For someone reason AoPS and Mathlinks have been really slow to load on my computer for the past two months or so. I haven't noticed any other sites that do this. Anyone have any ideas?

maybe it's because of all the posts and pics

I'm having this problem too. In particular, it takes forever for the latex to display. Weirdly enough, by quoting someone, I can scroll down and view the latex faster  :huh:

Weird. Do you have hard disk cache enabled?

Yes, but disabling it changes nothing.

Edit: Latency isn't too bad, but it seems like throughput is the problem. The things that take longest to load are the images... I'm going to go do a test on downloading a large-ish file.

Edit 2: A 1MiB file downloads pretty quickly, it seems. Can't upload anything larger, though.


a continuous map $ f : (X,d) \to (X,d)$ is called a contraction if there is $ 0 < \alpha < 1$ st : $ d(f(x),f(y))\leq \alpha d(x,y)$

let $ C$ be the union of line segments in $ \mathbb{R}^2$ from $ (0,1)$ to $ (1/n, 0)$ ,$ n \equal{} 1,2,3,\ldots$  and the line segment from $ (0,1)$ to $ (0,0)$,
 prove that there is no contraction map $ f : C\to C$ s.t :  $ f(0,0) \equal{} (0,0)$

thanks so much

How about $ f(x)\equal{}(0,0)$ for all $ x$? :?


Define the polynomials $P_0, P_1, P_2 \cdots$ by:
\[ P_0(x)=x^3+213x^2-67x-2000 \]
\[ P_n(x)=P_{n-1}(x-n), n \in N \]
Find the coefficient of $x$ in $P_{21}(x)$.

BUMP
How has this gone unanswered for two years!!! Should it be moved to a tougher forum? Right now I can't think of anything but its probably tractable.

what? how is this in the least bit difficult, you just need the x coefficient in: $P(x-\frac{21*22}{2})=P_{21}(x)$

What? THat was obvious. I just read it wrong... must be a pretty weak olympiad :P

[quote="t0rajir0u"][quote="me@home"]I just read it wrong... [/quote]

Me too.  Personally, I read the recursion as $P_{n}(x) = P_{n-1}(x) \cdot (x-n)$, which is a little more interesting, I think, although still not hard.  For one thing, quadratic and higher terms can be ignored entirely.[/quote]
if it was $p_{n}(x)=p_{n-1}(x)(x-n)$
$p_{n}(x)=p_{0}(x)\prod_{i=1}^{n}(x-i)=\sum_{k=0}^{n+3}c_{k}x^{k}$
the coefficient of $x$ is $c_{1}=(-1)^{n}(-67+2000\sum_{k=1}^{n}\frac{1}{k})n!$

this seems to be the same problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80691

[hide=induction]claim:- $P_{n}(x)=P_{0}\left( x-\sum_{i=1}^{n} i \right)$\\
pf:- base case:\\
$n=1$ ( trivial)\\
assume the claim is true for $n$\\
from definition we have $P_{n+1}=P_{n}(x-(n+1)) \qquad \qquad (\spadesuit)$\\
we also have $P_{n}(x)=P_{0}\left( x-\sum_{i=1}^{n} i \right)\implies P_{n}(x-(n+1))=P_{0}\left(x-(n+1)-\sum_{i=1}^{n} i \right)$\\
from $(\spadesuit)$ we conclude $P_{n+1}=P_{0}\left( x-\sum_{i=1}^{n+1} i \right)$, which makes the claim true $\blacksquare$\\
so from claim it follows $P_{21}(x)=P_{0}(x-231)$\\
coefficient of $x$ in $P_{21}(x)= 3(231)^2 -213(231)(2)-67=\boxed{61610}$ $\blacksquare$[/hide]


The rate at which water flows out of a pipe, in gallons per hour, is given by a differentiable function R of time t. The table below shows the rate as measured every 3 hours for a 24-hour period.

t (hours)            R(t) (gallons per hour)
0                      9.6
3                      10.4
6                      10.8
9                      11.2
12                     11.4
15                     11.3
18                     10.7
21                     10.2
24                     9.6

Is there some time t, 0<t<24, such that R'(t) = 0?


algebra
polynomial
complex numbers
factoring polynomials

hey

i know that all sums of the form $ a^{2n\plus{}1}\plus{}b^{2n\plus{}1}$ can be factorised (although i do not remember the name of this theorem). Is this true for three or more terms? For example can we factorize $ a^{5}\plus{}b^{5}\plus{}c^{5}$?

thanks

p.s. was not sure where to put this - please move if it is in the wrong place!

Well... I don't think $ a^{3}\plus{}b^{3}\plus{}c^{3}$ can be factorized... but $ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc \equal{} (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca)$.

Well the reason $ a^{2n\plus{}1}\plus{}b^{2n\plus{}1}$ is divisible by $ a\plus{}b$ is that $ a^{2n\plus{}1}\equiv\minus{}b^{2n\plus{}1}\mod{a\plus{}b}$.  But if we try to use a similar argument for three terms,

$ a^{2n\plus{}1}\equiv\minus{}(b\plus{}c)^{2n\plus{}1}\mod{a\plus{}b\plus{}c}\not\equiv\minus{}(b^{2n\plus{}1}\plus{}c^{2n\plus{}1})\mod{a\plus{}b\plus{}c}$

The third term makes it impossible.  Perhaps there's a factorization using complex numbers (say, third roots of unity), but that's no long an integer factorization.

Homogeneous polynomials in two variables are essentially polynomials in one variable; what I mean is that a factorization for $ a^{2n\plus{}1}\plus{}b^{2n\plus{}1}$ corresponds exactly to a factorization for $ x^{2n\plus{}1}\plus{}1$, where $ x \equal{}\frac{a}{b}$ (if you like).  There are well-known techniques for factoring polynomials in one variable.

When a third variable is introduced, the problem becomes one of factoring polynomials in two variables; $ a^{5}\plus{}b^{5}\plus{}c^{5}$ corresponds exactly to $ x^{5}\plus{}y^{5}\plus{}1$, where $ x \equal{}\frac{a}{c}, y \equal{}\frac{b}{c}$ (again, if you like).  Factoring polynomials in two variables is much more complicated; the zero set of a polynomial in two variables $ f(x, y)$ is not a discrete set of roots, but rather a curve, and a factorization of such a polynomial corresponds to a union of two curves, and I have found empirically that "most" polynomials in two variables are irreducible, that is, they are not a union of two curves.


induction
number theory unsolved
number theory

Solve the equation:

1/x1+1/x2+...+1/xn+1/(x1*x2*...*xn)=1;
xi are natural numbers!

... this problem is in connection with the problem 1 / romanian TST 2003 :)

I found out that the problem is solved in KVANT magazine number 7/1987,in an article about partitions of the unity,where it says that the solution is:2,3,7,1897,3263443,...
(1/2+1/3+1/6=1;
1/2+1/3+1/7+1/42=1
...),
but i don't have the magazine and cannot get it.
(I suppose you must use induction !)

Where can i find the problem 1 from romanian TST 2003?

Somehow I doubt that those are the only solutions. Look at this:

1/2+1/3+1/7+1/43+1/(2*3*7*43) = 1. I can't see any 43 in that sequence xxxxtt posted.

3=2+1; 7=2*3+1; 43=2*3*7+1; see the pattern? I think it can be proven by induction that (this is, of course, just a guess) if for n we have a soln x1, x2, .., xn then for n+1 the numbers x1, x2, .., xn, x1*..*xn+1 work. I think thias is the main idea (I recal seeing this problem before). Someone needs to check that since I'm really not sure it works.

You are right Grobber. The fact that for each n  \geq  1 there exists a solution seems to be due to Leo Moser who gave the same set of solutions as yours. According to a foot-note, there are infinitely many solutions when n  \geq 3. Thus, I doubt about solving it, but maybe...

Pierre.

For Grobber
Sorry,i missed the 43 from the set of solutions!
I assure you these are the right solutions!!

For pbornsztein
I dont know where did you find the fact that for  n>3 there are infinitely many solutions but i found somewhere demonstrated that the equation has only a finite number of solutions!

Grobber,you are right!
If for n we have the solutions x_1,x_2,...,x_n then for n+1 x_1,x_2,...,x_n,x_n+1 where x_n+1=x1*x2*...*xn works(Just make the calculations!)
So solutions exist for every n.
The question is now if these solutions are unique.
Can someone prove this?

But are we positive that these are the ONLY solutions? This seems somewhat nasty to prove. I've given it some thought,but still nothing.

[quote]But are we positive that these are the ONLY solutions?[/quote]

I doubt it.  The discussion in UPINT3 at D28 mentions an extensive analysis by Judith Longyear for finding all solutions, and reviews of some of the cited references refer to parametric solutions.


Actually, I'm even going to provide the answer and just ask for proof.

Show that

$\sum_{n=0}^\infty \frac{(-1)^n(4n+3)\prod_{m=1}^n(2m-1)}{2^n(n+1)!} = 2$

The notation is a little more compact if I use the gamma function:

$\sum_{n=0}^\infty \frac{(-1)^n(4n+3)\Gamma(n + \frac{1}{2})}{\sqrt{\pi}\Gamma(n+2)} = 2$

Enjoy!

My friend points out that it is nicer to write the sum as

$\sum_{n=0}^\infty \frac{(-1)^n(4n+3)(2n)!}{2^{2n}n!(n+1)!\sqrt{\pi}}$

which is then just

$\sum_{n=0}^\infty \frac{(-1)^n(4n+3)}{2^{2n}(n+1)\sqrt{\pi}}\binom{2n}{n}$

Any thoughts?


31.  A divided highway into the city passes under a number of bridges.  The arch over each lane is in the form of a semi-ellipse, where the height is equal to the width.  What is the lowest bridge under which a truck 6 feet wide and 12 feet high can theoretically pass?

NOTE:  This is multiple choice, but I did not include that part  :D

[hide]

let the equation of this ellipse be
$\frac{x^{2}}{\left(\frac{h}{2}\right)^{2}}+\frac{y^{2}}{h^{2}}= 1$

$\frac{4x^{2}}{h^{2}}+\frac{y^{2}}{h^{2}}= 1$
where h is the height

since all the possible bridges are similar, the lowest bridge must pass through the point (3,12)

$h^{2}= 4(3^{2})+12^{2}= 2^{2}3^{2}+2^{4}3^{2}= 2^{2}3^{2}(1+2^{2})$
$h = 6\sqrt{5}$
[/hide]

[hide]
Equation of ellipse is
$x^{2}+(\frac{y}{2})^{2}= r^{2}$
The point (3, 12) lies on the ellipse.
Subbing it in gives $r = 3\sqrt{5}$.
Set $x = 0$ which gives a height of $\boxed{6\sqrt{5}}$.
[/hide]


I just get little confused about this I see it easy but I'm just confused :huh:. If $ \begin{array}{l}a \plus{} b \equal{} 1 \\ a^2  \plus{} b^2  \equal{} 2 \\  \end{array}$ Find $ a$ and $ b$

hello, rewriting the equation $ a\plus{}b\equal{}1$ in the form $ b\equal{}1\minus{}a$ and plugging this in the second equation we have $ a^2\plus{}(1\minus{}a)^2\equal{}2$ or $ a^2\minus{}a\minus{}\frac{1}{2}\equal{}0$.
Solving this quadratic equation we get $ a_1\equal{}\frac{1\plus{}\sqrt{3}}{2}$ or $ a_2\equal{}\frac{1\minus{}\sqrt{3}}{2}$ and we find $ b_1\equal{}1\minus{}a_1$ and $ b_2\equal{}1\minus{}a_2$.
Sonnhard.


calculus
integration
function
real analysis
real analysis unsolved

determine the following integral
\[ \int_0^\infty\frac{dx}{1+x^n} \]
where n is of course positive integer greater than 1.

For a complex variable approach, see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39543[/url].

[quote="jmerry"]That's not the only place; there's also a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21316]thread I posted almost a year ago[/url] and one it was inspired by.

My problem was: Calculate $\sum_{n=-\infty}^\infty\frac{(-1)^n}{1+nk}$. The answer shouldn't depend on $k$ being an integer, although $k>1$ is useful.[/quote]
you can reduce the series to very nice result with the sinus function

Of course I know the answer- I proposed that problem in the first place because I thought it was nice. You could at least follow the link...


[quote="bakerbakura"]If $ f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous increasing function such that $ f^2(x) \equal{} x$ $ \forall$ $ x \in \mathbb{R}$, prove that $ f(x) \equal{} x$ $ \forall$ $ x \in \mathbb{R}$.[/quote]

You should modify your problem : We cant have both $ f^2(x)\equal{}x$ and $ f(x)\equal{}x$ $ \forall x$

Do you mean $ f\left( {f\left( x \right)} \right) \equal{} x$ implies $ f(x) \equal{} x$ ? So we can prove it easily by reductio ad absurdum. If $ f(x) > x$ then $ f(f(x)) > f(x) > x$, the contradiction. The case $ f(x) < x$ is similar. So $ f(x) \equal{} x$


prove that the equation x^{3}+y^{3}=z^{3} has no solutions.

Consider either Fermat's last theorem for $ n\equal{}3$ or

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=177026[/url]


geometry
rhombus
ratio
LaTeX
angle bisector

Consider rhombus $ABCD$ and point $E$ which lies on $AC$, the longest diagonal fo the rhombus. If angle $BCD=60$ degrees, and $CE=CD$, then what is the ratio of the area of quadrilateral $ABED$ to the area of quadrilateral $BCDE$. Hopefully, y'all won't need a diagram :P

Ohh, this one. :lol: I've solved it before, but I'll redo it. ;)

[hide="Answer"]Setting each side of the rhombus equal to $1$, we see that diagonal $AC=\sqrt{3}$, and $BD=1$, since the rhombus consists of two equilateral triangles. If $F$ is the intersection of the diagonals, we see that $CF=\frac{\sqrt{3}}{2}$, so $EF=1-\frac{\sqrt{3}}{2}$. Since $BF=\frac{1}{2}$, the area of $\triangle BDE=\frac{1}{2}\cdot (1-\frac{\sqrt{3}}{2})=\frac{1}{2}-\frac{\sqrt{3}}{4}$. Therefore, the ratio between the two quadrilaterals is $\frac{\frac{\sqrt{3}}{4}-(\frac{1}{2}-\frac{\sqrt{3}}{4})}{\frac{\sqrt{3}}{4}+(\frac{1}{2}-\frac{\sqrt{3}}{4})}= \frac{\frac{\sqrt{3}}{2}-\frac{1}{2}}{\frac{1}{2}}=\sqrt{3}-1$.[/hide]

[hide]By setting CE=CD, we are making two isosceles 75-75-30 triangles.  We can put the midpoint of AC as well and connect and see the two 15-75-90 triangles which are formed.  By the angle bisector theorem, we can find that tan15 degrees=2+sqrt[3].  Thus, setting a side of the rhombus as x, we can find that the sum of the areas of those two triangles is (x/2)(x/(4+2sqrt[3]))=x^2/8+4sqrt[3]=x^2(8-4sqrt[3])/(16)=x^2(2-sqrt[3])/(4).  The area of BDC is sqrt[3](x^2)/4.  Thus, we can add and get the area of BCDE as (x^2)/2.  We know that the original area of the rhombus is (x^2)(sqrt[3])/2.  Thus, by subtracting, we get the ratio between ABED and BCDE is $\sqrt{3}-1$.[/hide]
Sorry if it's a bit unclear, pretty void of latex and hard to follow, it's late over here and i'm sleepy...i might fix it in the morning....

