?

[quote="Chocoboom"]youaretheloser[/quote]

how nice of you to point that out... -___-

/sarcasm

I am not getting this at all. And as for the game it is stupid. I have seldom seen such horrid graphics!

I have a character on 90 level and i am guardian i give my guardian to another guardian if U want send me your pasword and user to free9jocker@yahoo.com :D


My question is: if an entire function $ g$ satisfies $ g(z)\equal{}g(2z)$ for all $ z \in \mathbb C$ then $ g$ is a constant. My solution is as follows: for each $ z$ fixed, one has $ g(z)\equal{}g\left( \frac{z}{2^k} \right)$ for every $ k$. Taking $ k \to \infty$ we obtain $ g(z) \equal{} g(0)$ which implies that $ g$ is a constant. So how about the approach  :oops: ?

[quote="ZetaX"]It's ok.[/quote]

Aha, thanks. Now I want to know where did I use entire property in the argument?

You used that it's defined at $ 0$ and the other points you used. That's weaker, but entire is easier to write down (and a lot of theorems/problems use stronger assertions than necessary for simplicity).

One consequence of using strong assumptions is that there are multiple ways to solve the problem.  For example, using the fact that $ g$ is bounded on, say, the unit disk, it follows that $ g$ is bounded everywhere and the result follows by Liouville's theorem, which was perhaps the intended argument.

The weakest assumption for the problem statement to be true, however, is being defined everywhere and being continuous in a neighborhood of zero, which is much weaker than being entire (it does not require that $ g$ is analytic).


[img]http://agutie.homestead.com/files/equilic/equilic_quadrilateral_04.gif[/img]

Please prove AJD is an equilateral.

[color=blue]Equilic quadrilateral is a quadrilateral ABCD with one pair of opposite sides equal (AB = CD)
and the sum of angles A and D equal to 120° [/color]

The original problem says that the triangle $JBC$ is equilateral and it is drawn externally of $ABCD$.
You should post the wording of the problem

[hide="Hint"]Try to prove that the triangles $ABJ,DCJ$ are congruent[/hide]

I think that in a book of Ross Hornberger I have seen 4-5 good properties of equilic quadrilaterals. I had proven all them years ago. This , proven by PONTIOS , is a simple and nice. Students who have the book can post some more properties(problems).

   [u]Babis[/u]

For those who has Honsberger's book [i]Gems III[/i] it is easy to check that the proof has a subtle flaw. For an explanation and a valid proof, see

http://www.cut-the-knot.org/Curriculum/Geometry/EquilicQuad1Tri.shtml

A. Bogomolny
http://www.cut-the-knot.org

P.S. I'd be grateful for more slips like that.


I'm from Fairfield County, and I just got back on this site!  (I made my account a long time ago, but haven't used it in a long time.)  I just wanted to introduce myself.

So, who here took the AMC?  I just did it yesterday.

@360 fan:do you do mathcounts

gl with ur wii (im so glad my bro didn't come early enough :) )
-jorian

Yeah, I've been doing Mathcounts since last year.  Got up to the state round individually.  This year I formed a team, and we're practicing.

What chapter are you in?  I'm in the SW CT chapter.

Fairfield University gave me some free courses for participating.  Did you get anything?  (We almost threw that letter away because it was in all the junk in my backpack. :P )

We don't give out any prizes in New Haven County


group theory
abstract algebra
superior algebra
superior algebra unsolved

This problem is from Martin Isaac's Algebra, chapter 5.

Let $ |G|\equal{}p(p\plus{}1)$, where $ p$ is prime. Show that $ G$ has either
a normal subgroup of order $ p$ or one of order $ p\plus{}1$.

There is a hint on the book saying that: if $ n_p (G)>1$ and
$ x\in G$ with $ o(x)\neq 1,p$, then $ |C_G (x)|\equal{}p\plus{}1$.
Here, $ n_p(G)$ denotes the number of $ p$-Sylow subgroups
of $ G$ and $ C_G (x)$ denotes the centralizer of $ x$
in $ G$.
I'm still confused of the connection between Sylow subgrups
and the centralizer of some elements.
I've tried using the action of $ G$ on the set $ \{x\in G | o(x)\neq 1\}$
by conjugation but still getting nowhere.
Can somebody help me?

You can try to use the action of $ G$ on the set $ \{x \in G|o(x) \ne 1,p\}$ by conjugation.

If $ n_p(G) \equal{} 1$, the $ p$-Sylow is normal and so the group has a normal subgroup of order $ p$.

Let's suppose $ n_p(G) > 1$.  $ n_p(G) \equiv 1\mod p$ and $ n_p | (p\plus{}1)$, then $ n_p \equal{} p\plus{}1$.

Let's consider the set $ X \equal{} \{x \in G\ |\ o(x) \neq 1,p\}$. $ |X| \equal{} p$ because if $ x \in X$ then it doesn't exist any $ p$-Sylow $ P$ such that $ x \in P$ and there are $ p(p\plus{}1) \minus{} [(p\minus{}1)(p\plus{}1) \plus{}1] \equal{} p^2 \plus{} p \minus{} p^2 \equal{} p$ elements that are not in a $ p$-Sylow.
Conjugate elements have the same order and so $ Cl(x) \subset X$. But it imply that $ Cl(X) \equal{} X$ and $ N_G(X) \equal{} G$. Now $ X \cup \{1\}$ is a subgroup, in fact, $ \forall x, y \in X$, $ p$ not divides $ o(xy^{\minus{}1})$ and so $ xy^{\minus{}1} \in X$. But $ N_G(X) \equal{} G$ and so $ X \cup \{1\}$ is normal and $ |X \cup \{1\}| \equal{} p\plus{}1$.

I hope my english is not too bad...  :blush:


a,b,c,d be real numbers in G.P. If u,v,w satisfy the equations u+2v+3w=6; 4u+5v+6w=12; 6u+9v=4 then show that roots of the equations (1/u+1/v+1/w)x^2 + [(b-c)^2 + (c-a)^2 + (d-b)^2]x + u+v+w=0 and 20x^2 + [10(a-d)^2]x - 9 =0 are reciprocal to each other.

plz give a detail solution...
thank you...

Solving the equation for $ u , v\ and\ w$ we get
$ 2u\plus{}v\equal{}0\ and\ 6u\plus{}9v\equal{}4\Rightarrow v\equal{}2/3,\ u\equal{}\minus{}1/3\ w\equal{}5/3$

$ 20x^2 \plus{} 10(a\minus{}d)^2x \minus{} 9 \equal{}0 .....(A)$
$ (1/u\plus{}1/v\plus{}1/w)x^2 \plus{} [(b\minus{}c)^2 \plus{} (c\minus{}a)^2 \plus{} (d\minus{}b)^2]x \plus{} u\plus{}v\plus{}w\equal{}0...(B)$
the equation having roots reciprocal of $ A$ is
$ 20\plus{}10(a\minus{}d)^2x\minus{}9x^2\equal{}0 ...(C)$
for the equation $ B$ to be same as $ C $
if we put the values for $ u,v,w$ in $ B$ we get
$ \minus{}9x^2\plus{}[(b\minus{}c)^2 \plus{} (c\minus{}a)^2 \plus{} (d\minus{}b)^2]x\plus{}20\equal{}0$
what remains is to prove
$ (a\minus{}d)^2\minus{}[(b\minus{}c)^2 \plus{} (c\minus{}a)^2 \plus{} (d\minus{}b)^2]\equal{}0$
$ a,b,c,d$ are in gp . let comon ratio be $ r$
$ (a\minus{}d)^2\minus{}[(b\minus{}c)^2 \plus{} (c\minus{}a)^2 \plus{} (d\minus{}b)^2]\equal{}a^2(1\minus{}r^3)^2\minus{}[a^2r^2(1\minus{}r)^2\plus{}a^2(1\minus{}r^2)^2\plus{}a^2r^2(1\minus{}r^2)^2] \equal{}a^2(1\minus{}r)^2[(1\plus{}r\plus{}r^2)^2\minus{}r^2\minus{}(1\plus{}r)^2\minus{}r^2(1\plus{}r)^2]\equal{}0$

hence done !


Prove that a number is divisible by 3 iff the sum of its digits are divisible by 3.
Also prove that a number is divisible by 9 iff the sum of its digits is a multiple of 9.

[hide="Rather simple"]Any natural no. can be expressed in the form
$ a_0 \plus{} 10a_1 \plus{} 10^2a_2 \plus{} 10^3a_3 \plus{} ...$
(where $ a_0, a_1, a_2, ...$ are the digits in respective positions)

seeing modulo 3 or 9, it is equivalent to
$ a_0 \plus{} a_1 \plus{} a_2 \plus{} ...$  which is the sum of the digits.

Hence, a number is divisible by 3 iff the sum of its digits are divisible by 3,
           a number is divisible by 9 iff the sum of its digits is a multiple of 9.[/hide]


I think it's equivalent to
$2 \sum \frac{a}{(b+c)^2} (a-b)(a-c) + \sum ab(a-b)\left(\frac{1}{(b+c)^2} - \frac{1}{(a+c)^2}\right) \ge 0$

[quote="Vasc"]Does this method solve the problem ?[/quote]

emm, I think first part by vornicu-schur, second part it's a sum of positive number * (a-b)^2

Indeed, it's valid. :)
I solved it using the following hint:
We can write the inequality as
$\frac{X}{(b+c)^3}+\frac{Y}{(c+a)^3}+\frac{Z}{(a+b)^3} \geq 0$  with $X+Y+Z=0$.


given a,b,c in R satisfies $a+b+c \geq abc$. Prove that
$a^{2}+b^{2}+c^{2}\geq abc$

I only consider the case $abc\geq 0$ (which implies $a+b+c\geq0$) otherwise this inequality is trivial.
So from the hypothesis $\rightarrow$ $1\geq \sqrt{\frac{abc}{a+b+c}}$ $\rightarrow$ $\sum a^{2}\geq \sqrt{\frac{abc}{a+b+c}}\sum a^{2}\geq abc$ $\leftrightarrow$ $\sqrt{\sum a^{3}b^{2}c^{2}}\leq \sum \sqrt{a^{5}bc}$ ,which is obviously true.


