quadratics
ratio
algebra
algebra unsolved

Let $ \alpha$ and $ \beta$ the roots of the quadratic equation (x – 2)(x – 3) + (x – 3)(x + 1) + (x + 1)(x – 2) = 0.
Determine the value of:

[img]http://img121.imageshack.us/img121/192/calsf.jpg[/img]

Can anyone explains me step by step how to do it ?
Thanks.

We can write $ (x \minus{} 2)(x \minus{} 3) \plus{} (x \minus{} 3)(x \plus{} 1) \plus{} (x \plus{} 1)(x \minus{} 2) \equal{} \boxed{3}(x \minus{} \alpha)(x \minus{} \beta)$, so plugging $ x \equal{} \minus{} 1,\ 2,\ 3$ into the equality gives

$ (\alpha \plus{} 1)(\beta \plus{} 1) \equal{} 4,\ (\alpha \minus{} 2)(\beta \minus{} 2) \equal{} \minus{} 1,\ (\alpha \minus{} 3)(\beta \minus{} 3) \equal{} \frac 43$, yielding the desired answer is $ \boxed{0}$.

Nice, kunny.

I found the roots of $ 3x^{2} \minus{}8x \plus{}1\equal{}0$ as $ \frac{4 \pm \sqrt{13}}{3}$ and then when I substituted it was easy because the difference of two squares gave $ \frac{1}{4} \minus{}1 \plus{} \frac{3}{4} \equal{} 0$. (Not as neat as yours and not as fast.)

[quote="kunny"]
$ (\alpha \plus{} 1)(\beta \plus{} 1) \equal{} 12,\ (\alpha \minus{} 2)(\beta \minus{} 2) \equal{} \minus{} 3,\ (\alpha \minus{} 3)(\beta \minus{} 3) \equal{} 4$, yielding the desired answer is $ \boxed{0}$.[/quote]

Hi kunny, I don't understand this line, how do you know that $ (\alpha \plus{} 1)(\beta \plus{} 1) \equal{} 12$  ?
And how do you do the orders ?
Thanks.

Hi Luis. As kunny's not on-line, I'll try to answer your question:

Substituting $ x\equal{}\minus{}1$, we get $ \minus{}3\times \minus{}4 \plus{} 0 \plus{}0 \equal{} 3(\minus{}1\minus{}\alpha)(\minus{}1\minus{}\beta)$, that is,

$ 4\equal{} (1\plus{}\alpha)(1\plus{}\beta)$.

Substituting $ x\equal{}2$, we get $ 0 \plus{} \minus{}1\times 3 \plus{}0 \equal{} 3(2\minus{}\alpha)(2\minus{}\beta)$

$ \minus{}1 \equal{} (\alpha \minus{} 2)(\beta \minus{}2)$.

Substituting $ x\equal{}3$, we get $ 0 \plus{} 0 \plus{} 4 \times 1 \equal{} 3(3\minus{}\alpha)(3\minus{}\beta)$

$ \frac{4}{3} \equal{} (\alpha \minus{} 3)(\beta \minus{} 3)$.

Substituting these, we get $ \frac{1}{4} \minus{}1 \plus{} \frac{3}{4} \equal{} 0$.

Hope this makes sense

Sorry, my first solution was wrong.

Thank you for the expalnation in place of me, Andrew Tom.

You're welcome, kunny. I'd hoped you wouldn't mind as you weren't on-line. After I'd posted, I noticed that you were back.

[quote="AndrewTom"]Hi Luis. As kunny's not on-line, I'll try to answer your question:

Substituting $ x \equal{} \minus{} 1$, we get $ \minus{} 3\times \minus{} 4 \plus{} 0 \plus{} 0 \equal{} 3( \minus{} 1 \minus{} \alpha)( \minus{} 1 \minus{} \beta)$, that is,

$ 4 \equal{} (1 \plus{} \alpha)(1 \plus{} \beta)$.

Substituting $ x \equal{} 2$, we get $ 0 \plus{} \minus{} 1\times 3 \plus{} 0 \equal{} 3(2 \minus{} \alpha)(2 \minus{} \beta)$

$ \minus{} 1 \equal{} (\alpha \minus{} 2)(\beta \minus{} 2)$.

Substituting $ x \equal{} 3$, we get $ 0 \plus{} 0 \plus{} 4 \times 1 \equal{} 3(3 \minus{} \alpha)(3 \minus{} \beta)$

$ \frac {4}{3} \equal{} (\alpha \minus{} 3)(\beta \minus{} 3)$.

Substituting these, we get $ \frac {1}{4} \minus{} 1 \plus{} \frac {3}{4} \equal{} 0$.

Hope this makes sense[/quote]

Thanks Andrew, the last question: why this line :

 $ \minus{} 3\times \minus{} 4 \plus{} 0 \plus{} 0 \equal{} 3( \minus{} 1 \minus{} \alpha)( \minus{} 1 \minus{} \beta)$, that is,

$ 4 \equal{} (1 \plus{} \alpha)(1 \plus{} \beta)$.

Why the division of 12 / 3 the ratio was positive and the equation has changed its sign?

$ (x \minus{} 2)(x \minus{} 3) \plus{} (x \minus{} 3)(x \plus{} 1) \plus{} (x \plus{} 1)(x \minus{} 2) \equal{} 3x^2 \minus{} 8x \plus{} 1 \equal{}0$.
So, $ \alpha \plus{} \beta \equal{} \frac{8}{3}$, $ \alpha \beta \equal{} \frac{1}{3}$.
Therefore, $ (\alpha \plus{} x)(\beta \plus{} x) \equal{} \alpha \beta \plus{} x(\alpha \plus{} \beta) \plus{} x^2 \equal{} \frac{1\plus{}8x\plus{}3x^2}{3}$  i.e.  $ \frac{1}{(\alpha \plus{} x)(\beta \plus{} x)} \equal{} \frac{3}{1\plus{}8x\plus{}3x^2}$.

Hence,
$ \frac{1}{(\alpha \plus{} 1)(\beta \plus{} 1)} \equal{} \frac{3}{1\plus{}8 \times 1\plus{}3 \times 1^2} \equal{} \frac{1}{4}$,
$ \frac{1}{(\alpha \minus{} 2)(\beta \minus{} 2)} \equal{} \frac{3}{1\plus{}8 \times (\minus{}2)\plus{}3 \times (\minus{}2)^2} \equal{} \minus{}1$,
$ \frac{1}{(\alpha \minus{} 3)(\beta \minus{} 3)} \equal{} \frac{3}{1\plus{}8 \times (\minus{}3)\plus{}3 \times (\minus{}3)^2} \minus{} \frac{3}{4}$.

Thus $ \frac{1}{(\alpha \plus{} 1)(\beta \plus{} 1)} \plus{} \frac{1}{(\alpha \minus{} 2)(\beta \minus{} 2)} \plus{} \frac{1}{(\alpha \minus{} 3)(\beta \minus{} 3)} \equal{} \frac{1}{4} \minus{} 1 \plus{} \frac{3}{4} \equal{} \boxed{0}$.


[quote="Sung-yoon Kim"]I think this is somewhat tough, but I have a nice proof.  :lol: Let me express $\sum_{sym}a^{p}b^{q}c^{r}$ as $(p,q,r)$.

For any positive $a,b,c$, prove that $(6,1,1)+(4,4,0)+3(3,3,2) \geq 2(5,2,1)+2(4,3,1)+(4,2,2)$.[/quote]

Hi Kim, try this one:

http://www.mathlinks.ro/Forum/viewtopic.php?p=618542

Let $f(a, b, c) = LHS-RHS$. Let $abc=1$. Then we need to show
$f\left(a, b, \frac{1}{ab}\right) \ge 0$
But
$a^{5}b^{5}f\left(a, b, \frac{1}{ab}\right) = 2S_{a}S_{b}$
where
$S_{a}= a^{5}b+a^{4}b^{5}+1-a^{2}b-a^{3}b^{3}-a^{4}b^{2}$
and $S_{b}$ the same with $a$ and $b$ switched. So we need only show that

$a^{5}b+a^{4}b^{5}+1 \ge a^{2}b+a^{3}b^{3}+a^{4}b^{2}$
what is done by
$a^{5}b+4a^{4}b^{5}+2\ge 7a^{3}b^{3}$
$4a^{5}b+2a^{4}b^{5}+1\ge 7a^{4}b^{2}$
$2a^{5}b+a^{4}b^{5}+4 \ge 7a^{2}b$

:D :D :D :D

I still struggle to find nice mixing variable solution. Sung-yoon Kim, what is your solution?


logarithms
trigonometry
calculus
calculus computations

$\begin{gathered} \left( {f \circ g} \right)^\prime \left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right) \hfill \\ \left( {f \cdot g} \right)^\prime \left( x \right) = f'\left( x \right) \cdot g\left( x \right) + f\left( x \right) \cdot g'\left( x \right) \hfill \\ \end{gathered}$
 :?


Let $ n \in \mathbb{N}$ and $ n > 6$ Then prove
$ n \minus{} 1$ and $ n \plus{} 1$ are primes $ \implies$  $ 720/n^2(n^2 \plus{} 16)$

Is the reciprocal true?

Morrocan olympiad  :blush: .

$ n \minus{} 1$ and $ n \plus{} 1$ are primes
$ \Rightarrow n\equiv 0[6],\ and\ n\equiv 2\ or\ \minus{} 2[5]$
$ \Rightarrow n\equiv 0[6],\ and\ n^2 \plus{} 16\equiv 0[5]$
$ \Rightarrow 6^2\times 4|n^2(n^2 \plus{} 16)\ and\ 5|n^2(n^2 \plus{} 16)$
$ \Rightarrow 720 \equal{} 6^2\times 4\times5|n^2(n^2 \plus{} 16)$


If $a , b , c \in \mathbb{R}$ and $a^{2}+b^{2}+c^{2}= 9$ , prove that :

$3min$ {$a,b,c$} $\le 1+abc$ .

                                                                                Author : Virgil Nicula

PS : I'm searching a solution for  the case $a , b , c > 0$ .

Posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=34669

And similar problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=127737


abstract algebra
group theory
superior algebra
superior algebra solved

Find all the irreducible $\mathbb{Z}$-modules

if irreducible module=simple module, then this is equivalent to finding all the abelian groups $G$ without proper subgroups

$Z_p$ with p prime would be some of them(if not all)

yes $Z_{p}$ with p prime is everything

take a random nontrivial in the group $G$
it will generate a cyclic nontrivial subgroup
thus it is the full group
but if its order is infinite, well then we have Z and that does have more subgroups
so the order is finite, but if it is not prime we would be able to construct nontrivial subgroups again


vector
analytic geometry
linear algebra
matrix
induction
linear algebra unsolved

Let $ M$ be the a set consisting of vectors in $ \mathbb{R}^n$, such that:
[b](1)[/b] every vector in $ M$ has all its coordinates from the set $ \left\{1,\minus{}1\right\}$;
[b](2)[/b] every two different vectors in $ M$ must be orthogonal with each other.

Find the maximum possible cardinality of $ M$ (depending on $ n$).

I can do the case when $ n$ is an odd number :) - the cardinality is $ 1$.

The Hadamard conjecture says that there are $ n$ such vectors when $ n$ is divisible by $ 4$, but this remains open as far as I know.  :o  See: [url=http://en.wikipedia.org/wiki/Hadamard_matrix]Hadamard matrix[/url].

I don't know about the case $ n\equiv 2\mod 4$.  :maybe:

What can definitely be said: the maximal cardinality of $ M$ is at least $ 2^q$, where $ q$ is the greatest integer $ s$ such that $ 2^s\mid n$.

The existence of a set $ M$ with $ \left|M\right| \equal{} 2^q$ is proved by induction (step $ n\mapsto 2n$). See http://www.mathlinks.ro/Forum/viewtopic.php?t=15359 for details in the case of $ n \equal{} 2^q$; for the case $ n \equal{} 2^qs$ with $ s$ being odd, either proceed similarly, or split the $ 2^qs$ coordinates of the space $ \mathbb{R}^n$ into blocks of $ s$ coordinates each, and apply the $ \mathbb{R}^{2^q}$-construction to them.

The bound $ 2^q$ is sharp (i. e. $ \left|M\right|\leq 2^q$ holds for any set $ M$ satisfying [b](1)[/b] and [b](2)[/b]) for $ n$ being a power of two, of course (since the vectors in the set $ M$ must be linearly independent), and for $ n$ being odd.

The maximal cardinality of $ M$ is $ 2$ for $ n\equiv 2\mod 4$. This is a very easy combinatorial exercise.

  Darij

Indeed, let $ n \equal{} 4k \plus{} 2$. The first two vectors may as well be $ (1,\dots,1)$ and $ (1,\dots,1, \minus{} 1,\dots, \minus{} 1)$. The third vector must have $ 2k \plus{} 1$ coordinates equal to $ 1$ out of which $ m$ will be in the first half. Thus, the scalar product of the second and third vectors is $ m \minus{} (2k \plus{} 1 \minus{} m) \minus{} (2k \plus{} 1 \minus{} m) \plus{} m \equal{} 4m \plus{} 4k \minus{} 2$, which is nonzero.

One is tempted to do the same for the case $ n\equal{}8k\plus{}4$ and obtain a contradiction using four vectors, but :P

Thank you,Mlok and Darij.
Frankly,I have shown the case you talked when I post the topic.I have get the problem from a case when $ n \equal{} 64$

And my solution for $ n \equal{} 2^q$ by induction is consider the matrix:
(A  A
 A  -A)
where $ A$ is the matrix satisfies the $ \equal{} 2^{q \minus{} 1}$ case.

But it surprised me that it was so hard.

Thanks again

So,The smallest case we haven't solved is $ n\equal{}12$.
Have anybody get the result for $ n\equal{}12$?

According to the Wikipedia article cited above, the smallest unknown case of the Hadamard conjecture is $ n\equal{}668$. See also this accessible [url=http://scholar.lib.vt.edu/theses/available/etd-05042004-120929/unrestricted/thesis_revised.pdf]survey[/url] (a master's thesis, actually).

Thank you linboll and mlok.
It is unbelievable to me that for n=12 the answer is 12. :P

And to mlok,I don't know why in China I can't open the Wikipedia Page,But I can download the article you offered,Thanks.

Zhaobin: does this work?
http://anonymouse.org/cgi-bin/anon-www.cgi/http://www.wikipedia.org

  dg

Hi,Darij,
It works,But a little lag.Thank you :)


What is the value of $ x$ if $ |x \minus{} 1| \equal{} |x \minus{} 2|$?