AMC daily contest?  Where is that?  Is this problem over yet, or should i edit my post?

The problem was over around a week ago, probably. It's on the AMC forums, but it's kind of on hold for now, since Silverfalcon, who's doing the practice, has computer problems.

[quote="xxreddevilzxx"]Wasnt this on the AMC daily contest thing going on right now????(I misunderstood that time and got it wrong :blush: )[/quote]

It was? :? I got it from a South Carolina Math competition


Slove this equation: $ tanx\plus{}tan^2x\plus{}tan^3x\plus{}cotx\plus{}cot^2x\plus{}cot^3x\equal{}6$

[quote="Brut3Forc3"][hide]\begin{eqnarray*}\tan^3 x + \cot^3 x & = & (\tan x + \cot x)^3 - 3\tan x \cot x (\tan x + \cot x) \\
& = & (\tan x + \cot x)^3 - 3(\tan x + \cot x) \\
\tan^2 x + \cot^2 x & = & (\tan x + \cot x)^2 - 2 \tan x \cot x \\
& = & (\tan x + \cot x)^2 - 2\end{eqnarray*}
So the original equation is equivalent to
$ 0 = (\tan x + \cot x)^3 + (\tan x + \cot x)^2 - 2(\tan x + \cot x) - 8$
Let $ y = \tan x + \cot x$, then
$ 0 = y^3 + y^2 - 2y - 8 = (y - 2)(y^2 + 3y + 4) \Rightarrow y = 2 \iff \tan x + \cot x = 2$
Both $ \tan x$ and $ \cot x$ must be positive for this to be true, so we can use AM-GM to find
$ \frac {\tan x + \cot x}{2} \geq \sqrt {\tan x \cot x} = 1 \Rightarrow \tan x + \cot x \geq 2$
Since equality holds, $ \tan x = \cot x = 1$, and $ x = \frac {\pi}{4} + n\pi , n\in \mathbb{Z}$
[/hide][/quote]
Thank you very much.  :P

in this case, i think AM-GM inequality with 6 variables works better, again this is my opinion

the 6 variables would be the 6 terms on the left side in the beginning equation
ahh... brut3forc3 just used brut3forc3
:rotfl:

[quote="stevenmeow"]in this case, i think AM-GM inequality with 6 variables works better, again this is my opinion

the 6 variables would be the 6 terms on the left side in the beginning equation[/quote]
......
*slaps self for taking long solution*


inequalities
trigonometry
inequalities proposed

I make "rude way"(That is, you do not like the method) to prove that:
Let $ x,y,z>0$,then

$ 6\,zy \plus{} 6\,xz \plus{} 6\,yx\leq 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( y \plus{} z \right) \left( z \plus{} 2\,x \plus{} y \right) }} \plus{} 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( z \plus{} x \right) \left( x \plus{} 2\,y \plus{} z \right) }} \plus{} 8\,{ \frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( x \plus{} y \right) \left( y \plus{} 2\,z \plus{} x \right) }} \plus{} 3\,{x}^{2} \plus{} 3\,{y}^{2} \plus{} 3\,{z}^{2}.$

[size=150]Tell me a wonderful solution,[/size]

and

[size=200]I also you a wonderful background[/size]

It's obviously trues by AM-GM and the well-known:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 2$
 :)
By this way, we can also prove the general problem:
\[ 6(xy\plus{}yz\plus{}zx) \le 2(k\plus{}2)xyz(x\plus{}y\plus{}z) \sum \frac{1}{(y\plus{}z)(kx\plus{}y\plus{}z)}\plus{}3(x^2\plus{}y^2\plus{}z^2)\]
With $ a, b, c, k > 0$

[quote="xzlbq"]I make "rude way"(That is, you do not like the method) to prove that:
Let $ x,y,z > 0$,then

$ 6\,zy \plus{} 6\,xz \plus{} 6\,yx\leq 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( y \plus{} z \right) \left( z \plus{} 2\,x \plus{} y \right) }} \plus{} 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( z \plus{} x \right) \left( x \plus{} 2\,y \plus{} z \right) }} \plus{} 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( x \plus{} y \right) \left( y \plus{} 2\,z \plus{} x \right) }} \plus{} 3\,{x}^{2} \plus{} 3\,{y}^{2} \plus{} 3\,{z}^{2}.$

[size=150]Tell me a wonderful solution,[/size]

and

[size=200]I also you a wonderful background[/size][/quote]

$ 6\,zy \plus{} 6\,xz \plus{} 6\,yx\leq 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( y \plus{} z \right) \left( z \plus{} 2\,x \plus{} y \right) }} \plus{} 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( z \plus{} x \right) \left( x \plus{} 2\,y \plus{} z \right) }} \plus{}$ $ 8\,{\frac {\left( x \plus{} y \plus{} z \right) xyz}{\left( x \plus{} y \right) \left( y \plus{} 2\,z \plus{} x \right) }} \plus{} 3\,{x}^{2} \plus{} 3\,{y}^{2} \plus{} 3\,{z}^{2}.$

$ \Longleftrightarrow \sum_{sym}(6x^7z \plus{} 15x^6y^2 \plus{} 19x^6yz \minus{} 6x^5y^3$ $ \plus{} 42x^5y^2z \minus{} 15x^4y^4 \minus{} 38x^4y^3z \minus{} 23x^3y^3z^2)\geq 0$

true due to Muirhead theorem.

 :lol:

[quote="nguoivn"]It's obviously trues by AM-GM and the well-known:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 2$
 :)
By this way, we can also prove the general problem:
\[ 6(xy \plus{} yz \plus{} zx) \le 2(k \plus{} 2)xyz(x \plus{} y \plus{} z) \sum \frac {1}{(y \plus{} z)(kx \plus{} y \plus{} z)} \plus{} 3(x^2 \plus{} y^2 \plus{} z^2)\]
With $ a, b, c, k > 0$[/quote]

wait a lot,I go home,copy background ,
thank you.
BQ

Let $ ABC$ is a triangle,

Note:
$ \cos\frac {A}{2}\le \frac {2h_a}{b \plus{} c} ,$

$ 4\sqrt {3}S \plus{} 3\sum{(b \minus{} c)^2} \geq a^2 \plus{} b^2 \plus{} c^2$(F-H ineq)
<=

$ 4(\frac {2}{3}\sum{\cos\frac {A}{2}})S \plus{} 3\sum{(b \minus{} c)^2} \geq a^2 \plus{} b^2 \plus{} c^2$

<=
$ 4(\frac {2}{3}\sum{\frac {2h_a}{b \plus{} c}})S \plus{} 3\sum{(b \minus{} c)^2} \geq a^2 \plus{} b^2 \plus{} c^2$

(a=y+z,b=z+x,c=x+y)
<=>

$ 6\,yz \plus{} 6\,zx \plus{} 6\,xy\leq 8\,{\frac {x \left( x \plus{} y \plus{} z \right) yz}{\left( y \plus{} z \right) \left( 2\,x \plus{} z \plus{} y \right) }} \plus{} 8\,{\frac {x \left( x \plus{} y \plus{} z \right) yz}{\left( z \plus{} x \right) \left( 2\,y \plus{} x \plus{} z \right) }} \plus{} 8\,{ \frac {x \left( x \plus{} y \plus{} z \right) yz}{\left( x \plus{} y \right) \left( 2\,z \plus{} y \plus{} x \right) }} \plus{} 3\,{x}^{2} \plus{} 3\,{y}^{2} \plus{} 3\,{z}^{2} .$
BQ

$ 4\sqrt {3}S \plus{} 3\sum{(b \minus{} c)^2} \geq a^2 \plus{} b^2 \plus{} c^2$(F-H ineq)----------------(*)
<=

$ 4(\frac {h_a \plus{} h_b \plus{} h_c}{s})S \plus{} 3\sum{(b \minus{} c)^2} \geq a^2 \plus{} b^2 \plus{} c^2$

(a=y+z,b=z+x,c=x+y)
<=>
$ 2xyz(\frac {1}{y \plus{} z } \plus{} \frac {1}{z \plus{} x} \plus{} \frac {1}{x \plus{} y}) \plus{} x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xy \minus{} 2zx \minus{} 2yz\geq 0.$-----------------（**）

<=>
$ 0\leq {x}^{4}y \minus{} {x}^{3}{y}^{2} \minus{} {x}^{3}{z}^{2} \plus{} {x}^{4}z \plus{} {y}^{4}z \minus{} {y}^{3} {x}^{2} \plus{} {y}^{4}x \minus{} {y}^{3}{z}^{2} \minus{} {z}^{3}{y}^{2} \plus{} {z}^{4}x \plus{} {z}^{4}y \minus{} {z}^{ 3}{x}^{2}$

<=>
$ {x}^{2} \left( y \plus{} z \right) \left( x \minus{} y \right) \left( x \minus{} z \right) \plus{} {y}^{2} \left( z \plus{} x \right) \left( y \minus{} z \right) \left( y \minus{} x \right) \plus{} {z}^{2} \left( x \plus{} y \right) \left( z \minus{} x \right) \left( z \minus{} y \right) \plus{} 2\,xyz \left( {x}^{2} \plus{} {y}^{2} \plus{} {z}^{2} \minus{} xy \minus{} yz \minus{} zx \right)\geq 0 .$

Clear from the above,
Ineq (*) Is not too strong。
BQ

[size=200]Last，for nice,Ineq(**),by CS？[/size]BQ

[size=150]The above facts give us an good idea:
If you feel empty in Algebra,
Went to the triangle to find。
BQ[/size]


function
algebra
polynomial
number theory
relatively prime
algebra unsolved

(i) Are there such rational functions $ f$ and $ g$ that $ (f(x))^3 \plus{} (g(x))^3 \equal{} x$

(ii) Are there such rational functions $ f,g,h$ that $ (f(x))^3 \plus{} (g(x))^3 \plus{} (h(x))^3\equal{} x$

i)  Equivalently, are there polynomials $ p, q, r$ such that $ p(x)^3 \plus{} q(x)^3 \equal{} x r(x)^3$?  Such polynomials can be taken to be pairwise relatively prime without loss of generality.  Now, by the [url=http://mathworld.wolfram.com/MasonsTheorem.html]Mason-Stothers theorem[/url], the number of distinct roots among $ p(x)^3, q(x)^3, x r(x)^3$ is at least $ 1 \plus{} 3 \text{max}( \text{deg}(p), \text{deg}(q), \text{deg}(r) \plus{} \frac {1}{3} )$, but the number of distinct roots is at most $ \deg{p} \plus{} \deg{q} \plus{} \deg{r} \plus{} 1$; contradiction.

ii)  Over what field?


geometry trigonometry ARML AMC AIME USA(J)MO USAMO

Does anyone know a good site that has good information on theorems/techniques that are good for amc/aime geometry?  I know that the AoPS are good but until I get those I would like to know if there are any good sites to learn from.

Or you guys could just post stuff here.  Either way I would appreciate it because I really need to get better with geometry.

AMC/AIME geometry usually uses the same geometry as any normal geometry class.  I don't think there's any one topic in geometry that is most important for AMC/AIME.  You might need to know some trig though.  Just basically try to label a few lengths/angles/whatever as variables, and using your knowledge, express other lengths/angles/whatever using previously established variables.

Well, I just need a ton of practice then.

joml88 you should try the formula sheet that was made by Jackpo and others. It doesn't give any proofs or motivations for the results, but it gives you an idea of some things that are good to know. You should see if you can try to prove some of these results (some are easier to prove than others). We put a copy of it on our ARML website here: http://155.225.48.46/statemat/stuf2004/formulas.pdf

And then you can try these Geometry Problems for practice. Most of them are former AIME and ARML problems that were discussed here. http://155.225.48.46/statemat/stuf2004/Geoprobs.htm

Bring a copy of these on the bus to ARML and we can talk about the ones that you have questions about on the way up there.

We will eventually offer a course in AMC/AIME level geometry.

Get this book: [url=http://www.amazon.com/exec/obidos/tg/detail/-/0486691543/qid=1084731527/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-0048912-6303167?v=glance&s=books&n=507846]Challenging Problems in Geometry[/url] (best value math book, only $10) and do all the problems in the book.

A friend of mine started working on the book when he is in 7th grade. After a year, He solved all the geometry problems (left out some easy algebra problems) in AMC12B/AIME and made USAMO.  And he got 7 points on the USAMO by essentially solving the first (geometry) problem!


calculus
integration
Putnam
trigonometry
calculus computations

sweet !!  :thumbup:

don't worry, i am not going to even try and attempt $\textbf{(ii)}$ ! LOL !!  :D

yesterday, my colleague asked me to find one similar integral for

$\boxed{20\;+\;\pi-e^{\pi}}$ and i already wasted half of my day today without any significant lead.  :(

 :rotfl:

misan, can't you just do

$\int_{e^{\pi}}^{20+\pi}dx$ ?

 :rotfl:


[b]Three sides of a quadrilateral are 1,2, and 5. The fourth side is an integer, $ x$,. What is the sum of all possible values of $ x$?[/b]

[hide]

let's see it has to be less than 1+2+5=8, but more or equal to than 2. Hence, 2+3+4+5+6+7=27

So, the answer is 27?

