trigonometry
inequalities
inequalities unsolved

For all positive reals $ a$, $ b$, $ c$, show that:

$ c\sqrt{a^2 \minus{}ab \plus{} b^2} \plus{} a\sqrt{b^2 \minus{}bc \plus{} c^2} \geq b\sqrt{a^2 \plus{}ac \plus{} c^2}$.

I am sure this was solved on ML before, but since I cannot find the link, let me post the solution.

[color=blue][b]Problem.[/b] If a, b, c are three positive reals, then show that

$c\sqrt{a^2-ab+b^2}+a\sqrt{b^2-bc+c^2}\geq b\sqrt{c^2+ca+a^2}$.[/color]

[i]Solution.[/i] Let O be a point, and A, B, C three other points in the plane such that < AOB = < BOC = 60° (and the angles AOB and BOC are equally directed) and OA = a, OB = b and OC = c. Then, < COA = < AOB + < BOC = 60° + 60° = 120°. Now, by the cosine law in triangles BOC, COA and AOB, we have

$BC=\sqrt{OB^2+OC^2-2\cdot OB\cdot OC\cdot\cos\measuredangle BOC}=\sqrt{b^2+c^2-2bc\cdot\cos 60^{\circ}}$
$=\sqrt{b^2+c^2-2bc\cdot\frac12}=\sqrt{b^2-bc+c^2}$;
$CA=\sqrt{OC^2+OA^2-2\cdot OC\cdot OA\cdot\cos\measuredangle COA}=\sqrt{c^2+a^2-2ca\cdot\cos 120^{\circ}}$
$=\sqrt{c^2+a^2-2ca\cdot\left(-\frac12\right)}=\sqrt{c^2+ca+a^2}$;
$AB=\sqrt{OA^2+OB^2-2\cdot OA\cdot OB\cdot\cos\measuredangle AOB}=\sqrt{a^2+b^2-2ab\cdot\cos 60^{\circ}}$
$=\sqrt{a^2+b^2-2ab\cdot\frac12}=\sqrt{a^2-ab+b^2}$.

Now, applying the Ptolemy inequality to the quadrilateral OABC, we get $OC\cdot AB+OA\cdot BC\geq OB\cdot CA$. Using these relations, this becomes $c\sqrt{a^2-ab+b^2}+a\sqrt{b^2-bc+c^2}\geq b\sqrt{c^2+ca+a^2}$, and the problem is solved.

  Darij

Yes, that is an old problem. We can make substitutions to have the following
\[ \sqrt{x^2-xy+y^2}+\sqrt{y^2-zy+z^2}\geq\sqrt{x^2+zx+x^2}.  \]


calculus
integration
trigonometry
calculus computations

whats the easiest way to integrate this

\[ \int {\frac{1}{x^4 + 1}}dx \]

and is this possible

 \[ \int {\frac{x \sin{x}} {1 + \cos^2{x}}} dx  \]

i know you can use partial fractions for the first one but is their an easier way.


integration
function
Functional Analysis
real analysis
real analysis unsolved

Is the weak convergence in $ L^P,1\le p<\infty,$ unique? How?

[quote="WWW"][quote="man.maiko"]Is the weak convergence in $ L^P,1\le p < \infty,$ unique? How?[/quote]

Let's ask a proper question:  "If $ f_n \rightarrow f$ weakly in $ L^p$ and $ f_n \rightarrow g$ weakly in $ L^p$,  is $ f \equal{} g$ (in the $ L^p$ sense)?" Is that what you mean?[/quote]

Yes. Thanks.

This is true for any Banach space. To prove this just observe that if $ f\ne g$ are both weak limits then $ l(f\minus{}g)\equal{}0$ for any linear bounded functional $ l$. Now just use Hahn-Banach theorem to construct the functional $ l$ such that $ l(f\minus{}g)\equal{}1$ :wink:

[quote="leshik"]This is true for any Banach space. To prove this just observe that if $ f\ne g$ are both weak limits then $ l(f \minus{} g) \equal{} 0$ for any linear bounded functional $ l$. Now just use Hahn-Banach theorem to construct the functional $ l$ such that $ l(f \minus{} g) \equal{} 1$ :wink:[/quote]

It seems complicated, at least for me.

Please check my arguments as follows:

It is clear
\[ \int (f \minus{} g)w \equal{} 0\quad (*)\]
for all $ w\in L^q.$ Choose $ w \equal{} |f \minus{} g|^{p/q}sgn(f\minus{}g).$ Then $ w\in L^q$. Therefore
\[ (*) \equal{} > \int |f \minus{} g|^p \equal{} 0.\]
The claim is proved.

The only real difference between that and what leshik said: instead of relying on a general existence theorem obtain the linear functional $ l,$ you used your concrete knowledge of $ L^p$ and its dual to explicitly construct $ l.$

Minor notational statement: in the case $ p \equal{} 1,q \equal{} \infty,$ you intend $ w \equal{} \text{sgn}(f \minus{} g).$ You're also tacitly assuming real-valued functions; for complex-valued functions you'd be using $ \overline{\text{sgn}(f \minus{} g)}.$

If I were to ask you whether this statement were true for $ L^{\infty},$ you'd be forced to use leshik's proof which quotes Hahn-Banach.

New question: if $ f_n\to f$ and $ f_n\to g$ in the weak* sense of $ L^{\infty},$ does $ f \equal{} g$ in the $ L^{\infty}$ sense?


Complex numbers (write in terms of a complex variable)

[hide="solution"]
Let $ r_1$, $ r_2$, ..., $ r_7$ be the 7th roots of unity, A.K.A the roots of the equation $ x^7\minus{}1\equal{}0$. Then $ r_n\equal{}\cos{\frac{n\pi}7}\plus{}i\sin{\frac{n\pi}7}$, with $ r_7\equal{}1$. Let $ S$ be the sum that we are trying to find. Then $ S\equal{}re[r_1\plus{}r_2\plus{}r_3]\equal{}.5re[r_1\plus{}r_2\plus{}...\plus{}r_6]\equal{}.5(r_1\plus{}r_2\plus{}...\plus{}r_6)\equal{}.5(r_1\plus{}r_2\plus{}...\plus{}r_7 \minus{}1)$. Now we can use the fact that $ r_1$,...,$ r_7$ are the roots of $ x^7\minus{}1\equal{}0$, and therefore $ r_1\plus{}r_2\plus{}...\plus{}r_7\equal{}0$. Thus, $ S\equal{}.5(0\minus{}1)\equal{}\boxed{\minus{}.5}$
[/hide]


under what conditions are you allowed to rearrange the terms of an infinite sum?

For example in the case of
$ \sum_{m \equal{} 0}^{\infty}\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^km^{2k}}{m^2(2k)!}$
both
$ \sum_{k \equal{} 0}^{\infty}\left|\frac{( \minus{} 1)^km^{2k}}{m^2(2k)!}\right|$
and
$ \sum_{m \equal{} 0}^{\infty}\left|\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^km^{2k}}{m^2(2k)!}\right|$ converge,

yet if you rearrange it to
$ \sum_{k \equal{} 0}^{\infty}\sum_{m \equal{} 0}^{\infty}\frac {( \minus{} 1)^km^{2k}}{m^2(2k)!}$
it doesn't converge

You are trying to use the associative property of the sum.  As far as I know, you can only be sure that this property may be used when the ENTIRE series is absolutely summable.  In your case, this means taking ALL terms of the form $ \frac {m^{2k}}{m^2\left(2k\right)!}$ and adding them, for ALL values of $ k$ and $ m$, and finding a finite number.  This is obviously not true in your case, so the associative property cannot be guaranteed to work.  As you may know, one of the most famous examples of this fact comes from the series $ 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} \frac {1}{5} \minus{} ...$.  Calling $ S \equal{} \frac {1}{1} \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} ...$, then $ \frac {S}{2} \equal{} \frac {1}{2} \plus{} \frac {1}{6} \plus{} \frac {1}{10} \plus{} ...$, $ \frac {S}{4} \equal{} \frac {1}{4} \plus{} \frac {1}{12} \plus{} \frac {1}{20} \plus{} ...$, and so on, so $ 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} \frac {1}{5} \minus{} ... \equal{} S \minus{} \frac {S}{2} \minus{} \frac {S}{4} \minus{} ... \equal{} S \minus{} S \equal{} 0$, which is false!

Whether there are other less exigent conditions than the absolute summability, I cannot recall, but the condition of absolute summability is definitely not satisfied by your series.

Usually, the absolute summability is not directly proved; instead, the absolute sum is shown to have a finite upper bound, typically by bounding each of the terms in the given series by a corresponding term in an absolutely summable series.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1928392338&t=89719]This topic[/url] is closely related.

In this particular case, the double sum does not converge absolutely - that is,

$ \sum_{k\equal{}0}^{\infty}\sum_{m\equal{}0}^{\infty}|a_{k,m}|\equal{}\infty.$

Hence we cannot justify switching the order of summation.


find all pairs (k,n) of positive integers for which 7^k - 3^n divides k^n + n^2

That problem is from IMO Shortlist 2007.

See here: http://www.mathlinks.ro/viewtopic.php?p=1106459#p1106459


Does anyone know how to draw a border around a chunk of text? I  have a figure that consists of some sample text, and I would like to have a border around it to distinguish it from the body of the document. I'm looking for something that is like [code]\boxedverbatim[/code] only without the verbatim part.

Here's what my figure looks like:
[code]
\begin{figure}[htp]
\small
The bordered text should be here.
\caption{\small
Some caption.}
\label{fig:myfig}
\end{figure}
[/code]

Thanks!

$\boxed{some random text}$

\boxed{TEXTHERE}

That doesn't seem to work. First, it doesn't allow math-mode inside of the brackets. Second, if even if there are no math commands, the spaces inside disappear (I get text like "hellothisistext" instead of "hello this is text") and the text stays on one line and goes off the edge of the page.

\boxed is for text that is [i]already[/i] in math mode such as $\boxed{\sin x}$. If you want a box round text mode then use \fbox as in $\fbox{this is framed text}$. You can also change to math mode inside the box.

But that is for just one line of text. If you want more than one line of text then use \parbox inside \fbox as in $\fbox{\parbox{2.5cm}{More than \\ one line of text}}$

Is there a way to make the text auto-wrap when using the fbox and parbox?

It does auto-wrap so I didn't need the  newline \\ in the example I gave :)
$\fbox{\parbox{2.5cm}{More than one line of text}}$
or
$\fbox{\parbox{1.2cm}{More than one line of text}}$
the difference just being the width of the parbox you set, which you will need to experiment with.
You can have double boxes
$\fbox{\fbox{\parbox{1.5cm}{More than one line of text}}}$
and you can change the thickness of the box lines
$\setlength{\fboxrule}{0.5mm}\fbox{\parbox{2.5cm}{More than one line of text}}$
and the amount of space between the frame and the text
$\setlength{\fboxsep}{3mm}\fbox{\parbox{2.5cm}{More than one line of text}}$


[quote="luiseduardo"](Peru/2007) Determine the value of x, so that the sum of the numbers of each row, column or diagonal is always the same:

[img]http://img219.imageshack.us/img219/6227/quadrado.jpg[/img][/quote]

Comparing upper line with leftmost row gives $ c\equal{}114$
Comparing middle line with rightmost row gives $ a\plus{}1\equal{}e\plus{}96$
Comparing upper line with diagonal containing $ x$ gives $ a\plus{}e\equal{}19\plus{}96$ and so, with previous line : $ a\equal{}105$
Comparing then upper line and the other diagonal gives $ x\plus{}19\equal{}a\plus{}c\equal{}105\plus{}114$ and so $ \boxed{x\equal{}200}$


The 1st, 2nd, and 3rd perfect squares are underlined in the following arithmetic sequence, [b]1[/b], 8, 15, 22, 29, [b]36[/b], 43, 50, 57, [b]64[/b], ... Find the 100th perfect square in the sequence.

[hide]The solutions to x^2=1mod7 are x=1 or 6 mod7. Thus the 1000th square is (50*7-1)^2=121801[/hide]

The general term of the sequence is $ 7n\minus{}6$.  We require:
$ 7n\minus{}6\equal{}k^2 \Rightarrow k\equiv 1, 6\pmod{7}$
The 100th square will be 6 (mod 7), so it will be the 50th term of the sequence 6, 13, 20, ...
$ \therefore k \equal{} 6 \plus{} 7 (50 \minus{} 1) \equal{} 349$
And so the 100th square in the sequence is $ 349^2 \equal{} 121801$

Why is the 100th perfect square $ 6 \mod7$?

Squares given by machodog's formula are... $ 1, 6, 8, 13, etc.$
as you can see, the 100th number will be equivalent to 6 mod 7.

Yeah sorry I should have made that clearer, I meant the hundredth square will have $ k \equiv 6 \pmod{7}$.


The $ 2N\equal{}2004$ real numbers $ x_1,x_2, \dots, x_{2004}$ are all equal to $ \sqrt2\minus{}1$ or $ \sqrt2\plus{}1$. Consider the sum $ M\equal{}\sum_{k\equal{}1}^{N} x_{2k\minus{}1}x_{2k}$.
a) Is $ M\equal{}2004$ possible?
b) How many different integer values can $ M$ attain?

$ x_{2k\minus{}1}x_{2k}$ can take three values : $ 1$, $ 3\minus{}2\sqrt{2}$ and $ 3\plus{}2\sqrt{2}$.

Let $ A$, $ B$ and $ C$ the number of terms $ x_{2k\minus{}1}x_{2k}$ in $ M$ which take values $ 1$, $ 3\minus{}2\sqrt{2}$ and $ 3\plus{}2\sqrt{2}$ respectively ($ A\plus{}B\plus{}C\equal{}N\equal{}1002$).

Then $ M\equal{}A\plus{}3(B\plus{}C)\plus{}2\sqrt{2}(C\minus{}B)$ is an integer if and only if $ B$=$ C$.

