It is problem 6 on page 318 of Algebraic inequalities of Vasile Cirtoaje.

Really? What is the solution proposed in this book? I prove it by SOS.

[quote="GacKiem"]Let $ a,b,c$ be non-negative real numbers. Prove that
\[ \frac {2a^{2} \minus{} bc}{b^{2} \minus{} bc \plus{} c^{2}} \plus{} \frac {2b^{2} \minus{} ca}{c^{2} \minus{} ca \plus{} a^{2}} \plus{} \frac {2c^{2} \minus{} ab}{a^{2} \minus{} ab \plus{} b^{2}}\ge 3.
\]
[/quote]
$ \frac {2a^{2} \minus{} bc}{b^{2} \minus{} bc \plus{} c^{2}} \plus{} \frac {2b^{2} \minus{} ca}{c^{2} \minus{} ca \plus{} a^{2}} \plus{} \frac {2c^{2} \minus{} ab}{a^{2} \minus{} ab \plus{} b^{2}}\ge 3$
<=>$ \sum \frac{2a^2\plus{}(b\minus{}c)^2}{b^2\minus{}bc\plus{}c^2} \geq 6$
By Cauchy -Sachwarz we have
$ \sum \frac{2a^2\plus{}(b\minus{}c)^2}{b^2\minus{}bc\plus{}c^2} \geq \frac{4(2\sum a^2\minus{}\sum bc)^2}{\sum (b^2\minus{}bc\plus{}c^2)(2a^2\plus{}(b\minus{}c)^2)}$
We need prove that $ 2(2\sum a^2\minus{}\sum bc)^2 \geq 3\sum(b^2\minus{}bc\plus{}c^2)(2a^2\plus{}(b\minus{}c)^2)$
or $ 2\sum a^4\plus{}2abc\sum a\plus{}\sum bc(b^2\plus{}c^2) \geq 6 \sum b^2c^2$
Notice that $ \sum a^4\plus{}abc\sum a \geq \sum bc(b^2\plus{}c^2)$
and $ \sum bc(b^2\plus{}c^2) \geq 2\sum b^2c^2$
Done!

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25551


I posted this earlier, but could not find the post anywhere in the forum... anyways, here it is again.

Can you form the number $2006$ using only zeros and simple arithmetic symbols? I could get it using $14$ zeros - can anyone better that?

[quote="julien_santini"]What is a simple arithmetic symbol, and what does "to form" really mean ?[/quote]

simple arithmetic symbol: $+$, $-$, $\times$, $\%$, $/$, may be $!$ also

to form: arrive at a figure of 2006 using above operators and only $0$s.

Had the question be arrive at $9$ using only $3$, one approach is: $9 = 3+3$... just to give an example.

No tricks like placing two $0$s and making a sign of $8$.

LATE EDIT: You can use any mathematical operator, but the simpler the better.

[quote="julien_santini"]Best I could do is with $15$ zeroes ...[/quote]

You can use any operator, but the simpler the better. and do you mind telling your solution?

So you can't do $\cos 0$ or anything like that? What about $\sqrt{\text{ }}$ and $\lfloor \text{ }\rfloor$?

[quote="arsemesomething"]i don't get it. why don't we just add 0! 2006 times?[/quote]

because that would require 2006 0s?

I did a similar way to pontios: i realized that 2006 = 2000 + 6 and worked from there, there might be a nicer way though.

but the question doesn't say anything about the number of zeros one cannot use. what are the conditions?

The point is to use the minimum number of 0s :)


Hi

I have a question: Do you know how to type Math signs (equations, root signs, ...) in a forum? Do I have to install Latex or anything like that? I mean I am the administrator, so now what should I do in order that everyone can type Math signs and stuff in the forum? This is a free forum, you can have a look at it here:
[url=http://p4.forumforfree.com/mathpassion.html]p4.forumforfree.com/mathpassion.html[/url]

Thanks in advance.

Download [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url] which has full instructions for installation on a phpBB forum. You will need LaTeX, ImageMagick and GhostScript installed on the server, though mimeTeX is an alternative if that's not possible.


Let $ ABC$ be a triangle. Denote the midpoints $ E$ , $ F$ of the sides $ [AC]$ , $ [AB]$ respectively. For a point $ D\in (BC)$ define the intersections $ X\in DE\cap AB$ , $ Y\in DF\cap AC$ , $ T\in XY\cap BC$ . Prove that $ TA\equal{}TD$ iff the ray $ [AD$ is the bisector of the angle $ \widehat {BAC}$ and in this case the line $ TA$ is tangent to the circumcircle of the triangle $ ABC$ .

[quote="Virgil Nicula"]Let $ ABC$ be a triangle. Denote the midpoints $ E$ , $ F$ of the sides $ [AC]$ , $ [AB]$ respectively.  $ AD$ is the bisector of the angle $ \widehat {BAC}$,define the intersections $ Y\in DE\cap AB$ , $ X\in DF\cap AC$. $ T\in BC$ such that TA=TD. Prove that X,T,Y are collinear.[/quote]
$ TA \equal{} TD$ if and only if the line $ TA$ is tangent to the circumcircle of the triangle $ ABC$. When this case,
we have $ \triangle TAB\sim\triangle TCA$ $ \Rightarrow\frac {TB}{TA} \equal{} \frac {TA}{TC} \equal{} \frac {AB}{AC}\Rightarrow\frac {TB}{TC} \equal{} (\frac {AB}{AC})^2$
By Menelaus theorem,
*)$ \triangle ABC$ with $ (XFD)$:
$ \frac {XC}{XA}.\frac {FA}{FB}.\frac {DB}{DC} \equal{} 1\Rightarrow\frac {XC}{XA} \equal{} \frac {DC}{DB} \equal{} \frac {AC}{AB}$
*)$ \triangle ABC$ with $ (YDE)$:
$ \frac {YA}{YB}.\frac {DB}{DC}.\frac {EC}{EA} \equal{} 1\Rightarrow\frac {YA}{YB} \equal{} \frac {DC}{DB} \equal{} \frac {AC}{AB}$
$ \Rightarrow\frac {TB}{TC}.\frac {XC}{XA}.\frac {YA}{YB} \equal{} 1$
Hence $ X,T,Y$ are collinear.


http://www.surveymonkey.com/s.aspx?sm=21rYpe_2faaeLZZoriFFuT9A_3d_3d

I need to poll people on these questions for class. We're collecting data on these questions, related to a book we're reading. If you can take it, thanks.

I did it!

 The message is too small. Please make the message longer before submitting.


please could someone help me find something about this problem, solutions or links are welcome

n|5^n-3^n

Could you post the complete statement of the problem?  :wink:

Obviosly $ n=1,2$ are solutions.
If n is solution, and $ m|\frac{5^n-3^n}{n}$, then $ nm$ is solution too.
$ n_0=1\to n_1=2\to n_{2,1}=4,n_{2,2}=8,n_{2,3=16}$.
$ n_{2,1,1}=n_{2,2}=8,n_{2,1,2}=68,...$
All solutions find in these wey.


For all of you that spend most of your time mathing  :) ...
  I want to become ready for the next IMO , so I was wondering, from what I should start ..Inequalities ,Probability or ... something like that .
  Anyone who want to help me : please post the List

P.S- Moderators : I'm not sure if this is  the  correct place to post this .  Move It if It is'nt .

I think its a good idea to start with Geometry!I may be wrong though!


geometry
circumcircle
incenter
geometric transformation
reflection
geometry proposed

Dear Mathlinkers,
let ABC be a triangle, O the centre of the circumcircle of ABC, (1) the incircle of ABC, I the incenter of (1), DEF the contact triangle of ABC, D' the foot of the D-altitude of DEF, A' the point of intersection of OI with BC and A" the midpoint of DA'.
Prove thet A"D' is parallel to AA'.
Sincerely
Jean-Louis

dear [b]jayme [/b]

from this problem [u]http://www.mathlinks.ro/viewtopic.php?t=291602[/u]
if we denote D1 is the reflection of D w.r.t EF then D1,A,A' are collinear
and thus A"D'  // A'D1(A'A)

Dera Plane Geometry,
thanks for your answer.
I know your link and I think that perhaps with my approach, it is possible to avoid the calculus...
I continue my research.
Sincerely
Jean-Louis


This year, will there be a math jam for each round like last year?

No -- because we're posting "official" solutions very quickly after the deadline date, we're not doing the Math Jams anymore.


integration
trigonometry
calculus
geometric series
calculus computations

How does one show that
\[\pi = \sqrt{12}\left(1-{1\over 3\cdot3}+{1\over5\cdot 3^{2}}-{1\over7\cdot 3^{3}}+\cdots\right) \]

That's $\arctan\left(\sqrt{\frac13}\right)$ inside the parentheses.

To be technical, what is inside the parentheses is $\sqrt{3}\arctan\left(\frac1{\sqrt{3}}\right).$

jmerry and Kent answered question perfectly.
Im just going to explain this formula in case the original poster doesnt know.

We know that,

$\frac{1}{1-y}= 1+y+y^{2}+y^{3}+...$ for $|x|<1$

So if $y=-x^{2}$ and $|x|<1$ then  $|y|<1$ so:

$\frac{1}{1+x^{2}}= 1-x^{2}+x^{4}-x^{6}+...$ for $|x|<1$

(If you studied some analysis this series is "uniformly convergent" because the interval is open).

So, we can integrate term by term,

$\int_{0}^{t}\frac{1}{1+x^{2}}dt = \int_{0}^{t}(1-x^{2}+x^{4}-... ) dt$ for $|t|<1$

And so,

$\tan^{-1}t = t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-...$ for $|t|<1$

About halfway down the the posts in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=141608]this topic[/url] I refer to my calculus class handout entitled "Seven Power Series to Memorize." The arctangent series is definitely one of the seven. Sylow_theory has presented the standard justification - that is more or less how I would develop this series in class, along with the closely related derivation of the power series for $\ln(1+x).$

One technicality: the geometric series for $\frac1{1+x^{2}}$ is [b]not[/b] uniformly convergent on the whole interval $(-1,1).$ It is uniformly convergent on compact subsets of that interval - which is sufficient to justify the interchange of sum and integral. And since the artangent series converges at the endpoints of that interval, it is actually uniformly convergent on $[-1,1].$ But even to claim you know the value of the series for $x=1$ requires the use of Abel's Theorem.

And that discussion - uniform convergence, Abel's Theorem, and so forth - has no place in the Calculus II course that the list of series to memorize belongs to.

[quote="Kent Merryfield"] And since the artangent series converges at the endpoints of that interval, it is actually uniformly convergent on $[-1,1].$ But even to claim you know the value of the series for $x=1$ requires the use of Abel's Theorem.
[/quote]

Im sure you know this, but I once so somebody develope the arctangent series for the endpoint without referring to Abel's theorem. We show that the error between the arctangent series and the truncated geometric series goes to zero.


modular arithmetic
number theory proposed
number theory

Determine the number of positive integers $a$, so that there exist nonnegative integers $x_0,x_1,\ldots,x_{2001}$ which satisfy the equation
\[ \displaystyle a^{x_0} = \sum_{i=1}^{2001} a^{x_i} \]

[quote="jin"]
And when $a-1|2000$,we can rewrite $a^n=a\times a^{n-1}$by $\frac{2000}{a-1}$ times.And we get a equation.
[/quote]

Hm, I'm somehow a bit dumb to understand this part of your solution :?

I do it so:

if $(a-1)\mid 2000$ then we have
\begin{eqnarray*} a^{\frac{2000}{a-1}} &=& (a-1) \frac{ a^\frac{2000}{a-1}-1}{a-1}+1\\ &=& (a-1)( 1+a+\ldots + a^{\frac{2000}{a-1}-1})+1\\ &=& 1+\ldots+1 +\\ &\ & a+\ldots+a +\\ &\ & \ldots\\ &\ & a^{\frac{2000}{a-1}-1} + \ldots+a^{\frac{2000}{a-1}-1} + \\ &\ & 1 \end{eqnarray*}

which obviously consists of 2001 summands

I mean you can choose $x_0$ enough and First rewrite $a^{x_0}=a_{x_0-1}+a_{x_0-1}+\cdots+a_{x_0-1}$
Then each time choose one $a^n$ in the right side and rewrite it like $a^{n-1}+a^{n-1}+\cdots+a^{n-1}$
So each time the number of the $a^i$ in the right side increase $a-1$.First it's 1 so by $\frac{2000}{a-1}$ times it's 2001. :P

[quote="Redriver"]why we have $a-1|2000$[/quote] As jin wrote, just consider $ a^{x_{0}}= \sum_{i=1}^{2001}a^{x_{i}}$ modulo $a-1$ :)

[quote=PaperMath]Wait how do we know that all solutions that satisfy $a-1|2000$ work?[/quote]

Because then we can represent $a^{x_0}$ like this: $a^{x_0}=a^{x_0-1}(a-1)+a^{x_0-2}(a-1)+...+a^{x_0-b}(a-1)+a^{x_0-b}$

When $a-1|2000,$ for some proper integer $b,$ the $\textrm{RHS}$ would have exactly $2001$ terms.


inequalities
calculus
derivative
inequalities proposed

[quote="xtar"]Let $x,y,z$ non-negative, such that $x+y+z=1$. Show that :
$\frac{1}{\sqrt{x^2+xy+y^2}}+\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}\geq 4+\frac{2\sqrt{3}}{3}$
.[/quote]

Looks hard....! :?
Can anyone beat it?? :D  :)

Actually, I think the following more general statement is valid:

Let $x,y,z$ be non-negative numbers, such that $x+y+z=3$.
(a) If $-2<p \leq \frac{17}{16}$, then
$\frac{1}{\sqrt{x^2+pxy+y^2}}+\frac{1}{\sqrt{y^2+pyz+z^2}}+\frac{1}{\sqrt{z^2+pzx+x^2}}\geq \frac{4}{3}+\frac{2}{3\sqrt{2+p}}$;
(b) If $p \geq \frac{17}{16}$, then
$\frac{1}{\sqrt{x^2+pxy+y^2}}+\frac{1}{\sqrt{y^2+pyz+z^2}}+\frac{1}{\sqrt{z^2+pzx+x^2}}\geq \frac{3}{\sqrt{2+p}}$.