Zelma is having a luncheon for herself and nine of the women in her tennis league. On the morning of the luncheon she places name cards at the ten places at her table and then leaves to run a last minute errand. Her husband, Herbert, comes home from his morning tennis match and unfortunately leaves the back door open. A gust of wind scatters the ten name cards. In how many ways can Herbert replace the ten cards at the places at the table so that exactly four of the ten women will be seated where Zelma had wanted them?

Maybe PIE? Don't know where to start.

hmmm i'd think:

[hide]
10 choose 4  {for the 4 correct spots}
multiplied by
the number of derangements for 6
[/hide]

[quote="mustafa"][hide]There has to be exactly 4 cards in the correct place, and there are 10 choose 4 ways to pick 4 people. There are 6 remaining people, who can be in any order that you want, so there are $6!$ ways to pick the rest. Multiply 10 choose 4 (210) by $6!$ (720), and you get 151200.[/hide][/quote]

The problem asks for arrangements where [b]exactly[/b] four people are sitting in the correct seat. $\binom{10}{4}\times 6!$ counts those and others.

[hide="Correct Solution"]The number of possibilities for at least 4 correct is $\binom{10}{4}\times 6! = 151200$. The number of possibilities for at least 5 correct is $\binom{10}{5}\times 5! = 30240$. Thus, the number of possibilities for exactly 4 correct is $151200-30240=\boxed{120960}$.[/hide]

I still dont' think thats quite correct.

[hide]$\binom{10}5 * 5!$ isn't the number of arrangements with at least 5 right. For example: $\binom 31 * 2! = 6 = 3!$ isn't the number of permutations of a 3-set with at least one fixed point. Some of the arrangements will be counted multiple times.
The correct solution (I think) is, as krustyteclown said, $\binom{10}4 * d_{6}$, where $d_{n}$ is the number of permutations of a n-set with no fixed points ("derangements"). Calculating $d_{n}$ is pretty straightforward: just use PIE.
There might be a slicker way, but it is not immediately obvious.[/hide]

Can you show us how to calculate $d_{n}$ by using PIE? I only know the recursive method.

Lets say we want the number of derangements of $T$. For a subset $S$ of $T$, let $M_{S}$ be the set of permutations $\sigma$ such that each point in $S$ is a fixed point of $\sigma$.

We calculate the number of permutations with at least one fixed point. This is the magnitude of the union of $M_{S}$ for each singleton $S$ that is a subset of $T$. To compute this magnitude, we just use PIE: \[\sum |M_{S}|(-1)^{|S|+1}\] Since $|M_{S}|$ is just $(|T|-|S|)!$, and there are exactly $\binom{|T|}{n}$ subsets of size $n$, this is straightforward to compute.


Prove that

$ \frac{1}{m_a\plus{}h_a}\plus{}\frac{1}{m_b\plus{}h_b}\plus{}\frac{1}{m_c\plus{}h_c}\leq\frac{1}{2r}$

[quote="Ligouras"][quote="Ligouras"]Prove that

$ \frac {1}{m_a \plus{} h_a} \plus{} \frac {1}{m_b \plus{} h_b} \plus{} \frac {1}{m_c \plus{} h_c}\leq\frac {1}{2r}$[/quote]

Who can solve this?   :(   :([/quote]

This is principally the same as Ineq-Ligouras-94 and works again exactly the way [b]xzlbq[/b] said (I think you just didn't unerstand what he meant...): $ m_a \geq h_a$ and we have to prove
\[ \sum \frac{1}{h_a}\leq\frac{1}{r}\mbox{,}\]
where we have equality again  :P


Is there a problem with the forum times this morning?  For example, I just posted a topic on the test forum.  The forum says the post was at 7:28 am, but here in New York it was 10:28 am.  Before today, the post times agreed with my local time.  I checked my profile, and my time zone setting is still correct (UTC minus 5 hours, appropriate for US Eastern time).

Thanks.

The server was rebooted last night, but the time not correctly set.  Fixed now.  Thanks for pointing this out - it was causing all sorts of weird bugs that I couldn't figure out.  When I scanned the forum, I saw your message about the times and I said "Aha!"

I don't know if this is related, but it's very, very weird.  In the thread http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135094 in the college forum, the final post is dated  Sun Jul 14, 2013 11:01 pm.  In fact, it was posted some time before my post with timestamp "Today, at 11:52 am."

Yep, that's related (the server had a time 6-7 years in the future for about 90 seconds).  Let me know if you see any other weirdly time-stamped posts.

[quote="b-flat"]Does the rebooting have anything to do with there being 866 people online at about 3:13 EST this morning?

If not, does anyone know why there were so many people at such an odd time?[/quote]

Insomniacs  :D

[quote="#H34N1"]Some times are still messed up.[/quote]

Details, please.

[quote="kstan013"]On the online status page. It will say "last updated July __ 2013"[/quote]

Yeah, we'll have to sort that out for a few users.  It's on the todo list . . .


[quote="X-man"]Let $a_1, ..., a_n > 0$ be integers such that:
 1. $a_1=1$
 2. $a_i \le a_{i+1} \le 2a_i$ for $i=1, 2, ..., n-1$
 3. $a_1 + a_2 + ... + a_n = 1992$
Is it possible to divide numbers in two groups, such that sum of the numbers in one group be equal to the sum in the other group.[/quote]
Lemma 1: $a_{i+1} \le a_1+a_2+\ldots+a_i+1$;for $i=1,2,\ldots,n-1$
It prove by induction.
Lemma 2: If $1\le x\le a_1+a_2+\ldots+a_j \Rightarrow x=a_{t_1}+a_{t_2}+\ldots+a_{t_k}$; $1\le t_1<t_2<\ldots<t_k\le j$ for j=1,2,...n-1
It prove by induction.
therefore $996=a_{t_1}+a_{t_2}+\ldots+a_{t_k}$; $1\le t_1<t_2<\ldots<t_k\le n$
hence $a_{t_1}+a_{t_2}+\ldots+a_{t_k}=( a_1+a_2+\ldots+a_n)-(a_{t_1}+a_{t_2}+\ldots+a_{t_k})=996$


Solve for $x$.

$\sqrt{x} - 1 = \sqrt{x - 2}$

[hide]square both sides:
(sqrt(x)-1)(sqrt(x)-1)=x-2
x-2sqrt(x)+3=x
2sqrt(x)=3
sqrt(x)=3/2
x=9/4[/hide]


inequalities Vieta calculus derivative trigonometry LaTeX algebra

Suppose $ a$ and $ b$ are real numbers such that the roots of the cubic equation $ ax^3\minus{}x^2\plus{}bx\minus{}1$ are positive real numbers. Prove that:
\[ (i)\ 0<3ab\le 1\text{ and }(i)\ b\ge \sqrt{3}
\]

[19 points out of 100 for the 6 problems]

[hide="My Solution"]Suppose the roots of the equation are $ p,q,r$
By Vieta's theorem:
$ \frac {1}{a} \equal{} p \plus{} q \plus{} r$
$ \frac {b}{a} \equal{} pq \plus{} qr \plus{} rp$
$ \frac {1}{a} \equal{} pqr$

$ (i)$ By AM-GM:
$ pq \plus{} qr \plus{} rp\le \frac {p^2 \plus{} q^2}{2} \plus{} \frac {q^2 \plus{} r^2}{2} \plus{} \frac {r^2 \plus{} p^2}{2}\iff 3(pq \plus{} qr \plus{} rp)\le (p \plus{} q \plus{} r)^2\iff$
$ \iff 3\left(\frac {b}{a}\right)\le \left(\frac {1}{a}\right)^2\iff 3ab\le 1$
[b]QED[/b]

$ (ii)$ By AM-GM:
$ \frac {p^2q^2 \plus{} q^2r^2}{2} \plus{} \frac {q^2r^2 \plus{} r^2p^2}{2} \plus{} \frac {r^2p^2 \plus{} p^2q^2}{2}\ge pqr(p \plus{} q \plus{} r)\iff$
$ \iff p^2q^2 \plus{} q^2r^2 \plus{} r^2p^2\ge pqr(p \plus{} q \plus{} r)\iff$
$ \iff p^2q^2 \plus{} q^2r^2 \plus{} r^2p^2 \plus{} 2((pq)(qr) \plus{} (qr)(rp) \plus{} (rp)(pq))\ge 3pqr(p \plus{} q \plus{} r)\iff$
$ \iff (pq \plus{} qr \plus{} rp)^2\ge 3pqr(p \plus{} q \plus{} r)\iff \left(\frac {b}{a}\right)\ge 3\left(\frac {1}{a}\right)\left(\frac {1}{a}\right)\iff$
$ \iff b^2\ge 3\iff b\ge \sqrt {3}$
[b]QED[/b][/hide]

first part ws easy the second dreadful...

let p,q r be the roots...