So the integer values that $ M$ can attain are $ A\plus{}6B$ with $ A\plus{}2B$ being equal to $ 1002$.

The integer values that $ M$ can attain are $ 1002\plus{}4B$ with $ B$ in the set $ \{0,1,...,501\}$.

Conclusion : $ M$ can attain $ 502$ integer values, but not 2004 because $ 2004\not \equiv 1002 \mod 4$.


inequalities
topology
function
IMO Shortlist
inequalities proposed

Prove that if $ n>2$ and $ a_1,...,a_n\geq 0$ verify $ (a_1)^n+(a_2)^n+...+(a_n)^n=n$ then we have the inequality $ (a_2\cdot\ a_3\cdot...\cdot a_n)^{n+1}+...
 +(a_1\cdot a_2\cdot...\cdot a_{n-1})^{n+1}\leq n $.  I have a terrible proof for this one and I'm not even sure it is good. What do you think?

I hope I didn't make any mistakes:

for $n=2$ we get $a_{1}^{2}+a_{2}^{2}=2$ and we take $a_{1}=\sqrt{2}$ and $a_{2}=0$ then we have:$a_{2}^{3}+a_{1}^{3} \leq 2$ which is $ (\sqrt{2})^{3} \leq 2$ which is false.

Please tell me if I didn't understand the inequality correctly!
Thank you very much!

Yes Harazi, the your claimed inequality is valid.
Using the known way, it reduces to $f(x)\leq n^2$, where

$f(x)=x_1x_2...x_n[(n+1)(1/x_1+1/x_2+...+1/x_n)-n]$

and $x_1+x_2+...+x_n=n$. This follows from

$f(x_1,x_2,x_3,...,x_n)\leq f(\frac{x_1+x_2}{2},\frac{x_1+x_2}{2},x_3,...,x_n)$,

which is true because of

$1/x_3+...+1/x_n\geq \frac{(n-2)^2}{x_3+...+x_n}= \frac{(n-2)^2}{n-x_1-x_2}\geq \frac{(n-2)^2}{n}\geq \frac{n}{n+1}$

for $n\geq4$.

I belive that the inequality can be refined as follows:
If  a_i are positive numbers such that $(a_1)^n+...+(a_n)^n=n$, then

$(a_2...a_n)^{2n-2}+...+(a_1...a_{n-1})^{2n-2}\leq n$.

This inequality is also valid for $n=2$.

It's not my inequality, it was just a question. But you're right, I should have said Vasc instead of Harazi. I apologise. And your solution is far more interesting and beautiful than mine.

You have correctly written "After a problem of Vasc-a general form" and
"Vasc problem generalized"
I don't understand which is the problem! :)
I haven't now a solution to the refined inequality. Try you?

No, I don't have but I will try to think, though I have tomorrow a stupid exam in informatics.

This is really a beautiful inequality and I would like to understand its beautiful proof.

Sorry, but I don't understand :

why the original inequality is equivalent to

$ 1)  $  $ f(x) \leq n^2 $

and why $1) $ follows from $2)$

$2)$ $f(x_1,x_2,..,x_n) \leq f(\frac{x_1+x_2}{2},\frac{x_1+x_2}{2},x_3,...x_n) $

Thank you very much.

Has anyone proved the stronger form of the inequality that Vasc suggested? I spent ne night on this inequality and couldn't prove anything interesting.

I propose to begin with the case n=4:
[quote]If $a,b,c,d$ are positive numbers such that $a^2+b^2+c^2+d^2=4$, then
$(abc)^3+(bcd)^3+(cda)^3+(dab)^3\leq4$.[/quote]
I wish you a beautiful week end.
 :)

Even the case n=4 is one of the most difficult inequalities I have ever solved. I have found ana allmost solution:
         Take $ d$ the minimum among the four numbers. And take $ f(a,b,c,d)=(abc)^3+(bcd)^3+(cda)^3+(dab)^3 $. We will prove that if $ u=\sqrt{\frac{a^2+b^2+c^2}{3}} $ then we have $ f(a,b,c,d)\leq f(u,u,u,d) $. But this comes down to the following monster (hi, Lagrangia!):
       $\sqrt{\frac{a^2+b^2+c^2}{3}}\cdot (\frac{a^2+b^2+c^2}{3})^4-(abc)^3\geq d^3\cdot ((ab)^3+(bc)^3+(ca)^3-\frac{(a^2+b^2+c^2)^3}{9}) $. Since $ d^3\leq abc$ it is enough to prove the monster when $ d^3=abc$ and in this case it becomes a homogenuous monster ( I think he is proud he is homogenuos monster!  :D  :D ) and thus we can assume that $ abc=1 $. Then we obviously have $\sqrt{\frac{a^2+b^2+c^2}{3}}\geq 1$ and it is enough to prove that $ (bc)^3+(ca)^3+(ab)^3-\frac{(a^2+b^2+c^2)^3}{9}\leq (\frac{(a^2+b^2+c^2)^4}{81}) -1 $. Now, let us study this separately. We have from Schur inequality and obvious identities and the fact that $ a^2b^2(a^2+b^2)+b^2c^2(b^2+c^2)+c^2a^2(c^2+a^2)\geq 2(a^3b^3+b^3c^3+c^3a^3) $ the inequality $ (ab)^3+(bc)^3+(ca)^3-\frac{(a^2+b^2+c^2)^3}{9}\leq\frac {(ab+bc+ca)( (ab)^2+(bc)^2+(ca)^2-a-b-c)}{9} $ and this is clearly at most
 $\frac{(ab+bc+ca)(\frac{(a^2+b^2+c^2)^2}{3}-3)}{9} $  and this is
 $\frac{(ab+bc+ca)((a^2+b^2+c^2)^2-9)}{27} $. And so we are left to proving that $\frac{((a^2+b^2+c^2)^2-9)((a^2+b^2+c^2)^2+9)}{81}\geq\frac{(ab+bc+ca)((a^2+b^2+c^2)^2-9)}{27} $ and this is really trivial.
      Thus, we are left to proving that for any $ d\leq 1$ we have the inequality $ f(t,t,t,d)\leq 4$ where $ t=\sqrt{\frac{4-d^2}{3}} $. This is equivalent if I'm not wrong to proving that for any $ a\geq 1$ we have $ a^6(a^3-9a^2+12)\cdot\sqrt{4-3a^2}\leq 4 $. Someone could please check that this is true. I'm not going to do this.
   By the way, a beautiful week-end to you too.
    Hope I didn't do major mistakes. By the way, Shrek 2 is a great film. It inspired me. That monster is similar to this one in the problem. Have a good night!

Harazi,you are really great.

I will check now your computations.

Harazi wrote
[quote]Hope I didn't made major mistakes.[/quote]
I think you have made a major mistake when you replaced $d^3$ with $abc$,
because of the factor of  $d^3$, which is not positive.
I suggest the following solution. Replacing $a, b, c, d$ with $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$, $\sqrt{d}$ respectively, the problem becomes
$a+b+c+d=4$   $\Rightarrow$   $\sum (abc)^{3/2}\leq4$.
From AM-GM inequality, we have
  $abc \leq (\frac{1+a+b+c}{4})^4=(\frac{5-d}{4})^4$,
$16\sqrt{abc}\leq25-10d+d^2$,
$16(abc)^{3/2}\leq25abc-10abcd+abcd^2$,
$16\sum{(abc)^{3/2}}\leq 25(\sum{abc})-36abcd$.
It remains to prove that
	$25(\sum{abc})-36abcd\leq64$.
I wait a nice solution for this one.

I don't think it was a major mistake. Just notice that if the factor of $ d^3 $ is not positive then the inequality is trivial, since LHS is always positive.

Your solution is much more beautiful. For the last inequality, I have a nice solution. Put $ a=4x, b=4y, c=4z, d=4t$ and it becomes $ xyz+yzt+ztx+txy\leq \frac{ 1+144xyzt}{25} $. But a problem from IMO SHORTLIST 1993 says that
 $ xyz+yzt+ztx+txy\leq\frac{1+176xyzt}{27} $ and your inequality follows directly from that one and AM-GM!

You are right, Harazi. [i]If the factor of  $d^3$ is not positive, then the inequality is trivial, since LHS is always positive.[/i]
I apologize to you for my affirmation. Your solution is monstruous, but correct.
And yet I wait a nice solution to the nice inequality

$64+36abcd \geq 25(abc+bcd+cda+dab)$.

OK.
I don't saw the last your message.
You have a great response speed.

Yes, Vasc, fortunately I have a good Internet connection. I really don't know what I will do in the summer, when I leave Bucharest and when I'll have to use dial up connection! Indeed, my proof is very very ugly, but I'm glad I could find it, since this inequality really is a difficult one. Your solution is magnificient, as usual.

I have another question, but no idea how to find or solve:
   What is the maximum value of $ (abc)^2+(bcd)^2+(cda)^2+(dab)^2 $ when $ a,b,c,d$ are non-negative with sum 4?

I propose to prove at first that if $a+b+c=const$ and $a\geq b\geq c \geq d$, then
$E(a,b,c)= a^2b^2+b^2c^2+c^2a^2$ is maximal for $a=b$. Using this result,
it follows that $F(a,b,c,d)=d^2E(a,b,c)+a^2b^2c^2$ is maximal for $a=b$ and also for $b=c$.

You really are incredible, really! So many ideas, so many strategies, absolutely incredible!

We can use the contra-positive way. Mainly, the idea is the following.
Consider that $x_1+...+x_n=na$, $a=constant$, and $f$ is maximal at $(a_1,...,a_n)$. If $a_i<>a_j$, then
$f$ decreases replacing $a_i$ and $a_j$ by $(a_j+a_j)/2$ and $(a_j+a_j)/2$; contradiction.

[quote="perfect_radio"]Why can the maximum of $f$ be reached?[/quote]

If $f$ is continuous, then since the set $S = \{(a_1,\dots,a_n)\,|\, a_1, \dots, a_n \ge 0, a_1+\cdots+a_n = k\}$ is a closed and bounded set in $\mathbb{R}^n$, the extreme value theorem specifies that a maximum is achieved. (One easy proof comes from topology: the closed interval $[-M,M]$ as a subspace of the real line is compact, $[-M,M]^n$ is a product of compact spaces, the closed set $S$ in $[-M,M]^n$ for some $M$ is compact, the continuous image of a compact space is compact, and the real line being a metric space, the compact image must be closed and bounded, from which the claim easily follows.)

[quote="arqady"][quote="Vasc"]Very nice! :lol:[/quote]
Thank you!
The next stronger:
$a>0,b>0,c>0,d>0,a+b+c+d=4.$ Prove that
$5(abc+abd+acd+bcd)\leq12+8abcd.$ ;)[/quote]
I think this is also in the preolympiad forum, right now. :)

Mildorf, I realised some time after I posted my question that the function Vasc used was continuous and that it must reach its extremal values. Anyway, thanks for the proof, as I didn't know how to make my argument very rigorous. (topology is not something I'm very familiar with)


Prove that if $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}k$, then $ ab\plus{}bc\plus{}ca \le \frac{k^2}{3} \le a^2\plus{}b^2\plus{}c^2.$

[quote="moldovan"]Prove that if $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} k$, then $ ab \plus{} bc \plus{} ca \le \frac {k^2}{3} \le a^2 \plus{} b^2 \plus{} c^2.$[/quote]

$ 3(a^2 \plus{} b^2 \plus{} c^2)\minus{}(a\plus{}b\plus{}c)^2\equal{}(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2$

$ 2(a\plus{}b\plus{}c)^2\minus{}6(ab\plus{}bc\plus{}ca)\equal{}(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2$

ineq holds for any reals a,b,c...


quadratics
percent
algebra
quadratic formula

2. Tub A contains 12% apple cider.
Tub B contains 55% apple cider.
How many ounces of cider from each tub
must be gotten to make a 64 ounce drink that
is 20% apple cider?

3.	Mary is running to school. She leaves home at
7:00 AM trotting at a speed of 12 mph.
Her little lamb leaves home at 7:15 AM
running after her at a speed of 17 mph.
What time of day will the lamb catch Mary?
Round your answer to the nearest minute.
(Example: 10:27 AM)

4.	Granny needs 3 gallons of 15% vinegar for
her pickles. Bud went to the store and bought
2 gallons of 50% vinegar. Granny asks:
How much pure spring water must I mix with
some of that 50% stuff to get my canner
filled with 3 gallons of 15% solution?

5.	Use the formulae:

Given: distance =16t^2 for a falling object,
And the fact that distance=1100 *t, where the speed of sound is 1100ft/sec in air.
If you drop a rock into a deep dark well and note that it takes 7 seconds to hear a splash after you drop the rock, how deep is the well? Hint: it takes some time for the rock to fall and then the remainder of the time for sound to travel back up the well for you to hear the splash. You will need the Quadratic formula to solve this one.

7.	Calculate the Medical Dosage.
Given that a saline medicine bottle contains 5 milligrams
of Tylenol for each 9 ml of saline
solution in which it is dissolved.
Grandma Wilson, who weighs 188 pounds needs
25 milligrams of Tylenol per 50 kilograms of body
weight for an I.V. How many ml of saline must she be administered?
Note: 1kilogram=2.2pounds

8.	I have a collection of nickels, dimes, and quarters
in a money bag worth 7.95. If the number of
quarters is one more than the number of nickels
and the number of dimes is five more than the number
of nickels, find the number of each type of coin.
# Nickels ________
# Dimes_________¬
# Quarters________

9.	Freezing weather is coming!
The radiator in my 450 John Deere track loader
is full with 18 quarts of fluid that is 15% antifreeze.
Question: How much must I drain out into a disposable container,
Then retighten drain plug, and pour back 100% pure
antifreeze into radiator to make 18 quarts of 40%
antifreeze.

i dunno if its rite but i got tub A has 52 oz and tub B has 12 oz. Im kinda sure on it.

These problems are quite elementary.