[/hide]

[quote="Nexmus"][hide]let's see it has to be less than 1+2+5=8, but more or equal to than 2. Hence, 2+3+4+5+6+7=27

So, the answer is 27?[/hide][/quote]
[hide]I dont think 2 would work

Its sides are 1,2,5,x then 8>x and 1+2+x>5, so x<8 and x>2 so [3,7] work so we have 3+4+5+6+7=25[/hide]

[color=darkred][size=75]Don't forget hide tags.
-nebula42[/size][/color]

[hide] Yes, two won't work, since then it would be $ 2+1+2=5$, which would result in a straight line. Also, please hide your answers. [/hide]

can you guys explain your reasoning? im completely lost in this problem. the first thing that i thought when i encountered it was triangle inequality which is stupid because its a quadrilateral. :blush:

It's the same logic as triangle inequality, If it is too little, it will be a straight line, so basically sum of 3 sides<1 side.

well, if we can find the longest possible side and the shortest possible side. if we add them together we get 8. we cant have 8 because then it would just be a straight line. now we take 5 and subtract the other two and we get 2. once again we have a straight line. so it must be the numbers between 3 and 7 or $ \sum_{k=3}^{k=7}{k}$


Let $ ABC$ be a triangle with $ \angle ABC$ being acute. Suppose that $ K$ be a point on the side $ AB$, and $ H$ be its orthogonal projection on the line $ BC$. A ray $ Bx$ cuts the segment $ KH$ at $ E$ and meets the line passing through $ K$ parallel to $ BC$ at $ F$. Prove that $ \angle ABC \equal{} 3\angle CBF$ if and only if $ EF \equal{} 2BK$.

For History, we must say that this problem ( easy to prove ), is the [b][size=100]Pappos method[/size][/b], for the trisection of arbitrary acute angle.

But, the point $ F$ on the line through $ K$ and parallel to $ BC$ such that $ EF \equal{} 2\cdot BK,$ it is not constuctable geometrically (with ruler and compass).

Kostas Vittas.

[quote="outback"]Can you show the solution of the problem as well?[/quote]
Let $ F$ be, the point on the line through $ K$ and parallel to $ BC,$ such that $ EF \equal{} 2\cdot BK,$ where $ E\equiv KH\cap BF$ and let $ M$ be, the midpoint of the segment $ EF.$

So, from $ BK \equal{} KM \equal{} MF$ and $ KF\parallel BC,$ we have $ \angle ABF \equal{} \angle KMB \equal{} \angle 2\angle KFB \equal{} 2\angle CBF$ $ ,(1)$

From $ (1)$ $ \Longrightarrow$ $ \angle ABC \equal{} 3\angle CBF$ and we are done.

Kostas Vittas.


A cube 4 units on each side is composed of 64 unit-cubes. 2 faces of the larger cube that share an edge are painted blue and then disassembled into the 64 cubes. Two of the cubes are selected randomly. What is the probability that exactly 1 cube will have 2 painted faces and the other cube has none?

A total of $ 28$ faces are painted, and only $ 2$ have $ 2$ painted faces. There are also $ 8$ cubes which have no paint. The required probability is $ \frac{2}{64}\times\frac{8}{63}\equal{}\boxed{\frac{1}{252}}$

ok, first of all, way more than 8 cubes have no paint, second of all, 2 faces of the large cube are painted, that means that there are 16*2= 32 small faces painted....
the number of cubes that arent painted is 4*3*3= 36
ok so the answer is $ 2*\frac{2*36}{64*63}$ we have to multiply by 2 because the one with no faces painted can be the first cube, or the second one...

the answer is $ \frac{1}{28}$

Wouldn't four unit cubes have two faces painted?

Yeah, that's what I was thinking, four unit cubes have two faces painted, those cubes are on the edge that the blue faces share.

So like RoFlLoLcOpT said, there are $16 + 16 - 4 = 28$ cubes which are painted.  Then there are $36$ cubes which do not have any paint on them.  So $\frac{36 \cdot 4}{\binom{64}{2}}$ would be the answer.  $\boxed{\frac{1}{14}}$.

Right. RoFlLoLcOpT missed a factor of 2.

Sorry for revive but shouldn't there be only 8 with no colored sides and 24 with 2 colored sides?

EDIT: Nvm I didn't read the question I'm stupid :P

[hide=Solution]Since there are four unit cubes with exactly two painted faces and 64-28=36 unit cubes with no painted faces, our answer is $$2\cdot \frac{4}{64} \cdot \frac{36}{63}= \boxed{\frac{1}{14}}.$$[/hide]

HEY! I PUT $\dfrac{\dbinom{4}{1} \cdot \dbinom{36}{1}}{\dbinom{64}{2}}$ and it didn't work even though it was right!!! :mad:

it's kind of common sense to simplify lmao calculators are allowed in c&p calculations like this tho

ouch lol

tbh I think natsuki trying to act like natsuki, angry at everything

[quote=HIA2020][quote=ilovepizza2020][quote=Natsuki]NOO i put that because i was too lazy to compute and ALCUMUS SAID IT WAS WRONG!

It could've atleast given me a warning that I HAD TO SIMPLIFY, right????? But it didn't!!![/quote]

Stop blaming the system for your mistake. You forgot to simplify, so the blame goes totally to you. (The formatting tips says to simplify everything)[/quote]

Let's all be aware of other's feelings.[/quote]

uh ok then

noooooooo i did 16/64*4/64=9/256 like an idiotttttgttgtgdfsgre


parameterization
vector
advanced fields
advanced fields unsolved

Let $ \gamma: (0, 1) \longrightarrow M$ be a natural parametrization of curve $ \Gamma$ which is on a smooth surface $ M \subset \mathbb{R}^3$ with the orientation given by $ N: M \longrightarrow \mathbb{R}^3$ (i. e. $ N$ is an unit vector such that $ N(p) \perp T_pM$). Prove that if $ N(\gamma(s)) \equal{} b(s)$, for $ s \in (0, 1)$, then the torsion $ \tau$ of $ \Gamma$ and the Gaussian curvature $ K$ of $ M$ satisfy the equation $ K(\gamma(s)) \equal{} \minus{}\tau(s)^2$, $ s \in (0, 1)$ ($ b(s)$ denotes the binormal vector of $ \Gamma$)

We have \[ \dot{\overrightarrow{N}}(s)=\pm{\dot{\overrightarrow{b}}(s)}\] then using frenet \[ \dot{\overrightarrow{N}}(s)=\pm{(-\tau(s)\overrightarrow{n}(s))}\] \[ (\dot{\overrightarrow{N}}(s))^{2}=(\pm{(-\tau(s)\overrightarrow{n}(s)})^{2}\] but \[ (\dot{\overrightarrow{N}}(s))^{2}=(\overrightarrow{N}_{u}\dot{u}(s)+\overrightarrow{N}_{v}\dot{v}(s))^{2}=III(du,dv)\] So \[ III(du,dv)={\tau}^{2}(s)\] because $ (\overrightarrow{n}(s))^{2}=1$. Using the identity ${ III-2HII+KI=0}$ where $ 2H$ the mean curvature and $ K$ the gaussian curvature we have \[ 2HII-KI={\tau}^{2}(s)\] but $ 2H=0$ because we have $ K_{N}=0$ and $ K_{N}=\frac{II}{I}$ where $ K_{N}$ the normal curvature. So we have \[ -KI={\tau}^{2}(s)\] but $ I=(\dot{x}(s))^{2}=1$ because $ s$ natural parameter and we have \[ -K(s)={\tau}^{2}(s)\]


$ a,b,c,d\in R$

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} ab \minus{} bc \minus{} cd \minus{} d \plus{} \frac {2}{5} \equal{} 0$

$ a \equal{} ?$

hello, writing your equation in the form
$ a^2\minus{}ab\plus{}b^2\plus{}c^2\plus{}d^2\minus{}bc\minus{}cd\minus{}d\plus{}\frac{2}{5}\equal{}0$ and applying the formula for a quadratic equation we got
$ a{_{1_2}}\equal{}\frac{b}{2}\pm\sqrt{\frac{b^2}{4}\minus{}b^2\minus{}c^2\minus{}d^2\plus{}bc\plus{}cd\plus{}d\minus{}\frac{2}{5}}$
$ a_{1_2}\equal{}\frac{b}{2}\pm\sqrt{\minus{}\frac{3}{4}b^2\minus{}c^2\minus{}d^2\plus{}bc\plus{}cd\plus{}d\minus{}\frac{2}{5}}$
Sonnhard.

[hide="solution"]
see the equation for $ a$,its discriminant $ D$ must be non-negative since $ a$ is real,
thus $ D \equal{} b^2 \minus{} 4(b^2 \plus{} c^2 \plus{} d^2 \minus{} bc \minus{} ca \minus{} d \minus{} \frac {2}{5}) \equal{} \minus{} 3(b \minus{} \frac {2c}{3})^2 \minus{} \frac {8}{3}(c \minus{} \frac {3d}{4})^2 \minus{} \frac {5}{2}(d \minus{} \frac {4}{5})^2\ge 0$,then because $ b,c,d$ are reals,
$ b \minus{} \frac {2c}{3} \equal{} c \minus{} \frac {3d}{4} \equal{} d \minus{} \frac {4}{5} \equal{} 0$ must hold,i.e.

$ b \equal{} \frac {2}{5},c \equal{} \frac {3}{5},d \equal{} \frac {4}{5}$.

then the solution is $ a \equal{} \frac {b}{2} \equal{} \frac {1}{5}$.
[/hide]


In the teacher's manual, it says the 10th or under student can choose which test to take - the AMC 10 or AMC 12, but I see several students here who have goals for both exams.  Do you have to take one on the first administration (a) and the other at the B time? Is that OK?

You can take both the AMC 10 and 12.  You are correct in that you would probably have to take one Test A and one Test B, as the AMC 10 and 12 are administered on the same day.

[quote="AMC website"]
AMC 12 Eligibility – A student in a program
leading to a high school diploma, and under 19.5
years of age on the day of the contest.

AMC 10 Eligibility – A student in a program
leading to a high school diploma, under 17.5
years of age on the day of the contest, and not
enrolled in grades 11 or 12 or equivalent.
[/quote]

Also, I'm not sure if this belongs in Middle School Contests.

Anyone in or below 12th grade can take two of the tests, as long as he/she can find a place to take it.

For 10th graders or below there are 4 options for two dates:
1) AMC10 on the A and B dates
2) AMC10 on the A date and AMC12 on the B date
3) AMC12 on the A date and AMC10 on the B date
4) AMC12 on the A and B dates
If you choose to take the test on only one date, it can either be AMC10 or AMC12, but not both.

For 12th graders and 11th graders, there is only 3 options:
1) AMC12 on both A and B dates
2) AMC12 on only A date
3) AMC12 on only B date

For more information, see [url=http://www.unl.edu/amc/f-miscellaneous/faq.shtml]here[/url] (near the bottom, where it says Q, AMC 10/12 #12). Also, this might better belong in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]AMC Forum[/url].

Thank you for all the thorough info.  I'm sorry I posted in the wrong place - the student I was inquiring about (has an interest in AMC 10 and maybe 12) is in middle school - that's how I ended up here.  I will try to pay better attention in the future! Thanks again.


In circle $ O$ chord $ AB$ is produced so that $ BC$ equals a radius of the circle.  $ CO$ is drawn and extended to $ D$.  $ AO$ is drawn.  Which of the following expresses the relationship between $ x$ and $ y$?

[asy]size(200);defaultpen(linewidth(0.7)+fontsize(10));
pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0);
draw(O--A--C--D);
dot(A^^B^^C^^D^^O);
pair point=O;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$O$", O, dir(285));
label("$x$", O+0.1*dir(172.5), dir(172.5));
label("$y$", C+0.4*dir(187.5), dir(187.5));
draw(Circle(O,1));
[/asy]

$ \textbf{(A)}\ x\equal{}3y \\
\textbf{(B)}\ x\equal{}2y \\
\textbf{(C)}\ x\equal{}60^\circ \\
\textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y \\
\textbf{(E)}\ x\equal{}2y \text{ or }x\equal{}3y\text{, depending upon the length of }AB$

[hide="sketching strongly advised"]
First let $ E$ be intersection of the circle and $ CD$.

$ \triangle AOD$ is isosceles $ (|OA|\equal{}|OD|)$ $ \implies \angle DAO\equal{}90^\circ\minus{}\frac{x}{2}, \ \angle ODA\equal{}90^\circ\minus{}\frac{x}{2}$
$ (90^\circ\minus{}\frac{x}{2}\plus{}90^\circ\minus{}\frac{x}{2}\plus{}x\equal{}180^\circ)$

$ \angle DAE\equal{}90^\circ$ (Thales' theorem) $ \implies \angle OAE\equal{} \angle DAE \minus{} \angle DAO \implies \angle OAE\equal{}\frac{x}{2}$

$ \triangle CBO$ is isosceles $ (|BO|\equal{}|BC|) \implies \angle BOE\equal{} \angle OCB \equal{} y$

$ \angle BAE\equal{} \frac{\angle BOE}{2}$ (because of the common chord $ BE$) $ \implies \angle BAE\equal{}\frac{y}{2}$

$ \angle AOE \plus{} \angle AOD\equal{}180^\circ \implies \angle AOE\equal{}180^\circ \minus{} x$

Now we have all the angles of $ \triangle AOC$:

$ \angle OAE \plus{}\angle BAE \plus{} \angle OCB \plus{} \angle AOE\equal{}\frac{x}{2}\plus{}\frac{y}{2}\plus{}y\plus{}(180^\circ \minus{} x)\equal{}180^\circ \implies \boxed{x\equal{}3y}$ or $ \textbf{(A)}$[/hide]

[hide="No sketching solution"]Just use the Exterior Angle Theorem (EAT) twice:
$ OB\equal{}BC$ implies that $ \angle BOC\equal{}\angle BCO\equal{}y$ (Angles of Isosceles Triangle Theorem) and so by EAT, $ \angle OBA\equal{}\angle BOC\plus{}\angle BCO\equal{}y\plus{}y\equal{}2y$.
$ OB\equal{}OA$ implies that $ \angle OBA\equal{}\angle OAB\equal{}2y$ (AITT) and so again EAT gives $ x\equal{}\angle OAC\plus{}\angle OCA\equal{}2y\plus{}y\equal{}3y$.[/hide]


with  very ugly way :blush:  I have proved a stronger one:

$ \sum{ \frac{1}{(a\plus{}b)^2}} \geq  \frac{ a^4\plus{}b^4\plus{}c^4\plus{}7(ab\plus{}ac\plus{}bc)^2}{(a\plus{}b)^2(a\plus{}c)^2(b\plus{}c)^2}$  one can prove that is stronger with pqr :wink:

This inequality is in my opinion very nice but also hard as we have equality only for $ a\equal{}0,b\equal{}c$ and cyclic. I tried to solve it by full expansion and p,q,r method but it is very ugly, but I cannot solve it by a nice trick :blush: so I am waiting for a nice solution. :)

Thank you very much to anyone will find a solution. :roll:

[quote="nguoivn"]Given $ a, b, c \ge 0$. Prove that: $ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {8(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^4} \ge \frac {11}{4(ab \plus{} bc \plus{} ca)}$
:)[/quote]
[quote="anas"]with  very ugly way :blush:  I have proved a stronger one:

$ \sum{\frac {1}{(a \plus{} b)^2}} \geq \frac { a^4 \plus{} b^4 \plus{} c^4 \plus{} 7(ab \plus{} ac \plus{} bc)^2}{(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2}$  one can prove that is stronger with pqr :wink:[/quote]
In the both inequalities enough check only two cases:
1) $ b\equal{}1,$ $ c\equal{}0$;
2) $ b\equal{}c\equal{}1.$

[quote="anas"]with  very ugly way :blush:  I have proved a stronger one:

$ \sum{\frac {1}{(a \plus{} b)^2}} \geq \frac { a^4 \plus{} b^4 \plus{} c^4 \plus{} 7(ab \plus{} ac \plus{} bc)^2}{(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2}$  one can prove that is stronger with pqr :wink:[/quote]
This is not true; try $ a\equal{}5, b\equal{}6, c\equal{}7.$

[quote="nguoivn"]Given $ a, b, c \ge 0$. Prove that: $ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {8(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^4} \ge \frac {11}{4(ab \plus{} bc \plus{} ca)}$
:)[/quote]
We can prove it by only Am-Gm and Cauchy Schwarts  :)

[quote="manlio"]Can you please post your proof, nguoivn?