$ \textbf{(A)}\ \minus{}\!\frac {1}{2}\qquad
\textbf{(B)}\ \frac {1}{2}\qquad
\textbf{(C)}\ 1\qquad
\textbf{(D)}\ \frac {3}{2}\qquad
\textbf{(E)}\ 2$

[quote="funcia"][quote="#H34N1"]What is the value of $x$ if $|x-1|=|x-2|$?

A. $-\frac{1}{2}$
B. $\frac{1}{2}$
C. $1$
D. $\frac{3}{2}$
E. $2$[/quote]
[hide]D, E and C didn't make sense negatives don't also so i tested, probably a better way[/hide][/quote]
Well, that is a good way to do it, but there is another way
[hide] Since |x-1|=|x-2|, we have that either x-1=x-2, which is never true, or x-1=-1(x-2).  Taking the second equation, we get x-1=2-x, so x=3/2.  Thus, the answer is D. [/hide]

[quote="#H34N1"]What is the value of $x$ if $|x-1|=|x-2|$?

A. $-\frac{1}{2}$
B. $\frac{1}{2}$
C. $1$
D. $\frac{3}{2}$
E. $2$[/quote]

[hide]We have either of the two equations:

$(1) x-1=x-2$
$(2) x-1=-(x-2)$

The first equation yields no solution but the second equation gives us $x=\frac{3}{2}$.

or you could just take the averafe of 1 and 2 to get $\frac{3}{2}$[/hide]

[quote="davidlizeng"]isn't that exactly what quatto did??[/quote]You can also see that $x$ is equidistant from 1 and 2 on the real line since the distance function in $\mathbb{R}^{1}$ is the absolute value :)


trigonometry
algebra
polynomial
calculus
calculus computations

the question is this:

y'' + 25y = 0, y(0) = 2, y'(0) = 5

then what I have is:

y = c1 + c2e^(-25t).
y' = -25c2e^(-25t)
5 = -25c2
so c2 = -1/5

and c1 is equal to 11/5

is this correct?

No, that's not correct, not as you've written the problem.

$ y\equal{}c_1\plus{}c_2e^{\minus{}25t}$ is the general solution of $ y''\plus{}25y'\equal{}0.$

But the equation you wrote is $ y''\plus{}25y\equal{}0.$

Different equation. Its general solution is $ y\equal{}c_1\cos 5t\plus{}c_2\sin 5t.$

[quote="Kent Merryfield"]No, that's not correct, not as you've written the problem.

$ y \equal{} c_1 \plus{} c_2e^{ \minus{} 25t}$ is the general solution of $ y'' \plus{} 25y' \equal{} 0.$

But the equation you wrote is $ y'' \plus{} 25y \equal{} 0.$

Different equation. Its general solution is $ y \equal{} c_1\cos 5t \plus{} c_2\sin 5t.$[/quote]

thanks, that was a silly mistake.. however can you guide me on how you get the sin and cos from?

We have a constant coefficient linear homogeneous ODE. For such equation, we guess that there is a solution of the form $ e^{rt}$ for some $ r.$ Put that back into the original equation as a check; the result will be a polynomial equation for $ r,$ with the degree of the polynomial equal to the order of the equation.

In this case, we get that $ r^2\plus{}25\equal{}0$ so $ r\equal{}\pm 5i.$ That means that $ y\equal{}e^{5it}$ and $ y\equal{}e^{\minus{}5it}$ are solutions, as are arbitrary linear combinations of those.

But the equation probably came from some physical source and there's a presumption that the solutions are real. Can we make real linear combinations of $ e^{5it}$ and $ e^{\minus{}5it}?$

One way would be $ y\equal{}\frac12e^{5it}\plus{}\frac12e^{\minus{}5it}\equal{}\cos 5t.$

Another way would be $ y\equal{}\frac1{2i}e^{5it}\minus{}\frac1{2i}e^{\minus{}5it}\equal{}\sin 5t.$

Since we can solve for all possible initial conditions by setting $ y\equal{}c_1\cos 5t\plus{}c_2\sin 5t,$ that is the complete general solution.

For another example: suppose the equation is $ y''\plus{}2y'\plus{}5y\equal{}0.$ Letting $ y\equal{}e^{rt}$ and checking gives us the polynomial equation $ r^2\plus{}2r\plus{}5\equal{}0$ which has roots $ r\equal{}\minus{}1\pm 2i.$ We could write the solution in terms of $ e^{(\minus{}1\plus{}2i)t}$ and $ e^{(\minus{}1\minus{}2i)t},$ but we usually recombine those to $ y\equal{}c_1e^{\minus{}t}\cos 2t\plus{}c_2e^{\minus{}t}\sin 2t.$

You get used to all of this appearing, in essentially the same way each time. Try looking in any standard (college) sophomore-level differential equations text, and you'll see all of this (along with how to handle nonhomogeneous equations and what to do in such exceptional cases as the polynomial equation having repeated roots.)


$\mathbb{Z}/ 6 \mathbb{Z}$
$S_3$

This was asked before.

If p and q are primes there are three possibilities :

p=q : there are exactly two groups of order $p q$ and they are abelian

$p\neq q$ but p does not divide q-1 nor does q divide p-1 : only the cyclic group

one divides the other minus one : exactly two groups , one cyclic and one nonabelian

[quote="spx2"]could you explain your solution zetax or maybe post a link to a solution[/quote]
If there is an element of order 6, it's $Z_6$.
By Cauchy there exists an element of order 3 and one of order 2. Let $a^3 = e$, $b^2 = e$, and consider $e,a,a^2,b,ab,a^2b,ba, ba^2$
They form a group, so some elements coincide. It suffices to consider $ab = ba$ and $ab = ba^2$ since $a \neq e$ and $b \neq e$.
For $ab = ba$, it's $Z_6$
For $ab = ba^2$, it's $S_3$


Let $ x$, $ y$ and $ z$ be positive real numbers.
Prove that
\[ \left(\dfrac{x \plus{} 2}{3x^2} \plus{} \dfrac{y \plus{} 3}{4y^2} \plus{} \dfrac{z \plus{} 4}{5y^2}\right)\left(\dfrac{3x}{x \plus{} 2} \plus{} \dfrac{4y}{y \plus{} 3} \plus{} \dfrac{5z}{z \plus{} 4}\right)\ge 3\left(\dfrac{1}{x} \plus{} \dfrac{1}{y} \plus{} \dfrac{1}{z}\right)
\]
:)

[quote="ElChapin"]Let $ x$, $ y$ and $ z$ be positive real numbers.
Prove that
\[ \left(\dfrac{x \plus{} 2}{3x^2} \plus{} \dfrac{y \plus{} 3}{4x^2} \plus{} \dfrac{z \plus{} 4}{5x^2}\right)\left(\dfrac{3x}{x \plus{} 2} \plus{} \dfrac{4y}{y \plus{} 3} \plus{} \dfrac{5z}{z \plus{} 4}\right)\ge 3\left(\dfrac{1}{x} \plus{} \dfrac{1}{y} \plus{} \dfrac{1}{z}\right)
\]
:)[/quote]
I think you have some mistakes.
I think it should be
[quote]Let $ x$, $ y$ and $ z$ be positive real numbers.
Prove that
\[ \left(\dfrac{x \plus{} 2}{3x^2} \plus{} \dfrac{y \plus{} 3}{4y^2} \plus{} \dfrac{z \plus{} 4}{5z^2}\right)\left(\dfrac{3x}{x \plus{} 2} \plus{} \dfrac{4y}{y \plus{} 3} \plus{} \dfrac{5z}{z \plus{} 4}\right)\ge 3\left(\dfrac{1}{x} \plus{} \dfrac{1}{y} \plus{} \dfrac{1}{z}\right)
\]
[/quote]

Yes, of course. Thank you, I have edited it.

:)

Er, trivial by Cauchy-Schwarz?

EDIT: No, I take that back. I fail at inequalities.

Then try with another one.  :wink:

You could say that this is "interesting" because it is "stronger" than cauchy when applied directly.

:)

But by Cauchy-Schwarz, $ \left(\dfrac{x \plus{} 2}{3x^2} \plus{} \dfrac{y \plus{} 3}{4y^2} \plus{} \dfrac{z \plus{} 4}{5z^2}\right)\left(\dfrac{3x}{x \plus{} 2} \plus{} \dfrac{4y}{y \plus{} 3} \plus{} \dfrac{5z}{z \plus{} 4}\right)\ge \left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)^2$, so it suffices to prove $ \left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)^2\geq 3\left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)$, which... isn't true.

 :huh: I probably have no idea what I'm talking about.

That just means that the inequality is too sharp for Cauchy to work (at least directly).

With respect to the description "almost direct application", [hide="I believe"][url=http://en.wikipedia.org/wiki/Chebyshev's_sum_inequality]this[/url] is what is being referred to. There's still work left though.[/hide]

Ahhh, so [i]that's[/i] what he meant by "it is 'stronger' than cauchy when applied directly."

[quote="grn_trtle"]But by Cauchy-Schwarz, $ \left(\dfrac{x \plus{} 2}{3x^2} \plus{} \dfrac{y \plus{} 3}{4y^2} \plus{} \dfrac{z \plus{} 4}{5z^2}\right)\left(\dfrac{3x}{x \plus{} 2} \plus{} \dfrac{4y}{y \plus{} 3} \plus{} \dfrac{5z}{z \plus{} 4}\right)\ge \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)^2$, so it suffices to prove $ \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)^2\geq 3\left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)$, which... isn't true.

 :huh: I probably have no idea what I'm talking about.[/quote]

By Cauchy,the inequality is true for $ 0 < x\leq y\leq z \leq 1$,but we need to use another way to prove for all positive numbers.

[quote="grn_trtle"]But by Cauchy-Schwarz, $ \left(\dfrac{x \plus{} 2}{3x^2} \plus{} \dfrac{y \plus{} 3}{4y^2} \plus{} \dfrac{z \plus{} 4}{5z^2}\right)\left(\dfrac{3x}{x \plus{} 2} \plus{} \dfrac{4y}{y \plus{} 3} \plus{} \dfrac{5z}{z \plus{} 4}\right)\ge \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)^2$, so it suffices to prove $ \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)^2\geq 3\left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)$, which... isn't true.

 :huh: I probably have no idea what I'm talking about.[/quote]

I think Cauchy gives you \[ \left(\dfrac{x \plus{} 2}{3x^2} \plus{} \dfrac{y \plus{} 3}{4y^2} \plus{} \dfrac{z \plus{} 4}{5z^2}\right)\left(\dfrac{3x}{x \plus{} 2} \plus{} \dfrac{4y}{y \plus{} 3} \plus{} \dfrac{5z}{z \plus{} 4}\right)\ge \left(\frac {1}{\sqrt x} \plus{} \frac {1}{\sqrt y} \plus{} \frac {1}{\sqrt z}\right)^2\]

Yes Differ, Cauchy gives $ \left(\dfrac{1}{\sqrt{x}}\plus{}\dfrac{1}{\sqrt{y}}\plus{}\dfrac{1}{\sqrt{z}}\right)^2$ and that is never greater than $ \dfrac{3}{x}\plus{}\dfrac{3}{y}\plus{}\dfrac{3}{z}$.

And yes MellowMelon, I was thinking about that.

:)


inequalities
quadratics
inequalities proposed

Prove that for all non-negative real numbers a,b,c holds the following inequality:

4((ab)^3/2 + (bc)^3/2 + (ac)^3/2) =< 4c^3 + (a+b)^3

Can this inequality be transformed into the quadratic inequality with the variable $ x\equal{}\sqrt{c^3}$?


function
integration
complex analysis
calculus
complex analysis theorems

Let $f$ be a complex function on the unit disc $D$ of the complex plane which is measurable and locally $\in L_{1,loc}$. Assume that for almost all pairs $(a,r)$ with $a \in C$ and $r > 0$  for which $C_r(a) \subset D$ we have

       \[  \int_{C_r\left(a\right)}fdz = 0 \].

Then there is a holomorphic function on $D$ which is almost everywhere equal to $f$.

  This one appears in Serge Lang - Complex Analysis as an exercise of Chapter V  :D   If anyone know the proof, would you please post here or indicate the source??!! :)

[quote="dickchimney"]Let $f$ be a complex function on the unit disc $D$ of the complex plane which is measurable and locally $\in L_{1,loc}$. Assume that for almost all pairs $(a,r)$ with $a \in C$ and $r > 0$  for which $C_r(a) \subset D$ we have
       \[  \int_{C_r\left(a\right)}fdz = 0 \].
Then there is a holomorphic function on $D$ which is almost everywhere equal to $f$.[/quote]
Is it contour integral or integral in sense $L_1(\mathbb{R^2})$?

That has to be a contour integral around a circle. The other interpretation leads only to the zero function and is uninteresting. With the contour integral, it makes sense as a problem. One possible path: show that $f$ must be a [b]weak[/b] solution of the differential equation $\frac{\partial f}{\partial \overline{z}}=0.$ From there, it is essentially a regularity result for an elliptic operator.

Actually, this exercise-theorem has a hint. It said that one can find the answer to this problem at p.412 of SL2. SL2 must be a book in the bibliography list ( In Serge Lang book). Unfortunately, my book lost some of the last pages, so I dont know what the book is  :(  :D

Here's a reference of some stronger results.

http://www.math.su.se/~fk04lho/public/springer%20-%20on%20moreras%20theorem.pdf

Was this exercise marked as harder or anything? This doesn't belong in Calculus - Analysis.

[quote="koo"]Here's a reference of some stronger results.

http://www.math.su.se/~fk04lho/public/springer%20-%20on%20moreras%20theorem.pdf

Was this exercise marked as harder or anything? [/quote]

 Of course!
 Thank u for the paper, it is really helpful!

There is a solutions manual to Lang's complex analysis text.


Να λυθεί η ανίσωση
$ 4(2\plus{}cot^2{x}\minus{}cos^2{x})^3\plus{}4(2\plus{}tan^2{x}\minus{}sin^2{x})^3\leq{5^3}$

Εξαιρετικη η κατασκευη της ανισοτητας.

Η ποσοτητα $ (2 \plus{} cot^{2}{x} \minus{} cos^{2}{x})^{3}$ καταληγει στην μορφη $ (\frac {1}{cos^{2}{x}} \plus{} cos^{2}{x})^3$.

Ομοιως ισχυει  $ (2 \plus{} tan^{2}{x} \minus{} sin^{2}{x})^{3} \equal{} (\frac {1}{sin^{2}{x}} \plus{} sin^{2}{x})^3$.

Θετω $ a \equal{} sin^{2}{x},b \equal{} cos^{2}{x}$. Ισχει $ a \plus{} b \equal{} 1$ και $ 0\leq {a,b}\leq {1}$.