p+q+r=1/a

pq+qr+pr=b/a

$ \rightarrow {ab}\equal{}\frac{(pq\plus{}qr\plus{}pr)}{{(p\plus{}q\plus{}r)}^2}$

since we have to prove that $ 3\le\frac{1}{ab}\\
\rightarrow\frac{p^2\plus{}q^2\plus{}r^2\plus{}2pq\plus{}2qr\plus{}2pr}{pq\plus{}qr\plus{}pr}\ge3$

solve and see we get the simple inequality

$ p^2\plus{}q^2\plus{}r^2\ge pq\plus{}qr\plus{}pr$

which holds true by AM-GM

hence proved

the second part i cud not get...what bout u...anyways how was it..on the whole?

i used calculus and rolles theorem. since all the roots are positive a and b are positive. so we get ab>0
For ab less than or equal to 1/3
i used rolles theorem
if a cubic has 3 real roots then its derivative has 2 real roots.
implies 3ax^2-2x+b has 2 real roots
implies discriminant greater than or equal to zero
So 4-12ab>0
implies ab less than or equal to1/3
for the second part let the roots be p,q,r
I gave a proof in calculus because this technique of Rolle's theorem is very useful for finding bounds on the real roots of an equation of higher degrees. :D

$ b\equal{}a.b/a$
it implies.....$ (\sum pq)^2 \ge 3pqr(\sum p)$

Okay , here is my proof for the second part. By vieta's theorem we know that the sum of the roots is equal to the product of the roots since both of them are equal to 1/a. Now let the roots be p,q and r. We proceed from our observation that p + q + r = pqr. Dividing throughout by pqr we know that $ 1/pq \plus{} 1/qr \plus{}1/pr \equal{}1$.
Now applying AM - HM inequality we get $ (pq \plus{} qr \plus{} rp)/3$>= 3/$ 1/pq \plus{} 1/qr \plus{}1/pr$.
But we know $ 1/pq \plus{} 1/qr \plus{}1/pr$ = 1. SO we get $ (pq \plus{} qr \plus{} rp)$ >= 9. But from the first part we proved that $ ab<\equal{} 1/3$. Substituting this in above inequality we get $ b^2>\equal{}3$. Implies $ b>\equal{}sqrt(3)$
$ QED$ :D
Whoo hoo! my first post in latex.

hey i got the first part correct and proved the second by contradiction i.e by assuming the contrary and obtaining a contradiction .

Hey can any one tell if i will get full credit?

a trigonometric substitution makes a nice work  :D

the from vieta's formulas we have that $ x \plus{} y \plus{} z \equal{} xyz$, so we can substitute $ (x,y,z) \equal{} (\tan A, \tan B, \tan C)$ for an acute triangle $ ABC$...

we have that $ \tan A\tan B\tan C \equal{} \frac {1}{a}$ and $ \tan A\tan B \plus{} \tan A\tan C \plus{} \tan B\tan C \equal{} \frac {b}{a} \equal{} b\tan A\tan B\tan C$...

so, $ a \equal{} \cot A\cot B\cot C$ and $ b \equal{} \cot A \plus{} \cot B \plus{} \cot C$... we have to prove that $ 0 < 3\cot A\cot B\cot C(\cot A \plus{} \cot B \plus{} \cot C)\leq 1$ and $ \sum \cot A\geq \sqrt 3$..

the lhs is obvious, and the rhs follows from the inequality
$ 3\cot A\cot B\cot C\sum \cot A\leq (\sum \cot A\cot B)^2 \equal{} 1$

the inequality $ \cot A \plus{} \cot B \plus{} \cot C\geq \sqrt 3$ follows from
$ (\cot A \plus{} \cot B \plus{} \cot C)^2\geq 3 (\cot A\cot B \plus{} \cot A\cot C \plus{} \cot B\cot C)^2 \equal{} 3$, and we're done  :D

ps: i know this is basically the same solution, but i didn't manage to "see" which inequalities i had to use till i did this  :D

Actually the solution is a lot simpler.

By Viete's formulae we have (x,y,z being roots)

x+y+z=1/a

xy+yz+zx = b/a

xyz=1/a

Now,

(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx)
1/a^2 = x^2 + y^2 +z^2 + 2b/a

Thus,

(1-2ab)/a^2 = x^2 + y^2 + z^2

Now by Q.M. >= A.M

we have

(1-2ab)/3a^2 >= 1/9a^2
3-6ab >= 1
3ab=<1
And as a and b are all positive 3ab > 0 is easily proved.

now (xy + yz + zx)/xyz = b
1/x + 1/y + 1/z = b
Again x+y+z / xyz = 1/xy + 1/yz + 1/zx = 1

Thus we have,

1/x^2 + 1/y^2 + 1/z^2 = b^2 - 2

Now by Cauchy Schwartz inequality

(1/x^2 + 1/y^2 + 1/z^2)(1^2 + 1^2 + 1^2)>= (1/x.1 + 1/y.1 + 1/z.1)^2
3(b^2 - 2)>= b^2
2b^2>=6
b>= root(3)

Proof completed

by sum of the roots and product of roots, let roots be x y z
x+y+z = 1/a ___ (1)
  xy+xz+zx = b/a ___(2)
 xyz=1/a___ (3)
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
  i did for first part , (1)^2 /(2)
  we get  (x+y+z)^2 /xy+xz+zy = 1 / ab
  as we know that (x+y+z)^2>=3(xy+yz+zx)___(4)  :idea:
  we get, 1/ab>=3, ie 3ab<=1
  as x y z are positive,( given), a, b are positive ie ab>0
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
2nd part
divide (1)  by  (3)
we get 1/xy +1/yz +1/xz = 1___(5)
by dividing (2) by(3) we get
1/x + 1/y + 1/z = b___(6)
  by  (6)^2 /(5) we get
 1/x+1/y+1/z ) ^2 / 1/xy + 1/yz + 1/zx  = b^2___(7) :idea:  :idea:
 let f(x,y,z) be (4) then (7) will be f(1/x, 1/y,1/z)
which implies by 4
           b^2>=3
           b>=sqrt 3    :D  :D  :D  :D  :D  :D

      my STRATEGY WAS TO ELIMINATE a and  get  b out of equations and i used the very basics

I got the first part all right; but I could not prove the second part - it was DREADFUL! :(

I only got this part and the "permutation" one all right (it was my first year giving RMO, and I was in class VIII)..........

Really, I was too nervous to write down anything correctly, I think my bad-proof-writing gave me less marks. :)

@Potla: Is it required to describe about how you solved it and how you did the RMO almost a year after the exam is over that too in the OLYMPIAD SECTION?
I feel people have already posted solutions to this very simple problem using 1. Viete Relatins and some suitable inequality
2. ROlles theorem application.
So I feel your post was not necessary.
{No offence :lol: }.

My solution(though similar to above I'm still posting it)
See we have to get an inequality on $ab$ so it's almost obvious that we have to calulate $ab$ by the VIETE'S RELATIONS..
IF the roots are $p,q,,r$ then $ab=\frac {pq+qr+rs}{(p+q+r)^2}$.
This is obviously less than $\frac {1}{3}$ as it converts into the well known $\dfrac {1}{2}$$[(p-q)^2 +(q-r)^2 +(r-p)^2]$.
Obviously $ab>0$ as roots are positive(that goes without saying :) )
The second part is better..
By AM-GM:
$ \frac {p^2q^2 + q^2r^2}{2} + \frac {q^2r^2 + r^2p^2}{2} + \frac {r^2p^2 + p^2q^2}{2}\ge pqr(p + q + r)\iff$
$ \iff p^2q^2 + q^2r^2 + r^2p^2\ge pqr(p + q + r)\iff$
$ \iff p^2q^2 + q^2r^2 + r^2p^2 + 2((pq)(qr) + (qr)(rp) + (rp)(pq))\ge 3pqr(p + q + r)\iff$
$ \iff (pq + qr + rp)^2\ge 3pqr(p + q + r)\iff \left(\frac {b}{a}\right)\ge 3\left(\frac {1}{a}\right)\left(\frac {1}{a}\right)\iff$
$ \iff b^2\ge 3\iff b\ge \sqrt {3}$
QED

[hide=Solution for (i)]Since the roots are positive, then their sum, sum of product taken twice, and product, all will be positive. So, $\frac{1}{a}>0, \frac{b}{a}>0$ which means that $a, b>0$.
Let the roots be $r, m, o$. So, $r+m+o=\frac{1}{a}, rm+mo+or=\frac{b}{a}$
Now, $(r+m+o)^2\geq3(rm+mo+or)$.
So $\frac{1}{a^2}\geq\frac{3b}{a}$ so $3ab\le1$.
So $(i)$ is proved. [/hide]


inequalities
geometry
inequalities proposed

[quote="hophinhan"][quote="alex2008"]Prove that , in any triangle , the following inequality holds :

$ \frac {1}{m_a^2(m_b \plus{} m_c \minus{} m_a)} \plus{} \frac {1}{m_b^2(m_c \plus{} m_a \minus{} m_b)} \plus{} \frac {1}{m_c^2(m_a \plus{} m_b \minus{} m_c)}\ge \frac {1}{S^2}$

where $ m_a,m_b,m_c$ are the medians .[/quote]

I think this geometry inequality is not true...Could you plz check your problem again ?[/quote]

Now i think it must be true  :)

[quote="alex2008"]Prove that , in any triangle , the following inequality holds :

$ \frac {1}{m_a^2(m_b \plus{} m_c \minus{} m_a)^2} \plus{} \frac {1}{m_b^2(m_c \plus{} m_a \minus{} m_b)^2} \plus{} \frac {1}{m_c^2(m_a \plus{} m_b \minus{} m_c)^2}\ge \frac {1}{S^2}$

where $ m_a,m_b,m_c$ are the medians .[/quote]