[hide="3"] There is no need to use algebra on this question. Positive difference in speeds is 5 mph, and by 7.15 a.m., Mary would have travelled $ 12 \times \frac{1}{4}\equal{}3$ miles. (Think about this carefully, and see by logic why this method works.) Thus the lamb needs $ 3/5$ hours. Time of day: 8.00 a.m.[/hide]

[hide="5"] Let time taken for rock to hit the bottom of the well be $ t_1$, and time taken for sound to travel up well from the bottom be $ t_2$.7

$ 16t_1^2 \equal{} 1100t_2$, and $ t_1 \plus{} t_2 \equal{} 7$. From here, you could solve for $ t_1$ in the second equation and plug this into the first equation, then solve.[/hide]

[hide="8"] Let there be $ q$ quarters, $ n$ nickels and $ d$ dimes. Then $ q \equal{} n \plus{} 1$, so $ n \equal{} q \minus{} 1$. Also, $ d \equal{} 5 \plus{} n \equal{} q \plus{} 4$. Then $ 5n \plus{} 25q \plus{} 10d \equal{} 795$, i.e. $ 5(q \minus{} 1) \plus{} 25q \plus{} 10(q \plus{} 4) \equal{} 795$. You simply expand out to get $ 5q \minus{} 5 \plus{} 25q \plus{} 10q \plus{} 40 \equal{} 795,40q \plus{} 35 \equal{} 795,q \equal{} 19$. Substitute $ q \equal{} 19$ to the appropriate formulas derived previously to evaluate $ n$ and $ d$.[/hide]

I have never been really good at proportions so I don't think I should answer them; maybe someone else.

ThetaPi, you did a little mistake in #3:
3/5 hours= 36 minutes.
7:15+36=7:51

[hide="2"]
$ 0,12x \plus{} 0,55y \equal{} 0,2*64$
$ x \plus{} y \equal{} 64$

$ x \equal{} 64 \minus{} y$

$ 0,12(64 \minus{} y) \plus{} 0,55y \equal{} 12,8$
$ 0,43y \equal{} 5,12$
$ y \equal{} 11,907$

$ x \equal{} 64 \minus{} 11,907 \equal{} 52,093$

Answer: 52,093 gallons of tube A and 11,907 gallons of tube B[/hide]

[hide="4"]This is similiar to problem 2:
x-vinegar

0,5x=0,15*3
x=0,9
3-0,9=2,1

EDIT: Answer: 0,9 gallons of vinegar and 2,1 gallons of water.[/hide]

[hide="7"]25 ml Tylenol is in 45 ml of saline solution.

50 kg = 110 gal.

110 gal. - 45 ml.
188 gal. - x ml.

$ x \equal{} \frac {188\cdot45}{110} \equal{} 76,9091 ml.$[/hide]

My english is to poor to understand #9, sorry.

[quote]2
Answer: 52,093 gallons of tube A and 11,907 gallons of tube B [/quote]

there are only 64 ounces in the final liquid,  how can there be gallons of liquid added in???

Sweetie, please don't post a bunch of problems in a thread like this. It makes it harder for a lot of people to read and answer the problems individually.

Also, if this is your homework, post you work (so far) as well, so that we know that we are actually helping you, and not just boosting your grade with the correct answers.

Thanks.

anyway, we learned how to solve problems like number 2 in math class about 2 weeks ago (and i'm in 8th grade)

anyway,
[quote]2. Tub A contains 12% apple cider.
Tub B contains 55% apple cider.
How many ounces of cider from each tub
must be gotten to make a 64 ounce drink that
is 20% apple cider?
[/quote]

let's set up a table first (which i have no idea how to do and i will make a very sad attempt at it)
............percent cider..ounces.......expression
tub A.....12%..............64-x..........0.12(64-x)
tub B......55%..............x.............0.55(x)
mixture...20%...............64...........0.2(64)

if you don't understand this table, i'll explain it
in tub A the liquid is 12% cider
in tub B, the liquid is 55% cider
in the mixture, we want a liquid with 20% cider
we want to find out how much of each liquid to put in, so we'll call the amount of liquid we put in from tub B x
since we want a total of 64 ounces, and we put in x ounces of liquid from tub B, we want to put in 64-x ounces of liquid from tub A

to find out how much cider is in tub A, we multiply the percent by the amount to get 0.12(64-x)
to find out how much cider is in tub B, we multiply the percent by the amount to get 0.55(x)
to find out how much cider is in the mixture, we multiplty the percent by the amount to get 0.2(64)

that was the hard part
now, all we do is make an equation and solve for the variable

we add together the amount from tub A and the amount from tub B to get the amount in the mixture
$ 0.12(64 \minus{} x) \plus{} 0.55(x) \equal{} 0.2(64)$
$ 7.68 \minus{} 0.12x \plus{} 0.55x \equal{} 12.8$
$ 7.68 \plus{} 0.43x \equal{} 12.8$
$ 0.43x \equal{} 5.12$
$ x \equal{} 11.9$

we put in 11.9 ounces of liquid from tub B

to find out how much liquid to put in from tub A, we solve the expression 64-x since 64 is the total and x is the amount we put in from tub B

$ 64 \minus{} x$
$ 64 \minus{} 11.9$
$ 52.1$

we put in 11.9 ounces of liquid from tub B and 52.1 ounces of liquid from tub A

this is not exact do to rounding done in the final step of finding x


There are 10 people in a classroom.  Find the number of ways to choose groups with 1,2,3,...10  people in a group.  If you solve this case try solving for the general case.

the number of ways you can make  a group of n people, out of K people (n <= k) is:
C(k,n) which equals to:
K!/(n!.(k-n)!).
so the number of ways to choose groups with 1,2,3,...10 people in a group of ten people is the number of way you can choose one person, plus two people, plus three people, etc. out of 10:
 (10!/9!)+(10!/8!2!)+(10!/3!7!)+....+(10!/9!)
so ingeneral you will have to chool 1,2,3,...,k people out of K people. you will have a total number of
(K!/(k-1)!)+(k!/(k-2)!2!)+(k!/(k-3)!3!)+(k!/(k-4)!4!)+....+(k!/(k-1)!)
now you can simplify this, and get a generalize into one formula.

There is an easier way.

[hide="Huge Hint"]Think: binomial expansion[/hide]

[hide]The sum of the numbers in each row is $2^n$ where $n$ is the row number. Since we are not counting picking zero people, the answer should be $2^n - 1$.[/hide]

Nice work tarquin.  If anybody doesn't understand what tarquin did because it was a little vague, here is a more in depth solution.

We know by that binomial coeffecients are defined by nC0..... nCn.  Which is the sum of the coefficients of the expansion (x+1)^n.  So plugging in one for x to find the sum of the coefficients we get 2^n.  ALso, since we are not including the group nC0 in the question, the answer is 2^n -1 as tarquin stated.

yeah, the answe is the sum of the $nth$ row of the pascal's triangle minus one.
i didn't want to just say the answer. i gave a hint. :D

Just a note, these can also be called proper subsets.

nah i still think that shoving 10Cx, where x is the number: 1,2,3,... is the easiest way, even though pascal's triangle is clever....

but another question is what does mC1+mC2+mC3+...+mCm equal??

[hide]$2^m$[/hide]

n.b. mCr mean m choose 1 ie. m!(m-r)!/r!

[quote="random_mathematician"]but another question is what does mC1+mC2+mC3+...+mCm equal??

[hide]$2^m$[/hide]

n.b. mCr mean m choose 1 ie. m!(m-r)!/r![/quote]I think it's actually [hide]$2^{m}-1$[/hide]like tarquin said.

the answer is that!!! :D
$2^m-1$, because you are not counting zero.

BTW your "other" question was exactly the same as mine. :P


search
number theory proposed
number theory

Let  x, a, b positive integers and  $ x^{a\plus{}b}\equal{}a^b b$. Prove  that  $ a\equal{}x$ and $ b\equal{}x^x$

this has been posted before it's iran 3-rd round,search in the resources...
http://www.mathlinks.ro/viewtopic.php?p=350349#p350349
it's amazing that i don't see any solutions ,anyway if u have any solutions plz post your solution there.or if u have seen this question anywhere plz post the site...
thx in advance.

It was posted on PEN forum and very famous. See here
http://www.mathlinks.ro/viewtopic.php?search_id=1219895255&t=150685


geometry
geometric transformation
homothety
ratio

S is a similarity. M, N, M', N' are points in a line (r) such that M<N<M'<N', where S(M)=M' and S(N)=N'.
Given a circle (C), construct S(C), ie, the image of (C) under S.

This is what I tried.

S must be a homothety with center $O\in MN$ and ratio k, where k=M'N'/MN and O belongs to Apollonius Circle of MM' and constant k. After I find O, I'll use a standard construction of S(C).

1) I must prove that S is actually a homothety with center O, $O\in MN$. I think it is clear but can't give a rigurous proof.

2) I wonder if there is an easier construction.

Thanx.

i presume that you mean homothety... do you know the center of homothety? that might be the purpose of the problem, here is a
[hide="hint"]what do you do with side length ratios?[/hide]

for your question, is this a homothety, of course it is, the way that you define it as similarity is awkward, and this is the way that homothety is defined, if you want a rigirous proof, then all you have to say is that homothety is a mapping of a point such that on a ray, you make a point that makes a constant length ratio with the center of homothety, and "a similarity" has to be such that you do this type of mapping


geometry
circumcircle
incenter
geometry proposed

Let $ \triangle ABC$,and $ D \equal{} (K) \cap (L)$ ; $ E \equal{} (L) \cap (M)$ and $ F \equal{} (M) \cap (K)$ where $ (K),(L)$ and $ (M)$ are the external bisectors of $ \angle BAC,\angle ABC$

and $ \angle CAB$ respectively.$ A',B'$ and $ C'$ are the circumcenters of $ \triangle EBC,\triangle AFC$ and $ \triangle ADB$ respectively.$ C'B'$ meets $ AB$ at $ E'$ and

$ AC$ at $ F',A'C'$ meets $ AB$ at $ J$and $ BC$ at $ I,A'B'$ meets $ AC$ at $ G$ and $ BC$ at $ H$.

Prove that $ B'HIC',E'B'A'J$ and $ F'GA'C'$ are cyclic.

[hide]
My first geometry problem  :)
[/hide]

It's not possible for $ C'B'$ to meet $ AB$ at $ E$, $ E$ being the intersection of $ L$ and $ M$. Same for $ C'B'$ meeting $ AC$ at $ F$.

Is there a mistake in the problem?

My work on the diagram thus far:
 [geogebra]2cdc82a783281498609fd43786dc7709d4d867dc[/geogebra]

I think you have two $ E$s and two $ F$s, which are making the statement that we need to prove ambiguous.

Yes,sorry for the typo,I meant $ C'B' \cap AB\equal{}E'$ and $ C'B' \cap AC\equal{}F'$.

Circles $\odot AFC, \odot CBE$ and $\odot BDA$ concur at the incenter $ I$ of $ \triangle ABC$ and $ \triangle A'B'C'$ has parallel sides to  the intouch triangle of  $ \triangle ABC$ $\Longrightarrow$ $ \angle E'B'A' \equal{}90 ^{\circ} \minus{}\frac{_1}{^2}A$ and $ \angle C'JE' \equal{}\angle BJA' \equal{}90^{\circ} \minus{}\frac{_1}{^2}A.$ Therefore, $\angle E'B'A'\equal{}\angle C'JE'$ $\Longrightarrow$ $ E'B'A'J$ is cyclic. The same reasoning is used for the remaning quadrilaterals.


$0 < b, c \leq  1$

prove that $2bc + 1 \geq  b + c$

Aren't proofs not allowed in the Getting Started forum?

no source

i just need to prove it to complete a question, but i think the inequality really does hold,

it seemed so trivial that i thought you guys can whizz bang.. get it just like that.. i dont know.. i cant seem to get it..

I think this problem is probably an appropriate level for Getting Started. It would be a tougher problem, but I don't think it needs to be bumped up to Intermediate.

the initial condition --> $0 < b, c \leq 1$

sorry if its ambiguous, meaning (0, 1]

b and c are both larger than 0 but both less than or equal to 1

2bc+1>=b+c
2bc+1-c>=b
2c+(1-c)/b>=1

if the following statements are true, we're done:
a) 2c+(1-c)/b>=2c+(1-c)
b) 2c+(1-c)>=1

Proof:
a)
2c+(1-c)/b>=2c+(1-c)
(1-c)/b>=1-c
1/b>=1
since 0=<b=<1, 1/b>=1 is a true statement

b)
2c+(1-c)>=1
c+1>=1, which is true as c>0

Therefore:
2c+(1-c)/b>=1 is a true statement, which proves our original statement 2bc+1>=b+c to be true.

Oh I'm dumb... :blush:

[hide="I couldnt find a better way"]
Assume $b\ge c$:

Suppose $b=c$
Then:
$(b-1)^2> 0$
$b^2-2b+1>0$
$b^2+1>2b$
$bc+1>b+c$
So since $bc$ is positive,
$2bc+1>b+c$

Suppose $b,c\le 0.5$.
Then:
$b+c\le 1$
$2bc>0$
$2bc+1>1$
$2bc+1>b+c$

Suppose $b\ge 0.5$,
$2b\ge 1$
$2bc\ge c$
$1\ge b$
$c+1\ge b+c$
Substituting,
$2bc+1\ge b+c$

And I think I'm done.
[/hide]

How about just $(b-1)(c-1) \ge 0 \Rightarrow bc+1 \ge b+c \Rightarrow 2bc+1 > b+c$... It's a pretty weak inequality (not to mention strict) since if you subtract $bc$ from the LHS it still holds...


Okay, those of you who don't know how to play, Bughouse is simple:

Instructions:

1. Have an even number of players, at least 4.
2. Group into teams.
3. If you are player 1 and your opponent is player 4, and your teammates are player 2, then if you are white and the other player is black, then player 2 must be black and his opponent must be white (you must have the opposite color of your teammate). every black piece you destroy would be sent to one of player 2's pieces (the piece must be in the position where it would actually be when the game starts, if the space is already taken, you may choose where to put it, but it has to be an empty space). The game goes on for quite awhile.