I think that the thechnique of solving such problems is like this
http://www.mathlinks.ro/Forum/viewtopic.php?t=13897

I am wrong ?

I don't think this technique helps us to solve these very hard inequalities.
Maybe EV-Theorem is a possible solution.  :(
Or maybe Xtar has a solution.

[quote="behzad"]Noone can prove this new problem?It looks hard... :([/quote]
NO,it is not new.I have seen the one which xtar post before,but I haven't seen the solution.

Can we use this result?
  Find the minimum of the expression:    $\sqrt{x^2+xy+y^2}.\sqrt{y^2+yz+z^2}+\sqrt{y^2+yz+z^2}\sqrt{z^2+zx+x^2}+\sqrt{z^2+zx+x^2}\sqrt{x^2+xy+y^2}.$

[quote="behzad"]I have proved the left term is minimum when $z=0$.so we get :
[b][color=blue]PROBLEM[/color].[/b]let $x,y$ be positive numbers such that $x+y=1$ prove that : $\frac{1}{\sqrt{x^2+xy+y^2}}+\frac{1}{x}+\frac{1}{y}\ge 4+\frac{2\sqrt3}{3}$[/quote]

The proof follows from the following inequality:
Denote $f(x,y)=\frac{1}{\sqrt{x^2+xy+y^2}}+\frac{1}{x}+\frac{1}{y}$
Now we will prove that $f(x,y)-f(\sqrt{xy},\sqrt{xy})\ge 0$ for all $x,y>0$
Proof:
$f(x,y)-f(\sqrt{xy},\sqrt{xy})=\left(\frac{1}{x}+\frac{1}{y}-\frac{2}{\sqrt{xy}}\right)-\left(\frac{1}{\sqrt{x^2+xy+y^2}}-\frac{2}{\sqrt{3xy}}\right)$

Now this can be rewritten as $\displaystyle\frac{(\sqrt{x}-\sqrt{y})^2}{xy}-\displaystyle\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})^2}{\sqrt{a}\sqrt{b}\cdot [x,y]}$ where $a=x^2+xy+y^2,b=\sqrt{3xy},[x,y]=\sqrt{a}+\sqrt{b}$
or

$f(x,y)-f(\sqrt{xy},\sqrt{xy})=\displaystyle\frac{(\sqrt{x}-\sqrt{y})^2}{xy}\left(1-\displaystyle\frac{\sqrt{xy}(\sqrt{x}+\sqrt{y})^2}{\sqrt{a}\sqrt{b}\cdot [x,y]}\right)$
Now using that $[x,y]\ge 2\sqrt{3xy}$ and $(\sqrt{x}+\sqrt{y})^2\leq 2x+2y$ we get

$f(x,y)-f(\sqrt{xy},\sqrt{xy})\ge \displaystyle\frac{(\sqrt{x}-\sqrt{y})^2}{xy}\left(1-\frac{x+y}{3\sqrt{a}}\right)\ge 0$   $\blacksquare$

Now to the proof of the original problem...
We have $f(x,y)\ge f(\sqrt{xy},\sqrt{xy})=\frac{\left(2+\frac{\sqrt{3}}{3}\right)}{\sqrt{xy}}\ge 4+\frac{2\sqrt{3}}{3}$  follows from  $1=x+y\ge 2\sqrt{xy}$

How did you prove that $z\longrightarrow 0$

[quote="leepakhin"][quote="hardsoul"]Now we will prove that $f(x,y)-f(\sqrt{xy},\sqrt{xy})\ge 0$ for all $x,y>0$[/quote]
But if you mix $(x,y)$ with $(\sqrt{xy},\sqrt{xy})$, the given condition $x+y=1$ is not preserved. :?[/quote]

You are right at this point,but I think we can mix $(x,y)$ with $(\frac{x+y}{2},\frac{x+y}{2})$ and if my calculations are correct then:

$f(x,y)-f\left(\frac{x+y}{2},\frac{x+y}{2}\right)=\displaystyle \frac{(x-y)^2}{2b}\left(\frac{2\sqrt{3}a(a+\sqrt{3}b)-xy}{2\sqrt{3}xya(\sqrt{3}b+a)}\right)\ge 0$ where $a=\sqrt{x^2+xy+y^2},b=\frac{x+y}{2}$

$\sqrt{xy}+\sqrt{xy}$ is not necessarily equal to $1$.

[quote="zhaobin"]hardsoul,
I think you proof with $f(x,y)-f(\sqrt{xy},\sqrt{xy})\ge 0$ is also right.
you proved for $x+y=1$
we have $f(x,y) \geq f(\sqrt{xy},\sqrt{xy})$,
then you proved $f(\sqrt{xy},\sqrt{xy}) \geq 4+\frac{2\sqrt{3}}{3}$
I don't know where you are wrong :)[/quote]

If I took $f(x,y)$  to be
$\frac{x+y}{\sqrt{x^2+xy+y^2}}+\frac{x+y}{x}+\frac{x+y}{y}\ge 4-\frac{2\sqrt3}{3}$
then showing $f(x,y) \geq f(\sqrt{xy},\sqrt{xy})$ would be correct...

[quote="zhaobin"][quote="behzad"][quote="hardsoul"]

How did you prove that $z\longrightarrow 0$[/quote]
Let (x,y,z) be a triple and $f(x,y,z)$= $\frac{1}{\sqrt{x^2+xy+y^2}}+\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}$. Let $(x,y,z)$ be triple such that $x,y,z\not=0$ and $x\gey\gez$.we have : $y^2+yz+z^2\le(y+z)^2$and $z^2+zx+x^2 \le(z+x)^2$and equality holds iff $z=0$.So
$\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}\ge \frac{1}{x+z}+\frac{1}{y+z}$so min of  $\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}$ is when $z=0$. Now note that $g(x)=x^2+xy+y^2$ is ascendant with $x,y$ .so min of $\frac{1}{\sqrt{x^2+xy+y^2}}$ is also when $z=0$. So the min of $f(x,y,z)$ is when $z=0$ , $x+y=1$.[/quote]
sorry,I don't think your "solution"is right...[/quote]
Really???
But I don't think so.
Write my mistake please ... ;)

I found a very useful lemma to solve this inequality. The following belongs to Vasc.

if $x,y,z>0$ then $\frac{1}{x^2+xy+y^2}+\frac{1}{y^2+zy+z^2}+\frac{1}{z^2+xz+x^2}\geq\frac{9}{(a+b+c)^2}=9$ in our case

[quote="behzad"][/quote][quote="zhaobin"][quote="behzad"][quote="hardsoul"]

How did you prove that $z\longrightarrow 0$[/quote]
Let (x,y,z) be a triple and $f(x,y,z)$= $\frac{1}{\sqrt{x^2+xy+y^2}}+\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}$. Let $(x,y,z)$ be triple such that $x,y,z\not=0$ and $x\gey\gez$.we have : $y^2+yz+z^2\le(y+z)^2$and $z^2+zx+x^2 \le(z+x)^2$and equality holds iff $z=0$.So
$\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}\ge \frac{1}{x+z}+\frac{1}{y+z}$so min of  $\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}$ is when $z=0$. Now note that $g(x)=x^2+xy+y^2$ is ascendant with $x,y$ .so min of $\frac{1}{\sqrt{x^2+xy+y^2}}$ is also when $z=0$. So the min of $f(x,y,z)$ is when $z=0$ , $x+y=1$.[/quote]
sorry,I don't think your "solution"is right...[/quote][quote="behzad"]
Really???
But I don't think so.
Write my mistake please ... ;)[/quote]
because when $z$ is changing , $x$ ,$y$ is changing,too.
so you can't say $\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}\ge \frac{1}{x+z}+\frac{1}{y+z}$...
maybe I can't explain well.but it is certainly wrong,can somebody help me to explain to behzad,thanks.

to silouan,can you detail you solution?

[quote="zhaobin"]maybe I made some mistakes.but I also think the following is right
if $x+y=1$   $x>0$ $y>0$
and $f(x,y)=\frac{1}{\sqrt{x^2+xy+y^2}}+\frac{1}{x}+\frac{1}{y}$
$f(x,y) \geq f(\sqrt{xy},\sqrt{xy})$

to leapakin:
why should $\sqrt{xy}+\sqrt{xy}$ must equal to 1?
I think it is not necessare. :)[/quote]
You are right... because in the first proof  I have showed that $f(x,y) \geq f(\sqrt{xy},\sqrt{xy})\ge f\left(\frac{x+y}{2},\frac{x+y}{2}\right)$ for all $x,y>0.$

[quote="zhaobin"]
because when $z$ is changing , $x$ ,$y$ is changing,too.
so you can't say $\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}\ge \frac{1}{x+z}+\frac{1}{y+z}$...
maybe I can't explain well.but it is certainly wrong,can somebody help me to explain to behzad,thanks.

[/quote]
Oh,excuse me "zhaobin",you are right  :blush: .

[quote="flavian"]I have a new idea for resolve this problem. But I can't finish the solution. Maybe u can help me. Let be a triangle ABC with the angles max 120. If M is Toriceli"s point we must prove this: If MA+MB+MC=1 (so for every point P PA+PB+PC>=1), we have 1/a+1/b+1/c>= than that number from right member of inequality. We have equality if the triangle has an angle = 120.    Can u resolve this?? :)[/quote]
I have thought about.but the result is the same to you :(

to xtar:
can you post a solution for your problem?

In my dream, God tell me that It can't be have a nice solution with elementary mathematical.  :lol:

OK
My solution with Lagrange multipliers
WLOG  $0\le x\le y\le z<1$
Case 1: $x=0$ we have to prove
$f(y,z)=\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{y}+\frac{1}{z}\ge 4+\frac{2\sqrt{3}}{3}$
solution [b]hardsoul[/b] page 1 post #12
$f(y,z)\ge f(\sqrt{yz},\sqrt{yz})\ge f(\frac{y+z}{2},\frac{y+z}{2})=4+\frac{2\sqrt{3}}{3}$
Case 2: $0< x\le y\le z<1$
$F(x,y,z,\lambda)=\sum \frac{1}{\sqrt{x^2+xy+y^2}}+\lambda (x+y+z-1)$
$\frac{\partial F}{\partial x}=0$ , $\frac{\partial F}{\partial y}=0$ , $\frac{\partial F}{\partial z}=0$ , then
$-\frac{1}{2}\left [\frac{2x+y}{\sqrt{(x^2+xy+y^2)^3}}+\frac{2x+z}{\sqrt{(z^2+zx+x^2)^3}}\right ]+\lambda =0$
$-\frac{1}{2}\left [\frac{2y+x}{\sqrt{(x^2+xy+y^2)^3}}+\frac{2y+z}{\sqrt{(y^2+yz+z^2)^3}}\right ]+\lambda =0$
$-\frac{1}{2}\left [\frac{2z+x}{\sqrt{(z^2+zx+x^2)^3}}+\frac{2z+y}{\sqrt{(y^2+yz+z^2)^3}}\right ]+\lambda =0$
Adding
$\lambda =\frac{1}{2}\left[ \frac{x+y}{\sqrt{(x^2+xy+y^2)^3}}+\frac{y+z}{\sqrt{(y^2+yz+z^2)^3}}+ \frac{z+x}{\sqrt{(z^2+zx+x^2)^3}}\right]$
Inserting $\lambda$ in first we get
$x\sum \frac{1}{\sqrt{(x^2+xy+y^2)^3}}=\frac{1}{\sqrt{(y^2+yz+z^2)^3}}$

Similarly
$y\sum \frac{1}{\sqrt{(x^2+xy+y^2)^3}}=\frac{1}{\sqrt{(z^2+zx+x^2)^3}}$

$z\sum \frac{1}{\sqrt{(x^2+xy+y^2)^3}}=\frac{1}{\sqrt{(x^2+xy+z^2)^3}}$
Hence
$y^2(z^2+zx+x^2)^3=z^2(x^2+xy+y^2)^3$
We put $y=ax,z=bx$, where $1\le a\le b$
$a^2(b^2+b+1)^3=b^2(a^2+a+1)^3$  we get $a=b$
Hence
$y=z$ as necessary for critical points in the interior of the region  $0<x,y,z<1$
We have to prove
$\frac{2}{\sqrt{x^2+xy+y^2}}+\frac{1}{y\sqrt{3}}\ge 4+\frac{2\sqrt{3}}{3}$ where $x+2y=1$
$g(y)= \frac{2}{\sqrt{3y^2-3y+1}}+\frac{1}{y\sqrt{3}}\ge 4+\frac{2\sqrt{3}}{3}$ where $\frac{1}{3}\le y\le \frac{1}{2}$ since $x\le y\le z$
By differentiation it is easily checked that the absolute minimum of $g(y)$ on $\left [1/3,1/2 \right]$ is $4+\frac{2\sqrt{3}}{3}=g(1/2)$

[quote="hungkhtn"]Oh... I'm right!  :lol:[/quote]
but it depends on how we define "nice" :lol: ...

Thank you for the solution xtar .Do you think about an elementary solution ?

[quote="xtar"]
Case 2: $0< x\le y\le z<1$
$F(x,y,z,\lambda)=\sum \frac{1}{\sqrt{x^2+xy+y^2}}+\lambda (x+y+z-1)$

What is $lamda$?

the number $l$ is the Langrange multiplier .

