function
induction
linear algebra
matrix
quadratics
algebra
polynomial

The sequence $a_{n}$ satisfies the recursion $a_{n}=3a_{n-1}+28a_{n-2}$ for integers $n>2$. If $a_{1}=3$ and $a_{2}=65$, write an explicit formula for $a_{n}$ in terms of $n$.

I used generating functions to obtain that [hide]$a_{n}= 7^{n}+(-4)^{n}$[/hide]Perhaps if one conjectures it, it's easy to prove by induction?

Yeah.

I'm not very good at generating functions..could you post your solution?

Yeah, I'm not good at characteristic equations  :(  I'd like to learn though. Is there a good explanation of it on the internet?

I'll post my step-by-step solution with generating functions. [hide]To keep it readable, let's use $S_{k}(\psi(x)) = \sum_{i=k}^{\infty}\psi(x)$. Let $f(x) = S_{1}\left(a_{i}x^{i}\right)$. We have that \[\begin{aligned}f(x) & = S_{1}\left(a_{i}x^{i}\right) = a_{1}x^{1}+a_{2}x^{2}+S_{3}\left(a_{i}x^{i}\right) \\ & = 3x+65x^{2}+S_{3}\left(3a_{i-1}x^{i}+28a_{i-2}x^{i}\right) \\ & = 3x+65x^{2}+3xS_{3}\left(a_{i-1}x^{i-1}\right)+28x^{2}S_{3}\left(a_{i-2}x^{i-2}\right) \\ & = 3x+65x^{2}+3xS_{2}\left(a_{i}x^{i}\right)+28x^{2}S_{1}\left(a_{i}x^{i}\right) \\ & = 3x+65x^{2}+3x\left(S_{1}\left(a_{i}x^{i}\right)-a_{1}x^{1}\right)+28x^{2}S_{1}\left(a_{i}x^{i}\right) \\ & = 3x+65x^{2}+3x\left(f(x)-3x\right)+28x^{2}f(x) \\ & = 3x+56x^{2}+f(x)\left(3x+28x^{2}\right).\end{aligned}\] Solving for $f(x)$, we get \[f(x) = \frac{3x+56x^{2}}{1-3x-28x^{2}}=-2-\frac{3x-2}{1-3x-28x^{2}}.\] But $1-3x-28x^{2}= (1-7x)(1+4x)$, so \[f(x) =-2-\frac{3x-2}{(1-7x)(1+4x)}.\] Split that last one into partial fractions: \[\frac{3x-2}{(1-7x)(1+4x)}=-\frac{1}{1-7x}-\frac{1}{1+4x}.\] We obtain that \[\begin{aligned}f(x) & =-2+\frac{1}{1-7x}+\frac{1}{1-(-4x)}\buildrel{\text{Sum of geo. series}}\over=-2+S_{0}\left((7x)^{i}\right)+S_{0}\left((-4x)^{i}\right) \\ & = S_{1}\left((7x)^{i}+(-4x)^{i}\right) = S_{1}\left((7^{i}+(-4)^{i})x^{i}\right).\end{aligned}\] From which, $a_{n}= 7^{n}+(-4)^{n}$. Tadaaaaaaaaaa  :P [/hide]

[quote="t0rajir0u"][hide="Solution"] We have the characteristic equation $x^{2}-3x-28 = 0$ with roots $-4, 7$.  Then

$a_{n}= \alpha \cdot 7^{n}+\beta \cdot (-4)^{n}$

And we calculate $\alpha = \beta = 1$. [/hide][/quote]

WHy do you get that characteristic equation...? then, why are the roots expressed as $r^{n}$ in $a_{n}=\alpha \cdot 7^{n}+\beta \cdot (-4)^{n}$? then why $\alpha=\beta=1$?   :blush:

How about this one?

Hosei University entrance exam/management 1987

Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=1,\ a_{2}=2,\ a_{n+2}-a_{n+1}-2a_{n}=2.$

Substitute $a_{n}=b_{n}-1$ and then you can use characteristic equations

[quote="Hokkage"]Yeah... I like the characteristic equation solution a bit more than that generating function madness...

Only when I google characteristic equation, I get matrix stuff. Explanation/link t0r?[/quote]

[hide="A theorem"] [b]Theorem:[/b]  Every sequence of the form $a_{n}= \alpha \gamma^{n}+\beta \delta^{n}$ is a second-order linear recurrence of the form

$a_{n}= A a_{n-1}+B a_{n-2}$

Proof:  The proof is actually very simple.  Let $A = \alpha+\beta, B =-\alpha \beta$.  Then

$\alpha \gamma^{n}+\beta \delta^{n}= A \left( \alpha \gamma^{n-1}+\beta \delta^{n-1}\right)+B \left( \alpha \gamma^{n-2}+\beta \delta^{n-2}\right)$

And we see that the quadratic

$x^{2}= Ax+B$

Has roots $\alpha, \beta$.

A very nonrigorous way of understanding this:  Let $a_{n}$ be a second-order linear recurrence.  We assume that $a_{n}$ is "approximately" exponential, that is, $a_{n}= \epsilon \zeta^{n}$ and then we simply divide out to get

$\zeta^{2}= A \zeta+B$

This is called the characteristic equation of the recurrence.

I claim that every recurrence such that its characteristic equation has distinct roots can be written in the form I gave above.  The proof uses generating functions and is very long, so I will not give it here.

From here it is a simple matter of calculation.  :) [/hide]

Something on recurrence equations and characteristic polynomials:

[url]http://en.wikipedia.org/wiki/Recurrence_relation[/url]
[url]http://www2.ics.hawaii.edu/~sugihara/course/ics141s98/note/04-23n24[/url]

[quote="Kurt Gödel"]

$a_{n}= \frac{1}{3}\left(5\cdot 2^{n-1}+(-1)^{n+1}-3\right)$.

Is this good, kunny?  :)[/quote]

Yes, your answer is correct. :)

[quote="SplashD"]I got $a_{n}=-\frac13 (-1)^{n}+\frac56 (2)^{n}-1$ using Elemennop's idea.[/quote]

That's the same as [b]Kurt Gödel[/b]'s answer, just with some exponents and common factors juggled around.


Solve for x
$ \sqrt{x\plus{}\sqrt{x^2\minus{}1}}\equal{}\frac{9\sqrt{2}}{4}(x\minus{}1)\sqrt{x\minus{}1}$

By using Maple, we get the result: $ x \equal{} \frac {5}{3}$. (Plugging in $ x \equal{} \frac {5}{3}$ in the equation $ \sqrt{x\plus{}\sqrt{x^2\minus{}1}}\equal{}\frac{9\sqrt{2}}{4}(x\minus{}1)\sqrt{x\minus{}1}$, it becomes $ \sqrt3\equal{}\sqrt3$.)

[quote="greenvert"]this is the problem from the newest Mathematics and Youth Magazine (7/2008). The dealine for submitting solutions is 15 September, 2008, so don't answer.See detail here:
[url]http://www.mathlinks.ro/viewtopic.php?p=1195403#1195403[/url][/quote]
Thank you. I didn't know that. Sorry for posting my comments. Moderators, please lock the topic.
And Doctor Graubner: Please don't post a solution in the meantime.


Three non-collinear points $A,B,C$ exist in the same plane. $P$ internally divides $AB$ into segments of $t : (1-t)$ and $Q$ divides the $BC$ into the same proportion internally as well. $R$ divides $PQ$ internally into segments of $t: (1-t)$. Find the locus of $R$ as $t$ varies if $A= (1,1)$, $B=(2,3)$ and $C=(3,1)$.

question:

when the segment $AB$ is divided, there are 2 possible points for $P$. This shouldn't matter, since t varies.

But what does matter is whether $Q$ divides $BC$ on the same side as $P$ or at the other point. Is the bigger side closer to B or closer to C?

I said that segment $AB$ is divided internally so for every $t$ there is only one possible $P$ (I think, correct me if I am wrong).

$BQ: QC = t: (1-t)$. Everything is divided internally (for simplicity) but you can try to prove the general case for if the points divide internally or externally.

[hide="what I was trying to differentiate between"]

I hope I'm not missing something obvious in your question

[img]http://img110.imageshack.us/img110/1310/mathvz6.jpg[/img]

[/hide]


It's easy to show that there are arbitrarily many composite consecutine positive integers but,

Is there a set of 1000 positive consecutive integers, with exactly 5 of them prime?

Well, consider the first 1000 numbers, which contains more than 5 primes, and a string of 1000 composite numbers, which contains 0 primes.
Every time we shift our string over 1, we either change the number of primes by 1,0, or -1, depending whether the next number is a prime, and whether the number you just left when you shifted is a prime.
Hence, starting with a number > 5, we end up with 0 after finitely many operations of +1,-1, and 0(do nothing). Therefore, logic says that every number between this number greater than 5 and 0 must have been reached, specifically, at some point 5 was reached. Hence we're done.

A question: Is there a number $ n > 1$ such that there exists a set of $ n$ consecutive numbers with more than $ \pi(n)\plus{}1$ primes? If so, what numbers? Are there infinitely many?