Θεωρω την συναρτηση $ f(x) \equal{} (\frac {1}{x} \plus{} x)^{3}$ η οποια ειναι παραγωγισιμη.

Τοτε $ f'(x) \equal{} 3(x \plus{} \frac {1}{x})^{2}(1 \minus{} \frac {1}{x^2})$. H $ f'(x)$ ειναι ομοιως παραγωγισιμη.

Τοτε μετα απο αρκετες πραξεις $ f"(x) \equal{} 6(x \plus{} \frac {1}{x})(\frac {x^4 \minus{} x^2 \plus{} 2}{x^4})$.

Αρα $ f"(x) > 0$ οποτε η $ f$ ειναι κυρτη και εφαρμοζω την ανισοτητα jensen.

$ f(a) \plus{} f(b)\geq 2f(\frac {a \plus{} b}{2}) \equal{} 2f(\frac {1}{2}) \equal{} \frac {125}{4}$.

Αρα $ LHS\geq \frac {125}{4}$.
Ομως $ LHS\leq \frac {125}{4}$.

Προφανως ισχυει η ισοτητα οταν $ a \equal{} b \equal{} \frac {1}{2}$, δηλαδη οταν $ tan{x} \equal{} 1, \minus{} 1$, αρα χ=κπ + π/4 ή χ=κπ - π/4.

Ωραία λύση ...
Η λυση μου είναι με  $ Jensen$ στην $ x^3$ και κατοπιν $ B\minus{}C\minus{}S$ ....


function
abstract algebra
group theory
linear algebra
linear algebra theorems

If f is any function which assigns to each permutation of degree n an integer, and if $ f(\sigma\tau ) \equal{} f(\sigma )f(\tau )$, then f is identically or f is the sign function.

Can't see how to prove that if is neither 0 nor 1 then f is the sign function for permutations.

This fits better as a group theory question. You're looking for a homomorphism from the symmetric group to the multiplicative group $ \{\minus{}1,1\}$ (the other integers are a red herring, since the image of any permutation has to be a root of unity).

That means that the kernel of this homomorphism has to be a normal subgroup of index 1 or 2 in $ S_{n}$. The full group and the alternating group $ S_{n}$ are the only such groups.

Actually, the only normal subgroups of $ S_{n}$ are the trivial group, $ S_{n}$, and $ A_{n}$. This means that there are essentially only three homomorphisms from $ S_{n}$ (for $ n\ge 3$); the identity, the trivial homomorphism, and sign.

sorry jmerry, this question was in my linear algebra book. consequently, i don't know what a homomorphism is and much of what you said was greek to me. i know the set of permutations of degree n forms a group under composition and thats about it.

The group theory approach is the insightful approach.

The "peasant" approach:
Suppose that $ f$ has no zeros. If $ a\neq b\neq c\neq a$, then $ (ab) (ac) \equal{} (acb)$ and $ (ac)(bc) \equal{} (acb)$, so $ f(ac) \equal{} f(bc)$.
Hence, for all $ a\neq b,d,\ c\neq d$ we have that $ f(ab) \equal{} f(ad) \equal{} f(cd)$, so $ f$ either sends each transposition into $ 1$ or into $ \minus{}1$.
In the first case, $ f$ is the identity, in the second, it's the sign.

How did you get (ab)(ac) = ac?

If a = (1,2,3), b = (2,1,3) and c=(3,2,1) it doesn't work. Also how do you know f is not the negative of the sign function?

You may misunderstand perfect_radio's meaning.
In his post $ a,b,c$ are integers,not  permutation.
A simple sence by $ f(23)f(12)f(23)\equal{}f(13)$ :wink:

Quite sorry but I did not understand because f was not defined. I thought f was a function over the permutations but now it is being used as a function over integers?

Actually, $ f$ was defined in your post and is over the permutations.

$ (ab)$ means the permutation that sends $ a$ into $ b,$ $ b$ into $ a$ and leaves the others fixed.
$ (abc)$ means the permutation that sends $ a$ into $ b,$ $ b$ into $ c,$ $ c$ into $ a$ and leaves the others fixed.

I should have written $ f((ab))$ instead of $ f(ab)$, but double parentheses are usually replaced by normal ones.

Key words that you can [i]Google[/i]: [b]cycle decomposition of a permutation[/b].

Alright thanks. I was not familiar with that notation


calculus derivative function quadratics algebra polynomial ratio

Perhaps this is a common problem, but I happened across it indirectly (Problem 263 in Art of Problem Solving 2) and cannot find a proof.
Given an infinite sequence S, define operation f(S) to return an infinite sequence whose terms are the difference between adjacent terms in S.  A well-known example is f({0,1,4,9,16,25,...})={1,3,5,7,9,11..}.
Show for any sequence S={a^n} and any positive integer  n , that f(f(..f(S)...)) becomes {n!} for some number of f's. (A consequence of this is that the infinite sum  :Sigma: k^n/2^k,k=1..:inf: can be found for all n).

Try an inductive approach.  I think that should do it.

The basic idea is that the difference operator is the discrete analogue of the derivative operator.  So (among other things) you will have to apply the
difference operator n times to the sequence {a^n} to end at the sequence {n!}.

In general, (I think) that applying the difference operator k times to the sequence
{f(a)} will give you the sequence {g(a)} where g(x) is the kth derivative of f(x).

That's true, give or take a little.  The give or take bit will come in depending how you define certain things -- I'm sure there's a standard definition, I just don't know it.  Basically, you have to choose whether f(2) - f(1) is equal to the value of the finite difference function (delta f) taken at 1 or 2, which might mess up whether or not it's exactly the derivative.

Anyhow, the fact that it is essentially a derivative doesn't help prove the statement he wants, which I think is best done inductively.

The first difference of f(x) is conventionally defined to be f(x+1) - f(x).  For example, if f(x) = x2, then the first difference is 2x + 1, whereas the derivative is 2x.  So they are not equal.  (Changing the definition to f(x) - f(x-1) doesn't help.  Changing the definition to f(x+1/2) - f(x-1/2) works for quadratics, but still doesn't equal the derivative for say cubics.)

So at best there is merely an analogy between differences and derivatives.

By the way, when the original post said {an}, did that mean the sequence 1n, 2n, ...?

"merely an analogy"

it's a pretty strong analogy, even if the specific numbers don't quite carry over.  For polynomials at least the ratio of the derivative to the difference should approach 1.  I bet that's general, in fact.

Yes, that was what he meant by {an}.

Yes, I agree it is a strong analogy.  You are right that for polynomials, the ratio approaches 1.  For non-polynomials, it's not true in general.  Take for example f(x) = ex.

Ah, but we have a discrete analogue of ex also. The forward difference of 2x is 2x. Umbral calculus is so neat.


In a Trinomial ax^2 + bx +c, if 'a' & 'c' are prime nos., and it can be factored,  how many values can 'b' have?

Um, please elaborate.  Do you mean for $ y$, where $ y\equal{}ax^2\plus{}bx\plus{}c$?  Do you mean $ x$?

[quote="AIME15"]Um, please elaborate.  Do you mean for $ y$, where $ y \equal{} ax^2 \plus{} bx \plus{} c$?  Do you mean $ x$?[/quote]

I think he means it for $ b$.

We assume that a,b,c and their factors have to be integral coefficients.

We can factor out in the following factors:

(ax+c)(x+1)=ax^2+(c+a)x+c
(ax+1)(x+c)=ax^2+(ac+1)x+c
(ax-c)(x-1)=ax^2-(c+a)x+c
(ax-1)(x-c)=ax^2-(ac+1)x+c

for a total of 4 different b.

But how do we know that all generate distinct numbers?
Well a,c are positive, so c+a and -(c+a) are distinct, and we only have to compare (c+a) and (ac+1).

Suppose equality occurs, i.e. c+a=ac+1, which is equivalent to 0=ac-a-c+1, or 0=(a-1)(c-1). Therefore either a or c = 1, but 1 is non-prime, thus completing our argument.

Note that the fact that there are only four possible factorings comes from the rational root theorem.


[quote="stevem"]It can be done. For example, $\begin{matrix}\sqrt{1\dots 1}\vspace{-2.5ex}\\ \hspace{0.7em}\underbrace{\hspace{2.6em}}_{n}\end{matrix}$ but it's a dreadful kludge.[/quote]
I see. And we should work a little more if we wanted to align it vertically too:
$s(n)=\begin{matrix}\sqrt{1\dots 1}\vspace{-2.5ex}\\ \hspace{0.7em}\underbrace{\hspace{2.6em}}_{n}\end{matrix}$
It seems too much work.
If only the square root ignored the underbrace to calculate its height.

$s(n)=\begin{matrix}\\ \sqrt{1\dots 1}\vspace{-2.5ex}\\ \hspace{0.5em}\underbrace{\hspace{2.6em}}_{n}\end{matrix}$
I agree though, it's too much work.

I'm reviving this thread because I have a better answer to[quote="lordWings"]I was posting an answer to a problem where we had to calculate
$\sqrt{\underbrace{1\dots 1}_{n}}$
(the square root of a number with $n$ ones).
The rendering is logical, but I would like to know if there's a way to avoid the square root to get bigger to include the underbrace, so that it would be more similar to this
$\underbrace{\sqrt{1\dots 1}}_{n}$[/quote]which is to use \smash. \smash[b] keeps the square root the normal size so doesn't extend downwards (\smash[t] does the same for upwards).
So [code]s(n)=\sqrt{\smash[b]{\underbrace{1\dots 1}_n}}[/code] gives
$s(n)=\sqrt{\smash[b]{\underbrace{1\dots 1}_{n}}}$


Just a sketch to miss boring computations:

Let $b_n=\frac{1}{a_n}$ then $b_{n+1} ^2=b_n ^2+b_{n-1}^2 +(b_nb_{n-1})^2$ (just multiply everything out and square). If $c_n=b_n ^2 +1$ then $c_{n+1}=c_n \cdot c_{n-1}$. Finally let $d_n = \ln c_n$. We have $d_{n+1}=d_n + d_{n-1}$. It is easy to count general term of $d_n$ and going backwards we can obtain general term for $a_n$.

Since all $a_n$ are obviously positive (the strict inductive proof is trivial), we can square the equation and substitute $b_n=a_n^2$:

$b_{n+1}=\frac{b_nb_{n-1}}{b_n+b_{n-1}+1}$, or

$\frac{1}{b_{n+1}}=\frac{1}{b_n}+\frac{1}{b_{n-1}}+\frac{1}{b_nb_{n-1}}$, or

$\frac{1}{b_{n+1}}+1=(\frac{1}{b_n}+1)(\frac{1}{b_{n-1}}+1)$.

Substituting $c_n=\frac{1}{b_n}+1$ we get

$c_{n+1}=c_nc_{n-1}$ where $c_1=\frac{1}{a_1^2}+1=2, c_2=\frac{1}{a_2^2}+1={26\over 25}$

From that we have $c_3=c_1c_2, c_4=c_1c_2^2, c_5=c_1^2c_2^3, c_6=c_1^3c_2^5, c_7=c_1^5c_2^8$ etc. Inductively it's very easy to show that $c_n=c_1^{F_n}c_2^{F_{n+1}}$ where $(F_n)_{n\geqslant 1}$ is Fibonacci (or Fibonacci-like) sequence beginning with $1,0$ (i.e. $1,0,1,1,2,3,5\dots$). It can be derived that

$F_n=\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n-2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-2}}{\sqrt{5}}, n\geqslant 1\qquad (*)$

Therefore $a_n=\frac{1}{\sqrt{c_n-1}}=\frac{1}{\sqrt{2^{F_n}\left({26\over 25}\right)^{F_{n+1}}-1}}$, where $F_n$ is given by $(*)$


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Let $ K$ be a field. Consider some homogeneous polynomial $ P(x_1,...,x_n)$ of degree 3 that has no zeroes over $ K$. Prove that it does not have a zero in any quadratic extension $ L$ of $ K$.

[Note: Of course, my proof from http://www.mathlinks.ro/viewtopic.php?t=130345 works here, too; however, it does not work in any other case and I suspect we can't have any more results of that type.]

Hi Peter,

I haven't read your solution for the other problem, but this result can't be true: how about taking $ K \equal{} \mathbb{Q}$ and identities such as $ \Big(1\plus{}\sqrt{\frac{1}{6}}\Big)^3 \plus{} \Big(1\minus{}\sqrt{\frac{1}{6}}\Big)^3 \equal{} 3$? [Accordingly, $ P(x,y,z) : \equal{} x^3 \plus{} y^3 \minus{} 3z^3$: this has no nontrivial zeros over $ \mathbb{Q}$ (e.g. modulo $ 7$), but has the zero $ \Big(1\plus{}\sqrt{\frac{1}{6}}, 1\minus{}\sqrt{\frac{1}{6}}, 1\Big)$ over $ \mathbb{Q}(\sqrt{6})$... ]

Well, you are wrong ;-). $ x\equal{}1,y\equal{}\minus{}1,z\equal{}0$ is of course a solution over $ \mathbb{Q}$ :-).

:) Indeed -- I was only thinking about the inhomogeneous equation $ x^3 \plus{} y^3 \equal{} 3$ -- which has a zero in $ \mathbb{Q}(\sqrt{6})$, but not in $ \mathbb{Q}$. Sorry for the spam. Meanwhile, I read your solution -- it is very nice and simple!