[quote="Vasc"]Actually, I think the following more general statement is valid:

Let $ x,y,z$ be non-negative numbers, such that $ x \plus{} y \plus{} z \equal{} 3$.
(a) If $ \minus{} 2 < p \leq \frac {17}{16}$, then
$ \frac {1}{\sqrt {x^2 \plus{} pxy \plus{} y^2}} \plus{} \frac {1}{\sqrt {y^2 \plus{} pyz \plus{} z^2}} \plus{} \frac {1}{\sqrt {z^2 \plus{} pzx \plus{} x^2}}\geq \frac {4}{3} \plus{} \frac {2}{3\sqrt {2 \plus{} p}}$;
[/quote]
This inequality is not always valid. Write itin the hormogeneous form
$ (x\plus{}y\plus{}z)\left( \frac {1}{\sqrt {x^2 \plus{} pxy \plus{} y^2}} \plus{} \frac {1}{\sqrt {y^2 \plus{} pyz \plus{} z^2}} \plus{} \frac {1}{\sqrt {z^2 \plus{} pzx \plus{} x^2}}\right) \ge 4\plus{}\frac{2}{\sqrt{p\plus{}2}}.$
and try $ x \equal{} 1,y \equal{} 0.7,z \equal{} 0,p\equal{} \minus{} \frac {5}{4}.$ ;)

[quote="xtar"]Let $x,y,z$ non-negative, such that $x+y+z=1$. Show that :
$\frac{1}{\sqrt{x^2+xy+y^2}}+\frac{1}{\sqrt{y^2+yz+z^2}}+\frac{1}{\sqrt{z^2+zx+x^2}}\geq 4+\frac{2\sqrt{3}}{3}$
.[/quote]
By Holder $\left(\sum_{cyc}\frac{1}{\sqrt{x^2+xy+y^2}}\right)^2\sum_{cyc}\left(x+y+(2\sqrt3-1)z\right)^3(x^2+xy+y^2)\geq\left(2\sqrt3+1\right)^3$.
Hence, it remains to prove that $3(2\sqrt3+1)(x+y+z)^5\geq4\sum_{cyc}\left(x+y+(2\sqrt3-1)z\right)^3(x^2+xy+y^2)$, which is fifth degree.
Hence, by $uvw$ it remains to prove the last inequality in two cases:
1. $y=1$, $z=0$, which gives $(x+1)((6\sqrt3-5)x^2+6(1+2\sqrt3)x+6\sqrt3-5)(x-1)^2\geq0$;
2. $y=z=1$, which gives $((6\sqrt3-5)x^4+2(11+6\sqrt3)x^3-4(42\sqrt3-67)x^2-24(41-22\sqrt3)x+96)x\geq0$, which is obvious.
Done!

[size=0]\(\textsc{Apologize for the necroposting.}\)[/size]
[quote=leonardg]What about a geometrical interpretation ?
Torricelli-Fermat .[/quote]

Nice approach. It is conceivable that [url=https://imgsa.baidu.com/forum/w%3D580/sign=4d1f10553c292df597c3ac1d8c305ce2/765dd409b3de9c8208a92cf56a81800a18d84386.jpg]１[/url] and [url=https://imgsa.baidu.com/forum/w%3D580/sign=e478826d64d0f703e6b295d438fb5148/c4a1432309f79052f64993440af3d7ca7bcbd578.jpg]２[/url] (both in [url=https://imgsa.baidu.com/forum/w%3D580/sign=3ee3c000a34bd11304cdb73a6aaea488/ee50b619ebc4b745b5df9365c1fc1e178a82157b.jpg]Chinese[/url]) may be useful.
The first two images are from “[color=#ADADAD]izzystar[/color]”, and the last one is from “[color=#B8E0F9]rosebaby时代[/color]”.
[hide=Note.]Another solution can be seen in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2132618]this post[/url].[/hide]


theres some trick i cant think of off the top of my head. its used for solving problems like this. try google

I couldn't find help to this in google. Please help me

oh. did you try AoPS wiki? sorry im 2 lazy to try searching myself.

$ a \equal{} b \equal{} c \equal{} 0,\pm1,\pm2$ are trivial solutions. I'm not sure whether there are more or not.

I know this solution, but I don't know how show, that it's all solutions

This is a problem from an ongoing math contest  :!:  :!:  :!:  (this year's Polish Mathematical Olympiad)


function
number theory
algebra
difference of squares
special factorizations

I am working through the number theory section of ACoPS and came to this problem.

Consider the set $F: =\{a+b\sqrt{-6},a,b\in%Error. "matbbl" is a bad command.
{Z}\}$. Define the norm, function $N$ by $N(a+b\sqrt{-6})=a^{2}+6b^{2}$.

I have already shown that for $\alpha,\beta\in%Error. "matbbl" is a bad command.
{Z}$ we have $N(\alpha\beta)=N(\alpha)N(\beta)$.
what I am having trouble with is this.

Show that 31 and 7 are not F-primes (an F-prime is an element in F which has no factors in F other than 1 and itself)

[hide="my ideas"]
well i know that $31=5^{2}+6(1)^{2}$. Then from a previous part of the problem i have the identity
$(ac-6bd)^{2}+6(ad+bc)^{2}=(a^{2}+6b^{2})(c^{2}+6d^{2})$ so i know that since 31 can be written in the form $a^{2}+6b^{2}$ it can also be written in the from $(a^{2}+6b^{2})(c^{2}+6d^{2})$ and therefore cannot be an F-prime. what i am having trouble with is finding the values of $a,b,c,d$
I know $5=ac-6bd$ and $1=ad+bc$ how would you solve this?[/hide]

[hide="hint"]$31 = 5^{2}-(1\sqrt{-6})^{2}$[/hide]
[hide="answer"]Use difference of squares factorization: 31 = $(5+1\sqrt{-6})(5-1\sqrt{-6})$[/hide]


So my math teacher had a problem on a test dealing with increase.
Basically it was a word problem and wanted me to increase something by 298%.
She thinks multiplying by 2.98 would result in the correct answer.

I've tried a lot of ways to explain to her that she should be multiplying by 3.98 instead.

I really can't find a good way to explain it. I mean an extremely simple way.

Your math teacher is saying that?

Was the problem asking for the increase amount or the final amount? What did other students say?

Maybe you could show her something off the internet. This would be my explanation:

Since we have an increase of 298 percent, the amount goes up by 298% of the original, or 100+298=398%.

If the question is asking for how much the increase is, then the math teacher is correct. If its the resulting price, you are right. To get the resulting price, you need to add the increase to the original, thus it will be 298%+100%=398% (multiplying by 3.98)


Is this number theory ?

anyway it is equivalent to prove that

$\sum_{k=1}^n\binom{k+p-1}{p}=\binom{n+p}{p+1}$

Combinatorial proof:

RHS is equivalent to choosing $p+1$ object from $n+p$ objects

LHS is assume you choose 1st object then it remains $\binom{n+p-1}{p}$ ways . If choose 2nd object then left $\binom{n+p-2}{p}$ (because 1st isnt included) . And so on you will get LHS .

And you found that LHS is actually = RHS .

This is a "wanna be" number theory exercice. Your combinatorial solution is nice. ;)

What does the $\bullet$ operator thing signify? It's a piece of notation I've never encountered before (unless it's just a really fat multiplication sign?)


This is a super weird question from my book of mensa quiz. I really don't get the solution they give. Can someone give a better solution?

A horse salesman went to a horse auction with a certain number of horses.  To his first customer, he sold half his horses plus half a horse.  To his second, he sold half of what he had left, plus half a horse.  To the last, he sold half of what he had left, plus half a horse.  Each of his customers received whole horses, however, not half horses.  How many horses did he start out with?  All help is needed!

Sorry for spamming, but this is the SAME EXACT question from my school that I have to do this week. My Mathcounts teacher probably will not like it if I saw this answer. Wow... Sorry again for spamming but I'm just amazed by the coincidence.    :lol:

name the people A, B, and C.

C received 1/2 of what was left, and 1/2 more. basically, the seller must have had one horse left, so that he gave up 1/2*1=1/2, and 1/2 more, leaving him with nothing.

the guy before, person B, had (1+.5)*2, or 3 horses from the seller.

the guy before had (3+1)*2=8 horses.

therefore, the seller had 8 horses. i don tknow if my reasoning makes much sense, but it makes sense in my head, and its the right answer (plug it in!), so, why not post.

@nikeballa:  Your reasoning is good, but I think you made a mistake at the end.  The seller has 7 horses, not 8.  If the seller had 8, he wouldn't be able to give the first person half a horse without chopping one in half.

Eureka!!!!!!!!

I solved the problem while i was in the shower, just like Archimedes!

Ok, so the number must be odd. Because... When you chop it in half you get a number of the form A.5. Then, when you add .5 to it, you get the number A+1 horses. Following this reasoning, the answer is that the seller started with 7 horses.

Here i will prove that it works.
(7/2)+.5= 4
7-4=3
(3/2)+.5=2
3-2=1
(1/2)+.5=1

1+2+4=7

Eureka! :rotfl: Hope that all of you can understand my reasoning especially those who asked me earlier in the day on For the Win.

Enjoy!


I just  have one question :

 how does  the  country  hosting  the imo  for a particular   year   affect  the imo?  for example  judging  from  ther   national  olympiads   vietnam   likes   lots  of   algebra problems they  seem to like locus problems  on geometry and in  number theory   they seem  to like sequence problems and things of the sort. are  we to expect  and  imo exam   similar  than  the  vietnam   mo  or tst?  how  can this answer  apply to any other  country  hosting the imo.  for  example  it seems to  that  the  2005  imo was very  suitable for mexico but  was this just  coincidence or  maybe not so much  but  always with the intention of playing a fair game?

As you might have noticed, the head of the problem-selection board in Slovenia this year was a Slovenian mathematician... Ofcourse he had a sort of influence... Using the same logic, and supposing that this is a tradition (a man from the host country in PS board), then the answer is yes, the host country has quite an influence...

Do you know who the other members of the problem selection committee are?

Robert Geretschläger from Austria
Marcin Kuczma from Poland
Geza Kos from Hungary
Svetoslav Savchev from Bulgaria

... so I guess that there is not too much influence :)

True. The problems are selected by a group of people, and they do not have the same nationality

Quite right... But a Slovenian mathematician would never have been a member of the comitee if Slovenia wasn't the host country- I mean: he's certainly making more influence when he's in the comitee than when he isn't... It doesn't have to be crucial influence, but...


geometry
geometric transformation
reflection
geometry solved

Let $A$, $B$, $C$ be three collinear points and $D$, $E$, $F$ three other collinear points. Let $G =BE\bigcap{CF}$, $H =AD\bigcap{CF}$, $I =AD\bigcap{CE}$. If $AI = HD$, and $CH = GF$, prove that $BI = GE$.

Indeed, I think that there is a typo in this problem: $I =AD\bigcap{CE}$ should actually be $I =AD\bigcap{BE}$. In other words, the problem should be:

[i]Let $A$, $B$, $C$ be three collinear points and $D$, $E$, $F$ three other collinear points. Let $G =BE\bigcap CF$, $H =AD\bigcap CF$, $I =AD\bigcap BE$. If $AI = HD$, and $CH = GF$, prove that $BI = GE$.[/i]

In fact, this problem follows from the following well-known theorem (the isotomic lines theorem):

[b]Theorem 1.[/b] If a line intersects the sidelines BC, CA, AB of a triangle ABC at the points X, Y, Z, then the reflections of these points X, Y, Z in the midpoints of the segments BC, CA, AB also lie on one line.

[i]Proof.[/i] Let X', Y', Z' be the reflections of the points X, Y, Z in the midpoints of the segments BC, CA, AB; then, with directed segments, we have BX' = XC, X'C = BX, CY' = YA, Y'A = CY, AZ' = ZB and Z'B = AZ, and thus

$\frac{BX^{\prime}}{X^{\prime}C}\cdot\frac{CY^{\prime}}{Y^{\prime}A}\cdot\frac{AZ^{\prime}}{Z^{\prime}B} = \frac{XC}{BX}\cdot\frac{YA}{CY}\cdot\frac{ZB}{AZ} =1:\left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)$.

Now, since the points X, Y, Z lie on one line, Menelaos yields

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$,

and hence,

$\frac{BX^{\prime}}{X^{\prime}C}\cdot\frac{CY^{\prime}}{Y^{\prime}A}\cdot\frac{AZ^{\prime}}{Z^{\prime}B} = 1:\left(-1\right)=-1$.

Consequently, after Menelaos, the points X', Y', Z' also lie on one line, and Theorem 1 is proven.

In order to solve our problem, we just apply Theorem 1 to the triangle GHI and the line ABC which intersects its sidelines HI, IG, GH at the points A, B, C, respectively; after Theorem 1, the reflections of these points A, B, C in the midpoints of the segments HI, IG, GH are collinear. But since AI = HD, the reflection of the point A in the midpoint of the segment HI is the point D, and since CH = GF, the reflection of the point C in the midpoint of the segment GH is the point F. Hence, the points D and F and the reflection of the point B in the midpoint of the segment IG are collinear. In other words: The point where the line DF meets the line IG is the reflection of the point B in the midpoint of the segment IG. But on the other hand, we know well that the point where the line DF meets the line IG is the point E. Hence, the point E is the reflection of the point B in the midpoint of the segment IG, and thus, BI = GE.

$\blacksquare$

  Darij

Sorry guys, I didn't know that there was a mistake in the problem. :blush: $I=AD\cap CE$ is what the source I had said. (maybe that's why I spent hours getting nowhere.) Sorry for any troubles caused.


I was just wondering...
How do you become a moderator?
do you have to post like a million posts or something?
It's not that i want to become a moderator, but i was just wondering...

You ask the administrators if you may become one. And NO! Never try to purposely reach a certain post count. People who post just for the sake of posting usually end of posting spam.

[quote="Nerd_of_the_Ages"]You ask the administrators if you may become one.[/quote]

No, you don't. The administrators [b]decide[/b] if you become one, without you asking. Asking to become a moderator will guarantee you not becoming a moderator.

[quote="Temperal"]No, you don't. The administrators [b]decide[/b] if you become one, without you asking. Asking to become a moderator will guarantee you not becoming a moderator.[/quote]
This isn't quite true. If there's a forum that needs more moderation, and you're an active member of the forum, you can offer to moderate that forum. If they think you'll be a good moderator and you are pretty active in that forum they will probably let you. But you can't just say "hey I want to be a moderator" and hope that they put you somewhere. It doesn't work like that.

If you want to be more involved with the forum here, the best thing to do is just to post a lot, but only with good reason and always trying to be helpful. When people post a problem that's been discussed before, provide the old link. When people ask for site support, help them. You don't have to be good at math to be a moderator, you have to be a good contributor to the community.

There's also the flip side- if you don't want to be a moderator, you won't be made one. I would probably be a moderator somewhere by now if I had indicated my intent- but I never did, and I'm comfortable as an "expert" without explicit powers.