Instructions for Monster Chess:

1. This game is basically normal chess, besides the pieces and rules. One player must only have 4 pawns and 1 king.
2. The other player gets all of his/her pieces (sounds like this guy is gonna win!)
3. The advantage for the person with 4 pawns and 1 king is that this player would be able to go [b]TWICE[/b] for every 1 turn of the other player. the ratio of the moves of the player with the 4 pawns and 1 king to the other player is 2:1.
4. (This also makes the player with 4 pawns and 1 king to have a greater advantage, for its range of checking the king is greater, and he can still change from a pawn to a queen if he gets to the end.)

Ninja Chess (the craziest of them all):

Intro: this is kinda like naruto, for those of you who watch the anime cartoon, but it's kinda crazy.
1. these are the pieces: rook: orochimaru, bishop: choji, knight: shikamaru, pawn: normal ninja, king: naruto, and queen: sasuke.
2. (it sounds pretty funny for queen:sasuke) Okay, just like the cartoon, choji can do choji meatroll, the problem is, that he can only destroy all pieces in one straight line, and in one direction (he may destroy his own pieces) On the last piece that he destroys is where he stays in.
3. shikamaru can do shadow attacks, so he can cast the shadow attack on a ninja in a straight line and only one, but he can do multi- numbers if he charges (takes one turn and wastes it). the other ninja may not move and would be in a shadow phase, so it may be destroyed by other ninjas. shikimaru may also control them (like the game-ex. if shik. throws kunai, all ninjas in shadow phase will also throw kunai. if he moves up 2 spaces, then they will also move up two spaces). ninjas that fall into his shadow max up to 3.
4. orochimaru can use his snake bite in this game, so he may reach out his head and bite the other side's character, and then let that character be on his side. the problem is that the character you bite, you must have the opposite color of that character in your graveyard. the character in your graveyard will be resurrected and be on your side (must be on the starting position, or empty space if the starting position is taken). then your opponent's piece dies. orochimaru can bite in a straight/curved line, but no other ninjas can be in his way.
5. normal ninjas can do shadow clone jutso only once in this game, and throw kunais and sharikens in straight lines.
6. naruto can do rasengan in a straight line and run to its opponent, kill it, and stand on his opponent's space. He may also destroy pieces (in a 3x3 square, with naruto in the middle) with the fox demon fire tornado thingy (in the first few episodes).
7. Sasuke may do fireball and chidori. Chidori is like Naruto's rasengan, but fireball is pretty much same, but he will stay in his same position for that turn, not on his opponent's spot.
8. Ninjas may travel in any line that is straight/curved, but no other ninjas may be in the way.
8. ALL PLAYERS MAY THROW KUNAIS, SHARIKENS, AND MAKE CLONES, KUNAIS AND SHARIKENS MAY ONLY BE THROWN IN STRAIGHT LINES. THE ATTACKER WOULD STAY ON HIS OWN POSTION, NOT ON HIS OPPONENT'S. CLONES CAN BE ONLY MADE ONCE IN THIS GAME, LIKE SHIKAMARU CAN MAKE ANOTHER SHIKIMARU, BUT THE NEW SHIKIMARU MUST STAY IN THE STARTING POSITION, IF TAKEN, THEN IN ANOTHER EMPTY SPOT. NORMAL NINJAS MAY MAKE AS MANY CLONES UNTIL THERE IS 8 NORMAL NINJAS.
9. If naruto dies, then the player with naruto dying loses. (this will be kind of tricky and cheat). As I said, this game is kind of crazy.

Alright, since I've posted 3 games on here, whoever wants to play one or both or all of these, please post here. The turns of each player (just like Tag Team Chess) are also going to have 2 to 3 days to make each move (for these chess games, we'll make different rooms for different players so it will be easier so we don't have to agree or anything). I want to make it a tournament, so it will be more easier, and not take for so long. Bughouse will also be in different rooms, but whenever one player destroyed the other player's piece, then go into their room and tell that person.

 :D

heard of bughouse, its fun. never heard of monster chess, but it seems fun. i dont really get how to play ninja chess. Anyone heard of suicide chess? its kinda lame, but the goal is to lose all your pieces first, but if a piece can be taken, then it must be

Like to play that one (the suicide one  :D )

http://en.wikipedia.org/wiki/Suicide_Chess

Just try it several times with friends.  :D

This kind of game ends quickly.

...bughouse. Or crazyhouse, but crazyhouse isn't as fun, I guess.

suicide chess, like BoesFX already said, ends quickly (although on forums itll still go for a while, i guess), and it doesnt take TOO much strategy. And of course, you dont want to be seen playing it in front of your chess instructor  :wink:

Bughouse is really good, you can get into some very interesting games with bughouse.

Simpler instructions for ninja chess:
       1. Stop reading if you don't watch naruto.
       2. For right now, you should know that king is naruto, queen is sasuke, bishop is choji, knight is shikimaru, rook is orochimaru, and pawn is a normal ninja.
       3. Cloning may be done only once in this game, and when you clone, you may make as many repricals of the same character until all of the same ninjas is used up in the graveyard. When the clones are made, they must be put it in the place where they would be when the game starts. If the spot is taken, switch to another empty spot.
        4. Sharikens and kunais may be thrown by all ninjas, in a straight line (diagonal, vertical, or horizontal) and kill ONE enemy.
        5. Attacks:
               a. naruto's attacks:
                        a1. rasengan-naruto's rasengan may be used in this game, but naruto must kill in a straight line (diagonal, horizontal, or vertical) and may destroy an enemy, but he will stand on the position that his opponent was standing on.
	        a2. fox demon-naruto may do fox demon in this game, like the one he did in the first few episodes. Anyway, he may destroy all enemies in a 3x3 square, with naruto in the middle square. If you don’t know what I mean, click the download and read the comment-there will be a picture (it’s at the bottom).
                b. choji's attacks:
                         b1. choji meatroll-choji may destroy all ninjas in one row/column in one direction. He can destroy ALL of the ninjas (even if they are his allies).
                c. shikimaru's attacks:
                         c1. shadow attack thingy-the jutso that shikimaru can do in the episodes. he can cast a shadow in a straight or swiggly line and make another opponent not able to move. then another player on your team may destroy that. he may also control the opponent. if shikimaru throws a kunai, then the opponent must also do it. if shikimaru walks forward 3 spaces, then the other person must also (both of them must have 3 spaces in front of them if this is done). shikimaru may only cast a shadow on more than one opponent unless he charges for one turn then he may cast it on 3 opponents at MAX.
                d. normal ninja's attacks:
                         d1. a normal ninja may only throw a kunai, shariken, or make clones.
                e. orochimaru's attacks:
                         e1. snake bite-orochimaru may bite in this game. he may bite in a straight or swiggly line on the board. then the opponent must be killed, then you find the same character, but the opposite color, in your graveyard. REMEMBER-BEFORE YOU BITE, YOU MUST CHECK TO SEE IF YOU HAVE THE SAME PIECE BUT OPPOSITE COLOR IN YOUR GRAVEYARD. then, after you receive the new piece, then you may put that piece in the position where it would be at the beginning of the game. if that position is taken, then put it at another empty spot.
                f. sasuke's attacks:
                         f1. chidori-same as naruto's rasengan.
                         f2. fireball-same as chidori, except you will stand in the position you are standing currently.

The point of the game is to destroy all of the opponent's pieces.

 :D

Bughouse is a good game. It can even help you in standard games. The idea of sacrifice is a difficult concept to grasp to new players who get in the habit of castling early and playing it safe. Playing bughouse can help you develop skills that will tell you when a piece should and should not be sacked.

Plus it's really fun if you're playing with a people 2x your rating  :D .

i like bughouse, crazyhouse, and suicide.

Ninja???? dude, you watch wayyy too much Naruto

i hate naruto, i havent even watched it, but its annoying to even hear those voices. anyway, you forgot to say that you cant place a pawn on the final rank, only on the 7th. (otherwise every  pawn would be the equivalent of a queen.)

All right, bughouse it is.......

All players who want to play please sign in this forum by 9/31/07.

The game [i]will[/i] start at 10/1/07. How about that?

By now, players are:

Nerd_of_the_Ages
davidyko
shaggy75
rem
7h3.D3m0n.117
mz94
toadoncart

Total amount of players: 7.

Alright, if we don't get this odd number changed to an even number soon, one player might have to be drawed out of the game and standing out of the tournament (6 people player, so 3 pairs, then 3 people win, so the last person will join and form 4 people).

Tournament pairs for RIGHT NOW:

[code]Nerd_of_the_Ages     V.S.      mz94
rem                                 V.S.      shaggy75
davidyko                          V.S.      7h3.D3m0n.117[/code]

(Sorry, but for some reason the code won't line it up straight. If I type, the spaces will be cancelled out.)  :blush:

Teams:

Nerd_of_the_Ages, shaggy75, and 7h3.D3m0n.117
mz94, rem, and davidyko.

draw: toadoncart

winners V.S. winners will be determined later.......

*sigh* guess i have to draw, unless one of you volunteer to or someone else joins the game.

 :D

toadoncart, i think your forgetting that you're playing also (or i assume so, because you started the topic). And im gonna die against mz94  :ninja:

I AM playing, but there must be an EVEN number to play.......

 :D

So are the teams 3 vs. 3? so when you take a piece, you choose which teammate you give the piece to? Maybe we should do two 2v2's at once by waiting for another person to come.

How about this: let's wait till more people post up and then pair groups of 4.....

 :D

[quote="mz94"]dude, if we wait for groups of 4, then we can have 4 groups of 2, have 2 separate bughouse tournaments, then have a championship![/quote]

A bughouse tournament is a good idea, but you have to deal with time problems. It won't really be bughouse if you have 2 days to play your move (not to mention games can get really confusing with pieces that are on/off the board).

Then we can just set a time and then each player plays (i.e) 30 minutes, then one person moves which takes 5 min., the other player moves, takes 5 min., etc.

Something like that........

 :D

I'll sign up for bughouse, and suicide sounds fun too. I haven't played chess for a while though, so I'm probably going to do something stupid in the beginning.

bughouse...my favorite!!!!!! but what about stalling for pieces?

We could make it so all moves don't count until the end of the time period given.

Players by now:

Nerd_of_the_Ages
davidyko
shaggy75
rem
7h3.D3m0n.117
mz94
toadoncart
Asiodea
Lawrence Wu

Total Players: 9.

*sigh* another person has to be taken out. o well, see about that later. ok, anybody else?

i can see a ton of you'll want to play suicide chess. maybe we can try that too sometime.........

 :D

Okay, kinda late, but it's starting

[code]Nerd_of_the_Ages      v.s.     davidyko [/code]
[code]shaggy75                  v.s.        rem[/code]
[code]7h3.D3m0n.117          v.s.     mz94 [/code]
[code]Asiodea                    v.s.     Lawrence Wu [/code]

(might appear a little funny).

[code]Pass: toadoncart[/code]

Teams:

[code]Nerd_of_the_Ages and shaggy75[/code]
[code]davidyko and rem[/code]
[code]7h3.D3m0n.117 and Aiodea[/code]
[code]mz94 and Lawrence Wu[/code]

(Rooms will be created tomorrow.)

(if changes want to be made, post them quickly!)

 :D

Lol, I'm gonna get OWNED by all of you guys. shaggy and I are DEAD


Since the equation is linear, any other solution $y_{2}$ will have the form $y_{2}=y_{1}+y$, where $y$ solves the homogeneous equation
$x^{2}y''+7xy'+5y=0$. The latter is just the Euler equation, whose solutions are known and can be found [url=http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc2.htm]here[/url].

(There may be a way to reduce the order of the ODE using $y_{1}$, but I don't remember it)

[quote="sinajackson"]Solve The following differential equation for $x>0$ with knowing that one of the answers is $y_{1}(x)=x^{-1}$,

                  $x^{2}y^{''}+7xy^{'}+5y=x$[/quote]

We get the following:
$y_{2}= u y_{1}$, where $y_{1}= x^{-1}$
$y'_{2}=-x^{-2}u+x^{-1}u'$
$y''_{2}= 2x^{-3}u-2x^{-2}u'+x^{-1}u''$.

Plugging them into the original equation you get $u''+5x^{-1}u'= 1$. Letting $z = u'$ and $z' = u''$,
we get $z'+5x^{-1}z = 1$ and solving this presents no difficulty.


Let $a,b,c >0$ such that $a+b+c = 1$. Prove that \[\frac1{a}+\frac1{b}+\frac1{c}\geq 3 \left( a^{2}+b^{2}+c^{2}\right)+8 .\] [hide="The idea of my ugly solution"]$f(x) = \frac1{x}-3 x^{2}$ is convex on some interval $\left( 0,\frac12 \right) \subset I \subset \left( 0,1 \right)$.[/hide]

[hide]Try to make a substitution $a^{2}+b^{2}+c^{2}=1-2(ab+ac+bc)$.[/hide]

if we let $a=\frac{x}{x+y+z}$
$b=\frac{y}{x+y+z}$
$c=\frac{z}{x+y+z}$

we end up having to prove:
$(x+y+z)^{3}(xy+xz+yz)\geq xyz(3(x^{2}+y^{2}+z^{2})+8(x+y+z)^{2})$

[quote="M4RI0"][quote="Lovasz"]
$f(a,b,c)-f\left(a,\frac{b+c}{2},\frac{b+c}{2}\right) \ge 0$
[/quote]

How does this help in the solution :?: . I don't understand.[/quote]
Then, you need to prove $f\left(a,\frac{b+c}{2},\frac{b+c}{2}\right) \ge 8$. Don't forget the condition $a+b+c=1$.