Heuristically the answer should be no, since primes get rarer as you move up. Note: The reason for the $ \plus{}1$ is that for any prime $ p$, the $ p\minus{}1$ numbers $ 2,3,\ldots,p$ will contain $ \pi(p\minus{}1)\plus{}1$ primes.

$ \pi \left( n \right)$ is the number of primes $ \leq n$.

[quote="Hamster1800"]A question: Is there a number $ n > 1$ such that there exists a set of $ n$ consecutive numbers with more than $ \pi(n) \plus{} 1$ primes? If so, what numbers? Are there infinitely many?
[/quote]

The number of primes between $ n\plus{}1$ and $ n\plus{}k$ is $ \pi(n\plus{}k)\minus{}\pi(n)$. The question of whether this is always no more that $ \pi(k)$, or in other words whether $ \pi$ is subadditive

$ \pi(n\plus{}k) \leq \pi(n) \plus{} \pi(k)$

is the (second) Hardy-Littlewood conjecture, which is unsettled but probably false, since it was shown to contradict other widely believed conjectures.


Find all positive real a,b,c such that
4(ab+bc+ac)-1>=a^2+b^2+c^2>=3(a^3+b^3+c^3)

$ 4(ab \plus{} bc \plus{} ac) \minus{} 1 \ge a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$, so $ ab \plus{} bc \plus{} ca \ge \frac {1}{3}$
$ a^2 \plus{} b^2 \plus{} c^2 \ge 3(a^3 \plus{} b^3 \plus{} c^3) \ge (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)$, so $ a \plus{} b \plus{} c \le 1$, $ ab \plus{} bc \plus{} ca \le \frac {(a \plus{} b \plus{} c)^2}{3} \le \frac {1}{3}$,
so $ \begin{cases}ab \plus{} bc \plus{} ca \equal{} \frac {1}{3} \\
a \equal{} b \equal{} c \end{cases}$
$ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ are the only solution.


In this topic you will see the current confirmed bugs

:arrow: LaTeX pop-up window doesn't show the code (IE, Firefox, Opera) (Valentin)

:arrow: Some lang entries need filling (lang) (Valentin, Dave)

:arrow: Highlight function doesn't work with LaTeX properly (Valentin)

:arrow: Smilies and other spam limits (Valentin)

:arrow: Topic shadow can't be erased (Valentin)

:arrow: Blogs page needs some more patching up (attachments box not displaying properly) (Valentin) reported @ http://www.mathlinks.ro/Forum/viewtopic.php?t=97878

Blogs auths now revised. Users can now post in blogs :)

Also, could the PDF in resource center bug also be added? Thanks.

It used to be soo awesome :)

[quote="anirudh"]Also, could the PDF in resource center bug also be added?[/quote]

I think it's not a bug but a decision because of copyright morons who don't want a considerable amount of problems of one competition posted at one place.

What would be much more awesome (and should not violate copyright): http://www.mathlinks.ro/Forum/viewtopic.php?t=125784 .

  darij


Let \[ a,b,c>0 \]Prove the inequality:
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c+\frac{4(a-b)^2}{a+b+c} \]

Man, i'm sick of posting solutions to this one. Anyway, here it goes:

Let us observe that $\frac{a^{2}}{b}=2a-b+\frac{(a-b)^{2}}{b}$. Writing the analogous and applying Cauchy-Schwarz inequality we have:

                                              \[ \frac{(a-b)^{2}}{b}+\frac{(b-c)^{2}}{c}+\frac{(c-a)^{2}}{a}\geq\frac{(2a-2b)^{2}}{a+b+c}  \].

  So from this inequality it follows that the inequality is done! ;)

[quote="yassin88"]I do not know how you get to:

\[ \frac{(a-b)^{2}}{b}+\frac{(b-c)^{2}}{c}+\frac{(c-a)^{2}}{a}\geq\frac{(2a-2b)^{2}}{a+b+c}  \][/quote]
cezar lupu has already stated that this is from Cauchy-Schwarz.
$(a+b+c)(\frac{(a-b)^{2}}{b}+\frac{(b-c)^{2}}{c}+\frac{(c-a)^{2}}{a})=(a+b+c)(\frac{(a-b)^{2}}{b}+\frac{(c-b)^{2}}{c}+\frac{(a-c)^{2}}{a})\geq (a-b+c-b+a-c)^2=(2a-2b)^2$

Thanks, this is exactly what I wanted to say. ;)

[quote="socrates"]This was given in BMO 2005(problem 1) if i am not wrong...

nice! :D  :lol:[/quote]

It was also given in Morocco National Olympiads 2005. :D


Let a,b,c>0 and $ a\plus{}b\plus{}c\equal{}1$. Prove that:
$ \frac{ab}{\sqrt{ab\plus{}c}}\plus{}\frac{bc}{\sqrt{bc\plus{}a}}\plus{}\frac{ca}{\sqrt{ca\plus{}b}} \le \frac{1}{2}$

I have just realized this ineq is very easy by AM-GM.  :blush:
We continiuos with another old lemma:
Let a,b,c>0. Prove that:
$ \frac {1}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} ca \plus{} a^2} \ge \frac {9}{(a \plus{} b \plus{} c)^2}$

[quote="mathstarofvn"]Let a,b,c>0 and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \frac {ab}{\sqrt {ab \plus{} c}} \plus{} \frac {bc}{\sqrt {bc \plus{} a}} \plus{} \frac {ca}{\sqrt {ca \plus{} b}} \le \frac {1}{2}$[/quote]
 $ ab\plus{}c\equal{}ab\plus{}c(a\plus{}b\plus{}c)\equal{}(c\plus{}a)(c\plus{}b)$
we have
$ LHS \le \frac{1}{2}(\sum \frac{ab}{c\plus{}b}\plus{}\sum \frac{ab}{c\plus{}a})\equal{}\frac{1}{2}$


calculus
integration
geometry
function
factorial
calculus computations

calculate the area betewen the graphe of the function f(x)=x^6/(1+x^4)^1/2
and x=0 and x=1

since the function $f(x)$ is always positive between $0$ and $1$ .so the problem is equivalent to this

 $\int_{0}^{1}\frac{x^{6}}{\sqrt{1+x^{4}}}$ $dx$

substitute $x^{2}$ $=$ $tan$ $\theta$

the inegral becomes
 $\int_{0}^{\frac{\pi}{4}}sin^{\frac{5}{2}}\theta cos^{-\frac{7}{2}}\theta$ $d\theta$

which can be solved using $gamma$ $functions$

[quote="drapt@"]waht is gamma function??[/quote]

$gamma$ $function$ is defined $\Gamma(x)$

$\Gamma(x)$ $=$ $\int_{0}^{\infty}e^{-t}t^{x-1}dt$

also it is the extension of factorial into the field of real no.s :)

is there an other way to fin the result???


from number theory excercises:
after solving Pellian, I get solutions given by
\[ w_n\plus{}y_n\sqrt{15} \equal{} \left(5\plus{}\sqrt{15}\right)\left(4\plus{}\sqrt{15}\right)^n\]
then $ w_0\equal{}5$.
how can I prove that $ w_0$ is the only power of 5 in it ?


function
floor function
ceiling function

lets say that $\lfloor{x}\rfloor=k$ if $y$ is the decimal part of x then. $\lfloor{2x}\rfloor=2k+\lfloor{2y}\rfloor$ also for the others then $17k\leq2005=17k+\lfloor{2y}\rfloor+\lfloor{4y}\rfloor+\lfloor{8y}\rfloor\leq17k+1+3+7=17k+11$ but 2005 is 17x+16 then that isnt true for any x.

You need to replace 17 with a 15.  And a solution does exist.  Try x=133.8

Stemming from joshua_mex we know
\[15k\leq 2005\leq 15k+\lfloor{2y}\rfloor+\lfloor{4y}\rfloor+\lfloor{8y}\rfloor\]
This implies k=133.
Our problem is then reduced to solving
\[\lfloor{2y}\rfloor+\lfloor{4y}\rfloor+\lfloor{8y}\rfloor=10\]
Substitute $y=1-p, p>0$ and using the identity $\lceil{u}\rceil=-\lfloor{-u}\rfloor$, we now must solve
\[\lceil{2p}\rceil +\lceil{4p}\rceil+\lceil{8p}\rceil=4\]
because $p>0$ all terms on the LHS will be at least 1 implying that the LHS is at least 3.  So we need for $\lceil{4p}\rceil=1$ and $\lceil{8p}\rceil=2$.  For all real $u$ it is known $u\leq\lceil{u}\rceil<u+1$.  For our first equation we have $4p\leq 1<4p+1$ yielding $0<p\leq \frac{1}{4}$.  Similarly we have $8p\leq 2<8p+1$ which yields $\frac{1}{8}<p\leq\frac{1}{4}$.  This last range is also the intersection of the two and thus gives all possible values for $p$.  After substitution we arrive at our solution set to the original problem: $133\frac{3}{4}\leq x <133\frac{7}{8}$


Find all solutions in integers to the system of equations \begin{eqnarray*}a^{3}+b^{3}+c^{3}&=&3 \\ a+b+c&=&3 \end{eqnarray*} edited to fix tex.


limit
trigonometry
inequalities
calculus
calculus computations

Find $\lim_{x \rightarrow 0}\frac{sinx-x}{x^{3}-2x^{2}}$ without using L'Hospital rule and without expanding series.