[quote="Xevarion"][quote="Temperal"]No, you don't. The administrators [b]decide[/b] if you become one, without you asking. Asking to become a moderator will guarantee you not becoming a moderator.[/quote]
This isn't quite true. If there's a forum that needs more moderation, and you're an active member of the forum, you can offer to moderate that forum. If they think you'll be a good moderator and you are pretty active in that forum they will probably let you. But you can't just say "hey I want to be a moderator" and hope that they put you somewhere. It doesn't work like that. [/quote]

Sorry, I just had an allergic reaction to the words, "ask to be a moderator".  :oops:

I thought you were a moderator, jmerry. I guess I was wrong there too.

Thanks everyone! You guys helped me a lot!


What is the sum of the digits of the decimal form of the product $ 2^{1999}\cdot 5^{2001}$?

$ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$

[hide]
$2^{1999} \cdot 5^{2001}=5^2 \cdot 10^{1999}$. Discarding the trailing zeros, we get 2+5=7. So the answer is D.
[/hide]

[size=75][color=green]edited by a mod to fix the latex problem mentioned below[/color][/size]

Good job. Just for sidenote, use {} when the power is two-digit or longer. So, $5^1999$ looks nicer as $5^{1999}$.


Given triangle $ABC$, let $D$ and $E$ be any two points on the side $BC$.
A circle through $A$ cuts the lines $AB$, $AC$, $AD$, $AE$ at the points $P$, $Q$, $R$, $S$, respectively. Prove that
$\frac{AP\cdot AB-AR\cdot AD}{AS\cdot AE-AQ\cdot AC}=\frac{BD}{CE}$.

Here's a proof using inversion (that KIND OF uses the power of a point...      :D):
Let d1, d2, d3, d4 be lines through B, D, E, C (I put them on BC in this order) resp. all of them perpendicular to the diameter of the circle which passes through A (the diameter passes through A). Then an inversion of pole A and power AP*AB transforms the circle (let's say it has radius r) in the line d1, the same for AR*AD and line d2 and so on. Then AP*AB-AR*AD= 2r*d(A,d1)-2r*d(A,d2), where d(A,d) is the distance between A and the line d. Also, AS*AE-AQ*AC=2r*d(A,d3)-2r*d(A,d4). Just replace these, and you get what you want (remember that d1, d2, d3, d4 are parallel). It's kind of nice, isn't it?

it's really kool.
 what about that vector solution? :D

thanks a lot grobber, very nice sol ;-)

well, my vector sol is rather ugly but let's go:
we will prove the following LEMMA:
===
let ABC be a triangle. A circle W through A cuts AB and AC
at D and E (resp.). The altitude AH of tr ABC cuts W at Q and
AP is a diameter of W (P on W). Then
abs( AD*AB-AE*AC)=PQ*BC
===
in the following, AB is vector AB , [AB] is lenght AB, and
AB.AC is the scalar product of vectors AB and AC:
***
[PQ]*[BC]=abs(PB.PC)   (BC and PQ parallel)
PQ.BC=(BA+AC).PQ=BA.PQ+AC.PQ=BA.(PD+DA+AQ)+AC.(PE+EA+AQ)
   but BA.PD=0 and AC.PE=0
then
PQ.BC=AQ.(BA+AC)-[AC]*[EA]+[AB]*[AD]
   moreover AQ.(BA+AC)=0
***
now AB is the lenght AB as usual, etc..
hence, abs( AD*AB-AE*AC)=PQ*BC  as desired
(i know it's not very elegant ...)

Then it's pretty easy to finish the problem:
You apply the LEMMA to triangle BAD and AEC
(hope it's correct)

[quote="alekk"]we will prove the following LEMMA:
===
let ABC be a triangle. A circle W through A cuts AB and AC
at D and E (resp.). The altitude AH of tr ABC cuts W at Q and
AP is a diameter of W (P on W). Then
abs( AD*AB-AE*AC)=PQ*BC[/quote]

Since the segment AP is a diameter of the circle W, while the points D, E and Q lie on this circle, we have < ADP = 90°, < AEP = 90° and < AQP = 90°. Thus, the points D, E and Q are the orthogonal projections of the point P on the lines AB, AC and AH, respectively. And thus, your lemma is equivalent to the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=23012 .

  Darij


limit
trigonometry
calculus
calculus computations

Hi,
I need help with this question:

Using $ \lim_{x \to 0} \frac{sinx}{x}\equal{}1$ find

$ a) \lim_{x \to 0}( \frac{2}{sin2xsinx} \minus{}\frac{1}{sin^2x})$ and

$ b) \lim_{x \to 0} \frac{sinx}{sin6x\minus{}sin7x}$

hello, to a) we have $ \displaystyle{\frac{2}{2\sin(x)^2\cos(x)}\minus{}\frac{1}{\sin^2(x)}\equal{}\frac{1\minus{}\cos(x)}{\sin^2(x)\cos(x)}\equal{}\frac{(1\minus{}\cos(x))(1\plus{}\cos(x))}{\sin^2(x)\cos(x)(1\plus{}\cos(x))}\equal{}\frac{1}{\cos(x)(1\plus{}\cos(x))}}$
so the searched limit for $ x$ tends to zero is $ \frac{1}{2}$.
Sonnhard.

hello, using the formula $ \sin(x)\minus{}\sin(y)\equal{}2\sin(\frac{x\plus{}y}{2})\cos(\frac{x\plus{}y}{2})$ we get
$ \frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{\minus{}2\sin(\frac{x}{2})\cos(\frac{13x}{2})}\equal{}\minus{}\frac{\cos(\frac{x}{2})}{\cos(\frac{13x}{2})}$ we get for the searched limit when $ x$ tends to zero $ \minus{}1$.
Sonnhard.


inequalities
LaTeX
real analysis
real analysis unsolved

N is the norm
N(x-y) is greater than 1/2[sup(N(x),N(y)][N(x/N(x) - y/N(y))]
N(x-y) is greater than 1/4[N(x)+N(y)][N(x/N(x) - y/N(y))]
Show also that 1/2 and 1/4 are the biggest possible values in each inequality.

Much easier to read in $ \text{\LaTeX}$:
$ \|x-y\|\ge \frac12 \max(\|x\|,\|y\|)\left\|\frac{x}{\|x\|}-\frac{y}{\|y\|}\right\|$
$ \|x-y\|\ge \frac14(\|x\|+\|y\|) \left\|\frac{x}{\|x\|}-\frac{y}{\|y\|}\right\|$
The second inequality follows from the first one.

Hi and happy new year
I dont think it difficult to solve it.We only use norm properties and elementary tricks.

Which is why this is not a very interesting problem.  :|
Assuming $ \|y\|\le \|x\|\equal{}1$, we have $ \left\|x\minus{}\frac{y}{\|y\|}\right\|\le \|x\minus{}y\|\plus{}\left\|y\minus{}\frac{y}{\|y\|}\right\|\equal{}\|x\minus{}y\|\plus{}(1\minus{}\|y\|)\le 2\|x\minus{}y\|$.
The two-dimensional $ \ell_1$ or $ \ell_{\infty}$ is the worst case as usual.


[quote="Algadin"]Find all $ f: R\to R,g: R\to R,h: R\to R$ and $ f,g,h$ is continuous on $ R$ and satisfying

$ f(x^2) \minus{} f(y^2) \equal{} [h(x) \minus{} h(y)]g(x \plus{} y)$ with $ \forall x,y\in R$[/quote]
Posted here http://www.mathlinks.ro/viewtopic.php?t=237447


Simple Probability

--------------------------------------------------------------------------------
Yep, it's simple but i'd like to know

There are 3 coloured balls that can be spun in a lotto motion with a funnel for the ball to dropout
The red ball comes out which is a one third probabilty.
What's the formulae for the red ball to come out a 2nd time and a 3rd and 4th and so on.
Is it correct that 2 times would be 1/2 x1/3 equals 1/6 or 0.16 ththe
3 times 1/3 X 1/6 equals 1/18 or .05
4 times 1/4 x 1/18 equals 1/72 0.013 etc etc
Whats the formalue? wcosmos@dodo.com.au

assuming the balls are replaced, wouldnt the red ball have a 1/3 chance on all of the tries, and since probabilities are multiplied when mutually exclusive(if i got a red last time, it wont affect how i do this time) so it is 1/3^k. if i understand the problem?


calculus
derivative
trigonometry
real analysis
real analysis unsolved

Hi everyone,

Is the derivative on Sin^2x -cos^2x?

Thank you very much

Are you asking for the derivative of $ \sin^{2}{x}$?  It's derivative is $ 2\sin{x}\cos{x}\equal{}\sin{2x}$.


group theory
abstract algebra
superior algebra
superior algebra solved

[quote="yannick"]We know that any finite subgroup of the multiplicative group of a field is cyclic. Counterexamples for division rings that are not commutative are known: $\mathbb{H}$, the quaterions over $\mathbb{R}$. Are there counterexamples  for finite division rings?[/quote]

Never mind. Every finite division ring is a field. :)


Salut à tous,

J'ai commencé à m'intéresser aux résolutions de problèmes, J'ADORE CA!

Le seul ennui c'est qu'il n'est pas facile de progresser et de naviguer sans rupture. Quelqu'un pourrait-il communiquer des pistes des chemins à suivre au travers de ses propres expériences?

Je suis en seconde et je sais lire l'anglais.

Merci beaucoup pour le coup de pouce :roll:  :D

Commence déjà par lire les documents disponibles en français  :P
http://www.animath.fr

Va voir l'entrainement et les cours de l'OFM.

Pierre.

Pas mal du tout, en particulier le "petit" cours d'arithmétique...!

Eh bien il va falloir se lancer maintenant..... :roll:

En attendant je suggère de continuer à alimenter ce fil avec des infos et des commentaires sur vos lectures...

Pour ma part je suggère un livre terrible d'exercices de géométrie de Posamentier intitulé Challenging Problems in Geometry. :roll:

[quote="JohnDifool"]le "petit" cours d'arithmétique...!
[/quote]

Bah..c'est juste la première partie, hein?  :D

Pierre, qui se demande si d'ailleurs la deuxième partie verra effectivement le jour.


Let f twice diferentiable and f(f΄(x))+f(x)=0 for x>0
Prove that:
(i). f 1-1
(ii). f΄(f΄(x))=x, x>0
(iii). If f(1)=0 then f(x)=lnx

Hello, what is "f 1-1" ? Try to use LaTex. Just write \$f(x) = ...\$ (without \) in the body of the massage and you will see the result. $ f(f'(x))\plus{}f(x)\equal{}0$...

[quote="Yuriy Solovyov"]Hello, what is "f 1-1" ?[/quote]
one to one

[quote="Yuriy Solovyov"]Hello, what is "f 1-1" ? Try to use LaTex. Just write \$f(x) = ...\$ (without \) in the body of the massage and you will see the result. $ f(f'(x)) \plus{} f(x) \equal{} 0$...[/quote]
Let f twice diferentiable and f(f΄(x))+f(x)=0 for x>0
Prove that:
(i). f one to one
(ii). f΄(f΄(x))=x, x>0
(iii). If f(1)=0 then f(x)=lnx


[b] Generalize an equation [/b]

Dear friends.
Today, I attend a class of Development Students’ thought through mathematics problem solving. I have something to ask you for help.

My instructor gives us a problem like this: “Solve the equation   $ \displaystyle 3^x + 4^x =5^x $  and generalize that problem”. I can solve the equation easily but I have a difficulty in generalizing that problem. Can you help me?

After that, she gives us one more problem: “Let $ a, b, c $ be positive real numbers and  $x$  be a natural number. Prove that there exists no more one  $ x $ such that  $ \displaystyle a^x + b^x = c^x $”. I can’t prove this. I wonder if this problem is right. If it’s right, how can we solve it?

Thank you very much for your help!

This problem seems very easy to me :

Suppose we have $a^x+b^x=c^x$. Then $a<c$ and $b<c$.

If $y>x$ then $a^y+b^y<(a^x+b^x)c^{y-x}=c^y$ , so y isn't solution.
If $y<x$ then $a^y+b^y>(a^x+b^x)c^{y-x}=c^y$ , so y isn't solution either.

Thank you very much for your reply! Now I have some questions to ask you.

1/ Can you generalize the problem “Solve the equation   $ \displaystyle 3^x + 4^x =5^x $?

2/ What can you do if $ x $ is a real number in the problem that you have solved?


David borrowed $ \$10,000$ to buy a new car. He had to pay
back $ 160\%$ of what he borrowed. What was the total cost of
the car to David?


arithmetic sequence
combinatorics proposed
combinatorics

[quote="moldovan"]Prove that among $ 12$ different positive integers of two digits, there are two of whose difference is a number with two equal digits.[/quote]

[hide="Hint"]
[color=blue]
Consider 11 different arithmetic progression each having common difference 11 and first term 10 or 11 or 12 ... or 20.
[/color]
[/hide]

[hide="Complete Solution"]
[color=blue]
Arithmetic progression starting with 10 or 11 will have 9 terms each and remaining  9 arithmetic progressions will have 8 terms each.Total number of term of all arithmetic progression is 2x9+9x8=90= Number of two digit numbers.

No two arithmetic progression has common term and all the two digit numbers appears exactly once in any one of the arithmetic progression. As common difference is 11 in each of the arithmetic progression , difference of any two term of an arithmetic progression is a number with two equal digits .

Since we have to select 12 numbers and their are only 11 arithmetic progression, atleast two of the selected numbers must belong to same arithmetic progression.

Hence among 12 different positive integers of two digits, there are two of whose difference is a number with two equal digits.
[/color]
[/hide]


induction
logarithms
inequalities proposed
inequalities

$ a_n^2 \leq a_n\minus{}a_{n\plus{}1},\Longrightarrow a_n<\frac1n.$

[quote="jcc0107"]$ a_n^2 \leq a_n \minus{} a_{n \plus{} 1},\Longrightarrow a_n < \frac1n.$[/quote]
Need more condition Example $ a_1 < 1$
We prove by induction
$ a_{n \plus{} 1}\le a_n \minus{} a_n^2\le \frac {1}{n} \minus{} \frac {1}{n^2}\le \frac {1}{n \plus{} 1}$
Done

[quote="Allnames"]Need more condition Example $ a_1 < 1.$ We prove by induction[/quote]$ a_{1}\leq a_{0} \minus{} a_{0}^2 \equal{} \frac {1}{4} \minus{} \left(\frac {1}{2} \minus{} a_0\right)^2 < 1.$

By the same way, we have

$ a_{n}^2 \leq a_{n} \minus{} a_{n \plus{} 1}(n\geq0)\Longrightarrow a_n < \frac {1}{n} \minus{} \frac {\ln{n}}{n^2}(n\geq1).$


why is it that when I click memberlist I see this user have like 16777215 posts?