Thank you very much. :)[/quote]
I am now very ill, but I will help you...

Write the inequality as
$ \sum \frac{ab\plus{}bc\plus{}ca}{(b\plus{}c)^2}\plus{}\frac{8(ab\plus{}bc\plus{}ca)^2}{(a\plus{}b\plus{}c)^4} \ge \frac{11}{4},$
or
$ \sum \frac{a}{b\plus{}c}\plus{}\sum \frac{bc}{(b\plus{}c)^2} \plus{}\frac{8(ab\plus{}bc\plus{}ca)^2}{(a\plus{}b\plus{}c)^4} \ge \frac{11}{4}.$
It is easy to check that (by applying the Cauchy-Schwarz Inequality)
$ \sum \frac{a}{b\plus{}c} \ge \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}.$
In addition, one has
$ \sum \frac{bc}{(b\plus{}c)^2} \ge \frac{1}{2}\sum \frac{bc}{b^2\plus{}c^2} \equal{}\frac{1}{4}\sum \frac{(b\plus{}c)^2}{b^2\plus{}c^2}\minus{}\frac{3}{4} \ge \frac{(a\plus{}b\plus{}c)^2}{2(a^2\plus{}b^2\plus{}c^2)}\minus{}\frac{3}{4}.$
By using this and the above inequality, we may reduce our inequality to the following
$ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}\plus{}\frac{(a\plus{}b\plus{}c)^2}{2(a^2\plus{}b^2\plus{}c^2)}\plus{}\frac{8(ab\plus{}bc\plus{}ca)^2}{(a\plus{}b\plus{}c)^4} \ge \frac{7}{2}.$
Note that
$ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)} \equal{}\frac{(a\plus{}b\plus{}c)^2}{8(ab\plus{}bc\plus{}ca)}\plus{}\frac{(a\plus{}b\plus{}c)^2}{8(ab\plus{}bc\plus{}ca)}\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{4(ab\plus{}bc\plus{}ca)}\plus{}\frac{1}{2},$
and
$ \frac{(a\plus{}b\plus{}c)^2}{2(a^2\plus{}b^2\plus{}c^2)}\equal{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{1}{2}.$
Therefore, the last inequality can be written as
$ \frac{(a\plus{}b\plus{}c)^2}{8(ab\plus{}bc\plus{}ca)}\plus{}\frac{(a\plus{}b\plus{}c)^2}{8(ab\plus{}bc\plus{}ca)}\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{4(ab\plus{}bc\plus{}ca)} \plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{8(ab\plus{}bc\plus{}ca)^2}{(a\plus{}b\plus{}c)^4} \ge \frac{5}{2}.$
In this form, we can see immediately that it is a direct consequence of the AM-GM Inequality.

My dear friend can_hang2007,

I regret that you are ill. :(

Your proof is really beautiful. :)

Thank you very much. :)


geometry
rhombus
angle bisector
perpendicular bisector

Hello everyone,

   I would appreciate it greatly again if you could lend me a hand on this problem: without the solution please!

---

Question: Consider the triangle ABC with AC = 6, AB = 12 and BC = 8. A rhombus is inscribed in triangle ABC in such a way that one of its vertices is A and two of its sides lie along AB and AC. Find the length of the side of the rhombus.

(Hint: denote the side of the rhombus x and write similarity ratios.)

---

Here is my diagram of how I visually interpreted the question:

[img]http://img95.imageshack.us/img95/6540/problem5mv8.jpg[/img]

Thank you very much!

[hide]Here's how you construct the rhombus:

The diagonals of a rhombus bisect the angles, so construct the angle bisector of $\angle A.$ The intersection of the the angle bisector and side $BC$ is one of the vertices of the rhombus. Then, construct the perpendicular bisector of the diagonal. The intersections of the perpendicular bisector with the sides of the triangle are the other two vertices.[/hide]

[hide="hint"]One of the sides of the rhombus makes a similar triangle to the bigger triangle.[/hide]

[hide="bigger hint"]The side that does this is parallel to AC.[/hide]

[hide="solution"]Let the other 3 vertices, D, E, and F be on the sides AB, BC, and AC respectively.  Since opposite sides of a rhombus are parallel, DE is parallel to AF and therefore, DE is also parallel to AC.  Since triangle BED is similar to triangle BCA, we have:
$\frac{DE}{AC}= \frac{BD}{BA}$
Since AD = x, BD = 12 - x.  AC = 6, DE = x, AB = 12...
$\frac{x}{6}= \frac{12-x}{12}$
Multiplying by 12 on both sides,
$2x = 12-x$
$3x = 12$
$x = 4$[/hide]

He asked for people to NOT post a solution.

Please honor his wishes  :)


function
inequalities
integration
real analysis
real analysis unsolved

Given this question, but I'm stuck need some help. Thanks.

(1). Show that if $x_1, ...x_n$ are positive real numbers, $\eta_1, ... \eta_n \in [0,1]$, $\sum_{i=1}^{n} \eta_i = 1$ and $0<p<q$ then  $(\sum_{i=1}^{n} \eta_i x_i^p)^{1/p}\leq (\sum_{i=1}^{n} \eta_i x_i^q)^{1/q}$

I guess that I need to use a convex function, or Jensen's Inequality, but I dont know how to, please help.

(2). Let $f: [a,b] \rightarrow R$ be an integrable function and suppose that there is some $q \geq 1$ such that $\int_{a}^{b}| f |^q=0$. Show that $\int_{a}^{b} f=0$

Well, I guess I might need use a result from part (1), right? Or something esle, please help to approach.

I guess I still dont know how it related to the link you sent. Could you explain? Thx

The link is correct.  However, if it's still not working, google 'planetmath power mean inequality', click on the first result, and click on the proof.

Anyways, for the second, use Holder's inequality.

For the first, the proof contained in the link I posted is precisely what you want to prove, only in different notation.  The parent object is a weaker version of it, but it is easily modified.

Holder inequality... umm interesting!! What does it say to provide showing the part3?

You could just google it, or look in your textbook (it's almost certainly there)

Anyways, Holder's inequality in this case says that if $1/q+1/p=1$, then

$0=\left(\int^b_a |f|^q\right)^{\frac 1q} \left(\int^b_a |1|^p\right)^{\frac 1p} \geq \int^b_a |f|$

But if $\int^b_a |f|=0$, then clearly $\int^b_a f=0$.


Find the sum of all quotients $ \frac{a}{b}$ such that $ a$ and $ b$ are relatively prime divisors of $ 27000$.

$ 27000 \equal{} 10^3 * 3^3 \implies 2^3 * 3^3 *5^3.$
Now the relatively prime divisors will either be ratios of $ 3,5,2$ 's by themselves or the combination of 2 of them divided by the remaining one.
$ \sum \frac {3^i}{5^i} \plus{} \sum\frac {5^i}{3^i} \plus{} \sum\frac {3^i}{2^i} \plus{} \sum\frac {2^i}{3^i} \plus{} \sum\frac {5^i}{3^i} \plus{} \sum\frac {3^i}{5^i} \plus{} \sum\frac {6^i}{5^i} \plus{} \sum\frac {5^i}{6^i} \plus{} \sum\frac {10^i}{3^i} \plus{} \sum\frac{3^i}{10^i} \plus{} \sum\frac {15^i}{2^i} \plus{} \sum\frac {2^i}{15^i}$

Evaluating would be messy. One technique would be to find the sum (1 + 2 + 4 + 8)(1+3+9+27)(1+5+25+125) then subtract the ones that aren't relatively prime.

[hide="How about we extend that a bit further.."]
$ S \equal{} \left(\frac{1}{8} \plus{} \frac{1}{4} \plus{} \frac 12 \plus{} 1 \plus{} 2 \plus{} 4 \plus{} 8\right)\left(\frac{1}{27} \plus{} \frac{1}{9} \plus{} \frac 13 \plus{} 1 \plus{} 3 \plus{} 9 \plus{} 27\right)\left(\frac{1}{125} \plus{} \frac{1}{25} \plus{} \frac 15 \plus{} 1 \plus{} 5 \plus{} 25 \plus{} 125\right)$
[/hide]

And multiplying out gives the above sum...How did you think of such a clever factorization?

[quote="mathemonster"]And multiplying out gives the above sum[/quote]  Not quite, actually -- the sum you proposed misses terms like $ \frac{5}{24}$, while azjps's expression includes it.

:blush: I forgot to include when the degrees weren't the same, Sorry. Still how can you prove this?

[quote="mathemonster"]How did you think of such a clever factorization?[/quote]
Well, for one thing, a very similar problem was on the AIME.  Here's another way to think about it:  take an arbitrary divisor of $ n^2$, divide it by $ n$, and reduce the fraction.  What do you get?

You didn't reduce the fraction.  If $ d$ is an arbitrary divisor of $ n^2$, then $ \frac {d}{n}$ reduces to a fraction $ \frac {a}{b}$ with the following properties:

- $ b | n$ (from the properties of reduction)
- $ bd \equal{} an$ where $ \gcd(a, b) \equal{} 1$.

On the other hand, there exists an integer $ e$ such that $ de \equal{} n^2$.  This means that

$ bde \equal{} bn^2 \equal{} ane \Leftrightarrow bn \equal{} ae$,

but since we still have $ \gcd(a, b) \equal{} 1$ it follows that $ a | n$.  So $ a, b$ are two relatively prime divisors of $ n$.  Now you only have to prove that we can get every pair of relatively prime divisors in this way, and it will then follow that the desired sum is

$ \frac {\sigma(n^2)}{n}$

where $ \sigma(n)$ is the sum of the divisors of $ n$ - and this is exactly what azjps wrote down.  (To prove that last fact you might want to write down this process explicitly in terms of prime factorization.)

Alternately, all the original problem is asking for is the sum of all terms of the form $ 2^a 3^b 5^c$ where $ \minus{}3 \le a, b, c \le 3$, and now it's obvious what to do if you know how to find the sum of the divisors of a number.


ye moammaye sadeh,mitoonin raveshetoono ham inja benevisin

man natonestam in soalo download konam bayad chi kar konam? :(  :?:

aval haftaeera por meekone, bad merezesh tooye panshtaee ta panshtaee por beshe. bad do litre memone tooye haftaee. inra mereze rooye daste. Hameen kara se bar anjam meede ta sheesh litre rooye daste baz she. bayad daste baz she deege.

yek joore deegeh ham man peida kardam. aval panjtaeera por meekone. bad meerezesh tooye haftaee. dobare panjtaeera por meekone va dobare meerezesh tooye haftaee. hala haftaeera khalee meekone. se litre tooye panjtaee hast ke oono meereze tooye haftaee. panjtaeera dobare por meekone. dobare meerezesh tooye haftaee. haftaeera khalee mekone. yek litre tooye panjtaee hast ke meerezesh tooye haftaee. panjtaeera por meekone. inra meereze tooye haftaee. hala haftaee toosh 6 litre hast. haftaeera meezare rooye daste ta dar baz beshe.

man in barname ro tonestan download konam va halesham kardam vali didam ke  [u]power[/u] gablan hallesh karde albate hale avali sorost nist chon bayed dagigan shish litr ab bashe na se ta se litr ama dovomi doroste ke manam az on rah hallesh kardam.
ba tashakor :)


A diagonal of a rectangle is 41 inches, and the perimeter is 98 inches. What is the number of square inches in the area of the rectangle?


modular arithmetic
Divisibility Theory
pen

Suppose that $p$ is a prime number and is greater than $3$. Prove that $7^{p}-6^{p}-1$ is divisible by $43$.

[quote="Robert Gerbicz"]If $ p > 3$ is a prime number then $ p \equal{} 6*k \plus{} 1$ or $ p \equal{} 6*k \minus{} 1$ for some positive integer k. In the first case:
\[ {7}^{p} \minus{} {6}^{p} \minus{} 1 \equal{} {7}^{6*k \plus{} 1} \minus{} {6}^{6*k \plus{} 1} \minus{} 1 \equiv 7*({7}^{6})^{k} \minus{} 6*({6}^{6})^{k} \minus{} 1 \equiv 7 \minus{} 6 \minus{} 1 \equiv 0\pmod{43}
\]
using : $ 7^{6}\equiv 6^{6}\equiv 1\pmod{43}$.
In the second case:
\[ {7}^{p} \minus{} {6}^{p} \minus{} 1 \equal{} {7}^{6*k \minus{} 1} \minus{} {6}^{6*k \minus{} 1} \minus{} 1 \equiv{7}^{5}*({7}^{6})^{k \minus{} 1} \minus{} {6}^{5}*({6}^{6})^{k \minus{} 1} \minus{} 1 \equiv 37 \minus{} 36 \minus{} 1 \equiv 0\pmod{43}
\]
This proves the statment.[/quote]

By the well-known fact that $ x^2 \plus{} x \plus{} 1|x^{2k} \plus{} x^k \plus{} 1$ for every $ k\equiv 1,2 \pmod{3}$ we have that $ 6^2 \plus{} 6 \plus{} 1|6^{2p} \plus{} 6^p \plus{} 1$ because 3 does not divide p. So $ 6^{2p} \plus{} 6^p \plus{} 1\equiv 0 \pmod{43}$.
Also we have that $ 6\cdot 7\equiv \minus{} 1\pmod{43}$ so $ (6\cdot 7)^p\equiv \minus{} 1\pmod{43}$ because p is odd.
$ 6^{2p} \plus{} 6^p \plus{} 1\equiv6^{2p} \plus{} 6^p \minus{} (6*7)^p\equiv6^p*(6^p \plus{} 1 \minus{} 7^p)\equiv0\pmod{43}$
Because $ \gcd(6,43) \equal{} 1$ we have that $ 43|7^p \minus{} 6^p \minus{} 1$

It is spesialiast problem about $ A_n\equal{}a^n\plus{}b^n\plus{}c^n, a\equal{}7, b\equal{}\minus{}6,c\equal{}\minus{}1,a\plus{}b\plus{}c\equal{}0$.