Trivial by the two things that are forbidden. Forbidding the usage of any theorems is useles and not mathematics, especially if one proof is archieved by directly mimicing the proof of L'Hospitals rule.
Any good reason for not using them¿

[quote="mayhem"]Find $\lim ...$ without using L'Hospital rule and without expanding series.[/quote]
There is no reason to apply L'Hospital rule and I suppose that Tailor formula isn't expanding serie. It is evident that it's applying is the best. So I am agree with [b]ZetaX[/b] in this part.

Let us consider $L=\lim_{x \rightarrow 0}\frac{sinx-x}{x^{3}-2x^{n}}$
Tailor formula gives: $\sin x-x=\frac{x^{3}}{6}+o(x^{4})$
and so $L=\lim_{x \rightarrow 0}\frac{\frac{1}{6}+o(x)}{1-2x^{n-3}}$
Now we have
$L=\frac{1}{6}$ for $n>3$
$L=-\frac{1}{6}$ for $n=3$
$L=0$ for $n<3$ (your case)

Perhaps a geometric argument would qualify?

It's enough to consider $0<x<\pi/2$. Mark the point $A(\cos x,\sin x)$ on the unit circle. Let $B$ be its projection onto the $x$-axis. Draw the tangent at $A$ and let $C$ be the point of its intersection with the $x$-axis. Clearly $|AB|=\sin x$ and $|AC|=\tan x$. Recalling that projection onto a convex region does not increase the length of any curve, we conclude that $\sin x\le x\le \tan x$. (There may be an easier way to prove this inequality, but this is what came to mind right now).

The rest is easy: $0\le x-\sin x\le \tan x-\sin x=\sin x \frac{1-\cos x}{\cos x}=\sin x \frac{1-\cos^{2}x}{\cos x(1+\cos x)}=\frac{\sin^{3}x}{\cos x(1+\cos x)}\le \frac{x^{3}}{\cos x}$.
This squeezes $x-\sin x$ hard enough to yield $L=0$.

[quote="mlok"]Perhaps a geometric argument would qualify? [/quote]
Yes, I think.

By the way I made mistake in sign.  :D
$\sin x-x=-\frac{x^{3}}{6}+o(x^{4})$
which produces corresponding change of the sign in the cases $n>3$ and  $n=3$


function
integration
limit
induction
real analysis
real analysis solved

$\{f_n(x)\}$ are defined on $[0,1]$. $f_0(x)$ is Riemann-integrable, and $f_0(x) > 0$. The sequence of functions satisfies the following recurrence:
\[f_n(x) = \sqrt{\int_0^x f_{n-1}(x) dx}\]
Find $\lim\limits_{n\to\infty} f_n(x)$.

That's what I found in all the cases I've tried. It seems amazing, doesn't it?

I can prove it. We just need to show that for arbitrary $a>0$ and sufficiently large $n$ we have
\[(1+a)\frac{x}{2}\geq f_n(x)\geq \frac{x-a}{2}.\]
Indeed, consider $f_1(x)$. It is increasing positive function. Let $M=\max f_1(x)=f_1(1)$. One can easy prove by induction that for all $x\in[0,1]$:
\[f_{n+1}(x)\leq M^{1/2^n}\alpha_n\cdot x^{\frac{2^n-1}{2^n}},\]
where $\alpha_n\to\frac{1}{2}$. So upper estimate is true.
For second estimate take $m=\min_{x\geq\delta} f_1(x)=f_1(\delta)>0$, then for all $x\in[\delta,1]$:
\[f_{n+1}(x)\geq m^{1/2^n}\alpha_n\cdot (x-\delta)^{\frac{2^n-1}{2^n}}.\]
And from here we obtain lower estimate.

[quote="Myth"]Indeed, consider $f_1(x)$. It is increasing positive function. Let $M=\max f_1(x)=f_1(1)$. One can easy prove by induction that for all $x\in[0,1]$:
\[f_{n+1}(x)\leq M^{1/2^n}\alpha_n\cdot x^{\frac{2^n-1}{2^n}},\]
where $\alpha_n\to\frac{1}{2}$. So upper estimate is true.
[/quote]
The difficulty for me is to prove that $\alpha_n\to\frac{1}{2}$  :(

Ok!
Let's reconstruct $\alpha_n$. What we have?
Denote $u_k=\frac{2^{k-1}}{2^k-1}$, then
\[\alpha_n=u_n^{1/2}u_{n-1}^{1/2^2}...u_1^{1/2^n}\]
(am I right?)
My statement follows from the fact that $u_k\to \frac{1}{2}$ if $k\to\infty$. Indeed, it easy to prove that $\alpha_n$ is decreasing bounded sequence, so it has limit, and
\[\alpha_n=\sqrt{\alpha_{n-1}}\cdot\sqrt{u_n},\]
therefore $a=\sqrt{a}\cdot\sqrt{\frac{1}{2}}$, where $a=\lim_{n\to\infty}a_n$. Thus $a=\frac{1}{2}$.

Assuming that the sequence converges and the limit exists, why not do it this way:

let [tex]f(x) = \lim\limits_{n\to\infty} f_n(x)[/tex]

so [tex]f(x)^2 = \int_0^x f(x) dx[/tex]

and after differentiating both sides [tex]2*f(x)*\frac{d f(x)}{dx} = f(x)[/tex]

this leads to [tex]\frac{d f(x)}{dx} = 1/2[/tex] or [tex]0[/tex], therefore [tex]f(x) = \frac{x}{2}[/tex], since f(x) cannot converge to 0.

And what? It gives nothing, but only possible answer. I am sure that Moubinool found it in the same way. Main part of the problem is convergence itself.

Ok now we know that $f_n$ converge simply to$x/2$.

Do we have uniform convergence on $[0,1]$ ?

Hmm... Moubinool is kidding  :lol:  :lol:  :lol:
Uniform convergence is a direct consequence of my proof, isn't it?

:huh: looks like the problem i had to do in my talk at the 3rd of BWM last year...

And what?  :?
Do you have another proof?

For uniform convergence we need to prove that

$||f_n - f||_{\infty}= \sup_{x\in [0,1]}|f_n(x)- x/2|$ tends to $0$

Moubinool!
[quote="Myth"]Let $M=\max f_1(x)=f_1(1)$. One can easy prove by induction that for all $x\in[0,1]$:
\[f_{n+1}(x)\leq M^{1/2^n}\alpha_n\cdot x^{\frac{2^n-1}{2^n}},\]
where $\alpha_n\to\frac{1}{2}$.[/quote]
[quote]$m=\min_{x\geq\delta} f_1(x)=f_1(\delta)>0$, then for all $x\in[\delta,1]$:
\[f_{n+1}(x)\geq m^{1/2^n}\alpha_n\cdot (x-\delta)^{\frac{2^n-1}{2^n}}.\][/quote]
Fix $\epsilon>0$. Take $\delta=\epsilon/2$ then
\[\sup |f_{n+1}(x)-x/2|\leq \sup \left|M^{1/2^n}\alpha_n\cdot x^{\frac{2^n-1}{2^n}}-m^{1/2^n}\alpha_n\cdot (x-\delta)\right|< \epsilon\]
for all sufficiently large $n$.


LaTeX
trigonometry
induction
algebra unsolved
algebra

yay i learned latex! its so much easier now..

Let  $x_1, x_2, ..., x_n$  be  $n$  positive real numbers such that

$x_1=\frac{1}{x_1}+x_2=\frac{1}{x_2}+x_3=...=\frac{1}{x_{n-1}}+x_n=\frac{1}{x_n}$.