Silver Medals

Name	Rank	Year
1. A. Melas	30	1984
2. Christos Athanasiadis	32	1989
3. Paul Tzermias	40	1985
4. Petros Bregiannis  58	1998
5. Nikolaos Romanidis	68	2007
6. Michail Rassias	69	2003
7. Romanos - Diogenis Malikiosis	73	1998
8. Minas Pagonakis	81	2004
9. Theofanis Matsoukas	85	1999
10. Spyridon Antonakopoulos	95	1999
11. Georgios Raptis	97	2001
12. Stergios Antonakoudis	113	2004

Source :  http://www.imo-official.com

Bronze Medals

Name	Rank	Year
Georges Panagopoulos	37	1987
Stavros Garoufalidis	37	1983
Vasilios Missoulis	60	1985
Ioannis Ivrissimtzis	72	1987
Haralampos Tamvakis	83	1987
Efthimios Anagnostou      	83	1983
Dimitrios Kodokostas 	91	1987
Georgios Papoutsi 	98	1989
Evangelos Kobotis  	101	1987
Prodromos Symeonidis 	108	1989
Georgios Takos	111	1996
Nikolaos Michalakis 	111	1996
Michail-Romanos Papalambros 	113	2002
Alexandros Karelas 	117	1991
Marios Papamichalis 	124	2004
Theodoros Evgeniou  	133	1991
Dimitrios Statopoulos 	134	1989
Argiriou Andreas 	139	1990
Dimitrios Poulopoulos 	140	2001
Ioannis Mavroidis 	143	1994
Efstratios Rappos 	148	1995
Anastasios Dimou 	148	2001
Theodosios Douvropoulos 	156	2005
Eleftherios Tsiokos 	156	2005
Stefanos Aretakis  	159	2004
Konstantinos Psychas  	161	2005
Andreas Galanis 	163	2003
Spyridon Alexakis 	164	1996
Theofanis Papamichalis 	169	2004
Athanasios Papaioannou	179	2003
Stavros Routzounis 	180	1996
Rallis Karamichalis 	189	2006
Dimitrios Papadimitriou 	197	2007
Ioannis Anapolitanos 	204	2000
Anastasios Vafeidis 	226	2007
Michail Savvas 	226	2007
Tilemahos Stamkopuluos	?	1986

Source :  http://www.imo-official.com


Sa se rezolve in $ \mathbb{R}$ ecuatia :

$ \{x\}^{2007}\minus{}[x]^{2007}\minus{}x^{2007}\equal{}0$

Pentru x>1 ecuatia clar nu are solutii deoarece $ [x]^{2007}\plus{}x^{2007}>1>{x}^{2007}$
pentru $ x \in [0,1) \rightarrow x^{2007}\equal{}{x}^{2007}$ adevarat iar pentru x<0 $ [x]^ {2007}\plus{}x^ {2007}<0\le {x}^{2007}$


vector
trigonometry
geometry
parallelogram
rhombus
absolute value

Let [b]u[/b] and [b]v[/b] be nonzero vectors on a plane or in 3-space.  Show that the vector
[b]w[/b] = ||[b]v[/b]||[b]u[/b] + ||[b]u[/b]||[b]v[/b]
bisects the angle between [b]u[/b] and [b]v[/b].

I don't get your notation. What are ||u and ||v? If they should be |u| and |v| instead, w is then a scalar, not a vector?

||v|| means |v|, and ||u|| means |u|.  Some authors use "double absolute value" for the norm (length) of a vector, whereas you are used to single absolute value.

Oh. Ah. Yes, that'd make sense... let me type it up in TeX:

$\mathbf{w} = |\mathbf{u}|\,\mathbf{v} + |\mathbf{v}|\,\mathbf{u}$

Proof: See diagram.

We have the following equations:

[tex]\begin{eqnarray}
\mathbf{w}&=&|\mathbf{u}|\,\mathbf{v}+|\mathbf{v}|\,\mathbf{u}\\
\mathbf{w}\cdot\mathbf{u}&=&|\mathbf{w}|\,|\mathbf{u}|\,\cos\alpha\\
\mathbf{w}\cdot\mathbf{v}&=&|\mathbf{w}|\,|\mathbf{v}|\,\cos\beta
\end{eqnarray}[/tex]

Substituting (1) into L.S. of (2) yields

[tex]\begin{eqnarray}
&&\mathbf{w}\cdot\mathbf{u}\nonumber\\
&=&\left(|\mathbf{u}|\,\mathbf{v}+|\mathbf{v}|\,\mathbf{u}\right)\,\mathbf{u}\nonumber\\
&=&|\mathbf{u}|^2\,|\mathbf{v}|\,\cos(\alpha+\beta)+|\mathbf{u}|^2|\mathbf{v}|\nonumber\\
&=&|\mathbf{u}|^2\,|\mathbf{v}|\left[1 + \cos(\alpha+\beta)\right]\nonumber
\end{eqnarray}[/tex]

This equals $|\mathbf{w}|\,|\mathbf{u}|\,\cos\alpha$, so

[tex]\displaystyle\cos\alpha = {{|\mathbf{u}|\,|\mathbf{v}|\left[1+\cos(\alpha+\beta)\right]}\over{|\mathbf{w}|}}[/tex].

By substituting (1) into the L.S. of (3), and equating it with the R.S. of (3), we see that

[tex]\displaystyle\cos\beta = {{|\mathbf{u}|\,|\mathbf{v}|\left[1+\cos(\alpha+\beta)\right]}\over{|\mathbf{w}|}}[/tex].

So $\alpha=\beta$ and our proof is complete.

You really don't need any trig to prove that the vector

$\mathbf{w} = |\mathbf{u}|\,\mathbf{v} + |\mathbf{v}|\,\mathbf{u}$

bisects the angle between the vectors $\mathbf{u}$ and $\mathbf{v}$. In fact, if you consider the vectors $\displaystyle \frac{\mathbf{u}}{\left| \mathbf{u} \right| }$ and $\displaystyle \frac{\mathbf{v}}{\left| \mathbf{v} \right| }$, then you see that these two vectors are unit vectors. Hence, they have the same length. In other words, if you take a point O somewhere in plane or in 3d space and two other points A and B such that $\displaystyle \overrightarrow{OA} = \frac{\mathbf{u}}{\left| \mathbf{u} \right| }$ and $\displaystyle \overrightarrow{OB} = \frac{\mathbf{v}}{\left| \mathbf{v} \right| }$, then the segments OA and OB have the same length. If we now construct another point D such that $\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{OB}$, then the quadrilateral OADB is a parallelogram. Since the sides OA and OB of this parallelogram have equal length, the parallelogram must be a rhombus. But in a rhombus, any diagonal bisects the angles at the two vertices through which the diagonal passes. Hence, for instance, the diagonal AD of our rhombus OADB bisects the angle AOB. In other words, the vector $\overrightarrow{OD}$ bisects the angle between the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$. But the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ equal $\overrightarrow{OA} = \mathbf{u}$ and $\overrightarrow{OB} = \mathbf{v}$, respectively, and the vector $\overrightarrow{OD}$ equals

  $\displaystyle \overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{OB} = \frac{\mathbf{u}}{\left| \mathbf{u} \right| } + \frac{\mathbf{v}}{\left| \mathbf{v} \right| } = \frac{\left| \mathbf{u} \right| \cdot \mathbf{v} + \left| \mathbf{v} \right| \cdot \mathbf{u}}{\left| \mathbf{u} \right| \cdot \left| \mathbf{v} \right| }$.

Hence we obtain that the vector $\overrightarrow{OD}$ bisects the angle between the vectors $\mathbf{u}$ and $\mathbf{v}$. Now, the vector $\mathbf{w} = \left| \mathbf{u} \right| \cdot \mathbf{v} + \left| \mathbf{v} \right| \cdot \mathbf{u}$ is just the vector $\displaystyle \overrightarrow{OD} = \frac{\left| \mathbf{u} \right| \cdot \mathbf{v} + \left| \mathbf{v} \right| \cdot \mathbf{u}}{\left| \mathbf{u} \right| \cdot \left| \mathbf{v} \right| }$ multiplied with the scalar $\left| \mathbf{u} \right| \cdot \left| \mathbf{v} \right|$; hence, the vector $\mathbf{w}$ is parallel to $\overrightarrow{OD}$, and consequently it also bisects the angle between the vectors $\mathbf{u}$ and $\mathbf{v}$. This completes the proof.

  Darij


[hide]Μήπως με MIQUEL????[/hide]αλλα πες για ποια ακριβώς ευθεία μιλάς!

Είναι η ευθεία CD .Δεν ξέρω αν λύνεται με αυτό που είπες πάντως η λύση μου είναι αρκετά μεγάλη.... :(  :)
Ελπίζω να βρεις κάτι σύντομο .

Ωραία πρόταση Σιλουανέ και σε ευχαριστώ για την αφιέρωση.

Εκεί που νομίζεις ότι κάτι καταφέρνεις στη Γεωμετρία, αρκεί μία πρόταση σαν κι αυτή, για να σε κάνει να αισθανθείς σαν αρχάριος...

Κώστας Βήττας.

[quote="silouan"]Βάζω μια ενδιαφέρουσα άσκηση (όχι εύκολη νομίζω ) και την αφιερώνω στους κ.Βήττα , Νίκο Ράπανο που αγαπούν τη γεωμετρία .

Έστω οξυγώνιο τρίγωνο $ABC$ . Τα σημεία $M,N$ είναι εσωτερικά σημεία των πλευρών $AC,\ BC$ αντιστοίχως , ενώ το $K$ είναι το μέσον του ευθυγράμμου τμήματος $MN$ . Οι περιγεγραμμένοι κύκλοι των τριγώνων $CAN$ και $BCM$ τέμνονται στα σημεία $C$ και $D$ .
Να δείξετε ότι η ευθεία $CD$ διέρχεται από το περίκεντρο $O$ του τριγώνου $ABC$ , [b]αν και μόνο αν[/b] ,η μεσοκάθετος της $AB$ διερχέται από το $K$ .

Δείξτε και το ευθύ και το αντίστροφο .
Περιμένω τις λύσεις σας .[/quote]

Σιλουανέ ήλπιζα ότι θα μου αφιέρωνες κάτι πιο δύσκολο... :wink: :P
Η άσκηση μπορεί να είναι θέμα της Μεσογειάδας του 2002 αλλά όλες οι λύσεις που κυκλοφορούν είναι υπερβολικά [b]ungly[/b] όπως συνηθίζεις να λες...
Σήμερα λοιπόν που είχα λίγο χρόνο και εν' όψει της προετοιμασίας μου για τη Μεσογειάδα βρήκα μία πολύ σύντομη και elegant λύση!
Προς το παρόν παραθέτω την εξαιρετική λύση σου (φαντάζομαι μου επιτρέπεις) για να μελετηθεί από τους διαγωνιζόμενους (το Σάββατο) και την επόμενη Δευτέρα θα δημοσιεύσω και τη δική μου.

[quote="silouan"]Μόνο που δε μου ζήτησες το copy -write για τις διορθώσεις πάνω στις λύσεις . :D
Πάντως από μία παρατήρηση που έκανες για ισοτομικά καταλαβαίνω που το πας  :wink:[/quote]

Με το Silouan τα είπαμε και στο τηλέφωνο...Η λύση μου δεν έχει να κάνει με ισοτομικά :o
Οι παρατηρήσεις εξ'άλλου έγιναν πριν λύσω την άσκηση :P
Ας διατηρήσουμε το saspens μέχρι τη Δευτέρα ([u]την επόμενη[/u] Δευτέρα)

[quote="galeos"]H askhsh auth vgainei poli grhgora xrhsimopoiwntas oti h eutheia twn meswn enos plhrous tetrapleurou einai katheti sthn eutheia Simpson tou tetrapleurou. :wink:[/quote]

Ara den xreiazetai na miliso katholou...geia sou re negale desko!!!

[quote="silouan"][quote="Nick Rapanos"]
geia sou re negale desko!!![/quote]

Αφού δεν την έλυσε ο Δέσκος !!! :P
galeos λέει όχι gdeskos . Αν δεν κάνω λάθος
galeos =Γαλάνης Ανδρέας   :wink:[/quote]

sorry-den eida kala!!

[quote="Nick Rapanos"]
sorry-den eida kala!![/quote]

Καλά δεν έγινε και τίποτα  :D   . Πάντως ωραία λύση μπράβο και στους δύο σας που τη βρήκατε.

[quote]Πάντως ωραία λύση μπράβο και στους δύο σας που τη βρήκατε.[/quote]
egw den thn vrhka akribws xaaxxa apla thimomoun meta to diagwnismo, to stefano aretakh na leei kati gia eutheies simpson tetrapleurou pou sthn ousia einai kai to 90% ths lushs. :P


analytic geometry
probability
AMC 12 A
AMC

The AMC-12 A 2003 #22

Objects A and B move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object A starts at (0,0) and each of its steps is either right or up, both equally likely. Objust B starts at (5,7) and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet?

A) 0.10
B) 0.15
C) 0.20
D) 0.25
E) 0.30

[hide="solution"]clearly, they have to meet after 6 moves, since they are 5+7 units away, also if we do the total probabilitiy of moves over the ways that they can meet, we will not overcount since they can only meet each other once in a sequence

consider their joined path, it is 5R7U in some order, and the number of arrangements for that word is $\binom{12}{5}$

then the total number of possible moves that they can do is 2^6*2^6, so $\frac{99}{512}\approx .193 \implies \boxed{C}$[/hide]

[quote="MysticTerminator"][quote="WarpedKlown1335"]I'm just curious, in their solution to this problem, they use (2^6)^2 = 2^12 for the total number of paths after 6 steps, but shouldn't it be (2^6-1)^2 = 3969?  Because you can't travel 6 steps to the right from point A or to the left coming from B.[/quote]

why not? does it say you have to stay in that little box thing?[/quote]

Uh...I don't think you can go all the way to the left....I think it says $2^6$ because you can only meet after each has taken 6 steps, after that, you pass each other and you can't meet at all, not because you have to stay in the little box.

But one of the possibilities covered by 2^6 is going all the way to the left.


geometry
perimeter
inradius
circumcircle
geometry proposed

An acute triangle $ ABC$ has three heights $ AD$, $ BE$ and $ CF$ respectively. Prove that the perimeter of triangle $ DEF$ is not over half of the perimeter of triangle $ ABC$.

[hide]it's very old: call A, p, r, R the area, the semiperimeter, the inradius and the circumradius of ABC, then the perimeter of the orthic triangle when ABC is acute, as know,  is $ \frac{2A}{R}$ so we have to prove $ \frac{2A}{R} \le p \ \Longleftrightarrow \ 2p \cdot r \le p \cdot R \ \Longleftrightarrow \ R \ge 2r$ that is know.[/hide]


Show an example of four infinite subsets of the set of all positive integers so that the intersection of any three of them is an infinite set, but the intersection of all four of them is empty.

Well,...take any four infinite and pairwise disjoint subsets of $ \mathbb {N}^*$, say $ A,B,C,D$.
Now consider the four sets $ A \cup B \cup C$, $ B \cup C \cup D$, $ C \cup D \cup A$ and $ D \cup A \cup B$. These four sets have the desired property. :wink:

Pierre.


Geometry
ASMC 2007 geometry
Two lines y and z form a 60æ angle at the point X , and Y1 is a point on y. From Y1 draw a line perpendicular to the line y meeting the line z at the point Z1. From Z1 draw a line perpendicular to z meeting the line y at Y2. Continue in this manner giving points Z2,Y2,Z3,Y3 and so on. These points are the vertices of right triangles X Y1 Z1 , X Y2 Z2,•••. If the area of X Y1 Z1 = 1, determine
areaX Y1 Z1 + areaX Y2 Z2 + ••• + areaX Y2007 Z2007.

Options:

1.) ((16^{2007}) -1)/7
2.) ((16^{2006}) -1)/15
3.) ((16^{2007}) -1)/15
4.) 16^{2007}/7

You've now posted this problem in three places.  Generally, just post it in the appropriate forum.  I think in this case that would be Intermediate Topics.