Thanks for edition. Now it became a easy problem :) :pilot:

Proof:

There is a triangle $ ABC$ with: $ BC \equal{} m_a ; CA \equal{} m_b$ and $ AB \equal{} m_c$ ; than $ S_{ABC} \equal{} S' \equal{} \frac {3}{4}.S$

So we need to prove :

$ \sum_{cyc} \frac {1}{a^2(b \plus{} c \minus{} a)^2} \ge \frac {9}{16S'^2}$

$ < \equal{} > \sum_{cyc} \frac {1}{a^2(b \plus{} c \minus{} a)^2} \ge \frac {9}{(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)} (*)$

Set : $ x \equal{} b \plus{} c \minus{} a\ ;\ y \equal{} c \plus{} a \minus{} b$ and $ z \equal{} a \plus{} b \minus{} c \ ( \equal{} > x,y,z \ge 0)$

$ (*) < \equal{} > \sum_{cyc} \frac {4}{(y \plus{} z)^2x^2} \ge \frac {9}{(x \plus{} y \plus{} z)xyz}$

$ < \equal{} > 0\le \sum_{cyc} \frac {(x \minus{} y)^2[3z^2(2xy \plus{} yz \plus{} zx) \plus{} x^2y^2 \plus{} 2xyz(x \plus{} y) \plus{} z^2(x^2 \plus{} xy \plus{} y^2)]}{(x\plus{}y\plus{}z)x^2y^2(x \plus{} z)^2(y \plus{} z)^2}$

Which is obviously true.  :starwars:


irrational number
advanced fields
advanced fields unsolved

In a set $ H$, whose elements are certain subsets of the set of all positive integers, any two distinct elements have a finite intersection. Is H necessarily countable (that is, does there exist a bijection between $ H$ and the set of natural numbers $ \mathbb{N}$?)

Suppose that $ f: Q\to N$ is a bijection
for every irrational number $ \alpha$ there is a sequence of rational number $ \{q_n\}_\alpha$ such that $ q_n\to \alpha$
let $ A_\alpha \equal{} \{\{q_n\}_\alpha\vert \alpha\;\rm is\; irrational\}$

let $ B \equal{} \{f(A_\alpha)\vert \alpha\;\rm is\; irrational\}$ so $ B$ has the desired property and is uncountable.

It's proper for the wise to say right so his words are reliable and it's proper for him to thank God,
 so His graces will be increased for him.


function
calculus
derivative
induction
logarithms
real analysis
real analysis unsolved

Let $f:\mathbb R\to\mathbb R$ be an additive function, and let $P$ be a non-constant polynomial. Show that if either one of $P\circ f$ or $\exp(f)$ is a derivative, then $f$ is continuous.

There is approach via IVP and general represantation of additive functions.

[quote="fedja"]If you know that a derivative has a point of continuity (as a pointwise limit of continuous functions), the problem immediately reduces to the well-known one-sided boundedness condition on an open interval. You can also use the fact that a derivative is a measurable function and therefore $f$ is bounded from above on a set $E$ of positive measure from where it follows that it is bounded from above on $E+E$, which contains an interval.
Anyway, in my humble opinion, this is a strange problem to propose for an olympiad: those who know the standard conditions can solve it in no time and those who don't may get stuck on it forever and a good olympiad should be about something else than knowledge of classical theorems... Or am I missing some really nice and original solution?  :?[/quote]

I don't know, maybe there are nice solutions, but the ones who decide whether a problem gets accepted for NMO probably didn't think so either :). As a result, this was not part of the contest.

Actually there is an elementary solution. I had it in contest two weeks ago. I solved it using baire th :D ,and later I realised that it can be done like this:

Proceed by induction on degree of $P$. Denote $Pof$ by $g$. Since $g$ is derivative we have that $g(x+a)-g(x)$is derivative. Hence $Q(f(x))=P(f(x)+f(a))-P(f(x))$ is derivative and $deg(Q)=deg (P)-1$.

From polinomials of degree $1$ it is obvious.

For the second part I have a solution but it is longer.

Xirti can you post your solution using Baire theorem?

Now I will write a proof for the secon question.

That probem is equiv with finding $f: [1,\infty)\to [1,\infty)$ such that $f(xy)=f(x)f(y)$, $f$ is derivative. It can easily be shown using the fact that $f\circ \ln$ is derivative iff $f$ is derivative.
Denote by $F$ the antiderivative of $f$ s.t F(1)=0.

We prove that  $G(x)=F(xy)-yF(x)f(y)-F(y)=0$. Indeed, differentiating we get that $G'(x)=yf(xy)-yf(x)f(y)=0$, and since $G(1)=0$, we have that $F(xy)=yF(x)f(y)+F(y),\forall x,y$. Intercharging $x,y$ we get that $F(yx)=xF(y)f(x)+F(x)$. Hence, $yF(x)f(y)+F(y)=xF(y)f(x)+F(x)\Rightarrow F(x)(yf(y)-1)=F(y)(xf(x)-1)$.

We observe that $f(x)=\frac 1 x$ is a solution and we look for others. Let $\alpha$ be  real number such that $\frac 1 \alpha\neq f(\alpha)$ and denote by $b=\frac{(\alpha f(\alpha)-1 )}{F(\alpha)}$. We have that $F(x)b=xf(x)-1$, which can be soved easily.

We get the solutions $f(x)=x^c$.

For, Cezar Lupu, I firstly proved that a derivative function is bounded on an interval (using Baire th). For a proof you can see:  http://www.mathlinks.ro/Forum/viewtopic.php?p=413091#p413091 for the proof. And then it becomes classic.


Point $ M(\minus{}2,\minus{}1,3) \in$ of line $ l$
Write the equation of $ l$  if $ l$ is parallel with plane $ \pi': 5x\minus{}y\minus{}z\minus{}4\equal{}0$ also $ l$ is perpendicular with  line $ d$ that is given with the system of two equation (equation of two planes) $ \pi1: 3x\minus{}y\minus{}5\equal{}0$  and $ \pi2 : 2x\minus{}z\minus{}2\equal{}0$

[quote="castigioni"]The equation of l=[t,-11t-23,16t+35][/quote]
how did you get this result ?

The line d is paralell to given plane .So line l must be orthogonal with d not perpendicular .Is it wright?


the formula is

                      2^n

Proof, please :)  In fact, give us 3 or 4 proofs.

the expansion of

    (1+x)^n = nC0 + nC1x + nC2x^2+ ..... nCnx^n

put x = 1

you will get your series = 2^n

proof 2

if you have n different object and you want to select them then the number of ways is

nC0 + nC1 + nC2 + .... + nCn      i.e ways nothing is select + one is select + 2 is select + ..... + n obj are select.

in the other way you can think that it is 2^n
i.e for 1st object 2 option wheather to select or not. similarly for second two option and so on. all possible cases are multiplied hence we can say

   your series = 2^n

please reply me if you have any other proof


A line splits a triangle into two new figures with equal perimeters
and areas. Prove that the center of the inscribed circle lies on this line.

It was posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=133939]here[/url]


geometry
3D geometry
Putnam
college
similar triangles
college contests

A right circular cone has base of radius $1$ and height $3$.  A cube is inscribed in the cone so that one face of the cube is contained in the base of the cone.  What is the side length of the cube?

[hide="Is this the Solution?"]
I set it up as a square of height $x$ and width $\sqrt{2}x$ inside a triangle of base $2$ and height $3$.  Using similar triangles I came up with

\[ \frac{1}{3}=\frac{1-\frac{\sqrt{2}x}{2}}{x}\implies x=\frac{6}{2+3\sqrt{2}}. \]

Does this seem right?  It just seemed like a really easy problem compared to many of the other Putnam problems (many I can't even start).

Thanks.
[/hide]

The Putnam has a very wide range of difficulty -- in general, questions A1 and B1 are very approachable.  (This isn't always true -- I think this year's A section lacked such an easy question.)

A few years ago, one of the national newsmagazines (I don't remember which one) did a story on the Putnam. They sent a photographer to Berkeley, where (after not being allowed in until the session was over) he got a picture of Austin Shapiro discussing one of the problems afterwards.

Anyway, the magazine decided that they should present one Putnam problem in a sidebar to the story - and they chose this exact problem. Hardly a representative choice.

Just for contrast, I just placed the following problem on a high school contest, on a team round:

A right circular cone has diameter $6$ and altitude $9$. Find the volume of the largest (in volume) rectangular box with a square base that fits inside the cone, with its square base flat against the base of the cone.

This was one of 8 questions that a six-member team had 30 minutes to answer. It was a fairly hard question - but not the hardest one on that round, and we did get three teams answering it correctly, with five other teams giving exactly twice the correct answer.

The magazine was Time; I remember reading the article before I got to university, solving the problem, and thinking 'man, this is easy'.


Propose a synthesis of [i]diethylpropion[/i] from benzene. Relevant information about this compound can be found [url=http://en.wikipedia.org/wiki/Diethylpropion]here[/url].

Was thinking of mannich reaction, but it doesn’t seem to give the req. product over here. so,
benzene to ethyl phenyl ketone(usual methods)l
Ethyl phenyl ketone + nh2oh ->oxime
Oxime + TsCl ->etoxime tosylate.
Tosylate + base\water to give alpha amino ketone.(neber rearrangement).
Now treat this wid dimethyl sulfate.

And what is the purpose of dimethyl sulphate?

sorry it shud be diethyl sulfate. doesnt nh2 react with diethyl sulfate to get ethylated in this case? also, SO42- or HSO4- is a gud leaving group.