Prove that $x_1=2 \cos \left( \frac{\pi}{n+2} \right)$.

i finding this very hard and pleas if you could help me i would be very glad. note that this problem work in radians not in degrees

[hide="hint"]notice that since the equation is read towards the left hand we have that changing $x_i$ with $\frac{1}{x_{n+1-i}}$ the equation is still true
$x_1=\frac{1}{x_n}$ or
$\frac{1}{x_1}+x_2=x_2+x_n=\frac{1}{x_{n-1}}+x_n\to x_2=\frac{1}{x_{n-1}}$
By induction $x_v=x_{n+1-v}$
Let's take the median term(s) $x_m$ and find their value using that fact...maybe use complex numbers... :D [/hide]

i be sorry but i still can not do it so if you would show me the solusion i would be very glad. thank you very much, in advance


$a,b,c$ are positive numbers such as ${a b c = 1}$

Proove that
${\frac{1}{c+\frac{1}{a+\frac{1}{b}}}+\frac{1}{a+\frac{1}{b+\frac{1}{c}}}+\frac{1}{b+\frac{1}{c+\frac{1}{a}}}\geq 2}$

without "brute force"

[quote="yagaron"]$a,b,c$ are positive numbers such as ${a b c = 1}$

Proove that
${\frac{1}{c+\frac{1}{a+\frac{1}{b}}}+\frac{1}{a+\frac{1}{b+\frac{1}{c}}}+\frac{1}{b+\frac{1}{c+\frac{1}{a}}}\geq 2}$
[/quote]
My ugly proof: ${\frac{1}{c+\frac{1}{a+\frac{1}{b}}}+\frac{1}{a+\frac{1}{b+\frac{1}{c}}}+\frac{1}{b+\frac{1}{c+\frac{1}{a}}}\geq 2}\Leftrightarrow\sum_{cyc}\frac{ab+1}{b+c+1}\geq2.$
Let $a=\frac{x}{y},$ $b=\frac{y}{z}$ and $c=\frac{z}{x},$ where $x,$ $y$ and $z$ are positive numbers.
Hence, $\sum_{cyc}\frac{ab+1}{b+c+1}=\sum_{cyc}\frac{x^{2}+xz}{z^{2}+xy+xz}=\sum_{cyc}\frac{x}{z+\frac{xy}{x+z}}=\sum_{cyc}\frac{x^{2}}{xz+\frac{x^{2}y}{x+z}}\geq$
$\geq\frac{(x+y+z)^{2}}{xy+xz+yz+\frac{x^{2}y}{x+z}+\frac{y^{2}z}{x+y}+\frac{z^{2}x}{y+z}}.$
Id est, it remains to prove that ( my old inequality  :D  )
 $x^{2}+y^{2}+z^{2}\geq2\left(\frac{x^{2}y}{x+z}+\frac{y^{2}z}{x+y}+\frac{z^{2}x}{y+z}\right),$
which equivalent to the following inequality: $\sum_{sym}(x^{4}y-x^{3}y^{2})\geq0.$ :)


[hide]TI-89 gives an approximation of $\boxed{x \approx .511312}$ but I don't know how to solve it long hand really.[/hide]

Well, obviously RRT won't work, seeing as there won't be any rational roots.
Hmm... :|


trigonometry
real analysis
real analysis unsolved

Determine whether $ \Sigma _{n\equal{}1} ^\infty \frac{\tan n}{2^n}$ is convergent or not.

You can use the command \sum instead of \Sigma to get the nice looking $ \sum$.


linear algebra
matrix
induction
vector
Gauss
algorithm
superior algebra

This problem came up in another forum, and I can't make any progress on it.

Show that the natural quotient map from $ \textsf{SL}_n(\mathbb{Z})$ to $ \textsf{SL}_n(\mathbb{Z}_m)$ is surjective.

Just to be clear about notation, $ \textsf{SL}_n(R)$ is the group of all n×n matrices over the (commutative, unital) ring R that have determinant 1, and $ \mathbb{Z}_m$ is the quotient ring $ \mathbb{Z}/m\mathbb{Z}$.

Even in the very simplest possible case, where n=2, m is prime, and the matrix to be lifted is diagonal, this result is less than obvious. For example, the natural lifting of the matrix $ \begin{bmatrix}4&0\\0&2\end{bmatrix}\in\textsf{SL}_2(\mathbb{Z}_7)$ to a 2×2 matrix over $ \mathbb{Z}$ has determinant 8. But there is a lifting of this matrix to $ \textsf{SL}_2(\mathbb{Z}),$ namely $ \begin{bmatrix}25&7\\7&2\end{bmatrix}.$

As far as I can tell by looking at examples, the claimed result is true for all positive integers m and n. But I have no idea how to prove it.

[quote="Opalg"]Show that the natural quotient map from $ \textsf{SL}_n(\mathbb{Z})$ to $ \textsf{SL}_n(\mathbb{Z}_m)$ is surjective.[/quote][hide="Hint"]Use induction on n. Given some matrix $ M\in\textsf{SL}_n(\mathbb{Z}_m)$ lift the first column of $ M$ to a vector $ x_0\in\mathbb Z^n$ whose components generate the unit ideal in $ \mathbb Z$. Then $ \mathbb Z^n/\mathbb Zx_0$ is torsion free, hence free of rank $ n\minus{}1$. This implies that there is a matrix $ M_0\in \textsf{SL}_n(\mathbb{Z})$ such that $ M_0e_1\equal{}x_0\equiv Me_1 \mod m$. The first column of the matrix $ N\equal{}M_0^{\minus{}1}M \in \textsf{SL}_n(\mathbb{Z}_m)$ is the first unit vector. Now build appropriate blocks in $ N$ and lift them. ...[/hide]

This can (even in a more general setting) be reduced to $ A$ diagonal and then $ n \equal{} 2$. Then, we have $ \begin{pmatrix}a & 0 \\
0 & a^{ \minus{} 1}\end{pmatrix} \equal{} \begin{pmatrix}1 & a \\
0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 \\
\minus{} a^{ \minus{} 1} & 1\end{pmatrix}\begin{pmatrix}1 & a \\
0 & 1\end{pmatrix}\begin{pmatrix}1 & \minus{} 1 \\
0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 \\
1 & 1\end{pmatrix}\begin{pmatrix}1 & \minus{} 1 \\
0 & 1\end{pmatrix}$.

[quote="myan"]This can (even in a more general setting) be reduced to $ A$ diagonal and then $ n \equal{} 2$.[/quote]
how does the reduction work?

We want to show that $ SL_n(R) \equal{} E_n(R)$ ($ E_n$: generated by elementary matrices) for a ring $ R$ with finitely many maximal ideals.

Reduce to
1. upper triangular
2. diagonal
3. $ n \equal{} 2$

For the first steps use something like the Gauss algorithm together with prime avoidance (here we use that $ R$ has finitely many maximal ideals) in the refinement given e.g. in Exercise 3.19b of Eisenbud, Commutative Algebra.

There is also a generalisation of the form "some condition on $ R,n$ $ \implies$ $ SL_m(R) \equal{} E_m(R)$ for $ m \geq n$" (which is satisfied for $ n \equal{} 2$, $ R$ finitely many maximal ideals), but the proof is quite lengthy.


geometry
3D geometry
pyramid
Support
calculus
trigonometry
quadratics

You're trying to build a bigger pyramid than your father, Aphluhent III, but you're on a tight budget.  The pyramid will cost you $20 per cubic meter of interior.  However, each of the sides will cost you $10 per square meter to finish.  The greatest incline your architects can support is 45 degrees.  If your budget is $10000, what is the tallest pyramid you can build?

(The pyramid has a square base and the surface area of the base is not included in the cost.)

Is this required to be a square-based pyramid?

Yes, though it should be solvable without calculus.

I believe the greatest height possible is approximately 8.774 feet. All you really need is right triangle trig and algebra.

The volume of the interior of a 45-degree angled square pyramid is
[hide] $\frac{4x^{3}}{3}$ - that was easy (x is the height, also half the length of the base) [/hide]
and the surface area (not counting the base) is
[hide] $8x^{2}$ - 8 45-45-90 triangles with side length $x\sqrt2$[/hide]
so our price formula is
[hide] $\frac{80x^{3}}{3}+80x^{2}=(80x^{2})(\frac{x}{3}+1)$[/hide]
our max price is 10000, so
[hide] $(80x^{2})(\frac{x}{3}+1)=10000$
$(x^{2})(\frac{x}{3}+1)=125$
$(x^{2})(x+3)=375$
$x^{3}+3x^{2}-375=0$
$x=\frac{\sqrt{1509}-3}{2}=17.923$ $feet$[/hide]

EDIT: I screwed that up - quadratic in the wrong place
I dont feel like solving the cubic


Hi guys...
I "stumbled" upon during one of my school's math team meetings.

The problems states something along these lines:

9*9=81
9+9=18

3*24=72
3+24=27
Find another pair where their product is the reverse of their sum.

My mom said that the digits in the product/sum should add up to 9, but even then, I wasn't able to find another pair.

Help...please?

 :huh:

Guess and check? The algebra seems far too tedious...

x + y = 10a + b
xy = 10b + a

[hide="One answer"]
Zero and zero =D .
[/hide]

I don't know if 0 and 0 would count.  I think the problem really wants another solution like 9 and 9.  Besides, does reversing a one digit number work.  That would have to be dancingblob's say because s/he posted the problem.

No, I don't think two one-digit numbers would work.
And I'm a she.  :D

This is proving to be quite a problem.

[quote="JEngel"]What about 2 and 2? 4 is the reverse of 4.[/quote]
I think that dancingblob said that two one-digit numbers don't work.  ;)

[quote="joseluis"]other pair of number are

[hide]2 and 47[/hide][/quote]
Nice!  :D How did you find it? I can't seem to find any... :(

$xy + x + y = 11(a+b)$
$(x+1)(y+1) = 11(a+b) + 1$

$xy - x - y = 9(a-b)$
$(x-1)(y-1) = 9(a-b) + 1$

This is about as simple as I can get it.  The two cases you propose have $a + b = 9$, which (and I have checked) has no other solutions.