Express $ \sqrt{12 \plus{} 6\sqrt{3}}$ in the form $ x \plus{} \sqrt{y}$.

$ \sqrt{12\plus{}6\sqrt 3}\equal{}\sqrt{9\plus{}6\sqrt 3\plus{}3}\equal{}3\plus{}\sqrt 3$


Dear All

When I insert a figure (ps) in latex, it shows/inserts  a "S" at the top of the figure. For some figures i have "SS".  It is not there when the ps is opened in GV. can someone help me with that?

cheers.
prasad..

Can you post a portion of code you used?


probability
Mafia
function
geometry
logarithms
combinatorics open
combinatorics

The function $ S(m, n)$ is defined on non-negative integers as follows:

- $ S(0, n) \equal{} 0$ whenever $ n \ge 0$,
- $ S(m, n) \equal{} 1$ whenever $ m \ge n$,
- $ S(m, n) \equal{} \frac{m}{m\plus{}n} S(m\minus{}1, n\minus{}1) \plus{} \frac{n}{m\plus{}n} S(m, n\minus{}1)$.

Compute either an ordinary or an exponential generating function for $ S$.

This recurrence is similar to (or the same as) a recurrence in theoretical computer science I once saw (in the areas of leader election or collective coin flipping).  If memory serves, we can approximate what's going on by passing to the continuous version, which leads to a partial differential equation that can be solved.  If I recall correctly, the character of the solution (meaning the value is near 0 or 1) is different depending on whether $ m$ is less than or greater than $ \frac {n}{\log n}$, or something like that.

Searching around,  I remembered that the work I was describing is a "baton-passing" protocol that was first proposed by Michael Saks and later analyzed more fully by Miki Ajtai and Nati Linial.  Here is a copy of the Ajtai-Linial paper (warning: the file is more than 10 MB):
[url]http://www.cs.huji.ac.il/~nati/PAPERS/ajtai_coalitions.pdf[/url].
See their Section 3.


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
Putnam
AMC 8

For the most part of my short life, I have had no mathematical experience AT ALL. The first time I had any contact at all was in the middle of January of eighth grade, 2 weeks before MATHCOUNTS. A friend of mine, Ethan Sterling, convinced me to try our for the school team. Amazingly, even with 2 weeks left, I was able to make it, albeit barely, to the team as an alternate. At the chapter rounds, I got second place, and thus I was "upgraded" to team member at states, where i got a pathetic 8th, 3 careless questions away from the nationals. However, our school, Hunter College High School, got 1st. I only regret i didn't join math team in 7th grade, when i could ahve made it to MATHCOUNTS as a seventh grader. Anyway, Ethan Sterling also forced me to take a weird exam, (at least at the time) called the AMC 8. I took it, passed it, and Mr. Hankin convinced me to take the AMC 10. I passed that too, and I took the AIME as an eighth grader, and did rather poorly, a 4. However, by that time, i was really interested in math, and thus began learning mathematics, but since i had such a late start, I was always a little behind.

ANYWAY, now that I'm done blabbing, I'm just wondering whether any of you guys have had similar experiences, and whether the truly good math people started out early, and what your experiences were like.

I definitely know what you mean when you talk about starting late. Luckily, I started a little younger than you did, in winter of 8th grade, so I managed to get just enough practice in to make it to nationals that year (where I didn't do very well). Since then I've made very rapid progress; by AIME score in 9th grade was 4, and this year (10th grade) my score was 13, and then I proceeded to take the USAMO and squeeze into MOP. I'm hoping my rate of progress continues for the next two years...

So definitely you can still go far if you start late. Just be glad you started before high school, so you still have plenty of time to rise up in the AMC competitions.

It still bothers me to this day, however, that I started as late as I did, because I never really had the chance to experience success at National Mathcounts. I think had I started doing math team in 6th grade, or even 7th, I could have gone quite far at National Mathcounts. I did feel a little left out at MOP because almost everyone there placed very well at National Mathcounts I think the reason is that almost all of them have been interesting in math their whole lives, whereas I hated it until 7th grade (and, as I said, didn't start doing competition-level stuff until 8th).

But keep working, and stay optimistic. It may be too late for Mathcounts, but you still have four more years to do the USAMO, and four years to do the Putnam after that.

LoL. Unfortunately, I only have 3 yrs left, and careless mistakes SLAUGHTERED me at this year's AIME.

In elementary school, we had a few contests, and I did quite well on them (almost always the highest in the school), but I wasn't aware of the existence of Saturday morning math competitions. In junior high school, we had occasional math competitions. Each year we had a class competition, and in sixth grade I finished second, and in seventh and eighth grade I finished first. Other than that, we did the grade level California Math League each year, and in eighth grade I got a perfect score, but the questinos were so routine and easy. They were not much more difficult than the problems we did in class, so I didn't think much of competitions. I wasn't informed of the existence of MathCounts until I got into high school, which wasn't very useful. In high school, I took tons of math contests (at least twenty-five per year I'd guess), and that was great, but I wasn't so good at first compared to all these other people who had had years of experience under their belts. I was probably the second best at my school when I got to high school, but second best wasn't what I had planned for, especially after my mother told all the math people at my school that I would be the future of mathematics at my school, which put unnecessary and unreasonable pressure on me, at least for a little while. I guess I'm a bit better now than I was in 9th grade, when I scored 86 on the AMC12.

Hmm, you're talking about starting in 8th grade. I started in second grade of High School. (We have three high school grades here in Sweden, and the last one starts one year "later" than yours). But I don't think It's fatal. If you have mathematical abilities you have them from the day you were born, they don't get lost that easily.

i participated in PA math league in 5-7 grades, then MC in 7 and 8, but i never did well...and then i found aops in march (while i was in 8). and i see the light! i did really well on practice amc8s, but for some reason my school stopped doing amc8.

I had a pretty late start in math.  In fifth grade, I was in this 7/8 math course thing that taught a little bit of algebra and such.

In the middle of my 7th grade, I was introduced to Mathcounts.  I got 11th regionals, 39th state that year.  So I was like "I'm so going to nationals next year!".  So after a year of work, I got 1st in regionals, 4th in state, and 53rd in nationals.

And now I will go into my sophmore year.  I really didn't prepare much math for math competitions during my freshman year because I had a lot of homework and such.  I did the AMC 10 and AIME that year but didn't make it to USAMO :cry: .  I hope I can squeeze in more preparation time!

-interesting_move

i did mc in 7th and 8th grades with very litte interest in it. i think i ended up around 13 regionals then 35 state in 8th grade. i started being really interested in math in january of my freshman year (last year). i was in a c++ class where i finished all my programs, so i started the unl.edu/amc sites and the like. i had a goal of making usamo freshman year, but with only 2 months till the amc12 and another for aime, i couldnt make it. i got a 6 on aime, with some stupid mistakes, so i believe i could have made usamo, but no point in looking back too much. instead im practicing now and hope to make usamo, and who knows? maybe even mop (i hear its being expanded) next year. good luck to all.

MOP is expanding next year!?  Yes!

-interesting_move

sigh. if only it were 2 years from now. super sigh

Well I started in the middle of seventh grade but then I rarely practiced.  I managed to get 2nd in chapter and well 38th in state that year though, but I thought I was good enough on ability alone.  Then in my eighth grade year, I took it a lot more seriously and I got 1st at chapter, 4th at state, and 59th at nationals.  And then on the AMC 10, I had a 138 and on the AIME, I had a 7, not good enough for the USAMO.  Sadly enough now, I'm cut off from most math competitions for a year because I am in Tokyo for my freshman year.  At least we still have the AMCs.

[quote]I missed 2 and on the AIME, I had a 7, not good enough for the USAMO[/quote]

shouldnt you get a 13 then?

ah. i misread it. he meant that he missed 2 on the amc. and he got a 7 on the aime. ah. sigh

[color=cyan]I started doing math competitions in 4th grade.  I started doing high school math competitions in 7th grade.  Yeah.  Starting early really helps, because very much of it just has to do with practice and familiarity with types of problems and methods of solution.[/color]

Hmm, I "started" in the middle of my junior year of high school where I signed up for the AMC.  I really had fun, though I didn't make AIME I did fairly well for my school (mid-upper 90's) and then I did the remaining 3 state math league contests.  In 8th grade, I participated in a Saturday morning math contests, placed 3rd but because of the situation I was in then, I did not pursue Mathematics farther.  I hope to make USAMO this spring  :D  but we'll see what happens.

I understand what it feels like to start extremely late. The most I did in middle school was the UIL state math contest (which I didn't place at.) In my 8th grade year my interested was piqued by an enthusiastic, if unorginazed teacher. I learned a lot of math on my own but had no outlet for it since I hadn't heard of any serious math contests. This summer I went to Texas Mathworks and put all my effort into learning as much as I could. The professor and the couselors encouraged me as well. Even though I am familar with a lot of advanced math, I have nothing to show for it yet. My sophmore year I would like to pass the AIME but considering it's my first year taking the AMC offically, not for practice, (not to mention my school's lack of a math team) I don't expect the world. I don't think there's a huge problem with starting late. When you grow up I doubt it makes a phenomenal difference if you started doing contests in the 7th or 10th grade. What matters is what you are doing NOW. I have a little book of Mathematician biographies, and many of them started relatively late. Though what made them genius was how quickly they caught up with their peers.   :D

wow. almost everyone started in grade school and said they started late. i've only been studying math (seriously) for the last couple of months, and i'm in grade 12.
   guess that's why i'm so not good. :(

[quote]i've only been studying math (seriously) for the last couple of months, and i'm in grade 12.
[/quote]

Don't worry, arsemesomething, im kind of in a similar situation as you.
I am currently in grade 11. Since grade 8 , i had been participating in the Australian Math Comp, but i never thought this as serious stuff.
Then,, this year, i learned about this forum and learned about what the math comp is about. I got interested and tried to prepare myself for the AUstralian MC this year. Well, i did not do as well as i expected, even worse than last year, maybe because i was too ambitious, oh well~

Anyway, i just think that winning or being placed at the top level in the maths competition can be very meaningful experience, but winning is not everything. If you can enjoy doing maths,, i reckon that can be just as awesome as winning the prize or somethin.

Don't give up!  i won't. I will study and try out next year again. Hopefully, i do my best as i can next year.

Good luck to everybody~,, whereever you live,

Mine started all of the way down... say 7 years old. Instead of doing 2nd grade math, I wiped out 3rd and 4th, and maybe 5th. 8 years old, I wiped out 6th, 7th, and I think 8th. At 9 years old, I was unfortunately an official 4th grader. Then, I started Algebra. Doing that, I finished in 3 months. I finished Algebra two in 5 or 7 months. Then I was 10 years old. I skipped 5th grade because of my outstandng achivements in the standardized test given out every other year before the SAT. Then I took pre-calc, of Advanced Mathematics, as it was called. I finished that just as I turned 11 years old.

When I was 9, I learned about Mathcounts. I went there, and improved some of my skills a little. I would've gotten onto the team, but I was too young. I als learned about the AMC 8. The outcome was similar. But, hen I was 10, I learned about AoPS. Correspondingly,  I learned about the AMC 10, AMC 12, AIME, USAMO, and IMO. I was already selected as a person on the scoring team in Mathcounts. I took the AMC 12 A and B, got 99.5 and 92.5, respectively. Then I boosted up my work, and the highest I got was 25th at states.(my first year in competing.)

Before states, I qualified for first alternate in NYSML. Some people dropped out, and I became a team member. I went to NYSML, my team got first place in division B, but I was too young to go to ARML. :noo:

I'm also going to take the December 2nd SAT this year.

And that's my experience.


:D  Solve in the  equation   $ [ \sin x] +[ \tan x] =[ \cos x] +[ \cot x ]$   :?:

_____________________________________
Azerbaijan Land of the Fire  :rotfl:

[quote="Pirkuliyev Rovsen"]:D  Solve in the  equation   $ [ \sin x] + [ \tan x] = [ \cos x] + [ \cot x ]$   :?: [/quote]

We can look for solutions in $ [0,2\pi)$. Then, in order to have $ \tan(x)$ and $ \cot(x)$ defined, $ x\notin\{0,\frac{\pi}2,\pi,\frac{2\pi}2\}$. Then :

1) ${ x\in(0,\frac{\pi}2})$ :
$ [\sin(x)]=[\cos(x)]=0$ and the equation is $ [\tan(x)]=[\cot(x)]$
But, if ${ x<\frac{\pi}4}$, then $ [\tan(x)]=0$ while $ [\cot(x)]\ge 1$
Same,  if ${ x>\frac{\pi}4}$, then $ [\tan(x)]\ge 1$ while $ [\cot(x)]=0$
Hence the unique solution $ x=\frac{\pi}4$

2) ${ x\in(\frac{\pi}2},\pi)$ :
$ [\sin(x)]=0$ and $ [\cos(x)]=-1$ and the equation is $ [\tan(x)]=-1+[\cot(x)]$
But, if ${ x<\frac{3\pi}4}$, then $ [\cot(x)]=-1$ and we have $ [\tan(x)]=-2$ and so ${ x\in[\pi-\arctan(2),\frac{3\pi}4})$
Same, if ${ x\ge\frac{3\pi}4}$, then $ [\tan(x)]=-1$ and we have  $ [\cot(x)]=0$, impossible

3) ${ x\in(\pi,\frac{3\pi}2})$ :
$ [\sin(x)]=-1$ and $ [\cos(x)]=-1$ and the equation is $ [\tan(x)]=[\cot(x)]$
But, if ${ x<\frac{5\pi}4}$, then $ [\tan(x)]=0$ while $ [\cot(x)]\ge 1$
Same,  if ${ x>\frac{5\pi}4}$, then $ [\tan(x)]\ge 1$ while $ [\cot(x)]=0$
Hence the unique solution $ x=\frac{5\pi}4$

4) ${ x\in(\frac{3\pi}2},2\pi)$ :
$ [\sin(x)]=-1$ and $ [\cos(x)]=0$ and the equation is $ -1+[\tan(x)]=[\cot(x)]$
But, if ${ x\le\frac{7\pi}4}$, then $ [\cot(x)]=-1$ and we have $ [\tan(x)]=0$, impossible
Same, if ${ x>\frac{7\pi}4}$, then $ [\tan(x)]=-1$ and we have  $ [\cot(x)]=-2$  and so ${ x\in(\frac{7\pi}4},2\pi-\arctan(\frac 12)]$

Synthesis of solutions :
=======================

Set of solution (modulo $ 2\pi$) is $ S=\{\frac{\pi}4\}\cup$ ${ [\pi-\arctan(2),\frac{3\pi}4})\cup$ $ \{\frac{5\pi}4\}\cup$ ${ (\frac{7\pi}4},2\pi-\arctan(\frac 12)]$


[quote="harazi"]Find good upper bounds of the form $N^{p}$ for the number of solutions of the equation $x_{1}^{r}+x_{2}^{r}+...+x_{k}^{r}=y_{1}^{r}+y_{2}^{r}+...+y_{k}^{r}$ in positive integers $x_{1},y_{1}$ between $1$ and $N$. Are there nontrivial lower bounds?[/quote]
Maximal value give $x_{i}=y_{i}=N$.


inequalities
trigonometry
geometry solved
geometry

Given a triangle $ ABC$ with angle $ C \geq 60^{\circ}$. Prove that:

$ \left(a \plus{} b\right) \cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \geq 4 \plus{} \frac {1}{\sin\left(\frac {C}{2}\right)}.$

Attempt of a halfways nice solution.