[quote="Peter"]Can you be more specific, Rust? Do you have a generalisation (with proof)?[/quote]
Let $ x_1,x_2$ are roots $ x^2 \plus{} ax \plus{} b \equal{} 0,a,b\in Z$ and ab -even (if $ x_1,x_2$ are integers, then always $ ab \equal{} \minus{} (x_1 \plus{} x_2)x_1x_2$ even). Let $ A_n \equal{} x_1^n \plus{} x_2^n \plus{} a^n, P \equal{} a^2 \minus{} b$. Then $ (2P)|A_{3k \minus{} 1}, \ (2P^2)|A_{3k \plus{} 1}.$
In your case $ P \equal{} 43$ and prime $ p > 3$ is $ p \equal{} \pm 1\mod 3$.
If $ ab$ odd we can prove only $ P|A_{3k \minus{} 1}, \ P^2|A_{3k \plus{} 1}$.
Generalized. Let $ x_1,x_2,...,x_k$ roots of $ x^k\equal{}ax\plus{}b, \ a,b\in Z$ and $ A_n\equal{}x_1^n\plus{}x_2^n\plus{}...\plus{}x_k^n$.
Then
$ A_{n\plus{}k}\equal{}aA_{n\plus{}1}\plus{}bA_n,A_0\equal{}k,A_1\equal{}A_2\equal{}...\equal{}A_{k\minus{}2}\equal{}0,$ and
$ A_{k\minus{}1}\equal{}(k\minus{}1)!(\minus{}1)^{k}a, A_k\equal{}bk.$
Therefore
$ A_{nk}\equal{}kc_nb^n\plus{}k!c_{n\plus{}1\minus{}k}b^{n\plus{}1\minus{}k}a^k\plus{}k!c_{n\plus{}2\minus{}2k}b^{n\plus{}2\minus{}2k}a^{2k}\plus{}...$
$ A_{nk\minus{}i}\equal{}k!c_{n\minus{}i}b^{n\minus{}i}a^i\plus{}k!c_{n\plus{}1\minus{}i\minus{}k}b^{n\plus{}1\minus{}i\minus{}k}a^{i\plus{}k}\plus{}...$
Therefore $ a^i|A_{nk\minus{}i}$.

Rust could you make you generalization clearer?I get confused with the expressions you name $ A_n$.. Thank you.

If $ A_n = x_1^n + x_2^n + .. + x_k^n$, then it may be defineed by recurent
$ A_{n + k} = \sigma_1 A_{n + k - 1} - \sigma_2 A_{n + k - 2} + ... + ( - 1)^{k - 1}\sigma_kA_n$, were $ \sigma_1 = x_1 + x_2 + ... + x_k,\sigma_2 = \sum_{i < j}x_ix_j,...$
$ x_1,x_2,..,x_k$ roots of $ x^k - \sigma_1x^{k - 1} + \sigma_2x^{k - 2} - ... + ( - 1)^k\sigma_k = 0$.
If all $ \sigma_i$ are integer, then $ A_n = P_n(\sigma_1,...,\sigma_k )$ are integer for all natural n.
Consider case, when ${ \sigma_1 = \sigma_2 = \sigma_{k - 2} = 0,\sigma_{k - 1}( - 1)^k = a,\sigma_k ( - 1)^k - 1} = b$.
Then $ A_{n + k} = aA_{n + 1} + bA_n$ and $ A_1 = ... = A_{k - 2} = 0, A_{k - 1} = aA_0 + bA_{ - 1} = (k - 1)a, A_k = bk.$
In these case $ A_n = P_n(a,b)$, were $ P_n$ is polinom with common degree n for all members, $ deg(a) = k - 1,deg(b) = k$.
$ P_n(a,b) = aP_{n - k + 1}(a,b) + P_{n - k}b,P_{k - 1}(a,b) = (k - 1)a,P_k(a,b) = kb$.
Therefore $ A_{n} = P_n(a,b) = \sum_{i,ik + j(k - 1) = n} B(n,i,j)b^ia^j$.
For coefficients $ B(n,i,j)$ we had $ B((k - 1)j,0,j) = k - 1,B(ki,i,0) = k,$
$ B(ki + (k - 1)j,i,j) = k\binom{i - 1 + j}{j} + (k - 1)\binom{i + j - 1}{i} = \frac {n(i + j - 1)!}{i!j!}$.
For case $ k = 3,x_1 = 7,x_2 = - 6,x_3 = - 1,\sigma_1 = 0,\sigma_2 = - 43,a = 43,\sigma_3 = 42 = b$ we had
$ 7^n + ( - 6)^n + ( - 1)^n = n\sum_{3i + 2j = n}\frac {(i + j - 1)!}{i!j!}a^ib^j$
If $ n$ odd, then $ i$ -odd, therefore $ a|P_n(a,b)$.

Such heavy machinery is not needed.

The powers of 7 mod 43 follow the pattern $7, 6, -1, -7, -6, 1, 7$, ...  The powers of 6 mod 43 follow the pattern $6, -7, 1, 6$, ...  All primes greater than 3 are 1 or 5 mod 6. Thus the expression $7^p - 6^p - 1$ must be either $(7 - 6 - 1)$ or $(-6 -(-7) - 1)$ mod 43.

Since $7 \equiv -6^{-1} \pmod{43}$, hence it suffices to show $6^{p}+6^{-p}+1 \equiv 0 \pmod{43} \Leftrightarrow 6^{2p}+6^p+1 \equiv 0 \pmod{43}$ .
Now since $ord_{43}(6)=3$ and $p>3$, hence $6^{p} \not\equiv 1 \pmod{43}$.
Hence $(6^p-1)\left(6^{2p}+6^p+1\right)=6^{3p}-1=216^p-1 \equiv 0 \pmod{43}$ yields the result. $\square$


Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q\minus{}q^p\equal{}7$

[quote="thaithuan_GC"]Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q \minus{} q^p \equal{} 7$[/quote]
[hide=" I think it can be helped you"]http://www.mathlinks.ro/Forum/viewtopic.php?t=217641[/hide][/hide]

[hide="Solution"][quote="thaithuan_GC"]Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q \minus{} q^p \equal{} 7$[/quote]
According Feramt's theorem, we easily show that 2p-q=7
So, $ 2p^q\minus{}q^p\equal{}2p\minus{}q$
We know that $ x^y>y^x$ for every $ y>x>3$, it help us to solve this problem.[/hide]

[quote="mr.danh"]According Feramt's theorem, we easily show that 2p-q=7[/quote]
I don't think so :blush:

Wow, if 2p-7 >0 then 2p-q=7. If 2p-7<0, we get p=2,q=3
Answer: two solutions (p,q) = (2,3), (5,3)

[quote="thaithuan_GC"]Find all pairs of prime numbers $ (p,q)$ satisfying $ 2p^q \minus{} q^p \equal{} 7$[/quote]
See that $ q \mid 2p\minus{}7$ and $ p \mid q\plus{}7$ This implies $ pq \mid (2p\minus{}7)(q\plus{}7) \Rightarrow pq \mid 14p \minus{}7q \minus{} 49$
If $ p \equal{} 7$ we have $ q \mid 14\minus{}7 \equal{} 7$ And so $ q \equal{} 7$ But $ (7,7)$ is not a solution.
If $ q \equal{} 7$ we have $ p \mid 7\plus{}7 \equal{} 7$ but neither $ (7,7)$ or $ (2,7)$ are solutions.

This means that $ pq \mid 2p \minus{} q \minus{} 7$ and therefore $ pq \le | 2p\minus{}q\minus{}7 |$ or $ 2p\minus{}q\minus{}7\equal{}0$
So $ pq \le 2p\minus{}q\minus{}7$ or $ pq \le q\plus{}7\minus{}2p$ or $ 2p\minus{}q\minus{}7 \equal{} 0$

$ pq \le 2p\minus{}q\minus{}7 \iff (p\plus{}1)(q\minus{}2) \le \minus{}9$ which is obviously false.
So $ pq \le q\plus{}7\minus{}2p \iff (p\minus{}1)(q\plus{}2) \le 5 \Rightarrow q \le 3, p \equal{} 2$

Obviously $ p \equal{} q$ is not a solution, So from $ pq \le q\plus{}7\minus{}2p$ we only get $ (p,q) \equal{} (2,3)$

Mr.Danh have taken care of $ 2p\minus{}q \equal{} 2p^q\minus{}q^p \equal{} 7$.

The key is to first see that $ q \mid 2p\minus{}7$ and $ p \mid q\plus{}7$ and after that multiply them. This means we have product is at most a sum or a sum is zero, which leaves very few opportunities for $ p$ and $ q$ :idea:


What product is obtained when allyl vinyl sulfide goes through the following steps?

1. $ sec\minus{}BuLi$

2. $ \ce{PhCH2Br}$

3. reflux

4. $ \ce{H3O\plus{}}$

isnt allyl vinyl sulfide unreactive????

It is nothing but  a thio ether isnt it????

SO why shloud it be reactive???? :maybe:  :maybe:  :maybe:

Of course it is reactive.

[quote="Valeriummaximum"]SO why shloud it be reactive????[/quote]

It is your job to explain that.

the hydrogen alpha to sulphur is quite acidic, so first step is an acid base reaction
second probably sn
third, god knows
fourth, hydrolysis
sorry for being so brief, no time to think abt this problem as im having a tight schedule rite now :D

the third step probably involves claisen rearrangent and in the fourth(assuming third to be correct) you get ketone.
the intermediate for the third step is probably this(forgot to add benzene.)

Very good. This is called the thio-Claisen rearrangement.


category theory
advanced fields
advanced fields unsolved

let $ A,B$ be rings (commutative, unital) and $ f,g: A \to B$ homomorphisms with $ Spec(f) \equal{} Spec(g)$, i.e. for every $ x \in A$ the prime ideals containing $ f(x)$ coincide with the ones containing $ g(x)$. how does it follow, that $ f \equal{} g$?

I just can show the existence of natural numbers $ n_x , m_x$, depending on $ x$, with $ f(x)\equal{}g(x)^{n_x}$ and $ g(x)\equal{}f(x)^{m_x}$.

hm, probably it's wrong. spec is faithful as a functor rings -> locally ringed spaces, but not -> topological spaces.

I think that a field and two different automorphism will do for a counterexample.


[i]A London telephone number consisted of three letters followed by four figures. How many telephone numbers could be made by
arranging the symbols in [b]BYR[/b] [b]2556[/b]?[/i]

What i think is that you would do the $ 3!$ for the letters and $ 4!$ for the numbers and multiply them, this gives $ 144$ but the the answer is ment to be $ 72$. What did i do wrong?

[quote="i_like_pie"]There are two $ 5$'s, so you must account for over-counting by dividing the result by $ 2!$.[/quote]

I noticed the two $ 5$'s, but should not they be counted as two separate entities?

There are many ways to look at this.

my advice: when in doubt, try some examples!

you start out with BYR 2556.
Now switch the two 5's around. You get: BYR 2556, the same thing!

Try YBR 2655.
Now switch the two 5's around. You get: YBR 2655, the same thing!

In permutations, the order does matter. But I think that you are getting confused here. The question asks you to find how many telephone numbers are possible. So the one way is to set it up as a permutation question. But when you get 144, you are really getting the number of telephone numbers if either one of the 5's was not 5, 2, or 6. So there would be 144 telephone numbers for a number like BYR 2956, and there are six more possibilities. But then you could switch the second 5, and that would give you another 7 different possibilities. But I'm getting off topic - you should understand my point.

At this point, if you're really good, you could almost start to conjecture that you would have half of the  possibilities if you switched one of the 5's to something else rather than 2, 5, or 6. Or, you could use the formula. "If a set of n elements has n1 elements of one kind alike [here, alike means the same], n2 of another kind alike, and so on, then the number of permuations, P, of the n elements taken n at a time is given by P = n!/(n1!n2!...) [here, n1, n2, ... the one and two after the n is simply the subscript but I'm not very good at latex]" (Algebra and Trigonometry Structure and Method Book 2 by McDougal Littell Houghton Mifflin). So you would divide 144 by 2!, as i_like_pie pointed out.

Or, you could set up the permutation "procedure" - as I like to call it - and say: _*_*_*_*_*_*_ (where * denotes multiplication). Then fill in the blanks to get 3*2*1*4*2*2*1. Note the first 2 in the pair of twos and why it was 2 instead of 3. This is very important!

Another way of looking at it would be to think: How many telephone numbers could be made by arranging the symbols in 555 - 5555. It's obviously 1 (this matches what you would get if you used the formula, too. Try it!). This example makes it clear. (You could also use the Reflexive Axiom and possibly Closure Axiom, too).

Hope this was helpful.

BYR can be arranged in 3! ways, like you said.

2556, however, can only be arranged in 4!/2 ways, because
given every single combination, you are overcounting twice.
In other words, you are counting 2[u]5[/u][b]5[/b]6 and
2[b]5[/b][u]5[/u]6 as different things, as if 5 and 5 were
separate numbers.  But they are, obviously, not.

So, the answer is simply:
3!*4!/2=[b]72[/b], as the book said.

thanks everyone for your time and help, i think i am getting the hang of things.


algebra
polynomial
quadratics
complex numbers
quadratic formula

At how many points do the graphs of $ y\equal{}x^{3}$ and $ y\equal{}x^{2}\plus{}4x$ intersect?

Please submit explanation and example.

At how many points do the graphs
of the equations y = x3 and y = x2 +
4x intersect? They intersect when
the value of y is the same for both.

[hide="Solution"] x3 = x2 + 4x
x3 – x2 – 4x = 0
x(x2 – x – 4) = 0
Now we set each part equal to zero
and solve.
x = 0 is one solution
There are two more solutions which
occur when x2 – x – 4 = 0
x2 = x + 4
x = x + 4 is one solution.
x = - x + 4 is the other solution.
1 + 2 = 3 Ans.      [/hide]

[hide="Click this"] :ddr:
Awesome, isn't it![/hide]

[quote="gogators"]At how many points do the graphs of $ y \equal{} x^{3}$ and $ y \equal{} x^{2} \plus{} 4x$ intersect?

Please submit explanation and example.[/quote]

The graphs intersect when the y values are the same. Thus, we have $ x^3\equal{}x^2\plus{}4x$.