I've been picking different values of $a+b$ and $a-b$ and have still not found another two-digit solution.    :huh:

Edit:  Joselius has found one!  Plugging in $a-b = 5$ gives $(x-1)(y-1) = 46$.  Choosing $x = 3, y = 24$ gives a solution already given, but choosing $x = 2, y = 47$ produces $x+y = 49, xy = 94$  :o  (I admit I had not bothered to check this one  :| )

Alternately, from the top equation, $a+b = 13$ gives $(x+1)(y+1) = 144$ (which, again, I had checked and not found solutions for  :| ), and here again, choosing $x = 2, y = 47$ produces a valid solution :o

really the number came to me keep traing but i think that there is not other number with that characteristic in thet the product of the number is less tha 100.

you guys never cease to amaze me.
thanks for your help.
 :lol:


trigonometry
inequalities
inequalities proposed

Prove if a,b,c>0 and ab+bc+ca=3 then
$ \sum\sqrt{a^2\plus{}3} >\equal{}a\plus{}b\plus{}c\plus{}3$

[quote="tranquoc"]Prove if $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ then
$ \sum\sqrt {a^2 \plus{} 3} \geq a \plus{} b \plus{} c \plus{} 3$[/quote]
The idea $ a \equal{} \sqrt3\tan\frac {\alpha}{2},$ $ b \equal{} \sqrt3\tan\frac {\beta}{2}$ and $ c \equal{} \sqrt3\tan\frac {\gamma}{2}$ works here.

Here is my proof:
Let $ a \equal{} \sqrt3\tan\frac {\alpha}{2},$ $ b \equal{} \sqrt3\tan\frac {\beta}{2}$ and $ c \equal{} \sqrt3\tan\frac {\gamma}{2},$ where $ \{\alpha,\beta,\gamma\}\subset[1,\pi).$
Then $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$ and our inequality equivalent to $ \sum_{cyc}\left(\frac{1}{\cos\frac{\alpha}{2}}\minus{}\tan\frac{\alpha}{2}\right)\geq\sqrt3.$
Let $ f(x)\equal{}\frac{1}{\cos\frac{x}{2}}\minus{}\tan\frac{x}{2}.$ Then $ f'(x)\equal{}\frac{\sin\frac{x}{2}}{2\cos^2\frac{x}{2}}\minus{}\frac{1}{2\cos^2\frac{x}{2}}\equal{}\minus{}\frac{1}{2(1\plus{}\sin\frac{x}{2})}.$
$ f''(x)\equal{}\frac{\cos\frac{x}{2}}{4(1\plus{}\sin\frac{x}{2})^2}\geq0.$
Thus, $ \sum_{cyc}f(\alpha)\geq3f\left(\frac{\alpha\plus{}\beta\plus{}\gamma}{3}\right)\equal{}\sqrt3.$
I don't see another way. :(

[quote="arqady"][quote="tranquoc"]$ a^2 \plus{} 3 \equal{} (a \plus{} b)(a \plus{} c)$......->  done[/quote]
What do you mean? Why it helps? Explain pleas. :wink:  Thank you![/quote]

It helps because it reduces the problem to http://www.mathlinks.ro/viewtopic.php?t=133953 .

  darij

Prove if $a,b,c>0$ and $a+b+c\leq abc$ . then
$$\sqrt{a^2\plus{}1} +\sqrt{b^2\plus{}1} +\sqrt{c^2\plus{}1} +3\leq ab+bc+ca$$


Does anyone here use books from the REA series for math? I have REA geometry and pre-calculus and i wanted to know if they were a good preparation for any AMC's(AMC 12, AIME, etc.). the problems seem easier than the ones in AOPS volume 1, at least in geometry, yet volume 1 geometry is challenging for me. are the REA books a good prep for volume 1 geometry? as for algebra, volume 1 is okay for me, not much of a challenge, but its not fantastically simple either. however, volume 2 algebra is way beyond my level. what books should i use for algebra in that case?

i have the geometry and algebra/trigonometry books.  i guess it depends what your goals are, I used the algebra/trigonometry book to prepare for class, and i think it might have helped for contests too.  overall i thought it was really good.  i never used the geometry book though so i cant say much about it.  I think i flipped through it and it looked pretty easy.

the REA algebra/trig book might help for AMC, but the best thing is just to stick with AOPS.  if volume 1 is too hard, maybe you can use the REA books to reinforce basic ideas and then go back to aops.


the number of permutations of letters of the word PARALLEL taken four at a time are ......................

pl suggest a shorter way

1 .A music academy has to send 10 members to 5 shows ,two in each show. two of the shows are in city and the 3 are outside .two of the five members prefer to work in city while 3 prefer to work outside city..the number of ways in which five shows can be held if preference are to be satisfied .
options are 5200, 5400,5472,5250

Q2 There are two students A and B having a and b numbers of books respectively.the number of ways in which they can exchange their their books so that after exchange they have same number of books with them but not he same books is .............

options are a+b^(C)a ,(a+b^C a)-1,a+b^(P)a,( a+b^P a)-1

[quote="ayush_2008"]
1 .A music academy has to send 10 members to 5 shows ,two in each show. two of the shows are in city and the 3 are outside .two of the five members prefer to work in city while 3 prefer to work outside city..the number of ways in which five shows can be held if preference are to be satisfied .
options are 5200, 5400,5472,5250
[/quote]

Since 2 members of internal show are fixed, the other 2 can be selected from 5 in $ {5\choose 2}$ ways. This automaticaly fixes the selection of the external show members. Now, the internal show guys have to be distributed into 2 distinct groups of 2 each. This is done in $ {4\choose 2} \cdot {2\choose 2}$ ways. The external show guys have to be distributed into 3 distinct groups of 2 each. This is done in $ {6\choose 2} \cdot {4\choose 2} \cdot {2\choose 2}\equal{}\frac {6!}{2!2!2!}$ ways.

So, total=$ {5\choose 2} \cdot {4\choose 2}\cdot {2\choose 2} \cdot \frac {6!}{2!2!2!}\equal{}5400$ ways.


inequalities
geometry
circumcircle
incenter
geometry proposed

[color=purple] Let $ O$and $ I$  are circumcenter and incenter of triangle $ ABC$ which has area $ S$ . Prove that
                 $ 2\sqrt {2}OI^2\ge S[|\frac {b \minus{} c}{a}| \plus{} |\frac {c \minus{} a}{b} \plus{} |\frac {a \minus{} b}{c}|]$
 :) [/color]

[color=purple] I have a mistake when I created the problem in previous post :|  but any one can show me the truly of it ?
  The following is true : $ 2OI^2\ge S[|\frac {b \minus{} c}{a}\plus{}\frac {c \minus{} a}{b}\plus{}\frac {a \minus{} b}{c}|]$
 :)  :) [/color]


By Cauchy Schwarz Inequality:
$\left(\sum a^2b\right)\left(\sum ab^2\right)\ge\left(\sum abc\right)^2$,
$\left(\sum a^2b\right)\left(\sum acd\right)\ge abcd\left(\sum a\right)^2$,
$\left(\sum ab^2\right)\left(\sum bcd\right)\ge abcd\left(\sum b\right)^2$.


Determine the sum of all the positive integers from $1$ to $2006$, inclusive, that do not contain the digit $7$.

[hide="flawed first attempt"]
First consider the sum of integers from $1$ to $999$ without any $7$s.

The sum of units digits $0$ to $9$ without $7$ is $\frac{9(9+1)}2-7=38$
And this sum of units digits appears 100 times.

The sum of tens digits $00$ to $90$ without $70$ is $38\cdot10=380$
And this sum of tens digits appears 10 times.

The sum of hundreds digits $000$ to $900$ without $700$ is $38\cdot100=3800$
And this sum of hundreds digits appears once.

The sum from $1000$ to $1999$ is the above sum plus $1000\cdot1000$

Therefore the sum is
$2\cdot(3\cdot3800)+1000\cdot1000+6\cdot(\frac{2001+2006}2)=\boxed{1,034,821}$[/hide]

[hide="Correct Solution"]
First consider the sum of integers from $1$ to $999$ without any $7$s.

The sum of units digits $0$ to $9$ without $7$ is $\frac{9(9+1)}2-7=38$
The sum of tens digits $00$ to $90$ without $70$ is $38\cdot10=380$
The sum of hundreds digits $000$ to $900$ without $700$ is $38\cdot100=3800$
So the total sum is $3800+380+38=4218$ and this set of sums appears $\frac{1000}{10}=100$ times from $1$ to $999$.

Now let's consider the sum from $1000$ to $1999$.

By PIE, the # of numbers with a $7$ in them from $1000$ to $1999$ is equal to $100+100+100-10-10-10+1=271$. So the # of numbers that don't have $7$ is $1000-271=729$. So the sum is equal to the sum from $1$ to $999$ plus $729\cdot1000$

Therefore, the total sum is:

$2(100\cdot4218)+729\cdot1000+6\left(\frac{2001+2006}2\right)=\boxed{1,584,621}$[/hide]

Whoops, I just realized my solution was totally flawed!  :blush:

[quote="surge"]What is PIE?[/quote]

PIE stands for Principle of Inclusion-Exclusion.