I have another solution , I hope I am not mistaken .
We have to prove that
$\sum\frac{1}{a}-3a^{2}-\frac{8}{3}\geq 0$  or equivalent
$\sum\frac{1}{a}-3a^{2}-\frac{19}{3}+11a\geq 0$ or equivalent
$\sum\frac{3-9a^{3}-19a+33a^{2}}{3a}\geq 0$ or equivalent
$\sum\frac{(3-a)(3a-1)^{2}}{3a}\geq 0$  which is trivial  :)

The best solution, \[(*) \iff \sum_{cyc}{(a+b)c \over ab}(a-b)^{2}\ge 0.\]


inequalities
trigonometry
inequalities unsolved

Let be given a triangle $ ABC$. Prove that \[ \sum\sin\frac{A}{2}\le 1\plus{}\frac{1}{6}\sum\cos^2\frac{B\minus{}C}{4}\]

note that the inequality is equivalent to $ \sum \cos \alpha\leq 1 \plus{} \dfrac{1}{6}\sum \cos^2\dfrac{\beta \minus{} \gamma}{2}$ (*), where $ \alpha,\beta,\gamma$ are the angles of an acute triangle...

rewrite the inequality as

$ 8\prod\sin\dfrac{\alpha}{2}\leq \dfrac{1}{3}\sum \cos^2\dfrac{\beta \minus{} \gamma}{2}$

from am-gm we have that

$ \dfrac{1}{3}\sum \cos^2\dfrac{\beta \minus{} \gamma}{2}\geq \prod\sqrt [3]{\cos^2\dfrac{\beta \minus{} \gamma}{2}}\geq \prod\cos\dfrac{\beta \minus{} \gamma}{2}$

recall that $ \cos\dfrac{\beta \minus{} \gamma}{2} \equal{} \dfrac{b \plus{} c}{a}\sin\dfrac{\alpha}{2}\geq \dfrac{2\sqrt {bc}}{a}\sin\dfrac{\alpha}{2}$, where $ a,b,c$ are the sides of the triangle corresponding to $ \alpha,\beta,\gamma$...

from this it is easy to conclude...

[in fact (*) holds still when it is not an acute triangle... this is problem 335.1 from the vietnamese journal the mathscope]

you can try this approach too: again rewrite is as $ \sum \cos \alpha\leq 1 \plus{} \dfrac{1}{6}\sum \cos^2\dfrac{\beta \minus{} \gamma}{2}$

you can use that $ \cos^2\dfrac{\beta \minus{} \gamma}{2}\geq \dfrac{2r}{R}$(**), and the result follows immediately since $ \sum \cos\alpha \equal{} 1 \plus{} \dfrac{r}{R}$.

[hide="proof of (**)"]
recall that $ \dfrac{2r}{R} \equal{} 8\prod\sin\dfrac{\alpha}{2} \equal{} 8\sin^2\dfrac{\alpha}{2}\cdot\dfrac{s \minus{} a}{a}$

then, we have to prove that $ \dfrac{\cos^2\frac {\beta \minus{} \gamma}{2}}{\sin^2\frac {\alpha}{2}}\geq \dfrac{8(s \minus{} a)}{a}$...

but we have that

$ \dfrac{\cos^2\frac {\beta \minus{} \gamma}{2}}{\sin^2\frac {\alpha}{2}} \equal{} \dfrac{(b \plus{} c)^2}{a^2} \equal{} \dfrac{(b \plus{} c \minus{} a \plus{} a)^2}{a^2}\geq \dfrac{4(b \plus{} c \minus{} a)}{a} \equal{} \dfrac{8(s \minus{} a)}{a}$

as we wanted to prove  :D

this is problem 2382 from crux
[/hide]


Fred has 10 identical oranges. In how many ways can he distribute these oranges among his 5 friends? (It is possible that a friend will receive no oranges.)

You have been told before that asking for solutions to problems belongs in the second forum. Click "Click here to discuss this problem on the forum" underneath a problem's solution to go to the thread.

Anyway, here's the solution.

This is equivalent to placing four dividers somewhere in the 11 slots between and on the ends of a row of oranges, the spaces between dividers being distributed to the friends. Thus, the answer is $ \binom{10\plus{}5\minus{}1}{5\minus{}1}\equal{}\binom{14}{4}\equal{}\frac{14\cdot13\cdot12\cdot11}{4\cdot3\cdot2\cdot1}\equal{}\fbox{1001}$.

[quote="smellydabian"]i posted here because nobody looks in the problem discussion[/quote]

Usually that's not the case; perhaps we're in a low right now, but there usually is someone, typically PowerOfPi or AIME15, who is browsing that forum.

Heh, I haven't been to the Problem Discussion forum for months :P

This forum is not the right place. It is like posting a game in the middle school classroom forum.

Also, one could be posting a current problem rather than a problem already answered.

[quote="smellydabian"]what is so bad about posting on the alcumus forum instead of the prob discussion? :?:[/quote]
Also, it keeps things organized. :wink:


1. A circle of radius r inches is inscribed in a right triangle with side lengths 6, 8, and 10 inches.
Find the value of r.

2. Simplify $ \frac{3}{1\minus{}\frac{3}{3\plus{}y}}\minus{}\frac{3}{\frac{3}{3\minus{}y}\minus{}1}$

3. The complex number 13 + i can be factored into the product of 1 + 2i and what other complex
number?

[hide="1"]
Since $ A\equal{}rs\equal{}\frac12bh$, this is $ \frac{\frac12\cdot6\cdot8}{\frac12(6\plus{}8\plus{}10)}\equal{}\frac{48}{24}\equal{}\boxed2$.
[/hide]
[hide="2"]
$ \frac3{1\minus{}\frac3{3\plus{}y}}\minus{}\frac3{\frac3{3\minus{}y}\minus{}1}\equal{}\frac{9\plus{}3y}y\minus{}\frac{9\minus{}3y}y\equal{}\frac{6y}y\equal{}\boxed6$.
[/hide]
[hide="3"]
$ \frac{13\plus{}i}{1\plus{}2i}\cdot\frac{1\minus{}2i}{1\minus{}2i}\equal{}\frac{15\minus{}25i}{5}\equal{}\boxed{3\minus{}5i}$.
[/hide]


If all the logarithms are real numbers, the equality
\[ \log{(x+3)} + \log{(x-1)} = \log{(x^2-2x-3)} \]
is satisfied for:

(A) all real values of $x$
(B) no real values of $x$
(C) all real values of $x$ except $x=0$
(D) no real values of $x$ except $x=0$
(E) all real values of $x$ except $x=1$

Do you mean the expression on the right to be $x^2+2x-3$?

Or is it meant to be the way it is?

If so, [hide](D)? Does the equality still work when an argument is negative?[/hide]

[quote="joml88"][hide]
Either way $x=0$ is not a solution...you have to remember to check your answers, i.e. plug them in!  :P

That being said, eliminating answer choices would crush this problem :D  [/hide][/quote]

Ok - thanks for the clarification.

[hide](x+3)(x-1) = x^2 + 2x - 3.  Since x^2 + 2x - 3 = x^2 - 2x - 3 is only satisfied by x = 0 and x = 0 is an extraneous solution, the answer is therefore B.[/hide]


[quote="arqady"][quote="VIETNAM"]Let a,b,c  be  positive  numbers  such  that $a^{2}+b^{2}+c^{2}=3$.Find the minimum of:
$A= \frac{a^{2}}{a+b}$+$\frac{b^{2}}{a+b}$+$\frac{c^{2}}{c+a}$[/quote]
Maybe $A= \frac{a^{2}}{a+b}$+$\frac{b^{2}}{b+c}$+$\frac{c^{2}}{c+a}$  $?$ :wink:
If so $\min A=\frac{3}{2}.$ SOS kills it.[/quote]
Can you type your solution sufficiently :) ?

[hide="I probably have inequality sign reversed somewhere but here it is"]
\[\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}=3\sum \left(\left(\frac{a^{2}}{a^{2}+b^{2}+c^{2}}\right)\left(\frac{1}{a+b}\right)\right)\geq 9\sum \frac{1}{a+b}\\ \geq 9\frac{1}{2(a+b+c)}\geq \frac96=\frac32\]
Where we used $a+b+c\leq 3\Longleftrightarrow \frac{a+b+c}{3}\leq \frac{\sqrt{a^{2}+b^{2}+c^{2}}}{3}$[/hide]

[quote="arqady"][quote="VIETNAM"]
Can you type your solution...  ?[/quote]
$\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}\left(\frac{a^{2}}{a+b}-\frac{a}{2}\right)\geq\sum_{cyc}\left(\frac{1}{2}-\frac{a}{2}\right)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{a(a-b)}{a+b}\geq\sum_{cyc}(1-a)\Leftrightarrow\sum_{cyc}\left(\frac{a(a-b)}{a+b}-\frac{a-b}{2}\right)\geq3-(a+b+c)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-b)^{2}}{2(a+b)}\geq\frac{3(a^{2}+b^{2}+c^{2})-(a+b+c)^{2}}{3+a+b+c}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a-b)^{2}\left(\frac{1}{2(a+b)}-\frac{1}{3+a+b+c}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-b)^{2}(3+c-a-b)}{a+b}\geq0,$ which obviously true. :)[/quote]
Nice,arqady! :)
And another problem.It's easy,too! :roll:
Let a,b,c be positive numbers such that $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq a^{2}b^{2}c^{2}$
Find the minimum of A:
$A= \frac{a^{2}b^{2}}{c^{3}(a^{2}+b^{2})}+\frac{b^{2}c^{2}}{a^{3}(b^{2}+c^{2})}+\frac{c^{2}a^{2}}{b^{3}(c^{2}+a^{2})}$.

[quote="VIETNAM"]
And another problem.It's easy,too! :roll:
Let a,b,c be positive numbers such that $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq a^{2}b^{2}c^{2}$
Find the minimum of A:
$A= \frac{a^{2}b^{2}}{c^{3}(a^{2}+b^{2})}+\frac{b^{2}c^{2}}{a^{3}(b^{2}+c^{2})}+\frac{c^{2}a^{2}}{b^{3}(c^{2}+a^{2})}$.[/quote]
No one like this problem? :(

[quote="VIETNAM"][quote="VIETNAM"]
And another problem.It's easy,too! :roll:
Let a,b,c be positive numbers such that $ a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2}\geq a^{2}b^{2}c^{2}$
Find the minimum of A:
$ A \equal{} \frac {a^{2}b^{2}}{c^{3}(a^{2} \plus{} b^{2})} \plus{} \frac {b^{2}c^{2}}{a^{3}(b^{2} \plus{} c^{2})} \plus{} \frac {c^{2}a^{2}}{b^{3}(c^{2} \plus{} a^{2})}$.[/quote]
No one like this problem? :([/quote]
Setting $ x \equal{} \frac {1}{a},y \equal{} \frac {1}{b},z \equal{} \frac {1}{c}$ We have: $ x^2 \plus{} y^2 \plus{} z^2 \ge 1$
We will prove that:
$ \frac {x^3}{y^2 \plus{} z^2} \plus{} \frac {y^3}{x^2 \plus{} z^2} \plus{} \frac {z^3}{x^2 \plus{} y^2} \ge \frac {\sqrt {3}}{2}$
Using Cauchy-Schwarz:
$ LHS \ge \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{x(y^2 \plus{} z^2) \plus{} y(x^2 \plus{} z^2) \plus{} z(x^2 \plus{} y^2)}$
By AM-GM we have:
$ x(y^2 \plus{} z^2) \plus{} y(x^2 \plus{} z^2) \plus{} z(x^2 \plus{} y^2) \le \frac {2}{3}(x^2 \plus{} y^2 \plus{} z^2)(x \plus{} y \plus{} z) \le \frac {2}{\sqrt {3}}(x^2 \plus{} y^2 \plus{} z^2)\sqrt {x^2 \plus{} y^2 \plus{} z^2}$
Because $ x^2 \plus{} y^2 \plus{} z^2 \ge 1$
So:
$ \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\frac {2}{\sqrt {3}}(x^2 \plus{} y^2 \plus{} z^2)\sqrt {x^2 \plus{} y^2 \plus{} z^2}} \ge \frac {\sqrt {3}}{2}$
We done!


Henry's Hamburger Heaven offers its hamburgers with the following condiments:
ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions.
A customer can choose one, two, or three meat patties, and any
collection of condiments.
How many different kinds of hamburgers can be ordered?

[quote="pinkmuskrat"]${ (1 + 1 + 1 + 1) ( \binom{8}{0} + \binom{8}{1} + \binom{8}{2} + ... + \binom{8}8} )$

$ 4 \cdot 2^8$

$ 2^2 \cdot 2^8$

$ 2^{10}$

$ \boxed{1024}$[/quote]
You have everything correct except we do not multiply $2^8$ by $4$.  Remember that there are only $3$ choices for patties--one patty, two patties, or three patties. Therefore, the number of different kinds of hamburgers that can be ordered is $3\times2^8=3\times256=\boxed{768}$.

[quote=pinkmuskrat]${ (1 + 1 + 1 + 1) ( \binom{8}{0} + \binom{8}{1} + \binom{8}{2} + ... + \binom{8}8} )$

$ 4 \cdot 2^8$

$ 2^2 \cdot 2^8$

$ 2^{10}$

$ \boxed{1024}$[/quote]

Can you elaborate on this a little bit please?

He posted that on 2010... they won't respond

im still confused... why do we do 2^8 for the total condiment combinations?

Each of the condiments could either be in the hamburger or not in the hamburger.