[color=blue][b]Problem.[/b] Let ABC be a triangle with $C\geq 60^{\circ}$. Prove the inequality

$\left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 4+\frac{1}{\sin\frac{C}{2}}$.[/color]

[i]Solution.[/i] First, we equivalently transform the inequality in question:

$\left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 4+\frac{1}{\sin\frac{C}{2}}$
$\Longleftrightarrow\ \ \ \ \ \left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{a+b}{c}\geq 4+\frac{1}{\sin\frac{C}{2}}$
$\Longleftrightarrow\ \ \ \ \ \left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}\right)-4\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$
$\Longleftrightarrow\ \ \ \ \ \frac{\left(a-b\right)^2}{ab}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$.

Now, by the Mollweide formulas,

$\frac{a+b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}$ and $\frac{a-b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}$, so that
$\frac{a-b}{a+b}=\frac{a-b}{c} : \frac{a+b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}} : \frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{\sin\frac{A-B}{2}\sin\frac{C}{2}}{\cos\frac{A-B}{2}\cos\frac{C}{2}}$.

Now, $\cos^2\frac{C}{2}\leq 1$ (as the square of every cosine is $\leq 1$). On the other hand, the AM-GM inequality yields $ab\leq\frac14\left(a+b\right)^2$. Hence,

$\frac{\left(a-b\right)^2}{ab}\geq\frac{\left(a-b\right)^2}{\frac14\left(a+b\right)^2}$       (since $ab\leq\frac14\left(a+b\right)^2$)
$=4\left(\frac{a-b}{a+b}\right)^2=4\left(\frac{\sin\frac{A-B}{2}\sin\frac{C}{2}}{\cos\frac{A-B}{2}\cos\frac{C}{2}}\right)^2=\frac{4\sin^2\frac{A-B}{2}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}\cos^2\frac{C}{2}}$
$\geq\frac{4\sin^2\frac{A-B}{2}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}$         (since $\cos^2\frac{C}{2}\leq 1$).
$=\frac{4\left(2\sin\frac{A-B}{4}\cos\frac{A-B}{4}\right)^2\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}=\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}$.

Thus, instead of proving the inequality $\frac{\left(a-b\right)^2}{ab}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$, it will be enough to show the stronger inequality

$\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$.

Noting that

$\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}=\frac{1}{\sin\frac{C}{2}}-\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{1-\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{2\sin^2\frac{A-B}{4}}{\sin\frac{C}{2}}$,

we transform this inequality into

$\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}\geq\frac{2\sin^2\frac{A-B}{4}}{\sin\frac{C}{2}}$,

what, upon multiplication by $\frac{\cos^2\frac{A-B}{2}\sin\frac{C}{2}}{16\sin^2\frac{A-B}{4}}$ and rearrangement of terms, becomes

$\sin^3\frac{C}{2}\cos^2\frac{A-B}{4}\geq\frac18\cos^2\frac{A-B}{2}$.

But this trivially follows by multiplying the two inequalities

$\sin^3\frac{C}{2}\geq\frac18$      (equivalent to $\sin\frac{C}{2}\geq\frac12$, what is true because $60^{\circ}\leq C\leq 180^{\circ}$ yields $30^{\circ}\leq\frac{C}{2}\leq 90^{\circ}$) and
$\cos^2\frac{A-B}{4}\geq\cos^2\frac{A-B}{2}$    (follows from the obvious fact that $\left|\frac{A-B}{4}\right|\leq\left|\frac{A-B}{2}\right|$ since $\left|\frac{A-B}{2}\right|<90^{\circ}$, what is true because $\left|A-B\right|<180^{\circ}$, as the angles A and B, being angles of a triangle, lie between 0° and 180°).

Hence, the problem is solved.

  Darij

Suppose w.l.o.g. $a\le b$. Because $C\ge 60^{\circ}$ there are only two cases:
$\blacktriangleright\ 1^{\circ}.\ a\le b\le c$. In the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=72226
I offered a strengthening of this inequality: \[ \boxed {\ (a+b)\left(\frac 1a+\frac 1b+\frac 1c\right)\ge 4+\frac{1}{\sin \frac C2}\cdot \sqrt{1+\left(\frac{a-b}{2c}\right)^2}\ }\ge 4+\frac{1}{\sin \frac C2}\ .  \] $\blacktriangleright\ 2^{\circ}.\ a\le c< b$.

[b]Proof.[/b] Remark two cases :
$\boxed {\ \bullet\ 1.1.\ }\ a\le c\le \frac{a+b}{2}\le b\ .$
$\frac{(a-b)^2}{ab}+\frac{a+b}{c}\ge \frac{(a-b)^2}{ab}+2\ge 2\ge \frac{1}{\sin \frac C2}\Longrightarrow$$\frac{(a+b)^2}{ab}+\frac{a+b}{c}\ge 4+\frac{1}{\sin \frac C2}\ .$
$\boxed {\ \bullet\ 1.2.\ }\ a\le \frac{a+b}{2}<c<b\ .$ [color=blue]???[/color]
[color=red]I don't not know how can continue !. Now and here I beg a little help.[/color]
I shall come back soon.

[hide="Please allow me to introduce you to ... the my darling !"] [quote="Virgil Nicula"]
$\blacktriangleright\ 1^{\circ}.\ a\le b\le c\Longrightarrow\boxed {\ (a+b)\left(\frac 1a+\frac 1b+\frac 1c\right)\ge 4+\frac{1}{\sin \frac C2}\cdot \sqrt{1+\left(\frac{a-b}{2c}\right)^2}\ }\ \ (\bigstar )$ [/quote]
[b]Proof.[/b] I shall use the following identities: $c^2=(a-b)^2+4(p-a)(p-b)\ ;$
$ab=p(p-c)+(p-a)(p-b)\ ;\ \sin \frac C2=\sqrt{\frac{(p-a)(p-b)}{ab}}\ ;\ \tan$ $\frac C2=\sqrt{\frac{(p-a)(p-b)}{p(p-c)}}\ .$ Thus,

$2c\ge a+b\Longleftrightarrow 2c(a+b)\ge (a+b)^2\ge 4ab\Longleftrightarrow$

$2c(a+b)+ab\ge \frac{a^2b^2}{(p-a)(p-b)}+4ab+ab-\frac{a^2b^2}{(p-a)(p-b)}\Longleftrightarrow$

$2c(a+b)+ab\ge \frac{a^2b^2}{(p-a)(p-b)}+4ab-\frac{abp(p-c)}{(p-a)(p-b)}$$\Longleftrightarrow 2c(a+b)(a-b)^2+ab(a-b)^2\ge$

$\frac{a^2b^2}{(p-a)(p-b)}\cdot \left[c^2-4(p-a)(p-b)\right]+4ab(a-b)^2-\frac{ab(a-b)^2p(p-c)}{(p-a)(p-b)}\Longleftrightarrow$

$2c(a+b)(a-b)^2+ab(a+b)^2\ge \frac{a^2b^2c^2}{(p-a)(p-b)}+ab(a-b)^2\left(4-\cot^2\frac C2\right)\Longleftrightarrow$

$\frac{2(a+b)(a-b)^2}{abc}+\left(\frac{a+b}{c}\right)^2\ge \frac{ab}{(p-a)(p-b)}+\left(\frac{a-b}{c}\right)^2\cdot \frac{3-5\cos C}{2\sin ^2\frac C2}\Longleftrightarrow$

$\left[\frac{(a-b)^2}{ab}+\frac{a+b}{c}\right]^2-\left[\frac{(a-b)^2}{ab}\right]^2\ge \frac{1}{\sin ^2\frac C2}\left[1+\left(\frac{a-b)}{c}\right)^2\cdot \frac{3-5\cos C}{2}\right]\ .$

But $C\ge 60^{\circ}\Longleftrightarrow \frac{3-5\cos C}{2}\ge \frac 14$. Therefore, $\frac{(a-b)^2}{ab}+\frac{a+b}{c}\ge \frac{1}{\sin \frac C2}\cdot \sqrt{1+\left(\frac{a-b}{2c}\right)^2}\Longleftrightarrow$

$\frac{(a+b)^2}{ab}+\frac{a+b}{c}\ge 4+\frac{1}{\sin \frac C2}\cdot \sqrt{1+\left(\frac{a-b}{2c}\right)^2}\Longleftrightarrow (\bigstar )\ .$[/hide]


What is the area of the intersection of the regions $x^{2}+y^{2}\le9$ and $x^{2}+y^{2}-10x+9\le0$?

[hide="Hint"]The first circle has a radius of 3 and is centered at (0,0)

With the second circle, we can add 16 to both sides and complete the square:

$(x-5)^{2}+y^{2}\le 4^{2}$

So we have a circle centered at (5,0) with radius 4.[/hide]


(Vasc) Let $ a,b,c$ be nonnegative real numbers and no two of them are zero. Prove that
$ \sqrt{1\plus{}\frac{48a}{b\plus{}c}}\plus{}\sqrt{1\plus{}\frac{48b}{c\plus{}a}}\plus{}\sqrt{1\plus{}\frac{48c}{a\plus{}b}} \ge 15.$

I wait for some nice solutions.

[quote="arqady"][quote="can_hang2007"](Vasc) Let $ a,b,c$ be nonnegative real numbers and no two of them are zero. Prove that
$ \sqrt {1 \plus{} \frac {48a}{b \plus{} c}} \plus{} \sqrt {1 \plus{} \frac {48b}{c \plus{} a}} \plus{} \sqrt {1 \plus{} \frac {48c}{a \plus{} b}} \ge 15.$

I wait for some nice solutions.[/quote]
For beginning:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64122[/quote]
Dear friend, see also that link for my solution. By the way, the lemma
$ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}} \ge 2\sqrt{\frac{a\plus{}b}{a\plus{}b\plus{}2c}},$
with $ c \equal{}\min\{a,b,c\},$ has also allowed to solve the following nice problem (of Vasc, too)
$ \left( \frac{a}{b\plus{}c}\right)^k\plus{}\left(\frac{b}{c\plus{}a}\right)^k\plus{}\left(\frac{c}{a\plus{}b}\right)^k \ge \frac{3}{2^k},$
for $ k \ge k_0$, where $ k_0$ is the root of the equation $ \frac{3}{2^k}\equal{}2.$

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers and no two of them are zero. Prove that

$ \sqrt {1 \plus{} \frac {48a}{b \plus{} c}} \plus{} \sqrt {1 \plus{} \frac {48b}{c \plus{} a}} \plus{} \sqrt {1 \plus{} \frac {48c}{a \plus{} b}} \ge 15.$[/quote]Without loss of generality, we can assume $ a \plus{} b \plus{} c \equal{} 1$ by the homogeneity.

$ 1 \plus{} \frac {48a}{1 \minus{} a} \minus{} \left(\frac {35 \plus{} 558a \minus{} 726a^2 \plus{} 1008a^3 \minus{} 216a^4}{35}\right)^2$

$ \equal{} \frac {12a(1 \minus{} 2a)^2(1 \minus{} 3a)^2(1645 \minus{} 2007a \plus{} 2415a^2 \minus{} 936a^3 \plus{} 108a^4)}{1225(1 \minus{} a)}\geq0\Longrightarrow$

$ \sum{\sqrt {1 \plus{} \frac {48a}{b \plus{} c}}} \equal{} \sum{\sqrt {1 \plus{} \frac {48a}{1 \minus{} a}}} \geq\sum{\frac {35 \plus{} 558a \minus{} 726a^2 \plus{} 1008a^3 \minus{} 216a^4}{35}}$

$ \equal{} \sum{\frac {35(a \plus{} b \plus{} c)^4 \plus{} 558a(a \plus{} b \plus{} c)^3 \minus{} 726a^2(a \plus{} b \plus{} c)^2 \plus{} 1008a^3(a \plus{} b \plus{} c) \minus{} 216a^4}{35}}$

$ \equal{} 15 (a \plus{} b \plus{} c)^4 \plus{} \frac {6}{35}\sum{(b \minus{} c)^2\left[17(b \plus{} c \minus{} a)^2 \plus{} 52bc\right]}\geq 15.$

[b]Remark[/b] $ 108a^4 \minus{} 936a^3 \plus{} 2415a^2 \minus{} 2007a \plus{} 1645$

$ \equal{} \frac {(18a^2 \minus{} 78a \plus{} 25)^2 \plus{} 261a^2 \minus{} 2121a \plus{} 4310}{3} > 0.$


In triangle $ ABC$, let $ A_1,A_2$ be points on $ BC$ such that $ AA_1$ is the median and $ AA_2$ is the angle bisector. Let $ K$ be a point on $ AA_1$ such that $ KA_2\parallel{}AC$. Prove that $ AA_2$ and $ KC$ are perpendicular.

[quote][color=darkred] Let $ ABC$ be a triangle in which $ AM$ and $ AD$ are the $ A$ - median and the $ A$ - bisector, where

$ \{M,D\}\subset BC$ . Take a point $ K\in AD$ such that $ MK\ \parallel\ AB$ . Prove that $ KA \perp KC$ .[/color][/quote][hide="Proof (very easy problem !)"]$ S\in AC\cap MK\implies$  $ \left\|\begin{array}{ccc}
MS\parallel\ AB & \Longrightarrow & SA\equal{}SC\\\\
\widehat {SAK}\equiv \widehat {BAK}\equiv\widehat {SKA} & \Longrightarrow & SA\equal{}SK\end{array}\right\|\ \implies\ SA$ $ \equal{}SC\equal{}SK\ \implies\ KA\perp KC$ .[/hide]


Let $ n$ coins be put into several (maybe one) piles. What is the greatest possible value of the product of the number of coins in the piles?

having piles of size 1 is clearly sub-optimal: integrating any singleton into any other pile increases the product. moreover, breaking any pile of size $ \geq 4$ into two piles of sizes both bigger than 1 makes the product at least as big (actually, always strictly bigger, except in the case of 4). so in a maximal product setting, all piles should have sizes 2 or 3. now, since $ 2 \plus{} 2 \plus{} 2 \equal{} 3 \plus{} 3$, and $ 3^2 > 2^3$, there should be no more than 2 $ 2$'s. it thus follows that

(1) if $ n \equal{} 3k$, the optimal product is $ 3^k$.
(2) if $ n \equal{} 3k \plus{} 1$, the optimal product is $ 2^2\cdot 3^{k \minus{} 1}$
(3) if $ n \equal{} 3k \plus{} 2$, the optimal product is $ 2\cdot 3^k$


Since $ f(x) \equal{} x^2 \minus{} 2x$, $ \frac{f(f(f(x)))\minus{}x}{f(x)\minus{}x}\equal{}(x^3\minus{}3x^2\plus{}3)(x^3\minus{}2x^2\minus{}x\plus{}1).$ Is there another function [i]f(x)[/i] of the form $ f(x)\equal{}x^2\plus{}ax$ where [i]a[/i] is integer such that polynom $ \frac{f(f(f(x)))\minus{}x}{f(x)\minus{}x}$ can be factorized into the terms with integer coefficients?