Let $ a_{i,j}$ be the sum of the common divisors to $ i$ and $ j$, and $ A\equal{}(a_{i,j})$.
Show that $ \det A\equal{}n!$.

Here's a simple way. Let $ \sigma(n)$ be the sum of divisors of $ n$ and $ \circ$ be Dirichlet convolution. Also let
\[ b_{nk}\equal{}\sum_{d|n}\sigma [(d,k)]\mu\left(\frac{n}{d}\right)\equal{}\sum_{d|n}a_{d,k}\mu\left(\frac{n}{d}\right)\]
because $ \sigma \circ \mu (n)\equal{} n$ we have $ b_{nn}\equal{}n$
for $ k<n$ we have
\[ b_{nk}\equal{}\sum_{d|n}\sigma[(d,k)]\mu\left(\frac{n}{d}\right)\equal{}\sum_{d|(n,k)}\sigma(d)\sum_{s|\frac{n}{(n,k)}}\mu\left(\frac{n}{sd}\right)\equal{}0\]
as $ \frac{n}{(n,k)}>1$. by substituting row $ n$ with $ \vec{b_n}$ we obtain $ \det (A_n)\equal{}n\det(A_{n\minus{}1})$ and we are done by induction.


limit integration calculus function algebra polynomial trigonometry

Let $ f: [0,1]\rightarrow \mathbb{R}$ a continuous function. Find
\[ \lim_n\int_0^1\dots \int_0^1f(\frac {x_1 \plus{} \dots \plus{} x_n}{n})dx_1\dots dx_n.
\]
[hide="click"] Limit equals $ f(\frac {1}{2})$ and the proof involves some well know facts from measure/probability theory, such as strong law of large numbers, Fubini theorem, LDKT...    Enjoy! [/hide]

Edit: added limit in front of integral

[hide="My suggestion"]
First prove the conclusion for $ e^{2 \pi i k x}$. Then, extend the result to $ f$'s such that $ f(0) \equal{} f(1)$ by approximating it uniformly by trigonometric polynomials.
Secondly, we should also prove the conclusion for linear functions to be able to assume that $ f(0) \equal{} f(1)$.
[/hide]

[quote="perfect_radio"] extend the result to $ f$'s such that $ f(0) \equal{} f(1)$ by [i]approximating it uniformly by trigonometric polynomials[/i].
[/quote]

How would you justify this, is this a consequence of Stone-Weierstrass theorem?

[quote="Mr.Doe"]How would you justify this, is this a consequence of Stone-Weierstrass theorem?[/quote]
It is a consequence of Stone-Weierstrass. However, I personally prefer to think of it as a specific theorem in its own right: the trigonometric version of the Weierstrass theorem. It has its own concrete proof, and I know that concrete proof better than I know the proof of Weierstrass for ordinary polynomials.

Suppose $ f$ is continuous and periodic of period $ 1.$

Define $ \widehat{f}(k)\equal{}\int_0^1f(x)e^{\minus{}2\pi ikx}\,dx$ and $ S_n(f)(x)\equal{}\sum_{k\equal{}\minus{}n}^n\widehat{f}(k)e^{2\pi ikx}.$

$ S_n(f)$ is the $ n$th partial sum of the Fourier series for $ f.$ It does [i]not[/i] behave nicely; in particular, we cannot claim that $ S_n(f)$ converges uniformly to $ f$ (or even that it converges at all). But we can form the Cesaro means of this series:

$ \sigma_N(f)(x)\equal{}\frac1{N\plus{}1}\sum_{n\equal{}0}^NS_N(f)(x)\equal{} \sum_{k\equal{}\minus{}N}^N\left(1\minus{}\frac{|k|}{N\plus{}1}\right)\widehat{f}(k)e^{2\pi ikx}.$

We then show that $ \sigma_N(f)(x)\equal{}K_N*f(x),$ where $ K_N$ is the so-called Fejer kernel. $ K_N\ge0$ and satisfies all of the properties we would want it to satisfy to be a nicely behaved approximation of the identity; this gives us the tools we need to show that $ \sigma_N(f)$ converges uniformly to $ f.$ But then each $ \sigma_N$ is a trigonometric polynomial, which shows that the trig polynomials are dense in $ C(\mathbb{T}).$

There are details and computations to be filled in there, including the specific formula for $ K_N$ and the details of the argument about convolving by an approximate identity. But it does all work.

This will work (without statistics):

$ f(x) \equal{} \sum_{i \equal{} 1}^{\infty} a_i e_i$,
where $ (e_i)_{i \in \mathbb{N}}$ is a Schauder basis in $ C[0,1]$.
$ e_i \equal{} f_{k,n} \plus{} g_{k,n}$, ($ k \equal{} k(i)$ and $ n \equal{} n(i)$) where e.g.
$ f_{k,n}(x) \equal{} 2^{n \plus{} 1}x \minus{} 2k$ for $ x \in [\frac {k}{2^n},\frac {2k \plus{} 1}{2^{n \plus{} 1}}]$
and zero outside this area.
So $ f_i$ (and also $ g_i$) are linear and so easy to integrate.
It's the integral over the complete domain minus 2 pieces, that can be computed inductively.
Only the functions $ f_{0,0}$ and $ g_{0,0}$ tend to a positive number, namely $ \frac12$.

and the Law of Large number implies that if $ U_1, U_, \ldots, U_n, \ldots$ is a sequence of iid random variable distributed uniformly on $ [0,1]$. then
\[ \frac{U_1\plus{}U_2\plus{} \ldots \plus{} U_n}{n} \to \frac{1}{2}\]
almost surely.Dominated convergence gives the result.

Mr. Doe: why did you almost give the answer with the problem ?

Well, I had a dilemma should I put it here or in probability forum ( but I thought it might be to easy for people interested in probability ) and I was not sure if people would have an idea how to solve it, but as you stated it, the proof is easy if you know probability.


[quote="mr. math"]how do you do this problem:

There are 20 members in the MAO club who each play at least one of three sports: tennis, ping-pong, or badmiton. Twelve play tennis, six play ping-pong, and fifteen play badmiton. There is only one person who plays all three sports. Two people play tennis and ping-pong, but not badmiton, and two play ping-pong and badmiton, but not tennis. How many people play tennis and badmiton, but not ping-pong?

Please include an explanation.

 :bomb:  :ninja:  :wallbash:[/quote]

[hide]You can use a "formula" instead of venn diagrams...I personally think venn diagrams are a waste of time. Add the ones who do individual sports, take away the ones who do two, and since you subtracted those out three times and added them in twice, you add the one person in. $12+6+15-2-2-x+1=20, \boxed{x=10}$[/hide]

[quote="mr. math"]I also got ten but the correct answer is 7. How can that be?

 :rotfl:  :spam:[/quote]

Really? Who did the instruction manual...?
I might be wrong, but i'm pretty sure if there's individual events A,B,C it's TOTAL=a+b+c+(a, b, and c)-(a and b)-(a and c)-(b and c)

[hide]20 people-19 don't play all.
11 tennis players, 5 ping-pong, and 14 badmitton players left.
2 play tennis and ping-pong, and two play ping pong and badmitton.
15 people-9 tennis players, 1 ping-pong player, and 12 badmitton.
9+1+12=22
22-15=$\boxed{7}$[/hide]

When 3 or fewer categories are involved, a Venn diagram usually works well.

Go through the clues filling in the items you know exactly (1 person plays all three sports, 2 people play tennis and ping-pong but not badminton, 2 people play ping-pong and badminton but not tennis).   [See top figure.]

Go through the clues again, filling in the items you know exactly (6 people play ping-pong).  [See middle figure.]

Repeat.

If there are still some unknowns, use logical reasoning or algebra to finish the problem.  [See bottom figure.]

When there are two categories, venn diagrams seem to make the problem unnecessarily complicated.  I suggest using them on some three catergory problems that don't have a simple solution using PIE.


[hide="Solution"]I wouldn't bother solving, because E obviously works.  :P [/hide]

[hide]$\frac{x/4}{2}=\frac{4}{x/2}$
divide each side by 2:
$\frac{x/4}{4}=\frac{4}{x}$
Cross multiply:
$\frac{x^2}{4}=16$
Multiply each side by four:
$x^2=64$
Square root each side:
$x=\pm 8\implies\boxed E$[/hide]

[quote="4everwise"]If $\frac{x/4}{2}=\frac{4}{x/2}$, then $x=$

$\text{(A)}\ \pm \frac 12 \qquad \text{(B)}\ \pm 1 \qquad \text{(C)}\ \pm 2 \qquad \text{(D)}\ \pm 4 \qquad \text{(E)}\ \pm 8$[/quote]

[hide]
(x/4)/2 then that equals x/8
4/(x/2) then that equals 8/x

Therefore
x/8=8/x
By cross multiplication,
x^2=64
x=plus minus 8[/hide]

[quote="pascal_1623"][hide]
Uhh... you can see that obviously, $E$ works.[/hide][/quote]

You're right, Pascal.

[hide]simplifi, get x/8=8/x
easy enough from here :P [/hide]

You don't really even need to work it out, option E is relatively obvious.

[hide]cross multiply to get $\frac{x^2}{8}=8$ multiply both sides by 8 $x^2=64$ so $x=\pm{8}$ [/hide]

[quote="shifty_pi"]E! at this stage of simplicity u can just punch the numbers in and see which works :P[/quote]

Yes, but is always better to show why E is the correct one......