[hide=Solution] A customer makes one of two choices for each condiment, to include it or not to include it. The choices are made independently, so there are $2^8 = 256$ possible combinations of condiments. For each of those combinations there are three choices regarding the number of meat patties, so there are altogether $(3)(256)=\boxed{768}$ different kinds of hamburgers.[/hide]

[hide=Combinatorics Solution]
We get $3$ choices for patty. For the condiments, the customer can choose up to $8$ options (including $0$), meaning the total number of possibilities is $8$ choose $0 + 8$ choose $2+\cdots+8$ choose $8$. That's $\binom{8}{0}+\binom{8}{1}+\binom{8}{2}+\binom{8}{3}+\binom{8}{4}+\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}$. Calculating gives $256$. So the total number of hamburgers that can be ordered is patty $\times$ condiments so $3\times256=\boxed{768}$.
[/hide]

[hide]We can note that the sum of the $n$th row of Pascal's Triangle is $2^n,$ so the number of possible choices for the condiments is $\binom{8}{0}+\binom{8}{1}+\binom{8}{2}+\binom{8}{3}+\binom{8}{4}+\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8} =2^8 = 256.$ Multiplying this by the number of choices for the patties, we have the desired $256\cdot3=768.$[/hide]

Sorry, but why is overcomplicating? First [i]original[/i] solution that does not overcomplicate.

[hide]
Consider the patties and condiments separately. There are 3 ways to choose how many patties there are, and for each of the 8 condiments we have 2 options - we can either include or not include that condiment. By the multiplication principle, this yields a final answer of $3 \cdot 2^8 = \boxed{768}.$
[/hide]


AMC
AIME
geometry
3D geometry
LaTeX
probability
tetrahedron

Hi guys,

This is similar to the AMC 10-12 marathon except it should be AIME problems. I figure it will be fun and good preparation for next year's AIME. The rules are similar to the AMC 10-12 Marathon. Have fun!

I'll start it off with a relatively easy problem from the 2009 AIME I:

There is a complex number z  with imaginary part 164 and a positive integer n such that z/(z+n)=4i

Find n.

[hide]Let $ z \equal{} a \plus{} 164i$. Then:
$ \frac {a \plus{} 164i}{(a \plus{} n) \plus{} 164i} \equal{} 4i$
Rearranging finds $ n \equal{} 697$.[/hide]

A right circular cone has base radius $ r$ and height $ h$. The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making $ 17$ complete rotations. The value of $ \frac {h}{r}$ can be written in the form $ m\sqrt{n}$, where $ m$ and $ n$ are positive integers and $ n$  is not divisible by the square of any prime. Find $ m \plus{} n$.

[quote="Ihatepie"]Because the cone is making a circle with the lateral edge as the radius, the circumference must be 2pi*l

and since the cone is rolling on it's circumference, and it turns 17 complete times, you get 34pi*r=2pi*l or 17r=l.

The lateral edge is just sqrt(r^2+h^2) due to the Pythagorean theorem, so 17r=sqrt(r^2+h^2)

289r^2=r^2+h^2   288r^2=h^2 You then get h^2/r^2=288, or h/r=12sqrt2. So m+n=12+2=14.

A relatively easy one now:
A is the probability of throwing more than 4 on one die.
B is the probability of throwing more than 8 on two dice. Find A/B.[/quote]

Assuming six-sided dice:

A = $ \frac {1}{3}$, B = $ \frac {5}{18}$.

So A/B = $ \frac {18}{15} \equal{} \frac{6}{5}$.

sdkudrgn88

Can you post another problem since you solved the previous one?Thanks.

-rts2007

note: there is/was an AIME marathon in the intermediate forum...

And we can't have one here why?

NP:
How many different 4x4 arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?

Just by glancing at the problem, I'd say brute force it and see whether there is a pattern. But, I'm sure there's a faster way. Am I right?

-rts2007

[quote="Ihatepie"][hide]Each row and column has two ones and two negative ones.[/hide][/quote]

[hide]Building on this information, we can create a 4x4 grid and look at what possible transformations there are.

For example, the checkerboard grid (for the sake of equal spacing and not having to much with LaTeX for half an hour, I'll say that 0 represents -1, and that the sum for each row and column is instead 2):

1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1

Applying the transformation:

1 0
0 1

becomes

0 1
1 0

Gives us 18 different ways to change the data around.

Not sure if there are others, but I'll say [b]018[/b] is my final answer.

[b]EDIT[/b] As it seems, I was wrong for the reason that you can wrap around as well (the transformation that I mentioned, that is). there are actually 36 different ways.

[/hide]

[quote="rts2007"]Now can you post another problem since you solved this one?

-rts2007[/quote]

Then again, I'm not even sure if my solution is right. Is it?

But all right. I'll make another one up.

A regular tetrahedron ABCD, with sidelength 1, sits on an arbitrary plane, where ABC is the triangular face touching the plane. The vertex D is moved so that the volume of the tetrahedron is the same, but the line AD is perpendicular to triangle ABC. The sum AB + AC + AD + BC + BD + CD is expressible as $ \frac {p\sqrt {q}\plus{}r \sqrt{t}}{u}$, where $ q$ and $ t$ are not divisible by the square of any integer, and $ r$, $ p$, and $ u$ are relatively prime. Find $ p \plus{} q \plus{} r \plus{} t \plus{} u$.

Alright, since I can't edit my previous post, I need to repost the problem to make corrections...

A regular tetrahedron ABCD has edge length 1. The vertex D is moved to D' so that the volume of the tetrahedron is the same, but the line AD is perpendicular to triangle ABC. The sum AB + AC + AD' + BC + BD' + CD' of this new tetrahedron is expressible as $ a \plus{} \frac {b\sqrt {c} \plus{} d \sqrt {e}}{f}$, where all terms are integers, $ c$ and $ e$ are not divisible by the square of any integer, and $ b$ and $ d$ are relatively prime. Find $ abcd \plus{} ef$.

[quote="sup3rcrash3r"]Ummm... The whole point of the marathon is 1 problem at a time...[/quote]

It is, which is why a person only needs to solve one out of three problems at a time.

I'd solve them, but I've been such a lazy bum recently...

Problem 1
[hide]The cones each have radius $ 2\sqrt{3}$ and height 2, thus volume $ 8\pi$, so 8.[/hide]
Problem 2
[hide]The lengths from A' to the other points are all $ \frac{1}{\sqrt{2}}$. The original height of the tetrahedron was $ \sqrt{\frac{2}{3}}$. The new height, using the Pythagorean theorem, is $ \sqrt{\frac{1}{6}}$ (coincidentally half the original height), so AA' is $ \sqrt{\frac{1}{6}}$, so 36.[/hide]
Problem 3 seems substantially harder, I may solve it later though.
Please correct me if I'm wrong; barring that, here is a modified version of a problem I saw on a GPML sprint, so it's fairly easy:
Given set $ H$ containing 4 elements, how many different ways are there to choose sets $ A$, $ B$, $ C$, $ D$, $ E$, and $ F$ such that $ A \subseteq B \subseteq C \subseteq D \subseteq E \subseteq F \subseteq H$ and $ D$ contains at least 2 elements? Sets may be empty.
I believe the answer fits in the range 0-999, I may have miscalculated though. In case you do not understand the notation, $ X \subseteq Y$ denotes that $ X$ is a nonstrict subset of $ Y$.

Hey guys! :)

Um, sdkgurn88, you're only supposed to post one problem at a time.

I thought this died out, but since it's still here...:)

What is the largest 2-digit prime factor of the integer $ \binom{200}{100}$?

[hide="Solution"]
We need an extra factor of a prime in the numerator.  Primes above $ 61$ have one on the top, one on the bottom.  However, $ 61$ appears twice in the numerator and once in the denominator, so the answer is $ \boxed{61}$.
[/hide]

[b]New Problem:[/b] Consider the two functions $ f(x)\equal{}x^2\plus{}2bx\plus{}1$ and $ g(x)\equal{}2a(x\plus{}b)$ where the constants $ a$ and $ b$ may be considered as a point $ (a,b)$ in the Cartesian plane.  Let $ S$ be the set of such points $ (a,b)$ for which the graphs of $ y\equal{}f(x)$ and $ y\equal{}g(x)$ do not intersect.  What is the area of $ S$?

[hide="Solution"]$ f(x) \equal{} g(x)$

$ x^2 \plus{} 2bx \plus{} 1 \equal{} 2ax \plus{} 2ab$

$ x^2 \plus{} (2b \minus{} 2a)x \plus{} (1 \minus{} 2ab) \equal{} 0$

We want there to be no solutionsto this so we want to the discriminant of this quadratic to be less than zero.

$ b^2 \minus{} 4ac < 0$

$ (2b \minus{} 2a)^2 \minus{} 4(1 \minus{} 2ab) < 0$

$ a^2 \plus{} b^2 < 1$

The area is just he area of a circle centered at the origin of radius 1 so the area is just $ \boxed{\pi}$.[/hide]

[b]New Problem[/b]

Given that $ A_k \equal{} \frac {k(k \minus{} 1)}{2}\cos{\frac {k(k \minus{} 1)\pi}{2}}$, find $ |A_{19} \plus{} A_{20} \plus{} ... \plus{} A_{98} |$

[hide]
This can be written as
$ \displaystyle\sum\limits_{i=19}^{98} {k\choose2} (-1)^{k\choose2}$
The terms are positive or negative depending on whether $ \frac{k(k-1)}{2}$ is even or odd; since every fourth $ k$ is divisible by $ 4$, the pattern goes $ -,+,+,-,-,+,+...$
Then, by factoring,

$ \sum\limits_{k=19}^{98} {k\choose2} (-1)^{k\choose2}\&=\frac{(-19*18+20*19)+(21*20-22*21)+...+(97*96-98*97)}{2}\\
\&= \frac{2*19-2*21+2*23-...-2*97}{2}\\
\&= (19-21)+(23-25)+...+(95-97)$
there are $ \frac{98-19+1}{4}=20$ of these pairs, so $ -2*20=-40$ and the absolute value of that is $ 040$ [/hide]

that was my first ever use of latex.  I hope it didn't go too badly.

Bill and Joe enter a cafeteria at random times between 4:00 and 5:00.  They each stay for 5 minutes.  What is the probability that they see each other?

I may have epicly made a serious mistake somewhere because I don't think an AIME problem can't be as easily solved as I did. But [hide="here"]
Since Bill and Joe arrive at different times and each stays for 5 mins, we wish to find the probability that they will see each other. So, any one of them, let's say Bill could arrive at any time between 4 and 5, now the probability that Joe will also arrive in that time span is $ 5\frac{1}{60} \implies \boxed{\frac{1}{12}}.$ I believe that is the answer, but that is not an integer from 0 to 999...[/hide]

[b]New Problem:[/b] (Please don't answer if you have seen this problem before)

$ \triangle ABC$ is isosceles with $ AC\equal{}BC$ and $ \angle ACB \equal{} 106^\circ.$ Point $ M$ is in the interior of the triangle so that $ \angle MAC \equal{} 7^\circ$ and $ \angle MCA \equal{} 23^\circ.$ Find the number of degrees in $ \angle CMB.$

The answer to Infrared's problem is wrong.

[hide="Solution"]Basically, if Joe arrives from 4:05 to 4:55, then Bill has a 10 minute span to arrive (4:00 to 4:10 for 4:05), which has a 5/6 * 1/6 = 5/36 probability.

If Joe arrives from 4:00 to 4:05, the "average" period Bill has to arrive is 7.5 minutes - same for 4:55 to 5:00.  The probability is 1/8 * 1/6 = 1/48.

Adding these yields 23/144.[/hide]

New problem is limac's.

[quote="AIME15"]You can use angle chasing and such to solve this without trig.[/quote]

I tried angle chasing, but it didn't work.

It's similar to this problem:

An isosceles triangle $ ABC$ has apex $ A$ (that is, $ AB$ = $ AC$) and $ \angle BAC \equal{} 20^\circ$. Points $ D$ and $ E$ are drawn on $ AB$ and $ AC$ respectively, such that $ \angle BCD \equal{} 50^\circ$ and $ \angle CBE \equal{} 60^\circ$. Calculate $ \angle CDE$.

: which I also don't know how to solve.

Here's the official solution (2003 AIME I #10):

[hide="Solution"]
Let $ \overline{CP}$ be the altitude to side $ \overline{AB}$.  Extend $ \overline{AM}$ to meet $ \overline{CP}$ at point $ L$, as shown.  Since $ \angle ACL\equal{}53^\circ$, conclude that $ \angle MCL\equal{}30^\circ$.  Also $ \angle LMC\equal{}\angle MAC \plus{} \angle ACM\equal{}30^\circ$.  Thus $ \triangle MLC$ is isosceles with $ LM\equal{}LC$ and $ \angle MLC\equal{}120^\circ$.  Because $ L$ is on the perpendicular bisector of $ \overline{AB}$, $ \angle LBA\equal{}\angle LAB\equal{}30^\circ$ and $ \angle MLB\equal{}120^\circ$.  It follows that $ \angle BLC\equal{}120^\circ$.  Now consider $ \triangle BLM$ and $ \triangle BLC$.  They share $ \overline{BL}, ML\equal{}LC$, and $ \angle MLC\equal{}\angle CLB\equal{}120^\circ$.  Therefore they are congruent, and $ \angle LMB\equal{}\angle LCB\equal{}53^\circ$.  Hence $ \angle CMB\equal{}\angle CML\plus{}\angle LMB\equal{}30^\circ\plus{}53^\circ\equal{}\boxed{83^\circ}$.
[/hide]

Alternate solution (although it does use trig)

[hide="Solution"]
WLOG, assume that $ AC\equal{}BC\equal{}1$.  Apply the Law of Sines in $ \triangle AMC$ to get $ \frac{\sin 150^\circ}{1}\equal{}\frac{\sin 7^\circ}{CM}$, from which $ CM\equal{}2 \sin 7^\circ$.  Apply the Law of Cosines in $ \triangle BMC$ to obtain $ MB^2\equal{}4\sin ^2 7^\circ\plus{}1 \minus{} 2 \cdot 2 \sin 7^\circ \cos 83^\circ \equal{} 4 \sin ^2 7 ^\circ \plus{} 1 \minus{} 4 \sin ^2 7^\circ\equal{}1$.  Thus $ CB\equal{}MB$, and $ \angle CMB\equal{}\boxed{83^\circ}$.
[/hide]

The problem I posted:

An isosceles triangle $ ABC$ has apex $ A$ (that is, $ AB$ = $ AC$) and $ \angle BAC \equal{} 20^\circ$. Points $ D$ and $ E$ are drawn on $ AB$ and $ AC$ respectively, such that $ \angle BCD \equal{} 50^\circ$ and $ \angle CBE \equal{} 60^\circ$. Without using trigonometry, calculate $ \angle CDE$.


i have a question,not being able to think out a answer

a+b+c = 10
find number of whole number solutions to the above equation

i can find out the answer through balls n urns concept ie no of various solutions is equal to 12C2.

but suppose i dont make any distinction between a,b,c
and just wanna know the no of 3 numbers which add up to 10
for example
for the above equation 1+2+7 = 10 is a solution
n so is 2+1+7 = 10
but i dont wanna distinguish between the two cases n wanna treat them as 1 case only {1,2,7}
how many more sets are there..

plz help

Well, for each of these solutions, a permutation of it gets repeated three times.
So if we divide by the amount of distinct permutations of each set, then we the amount of distinct solutions.
So we need to divide by EDIT: 3.
Try that.