$ x^3\minus{}x^2\minus{}4x\equal{}0 \implies x(x^2\minus{}x\minus{}4)\equal{}0$. Thus there are 3 roots.

Once we have $ x^3\minus{}x^2\minus{}4x\equal{}0$ we know there are $ 3$ points because all roots are real and this polynomial has $ \text{deg3}$

To check to see if none of the roots are the same, the discriminant ($ b^2\minus{}4ac$) of the quadratic BOGTRO gave must be positive (negative gives complex numbers and zero gives three equal roots).

[quote="gogators"]At how many points do the graphs of $ y \equal{} x^{3}$ and $ y \equal{} x^{2} \plus{} 4x$ intersect?

Please submit explanation and example.[/quote]

$ x^3 \equal{} x^2 \plus{} 4x$

There will be a maximum of three intersection points, because if there were any more then both of the graphs would have to be the same, and obviously they aren't. A trivial solution to this equation is $ x \equal{} 0$. We can find more points by dividing both sides by x;

$ x^2 \equal{} x \plus{} 4$
$ x^2 \minus{} x \minus{} 4 \equal{} 0$

Using the quadratic formula we find that the solutions to this equation are;

$ \frac{1 \plus{} \sqrt{17}}{2}$ and $ \frac{1 \minus{} \sqrt{17}}{2}$

So there are three intersection points.


ratio
geometry
circumcircle
geometry proposed

Let $ ABCD$ be a bicentric quadrilateral, Diagonals $ AC$ and $ BD$ meet at $ T$. Let $ EF$ be the diameter of its circumcircle such that $ EF$ is perpendicular to $ BD,BD$ intersects $ EF$ at $ S$. $ A,E$ are on the same side of $ BD$. Prove that $ \frac {ES}{SF} \equal{} \frac {AT}{TC}$


find all natural nos. a,b,c
 s.t.
 2^n*a+b divides c^n+1 for all natural nos. n.

Please, stop posting problems from the new issue of Mathematical Reflections.

Don't you understand that it is from a current issue?  :mad:
Someone should lock or delete this topic.

@who: explain why you posted this!
@all others: please tell me if there are more to be locked in Number Theory or if it is better to delete them.
(and locked now)

Now unlocked again...


geometry
3D geometry
algebra unsolved
algebra

Find the sum of 100 numbers after the comma of this number{(2)^1/2  +  1 }^2001
*x^y = y times x

Hi;

If you mean after the decimal point, then the sum of the first hundred digits after the decimal point is 0.

Hello ,,, yes i mean after the decimal point .. but why can you explain me this .?

Hi;

Here is how I did it. Take your number and convert it to decimal.

$ \sqrt {2} \plus{} 1 \approx 2.414213562$

Now cube it:

$ (\sqrt {2} \plus{} 1)^3 \approx 14.071067811$

You are interested in the part after the decimal. when you raise a number that has an integer part and a fractional part ( digits after the decimal ) the number grows in a particular way when being raised to a power. To the left of the decimal point grows to the left and to the right of the decimal point grows to the right by adding leading zeros. We only have to raise the right side up to some power to deduce the answer. Here is how.

Take the fractional part of the number cubed and raise it to the 19 th power.

$ .071067811^{19} \equal{} 1.51979788* 10^{ \minus{} 22}$

We already have a number with 21 leading zeros. Raise the new number to  the 5 th power again.

$ ( 1.51979788* 10^{ \minus{} 22})^{5} \equal{} 8.10828811*10^{ \minus{} 110}$

Since ((x^3)^19)^5) is equal to x^285 we have just raised the fractional part of your number to the 285 th power. As you can see their is already 109 leading zeros, That means their are 109 zeros after the decimal point. The number of leading zeros is only going to get larger than this as we raise the number up to 2001. So you already know that the first 100 digits after the decimal point of:

$ (\sqrt {2} \plus{} 1)^{285}$

are zeros, so the sum of them is 0.

There is another way :

Consider the sequence $ a_n$ defined as $ a_0 \equal{} a_1 \equal{} 2$ and $ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} \plus{} a_n$. It's immediate to establish $ a_n \equal{} (\sqrt 2 \plus{} 1)^n \plus{} (1 \minus{} \sqrt 2)^n$

So $ (\sqrt 2 \plus{} 1)^{2001} \equal{} a_{2001} \plus{} (\sqrt 2 \minus{} 1)^{2001}$ and, since $ a_n$ is integer and $ \sqrt 2 \minus{} 1 < \frac 12$, we get  $ \{(\sqrt 2 \plus{} 1)^{2001}\} \equal{} (\sqrt 2 \minus{} 1)^{2001} < \frac 1{2^{2001}} < 10^{ \minus{} 600}$

Hence the result $ 0$


function
integration
calculus
linear algebra
matrix
inequalities
derivative

a nice (and hard) one:
let $f$ be a $C^1$ function $f:R^n -> R^n$ such that
$df(x)$ always belongs to the orthogonal group of $R^n$ with the usual scalar product.
Prove that $f$ is an isometry.

We have first of all: $f(x)-f(y)=\int_{0}^{1}Df(x+t(y-x))dt(x-y)$ for all $x,y\in\Bbb R^n$ where the integration of the matrix means integration of the elements of it. Now we have to show that $\int_{0}^{1}Df(x+t(y-x))dt$ is orthogonal which can be proven by standard computations (I will complete it later when I have more time).

Misha

Alekk are you sure $f$ is $C^{1}$ ?

There is a version with $f$ $C^{2}$ using the trilinear linear form

$g(h,k,z)=(f''(x).h.k|f'(x)z)$

I was randomly browsing the forum for interesting problems, and I found this. I think this should work:

Let $x,y\in\mathbb R^n$ be two vectors. Lagrange's inequality gives us $\|f(x)-f(y)\|\le\sup_{[x,y]}\|f'\|\cdot\|x-y\|$. Since $\|f'\|=1$ everywhere, we find that a function $f$ defined on a convex open subset $D$ of $\mathbb R^n$ with values in $\mathbb R^n$ for which $f'$ is always orthogonal decreases the distances: $\|f(x)-f(y)\|\le\|x-y\|,\ \forall x,y\in D$.

Now, the Inverse Function Theorem says that for every $x\in\mathbb R^n$ there are open subsets $x\in V\subset\mathbb R^n,W\in\mathbb R^n$ s.t. $f|_V: V\to W$ is invertible, and its inverse is differentiable, it's derivative in $f(a)$ being the inverse of the derivative of $f$ in $a,\ \forall a\in V$, thus also orthogonal. This means (from the observations in the paragraph above) that the restriction of $(f|_V)^{-1}$ to some open ball in $W$ centered around $f(x)$ decreases the distances, and, since its inverse does the same, $f$ is a local isometry, i.e. for every $x$ we can find a neighbourhood $V_x$ of $x$ s.t. the restriction of $f$ to $V_x$ is an isometry. The conclusion that $f$ is an isometry is easy to draw from here.

I remember I did the same, and said to my teacher the conclusion was straighforward from here. But it is not very straightforward. You have to write a few lines to conclude if my memories are correct ...

Oh, I know it's not [b]that[/b] obvious :). Anyway, here's what I did (I don't know if it works, as I haven't checked it properly; I was walking around my university waiting for someone and just sort of mentally playing around with this :)):

Take a hypercube having two prescribed points as opposite vertices, and cover each point of this hypercube with an open neighbourhood for which the restriction of $f$ is an isometry. By Lebesgue's Number Lemma, if we divide this hypercube into small enough hypercubes (in the obvious manner), then any two adjacent hypercubes will lie in some common open set belonging to the described covering. Now go from one of the corners of the hypercube to the opposite corner passing through the vertices of adjacent small hypercubes one by one. Each time, the structure is preserved isometrically, because at each step, the distances from the next point we take into consideration to the vertices of the previous small hypercube are exactly what we want them to be, and the position of a point in $n$-space is determined by its distances to the vertices of an $n$-hypercube (actually, it's determined by a lot fewer distances). This means that the whole hypercube is transformed isometrically by $f$ so, in particular, the distance between the two opposite corners we considered is not changed by $f$.

I'm terribly sorry if it sounds very unclear.


linear algebra
matrix
algebra
polynomial
quadratics
complex numbers
linear algebra unsolved

[b]a)[/b] Let $ A$ be a skew-symmetric matrix (this means $ A^T \equal{} \minus{} A$, where $ A^T$ denotes the transpose of $ A$) with real entries.
Show that its determinant $ \det A$ is nonnegative.
[b]b)[/b] If $ A$ is a skew-symmetric matrix with integer entries, then show that its determinant $ \det A$ is the square of an integer!

You mean "nonnegative", not "positive", since zero matrix is skew symmetric and has det=0 :).
[Moderator edit: I have just edited Arline's post #1 according to this.]

I always like to see just how much mileage I can get out of Schur's lemma: every complex square matrix can be unitarily upper triangularized. That is, there exists $U$ with $U^{*}U=I$ such that $U^{*}AU=T$ with $T$ upper triangular.

Take the $*$ of that: $U^{*}A^{*}U=T^{*}.$ Since $A^{*}=A^{T}=-A,$ it follows that $T^{*}=-T.$ This in turn implies that all of the off-diagonal elements of $T$ are zero and all of the elements of $T$ on the main diagonal (the eigenvalues of $A$) are pure imaginary.

The eigenvalues of $A$ are pure imaginary. But since $A$ is real, the eigevalues come in complex conjugate pairs. The eigenvalues of $A$ (and hence the diagonal entries of $T$) are $i\alpha,-i\alpha, i\beta,-i\beta,\dots i\omega,-i\omega,0,\dots,0.$

Note that if $A$ is of odd size, there must be at least one zero in that list of eigenvalues and $\det A=0.$ To make the rest of the problem interesting, assume that $A$ is of even size and nonsingular. The characteristic polynomial of $A$ is $(x^{2}+\alpha^{2})(x^{2}+\beta^{2})\cdots(x^{2}+\omega^{2}).$ If $A$ has only integer entries, this characteristic polynomal has only integer coefficients.

I'd like to get from that polynomial having integer coefficients to $\alpha,\beta,\dots$ being themselves integers, but I don't have that argument yet.

The determinant of a skew-symmetric matrix (for even n) is equal to square of its Pfaffian, thus nonnegative. See [url]http://en.wikipedia.org/wiki/Pfaffian[/url]. You can also try here: [url]http://www.math.harvard.edu/~elkies/M55a.99/pfaff.html[/url].

[quote="I"]I'd like to get from that polynomial having integer coefficients to $\alpha,\beta,\dots$ being themselves integers, but I don't have that argument yet.[/quote]
Two points: first, we can get the determinant to be the square of an integer without this. Second, this is false anyway.

For a cheap example, if

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

Then the characteristic polynomial of $A$ is $x^{4}+3x^{2}+1,$ which cannot be factored into integer-coefficient quadratic polynomials.


algorithm
combinatorics unsolved
combinatorics

Arutyun and Amayak show another effective trick. A spectator writes down on a board a sequence of $N$ (decimal) digits. Amayak closes two adjacent digits by a black disc. Then Arutyun comes and says both closed digits (and their order). For which minimal $N$ they may show such a trick?
[i]K. Knop, O. Leontieva[/i]

It's clear that when Arutyun sees the sequence with the closed digits, he has to identify it in an unique way. So the possibilities of covering the digits by Amayak in the initial sequence has to be at least equal to the 100 possibilities that the sequence can be for Arutyun. The possibilities of covering is $N-1$, so $N-1\geq 100$. So the minimum is $101$. In this case a bijection can be constructed between the "complete sequences" and the "covered sequences". This is of course a silly game which demands a huge memory :P , and let's not forget that N can be anything larger then 101 so as it tends to infinite the capacity of our memory tends to infinite. This problem could be solved if we found an algorithm which can be applicable for larger N-s.

[quote="N.T.TUAN"][i]K. Knop, O. Leontieva[/i]
Arutyun and Amayak show another effective trick.
A spectator writes down on a board a sequence of $N$ (decimal) digits.
Amayak closes two adjacent digits by a black disc. Then Arutyun comes
and says both closed digits (and their order). For which minimal $N$
they may show such a trick?[/quote]

Very nice problem. I searched for a solution a long time.

1) $N$ must be $\geq 101$
Arutyun may have in front of him $(N-1)10^{N-2}$ different configurations ($N-1$ positions for the disc and $10^{N-2}$  for the reminder).
This number must be great enough to be mapped on any of the $10^{N}$ configurations the spectator may have choosen.
So $(N-1)10^{N-2}\geq 10^{N}$ and $N\geq 101$

2) there exists a $101$ strategy :

Let $a_{0},a_{1}, \cdots,a_{99},a_{100}$ be the first $101$ ordered numbers written by the spectator.

Amayak computes $S_{1}=10a_{0}+a_{1}+10a_{2}+a_{3}+\cdots+10a_{100}[100]$ (where $[100]$ means modulo $100$).
Amayak put then the black disc on numbers $a_{S_{1}}$ and $a_{(S_{1}+1)}$

Aruyun first looks at the position of the black disc and knows $S_{1}$
Aruyun then computes $S_{2}=10a_{0}+a_{1}+10a_{2}+a_{3}+\cdots+10a_{100}[100]$, using $0$ and $0$ for the two hidden positions.
Aruyun computes $S_{3}=S_{1}-S_{2}[100]$ and the final step is elementary :
$S_{3}$ is a number between $0$ and $99$ and so $S_{3}=10a+b$

Then, if $S_{1}$ is even, the ordered two hidden numbers are $(a,b)$
And if $S_{1}$ is odd, the ordered two hidden numbers are $(b,a)$

And minimal $N$ is $101$

Very nice strategy. I'd been trying to find it too, but i gave it up and thought there isn't a good strategy.

[quote="pco"]

Then, if $S_{1}$ is even, the ordered two hidden numbers are $(a,b)$
And if $S_{1}$ is odd, the ordered two hidden numbers are $(b,a)$

And minimal $N$ is $101$[/quote]
Shouldn't it be $S_{3}$?

if N=101 there's also strategy by marriage theorem


geometry
algebra
function
domain
geometry solved

Let ABC be a triangle. On it's sides are constructed 3 regular n - gons.
Let O1,O2,O3 be their centers, find all n for which AO1, BO2, CO3 are collinear.
     It is clear for n = 3 - Torricelli and for n = 4 - Vecten.
If this problem is anywhere appeared, i am sorry for giving it, but i don't know
the solution.

This is called Kiepert's theorem: if we construct similar isosceles triangles having their bases on the sides of $ABC$, then the lines joining the vertices of $ABC$ with the respective vertices of the triangles are concurrent, so, to answer your question, the concurrence occurs for all $n$ as a particular case of Kiepert's theorem.