[quote="math92"][hide]
i see how everyone else's solution works, but i tried to find a differnet one, can you help find whats wrong here:
multiply everything by two to get:
$\frac{x}{4}=\frac{4}{x}\\\\x^2=16\\\\x=4$[/hide][/quote]
[quote="biffanddoc"]If you multiply the left by two you get $\frac{x/2}{2}$[/quote]
Just to clarify, $\frac{x/2}{2}=\frac{x}{4}$, so that wasn't math92's mistake.  However, if you multiply the righthand side, you get $\frac{8}{x/2}=\frac{16}{x}$; mathman92 multiplied the denominator, rather than the numerator, by $2$.

I think we should have condition x is different from 0  :rotfl:

[quote="4everwise"]If $\frac{x/4}{2}=\frac{4}{x/2}$, then $x=$

$\text{(A)}\ \pm \frac 12 \qquad \text{(B)}\ \pm 1 \qquad \text{(C)}\ \pm 2 \qquad \text{(D)}\ \pm 4 \qquad \text{(E)}\ \pm 8$[/quote]

Well this is pretty simple.
[hide]I solved for a positive value. Set up a proportion.  $\displaystyle{2\cdot\14=\frac{x}{4}\cdot\frac{x}{2}}$
Simplify. $\displaystyle{8=\frac{x^2}{8}}$
Multiply both sides by 8. $\displaystyle{64=x^2}$
Find both square roots.  $\displaystyle{8=x}$

Your final answer is $\displaystyle{\boxed{x=\pm\18}}$, $\displaystyle{Answer\boxed{E}}$[/hide]

[quote="José"][quote="Ignite168"]And this topic is over two months old.[/quote]

Yes..... really, what you did is non-sense[/quote]
i would stop posting here

[quote="José"][quote="daermon"]its pretty easy  ;)
[hide="solution"]the equation simplifies to x/8=8/x so x is obviously 8  :D [/hide][/quote]

Stop posting!!!!!!!!![/quote]I second that, but calm down a little, Jose, it's not that big a deal.

[quote="4everwise"]If $\frac{x/4}{2}=\frac{4}{x/2}$, then $x=$

$\text{(A)}\ \pm \frac 12 \qquad \text{(B)}\ \pm 1 \qquad \text{(C)}\ \pm 2 \qquad \text{(D)}\ \pm 4 \qquad \text{(E)}\ \pm 8$[/quote]

[hide="solution"]simplifying gets us to x/8=8/x
64=x^2
x=8[/hide]

[quote="daermon"]oh oops I forgot i already gave an answer for this  :blush:
i [i]knew[/i] it looked familiar![/quote]

Wow, solution #233 for this problem (not really).

And it's supposed to be $\pm8$,

[quote="Knuth's_check"]Guys, guys! It is x^2=64, which is 8. Can you believe that?  :rotfl:[/quote]

:spam:  Lock it

[quote="pascal_1623"][hide]
$E$
Cross multiply and you get $\frac{x^2}{8}=8$
$x^2=64$
$x=\pm8$[/hide][/quote]

Hey mods lock it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Every one is spamming here!


superior algebra
superior algebra unsolved

Let $ G$ be a group of order $ 6$, and assume that $ G$ is not commutative. Show that $ G$ is isomorphic to $ S_3$.

Given one of your posts above, I'm assuming that you at least know of Cauchy's Theorem (if $ p$ divides the order of a finite group $ G$, then $ G$ has an element of order $ p$). So take $ a$ of order 3, $ b$ of order 2. As $ <a>$ has index 2, $ <a>$ is normal, so $ bab^{\minus{}1}\in <b>$. We can't have $ bab^{\minus{}1}\equal{}1$ because $ a\neq 1$, nor $ bab^{\minus{}1}\equal{}a$ because then $ ab$ would have order 6 making $ G$ cyclic.

So $ bab^{\minus{}1}\equal{}a^2$. In this case, $ G\equal{}<a><b>$ so we can define $ \varphi$ from $ G$ to $ S_3$ by $ \varphi (a^mb^n)\equal{}(123)^m(12)^n$. Check that this is a well-defined isomorphism.

There are other ways to do this: both simpler (showing that elements of orders 2,3 exist without Cauchy's Theorem) and more advanced ($ G$ acts on cosets of $ <b>$ and this action is 1-1 unless $ <b>$ is normal).


induction
number theory unsolved
number theory

[color=darkblue]Let $ n\in N$ and $ n\ge 2$. Prove that $ (n^3\plus{}20n)\vdots 48$[/color]

It is wrong,conterexample:$ n\equal{}3$
actually $ 48|n^3\plus{}20n \Leftrightarrow 2|n$

[color=darkblue]I'm sorry. I post not full my problem.[/color]
[color=darkblue]Let $ n \equal{} 2k, k\in N$ . Prove that $ (n^3 \plus{} 20n)\vdots 48$[/color]

n^3+20n=n(n^2+20)
As n=2k,n|2.n^2+20=4k^2+20=0(mod4).
An even number is congruent to2,4,6 (mod6).(all multiples of 6 are even so there difference must be even).
Considering this cxases we can cnclude that n^3+20n|48. :D

Here's something maybe even simpler:
$ n^3\plus{}20n\equal{}8k^3\plus{}40k.$
$ 48|8k^3\plus{}40k$ is equivalent to $ 6|k^3\plus{}5k$, which is obviously true since
$ k^3\plus{}5k\equiv k\plus{}5k\equiv 6k\equiv 0 \ (mod 2)$ and
$ k^3\plus{}5k\equiv k\minus{}k\equiv 0 \ (mod 3)$.

[quote="mornik"]Here's something maybe even simpler:
$ n^3 \plus{} 20n \equal{} 8k^3 \plus{} 40k.$
$ 48|8k^3 \plus{} 40k$ is equivalent to $ 6|k^3 \plus{} 5k$, which is obviously true since
$ k^3 \plus{} 5k\equiv k \plus{} 5k\equiv 6k\equiv 0 \ (mod 2)$ and
$ k^3 \plus{} 5k\equiv k \minus{} k\equiv 0 \ (mod 3)$.[/quote]
Thank you! :P

[quote="Cronus"]I think that this might also be solved by induction.Can anyone prove it with induction?[/quote]
$ (n\plus{}2)^3\plus{}20(n\plus{}2)\minus{}n^3\minus{}20n \equal{} n^3\plus{}6n^2\plus{}12n\plus{}8\plus{}20n\plus{}40\minus{}n^3\minus{}20n \equal{} 6n^2\plus{}12n\plus{}48 \equal{} 6n(n\plus{}2) \plus{} 48$
and $ n(n\plus{}2)$ is clearly divisible by 8 since $ 2 \mid n$.

That proves that the difference of (n+2)3+20(n+2)-n^3-20n is divisible by 48.How does it prove that n^3+20n|48? :?:

By checking $ n\equal{}0,1$.

thanhnam2902: this is the very last warning, don't ever again use such a useless title again. See http://www.mathlinks.ro/viewtopic.php?t=135914 .

[color=darkblue]$ \bullet$ Because $ n \equal{} 2k\Longrightarrow n^3 \plus{} 20n \equal{} 8k^3 \plus{} 40k \equal{} 8k(k^2 \plus{} 5)$

$ \equal{} 8k[(k^2 \minus{} 1) \plus{} 6] \equal{} 8k(k \minus{} 1)(k \plus{} 1) \plus{} 48k$

$ \bullet$ Because $ k(k \minus{} 1)(k \plus{} 1)\vdots 6 \Longrightarrow [8k(k \minus{} 1)(k \plus{} 1)]\vdots 48$ and $ (48k)\vdots 48$

$ \Longrightarrow (n^3 \plus{} 20n)\vdots 48$[/color]


Prove that

(10101)

is a composite integer for any natural number base >=2.

Thank you

Your number is $b^{4}+b^{2}+1=(b^{2}+b+1)(b^{2}-b+1)$ where $b=\text{base}$.

The latter is the lower of the 2 factors, and is only equal to $1$ when $b=0$ or $b=1$, both outside the scope of the question, so we know we have two distinct proper factors $> 1$.


calculus
integration
limit
function
calculus computations

First notational step in thinking about this:

Define $g_{k}(x)=\begin{cases}\left(1-\frac{x}{k}\right)^{k}& 0\le x\le k\\ 0& x>k\end{cases}.$

Then we're looking at $\lim_{k\to\infty}\int_{0}^{\infty}g_{k}(x)e^{x/3}\,dx.$

Now, we can see that pointwise, $g_{k}(x)=e^{-x}$ for all $x\in[0,\infty).$ If we could only somehow justify the exchange of limit and integral, then we would arrive at

$\int_{0}^{\infty}e^{-2x/3}\,dx=\frac{3}{2}.$

Now: what theorems on the subject of interchanging limits and integrals do you know? What hypotheses are needed?

But $k$ appears as an integral bound and also in the integrant  :maybe: ..
does it is a dominated function in there??

That's why I rewrote it, to make all of the integrals over $[0,\infty).$ Yes, you can say that $g_{k}(x)\le e^{-x}.$


how many amsp qs did you answer? i did 8.

I got all of them but not sure for one problem(prob.10). However, I haven't finished writing my proof!I'm always sucked at combinatorial :oops:

Eh, I saw the test for the first time two days ago at like 2 PM, then I wrote up 1, 2, 3, 4, 5, 7, 9, and solved 10. Figured out an easier way to do 10 the next day, bashed 8 and wrote them up. Then I solved 6 when trying to fall asleep. So yeah, I got all 10.

Really? Wow...I started doing the test like the day it came out.

I did 9...the order I did them in: 1, 2, 3, 4, 7, 10, 6, 8, 5...interesting...

Think I made it?