My reasoning for the problem is follows, if you can understand it...I'm really bad at explaining things (sorry :blush:)

[hide]We'll take care of the numbers 0-9 first. They sum up to 38 without 7. Now, we take care of the two digit numbers. If we break them up into tens and units (such as 17 = 10 + 7), for each tens digit x, we have the sum of numbers 10x + 0 through 10x + 9 as 9(10x) + 38. If we add them all up for x's 0 - 9 excluding 7, we have (9)(10 + 20 + 30 + ... + 90 - 70) + (9)(38) = (9)(418) = 3762, so the sum of the numbers between 0-99 without a 7 is 3762.

Now, we take care of the three digit numbers. If we split them up as a hundreds digit and a two digit part, we can call the hundreds digit x. Then the sum of numbers 100x + 0 through 100x + 99 is (81)(100x) + 3762 where we use 81 because there are (9 numbers w/o 7 in units digit)(9 numbers w/o 7 in tens digit) = 81 numbers between (0-99) w/o 7. For all numbers 0-999 ignoring the 700s, we have (81)(3800) + (9)(3762) = 341658.

Finally, we work on the thousands digit. We split the four digit numbers up into thousands digit, x, and a three digit number. Once again, the sum of all the thousands digits together for the numbers 1000x to 1000x + 999 is (9)(9)(9)(1000x) = (729)(1000x), with 729 numbers total since there are (9 # w/o 7 in units digit)(9 # w/o 7 in tens digit)(9 # w/o 7 in hundreds digit) = 729 numbers for every thousand without a 7. To add all the numbers for 0-999 and 1000-1999 we then have to do (729)(1000) + (2)(341658) = 1412316. Then, just simply add 2000 through 2006 = 14021 and the final answer is:

$\boxed{1426337}$[/hide]


Another one but easier
Let $ a,b,c,d$ are real numbers such that $ a+b+c+d=4$. Find the best constant $ k$ such that
$ \sqrt{a+b+k(a+b-c-d)^{2}}+\sqrt{b+c+k(b+c-d-a)^{2}}+\sqrt{c+d+k(c+d-a-b)^{2}}+\sqrt{d+a+k(d+a-b-c)^{2}}\ge 4\sqrt{2}$

Another one and may be the hint for the second problem
Let $ a,b,c,d$ are real numbers such that $ a\plus{}b\plus{}c\plus{}d \equal{} 4$. Find the best constant $ k$ such that
$ \sqrt{a\plus{}k(a\minus{}b\minus{}c\minus{}d)^{2}}\plus{}\sqrt{b\plus{}k(b\minus{}c\minus{}d\minus{}a)^{2}}\plus{}\sqrt{c\plus{}k(c\minus{}d\minus{}a\minus{}b)^{2}}\plus{}\sqrt{d\plus{}k(d\minus{}a\minus{}b\minus{}c)^{2}}\ge 3\sqrt2$

The last hint
Let $ x_{1},x_{2},...,x_{n}$ are real numbers such that $ \sum_{i \equal{} 1}^{n}x_{i}\equal{} m$. Find the best constant $ k$ such that
$ \sum_{i \equal{} 1}^{n}\sqrt{x_{i}\plus{}k\left(\sum_{j\neq i}^{n}x_{j}\minus{}x_{i}\right)^{2}}\ge (n\plus{}2)\sqrt{\frac{m}{8}}$


geometry
3D geometry
logarithms
algebra unsolved
algebra

For all positive integer $n$ prove the equality
$[\sqrt{n}]+[\sqrt[3]{n}]+...+[\sqrt[n]{n}]=[log_2n]+[log_3n]+...+[log_nn]$

LHS counts the number os squares, cubes, ..., $n$-th powers $\leq n$.
RHS counts the number of powers of $2$, $3$, ..., $n$ $\leq n$.

They are clearly the same.

huh..clever...
But then again, aren't the first powers counted on the right as well?  For example, doesn't $\log_{[2]}{n}$ count 2 as well?

Yes, I forgot some of them to count...

It's important to note that some powers are counted more than one time, e.g. $64$ is counted as $2^6, 4^3 , 8^2$ and $64^1$.

Now the full and correct proof:
On the LHS, also $1$ is counted everytime, thus $n-1$ times in total. On the RHS, also the first powers $\neq 1$ are counted, also $n-1$ of them. Thus we can now neglect $1$ and all first powers.
A number of type $s^t$ (by the above $s,t>1$) with $s$ being no power of degree $>1$ is counted as $\frac{t}{d}$-th power of $s^d$ for any divisor $d \neq t$ of $t$ on the LHS and as power of $s^d$ for all such $d \neq 1$ on the RHS. Thus both count the same.
By "changing" the counted $1$es into counted first powers, the argument posted before ("#(squares, cubes,...) = #(powers of 2,3,...)") already works and gives a similar proof.


Let $ C_1$ be a circle centered at $ O_1$. A circle $ C_2$, centered at $ O_2$, passing through $ O_1$ and meets $ C_1$ at $ A$ and $ B$. Let $ l$ be a line passing through $ O_1$. Points $ P$ and $ Q$ are on $ l$ such that $ AP$ and $ BQ$ are perpendicular to $ l$.
Prove that if $ R$ is the midpoint of $ AB$, then $ PQR$ is an isoceles triangle.

Because $ O_1O_2 \perp AB$ and $ R$ is the intersection of $ O_1O_2$ and $ AB$, so $ O_1, R, A, P$ and $ O_1, R, B, Q$ are concyclic. So:

$ \angle{RPQ}\equal{}\angle{RAO_1}\equal{}\angle{RBO_1}\equal{}\angle{RQP}$

Hence $ \triangle{PQR}$ is an isosceles triangle.

PQ is the projection of AB onto l, therefore the middle of AB is equally apart of P and Q (the projection of R will be the middle of PQ).

Best regards,
sunken rock


factorial
number theory
prime factorization

How many perfect squares are divisors of the product (1!)(2!)(3!)(4!).....(10!)?

It is well known that the exponent of a prime $ p$ in the prime factorization of $ n!$ equals $ \sum_{k=1}^{\infty} \left[{\frac{n}{p^k}\right]}$, where $ [x]$ denotes the integer part of $ x$.

Using this formula we can get the exponent of the primes $ 2$, $ 3$, $ 5$, and $ 7$ (they are the only ones to appear) in each of $ 1!, 2!, ..., 10!$, and then we add them up to get the prime factorization of $ P=(1!)(2!)...(10!)$.

Indeed, $ P=2^{38} \cdot 3^{17} \cdot 5^{7} \cdot 7^{4}$.

The perfect squares divisors of $ P$ are of the form $ 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta}$, with $ \alpha, \beta, \gamma, \delta$ even nonnegative integers less or equal than $ 38, 17, 7$ and $ 4$ respectively.

We have $ 20 \times 9 \times 4 \times 3 = 2160$ ways to choose $ \alpha, \beta, \gamma, \delta$, so there are $ 2160$ divisors of $ P$ which are perfect squares.

@uglysolutions, your answer can't be correct.

10 factorial= 3628800 and the highest square below or equal to this is 1904^2=3625216

This means there are AT MOST 1904 perfect squares which divide 10 factorial, so your answer is clearly wrong.

[b]EDIT: Ignore this post, I misread the question.[/b]

He's not counting perfect squares dividing 10!, he's counting perfect squares dividing the (substantially larger) number the question asked about ;)

Oops, misread the question. Oh well, lesson learnt.


function
analytic geometry
calculus
calculus computations

Sketch the level curves in the xy-plane of f(x,y)=3-y-x^2 for k=0,2,4.

Anyone mind helping me how to do these type of problems? Thank you. :)

A level curve is formed by all points with coordinates $ (x,y)$ such that $ f(x,y)=k$. Write this equation as $ y=3-k-x^{2}$: do you recognize the curve?


abstract algebra
superior algebra
superior algebra unsolved

Assume that $ L$ is a finite extension of a field $ K$. Let $ E_{1},\cdots,E_{n}$ (resp. $ F_{1},\cdots,F_{n}$) be linearly disjoint field extensions of $ K$ (resp. $ L$). Let $ \sigma_{i}: E_{i}\to F_{i}$ be isomorphism which coincide on $ K$ and for which $ \sigma_{i}(K)=L$ for every $ i=1,\cdots,n$. Then
\[ \sigma_{1}\otimes\cdots\otimes \sigma_{n}: E_{1}\otimes_{K}\cdots\otimes_{K}E_{n}\to F_{1}\otimes_{L}\cdots\otimes_{L}F_{n}\]
is a well defined isomorphism.

$ \sigma_{i}(K) \subseteq L$ implies that the map $ (x_{1},...,x_{n}) \mapsto \sigma_{1}(x_{1}) \otimes ... \otimes \sigma_{n}(x_{n})$ is $ K$-linear in every variable, so it extends to a unique $ K$-linear map of the tensor products. since the $ \sigma_{i}$ are multiplicative, their tensor product is also multiplicative. the $ E_{1},...,E_{n}$ are linearly disjoint, i.e. their tensor product is a  field. thus the map is injective. since the $ \sigma_{i}$ are surjective, it can be easily seen, that the map is also surjective.