[quote="mysmartmouth"]Out of curiosity, what was the problem?  I've never seen that bug in phpBB before.[/quote]This is not phpBB anymore. I've rewritten large parts of the code myself. The bug happened when the computer made in the DB $--0$ in a non-negative defined field. Thus the result was taken modulo $2^{24}$.


Find all values of m which will make $x+2$ a factor of $x^{3}+3m^{2}x^{2}+mx+4$.

or you can do it without synthetic division by plugging in $x=-2$
and solving the quadratic for $m$

I prefer anyone's method.
[hide] $-8+12m^{2}-2m+4=0\Rightarrow 12m^{2}-2m-4=0\Rightarrow 6m^{2}-m-2=0\Rightarrow (3m-2)(2m+1)=0\Rightarrow \boxed{m=\frac{-1}{2}, \frac{2}{3}}$ [/hide]


I've finished Interm Counting and Probability. What is a good book to move on to? I don't necessarily want to limit myself to things on found in competitions. I want to learn about how to count, so that I could eventually tackle a book like enumerative combinatorics by Richard Stanley.

I highly recommend [url=http://www.amazon.com/Proofs-that-Really-Count-Combinatorial/dp/0883853337]Proofs that Really Count[/url].  If you can master the concepts in that book, you'll be very well prepared for the arguments in EC (well, the combinatorial ones anyway).

You might also try [url=http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf]Generatingfunctology[/url] by Herbert Wilf, or [url=http://www.amazon.com/102-Combinatorial-Problems-Titu-Andreescu/dp/0817643176/ref=pd_sim_b_4]102 Combinatorial Problems[/url] by Titu Andeescu and Zuming Fang


Given $ a, b, c \ge 0$ satisfy $ a^2\plus{}b^2\plus{}c^2\equal{}2$.

Prove that: $ \frac{(a\plus{}b)^3}{a^2\plus{}6ab\plus{}b^2}\plus{}\frac{(b\plus{}c)^3}{b^2\plus{}6bc\plus{}c^2}\plus{}\frac{(c\plus{}a)^3}{c^2\plus{}6ca\plus{}a^2} \le 3$  :)

[quote="nguoivn"]Given $ a, b, c \ge 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$.

Prove that: $ \frac {(a \plus{} b)^3}{a^2 \plus{} 6ab \plus{} b^2} \plus{} \frac {(b \plus{} c)^3}{b^2 \plus{} 6bc \plus{} c^2} \plus{} \frac {(c \plus{} a)^3}{c^2 \plus{} 6ca \plus{} a^2} \le 3$  :)[/quote]

Assume: c min
So:
$ \frac {(b \plus{} c)^3}{b^2 \plus{} 6bc \plus{} c^2} \le b$
$ \frac {(c \plus{} a)^3}{c^2 \plus{} 6ca \plus{} a^2} \le a$
We need prove that:
$ \frac {(a \plus{} b)^3}{a^2 \plus{} 6ab \plus{} b^2} \plus{} a \plus{} b \le \frac {3}{2}\sqrt {2(a^2 \plus{} b^2)}$
$ < \equal{} > \frac {(a \minus{} b)^2(a \plus{} b)}{a^2 \plus{} 6ab \plus{} b^2} \le \frac {3(a \minus{} b)^2}{a \plus{} b \plus{} \sqrt {2(a^2 \plus{} b^2)}}$
It's true :)
Done!

[quote="nguoivn"]Yes, that's it. Now, you can try with the stronger  :) :
 $ \frac {(a \plus{} b)^{3}}{a^{2} \plus{} 6ab \plus{} b^{2}} \plus{} \frac {(b \plus{} c)^{3}}{b^{2} \plus{} 6bc \plus{} c^{2}} \plus{} \frac {(c \plus{} a)^{3}}{c^{2} \plus{} 6ca \plus{} a^{2}}$$ \le \frac {6(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)}{3(a^2 \plus{} b^2 \plus{} c^2) \plus{} 2(ab \plus{} bc \plus{} ca)}$
With $ a, b, c \ge 0$[/quote]

Assume:c min.
We have:
$ \frac {6(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)}{3(a^2 \plus{} b^2 \plus{} c^2) \plus{} 2(ab \plus{} bc \plus{} ca)} \ge \frac{6(a^2\plus{}b^2)(a\plus{}b)}{3(a^2\plus{}b^2)\plus{}2ab}$
And:$ LHS \le \frac {(a \plus{} b)^{3}}{a^{2} \plus{} 6ab \plus{} b^{2}} \plus{}a\plus{}b$
We need prove that:
$ \frac {(a \plus{} b)^{3}}{a^{2} \plus{} 6ab \plus{} b^{2}}\plus{}a\plus{}b \le \frac{6(a^2\plus{}b^2)(a\plus{}b)}{3(a^2\plus{}b^2)\plus{}2ab}$
$ <\equal{}>\frac{(a\minus{}b)^2(a\plus{}b)}{a^2\plus{}6ab\plus{}b^2} \le \frac{3(a\minus{}b)^2(a\plus{}b)}{3(a^2\plus{}b^2)\plus{}2ab}$.It's true :D
Done!


calculus
integration
trigonometry
calculus computations

Let $ a$ be positive constant number. Evaluate $ \int_{ \minus{} a}^a \frac {x^2\cos x \plus{} e^{x}}{e^{x} \plus{} 1}\ dx.$

Here is one solution
$ \begin{array}{l}
 I \equal{} \int\limits_{ \minus{} a}^a {\frac{{{x^2}\cos x \plus{} {e^x}}}{{{e^x} \plus{} 1}}} dx\mathop  \equal{} \limits_{dx \equal{}  \minus{} du}^{x \equal{}  \minus{} u} \int\limits_a^{ \minus{} a} { \minus{} \frac{{{u^2}\cos u \plus{} {e^{ \minus{} u}}}}{{{e^{ \minus{} u}} \plus{} 1}}du \equal{} }  \\
  \equal{} \int\limits_{ \minus{} a}^a {\frac{{{u^2}\cos u \cdot {e^u} \plus{} 1}}{{{e^u} \plus{} 1}}} du \\
 I \plus{} I \equal{} \int\limits_{ \minus{} a}^a {\frac{{{x^2}\cos x \plus{} {e^x}}}{{{e^x} \plus{} 1}}} dx \plus{} \int\limits_{ \minus{} a}^a {\frac{{{x^2}\cos x \cdot {e^x} \plus{} 1}}{{{e^x} \plus{} 1}}} dx \equal{}  \\
  \equal{} \int\limits_{ \minus{} a}^a {\frac{{\left( {{e^x} \plus{} 1} \right)\left( {{x^2}\cos x \plus{} 1} \right)}}{{{e^x} \plus{} 1}}} dx \equal{} \int\limits_{ \minus{} a}^a {\left( {{x^2}\cos x \plus{} 1} \right)} dx \equal{}  \\
  \equal{} \left[ {{x^2}\sin x} \right]_{ \minus{} a}^a \minus{} 2\int\limits_{ \minus{} a}^a {x\sin xdx}  \plus{} 2a \equal{}  \\
  \equal{} 2{a^2}\sin a \plus{} 2a \minus{} 2\left( {\left[ { \minus{} x\cos x} \right]_{ \minus{} a}^a \plus{} \int\limits_{ \minus{} a}^a {\cos xdx} } \right) \equal{}  \\
  \equal{} 2{a^2}\sin a \plus{} 2a \plus{} 4a\cos a \minus{} 4\sin a \\
 I \equal{} {a^2}\sin a \plus{} a \plus{} 2a\cos a \minus{} 2\sin a \\
 \end{array}$

hello
[hide]
We can directly use the property.
$ I \equal{} \int^{a}_{0} \frac {x^2cos(x) \plus{} e^{x}}{e^{x} \plus{} 1}dx \plus{} \frac {x^2cos(x) \plus{} e^{ \minus{} x}}{e^{ \minus{} x} \plus{} 1}dx$
$ I \equal{} \int^{a}_{0} (x^2cos(x) \plus{} 1)dx$
$ I \equal{} 2acos(a) \plus{} a \plus{} (a^2 \minus{} 2)sin(a)$[/hide]
thank u


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ a^3\plus{}b^3\plus{}c^3 \ge \frac{a^2(b^3\plus{}c^3)}{a^2\plus{}bc}\plus{}\frac{b^2(c^3\plus{}a^3)}{b^2\plus{}ca} \plus{}\frac{c^2(a^3\plus{}b^3)}{c^2\plus{}ab}.\]
:)

This inequality isn't so hard. It's equivalent to:

$ \sum_{\textrm{cyc}} a^3\left(1\minus{}\frac{b^2}{b^2\plus{}ca}\minus{}\frac{c^2}{c^2\plus{}ab}\right)\ge 0$
$ \Leftrightarrow  \sum_{\textrm{cyc}} a^3\frac{bc(a^2\minus{}bc)}{(b^2\plus{}ca)(c^2\plus{}ab)}\ge 0$

$ \Leftrightarrow \sum_{\textrm{cyc}}a^2\frac{a^2\minus{}bc}{(b^2\plus{}ca)(c^2\plus{}ab)}\ge 0$
$ \Leftrightarrow  \sum_{\textrm{cyc}}\frac{a^2(a^4\minus{}b^2c^2)}{(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}\ge 0$
$ \Leftrightarrow  \sum_{\textrm{cyc}} A^3\plus{}B^3\plus{}C^3\minus{}3ABC\ge 0$, where $ A\equal{}a^2$, $ B\equal{}b^2$ and $ C\equal{}c^2$.

Deservedly, you didn't choose it for the real contest.  :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \ge \frac {a^2(b^3 \plus{} c^3)}{a^2 \plus{} bc} \plus{} \frac {b^2(c^3 \plus{} a^3)}{b^2 \plus{} ca} \plus{} \frac {c^2(a^3 \plus{} b^3)}{c^2 \plus{} ab}.
\]
:)[/quote]

This inequaltiy is equivolent to:
$ \sum_{sym}{a^5b^2c^2}\plus{}\frac{1}{2}\sum_{sym}{a^3b^3c^3}\plus{}\sum_{sym}{a^6b^3}\plus{}\sum_{sym}{a^4b^4c}\plus{}\frac{1}{2}\sum_{sym}{a^7bc} \ge \sum_{sym}{a^5b^2c^2}\plus{}\sum_{sym}{a^3b^3c^3}\plus{}\sum_{sym}{a^6b^3}\plus{}\sum_{sym}{a^4b^4c}$
or
$ \sum_{sym}{a^7bc} \ge \sum_{sym}{a^3b^3c^3}$
 which is right.


geometry
trigonometry
circumcircle
analytic geometry
geometric transformation
reflection
cyclic quadrilateral

By an angle chase, we can see that $P$ moves on a circle tangent in $A,B$ to $CA,CB$ respectively. Let $Q$ be the second intersection of $CP$ with this circle. $PAQB$ is a harmonic cyclic quadrilateral, so, if $C'$ is the intersection between the tangent at $P$ to the circle and $AB$, we find $(PA,PB;PQ,PC')=-1$, meaning that $PQ$ is the symmedian of $PAB$, which is precisely what we want.

I was in a hurry while writing this, but I'll clarify if needed.

Rewriting your problem with different notations:

[i]Let ABC be an isosceles triangle with AB = AC. Let P be a point inside the triangle ABC such that < PBC = < PCA. Let M be the midpoint of the segment BC. Prove that < BPM + < APC = 180°.[/i]

Now, this problem was solved in posts #8 and #9 of http://www.mathlinks.ro/Forum/viewtopic.php?t=18258 (see the Lemma in post #8), with the only difference being that an additional assumption was used, namely the assumption that < PCB = < PBA; but this assumption is actually unnecessary, since it follows from the other assumptions of the problem (namely, since the triangle ABC is isosceles with AB = AC, we have < ACB = < ABC, and since < PBC = < PCA, we have < PCB = < ACB - < PCA = < ABC - < PBC = < PBA, so that < PCB = < PBA comes out as a consequence of the other assumptions of the problem).

  Darij

[quote="warut_suk"]Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.[/quote]

Take point Q such that APBQ is a parallellogram.

Take point R on AQ such that ACB, APR are similar. (I assume R lies between A and Q, the other case is similar) Then BPRQ is cyclic, and ACP, ABR are similar.

Hence BPQ = BRQ = 180 - ARB = 180 - APC. Hence result.

$S\in AB\cap CP$, $CA=CB=a$, $AB=b$, $m(\widehat {PCA})=u$, $m(\widehat {PCB})=v$,
$m(\widehat {PAB})=m(\widehat {PBC})=x$, $m(\widehat {PAC})=m(\widehat {PBA})=y$.