[quote="distracted523"]I did the test in one day, saw the problems Sunday morning. Got all of them except for #6  :( . I thought #10 was easy (maybe too easy...?)[/quote]
as did i... i did all except 8 and 9 b/c i am very bad at proving.

but w/e i also did the test in 1 day, the week before on sunday :)

I totally forgot about AMP until about 5 days before the test was due... then I did 3-1-2-4-7-6-5. I got super sick Sunday/Monday so I didn't do 8/9/10...

I solved 1-7 in order, then 10, then put down all the progress I was able to make on 8 and 9.

So I "got" an 8-9, I think.

$ \{ x| 1\leq x\leq 10, x\neq 8\}$

I hate geometry.


There are three different values of N such that the mean of the 5 members of set S, where S={62,59,76,43,N}, equals the median of the members of set N. What is the sum of those 3 values of N?

Billy

[hide]The sum of the values of N is 185.

The value of the numbers you know is 240. For your sequences you can have

43  N  59  62  76. The value of N would be 55.

43   59   N   62   76. The value of N would be 60.

43   59   62   X   76. The value of N would be 70. [/hide]

[hide]43, 59, 62, 76
Sum of #s we know is 240

If 59 = median,  59*5=295, N=55
Check it, add 55 and 59 becomes median.

If N = median, 4n=240, N=60
Check it, add 60 and 60 becomes median.

If 62 = median, 62*5=310, N=70
Check it, add 70 and 62 becomes median.

55+60+70=185
>>185<<[/hide]


abstract algebra
group theory
search
superior algebra
superior algebra unsolved

Show that if every proper subgroup of a finite group G is abelian,
then G is solvable.

Is it not trivial by noting that $ [H: H] \equal{} 1$ for every proper subgroup of G, so the derive series for G must terminate in {1}? Or is there more to say?

We could have $ [G,G]\equal{}G$, in which case the derived series never terminates.

If $ G$ isn't simple, then we have some nontrivial normal $ N$, but $ N$ is abelian and $ G/N$ satisfies the same problem conditions, hence is solvable by induction... so $ G$ is solvable.

So we can assume $ G$ is simple.  From here, the problem is supposedly solved in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57911]this thread[/url], but I'm having some trouble following the proof.

[url=http://www.mathlinks.ro/viewtopic.php?search_id=1375508065&t=46641]Here[/url]'s another topic which might interest you, and there you'll find a link to yet another one.

[quote="LydianRain"]So we can assume $ G$ is simple.  From here, the problem is supposedly solved in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57911]this thread[/url], but I'm having some trouble following the proof.[/quote]
And [url=http://www.mathlinks.ro/viewtopic.php?t=215046]here[/url] is another thread, with some proofs (see especially fourth post) :)

Show that if every proper subgroup of a finite group G is abelian, then G is solvable.

Proof: If G is cyclic or non-simple, then it would be solvable. So suppose G is a non-cyclic simple group whose proper subgroups are abelian.

Since G is finite it must have a proper subgroup of maximal size [Just note that the sequence of indices:
$ [G: H_{1}] > [G: H_{2}] > [G: H_{3}] > ...$ is decreasing and must terminate since G is finite] .

Since G is non-cyclic G must have more than one proper subgroups of maximal sizes. If M were the only proper subgroup of maximal size, then pick $ x \neq 1 \in G \minus{} M$ and consider the subgroup $ < x >$, which must belong to another maximal subgroup but since there is no other one, we have $ < x > \equal{} G$, a contradiction since G is non-cyclic by assumption.

Using the same $ x$, consider $ M \cap xMx^{ \minus{} 1}$, which is the identity subgroup for if it weren't then choose $ y \in M \cap xMx^{ \minus{} 1}$ and then the centralizer of $ y$ in G would properly contained $ M$, and this makes $ y$ central in $ G$, contradicting simplicity.

Now we have $ M$ and $ K \equal{} xMx^{ \minus{} 1}$. The union of the conjugates of $ M$ and $ K$ contain non-identity elements $ |G| \minus{} |G: M|$ and $ |G| \minus{} |G: K|$ respectively.
We must have $ |G| \minus{} |G: M| \plus{} |G| \minus{} |G: K| \plus{} 1 \leq |G| \implies |G| \plus{} 1 \leq |G|$, a contradiction.
ok?


IT'S MY QUESTIONS . I know I am a amateurboy.. but please help me [/url][/code]

Wrong forum...

What do you mean by "get math on the blog"? If you're referring to $ \text{\LaTeX}$, it works the same as on the forums. If not, please elaborate.


I was looking at the solution for 1983 AIME #7, and I don't get a part of it.  Here's the original problem:
[b]<<[/b]   Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let P be the brobability that at least two of the three had been sitting next to each other. If P is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?   [b]>>[/b]

   And while getting the probability of the case in which all three knights are separated by "at least" 2 seats, the solution provider wrote:  [b](25/25)*(20/24)*(19/23) = (380/552).[/b]

   I get why it's 20 in the 2nd fraction, but I don't see why
the third fraction has 19 instead of 15.
   Don't you need to disregard the 4 surrounding seats of the 2nd knight as you did for the 1st one? (where you got 20 instead of 24)
   How is it OK to have 19 ?   Pleast help me! Thanks.

I don't really understand how they are doing that either, as it isn't necessarily independent, but how about this:

[hide]
Take any two next to each other, there's of course 25 ways to do that.

Then the last guy can be in any of 23 places, so 23*25 ways, except anything with all three next to each other is overcounted twice (as, supposing we call the initial two we pick to be next to each other x, and the last guy y, we could have xxy or yxx), so subtract that: 1*25. So the number of ways is

23*25-1*25 = 22*25. Now divide by $\binom{25}{3}$, the number of ways to pick any number, and you get:

$\frac{22*25*6}{25*24*23}= \frac{22*6}{24*23}= \frac{11}{46}$.
[/hide]

Your solution is much more understandable and simpler.
Thanks 100 times. =]

I like this problem a lot because you can do it in 4 essentially different ways, either combinatorically or probabilisitically, and either by calculating $P$ directly or through $1-P$.

e.g. We can count the number of triangles with vertices on a circle with $25$ points such that no two points are adjacent.

[hide="Solution"]We first choose two nonadjacent points, in one of $\binom{25}{2}-25$ ways.

If the two points are 2 away from each other (in one of $25$ ways), then there are $20$ choices for the third point.

If the two points are 3 or more away from each other (in one of $\binom{25}{2}-50$) ways, there are $19$ choices for the third point.

This counts each triangle three times so we divide by $3$. Finally we divide by the total number of triangles, $\binom{25}{3}$. Thus $1-P$ is

$\frac{(25)(20)+(\binom{25}{2}-50)(19)}{3\binom{25}{3}}= \frac{(25)(20)+(25 \cdot 12-50)(19)}{3 \cdot \frac{25 \cdot 24 \cdot 23}{3 \cdot 2}}= \frac{20+(12-2)(19)}{12 \cdot 23}= \frac{35}{46}.$[/hide]

Oh, and I don't have an answer for your original question.


inequalities
geometry
triangle inequality
inequalities unsolved

[i][u][size=150]can you solve this inequality by Minkowski.a,b,c>=0.
prouve that            rout(a²+b²-ab)+rout(b²+c²-bc)>=rout(c²+a²-ac)[/size][/u][/i]

[quote="bibobeo"]Am I wrong?[/quote]
You are right! :D

[quote="guas"]1. Draw three segments with lengths a,b,c starting from one point and forming three angles, 120 degrees each. The other ends of these segments are vertices of a triangle, the sides of this triangle are [code]$\sqrt{a^2+b^2-ab}$, $\sqrt{b^2+c^2-bc}$, $\sqrt{a^2+c^2-ac}$[/code]. So what you need is triangle inequality.
2. Why do you call it Minkowski's inequality? Have you any source supporting your reference?[/quote]
 In fact the sides are $\sqrt{a^{2}+b^{2}+ab}$, $\sqrt{b^{2}+c^{2}+bc}$, $\sqrt{a^{2}+c^{2}+ac}$.
If we consider three segments with lengths a,b,c starting from one point and forming two angles, 60 degrees each and one 120 degrees we get even stronger $\sqrt{a^{2}+b^{2}-ab}+\sqrt{b^{2}+c^{2}-bc}\ge \sqrt{a^{2}+c^{2}+ac}$.

so we have two solution for this stuff, one uses geometry and one uses Minkowski inequality  :lol:

[quote="guas"]1. Draw three segments with lengths a,b,c starting from one point and forming three angles, 120 degrees each. The other ends of these segments are vertices of a triangle, the sides of this triangle are [code]$\sqrt{a^2+b^2-ab}$, $\sqrt{b^2+c^2-bc}$, $\sqrt{a^2+c^2-ac}$[/code]. So what you need is triangle inequality.
2. Why do you call it Minkowski's inequality? Have you any source supporting your reference?[/quote]

Yes it's the triangle inequality but in some books they call it also Minkowski's inequality.


Let $ a$ be a given constant number such that $ |a|\leq 2$. For real numbers $ p,\ q,\ r$ with $ |p|\geq 1,\ |q|\geq 1,\ |r|\geq 1$ and $ p \plus{} q \plus{} r \equal{} 0$,
prove that $ 2(pq \plus{} qr \plus{} rp)^2 \plus{} 3apqr\geq 3$.