Sign up for Mewto55555 and Ernie's co-hosted [size=150][b]64 person[/b][/size] tournament. Will be double elimination. Please include your highest rating when signing up!

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nikeballa313
highest: 1325

ill get killed, why not...=]

bunch of tourneys... :rotfl:

ill join highest-1502

Put "izzy" on your sign-up list, please.  :)

Highest rating -- 1349

good luck with the bracket, mewto and ernie :D

COME ON PPLS JOIN ALREADY!!!!! I'M GOING 2 FORGET ABOUT THIS TOURNEY SOON!! :D

ernie are u cancelling this or not?
(i hope not)

The status of this tourney is undecided. (kinda sounds like the FBI  :ninja:, wh00ps, thats a ninja :D)

BTW, my highest rating is now 1464
no change in seeding, but still

I'm sorry to disappoint everyone, but this tournament is now [b]canceled[/b]

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO! :(
Why?  Just give the top couple of people byes to a later round (or just give randomly selected people byes if you prefer that).  There are lots of other solutions, too.  There's lots of people signed up, and it can still be a really successful tourney.

THIS HAS NOTHING TO DO WITH POEPLE SIGNED UP....

WHAT?!?!??!?!?! NOW U TELL US MEWTO!!?!?!? does ctc have do with this. =]

Oh... :oops_sign: Sorry, then.  What's the problem, though?

But you know what I find stupid?  Mewto can certify other tourneys, but not his own.

I find that kinda stupid, like this guy.

[hide="Stupid Guy"]:yankchain:[/hide]


Find all x,y be a positive numbers such that
$ m^n\minus{}n^m\equal{}3$ :?:
 :huh:

I suppose you meant "find $ m$ and $ n$".

Clearly $ m$ is odd or $ n$ is odd, and $ m \not\equiv n mod 2$. If $ m > 1$ is odd, $ m^n \equiv 1 mod 4$ (every odd square is 1 mod 4) and $ 4|n^m$, thus $ 3 \equal{} m^n \minus{} n^m \equiv 1 mod 4$, contradiction. Note that $ m \equal{} 1$ is obviously absurd.

If $ n$ is odd and $ n > 1$, $ 8|m^n$ and $ n^m \equiv 1 mod 8$ (every odd square is $ 1 mod 8$), thus $ m^n \minus{} n^m \equiv 7 mod 8$, then $ 3 \equiv 7 mod 8$, contradiction. Then $ n \equal{} 1 \Rightarrow m \equal{} 4$.


calculus
derivative
calculus computations

After substitution you should work out the derivatives using the product rule. You will get an overall factor $ e^{\minus{}(bx\plus{}ay)}$ on the left hand side which you can drop.

[hide="Hint"]
\[ a \frac{\partial u}{\partial x} \equal{} a \left( \frac{\partial w}{\partial x} \right) e^{\minus{}(bx\plus{}ay)} \plus{} a w \frac{\partial}{\partial x} e^{\minus{}(bx\plus{}ay)} \equal{} a \left(\frac{\partial w}{\partial x} \minus{} b w  \right) e^{\minus{}(bx\plus{}ay)}
\][/hide]


Outline a general method for making the homologation of nitro alkanes:

$ R\minus{}CH_2NO_2 \longrightarrow R\minus{}NO_2$.

R-CH2-NO2 + H3O+->RCOOH
FOLL BY SCHMIDT REACTION TO GIVE RNH2 FOLL BY OXIDATION.

why not. Hydrolysis of nitro alkanes in presence of acid gives the corr acid. not sure abt the mech but i think it involves tautomerisation to acinitro form. this is applicable to only primary compounds tho

[quote="Carcul"][quote="VARUNARASU"]R-CH2-NO2 + H3O+->RCOOH[/quote]

I don't think so.[/quote]

Is VARUNARASU referring to this:

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

not exactly. i know what nef reaction is but my book says acid catalysed hydrolysis of primary nitro compounds gives acids. there is no base involved. also aldehydes are not formed

[quote="VARUNARASU"]not exactly.[/quote]

Yes you are. The Nef reaction is the name given to the conversion of nitro compounds into aldehydes, ketones, or carboxilic acids. However, the conversion to carboxilic acids with acid requires the use of [i]concentrated mineral acids[/i], which, because often leads to side reactions, is a method seldom used.

Just wondering Carcul. How do you know which ones are often use in synthesis and which ones are seldom/never used? Is it just experience?

It's just experience, but any reaction with a strong mineral acid is a reaction in harsh conditions, and so side reactions are always to be expected.

Yes, Pardesi, are you trying to say something?

no just wondering about ur experience . :)

i don know if this is write....correct me

first reduce it with Lithium Aluminium Hydride to get the primary amine. Treat with Caro's acid,it undergoes a rearrangement  called LOSSEN REARRANGEMENT and we get the primary amine with 1 carbon less. Oxidise with CF3COOOH-triflouro peracetic acid

The Lossen rearrangement is the rearrangement of hydroxamic esthers, which are prepared from hydroxamic acids. So, how does a hydroxamic acid/esther forms from a primary amine and Caro's acid? Also, the treatment of aliphatic primary amines with Caro's acid yields nitroso compounds or oximes.

[quote="valeriummaximum"]sorry
another method of mine...........see if it's right
first reduce with Lithium Aluminium Hydride.treat with nitrous acid to get the alcohol.oxidise using Jones reagent to get the respective acid.Heat with ammonia to get the respective amide.
Amide undergoes Hoffmann Bromamide rearrangement to give you the amine.Amine with triflouro peracetic acid gives you the product.

Is this Correct??????? :)[/quote]
but when u add excess NH3 to the acid you mainly get RCOONH4+ AND not RCONH2 so what i think is that to get that RCONH2 we can convert RCOOH to RCOCl and then add NH3 to convert it into RCONH2?? am i right carcul coz this is the correction you made in one of my previous conversions????

i heat with ammonia srinitrs.
heating with ammonia eliminates a mole of water to give you the amide.............. :lol:

[quote="Valeriummaximum"]treat with nitrous acid to get the alcohol[/quote]

Wrong: the treatment of primary aliphatic amines with nitrous acid is not a synthetically usefull reaction, because the diazonium salt is not stable enough to survive in solution, and so it forms a carbocation which then undergoes several reactions: elimination to the alkene, nucleophilic addition to the alcohol and/or alkyl halide, etc.

[quote="Valeriummaximum"]Heat with ammonia to get the respective amide.[/quote]

That's not a good way of making an amide from an acid, because of the reasons pointed out by Srinitrs: the use of the acid chloride is preferred. However, if you want to convert directly the acid into the amide you can use ammonia and DCC.

What is DCC?

Anyway my method:-

[list]1.Nef reaction (to covert it into aldehyde)
2.Tollen's reagent (to convert it into acid)
3.Smidt reaction (Reaction of HN3) to get primary amine
4.CF3COOOH(to oxidise -NH2 --> -NO2[/list]

Yes, that's right. Here's another method:

1) $ R\minus{}CH_2NO_2 \plus{} NaNO_2/DMSO \longrightarrow R\minus{}CO_2H$

2) $ R\minus{}CO_2H \plus{} HN_3/heat \longrightarrow R\minus{}NH_2$

3) $ R\minus{}NH_2 \plus{} oxidation \longrightarrow R\minus{}NO_2$

[quote]
What is DCC?
[/quote]
DCC is dicyclohexyl carbodiimide.


Could someone please post a solution to this problem?

Find $\sum_{n=0}^{\infty}\cot^{-1}(n^{2}+n+1)$, where $\cot^{-1}(t)$ for $t\geq 0$ denotes the number $\theta$ in the interval $0\leq\theta\leq\frac{\pi}{2}$ with $\cot(\theta)=t$.

It is in the "Do you need this" PDF on trig. I currently can do all of it except this one. Any help would be appreciated.

[hide="Solution"]$S=\sum_{n=0}^{\infty}\cot^{-1}\left(n^{2}+n+1\right)=\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{n^{2}+n+1}\right)$

Note that $\frac{1}{n^{2}+n+1}=\frac{\left(n+1\right)-n}{1+n\left(n+1\right)}$.

So you have $S=\sum_{n=0}^{\infty}\tan^{-1}\left(n+1\right)-\sum_{n=0}^{\infty}\tan^{-1}\left(n\right)=\lim_{n\rightarrow\infty}\tan^{-1}\left(n\right)=\boxed{\frac{\pi}{2}}$.[/hide]
What is this "Do you need this?" that you speak of?

[quote="amcavoy"]What is this "Do you need this?" that you speak of?[/quote]

i think it's something like a pretest for the online class of this site...

[quote="mysmartmouth"]I would just like to note that lately, a bunch of people have been posting these problems from the pretest things (a trend started by me!).  That's probably good news (financially) for AoPS, because it means more people are considering taking classes.[/quote]
By the way, where do you find pretests? Or do you have to enroll in the class to get them?