$\blacktriangle$ The theorem of Sinus in the triangle $APB\Longrightarrow \frac{PA}{PB}=\frac{\sin y}{\sin x}\ \ (1)$.
$\blacktriangle$ The trigonometrical form of the Menelaus' theorem in
the triangle $ABC$ for the inner point $P\Longrightarrow$
$\sin u\sin^2x=\sin v\sin^2y$ $\Longrightarrow$ $\frac{\sin u}{\sin v}=\left(\frac{\sin y}{\sin x}\right)^2\ \ (2)$.
$\blacktriangle\ \frac{SA}{SB}=\frac{CA}{CB}\cdot \frac{\sin \widehat {ACS}}{\sin \widehat {BCS}}$ $\Longrightarrow$ $\frac{SB}{SC}=\frac{\sin u}{\sin v}\ \ (3).$

From the relations $(1)$, $(2)$, $(3)$ results: $\frac{SA}{SB}=\left(\frac{PA}{PB}\right)^2$, i.e. the ray $[PS$ is
the symmedian from the vertex $P$ in the triangle $APB$, meaning

$\widehat {APM}\equiv \widehat {BPS}\Longrightarrow$ $\boxed {\ m(\widehat {APM})+m(\widehat {BPC})=180^{\circ}\ }.$

The problem can be generalized as follow (which is well-known)
 Let $PAB$ be a triangle and $M$ be the midpoint of $AB$. Suppose $C$ is the intersection of the tangents of the circumcircle at $A$ and $B$. Then $\measuredangle{BPM}= \measuredangle{CPB}$.
[Moderator edit: This is basically the assertion of http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 .]
 Back to the problem , it is easy to see that $CA$ and $CB$ are the tangents of $(PAB)$ at $A$ and $B$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55581
http://www.mathlinks.ro/Forum/viewtopic.php?t=43202
where are mentioned the properties of the harmonical quadrilateral.

Let us see the circumcircle of APB. Let v=<ABC.
Use the center of the circumcircle as origin of coordinate and give the coordinates as:

C (0,1)
M (0, (cos v)^2)
P (cos v*cos u, cos v*sin u)

It is easy to obtain: PM/PC=cos v

So, we can get: BC/PC=AM/PM.
So, <BPC=<APM or <BPC+<APM=180.

When <BPC=<APM, then <AMP=<BCP. So, <BPC=<BMC=90. As a result, <BPC+<APM=180 too.

P is on circle (U) tangent to BC at B and passing through A(then $ \angle PAB \equal{} \angle PBC$). $ \triangle ABC$ is isosceles $ \Longrightarrow$ (U) is also tangent to CA at A, AB is polar of C WRT the circle (U). CP meets (U) at P and again at Q. If P' is a reflection of P in $ CM \perp AB,$ $ P' \in (U)$ and CP' meets (U) again at Q'. PP'Q'Q is isosceles trapezod symmetrical WRT CM, the diagonals PQ', P'Q meet on CM and on the polar AB of C, $ M \equiv PQ' \cap P'Q \Longrightarrow$ $ \angle APM \equiv \angle APQ' \equal{} \angle BP'Q \equal{} \angle BPQ \equal{} 180^\circ \minus{} \angle BPC.$

We have $ \angle PBA\equal{}\angle PAC$. Let $ X$ be on $ PM$ such $ PAXB$ is parallogram. Then $ \angle XAB\equal{}\angle PAC,\angle XBA\equal{}\angle PBC$. Hence $ C,X$ are isogonal points wrt $ \triangle PAB$. Hence $ PC,PX$ is isogonal line $ \triangle PAB$. This imply $ \angle APM\plus{}\angle BPC\equal{}180$.

[b]Lemma:[/b] [color=#0000FF]Let $ABC$ be a triangle and $\Gamma $ its circumcircle. Let the tangent to $\Gamma $ at $B$ and $C$ intersect at $D$. Then $AD$ coincides with the $A$- symmedian of $\triangle ABC$. [/color] The proof can be found [url=http://web.mit.edu/yufeiz/www/olympiad/geolemmas.pdf]here[/url].

By angle chasing we can show that $AC$ and $BC$ are tangents to the circumcircle of $\triangle PAB$ at points $A$ and $B$ respectively. The tangents $AC$ and $BC$ intersect at $C$. So $PC$ coincides with the $C$- symmedian of $\triangle ABC$. Since $PM$ is the median of $\triangle ABC$, $\angle APM = \angle BPE$. or  $\angle APM = 180^{\circ} - \angle BPC$. So $\angle APM +  \angle BPC = 180^{\circ} $.

[hide="Translation of Official Solution"]
Assume first, that the point $P$ lies inside the triangle $AMC$ or on the line $MC$.

Let $Q\ne C$ be the point of intersection of the circumcircle of the triangle $BCP$ with line $CM$ (Fig. 1 and 2). Since $\angle PAM=\angle PBC=\angle PQC$, so the points $A, P, M, Q$ lie on a circle. Also
\[(1)\qquad\angle BPC=\angle BQC=\angle AQC.\]

If the point $Q$ is located on the outside of the triangle $ABC$ (Fig. 1) or on the segment $AB$, then by the equation $(1)$ we obtain
\[\angle APM+\angle BPC=\angle APM+\angle AQC=180^\circ.\]

If, however, the point $Q$ is inside the triangle $ABC$ (Fig. 2), it is once again using the equation $(1)$ we obtain
\[\angle APM+\angle BPC=\angle AQM+\angle AQC=180^\circ.\]

Suppose now that the point $P$ is inside the triangle $BCM$ . Since $\angle PBA=\angle PAC$ , so by analogy with the above, we obtain the equality $\angle BPM+\angle APC=180^\circ$. Hence,
\[\angle APM+\angle BPC=180^\circ,\]
which completes the solution.
[/hide]
[url=http://www.om.edu.pl/sites/default/files/zadania/om/om51_3r.pdf]Official Solutions[/url]
[hide="Problem Significance"]
I was trying to find a minimal-construction proof of 2014 JMO 6b, and it reduced to a certain lemma. I asked Kenny_cz about it, and he replied
[quote="Kenny_cz"]Your Lemma 5 is a well-known problem coming from Poland (from year 2000). Nowadays it appears on just about every handout concerning symmedians. And indeed the insight coming from symmedians gives the most natural proof.[/quote]
This problem is significant. Q.E.D.
[/hide]

A nice problem for symmedians  :)
We see that AC and AB are tangents to the circumcircle of $\triangle{APB} \implies CP $ is the symmedian.
Then the rest easily follows from how symmedians are defined

Since $\triangle ABC$ is isosceles, we know that $\angle PAC=\angle PBA$. Consider $(APB)$. Since $\angle PAB=\angle PBC$ and $\angle PAC=\angle PBA$, lines $AC$ and $BC$ are tangent to $(APB)$. This implies that $PC$ lies on the $P$-symmedian. Let $CP$ intersect $AB$ at point $E$. By the definition of the symmedian, $\angle APM=\angle BPE$. Clearly $\angle BPE+\angle BPC=180^{\circ}$, so $\angle APM + \angle BPC = 180^{\circ}$.

Let $CP\cap AB=G$.  Then the problem is equivalent to $\angle APM=\angle BPG$, aka. $CG$ is a symmedian in $\triangle APG$. Notice that $CB$ and $CA$ are the tangents to $(APB),$ so we are done.

Since $\angle CBP=\angle BAP$ and $\angle PAC=\angle PBA,$ $C$ is the intersection of the tangents to $(PAB)$ at $A$ and $B.$ Hence, $\overline{PC}$ is the $P$-symmedian point of triangle $PAB.$ Notice that $$\measuredangle BPC=-\measuredangle(\overline{CP},\overline{PB})=-\angle APM.$$ $\square$

Nice problem!

It is well known that $P$ is the $A$ humpty point, and so $M_1:=AP\cap BC$ is the midpoint of $BC$. Furthermore, we have that $\angle PCB=\angle PAC=\angle PBA$ and $\angle PAB=\angle PBC$. Hence, we get that $\angle APM+\angle BPC=\angle APM+\angle CPM_1+\angle M_1PB=\angle APM+\angle MPB+\angle BPM_1=180^{\circ},$ where this follows from similar triangles $\triangle PMB$ and $\triangle PM_1C$.

EDIT: oops just realized I already solved this before. At least this is a slightly different/more motivated solution.

Let $(APB) \cap CP=L$
As well: $CA$ and $CB$ are tangent of $(APB)$
Then: $PQ$ is symedian of $(APB)$
$\implies \angle APM=\angle LPB \implies \angle APM + \angle BPC=180$

Let $CP$ meet circumcircle of $APB$ at S. Since $\angle PAB = \angle PBC$ then $CB$ is tangent to $APB$ and since $CB = CA$ then $SP$ is symmedian of $ASB$ so $\angle APM = \angle BPS$ and so $\angle APM + \angle BPC = \angle 180$.

Let $CP \cap AB = D$. The angle condition implies that $AC$ and $AB$ are tangent to the circumcircle of $APB$, then we know that $PD$ is the $P$-symmedian of $APB$, which implies that $\angle TPC = \angle APD = \angle MPB$ thus we have that $\angle BPC = \angle MPT$, which clearly gives us that $\angle APM + \angle BPC = 180$

Let $\overline{PM'}$ be a symmedian in triangle $BPC$, so it suffices to show that $A$ lies along the symmedian. But this is evident by the tangent intersection definition.

[hide=Complex]Clearly $BC$ is tangent to $(ABP)$, so $AC$ is also tangent to $(ABP)$, by symmetry. Use complex numbers with $(ABP)=\mathbb{S}^1$, then $c=\frac{2ab}{a+b}$ and $m=\frac{a+b}{2}$. Hence\begin{align*}\frac{m-p}{a-p}\frac{c-p}{b-p} & =\frac{\frac{a+b}{2}-p}{a-p}\frac{\frac{2ab}{a+b}-p}{b-p} \\
& =\frac{(a+b-2p)(2ab-ap-bp)}{2(a+b)(a-p)(b-p)}.
\end{align*}So\begin{align*}\overline{\left (\frac{m-p}{a-p}\frac{c-p}{b-p}\right )} & =\overline{\left (\frac{(a+b-2p)(2ab-ap-bp)}{2(a+b)(a-p)(b-p)}\right )} \\
& =\frac{\left (\frac{1}{a}+\frac{1}{b}-\frac{2}{p}\right )\left (\frac{2}{ab}-\frac{1}{ap}-\frac{1}{bp}\right )}{2\left (\frac{1}{a}+\frac{1}{b}\right )\left (\frac{1}{a}-\frac{1}{p}\right )\left (\frac{1}{b}-\frac{1}{p}\right )} \\
& =\frac{(a+b-2p)(2ab-ap-bp)}{2(a+b)(a-p)(b-p)} \\
& =\frac{m-p}{a-p}\frac{c-p}{b-p}.
\end{align*}Therefore $\frac{m-p}{a-p}\frac{c-p}{b-p}\in \mathbb{R}$, so $\angle APM+\angle BPC=180^\circ$, as wanted.[/hide]

[hide=Synthetic]Let $D$ be the circumcentre of $APB$, clearly $BC$ is tangent to $(ABP)$, so $AC$ is also tangent to $(ABP)$, by symmetry. Then $ACBD$ is cyclic and $M$ and $C$ are inverses with respect to $(ABP)$. Therefore\begin{align*}\angle APM+\angle BPC & =\angle APD-\angle MPD+\angle BPC \\
& =\angle APD-\angle DCP+\angle BPC \\
& =\angle DAP-\angle DCP+180^\circ -\angle CBP-\angle PCB \\
& =180^\circ +\angle DAP-\angle PAB-(\angle DCP+\angle PCB) \\
& =180^\circ +\angle DAB-\angle DCB \\
& =180^\circ +\angle DAB-\angle DAB \\
& =180^\circ .
\end{align*}[/hide]


1. What is the only pair of non-negative integers (m,n) for which (3^m) - 1 = (2^n) + 1?

2. What are all integers k for which x^2 + kx + k + 17 = 0?

[hide="1"](3^m) - 1 is always even and (2^n) + 1 is always odd unless n=0. Thus, the ordered pair is (1,0)[/hide]

I don't really get #2... There are two variables ($k$ and $x$) :?

[quote="airelemental135"]I understand that (3^m) + 1 is even and (2^n) -1 is odd unless n=0. But why does the order pair must be (1,0)? Where did it come from?[/quote]

He means $(m.n)=(1,0)$

[quote="airelemental135"]Sorry I typed #2 wrong. But here it is.

2. What are all integers k for which x^2 + kx + k + 17 = 0 has integral roots?[/quote]

For question 2 , let the roots be $a,b$ where $a\geq b$ then you have

$a+b=-k$ ; $ab=k+17$   :arrow:  $ab+a+b=17$  :arrow:  $(a+1)(b+1)=18$

from here $(a+1,b+1)=(18,1)$ , $(9,2)$ , $(6,3)$ and also the negative cases

so you get $(a,b)=(17,0)$ ,$(8,1)$ , $(5,2)$ ,$(-2,-19)$ , $(-3,-10)$ , $(-4,-7)$

So you get 6 set of $k$ which are $-17,-9,-7,21,13,11$


$ (x,y)\in{\mathbb{IN}}$ such: $ xy|x^{2}\plus{}y^{2}\minus{}x$
prove that $ x$ is a perfect square.

thanx.

Let $ x\equal{}ka$, $ y\equal{}kb$, where $ a,b$ coprime.

Then $ k^2ab|k^2a^2\plus{}k^2b^2\minus{}ka$ $ \Rightarrow k|a$, and $ a|kb^2 \Rightarrow a|k$

So $ a\equal{}k$, $ x\equal{}ka\equal{}k^2$ which is a perfect square


In how many different ways can 3 men and 4 women be placed into two groups of two people and one group of three people if there must be at least one man and one woman in each group?

I may be wrong but for the first group of two, we have

$ 3 \times 4 \equal{} 12$, since there are 3 men and 4 women.

Then

$ 2 \times 3 \equal{} 6$, for the remaining men and women in the second group.

There is only one way to complete the last group.

So $ 12 \times 6 \equal{} \boxed{72}$

I don't believe I'm right. Someone please correct me.

so that means 72 is not the answer
anyone have any other suggestions?

Since there are 3 groups and 3 men, there must be one man in each group, so there are $ 3!\equal{}6$ ways to do that. Then, there are 4 ways to put the first woman in the first group, 3 ways to put the second woman in the second group, and the rest doesn't matter for the third group because no matter what, theyre stuck with the third group. Thus, there are $ 6\cdot 4\cdot 3\equal{}72$ ways to group these people, so izzy, you are right.

[quote="mewto55555"]When I entered 72 last time I did it, it counted it wrong. I don't know why.[/quote]
a glitch. i've gotten that before.

You are overcounting...First one to figure out why wins a brownie point.

OK trying a new approach.

With 3 men, their groups are guaranteed.