Every such function has the property you request: $ f(f(f(x))) \minus{} x$ is clearly divisible by $ x$.  In addition, $ f(1 \minus{} a) \equal{} 1 \minus{} a$ so $ f(f(f(1 \minus{} a))) \minus{} (1 \minus{} a) \equal{} 0$ and thus $ f(f(f(x))) \minus{} x$ is divisible by $ x \minus{} (1 \minus{} a)$.  Thus $ f(f(f(x))) \minus{} x$ is divisible by $ x(x \minus{} (1 \minus{} a)) \equal{} f(x) \minus{} x$.

I meant that quotient has to be factorized into two or more polynoms   :)


geometry circumcircle induction Stanford college

Given n > 3 points on the plane, is it true that there exists a circle passing through at least 3 of the points, not containing any other points in its interior?

I'm not sure if this will work

[hide]
consider any 3 of the points. If all the other points are either on or outside the circumcircle of the triangle formed by the three points, then we have the required circle. If any of the other points lies within the circumcircle, take two of the original three points plus that one, and thier circumcircle should exclude the other third point. presumably, we should be able to repeat this process until no more points are contained within the circumcircle.

It's the part that the new circumcircle should necessarily exclude all the other points who have been previously excluded that get's me...
[/hide]
help?

The opposite, but similar question:  Given $n > 3$ points on the plane, is it true that there exists a circle passing through at least $3$ of the points containing all of the other points in its interior?

[hide="Solution to the given problem"] Consider the triplet of points whose circumcircle is of minimum radius.  If any of the other points is in its interior, we can construct a smaller circumcircle by replacing one of the points on the circle with a point in its interior; contradiction. [/hide]
The proof to the opposite question is almost identical, and is featured in Engel in the chapter on the extremal principle.

Hey, we did this at Stanford  Math Circle!!   :) . Anyway, the basic idea is that any three points determine a circle. So, we take three points and make a circle through them. Then, if there are any points in the interior, we pick one and replace the point from our three points that is closest to that interior point. We repeat this process until there are no interior points remaining. Something like that...

Okay, can anybody offer a proof as to why replacing that third point will result in a smaller circumcircle. I can certainly find a point that will produce one large circumcircle with two small circumcircles. To achieve any degree of rigor, the proof has to account for this.

[quote="Altheman"]Okay, can anybody offer a proof as to why replacing that third point will result in a smaller circumcircle. [/quote]

Whoops.  On second thought, this might not be true.  Consider four points given by the vertices and centroid of an equilateral triangle.  All the circumradii are, I think, the same, which makes this an interesting case to account for.

I have no idea why I know this...

[hide]
For each of the n points, there is a region in the plane such that every point in that region is closer to that point than any of the other n-1 points, and the union of the regions plus the boundry is the plane (this is called a vononoi diagram for anyone interested).  The boundries of these regions are segments of the perpendicular bisectors between each pair of points, and importantly they intersect at a points, and there are finitely many straight edges.  So when you have this diagram, the point where three regions (or more) come together is the center of a circle which goes through the points that define those regions, and because of how we defined these regions we are guaranteed to have no points on the interior.

Interestingly, given n >= 3 points such that no 4 points are on a circle, there is a triangulation such that the circumcircle of each triangle contains no other points on the interior, and it is generated in the same way as I (tried to) describe above. This is called a Delaunay triangulation.[/hide]

The proposition is trivially false for all n points collinear. There exists be no circle to begin with.

Barring that scenario, I think we can agree that we can find a biggest circle that contains all of the points. So can we consider an inversion of the given points through a random circle, find that biggest circle containing the inverted points, and un-invert, hoping that the biggest circle does not pass through the center of the inversion circle?

edit. I take back the random circle part. I think it works if the inversion circle itself contains all of the given points. It's a bit sketchy when the biggest circle intersects the inversion circle.

[quote="nateharman1234"]Interestingly, given n >= 3 points such that no 4 points are on a circle, there is a triangulation such that the circumcircle of each triangle contains no other points on the interior, and it is generated in the same way as I (tried to) describe above. This is called a Delaunay triangulation.[/quote]

Wow, so then this is a [i]much[/i] stronger result than we had originally set out (and failed) to prove. Very interesting, indeed. I've heard of Voronoi diagrams and Delaunay triangulations in a computer science context, but never really looked up what they were, but it seems like they are powerful tools/results.

As has been pointed out, selecting the smallest circumcircle does not actually work. The only solution I have found uses a double application of the extreme principle:

[hide]

Choose the two points with smallest distance between them. Call them $A$ and $B$. Connect these with the point $P$ which maximizes $\angle APB$. Now, let's examine the consequences of this choice:

If a point $Q$ lies on the same side of $AB$ as $P$ and is within the circumcircle, then $\angle AQB > \angle APB$. If $Q$ lies on the other side, then $\angle AQB > 90$ and this contradicts the minimality of length $AB$.

[/hide]


If $\sqrt[3]{a} +\sqrt[3]{b} +\sqrt[3]{c}= 0$
Find the value of $(a+b+c)^3$

Too simple for Getting started i guess.

[hide]Ans: 27abc[/hide]

[hide]a^1/3 + b^1/3 + c^1/3 = 0

we know....if a+b+c=0 then a^3+b^3+c^3 = 3abc

using this

a + b + c = 3(abc)^1/3

(a+b+c)^3 = 27abc[/hide]


calculus
integration
trigonometry
derivative
calculus computations

Find the minimum value of $a\ (0<a<1)$ for which the following definite integral is minimized.

\[ \int_0^{\pi} |\sin x-ax|\ dx \]

$\int_0^\pi |\sin x - a x| \ dx = \int_0^k \left(\sin x - ax\right) \ dx + \int_k^\pi \left(ax-\sin x\right) \ dx = \frac{a\pi^2}{2}-ak^2$ $-2\cos k, \ k\in \left(0,\pi\right)$

$\frac{\partial}{\partial a} \left(\frac{a\pi^2}{2}-ak^2-2\cos k\right) = \frac{\pi^2}{2}-k^2=0 \Rightarrow k=\frac{\pi}{\sqrt{2}}$

$\sin \frac{\pi}{\sqrt{2}} - a \frac{\pi}{\sqrt{2}} = 0 \Rightarrow a = \frac{\sqrt{2}}{\pi}\sin \frac{\pi}{\sqrt{2}}$ and the minimum value is $-2\cos\frac{\pi}{\sqrt{2}}$.

I think theres is something wrong.
the $k$ you choose must be dependent of $a$
which means that $\frac{\partial}{\partial a} \left(\frac{a\pi^2}{2}-ak^2-2\cos k\right)$ is not equal to $\frac{\pi^2}{2}-k^2$
am I right ?  :?

what is wrong?
He differentiate it smoothly.since we need to minimize $\frac{a {\pi}^2}{2} - ak^2 - 2 \cos{k}$. we need to find $a$ and $k$ that minimize it. in this case it is stationary point.

In Japan, Japanese high school students don't study the pertial differentiation. :)

To avoid using partial derivatives we could say $f(a) = \frac{a\pi^2}{2}-ak^2-2\cos k \Rightarrow f'(a) = \frac{\pi^2}{2}-k^2$.

jimmy,How can you show the exixtence of minimal value? :roll:

Well, back to the partial derivatives:

$f(a,k) = \frac{a\pi^2}{2}-ak^2-2\cos k$

$f_a = \frac{\pi^2}{2}-k^2 = 0 \Rightarrow f_{aa} = 0 \Rightarrow f_{ak} = -2k$

$f_k = -2ak+2\sin k \Rightarrow f_{kk} = -2a+2\cos k \Rightarrow f_{ka} = -2k$

$D=\left| \begin{array}{cc}f_{aa} & f_{ak} \\f_{ka} & f_{kk} \end{array} \right| = 4k = 4\left(\frac{\pi}{\sqrt{2}}\right) >0$, so it's a minimum.


inequalities
algebra
system of equations

[b]Solution.[/b]  Suppose that $x\geq y\geq z$.  It then follows that
\[y-x\geq z-y\geq x-z\implies 2y\geq x+z,\ 2z\geq x+y,\ 2x\leq y+z\implies x\leq y\leq z.\]
As a result, $x=y=z$. Hence,
\[x+\sqrt{x^{2}+1}=1\implies x=0.\]
Therefore, the only solution to this system of equations is $(x,y,z)=(0,0,0)$.$\Box$

You also have to consider the case when $x,y,z$ are negative.

If one of $x,y,z$ is negative then if $x\leq y\leq z$ and $x<0$ so $10^{y-x}\geq 1$ so $x+\sqrt{x^{2}+1}\geq 1$ so $\sqrt{x^{2}+1}\geq 1+|x|$ as $x=-|x|$ so $x^{2}+1\geq 1+2|x|+x^{2}$ so $|x|\leq 0$ so $x=0$. Get a contradiction so none of the numbers are negative.
Now, if $x,y,z\geq 0$ multiply out the three equations to get: $(x+\sqrt{x^{2}+1})(y+\sqrt{y^{2}+1})(z+\sqrt{z^{2}+1})=1$. Now, $x+\sqrt{x^{2}+1}\geq 0$ if $x\geq 0$ so $(x+\sqrt{x^{2}+1})(y+\sqrt{y^{2}+1})(z+\sqrt{z^{2}+1})\geq 1$ with equality iff:
$x+\sqrt{x^{2}+1}=1 \Rightarrow x=0$;
$y+\sqrt{y^{2}+1}=1 \Rightarrow y=0$;
$z+\sqrt{z^{2}+1}=1 \Rightarrow z=0$;
This gives the only solution of $(0,0,0)$.

I don't see how boxedoxe's solution doesn't consider negatives.

You have $1-x=\sqrt{x^{2}+1}\Longrightarrow-2x=0\Longrightarrow x=0$.

My WLOG statement has no problem with having negatives in it.  Also, the inequalities I used do not have problems with having some of $x,y,z$ as negative integers.  As a result, I do not see how I did not consider the case when $x,y,z$ are negatives.

I understand, but you kind off skipped a few steps and jumped to the conclusion that if $x\geq y\geq z$ then $y-x\geq z-y\geq x-z$ which is not true if for example $z=-1000,y=0,x=1000$ so I think you have to show why $y-x\geq z-y\geq x-z$ is true based on the given information.

I was in a hurry to go eat lunch, so I skipped the clear steps.  However, the beginning was not explained enough which led to a controversy.


inequalities
calculus
derivative
calculus computations

For $x > 0,$ show that $e^{x}\ge \frac{e^{2}x^{2}}{4}.$

I used Laplace Transform to show the above inequality, but I am not sure if it is justified. Could anyone let me know if such a method is OK to use?

I basically took the Laplace transforms of both the sides of the given inequality, which yielded another inequality that was easy to prove.

Are there other (elementary) techniques?

[quote="FieryHydra"]Are there other (elementary) techniques?[/quote]

What do you want for "elementary?"  The two curves are tangent at $x = 2$.  For $x \geq 2$ the second derivatives are $e^{x}> e^{2}/2$, so it's easy to argue from there that $e^{x}$ is larger.  I don't think it's too hard to extend it backwards to $[0, 2)$.


quadratics
modular arithmetic
algebra
polynomial
function

prove that for every quadratic non residue $ a$ of a prime $ p$
$ a^{\dfrac {p\minus{}1} 2} \equiv {\minus{}1} \pmod{p}$

[hide="Discussion"] The polynomial $ x^{p\minus{}1} \minus{} 1 \equiv 0 \bmod p$ has as its roots every nonzero residue $ \{ 1, 2, 3, ... p\minus{}1 \} \bmod p$ by Fermat's Little Theorem.  On the other hand, when $ p$ is odd it factors as

$ \left( x^{ \frac{p\minus{}1}{2} } \minus{} 1 \right) \left( x^{ \frac{p\minus{}1}{2} } \plus{} 1 \right) \equiv 0 \bmod p$.

The left factor contains precisely the quadratic residues $ \bmod p$ (since there are at most $ \frac{p\minus{}1}{2}$ of them and all of them must be roots of the left factor) and the right factor contains precisely the quadratic non-residues $ \bmod p$.  The conclusion follows.

Restatement:  $ a^{ \frac{p\minus{}1}{2} } \equiv \left( \frac{a}{p} \right) \bmod p$.  I believe this is known as Euler's lemma. [/hide]

it is clear that all the quadratic residues are contained in $ x^\dfrac{p-1}{2}-1$ part
could you please elaborate how all the quadratic non-residues are contained in $ x^\dfrac{p-1}{2}+1$ and not in $ x^\dfrac{p-1}{2}-1$

P.S:if $ p$ is odd then there are exactly $ \dfrac{p-1}{2}$ quadratic residues and non-residues

[hide="Elaboration"] If $ a^2 \equiv b^2 \bmod p$ then either $ a, b \equiv 0 \bmod p$ or $ \left( \frac{a}{b} \right)^2 \equiv 1 \bmod p \implies \frac{a}{b} \equiv 1, \minus{}1 \bmod p$.  In other words, the function $ f(x) \equiv x^2 \bmod p$ takes each of its values (except $ 0$) exactly twice, so there are exactly $ \frac{p\minus{}1}{2}$ quadratic residues $ \bmod p$.  This settles every factor of $ x^{ \frac{p\minus{}1}{2} } \minus{} 1$.