Have you got a prove for this theorem, where can it be found?

You can find something on this subject [url=http://www.cut-the-knot.org/Curriculum/Geometry/Kiepert.shtml]here[/url]. There are more links on [url=http://www.google.com/custom?domains=cut-the-knot.org&q=kiepert&sitesearch=cut-the-knot.org&sa=Search&client=pub-9118218894256794&forid=1&ie=ISO-8859-1&oe=ISO-8859-1&cof=GALT%3A%23003324%3BGL%3A1%3BDIV%3A%2366CC99%3BVLC%3AFF6600%3BAH%3Acenter%3BBGC%3AC5DBCF%3BLBGC%3A73B59C%3BALC%3A000000%3BLC%3A000000%3BT%3A330033%3BGFNT%3A333300%3BGIMP%3A333300%3BFORID%3A1%3B&hl=en]this[/url] page, but they are all links to articles on the same site: Cut the Knot. Try Goolge-ing it :).

I never took the time to actually write the proof, but it can be proven with the aid of projective geometry.


geometry
perimeter
calculus
integration
inequalities
triangle inequality

There is a triangle with side's length expressed by integer . One side is 3 time biger than the other side  and the third side is 6 .Find the highest value of the perimeter .

Let me restate the question as I understand it.

There is a triangle with integral side lengths.  One side of the triangle is 6, and the other two sides are in a 3:1 ratio. What is the greatest possible perimeter of this triangle?

Is this a correct interpretation of the problem?

[hide="If it is, then..."]I believe the answer is 14, which would be a triangle with sides of 2, 6, and 6.

Assume that a triangle exists that meets the requirements and has a perimeter of greater than 14.  Since the triangle has integral side lengths, the next larger triangle will be a 3-6-9 triangle.  However, such a triangle is not possible because the length of the third side is equal to the lengths of the other two sides.[/hide]

p=a+3a+6=4a+6. Because of triangle inequality we have a+6>3a--> a<3 and 4a>6--> a>1 so a=2--> p=14

edited for shorter answer.

[b]Follow-up:[/b] [i]Same setup, different concept[/i]
Suppose the side-lengths don't have to be integers, but the sides are still $ 6,x,3x$. What is the maximum area of this triangle?


If the three-digit number $ 2d2$ is divisible by 7, what is $ d$?

The rule for divisibility of $ 7$ is twice the units digit subtracted from the number with the units digit dropped is divisible by $ 7$.  Meaning, for a number $ 100a \plus{} 10b \plus{} c$ that is divisible by $ 7$, $ 10a \plus{} b \minus{} 2c$ is divisible by $ 7$.  So, this means that $ 2(10) \plus{} d \minus{} 2(2)$ is divisible by $ 7$, so we have that $ 16 \plus{} d$ is divisible by $ 7$.  The closest number that is divisible by $ 7$ is $ 21$, so $ d \equal{} \boxed{5}$.

[hide="Silly way"]Or you could just try out numbers one by one (202,212,222...252) I think that would be just slightly faster.[/hide]

Okay, just notice that 42 is divisible by 7, and it ends with a 2.  So ${d=\boxed[5}$ .


Find the biggest number of positive integers which we can choose among the integers $1,2,...,100$ such that there isn't $a,b,c,d$ with $a+b+c=d.$

67: 34,35,36,...,100.
Let's suppose we have n>67 numbers, say $x_{1}<x_{2}<\ldots<x_{n}$. Then numbers \[x_{2}-x_{1},x_{3}-x_{1},x_{4}-x_{1},\ldots,x_{n}-x_{1}\] are distinct, between 1 and 99 and there're $n-1\ge67$ of them.
We have $x_{68}\le100,x_{67}\le99,\ldots,x_{18}\le50,\ldots.$ Numbers \[x_{1}+x_{2},x_{2}+x_{3},x_{3}+x_{4},\ldots,x_{17}+x_{18},\]  \[x_{1}+x_{3},x_{2}+x_{4},x_{3}+x_{5},\ldots,x_{16}+x_{18}\] are all distinct (it's obvious), between 1 and 99 and there're 33 of them. So we have 100 numbers between 1& 99. Thus we have solution of equation d-c=a+b.


In an isosceles triangle ABC we have AB=AC and the angle (BAC) is 20°. Let D be the point on the edge AB such that AD=DC, and let E be the point on the edge AC such that BC=CE. Find the angle (CDE).

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=179738]here[/url]  :wink:


For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6, 8^8,$ and $k$?

Oops, there was a typo.  It should have said $8^8$ instead of $8^6.$  It doesn't really make the problem any harder though (gives a few more solutions though) :D

[hide="Solution"]Clearly, $k$ can only have products of powers of primes $2$ and $3$ in its prime factorization. That is, $k=2^a\cdot3^b$, for some $a,b$.

$6^6=2^6\cdot3^6$
$8^8=2^{24}$
$12^{12}=2^{24}\cdot3^{12}$

Thus, $2^{24}\cdot3^{12}=LCM(2^6\cdot3^6, 2^{24}, 2^a\cdot3^b)=2^{max(6,24,a)}\cdot3^{max(6,0,b)}$.
Hence, $0\le a\le24$, but $b=12$, so there are $\boxed{025}$ possible values for $k$, as $a$ ranges from $0$ to $24$.[/hide]


Then,can you give solution of this problem? :)

From Cauchy we have $ \left(\sum\sqrt{\frac{a(b+c)}{a^{2}+bc}}\right)^{2}\leq (\sum\sqrt{a})\left(\sum\frac{\sqrt{a}(b+c)}{a^{2}+bc}\right)$.Now all we have to prove is $ \sum\frac{\sqrt{a}(b+c)}{a^{2}+bc}\leq \sum\frac{1}{\sqrt{a}}$ which is equivalent $ \sum \frac{(a-b)(a-c)}{\sqrt{a}(a^{2}+bc)}\geq 0$,which is Vornicu Schur.

[quote="mateivld"]From Cauchy we have $\left(\sum\sqrt{\frac{a(b+c)}{a^{2}+bc}}\right)^{2}\leq (\sum\sqrt{a})\left(\sum\frac{\sqrt{a}(b+c)}{a^{2}+bc}\right)$.Now all we have to prove is $\sum\frac{\sqrt{a}(b+c)}{a^{2}+bc}\leq \sum\frac{1}{\sqrt{a}}$ which is equivalent $\sum \frac{(a-b)(a-c)}{\sqrt{a}(a^{2}+bc)}\geq 0$,which is Vornicu Schur.[/quote]

nice  :lol:

[quote="mateivld"]Now all we have to prove is
$\sum\frac{\sqrt{a}(b+c)}{a^{2}+bc}\leq \sum\frac{1}{\sqrt{a}}$[/quote]
Unfortunately, this [b]does not[/b] hold !!!

...because you can't apply vornicu schur to it.
this inequality can be solved like the other inequality of ductrung.very nice! :)

Like which one? I though about a very ugly proof.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=78204]here[/url] you will get the inspiration. :)

From this idea we should square the inequality and then use that $\frac{ab(c+a)(c+b)}{(a^{2}+bc)(b^{2}+ca)}\leq 1$
for example .Then you will have to prove that
$\frac{(a+b)(a+c)}{a^{2}+bc}+\frac{(b+a)(b+c)}{b^{2}+ca}+\frac{(c+a)(c+b)}{c^{2}+ab}\leq \sum (\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{b}}{\sqrt{a}})$

[quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[\sqrt{\frac{a(b+c)}{a^{2}+bc}}+\sqrt{\frac{b(c+a)}{b^{2}+ca}}+\sqrt{\frac{c(a+b)}{c^{2}+ab}}\le \sqrt{ \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right) }\]
[/quote]
Very nice, Ductrung. Here is my solution for it
We have the inequality is equivalent to
\[\left(\sum\sqrt{\frac{a(b+c)}{a^{2}+bc}}\right)^{2}\le \left(\sum\sqrt{a}\right)\left(\sum\frac{1}{\sqrt{a}}\right) \]

\[\Leftrightarrow \sum\frac{a(b+c)}{a^{2}+bc}+2\sum\sqrt{\frac{ab(a+c)(b+c)}{(a^{2}+bc)(b^{2}+ca)}}\le \left(\sum\sqrt{a}\right)\left(\sum\frac{1}{\sqrt{a}}\right) \]
We can easily prove that
\[\sum\sqrt{\frac{ab(a+c)(b+c)}{(a^{2}+bc)(b^{2}+ca)}}\le 3 \]
So, it suffices to prove that
\[\Leftrightarrow \sum\frac{a(b+c)}{a^{2}+bc}+6 \le \left(\sum\sqrt{a}\right)\left(\sum\frac{1}{\sqrt{a}}\right) \]
To prove this ineq, we only need to prove that
\[\frac{a+b}{\sqrt{ab}}-\frac{c(a+b)}{c^{2}+ab}-1 \ge 0 \]
But this is trivial, because
\[\frac{a+b}{\sqrt{ab}}-\frac{c(a+b)}{c^{2}+ab}-1 =(a+b)\left(\frac{1}{\sqrt{ab}}-\frac{c}{c^{2}+ab}\right)-1 \ge 2\sqrt{ab}\left(\frac{1}{\sqrt{ab}}-\frac{c}{c^{2}+ab}\right)-1 = \frac{\left(c-\sqrt{ab}\right)^{2}}{c^{2}+ab}\ge 0 \]
We are done. :)

[quote=sqing][b]Potla inequality[/b]
For all $ a,b,c > 0$, show that
\[ \sqrt {\frac {a^2 + bc}{b(c + a)}} + \sqrt {\frac {b^2 + ca}{c(b + a)}} + \sqrt {\frac {c^2 + ab}{a(b + c)}}\geq 3\][/quote]
It's just AM-GM.

[quote=sqing][quote=ductrung]For any positive real numbers $a,b$ and $c$,
\[\sqrt{\frac{a(b+c)}{a^{2}+bc}}+\sqrt{\frac{b(c+a)}{b^{2}+ca}}+\sqrt{\frac{c(a+b)}{c^{2}+ab}}\le \sqrt{ \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right) }\][/quote]
For any positive real numbers $a,b$ and $c$,
$$ \sqrt{\frac{a(b+c)}{a^2+bc}}+\sqrt{\frac{b(c+a)}{b^2+ca}}+\sqrt{\frac{c(a+b)}{c^2+ab}}\le \sqrt{2(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})}$$
Crux,Problems 3507[/quote]

[img]http://s4.sinaimg.cn/middle/006ptkjAzy75Kc0TuRd23&690[/img]

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

\[\sqrt {{\frac {x \left( y+z \right) }{{x}^{2}+yz}}}+\sqrt {{\frac {y
 \left( z+x \right) }{{y}^{2}+zx}}}+\sqrt {{\frac {z \left( x+y
 \right) }{{z}^{2}+xy}}}\geq 2\,\sqrt {1+20\,\sqrt {2} \left( {\frac {xyz}
{ \left( y+z \right)  \left( z+x \right)  \left( x+y \right) }}
 \right) ^{3/2}}\]

[quote=zdyzhj][quote=luofangxiang][quote=ductrung]For any positive real numbers $a,b$ and $c$,
\[\sqrt{\frac{a(b+c)}{a^{2}+bc}}+\sqrt{\frac{b(c+a)}{b^{2}+ca}}+\sqrt{\frac{c(a+b)}{c^{2}+ab}}\le \sqrt{ \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right) }\][/quote]

[img]http://s2.sinaimg.cn/middle/006ptkjAzy75KgXRPQ571&690[/img][/quote]

这个不等式已经非常强了，不过，还可以进一步加强，两个不等式都可以包力展杀。
$\sum{\frac{2\sqrt{a\left( b+c \right)}}{a+\sqrt{bc}}}\le \sqrt{9+\left( \sum{\sqrt{a}} \right)\left( \sum{\frac{1}{\sqrt{a}}} \right)}$[/quote]

could explain in English?


function
algebra
functional equation
algebra unsolved

I found a solution to this functional equation, but was wondering whether I made any flaws?

Find all Functions f from the Real Numbers to the Real Numbers such that

$ f(x\minus{}f(y))\equal{}f(f(y))\plus{}xf(y)\plus{}f(x)\minus{}1$ for all x,y in the real numbers.

Solution

[hide]First, let $ x\equal{}0$, then we have $ f(\minus{}f(y))\equal{}f(f(y))\plus{}f(0)\minus{}1$.

Then set $ f(y)\equal{}g(y^{2})$, to get $ f(\minus{}g(y^{2}))\equal{}f(g(y^{2}))\plus{}g(0)\minus{}1$, which implies that $ g((g(y^{2}))^{2})\equal{}g((g(y^{2}))^{2})\plus{}g(0)\minus{}1$, so $ g(0)\equal{}1$, and then $ f(0)\equal{}1$.

Next, we observe that plugging $ y\equal{}0$, we get that
$ f(x\minus{}1)\equal{}f(1)\plus{}x\plus{}f(x)\minus{}1$.  Upon plugging $ x\equal{}1$, we get that $ f(1)\equal{}\frac{1}{2}$.

Furthermore, we see that $ f(x\minus{}1)\equal{}f(1)\plus{}x\plus{}f(x)\minus{}1$(*) implies that $ g(x^{2}\minus{}2x\plus{}1)\equal{}x\minus{}1/2\plus{}g(x^{2})$ (**).

However, $ \frac{g(x^{2}\minus{}2x\plus{}1)\minus{}g(x^{2})}{\minus{}2x\plus{}1}\equal{}\minus{}\frac{1}{2}$, for all values of x, except $ x\equal{}\frac{1}{2}$.  So, for right now, for all values of x, except for $ x\equal{}\frac{1}{2}$, $ g(x)\equal{}1\minus{}\frac{x}{2}$ (***), for we know that $ g(0)\equal{}1$.

Now, we set $ x\equal{}\frac{\sqrt{2}}{2}$ into (**).

Then we get that $ g(\frac{3}{2}\minus{}\sqrt{2})\equal{}\frac{\sqrt{2}}{2}\minus{}\frac{1}{2}\plus{}g(\frac{1}{2})$.

This implies that $ g(\frac{1}{2})\equal{}\frac{3}{4}$ from knowing (***) for all $ x \neq \frac{1}{2}$.

Likewise, we plug in $ x\equal{}\minus{}\frac{\sqrt{2}}{2}$ into (**), and we see that we get $ g(\frac{1}{2})\equal{}\frac{3}{4}$.

So, in either case, we get that $ g(x)\equal{}1\minus{}\frac{x}{2}$, for all $ x$.