[quote="kunny"]Let $ a$ be a given constant number such that $ |a|\leq 2$. For real numbers $ p,\ q,\ r$ with $ |p|\geq 1,\ |q|\geq 1,\ |r|\geq 1$ and $ p + q + r = 0$,
prove that $ 2(pq + qr + rp)\2 + 3apqr\geq 3$.[/quote]
Obviously not all of $ p,q,r$ are either positive or negative.
Case two of them are negative. Assume wlog $ p,q < 0 < r$.
Then set $ p = -x$, $ q = -y$, $ r = z$, where $ z = x+y$, and $ x,y,z \ge 1$.
We get:
$ 2(pq + qr + rp) = 2(xy - xz - yz) = 2xy - 2(x+y)^2 = -2x^2 - 2y^2 - 2xy$.
$ 3apqr = 3axyz$.
$ 3 = 3$ :P
So we have to prove: $ -2x^2 - 2y^2 - 2xy + 3axyz \ge 3$. Since $ |a| \le 2$ we get $ a \ge -2$. So $ 3axyz \ge -6xyz$. It should also hold in this case. So we have $ -2x^2 - 2y^2 - 2xy - 6xyz \ge 3$ which is obviously wrong.
So the counterexample would be $ p=q=-1$, $ r=2$ and $ a = -2$. Then $ p+q+r = 0$. $ LHS = 2(1 - 2 - 2) + 3(-2)1 \cdot 1 \cdot2 = -6-12 = -18 < 3$. Could you fix the problem? :)

Sorry, I made a typo. I have just fixed it.:oops:

kunny


If I finish the book (I know it's not out yet!), will I be prepared to take multivariable calculus? Does the book strictly focus on single variable calculus, or does it also venture onto multivariable calculus? Thanks (I can't wait until it comes out).

the description said it prepares for calc bc, so i would assume it covers some aspects of multivariable calculus.  on th eother hand, it said it will focus a lot on the applications of calculus rather tan just spitting out formulas and telling u to repeatedly plug them in.  correct me if im wrong.

The book covers traditional single variable calculus, including everything you'll see on the BC Calculus exam.  If you complete the book, you should be ready for a university multivariable calculus class.  (The Precalculus book would help for that too, since it covers information about vectors that would be useful to know for a multivariable calculus class.)

There's an calculus book already?
is it going to be available for purchase anytime soon?

We expect it to be available for purchase in about 4 weeks.


limit
integration
real analysis
real analysis unsolved

Let $ \alpha>0$ be a real number. Find the limits:

1) $ \lim\limits_{n\rightarrow \infty}\left(\frac{1}{n}\right)^{n\alpha}\plus{}\left(\frac{2}{n}\right)^{n\alpha}\plus{}\cdots\plus{}\left(\frac{n\minus{}1}{n}\right)^{n\alpha}.$

2) $ \lim\limits_{n\rightarrow \infty}\left(\frac{2}{2n\plus{}1}\right)^{n\alpha}\plus{}\left(\frac{4}{2n\plus{}1}\right)^{n\alpha}\plus{}\cdots\plus{}\left(\frac{2n}{2n\plus{}1}\right)^{n\alpha}.$

3) $ \lim\limits_{n\rightarrow \infty}\left(\frac{1}{2n\plus{}1}\right)^{n\alpha}\plus{}\left(\frac{3}{2n\plus{}1}\right)^{n\alpha}\plus{}\cdots\plus{}\left(\frac{2n\minus{}1}{2n\plus{}1}\right)^{n\alpha}.$

it's enough to solve only the first: answer is $ \frac {1}{e^{\alpha}\alpha}$

or we just need to find approximation of $ \sum_{k \equal{} 1}^{n \minus{} 1}k^{n\alpha}$
2, 3 follows from 1, because we can represent $ 2k$ as $ 2*k$ :lol:
and in 3 we can sey that $ 2k \plus{} 1 \equal{} \left( 2k \plus{} (2k \plus{} 1) \right) \minus{} (2k)$  :)
.for first consider

[hide="hint"]$ \frac {\sum_{k \equal{} 1}^{n \minus{} 1}k^{n\alpha} }{\int_{1}^{n \minus{} 1}x^{n\alpha}dx}$[/hide]

Extremal,
I think that your answer is incorrect.

I'm sorry I don't calculate right approximation. answer is  :blush:
$ \frac {1}{\minus{}1 \plus{} e^{  \alpha}}$
Hint about calculation approximation was not correct.
question: how to prove : $ \sum_{1}^{n}k^{n\alpha}\approx \frac {n^{n\alpha}}{1 \minus{} e^{ \minus{} \alpha}}$
we need to consider last $ [n^{1/3}]$ terms. other terms is very small .and just consider $ \sum_{[n^{1/3}]}^{n} ...$


Calculate the following expression in term of $ n$ $ (n\in N$ excluding $ 0)$
$ 2^{n\minus{}1} \plus{} 2\times 2^{n\minus{}2} \plus{} 3\times 2^{n\minus{}3} \plus{} ... \plus{} (n\minus{}1)\times 2 \plus{}n$

[quote="DeathlyHallows"]Calculate the following expression in term of $ n$ $ (n\in N$ excluding $ 0)$
$ 2^{n \minus{} 1} \plus{} 2\times 2^{n \minus{} 2} \plus{} 3\times 2^{n \minus{} 3} \plus{} ... \plus{} (n \minus{} 1)\times 2 \plus{} n$[/quote]

$ A\equal{}2^{n \minus{} 1} \plus{} 2\times 2^{n \minus{} 2} \plus{} 3\times 2^{n \minus{} 3} \plus{} ... \plus{} (n \minus{} 1)\times 2 \plus{} n$----->
$ 2A\equal{}2^{n } \plus{} 2\times 2^{n \minus{} 1} \plus{} 3\times 2^{n \minus{} 2} \plus{} ... \plus{} (n \minus{} 1)\times 2^{2} \plus{}2 n$------>
$ 2A\minus{}A\equal{}2^{n}\plus{}2^{n\minus{}1}\plus{}...\plus{}2^{2}\plus{}2\minus{}n$------> $ A\equal{}2^{n\plus{}1}\minus{}2\minus{}n$

[quote="DeathlyHallows"]Calculate the following expression in term of $ n$ $ (n\in N$ excluding $ 0)$
$ A\equal{}2^{n \minus{} 1} \plus{} 2\times 2^{n \minus{} 2} \plus{} 3\times 2^{n \minus{} 3} \plus{} ... \plus{} (n \minus{} 1)\times 2 \plus{} n$[/quote]
$ A\equal{}\sum_{k\equal{}1}^{n}k2^{n\minus{}k}$
we have $ \minus{}A\plus{}n\sum_{k\equal{}1}^{n}2^{n\minus{}k}\equal{}\sum_{k\equal{}0}^{n\minus{}1}k2^{k}\equal{}p'(2)$ wehre $ p(x)\equal{}1\plus{}x\plus{}x^2\plus{}...\plus{}x^{n\minus{}1}\equal{}\frac{x^n\minus{}1}{x\minus{}1}$


Find all $ k \in Z$ such that $ \frac{5k^{2}-32}{1-2k}$ and $ \frac{5k^{2}-32}{1+2k}$ are  integers


How many integers and non negatives solutions  has $x+y+z+w=5$?

[hide]Combinations help very much[/hide]

i dissagree
the answer i get is 8 choose 3

$\left(\stackrel{n + r - 1}{r-1}\right)$ where $n = 5$ and $r = 4$ so maokid7 is right

Andreas, try \binom{n}{r} for binomial coefficients :D ...

$\binom{n+r-1}{r-1}$

Can anyone prove it?

Oh Thanks very much chess 64  :lol:
If anyone has a proof I'll post mine

sure its not that hard

lets say that  x1+x2+x3+x4+...+xn=A
 your want to know how many ways you can make n groups so that there are a total of A items in all.
we then use a "_" to sepperate the groups.  so now we have (n-1) _s and A items.  we then find how many ways we can look for the number of ways that u can arrange the (n-1) _s.

so we reach  (n+A-1) choose (n-1).

sorry that this is not in LaTex but i just dont know how to use it. i will learn as soon as i have time.

Consider it as a grid 5 * 3 (picture). We have to get from point A to point B. The number of pissible solutions is a number of all possible combinations
(we can only move upwards and rightwards)
In the picture exemplary solution is:
$x=1$
$y=2$
$z=2$
$w=0$
The number of all possible solutions is $\binom{8}{5}$
Please click on the picture. Then it will become more visible


Hi,

I'm assistant coaching Mathcounts this year, and since there is about one month left, I wanted to know what is the best way to prepare these students for Mathcounts. I pondered the thought for a while and then realized that the best way is probably to go over the most important concepts.

I've already taught them the concepts of mean, median, and mode, and some geometry. I know combinatorics is another important area but what else.

I'm specifically asking for concepts to increase the chance of them qualifying from Chapter. Thanks!

I'd say definitely probability and counting, since those problems tend to appear a lot.  Also some work on algebra would help.

Some number theory might be good too.
Basically, stuff out of the Intro series.

Dont forget to train them to combine the basic topics.     some problems might require algebra AND geometry while others use  geometry AND probability........

The most important concepts, imo, are the ones that people get wrong the most.  If there was a test that had a lot of geometry and a lot of people got all those questions right, would you go over geometry just because it appeared most often on that test?  Since mathcounts teams are small, it's probably also possible to work with some people individually.

I think you should teach them the very basics of all the subjects-then, you should give them problems that make them extend their knowledge.  A good math student shouldn't have to  go over problems of every sort; try to show that they can solve many problems by just using what they know and applying thinking to them.  For example, you should not have to teach them problems about how to apply geometry and algebra if you have already taught them how to correctly apply their knowledge to a select few problems.  It is important that you stress that memorization of problems is not good, that application of a few skills will always do the trick.