Further reasoning along these lines allows you to prove the existence of a primitive root $ \bmod p$. [/hide]


modular arithmetic
quadratics
number theory
Perfect Square
Discriminant
IMO Shortlist

Let $ a, b, c$ be integers and $ p$ an odd prime number. Prove that if $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$ is a perfect square for $ 2p \minus{} 1$ consecutive integer values of $ x,$ then $ p$ divides $ b^2 \minus{} 4ac.$

To simplify the notation, we use $ \equiv$ for equality in modulo $ p$.
Denote $ A$ the set of all modulo of perfect squares. We know $ |A|=\frac{p+1}2$.

Case 1) $ a\equiv0$
$ ax^2+bx+c\equiv bx+c$. If $ b\not\equiv0$, we notice $ bx+c$ will go through all the modulo for $ 2p-1$ consecutive $ x$. But those $ ax^2+bx+c$ are perfect squares. We get contradiction with $ |A|=\frac{p+1}2$. Hence, $ b\equiv0$ and $ b^2-4ac\equiv0$.

Case 2) $ a\not\equiv0$, $ b\not\equiv0$.
Since $ 2a$ and $ p$ are coprimes, we can always find $ d$ such that $ 2ad+b\equiv0$.
$ f(x+d)=a(x+d)^2+b(x+d)+c=ax^2+b^*x+c^*$, where $ b^*=2ad+b$ and $ c^*=ad^2+bd+c$. It is easy to see that $ b^*^2-4ac^*=b^2-4ac$. Hence, we can always assume $ b\equiv0$ without loss of generality.

Case 3) $ a\not\equiv0$, $ b\equiv0$, $ p>3$.
We claim there exists $ u$ such that $ a\equiv u^2$.
When $ x\equiv0$, $ c\equiv ax^2+c\equiv f(x)\equiv t^2$ for some $ t$. Assume $ t$ to be even by replacing $ t$ with $ t+p$ when necessary.
i) $ c\equiv t^2\not\equiv 4$.
Find $ x$ such that $ ax^2+c\equiv (1+\frac{t^2}4)^2$.
or $ ax^2\equiv (1+\frac{t^2}4)^2-c\equiv (1+\frac{t^2}4)^2-t^2\equiv (1-\frac{t^2}4)^2\not\equiv0$.
Hence $ x\not\equiv0$ and the claim holds.
ii) $ c\equiv4$, $ p>5$.
Find $ x$ such that $ ax^2+c\equiv (\frac{(1+p)^2}4+4)^2$,
or $ ax^2\equiv (\frac{(1+p)^2}4+4)^2-c\equiv (\frac{(1+p)^2}4+4)^2-(2+2p)^2=(\frac{(1+p)^2}4-4)^2$. Notice $ =(\frac{(1+p)^2}4-4)^2\not\equiv0$ when $ p>5$.
Again $ x\not\equiv0$ and the claim holds.
iii) $ c\equiv4$, $ p=5$.
Find $ x$ such that $ ax^2+c\equiv0$. $ ax^2\equiv-c\equiv-4\equiv1$. The claim holds.
Now $ f(x)\equiv ax^2+c\equiv (ux)^2+c$. Obviously $ f(x)$ goes through $ A+c$. We know $ f(x)$ goes through $ A$ and $ c\in A$. So $ 2c$ has to be in $ A$, so do $ 3c$, $ 4c$, etc. The only way to make sure $ |A|=\frac{p+1}2$ is to let $ c\equiv0$. Hence, $ b^2-4ac\equiv0$.

Case 4) $ a\not\equiv0$, $ b\equiv0$, $ p=3$.
Using similar argument above, $ b^2-4ac\equiv0$ holds if $ a\equiv1$.
Now consider $ a\equiv-1$. We either have $ c\equiv0$ or $ c\equiv1$. For $ c\equiv0$, we get $ ax^2+c\equiv-1$ when $ x\equiv1$, which contradicts with $ -1\not\in A$.
Now consider $ a\equiv-1$, $ b\equiv0$, and $ c\equiv1$. We can look at  all possible combination of (a,b,c) with modulo 9 and prove there is no way to find 5 consecutive numbers to give $ f(x)$ whose modulo with 9 all falls into $ \{0,1,4,7\}$. I wish someone gives a more efficient proof.

the case $ a \equiv 0 \pmod{p}$ is easy to check.
now suppose that $ p \nmid b^2\minus{}4ac$ then see [url=http://www.mathlinks.ro/viewtopic.php?t=150500]here[/url]...

Dear xxp2000, I think we can prove there exists $u$ such that $a \equiv u^2$ by proving that there exist $d$ such that $p \mid bd+c$. Hence, $ad^2 \equiv f(d) \equiv t^2$.

let the variables of x be element of {k,k+1,....k+2p-2}    S= [(ax^2+bx+c)/p]  ( let [ ] be the lagrange symbol)
it's clear that S=2p-1. for an integer a coprime to p it's well known (MONT page 220) that for x is elements of residue class modulo p
 [(ax^2+bx+c)/p]= (p-1) [a/p]  for p|b^2-4ac
                                 -[a/p]  otherwise
it's clear that  {k,k+1,....k+2p-2,k+2p-1} contains all residues modulo p twice.
 Situation 1 : if p and a coprime
 then  other hand  S=2( -[a/p]) +1,  2( -[a/p]) -1, 2(p-1) [a/p] +1 ,2(p-1) [a/p] -1.
But for p>3 only solution S=2p-1= 2(p-1) [a/p] +1  where [a/p]=1  so p|b^2-4ac.

Situation 2 :
if p|a then S= [(ax^2+bx+c)/p]= [(bx+c)/p] then if p|b p|b^2-4ac so assume that p and b coprime.then for value of x    bx+c will be residues modulo p and we know the number of quadratic rresidues in modulo p is (p-1/2) so S =2(p-1/2)+1 or 2(p-1/2)-1   contridiction .

QED.


probability
combinatorics unsolved
combinatorics

A box contains $ 10$ light bulbs, $ 5$ of wihch are bad and $ 5$ which are good. if $ r$ less than or equal to $ 5$ light bulbs are drawn at random [b]with[/b] replacement from the box, what is the epxected number of good light bulbs drawn?

I understand without replacement, I can just use linearity of expectation and it would end up being $ r/2$
but I'm curious what it is with replacement?

Each bulb has probability 1/2 of being good whether you draw with or without replacement.  The condition $ r \leq 5$ is also totally unnecessary (although in the without replacement case, you do need $ r \leq 10$).


Take a triangle and inscribe an ellipse in it.  Join each vertex with the point of tangency (of the ellipse with the triangle) of the opposite side.  Do the three lines meet at one point?  Always?

Yes, this is a degenerate case of the Brianchon theorem: A hexagon $ABCDEF$ is circumscribed around a conic section, if and only if the lines $AD,BE$ and $CF$ concur. Alternatively, for this particular case, project the conic into a circle by central projection. The subject lines obviously concur at the Gergonne point of the projected triangle, so the three primitive lines also concur.


Greetings,

I do apologise if I make any errors regarding posting etiquette and such. This is my first post. The situtation is as follows,

(1) Downloaded and installed Asymptote 1.43
(2) Set up TeXNicCenter as an editor for Asymptote
(3) Tested that the installation works (some examples from the Asymptote manual)
(4) Downloaded trunk.tar.gz from [url]http://svnweb.tuxfamily.org/listing.php?repname=piprim/asymptote&path=%2Ftrunk%2F&rev=0&sc=0[/url]

(5) I then copied the .asy files from here into the Asymptote installation folder.
(6) I entered the following into the editor
[code]
import base_pi;
size(8cm,0);

for (int i=-4; i<=4; ++i)
  {
    /* View the definition of string texfrac(int,int,string,bool,bool,bool,bool) */
    label(texfrac(i,4), (i*cm,0));
    label(texfrac(i,4,signin=true), (i*cm,-cm));
    label(texfrac(i,4,factor="\pi"), (i*cm,-2cm));
    label(texfrac(i,4,factor="\pi",factorin=false), (i*cm,-3cm));
    label(texfrac(i,4,factor="\pi",signin=true,factorin=true), (i*cm,-4cm));
    label(texfrac(i,4,factor="\pi",signin=true,factorin=false,displaystyle=true,zero=false), (i*cm,-5cm));
  }
[/code]

And upon compilation receive the following error

"
C:\Program Files\Asymptote/base_pi.asy: 171.23: cannot cast 'real[][]' to 'real[
]'
C:\Program Files\Asymptote/base_pi.asy: 182.23: cannot cast 'real[][]' to 'real[
]'
C:\Program Files\Asymptote/base_pi.asy: 193.6: cannot convert 'real[][]' to 'rea
l[]' in assignment
C:\Program Files\Asymptote/base_pi.asy: 240.16: no matching function 'points(pat
h, real[][])'
"

From searching the forum it seems that the versions of the .asy files I downloaded are correct.

Does anyone have an idea as to how I can sort out this error? I am working on a textbook (guess what topic -- see my username  :) ) so I;m eagerly waiting to complete some diagrams in Asymptote.

Thanks in advance.

[quote="Ph. Ivaldi"]Hi,

The SVN version of base_pi that you downloaded works with the SVN version of Asymptote.
Can you wait the incoming version of Asymptote (1.44) ?[/quote]

Hi,

Thank you for your very quick reply. If there isn't a way of getting a version of base_pi that works with v1.43 of Asymptote then I will have to wait. I have a two weeks at most after which I will have to start constructing the illustrations for the book.  :)  Would you by any chance know when  v1.44 of Asymptote is scheduled to be launched?

Thanks again.

[quote="matlabguy"]If there isn't a way of getting a version of base_pi that works with v1.43 of Asymptote then I will have to wait.[/quote]
With Asy 1.43 you may try [url=http://svnweb.tuxfamily.org/listing.php?repname=piprim/asymptote&path=%2Ftrunk%2F&rev=113&sc=0]these versions[/url] of packages.

[quote="matlabguy"]Would you by any chance know when  v1.44 of Asymptote is scheduled to be launched?[/quote]
For very soon but I do not know exactly when  :maybe:
An exceptional work was done on 3D (PRC imbendding in PDF file and OpenGL output).
It's the documentation to write and surely some trailing bugs to fix ...

[quote]With Asy 1.43 you may try [url=http://svnweb.tuxfamily.org/listing.php?repname=piprim/asymptote&path=%2Ftrunk%2F&rev=113&sc=0]these versions[/url] of packages.[/quote]

Thank you! I downloaded the .asy files and it worked perfectly!!


function
limit
trigonometry
calculus
calculus computations

can you guys help me to prove the this limit is 0, using the definition of limits.

lim (x,y) --> (0,0) [(x^2)(sin^2(y))]/[x^2+2y^2]

thanks

Here is one of the cleanest solutions:

[hide="Solutions"]Divide the numerator and denominator by $ x^2$ to get:

$ \lim_{(x,y) \to \mathbf{0}} \frac {\sin ^2 y}{1 \plus{} 2 \left(\frac {y}{x} \right)^2}$

As we approach the origin, the numerator goes to $ 0$ but the denominator is always at least $ 1$, so the limit is $ 0$.[/hide]

It's pretty bad if you want to show it [i]purely [/i]using the definition of a limit, since our input accuracy is not a single offset but a vector. However,

[hide="you can"]squeeze $ \sin^2 y$ between $ 0$ and $ y^2$, which gives:

$ 0 \leq L \leq \lim_{(x,y) \to \mathbf{0}} \frac {x^2 y^2}{x^2 \plus{} 2y^2} \equal{} \frac {1}{2} \lim_{(x,y) \to \mathbf{0}} \frac {x^2 (2y^2)}{x^2 \plus{} 2y^2} \equal{} \frac {1}{2} \lim_{(x,y) \to \mathbf{0}} \frac {1}{\frac {1}{x^2} \plus{} \frac {1}{2y^2}}$

You can now easily use the definition of a limit to show that the rightmost limit equals 0, completing the proof.[/hide]

Although, if you're going to use squeeze, you may as well

[hide="do this"]
(continued from the previous hide)

$ 0 \leq L \leq \lim_{(x,y) \to \mathbf{0}} \frac {x^2 y^2}{x^2 \plus{} 2y^2} \equal{} \frac {1}{2} \lim_{(x,y) \to \mathbf{0}} \frac {1}{\frac {1}{x^2} \plus{} \frac {1}{2y^2}} \leq \frac {1}{2} \lim_{(x,y) \to \mathbf{0}} x^2 \equal{} 0$
[/hide]

this is a common calc III problem. first one must establish that there is a limit .

$ \lim_{(x,0) \to 0}\frac{sin^2{y}}{1+2\left(\frac{y}{x}\right)^2}$ must equal $ \lim_{(0,y) \to 0}\frac{x^2sin^2{y}}{x^2+2y^2}$

 the two ways that I wrote it are the same but each in its own way is necessary to do the limit
 the limit on the left hand side reduces to $ \frac{sin^2 {0}}{1+2(0)^2}=\lim_{x\to 0} 0=0$
  the limit on the right hand side reduces to $ \frac{(0)sin^2 {y}}{0^2+2(y)^2}= \lim_{y\to 0}\frac{0}{2y^2} = 0$

so both limits are equal and converge to 0 so the limit is 0

dude but........ that's doesn't prove it... it only shows that along these two paths the limits are 0. I know that already...

and these two paths are converging to the origin which is (0,0)
which is what we want


function
algebra
polynomial
trigonometry
integration
complex analysis
real analysis

Find the limit of this sequence of numbers:
     $ u_n=\sum_{i=1}^n \frac1{i^2}$
I know that the limit is $ \frac{\pi^2}6$ but I can't prove it.

do you know fourier series ? there is one simple proof of this result if you know that.

There are Fourier series methods - several different Fourier series methods, using different functions and different principles from Fourier series. There are complex analysis methods, that involve creating a function with the right poles and evaluating it by the residue theorem. There is Euler's original method, which applies Viete's relationships for the roots and coefficients of a polynomial to the infinite product representation of $\sin x$. All of these methods are capable (with increased work at each step) of finding $\sum_{k=1}^{\infty}\frac1{k^2}, \sum_{k=1}^{\infty}\frac1{k^4}, \sum_{k=1}^{\infty}\frac1{k^6},$ etc. (Also known as $\zeta(2),\zeta(4),\zeta(6),\dots$.)

But here's a method just for the squares. I'll offer it only as a first step and a hint, and leave the finish as a challenge:

$\int_0^1\int_0^1\frac{dxdy}{1-xy}= \int_0^1\int_0^1\sum_{k=0}^{\infty}x^ky^kdxdy$

$=\sum_{k=0}^{\infty}\int_0^1\int_0^1x^ky^kdxdy= \sum_{k=0}^{\infty}\frac1{(k+1)^2}=\sum_{k=1}^{\infty}\frac1{k^2}$

Your challenge: find a two-dimensional change of variables that gives you a different way of computing that double integral.

On Robin Chapman web page there are 14 proof.