[quote="DiscreetFourierTransform"]Well, for each of these solutions, a permutation of it gets repeated three times.
So if we divide by the amount of distinct permutations of each set, then we the amount of distinct solutions.
So we need to divide by $3!$, or 6.

Try that.[/quote]

i tried it doesn't work,the total no of solution by 12c2 = 66 when i divide them by 3! as u said,i get 11
now i try to find solutions mechanically
0,0,10
0,1,9
0,2,8
0,3,7
0,4,6
0,5,5
1,3,6
1,4,5
1,1,8
1,2,7
2,2,6
3,2,5
i have already exceed 11 here only so apparently there are more and hence division by 3! wont work..

Solution:

[hide]

This can be thought of as the "stars and bars" problem or any other two suitable things.

We are given a+b+c=10. Think of the + signs as "bars" and the 10 integer units as "stars." Thus we need to fit 10 "stars" with 2 "bars" in as many combinations as possible. For instance:

*****|**|***

is an admissible combination that corresponds to the case (a,b,c)=(5,2,3)

And this is also admissible:

******||****

Which corresponds to (a,b,c)=(6,0,4).

As such, the problem becomes: How many distinct ways can you order 10 stars and 2 bars? The answer is just $\frac{(10+2)!}{10!\times{2!}}=66$. So the answer is $\boxed{66}$.

Follow-up: how might you answer the same question for positive integer solutions? (that is, none of a,b,c=0) [/hide]

EDIT: Aha, just got the difference in what you're looking for. Now I'm working on that.

Solution:

[hide]

Of the 66 combinations, none have a=b=c. The following are all cases that have two of a,b,c being equal:
(0,0,10)
(1,1,8)
(2,2,6)
(3,3,4)
(4,4,2)
(5,5,0)

There are 6 of these. In the generic balls-and-urns, each was counted 3 times, so there are 66-6(3) other combinations.

Then the other 66-18=48 solutions have distinct a,b,c and so these represent all combinations counted 3!=6 times each. Thus from these there are 48/6=8 unique combinations.

Thus there are 8 combinations with distinct a,b,c; there are 6 combinations with a=b and distinct c; and there are 0 combinations with a=b=c. There are 8+6=$\boxed{14}$ solutions.[/hide]

[hide]
We know there are $\binom{12}{2}=66$ total solutions to $a+b+c=10$ with $a,b,c$ nonnegative integers. We also know $a=b=c$ can't happen. So if we want to find solutions where order does not matter, then there are two cases.

Case 1: Two of $a,b,c$ equal. Here we can say WLOG $b=c$ and have $a+2b=10$. Now $a$ is even; let $a=2m$ so the equation is $m+b=5$. We get 6 solutions.

Case 2: $a,b,c$ distinct. The number of ordered solutions is $66-3\cdot 6=48$. But we overcount these 6 times because rearrangements do not matter. This makes $\frac{48}{6}=8$ of these.

The total is therefore $\boxed{14}$.
[/hide]

The general function is the two-argument version of the following, and does not have a closed form:

http://mathworld.wolfram.com/PartitionFunctionP.html

are such kind of questions answered through a case wise approach only??
or is there a specific formula as well..?


function
trigonometry
induction
algebra solved
algebra

Find all functions $ f:R->R$ satisfying these conditions:
       $ f$ is continous on $ R$
       $ f(x).f(y)-f(x+y)=sin(x).sin(y)$

Set $y=\pi$ and you'll get $f(x)f(\pi)=f(x+\pi),\ \forall x\in\mathbb R$. Now turn $y$ into $y+\pi$ to get $f(\pi)[f(x)f(y)-f(x+y)]=-\sin x\sin y$, so $f(\pi)=-1$. Now put $x=y=\frac {\pi}2$ to get $f(\frac {\pi}2)=0$.

If we put $y=nx$ we get $f(x)f(nx)-f((n+1)x)=\sin x\sin nx$, so we can perform induction: if $f(nx)\ne 0,\ f(nx)=\cos (nx),\ f((n+1)x)=\cos ((n+1)x)$, then $f(x)=\cos x$. On the other hand, if $f(x)=\cos x,\ f(nx)=\cos (nx)$, then $f((n+1)x)=\cos ((n+1)x)$. We do this for $x=\frac {\pi}2$ and we get $f(\frac pq\frac{\pi}2)=\cos(\frac pq\frac{\pi}2)$, and this, together with the continuity of $f$ gives the result $f(x)=\cos x$.


The purpose of the Energetically Autonomous Tactical Robot (EATR)™ (patent pending) project is to develop and demonstrate an autonomous robotic platform able to perform long-range, long-endurance missions without the need for manual or conventional re-fueling, which would otherwise preclude the ability of the robot to perform such missions. The system obtains its energy by foraging – engaging in biologically-inspired, organism-like, energy-harvesting behavior which is the equivalent of eating. [b]It can find, ingest, and extract energy from biomass in the environment (and other organically-based energy sources)[/b], as well as use conventional and alternative fuels (such as gasoline, heavy fuel, kerosene, diesel, propane, coal, cooking oil, and solar) when suitable.

http://www.robotictechnologyinc.com/index.php/EATR

I was just thinking about this when I saw it, and I realized that humans fit the criteria in bold. What do you guys think about this?

[quote="Ihatepie"]I for one welcome our new robot overlords.[/quote]

Even if they eat you for fuel?

If they come close I'll use [url=http://pics.livejournal.com/eska/pic/0007wfdq]this.[/url]

But to go back on topic, I doubt the robot is very big. I would assume the biggest animals it could take down are bugs and an occasional rodent.

LOL this got kicked out of the round table.


geometry
trigonometry
function
inequalities
inequalities unsolved

We know that in an acute triangle $ abc \equal{} a \plus{} b \plus{} c$. Prove that $ \frac {\sqrt {3}}2 < A < \frac 32$ (maybe $ \le$ somewhere) where $ A$ is the area of the triangle.

Let $ K$ the triangle's area.

$ abc \equal{} a \plus{} b \plus{} c \rightarrow 1 \equal{} \frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca}\equal{}\frac{\sin A}{2K}\plus{}\frac{\sin B}{2K}\plus{}\frac{\sin C}{2K}$

$ K\equal{} \frac{\sin A \plus{} \sin B \plus{} \sin C}{2}$

Since The sine function is concave, so we can use the Jense's inequality in the following way: $ \sin  \frac{A\plus{}B\plus{}C}{3}\equal{} \sin  \frac{\pi}{3}\equal{}\frac{\sqrt{3}}{2}\le \frac{\sin A\plus{}\sin B\plus{}\sin C}{3}$ so $ K \ge 3\frac{\sqrt{3}}{4}> \frac{\sqrt{3}}{2}$.

Now using the fact $ \sin x \le x$ if $ x\ge 0$

$ K\equal{} \frac{\sin A \plus{} \sin B \plus{} \sin C}{2}< \frac{A \plus{} B \plus{} C}{2} \equal{} \frac{\pi}{2}$

The bible says $ \pi\equal{}3$ so $ K< \frac{3}{2}$

[quote="asdrojas"]
Since the sine function is concave, so we can use the Jensen's inequality in the following way: $ \sin \frac {A \plus{} B \plus{} C}{3} \equal{} \sin \frac {\pi}{3} \equal{} \frac {\sqrt {3}}{2}\ge \frac {\sin A \plus{} \sin B \plus{} \sin C}{3}$ so $ K \le 3\frac {\sqrt {3}}{4} < \frac {3}{2}$.

$ K \equal{} \frac {\sin A \plus{} \sin B \plus{} \sin C}{2} < \frac {1\plus{}1\plus{}1}{2} \equal{} \frac {3}{2}$
[/quote]

Sorry for not replying sooner but unfortunately your solution appears to be incorrect. :(

Ok, so if $ abc\equal{}a\plus{}b\plus{}c$ in an acute triangle show that \[ 1<A\le\frac {3\sqrt 3}4\]

[hide="Solution for the second part"]$ A\equal{}\frac {abc}{4R}\equal{}\frac {\sin\alpha\plus{}\sin\beta\plus{}\sin\gamma}2\le\frac {\frac {3\sqrt 3}2}2$ (by Jensen).[/hide]

$ S \equal{} \frac {sinA \plus{} sinB \plus{} sinC}{2}$

$ 1 < S \Longleftrightarrow sinA \plus{} sinB \plus{} sinC > 2$

$ \Longleftrightarrow 2sinAsinB \plus{} 2sinAsinC \plus{} 2sinBsinC > 1 \plus{} cos^2{A} \plus{} cos^2{B} \plus{} cos^2{C}$

$ \Longleftrightarrow [cos(A \minus{} B) \plus{} cos(B \minus{} C) \plus{} cos(A \minus{} C)] \plus{} cosA \plus{} cosB \plus{} cosC > 1 \plus{} cos^2{A} \plus{} cos^2{B} \plus{} cos^2{C}$

$ cosx > cos^2{x}$ So;

If we prove that $ cos(A \minus{} B) \plus{} cos(B \minus{} C) \plus{} cos(A \minus{} C) > 1$ then the solution ends.

$ cos(A \minus{} B)\plus{}cos(B\minus{}C)\plus{}cos(A\minus{}C)>$

$ \sqrt {cos^2{(A \minus{} B)} \plus{} cos^2{(B \minus{} C)} \plus{} cos^2{(A \minus{} C)}}$

Without the loss of generality we can say $ A \geq B \geq C$

Let's prove that $ cos^2{(A \minus{} B)} \plus{} cos^2{(B \minus{} C)} > sin^2{(A \minus{} C)}$

$ cos^2{(A \minus{} B)} \plus{} cos^2{(B \minus{} C)}\geq 2cos^2{(\frac {A \minus{} C}{2})}$ (by the Cauchy-Schwarz and Jensen) $ A \minus{} B < 90$ and  $ B \minus{} C < 90$

$ 2cos^2{(\frac {A \minus{} C}{2})} > sin^2{(A \minus{} C)} \Longleftrightarrow$

$ sin{(\frac {A \minus{} C}{2})} < \frac {1}{\sqrt {2}}$ which is obviously true because of $ A \minus{} C < 90$  :wink:


Is the definiton of ABCD is in this order on a line, (ABCD)=-1 $ \implies \frac{CA}{CB}\equal{}\frac{DA}{DB}$ or is it $ \frac{CB}{CE}\equal{}\frac{BA}{AD}$?

can someoneve a proof of:
(ABCD)=-1 implies $ \frac{2}{AC}\equal{}\frac{1}{AC}\plus{}\frac{1}{AD}$

[quote="D.P.L"]can someoneve a proof of:
(ABCD)=-1 implies $ \frac2{AC}\equal{}\frac1{AC}\plus{}\frac1{AD}$[/quote]
first of all it should be $ \frac2{AB}\equal{}\frac1{AC}\plus{}\frac1{AD}$,second of all you asked the same question here [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=948614#948614[/url]

anyway:
assume all the segments directed:
$ (AB,CD)\equal{}\minus{}1\Rightarrow \frac{AC}{BC}\equal{}\minus{}\frac{AD}{BD}\Rightarrow AC.BD\equal{}\minus{}AD.BC$ (*)

now we want to prove that $ \frac2{AB}\equal{}\frac1{AC}\plus{}\frac1{AD}$

$ \Rightarrow 2AC.AD\equal{}AB.AC\plus{}AB.AD$
$ \Rightarrow (AC.AD\minus{}AB.AC)\plus{}(AC.AD\minus{}AB.AD)\equal{}0$
$ \Rightarrow AC(AD\minus{}AB)\plus{}AD(AC\minus{}AB)\equal{}0$
$ \Rightarrow AC.BD\plus{}AD.BC\equal{}0$
$ \Rightarrow AC.BD\equal{}\minus{}AD.BC$

which is true because of (*)

[quote="BaBaK Ghalebi"][quote="D.P.L"]can someoneve a proof of:
(ABCD)=-1 implies $ \frac2{AC} \equal{} \frac1{AC} \plus{} \frac1{AD}$[/quote]
first of all it should be $ \frac2{AB} \equal{} \frac1{AC} \plus{} \frac1{AD}$,second of all you asked the same question here [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=948614#948614[/url]

anyway:
assume all the segments directed:
$ (AB,CD) \equal{} \minus{} 1\Rightarrow \frac {AC}{BC} \equal{} \minus{} \frac {AD}{BD}\Rightarrow AC.BD \equal{} \minus{} AD.BC$ (*)

now we want to prove that $ \frac2{AB} \equal{} \frac1{AC} \plus{} \frac1{AD}$

$ \Rightarrow 2AC.AD \equal{} AB.AC \plus{} AB.AD$
$ \Rightarrow (AC.AD \minus{} AB.AC) \plus{} (AC.AD \minus{} AB.AD) \equal{} 0$
$ \Rightarrow AC(AD \minus{} AB) \plus{} AD(AC \minus{} AB) \equal{} 0$
$ \Rightarrow AC.BD \plus{} AD.BC \equal{} 0$
$ \Rightarrow AC.BD \equal{} \minus{} AD.BC$

which is true because of (*)[/quote]

does (AB,CD)=(ACBD)?

and in the proof of the thread soemone wrote: (B,X,K,E)=-1 implies KX/KE=XB/BE :huh:

thank you

as far as I know $ (AB,CD)\equiv (A,B,C,D)\equiv (ABCD)$


complex numbers
linear algebra
linear algebra unsolved

Le $ n$ and $ r$ be  positive integers such that $ n>1$ and $ 0<r<n$. Show that for a field $ \mathbb{F}$, the space $ \mathbb{F}^{n \times n}$ has a subspace $ U$ such that:

i) $ \text{Rank}(A)\leq r$ for any $ A \in U$.
ii) There exist $ B$ such that all entries of $ B$ is positive.