Say Man A is in group A
Man B group B
Man C group C

TO figure out which 2 women are grouped together, 4c2=6

6x3(whichever group has 3)=18

Other women can be arranged 2 ways.

18x2=36

I get 26 when doing this way. I wonder how we overcounted previously....

Or do isabella's way, noting that the two two person groups can switch, so you have to divide by 2 to compensate.

Example being.... Group 1: man A, woman A and Group 2: man B, woman B is the same as Group 1: man B, woman B and Group 2: man A, woman A.

Hope I make sense!

[quote="vallon22"]hmm...

let's see
group A could have either man A, B, or C
therefore, group B would have the two that did not get taken by the first group
therefore, there are 6 ways to arrange the men....

and then group A can have either woman A, B, C, or D
so group B would have either one of the 3 women that didn't get taken
and the last group wouldn't matter, so there would be 12 ways to arrange the women

but taking into account iin's statement, then we might have to divide by 2 if group A and B are flipped
so we would have 36 possibilities, the same as what mew got
is that right?[/quote]

I think that's right. I'm pretty sure the answer is $ 36$...what I did earlier seemed correct, I just didn't divide by 2 because if A and B are flipped as iin and you said.


Support
probability
advanced fields
advanced fields open

I have 5 pumps operating. I need to figure out the minimum number of spare pumps required to support the 5 pumps.

Each pump has a MTBF of 30000
The MTTR is 24 hours
The time to switch pumps is 1 hour (mean time to fail)
Once a spare is used the old pump is de-installed and repaired. This takes 48 hours

The pumps will operate 24 hours a day 7 days a week
The pumps have a known availability of 99.5%

I get that MTBF = Mean Time Between Failure and MTTR = Mean Time To Repair. I don't understand why the time needed to switch pumps is called "mean time to fail". I don't understand why MTTR=$24$ if it takes $48$ hours to repair a pump. I interpret $99.5\%$ as the probability of all pumps being operational at any given time.

A rough count: on average you'll have one failure every $30000/5=6000$ hours. If it takes you $24$ hours to fix a problem, then the availability is $6000/6024=99.6\%$ even with no spare devices. With $48$ hours to repair you can't get $99.5\%$ this way. But since it takes just $1$ hour to replace a pump with a spare, the availability will be $\approx 6000/6001>99.9\%$ even with only one spare. The chance of a pump failing while another one is under repair (be it either $24$ or $48$ hours) is so small that it should not seriously affect the estimate of availability.

Disclaimer: I am not a reliability specialist, I just play one on AoPS.  :)

How would I show that having 1 spare is enough. If I focus on using the availability number of 99.5 over the MTBF I see I can have a pump failure lasting 43.8 hours during a year, assuming the pumps run 24/7 8766 hours/year

If a pump has an uptime of 8722 (99.5% of 1 Year). It means the pump can be down almost 44 hours.

With one spare I can cover the repair time (48 hours) or almost. But this is one pump failing. If a second pump was to fail then I am out of commission. I would like to understand the low probability of a second pump failing at the same time.

Sorry, I don't get this. When a pump fails, you replace it with a spare (which takes $1$ hour if I understood you correctly). The repair time may be $48$ hours, but the downtime is just $1$ hour. Also, I don't understand why a $1$-year period is relevant to the problem.

To zzqv9p: I'm confused too: what exactly do you mean by "availability" of a pump? Is it true that the pump can be in just two states: operational (when it is available) and failed (when it is not available and requires repairs)? If so, the MTBF completely determines the entire process of failing, which is just the Poisson process with intensity 1/MTBF for one pump and 5/MTBF for five pumps. We could figure out a lot of things from here but what bothers me is that "availability" in the above sense can also be figured out from this data combined with MTTR and the result doesn't quite fit the $99.5\%$ figure you mentioned, so, unless "availability" means something else, your data are self-contradictory and I would hate to proceed any further in such situation.

To mlok: I think we should assume that the repair of a pump is also governed by a Poisson process with intensity $1/24$. This makes the event that the pump is not repaired in $48$ hours not very likely (the exact probability is $e^{-2}$) but quite possible. Anyway, to answer the question, we'll need some model of the repair process, not just the mean. As to the one hour period for switching pumps, I think it can be safely neglected (we get a worse situation if the switch is immediate and the mean time to repair is increased to $25$ hours). As far as I understand, running fewer than all $5$ pumps for an extended period of time is some kind of disaster but I still don't understand how bad it is: one will have to live with the one-hour periods time to time just to switch pumps, so why not with $24$, or $48$, or whatever else hours? Usually one tries to balance the loss one incurs because of pump failures against the cost of acquiring and maintaining spare pumps but here we are just not given the corresponding data at all.

Anyway, I'm not a reliability specialist either and I'll be out for the next few days. I just tried to throw in my 2 cents :).

I did some checking and the MTBF is for one part of the pump. The proper value I was told to use is the availability value. I can figure out the entire pump MTBF by using
Availibility = mtbf/(mtbf+mttr)

We must try to run all 5 pumps to ensure we provide enough pressure in the system. Our service level requires 5 pumps to run.

What we would like to show is we can keep x pumps spare to always be able to have 5 pumps running.

Hope this helps.


I use two index fingers and pop the bubbles as fast as I can. I think I got 175 with that strategy.

My high is 212, I pop all the bubbles first and then start mass tapping with eight fingers.

My friends just cheat by bringing in multiple people. They do it on the bus and high score is about 208.

I'm so sorry for reviving this, but I didn't feel like making a new thread.

My record is 290, using six fingers.


Equation: $\sin x+\sin 2x+\sin 3x=1$.
Find all $x$.

hello, this have been solved already.
Sonnhard.


I want to know the coding that you can do to any webpage to produce the same effect as [hide] this [/hide]

It's not that hard using JavaScript  :wink:

[code]
Click to reveal hidden content

here is the hidden content

[/code]

thanks pontios, I was hoping for a javascript solution since I can easily embed that.

[quote="myyellowducky82"]Do you have to use a special program?[/quote]

I was wondering that too

A special program? The only thing you need is a web browser and Notepad.

And not necessarily Notepad, g-edit works too.

How do you make some part of the text in the post hidden?

[code][hide]This is a hidden text[/hide][/code]

has the following effect:

[hide]Lorem ipsum[/hide]

and the code

[code][hide=Another hidden text]Lorem ipsum[/hide][/code]

produces

[hide="Another hidden text"]Lorem ipsum[/hide]

The latter is used when you want some text other than the default one in the link.


I'm doing exponential equation in my math class now and it's pretty challenging. I get them but I'm just technically not SO sure.
Here is a question and answers.

[tex]7^{x-2} = 5^x[/tex]
[tex]\log 7^{x-2} = \log 5^x[/tex]
[tex](x-2) \log 7 = x \log 5[/tex]
[tex]x \log 7 - 2 \log 7 = x \log 5 [/tex]
[tex]x \log 7 - x \log 5 = 2 \log 7[/tex]
[tex]x (\log 7 - \log 5 ) = 2 \log 7[/tex]
[tex]x = \frac {2 \log 7}{\log 7 - \log 5}[/tex]
[tex]x = \frac {2(0.8451)}{0.8451-0.6990}[/tex]
[tex]x = \frac {1.6902}{0.1461}[/tex]
[tex]x  \approx {11.5688} [/tex]


ratio
parameterization
geometry
exterior angle

hi,
i keep thinking about this....an angle has rayz for 2 sidez and a vertex. but if there were an angle 0 then there is one ray only, but how is that an angle if an angle is supposed to have 2 rays its sides???! also an angle of zero has no interior.

does there exist an angle 0? i say no, but i want some confirmation.

tnx and God bless

Sure there is a 0 degree angle.  You just need your two rays to coincide with each other.

but if 2 rays coincide doesnt that mean they r the same ray then only a single ray?

tnx and God bless

Yeah.  Draw a ray.  Label it $ \overrightarrow{AB}$.  Now draw point $ C$ on this ray.  Then $ \overrightarrow{AC}$ coincides with $ \overrightarrow{AB}$, does it not?  So $ \angle BAC\equal{}0^\circ$.

i'm still not convinced...

suppose that a ray is a set of points including the endpoint...your example says that AC coincides with AB. but by the definition of the ray, AC and AB has the same elements!!!!

then ray AB = ray AC (i am equalling sets of points)...

then ray AB and ray AC are the same ray.... how can an angle have two sides of the same ray? my definition of angle in my book says the angle is the union of two rays with the same vertex....

im sorry if i'm being a pain, but i only want to understand.....

Does your book say that angle is the union of two [b]different[/b] rays?  Sure, ray AB and ray AC have the same set of points.

no, it says two rays....so it is ok for an angle to have its 2 sides as one ray? that's weird, but ok.

i will tell you how this begun

i am trying to prove that an exterior angle is always greater than the nonadjacent interior angles. c is the measure of exterior angle, a is the measure of interior angle and b is the ange adjactnet to it

i showed that c=a+b, then i use partition property that c>a if b>0 ... but one thing that bothered me is what if the angle was 0? i have not showed that the angle of measure b is > 0, and it should be obvious since no interior angle of a triangel can be 180... but i am questioning the existesness of such an angel.

I think you're overthinking this.  Just think of a clock.  The hands are essentially rays.  When it's 12:00, the two rays overlap each other and form an angle of 0 degrees.

I returned to the definition, and it kind of makes sense now. Thx.

Hello!!! Find 3 points A,B and C such that (A-B-C) and then $ \angle {BAC}\equal{}0^\circ$

That's my sign for 3 collinear points in that order.

You could choose all three points to be the same point.  I'm not sure if that's what you have in mind but it works.

[quote="Bugi"]Hello!!! Find 3 points A,B and C such that (A-B-C) and then $ \angle {BAC} \equal{} 0^\circ$[/quote]

uh, if (A-B-C), then BAC has to be zero? or do you mean something else?

JRAV: actually, three coinciding points do not define an angle.

An angle is defined as the ratio of arc length (along a circle of radius $ r$), $ s$, to the length of the radius.

That is an angle $ \theta \equal{}\frac{s}{r}$.

For example for a complete circle $ s \equal{}2\pi r$. Consequently $ \theta \equal{}\frac{2\pi r}{r} \equal{} 2\pi$. Of course $ \theta$ is in units of radians when calculated this way.

Considering only the definition, you're asking if $ \theta \equal{} 0$ is possible?

Sure, because

$ \theta \equal{} 0\implies \frac{s}{r} \equal{} 0 \\ \therefore s \equal{} 0$

$ s \equal{} 0$ just means that if you start at an arbitrary point on a circle, you haven't moved any distance along that circle.

The two rays that coincide are not the same rays, since they are given different names, like AB and AC. Thus, since they are collinear, the angle is 0 degrees.

The definition of an angle is:

The canonical parameter of the one parameter compact Lie-Group.

So there must exist an angle 0...

I don't think that definition is of any use to high school students.

Basically, a straight line is an angle of zero because it is formed of an indefinite amount of rays.

What about $ 180^\circ$? Isn't that why it's called the straight angle?

If A-B-C then $ \angle ABC\equal{}180$ and $ \angle BAC\equal{}0$

I think that it really depends on what you define to be an angle.

For example, in the Moise-Downs Geometry book, an angle is defined as follows: "If two rays have the same end point, but do not lie on the same line, then their union is an angle. The two rays are called its sides, and their common end point is called its vertex. If the rays are ray AB and ray AC, then the angle is denoted by angle BAC or angle CAB."

Clearly, in the Moise-Downs book, there is no such thing as a straight angle or a 0 degree angle.

Now take Geometry by Jurgensen, Brown, and Jurgensen, published by McDougal Littell, an angle is defined as follows: "An angle is the figure formed by two rays that have the same endpoint. The two rays are called the sides of the angle, and their common endpoint is the vertex of the angle. ... Straight angle: measure 180"

So what's my opinion? I think that it is important here to recall that there is no set, one geometry. If you assume different postulates (like the assumption that Euclid's fifth postulate - through a point not contained by the line, there is one and only one line that can be drawn through the point parallel to the line - if you assume that there are more or 0 lines that can be drawn parallel to the given line through the point, you will stumble upon the existence of hyperbolic geometry and alternate geometries that are still applicable), you get different geometries. So it really depends on how you define an angle. When the Moise-Down book states that a straight line really isn't an angle, it doesn't necessarily mean that there isn't such a thing as a straight angle (in this case, there is such a thing as a straight angle), but it just means that there isn't a straight angle in their geometry book (as they can set the parameters because it is their geometry).

[quote="theax"]
Clearly, in the Moise-Downs book, there is no such thing as a straight angle or a 0 degree angle.
[/quote]

Moise-Downs is most definitely not standard; it's a highly controversial textbook. I doubt that a significant number of mathematicians would consider the definition of an angle given in it legitimate.

[quote="theax"]True. However, it is used to illustrate my point: very few geometry books use exactly the same theorems, the same postulates, or the same definitions. Yet they are not wrong because geometry is based on your assumptions - so if you assume different things, then you end up with results.[/quote]

If you assume different postulates, you assume an entirely different geometry.

High school cirricula in the US teach the geometry circumscribed by the postulates of Euclid's [i]Elements[/i].

In some contexts, we can reasonably talk about a zero angle, even if it's an angle between two different objects. That's the angle a circle makes with its tangent line, for example. It is an issue of definitions.

[quote="Dr. No"]High school curricula in the US teach the geometry circumscribed by the postulates of Euclid's [i]Elements[/i].[/quote]Not really. Hardly anyone uses the Euclidean form of the parallel postulate, instead using a simpler version from much later. The [i]Elements[/i] is a textbook, and it has been replaced by modern expositions of similar material. Its axioms are also incomplete for what it tries to do; we really need axioms on the notion of "between", without which even the first proposition in Euclid can't be proved.

[quote="jmerry"]
[quote="Dr. No"]High school curricula in the US teach the geometry circumscribed by the postulates of Euclid's [i]Elements[/i].[/quote]

Not really. Hardly anyone uses the Euclidean form of the parallel postulate, instead using a simpler version from much later. The [i]Elements[/i] is a textbook, and it has been replaced by modern expositions of similar material. Its axioms are also incomplete for what it tries to do; we really need axioms on the notion of "between", without which even the first proposition in Euclid can't be proved.[/quote]

Let me clarify my remark. I did not say that US high school students used [i]The Elements[/i] as their textbook. I said the geometry taught in US high schools was [b]circumscribed[/b] by the postulates of [i]The Elements[/i], i.e. the postulates of Euclid.  To circumscribe means to encircle, to contain, or to bound.  Thus what is taught in US high schools is a subset of the postulates of Euclid, postulates he published in [i]The Elements[/i].