Therefore, $ f(x)\equal{}1\minus{}\frac{x^{2}}{2}$ is the only function that works in this problem, which can be checked easily.[/hide]

[quote="mathgeniuse^ln(x)"]I found a solution to this functional equation, but was wondering whether I made any flaws?[/quote]

I think you made some.

[quote="mathgeniuse^ln(x)"]
...
Then set $ f(y) \equal{} g(y^{2})$,
...
[/quote]

You can't set $ f(y) \equal{} g(y^{2})$ except if you prove first that $ f(\minus{}x)\equal{}f(x)$

[quote="mathgeniuse^ln(x)"]
...
$ \frac {g(x^{2} \minus{} 2x \plus{} 1) \minus{} g(x^{2})}{ \minus{} 2x \plus{} 1} \equal{} \minus{} \frac {1}{2}$, for all values of x, except $ x \equal{} \frac {1}{2}$.  So, for right now, for all values of x, except for $ x \equal{} \frac {1}{2}$, $ g(x) \equal{} 1 \minus{} \frac {x}{2}$ (***)[/quote]

How could you conclude from $ \frac {g(x^{2} \minus{} 2x \plus{} 1) \minus{} g(x^{2})}{ \minus{} 2x \plus{} 1} \equal{} \minus{} \frac {1}{2}$ that $ g(x) \equal{} 1 \minus{} \frac {x}{2}$.
It can't be immediate.

[quote="mathgeniuse^ln(x)"]I found a solution to this functional equation, but was wondering whether I made any flaws?

Find all Functions f from the Real Numbers to the Real Numbers such that

$ f(x \minus{} f(y)) \equal{} f(f(y)) \plus{} xf(y) \plus{} f(x) \minus{} 1$ for all x,y in the real numbers.

[/quote]

It's a problem I searched for a long time on another forum without finding general solution. I have a solution with one constraint more :

"The set of discontinuity points of $ f(x)$ is included in an interval $ (a,b)$" (for example, $ f(x)$ has a finite set of discontinuity points).

With this condition, the only solution is $ f(x)\equal{}1\minus{}\frac{x^2}{2}$. Here is the solution I gave on this other forum :

Let $ P(x,y)$ the property $ f(x\minus{}f(y)) \equal{} f(f(y)) \plus{} xf(y) \plus{} f(x) \minus{} 1$.

$ 1)$ $ f(x)$ is not bounded (it does not exist $ u,v$ such that $ u<f(x)<v$ $ \forall x$)
Obviously $ \exists y_0$ such that $ f(y_0)\neq 0$.
Then, in $ P(x,y_0)$, let $ x\rightarrow\plus{}\infty$ : RHS is not bounded, and so LHS can't be bounded. Q.E.D.

$ 2)$ $ f(x) \equal{} \frac{f(0)\plus{}1}{2} \minus{} \frac{x^2}{2}$ $ \forall x\in Im(f)$ (I name $ Im(f)$ the set $ \{f(x)$, $ x\in\mathbb{R}\}$).
$ P(f(x),x)$ implies $ f(0) \equal{} 2f(f(x)) \plus{} f(x)^2 \minus{} 1$ and so
$ f(f(x))\equal{} \frac{f(0)\plus{}1}{2} \minus{} \frac{f(x)^2}{2}$
And hence the result.

$ 3)$ $ f(x) \equal{} 1 \minus{} \frac{x^2}{2}$ For all $ x$ such that $ \frac{x}{2}\in Im(f)$
$ P(2f(x),x)$ implies $ f(f(x)) \equal{} f(f(x)) \plus{} 2f(x)^2 \plus{} f(2f(x)) \minus{} 1$ and so
$ f(2f(x)) \equal{} 1 \minus{} \frac{(2f(x))^2}{2}$
And hence the result.

$ 4)$ $ f(0)\equal{}1$
Since $ f(x)$ is not bounded (see $ 1$ above), and since the set of discontinuity points of $ f(x)$ is included in an interval $ (a,b)$, it is possible to find two values $ u$ and $ v$ such that $ f(v)\equal{}2f(u)$
Then, point 2 above gives $ f(f(v)) \equal{} \frac{f(0)\plus{}1}{2} \minus{} \frac{f(v)^2}{2}$
And point 3 above gives $ f(f(v)) \equal{} 1 \minus{} \frac{f(v)^2}{2}$
And so $ \frac{f(0)\plus{}1}{2}\equal{} 1$
And hence the result.

$ 5)$ $ f(x) \equal{} 1 \minus{} \frac{x^2}{2}$ $ \forall x\in\mathbb{R}$
Let $ g(x) \equal{} f(x) \minus{} (1 \minus{} \frac{x^2}{2})$. $ P(x,y)$ becomes : $ g(x\minus{}f(y)) \equal{} g(f(y))\plus{} g(x)$
Let then $ x\equal{}f(y)$, we have $ g(f(y))\equal{}\frac{g(0)}{2} \equal{} 0$. And so
$ g(x \minus{} y) \equal{} g(x)$ $ \forall x\in\mathbb{R}$, $ \forall y\in Im(f)$

Since we know that it exists an intervall $ [\minus{}\infty,x_1)$ or $ (x_2,\plus{}\infty)$ included in $ Im(f)$, we can conclude that $ g(x)\equal{}$contant$ \equal{}0$.
And hence the result.

========
It remains to study the case of functions having infinitly many discontinuity points whose set is unbounded.

So, to be continued ...

1) $ f(0) \equal{} 1$ and $ f(f(y)) \equal{} 1 \minus{} \frac {f(y)^2}2$
Let $ a \equal{} f(0)$.
$ P(0,y): f(f(y)) \equal{} \frac {1 \plus{} a \minus{} f(y)^2}2$
We know $ f$ cannot be constant function. So there exists $ x \equal{} \frac1{f(y)}$ for some $ y$.
Then we have $ f(\frac1{f(y)} \minus{} f(y)) \equal{} f(f(y)) \plus{} f(\frac1{f(y)})$.
Now let $ u \equal{} f(\frac1{f(y)} \minus{} f(y))$ and $ v \equal{} f(y)$. We see $ u$ and $ u \minus{} f(v)$ are all in $ Im(f)$.
$ P(u,v): \frac {1 \plus{} a \minus{} (u \minus{} f(v))^2}2 \equal{} \frac {1 \plus{} a \minus{} f(v)^2}2 \plus{} uf(v) \plus{} \frac {1 \plus{} a \minus{} u^2}2 \minus{} 1\Rightarrow a \equal{} 1$

2) $ f(x) \equal{} 1 \minus{} \frac {x^2}2$
Let $ g(x) \equal{} f(x) \plus{} \frac {x^2}2$, the original equation can be rewritten as
$ g(x \plus{} \frac {y^2}2 \minus{} g(y)) \equal{} g(x)$.
Hence $ g$ is periodic function with period $ \frac {y^2}2 \minus{} g(y)$ for any $ y$
$ g(0) \equal{} 1\Rightarrow g(x \minus{} 1) \equal{} g(x)$
Let $ y$ be arbitrary, we see $ \frac {(y \minus{} 1)^2}2 \minus{} g(y \minus{} 1)$ is also a period. Since the difference of two periods is still period,
$ [\frac {y^2}2 \minus{} g(y)] \minus{} [\frac {(y \minus{} 1)^2}2 \minus{} g(y \minus{} 1)] \equal{} y \minus{} \frac12$ is another period.
Since $ y\minus{}\frac12$ can be any real number, $ g(y) \equal{} g(0) \equal{} 1$ for any $ y$.

Hence $ f(x) \equal{} g(x) \minus{} \frac {x^2}2 \equal{} 1 \minus{} \frac {x^2}2$.

When you rewrite the equation in terms of $ g$, it is incorrect. You dropped the $ g(f(x))$ term and the constant. It should be $ g(x\minus{}f(y))\equal{}g(f(y))\plus{}g(x)\minus{}1$ (then you can replace $ f(y)$ again, but it is a fundamentally different equation).


How can you add up all the numbers between 1 and 144, including 1 and 144, without just having to, like, add them all up? There must be an easier way!! :(  :(  :D

The formula is n(n+1)/2. Therefore, the answer is 144*145/2. You can take it from there.

I forgot to mention why this formula works. Since the numbers are in an arithmatic sequence, you just need to multiply the mean (which by the way is also the median) by the number of terms. For example, add 1+2+3. If you add them, you get six. If you formula, n(n+1)/2, then you get the same thing. To prove this formula, try this with five numbers, adding the greates and least, and dividing by two. If you notice, you just have to displace quantities to the terms on the opposite side. If that didn't make sense to you, then just know that:

For any arithmatic sequence, the sum is
n(a+z)/2
Where n is the number of terms (a-z+1), z is the lowest term, and a is the greatest term.
The sum of the first n numbers is:
n(n+1)/2
The sum of the first n odd numbers is:
n^2
The sum of the first n even numbers is
n(n+1)
Bonus: The sum of the first n squares is:
n(n+1)(2n+1)/6

Knowing all but the last one will save you eons of time... and the last one can't hurt...

You may not have seen these, but under the left side of the site, under the main navigation, there is a Flash movie of various formula's proofs. One of them is that proof.

Billy

Don't worry, I've seen those (many times)

[hide]144*145/2
72*145
>>10440<<[/hide]

Oh, and a formula you forgot:

The sum of the first n cubes:
[nth triangular number]^2
-in other words-
{[n(n+1)]/2}^2

Of course, it's easiest just to know the formula and it's proper applications rather than adding out... but similar to what basooner said, it's good to know that its derivation is from averaging and then multiplying.

[quote="mathlover"]well... i learned it as the first plus the last times how many there are divided by 2. because the first number plus the last number is 145 and as u go in, the sum stays the same unless the number of numbers is odd. u divide by two cause u cut the amount the number of number u have in half. i hope i explained well. im usually bad at that. which is y i hate masters round. how do u people do that?[/quote]
...which can be written as $\frac{n(a+z)}{2}$, where n is the number of terms ($n=a-z+1$), a is the greatest term, and z is the least term


Find a closed form for $1+22+333+....+n(11\ldots1)$ where in the last term 1 is written $n$ times.

[hide="Answer"]$S_{n}=\frac{2(9n-1)10^{n+1}-81n(n+1)+20}{1458}$[/hide]

(Will post the solution if no one else does.)

[quote="amcavoy"]I put it into series form, but I'm not sure how to evaluate the double series:

$1+22+333+\cdots+n\left(11\dots 11\right)=\sum_{k=0}^{n}\sum_{i=0}^{k}\left(k+1\right)10^{i}$[/quote]

Evaluate the series on the inside first, of course!  Didn't you learn the order of operations in elementary school?   :P

For some reason I was thinking that I couldn't.  A similar problem came up with a double sum where I had to use the identity \[\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{k,n-k}=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_{i,k}\] to evaluate the double series and I guess I just assumed I had to for this as well.  Anyways, thanks for the comment.


Let $a_1,a_2,....,a_n\in Z$ be different to each other.Prove that the polynomial
$f(x)=(x-a_1)(x-a_2)....(x-a_n)-1$ doesn't write as a product of polynomials $f(x)$ and $g(x)$ with integer coefficients and with degree $\geq 1$

[hide]By contradiction and use the theorem which says that every polynomial of $n$ degree which has $n+1$ roots it is the polynomial of $0$ degree.[/hide]
Could anyone now finish it?

Let's suppose $f(x) = g(x)h(x)$ with $g, h \in Z[X]$ with d°g and d°h > 0

$g(a_i)h(a_i) = -1$ so $g(a_i) + h(a_i) = 0$
But f and g have degree at most n-1 so we have $g+h = 0$
$f(x) = -g(x)^2$ is not possible since in f coefficient of $x^n$ is > 0.

What about $f(x)=(x-a_1)(x-a_2)....(x-a_n) + 1$ ?
Is there a general result ? x(x-1)+1 is irreducible but not x(x-2)+1  :huh:

[quote="tµtµ"]
But f and g have degree at most n-1 so we have $g+h = 0$
$f(x) = -g(x)^2$ is not possible since in f coefficient of $x^n$ is > 0.
[/quote]
How do you find this? Could you explain it with all the details?

Hi,

$g+h$ has degree at most n-1 (d°g, dh $\leq$ n-1, by hypothesis)  and is null at n different places, hence it's null.

f(x) can be written $f(x) = x^n + ...$ so coefficient of $x^n$ is 1 but the main coefficient of $-g(x)^2$ can't  be positive.

Perhaps I'm wrong it seems to make sense.

yes but my questions are:
Why $g+h=0$?
And the second why $f(x)=-g(x)^2$?

Silouan, did you READ the above post? :huh:

I don't understand. Don't you ever try to figure something out yourself?

Anyway...

Suppose that $f(x) = g(x)h(x)$ where $g(x),h(x) \in \mathbb{Z}[x]$ and $\deg{g},\deg{h} \geq 1.$ Then, $g(x)h(x) = -1$ for $x = a_1,a_2,\dots,a_n$. Hence for all of these $a_i$ we have $g(a_i) = 1$ and $h(a_i) = -1$ (or the other way around). Anyway, we get $g(a_i) + h(a_i) = 0$ for all $a_i$. Then $g(x) + h(x)$ is a polynomial in $x$ and has $a_1$, $a_2$, ..., $a_n$ as zeroes. So, that's $n$ zeroes. The degree of $g(x) + h(x)$ is at most $n - 1$, since $f(x) = g(x)h(x)$ is a polynomial of degree $n$. Hence, $g(x) + h(x)$ must vanish everywhere (which means that $g(x) + h(x) \equiv 0$ for all $x$). So, $g(x) = -h(x)$ for all $x$. Since we have $f(x) = g(x)h(x)$ by assumption, we get $f(x) = -g(x)^2$, contradiction (because the leading coefficient of $f$ is 1).

Yes but I cannot understand. Could you explain it to me Arne with details. Please, Please. (My English are awful and i cannot understand what is <<null>>,etc.)Please help

I typed a detailed explanation in my above post. Read it carefully.
Please use a dictionary if you don't understand.

[quote="tµtµ"]
What about $f(x)=(x-a_1)(x-a_2)....(x-a_n) + 1$ ?
Is there a general result ? x(x-1)+1 is irreducible but not x(x-2)+1  :huh:[/quote]

I think that this is noy always true but this it is:
$f(x)=[(x-a_1)(x-a_2)....(x-a_n)]^2+1$ with the conditions from my first post. Try to solve it

Silouan, I'm just wondering now - you named this problem "easy" and you posted it in the proposed problems section, but at first you didn't understand the solution. Does that mean that you posted it in the wrong section? Or did you find a different solution? If you found a different one, I'd like to see it.

I had a different one. And ok if I have time i will post it. Ok? But now try to solve the new problem.