[quote="Beginner"]Le $ n$ and $ r$ be  positive integers such that $ n > 1$ and $ 0 < r < n$. Show that for a field $ \mathbb{F}$, the space $ \mathbb{F}^{n \times n}$ has a subspace $ U$ such that:

i) $ \text{Rank}(A)\leq r$ for any $ A \in U$.
ii) There exist $ B$ such that all entries of $ B$ is positive.[/quote]

You can't say that entries are positive or negative if you have arbitrary field.For example
 which numbers are positive in field of complex numbers?

Ok, i suppose that we take the field to be set of real numbers (as when i solved this).


limit
real analysis
real analysis unsolved

I don't understand the answer for part (b).
How to know if it's first two nonzero terms?
Please explain this.

[/url]

Do you know what the Alternating Series Estimation Theorem says?

It says that given $ S \equal{} \sum_{k\equal{}0}^{\infty} (\minus{}1)^k a_k$ where $ a_k$ is a monotonically decreasing positive sequence with $ \lim_{k \to \infty} a_k \equal{} 0$,

$ |S \minus{} \sum_{k\equal{}0}^{n} (\minus{}1)^k a_k| < a_{n\plus{}1}$.

When we take $ n \equal{} 2$ the error is bounded by $ a_3 \equal{} \frac{1}{11 \cdot 5!} \equal{} \frac{1}{1320} < 0.001$.

[quote="t0rajir0u"]Do you know what the Alternating Series Estimation Theorem says?
It says that given $ S \equal{} \sum_{k \equal{} 0}^{\infty} ( \minus{} 1)^k a_k$ where $ a_k$ is a monotonically decreasing positive sequence with $ \lim_{k \to \infty} a_k \equal{} 0$,
$ |S \minus{} \sum_{k \equal{} 0}^{n} ( \minus{} 1)^k a_k| < a_{n \plus{} 1}$.
When we take $ n \equal{} 2$ the error is bounded by $ a_3 \equal{} \frac {1}{11 \cdot 5!} \equal{} \frac {1}{1320} < 0.001$.[/quote]

I finally got it.
Thank you for your help, t0rajir0u.


find all positive integers m,n such that
mn-1 | (m+n-1)^2

Your problem is equivalent to this nice problem
http://www.mathlinks.ro/Forum/viewtopic.php?p=1566051#1566051


Eight people seated around a table need to be split into four groups of two, in a way such that the two people in any one group must be seated next to each other.  Each person randomly chooses one of their two neighbors to be the person they would like to group with, and if two people both choose each other, they become a group.  What is the probability that after these groupings are made, it will be possible to group the remaining people?

[hide="Solution"]
It is not too hard to find that the only two basic patterns of groups that don't allow the remaining people to be divided into groups are the ones depicted in following diagrams. (In these diagrams, each person is represented by a dot, and a group is represented by a line segment.)

[asy]size(175);

picture pic1, pic2;

pair A = dir(22.5);
pair B = dir(3*22.5);
pair C = dir(5*22.5);
pair D = dir(7*22.5);
pair E = dir(9*22.5);
pair F = dir(11*22.5);
pair G = dir(13*22.5);
pair H = dir(15*22.5);

dot(pic1, A);
dot(pic1, B);
dot(pic1, C);
dot(pic1, D);
dot(pic1, E);
dot(pic1, F);
dot(pic1, G);
dot(pic1, H);

draw(pic1, B--C);
draw(pic1, G--H);

add(pic1);

dot(pic2, A);
dot(pic2, B);
dot(pic2, C);
dot(pic2, D);
dot(pic2, E);
dot(pic2, F);
dot(pic2, G);
dot(pic2, H);

draw(pic2, B--C);
draw(pic2, E--F);
draw(pic2, G--H);

add(shift(4.5*right)*pic2);

label("Basic Pattern 1", (0, -2));
label("Basic Pattern 2", (4.5, -2));[/asy]
Of course, the basic patterns can be rotated to form other patterns which don't allow the remaining people to be grouped, but these two patterns are the "basic patterns." For each of the two basic patterns, there are multiple ways that it could occur. For Basic Pattern 2, there are four combinations of choices that produce it: the people in groups must choose each other, but each of the remaining 2 people can choose either of his neighbors, so there are $ 2^2 \equal{} 4$ ways that Basic Pattern 2 can occur. Basic Pattern 1 is a little trickier: the single person can choose either of his neighbors, but there are exactly 4 combinations of choices for the cluster of three people. These combinations are diagramed below, with each arrow going from a person to the neighbor he chose.

[asy]size(300);

picture pic1, pic2, pic3, pic4;

pair A = dir(22.5);
pair B = dir(3*22.5);
pair C = dir(5*22.5);
pair D = dir(7*22.5);
pair E = dir(9*22.5);
pair F = dir(11*22.5);
pair G = dir(13*22.5);
pair H = dir(15*22.5);

dot(pic1, A);
dot(pic1, B);
dot(pic1, C);
dot(pic1, D);
dot(pic1, E);
dot(pic1, F);
dot(pic1, G);
dot(pic1, H);
dot(pic2, A);
dot(pic2, B);
dot(pic2, C);
dot(pic2, D);
dot(pic2, E);
dot(pic2, F);
dot(pic2, G);
dot(pic2, H);
dot(pic3, A);
dot(pic3, B);
dot(pic3, C);
dot(pic3, D);
dot(pic3, E);
dot(pic3, F);
dot(pic3, G);
dot(pic3, H);
dot(pic4, A);
dot(pic4, B);
dot(pic4, C);
dot(pic4, D);
dot(pic4, E);
dot(pic4, F);
dot(pic4, G);
dot(pic4, H);

draw(pic1, B--C);
draw(pic1, G--H);
draw(pic2, B--C);
draw(pic2, G--H);
draw(pic3, B--C);
draw(pic3, G--H);
draw(pic4, B--C);
draw(pic4, G--H);

real c = 1.35;
draw(pic1, c*dir(7*22.5)--c*dir(8*22.5), Arrow);
draw(pic1, c*dir(9*22.5)--c*dir(10*22.5), Arrow);
draw(pic1, c*dir(11*22.5)--c*dir(12*22.5), Arrow);
draw(pic2, c*dir(7*22.5)--c*dir(6*22.5), Arrow);
draw(pic2, c*dir(9*22.5)--c*dir(10*22.5), Arrow);
draw(pic2, c*dir(11*22.5)--c*dir(12*22.5), Arrow);
draw(pic3, c*dir(7*22.5)--c*dir(6*22.5), Arrow);
draw(pic3, c*dir(9*22.5)--c*dir(8*22.5), Arrow);
draw(pic3, c*dir(11*22.5)--c*dir(12*22.5), Arrow);
draw(pic4, c*dir(7*22.5)--c*dir(6*22.5), Arrow);
draw(pic4, c*dir(9*22.5)--c*dir(8*22.5), Arrow);
draw(pic4, c*dir(11*22.5)--c*dir(10*22.5), Arrow);

add(pic1);
add(shift(4.5*right)*pic2);
add(shift(9*right)*pic3);
add(shift(13.5*right)*pic4);

label("4 Possiblities of Choices in Basic Pattern 1", (6.75, -2.5));[/asy]
Thus, since there are 4 ways that the group of three people can choose neighbors and 2 ways the the isolated person can choose a neighbor, there are 8 ways that Basic Pattern 1 can occur.

Each basic pattern can be rotated 8 different ways. Thus, for Basic Pattern 1, since it can occur 8 different ways and be rotated 8 different ways, there are 64 ways that the remaining people cannot be grouped that follow Basic Pattern 1. Similarly, Basic Pattern 2 can occur 4 different ways and be rotated 8 different ways, so there are 32 ways that the remaining people cannot be grouped that follow Basic Pattern 2. This gives a total of 96 ways that the remaining people cannot be grouped.

Since each of the 8 people must choose either the person to his left or the person to his right, there are $ 2 ^ 8 \equal{} 256$ different ways that the eight people can choose neighbors for grouping. In $ 256 \minus{} 96 \equal{} 160$ of these ways, the remaining people can be grouped. Thus, the desired probability is $ \frac{160}{256} \equal{} \boxed{\frac{5}{8}}.$

EDIT: I thought that the cluster of 3 people could make choices in only 2 ways; fixed now.
[/hide]

Extension: in general, what is the probability, given $ 2n$ people seated around a table?

For the 3-cluster in the first basic pattern, is it really true that everyone must choose in the same direction?

You're right; I discovered that mistake shortly after I posted it! :oops: There are 4 ways the group of three people can choose their neighbors. I'm working on fixing it, and I'll edit my post.

One way to think about what we're doing:

Given a valid pattern of pairings, if the runs of unpaired people have length $ a_1, a_2, \ldots, a_m$, then the total number of ways people could have been pointing is $ (a_1 \plus{} 1)(a_2 \plus{} 1) \cdots (a_m \plus{} 1)$, unless we have no pairings at all (a single run of length $ 2n$) in which case there are two.  We're trying to sum this expression over all of the $ L_{2n}$ ways of drawing a valid pattern of pairings, and then separately sum it over those pairings in which all the $ a_i$ are even, and compare these two numbers.  ($ L_{2n}$ is a Lucas number; $ L_n$ count the ways to cut up a circle with $ n$ dots so that each arc has either one or two dots on it.)

It seems to me that this might be easier if we ignored the weightings and just asked for the fraction of all those diagrams that have all runs of unpaired people of even length.  (This is a completely different problem, but it looks like it might be more approachable.)

We could also ask the same question of $ 2n$ people seated in a line; then Lucas numbers will be replaced by Fibonacci numbers.


modular arithmetic
quadratics
number theory proposed
number theory

Let $x,y$ positive integers. If the numbers $3x+4y$ and $4x+3y$ are perfect squares
prove that the numbers $x,y$ has the form $7k,7l$ respectively, where $k,l$ positive integers.

Please I want from the moderators to send all the problems to Olympiad recousrces.
I will put all the for problems from 2004

Just a little different... Let $3x + 4y = a^2$ and $4x + 3y = b^2$, with $a, b \in \mathbb{Z}$. Hence $a^2 + b^2 = 7(x+y)$, and $7 \mid a$ iff $7 \mid b$. So suppose $7 \nmid a$. You argue $a^2 \equiv -b^2 \bmod 7$. But $b$ is invertible $\bmod\;\! 7$, so you're actually looking for an integer solution of $u^2 \equiv -1 \bmod 7$, which is impossible, since $-1$ isn't a quadratic residue $\bmod\;\! 7$.


Determine whether the set of intergers $ n>1$ satisfying $ \omega(n)<\omega(n\plus{}1)<\omega(n\plus{}2)$ is finite,where $ \omega(n)$ denotes
$ a.$ the largest prime divisor of $ n$
$ b.$ the number of prime divisors of $ n$


if
$ 1\plus{}(\minus{}2)x\plus{}\frac{(\minus{}2)(\minus{}3)}{2}x^{2}\plus{}\frac{(\minus{}2)(\minus{}3)(\minus{}4)}{6}x^{3}\plus{}\cdots\equal{}\frac{1}{(1\plus{}x)^2}$

what is the general rule ?
example: what does this equal to.
$ a\plus{}bx\plus{}cx^2\plus{}dx^3\plus{}ex^4\plus{}\cdots$

$ (1 \plus{} x)^r \equal{} \sum_{n \equal{} 0}^\infty \binom{r}{n} x^n$ for $ x > \minus{} 1$.

For your second question, there is no general answer other than "exactly what it is".


can some one tell me what IS P versus NP, poincare conjecture, riemann hypothesis , etc etc etc?

These are the 7 Millenium Problems. Visit http://www.claymath.org/millennium/

A question that has been posted and answered MANY MANY times. Search for it. Locked.


The sequence {$x_{n}$(n=1,2,...)satisfy $0\leq x_{n+m}\leq x_{n}+x_{m}$ $\forall m,n \in N^*$.
Prove that  $lim_{n\to\infty} \frac{x_{n}}{n}$  esists.

0<=x(n)/n<=x(1). Therefore exist up limit u and down limit d. Let eps>0, x(m)/m<d+eps,then
x(km+r)<=kx(m)+x(r)<=k*m*(d+eps)+r*x(1)<=k*m*(d+eps)+m*x(1) (r<m). Therefore
x(km+r)/(km+r)<=[k(d+eps)+x(1)]/k=d+eps+x(1)/k<d+2eps,when k>x(1)/eps. Therefore d=u.

I really can't read this. Can moderators intervene in non-LaTeXed messages?

[b]Rust[/b] wrote:

$0\leq \frac{x_n}{n}\leq x_1$. Therefore exist up limit $u$ and down limit $d$. Let $\epsilon>0, \frac{x_m}{m}<d+\epsilon$, then
$x_{km+r}\leq kx_{m}+x_{r}\leq km(d+\epsilon)+rx_1\leq km(d+\epsilon)+mx_1 (r<m)$. Therefore
$\frac{x_{km+r}}{km+r}\leq \frac{k(d+\epsilon)+x_1}{k}=d+\epsilon+\frac{x_1}{k}<d+2\epsilon$,when $k>\frac{x_1}{\epsilon}$. Therefore $d=u.$