The weakness of the parallel postulate is well known. Even excluding the parallel postulate, high school students learn geometry based upon Euclid's postulates. Almost no one is taught Lobachevsky's geometry for example.  My point however was not that we learn geometry from some wizened ancient in a toga, but that the postulates one uses define the geometry one learns.

You missed my other point- that Euclid's postulates are incomplete. We can build a geometry in which Euclid's postulates hold and there are no equilateral triangles. We teach more than those postulates, and even Euclid implicitly assumed more. If you want a set of postulates that actually completely describe the geometry we call Euclidean, look to the late 19th century.

OK, this is now turned into discussion of books, and not the question of 0 degree angle.


If you are a parent, teacher, or student in New Jersey involved in high school math stuff, you'll surely be interested in the new Princeton Math Competition (read about it [url=http://www.artofproblemsolving.com/Forum/index.php?f=414]here[/url]!). :D It's going to be awesome. I hope you can come.


function
calculus
derivative
limit
algebra
polynomial
real analysis

1) Let $ f$ be four times differentiable even function on $ \mathbb{R},$ $ g(x) \equal{} f(\sqrt {x}\,), x\ge 0.$ Prove that $ g$ is twice differentiable at $ x \equal{} 0$ and find $ g''(0)$ in terms of derivatives of $ f$ at zero. (A. Kukush)

2) Let $ f$ be $ 2k$ times differentiable even function on $ \mathbb{R},$ $ g(x) \equal{} f(\sqrt {x}\,), x\ge 0.$ Prove that $ g$ is $ k$ times differentiable at $ x \equal{} 0$ and find $ g^{(k)}(0)$ in terms of derivatives of $ f$ at zero. (A. Kukush, V. Brayman)

The first is not so difficult, with a little care.

We compute $ g'(x)\equal{}{1\over 2} x^{\minus{}{1\over 2}} f'(\sqrt{x})$,
$ g''(x)\equal{}{1\over 4} x^{\minus{}1} f''(\sqrt{x}) \minus{} {1\over 4} x^{\minus{}{3\over2}}f'(\sqrt{x})$.

We write this as $ g''(y^2)\equal{}{{yf''(y)\minus{}f'(y)}\over 4y^3}$.  Now we apply L'Hopital's rule twice:

$ g''(0)\equal{}\lim_{y\minus{}>0} {{yf''(y)\minus{}f'(y)}\over 4y^3}\equal{}\lim_{y\minus{}>0} {{yf'''(y)}\over 12y^2}\equal{}\lim_{y\minus{}>0} {{f'''(y)}\over 12y}\equal{}\lim_{y\minus{}>0} {{f''''(y)}\over 12}$ $ \equal{}{1\over 12}f^{(4)}(0)$.

The generalization of this is below:

What we should note is that the above process for calculating the derivatives of $ g$ at $ 0$ is entirely dependent only on the first $ 2k$ derivatives of $ f$.  Specifically, I mean that we can write $ g^{(k)}(x^2)$ as $ x^{\minus{}N}\sum_{j\equal{}1}^{k} P_j(x) f^{(j)}(x)$, where the $ P_j$ are polynomials, independent of the choice of $ f$.

Let $ p(x)\equal{}\sum_{i\equal{}0}^{k} a_k x^{2k}$, so $ f$ is even.  Set $ q(x)\equal{}p(\sqrt{x})$.  Then we can write $ q(x)\equal{}\sum_{i\equal{}0}^{k} a_k x^k$, so that $ q^{(k)}(0)\equal{}k! a_k\equal{}{k!\over{(2k)!}}p^{(2k)}(0)$.  But since we can also compute $ q^{(k)}(0)$ by successive applications of L'Hopital's rule to the limit $ \lim_{x\minus{}>0} {x^{\minus{}N}\sum_{j\equal{}1}^{k} P_j(x) p^{(j)}}(x)$, we must get the same result.  We can set $ a_k$ so that the first $ 2k$ derivatives of $ f$ and $ p$ agree at $ 0$, (the odd derivatives of an even function must vanish at $ 0$) then, by the above, we have:

$ g^{(k)}(0)\equal{}\lim_{x\minus{}>0} {x^{\minus{}N}\sum_{j\equal{}1}^{k} P_j(x) f^{(j)}(x)}\equal{}\lim_{x\minus{}>0} {x^{\minus{}N}\sum_{j\equal{}1}^{k} P_j(x) p^{(j)}(x)}\equal{}q^{(k)}(0)$

$ \equal{}{k!\over{(2k)!}}p^{(2k)}(0)\equal{}{k!\over{(2k)!}}f^{(2k)}(0)$, and we are done.

The above proof has a serious flaw.  From what I've said, we are not guaranteed that we will only see the first $ 2k$ derivatives of $ f$--we may see up to $ k \plus{} N$ of them.  (in general $ N \equal{} 2k\minus{}1 > k$)  To be clear, this proof still works if all of these derivatives exist, but we are not guaranteed of this fact.

Perhaps there is a simple way to fix it?

for 2/

Use Faa Di Bruno's formula  you will find it in

http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno's_formula


algebra
polynomial
function
Rational Root Theorem
algebra unsolved

There is a problem from 6th edition in this site:
factor x^2005-2005x+2004 as a product of irreducible polynomial in Z

I'm new to this site so I dont now where the solution locates
Can anyone help me?

thanks

Sorry, I just notice this problem was discussed several days ago.
But it still didn't yield a solution

$ x^{2005}\minus{}2005x\plus{}2004$

Consider $ x^{n}\minus{}nx\plus{}n\minus{}1$

Clearly, $ x\equal{}1$ is a factor, since the coefficients sum to 1.

Hence, we get

$ (x\minus{}1)(x^{n\minus{}1}\plus{}x^{n\minus{}2}\plus{}x^{n\minus{}3}\plus{}\ldots\plus{}x^2\plus{}x\minus{}(n\minus{}1))$

The 2nd polynomial also factors, seeing as there are $ n\minus{}1$ cases of $ 1x^j$ and an $ \minus{}(n\minus{}1)$

We then get (assuming the degree > 2)

$ (x\minus{}1)^2(x^{n\minus{}2}\plus{}2x^{n\minus{}3}\plus{}\ldots)$

Finally, we need to show that this is irreducible.

By rational root theorem, the root should be of the form $ \pm \frac{n\minus{}1}{1}$

Since all the terms are positive, the root must be negative. I'm not sure where to go from here. Help would be appreciated.

Consider the tangent line of $ y\equal{}x^n$ at the tangency of point $ (1,\ 1)$.

[quote="kunny"]Consider the tangent line of $ y \equal{} x^n$ at the tangency of point $ (1,\ 1)$.[/quote]

Errmm... $ y\equal{}nx^{n\minus{}1}\minus{}n$?

I don't understand. Sorry

[quote="silouan"]The problem is quite difficult . For a proof see here
http://www.mathlinks.ro/viewtopic.php?t=124596[/quote]

I'm afraid the site is down:

The requested URL /borisov/Documents/papers/abc.pdf was not found on this server.

Could you give us an outline please

Why a polynomial is irreducible if it doesn't have integer roots
it may still be factored as two polynomial where each one doesn't have a integer root

Sorry, I can not open the link for the lemma.

[url]http://www.pitt.edu/~borisov/Documents/papers/abc.pdf[/url]

i believe this is the link we are talking about :) and this is working

Is there a document similar to that polynomial one contains details about methods of solving function equations?(With example, exercises, and solutions...)


geometry
parallelogram
geometry proposed

Let $ ABCD$ be a parallelogram. $ E$ and $ F$ be points on $ AD$ and $ CD$ respectively, such that $ BE \perp AD$ and $ AF \perp FB$. A line through point $ E$ parallel to $ CD$ intersects $ BF$ at point $ G$. $ H \equal{} AF \cap BE$ and $ I \equal{} DH \cap BC$. Prove that $ FI \perp GH$

[color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABC$ be a triangle with the orthocenter $ H$.

Denote $ \{\begin{array}{c}
E\in BH\cap AC\\\
F\in CH\cap AB\end{array}$ , $ \{\begin {array}{c}
X\in AC\ ,\ FX\parallel BC\\\
Y\in AB\ ,\ EY\parallel BC\\\
Z\in YH\ ,\ ZC\parallel AB\end{array}$. Prove that $ EZ\perp HX\ .$[/color]

[color=darkblue]Maybe such the our nice problem becomes more easily. Good luck ![/color]


1) Engineers want to connect two parts of an unfinished road. The two parts are parallel, but end at points A and B respectively. We draw line segments AC and BC are perpendicular at point C with AC perpendicular to the two unfinished roads (BC is parallel). The engineers join the two parts of the road with two congruent circular curves. Each curve is tangent to both one part of the road where the other curve (at the midpoint of line segment AB). Given that AC is 1200 meters and BC is 900 meters, what is the radius of the two curves in meters?

2) The sum and product of five positive integers x,y,z,u,v are equal. If
   x ≤ y ≤ z ≤ u ≤ v, how many distinct solutions (x,y,z,u,v) are there?

[hide="Slution of the second"]

$ xyzuv\equal{}x\plus{}y\plus{}z\plus{}u\plus{}v\le v\plus{}v\plus{}v\plus{}v\plus{}v\equal{}5v$,so we have $ xyzu\le 5$.

(1)$ xyzu\equal{}1$
only $ x\equal{}y\equal{}z\equal{}u\equal{}1$ is possible and $ 4\plus{}v\equal{}v$ doesnt work.

(2)$ xyzu\equal{}2$
only $ x\equal{}y\equal{}z\equal{}1,u\equal{}2$ is possible and $ 5\plus{}v\equal{}2v$ i.e.$ v\equal{}5$ ok.

(3)$ xyzu\equal{}3$
only $ x\equal{}y\equal{}z\equal{}1,u\equal{}3$ is possible and $ 6\plus{}v\equal{}3v$ i.e.$ v\equal{}3$ ok.

(4)$ xyzu\equal{}4$
there are two case:$ (x,y,z,u)\equal{}(1,1,1,4)(1,1,2,2)$

when the first case,$ 7\plus{}v\equal{}4v$ doesn't work.
when the second case,$ 6\plus{}v\equal{}4v$i.e. $ v\equal{}2$ ok.

(5)$ xyzu\equal{}5$
only $ x\equal{}y\equal{}z\equal{}1,u\equal{}5$ is possible but $ 8\plus{}v\equal{}5v$ i.e. $ v\equal{}2<u\equal{}5$ so doesn't work.

Therefore we have $ (x,y,z,u,v)\equal{}(1,1,1,2,5)(1,1,1,3,3)(1,1,2,2,2)$,so there are $ \boxed{3}$ solutions.

[/hide]

[hide="solution 1"]
Ive never used geogebra before,but this was way too hard to explain without a diagram
[note: diagram not to scale]

 [geogebra]04f5bc72d54708e4dfc487643ed329d5634e17cf[/geogebra]

Due to the fact that the circles meet at the midpoint of $ ab$ and are both tangent then both circles must have the same radius
The distance between the two parallel lines is 900 and the distance between the radius of the circle is 1200
After constructing the lines in the diagram we get a right angle triangle with side lengths; $ 1200, 900\minus{}2r$ and $ 2r$, therefore:
$ (2r)^2 \equal{} 1200^2 \plus{} (900\minus{}2r)^2$
$ r\equal{}625$
 [/hide]


@above,how did you get $\frac{4^x}{10^x}???$

@above basically you divide everything by $25^x$ and get $$\frac{4^x}{25^x} + \frac{10^x}{25^x} = 1$$
$$\implies \left(\frac{2}{5}\right)^{2x} + \left( \frac{2\cdot 5}{5\cdot 5}\right)^{x} = 1$$
$$\implies \left(\frac{2}{5}\right)^{2x} + \left(\frac{2}{5}\right)^{x} = 1$$
Which is the same equation as in #2 post

[quote=inom]$4^x+10^x=25^x$
find x.[/quote]

$x \in \mathbb{Z}$ or $x \in \mathbb{R}$ ?

[quote=TuZo][quote=starchan][hide = Just a Remark]The reason this works is because $4 \times 25 = 10^2$ [/hide][/quote]

Else after the division, the LHS it is descreasing, the RHS is constant, so we have maximum 1 solution.[/quote]

Yeah but I think in that case, the problem can't be solved within the range of Elementary Mathematics.

Yes, because in this case we can not quess the solution.  :(

[quote=HamstPan38825]Isn't it extremely simple to substitute $a=2^x$, $b=5^x$ to yield $$a^2+ab-b^2=0$$ which is homogeneous?[/quote]

Yes, and the solution for this it is: ${{a}^{2}}+ab-{{b}^{2}}=0/:{{b}^{2}}\Leftrightarrow {{\left( \frac{a}{b} \right)}^{2}}+\frac{a}{b}-1=0\Leftrightarrow {{x}^{2}}+x-1=0$

[quote=Eat314]Please put solutions in hide tags.[/quote]

It is to late!

@TuZo: You can edit your post by clicking the pencil shaped icon near your post. Then, click on the eye with a line through it to create a hide tag. Put your text inside of the hide tag.

I know, but I reffered to the older posts, because in my post do not exist complete solution, only an idea, and what's the point to hidde my idea, when all the previouse complete solutions there are not hiddens?  ;)


Let $ D$ be a point on the side $ BC$ of $ \triangle{ABC}$.
and let $ E$ and $ F$ be the incenters of triangles $ ABD$ and $ ACD$,respectively
Suppose that $ B,C,E,F$ are concyclic.
Prove that $ \frac {AD \plus{} BD}{AD \plus{} CD} \equal{} \frac {AB}{AC}$

Posted before, see
[url]http://www.mathlinks.ro/viewtopic.php?t=62431[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=172995[/url]