[quote="towersfreak2006"]http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php[/quote]
So... I guess WOOT doesn't have a pretest? Or is it on another page?  :|
Wow, my 200th post!  :)  :P

WOOT! does not have any of those. I believe its Week 0 class (Orientation) is designed to place you into the correct level though.

Note that with WOOT, you can do it, and if you find it too hard, you can drop out and get your money back within x number of days.  The pretests are designed to see if you are ready for the class.  And Neal, I think you are ready for WOOT, despite that drastically low AIME score *sarcasm*. :rotfl:

I can remember seeing this problem in a GRE practice book...


1.Let $ a,b,c \in [3,4].$
Prove that
$ \sum  \sqrt{a^2\plus{}b^2\minus{}c^2}  \leq a\plus{}b\plus{}c$
2.Let $ a,b,c \in [3,4].$
Prove that:
   $ \sqrt[3]{a(a^2\plus{}b^2\minus{}c^2)}\plus{}\sqrt[3]{b(b^2\plus{}c^2\minus{}a^2)}\plus{}\sqrt[3]{c(c^2\plus{}a^2\minus{}b^2)}\leq a\plus{}b\plus{}c$

My idea is let $ a,b,c$ be sides of a triangle  :)

(Since $ a,b,c \in [3,4]$, it is possible)

Then $ \sum \sqrt{a^2\plus{}b^2\minus{}c^2} \equal{} \sum \sqrt{2abcosA} \leq \sqrt{(\sum 2ab)(\sum cosA)}$

But we have $ \sum cosA \leq \frac{3}{2}$, $ 3 \sum ab \leq (a\plus{}b\plus{}c)^2$.

Also, $ \sum \sqrt[3]{a(a^2\plus{}b^2\minus{}c^2)} \equal{} \sum \sqrt[3]{2a^2bcosA} \leq \sqrt[3]{(\sum a)(\sum 2ab)(\sum cosA)}$.

The remaining parts should be trivial  :P


A group of jealous professors are locked in a room. There is nothing else in the room but pencils and one tiny scrap of paper per person. The professors want to determine their average (mean, not median) salary so that each one can gloat or griece over their personal situation compared to their peers. However, they are secretive people and do not want to give away any personal salary information to anyone else. Can they determine the average salary in such a way that no professor can discover [i]any[/i] fact about the salary of anyone but herself? For example, even facts such as "3 people earn more than $\$$40,000" or "no one earns more than $\$$90,000" are not allowed.

Anyone? I decided that they *could* have the pencil represent like $\$$20,000 or some other base amount, then rip up the pieces of paper to represent other increments and toss it all into a pot, but you would need to be able to rip the paper into some pretty small pieces...

Every professor can do the following process for himself or herself.

Prof#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues and every one can know just the mean salary without any further details.

Suppose, there are 5 professors.

Prof#1 writes 100 (dummy value) and passes it to Prof # 2.  Prof # 2 adds his own salary, say 300, and passes just the total value of 400 to Prof#3. Prof # 3 adds his salary (say 500) and lets Prof # 4 know the total as 900. Prof # 4 passes a total of 1600 to Prof # 5. Prof # 5 returns back a total of 2500 to Prof # 1.

Now, prof # 1 can determine the average salary. Say, his actual salary was 350.

Average salary = {(2500 - 100) + 350}/5 = 550. All other professors can also follow a similar process.

[quote]
Prof#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues and every one can know just the mean salary without any further details.
[/quote]

Prof #3 now knows that that value passed to him is the upper limit of what prof#2 makes.  Isn't this one of the "facts" that the problem statement doesn't allow? Moreover, if the group consists of only 2 profs (the problem doesn't specify the group's minimum size), the first one can easily determine what the second prof makes.

I don't think there's a solution. Here's why:

[hide]
Consider a group that consists of only two profs.  Suppose there is some elaborate protocol that allows them to determine their average salary without revealing any personal salary info to the other prof. At the end of this, each will know the average salary. If prof #1 makes the same as the average salary, he's also just determined that prof #2 must make exactly the same amount. If prof #1 makes _more_ than the average salary, he's just determined prof #2 makes _less_ than the average salary. And if prof #1 makes _less_ than the average, then prof #2 must make more. I assume these are the type of "facts" that the problem statement doesn't want you to be able to determine.

If you claim that "group" implies 3 or more people, then consider the case with 3 people. Once everyone knows the average, they can simply compare it to their own salary to conclude some facts about the salaries of the others: If a prof makes the same as the average, then either everyone makes exactly the same or there's one person who makes more and one person who makes less. If a prof makes more than the average, then there must be at least one prof who makes less than the average.
[/hide]

[quote="gubd"]
Prof #3 now knows that that value passed to him is the upper limit of what prof#2 makes.  Isn't this one of the "facts" that the problem statement doesn't allow? Moreover, if the group consists of only 2 profs (the problem doesn't specify the group's minimum size), the first one can easily determine what the second prof makes.

I don't think there's a solution. Here's why:

[hide]
Consider a group that consists of only two profs.  Suppose there is some elaborate protocol that allows them to determine their average salary without revealing any personal salary info to the other prof. At the end of this, each will know the average salary. If prof #1 makes the same as the average salary, he's also just determined that prof #2 must make exactly the same amount. If prof #1 makes _more_ than the average salary, he's just determined prof #2 makes _less_ than the average salary. And if prof #1 makes _less_ than the average, then prof #2 must make more. I assume these are the type of "facts" that the problem statement doesn't want you to be able to determine.

If you claim that "group" implies 3 or more people, then consider the case with 3 people. Once everyone knows the average, they can simply compare it to their own salary to conclude some facts about the salaries of the others: If a prof makes the same as the average, then either everyone makes exactly the same or there's one person who makes more and one person who makes less. If a prof makes more than the average, then there must be at least one prof who makes less than the average.
[/hide][/quote]

First of all, Shamik's procedure does work. If the initial number prof#1 writes down is large enough, the maximum value prof#3 recieves will say almost nothing about prof#2's salary. For example, if prof#1 writes down 1,002, 345 and prof#2's salary is 3,000, then the maximum value that prof#3 gets is 1, 005, 345, and isn't very revealing at all about prof#2's salary.

Also, I agree that such a procedure would be impossible for a group of 2 professors because knowing the mean and their own salary they can figure out the other's salary. However the argument for 3 or more professors is faulty. Knowing the mean and their own salary they can figure out a fact about [b]someone's[/b] salary, but not a fact about the salary of [b]someone in particular[/b].

[quote]
However the argument for 3 or more professors is faulty. Knowing the mean and their own salary they can figure out a fact about [b]someone's salary[/b], but not a fact about the salary of [b]someone in particular[/b].
[/quote]

From the problem statement: [i]even facts such as "3 people earn more than \$40,000" ... are not allowed[/i]. This is a fact about [b]someone's salary[/b], not the salary of [b]someone in particular[/b], as you put it. My interpretation of this is that learning either type of fact is ruled out by the problem statement.

If the problem statement didn't rule out learning such general facts, then I agree, Shamik's solution would work in a group of 3 or more people.

Gubd's right. I guess I didn't read the problem very carefully :blush: .

Halluci gave me an idea though:

[hide]They could rip their scraps into pieces and write on each scrap a portion of their salary in different handwritings. For example, if prof#1 earned 500 dollars, he could rip his scrap into three pieces and write 400, 200, and -100 on different scraps. Then all pieces would be dumped into the middle of the room , and they would then find the mean. They need to write down the portions in different handwritings so that no one could piece the scraps together.[/hide]

This still gives away information after the pofessors know the mean though, as gubd said. Perhaps the problem wasn't worded correctly? :huh:  :huh:

I was thinking, could gubd's problem be solved if professor #1 is randomly determined? Then they can't make any assumptions about how much anyone makes. And making professor #1 randomly determined can be done if they setup a system like in mafia games(shuffling cards).

I don't think I quite understand what you mean. After any professor knows the mean and his own salary, he can compare his salary to the mean and find out a fact about someone else in the group. (For example, if the mean was 500 and prof#1 made 100, he would know that at least on person in the group made more than 500.) Also, I have just noticed that knowing the mean, a professor can also find the total of everyone else. I still think the problem was misworded.

[quote="gubd"]Prof #3 now knows that that value passed to him is the upper limit of what prof#2 makes.  Isn't this one of the "facts" that the problem statement doesn't allow?[/quote]

Seconding laughinghead: this objection is a good objection to the statement of the problem, not to the given solution.  If a professor is to know the mean salary, he will also learn the total salary, and thus an upper bound on all salaries and in particular an upper bound on the salary of any particular professor.  So any solution to the problem is a violation of that restriction.  What this suggests is that the restriction in the problem is flawed and should be re-written.  Under a reasonable re-writing of the problem, the given solution would be correct.

In any case, the original objection about the third professor knowing upper limits is easily removed if the first arbitrary value passed is allowed to be negative.

[quote="Shamik Banerjee"]Every professor can do the following process for himself or herself.

Prof#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues and every one can know just the mean salary without any further details.
[/quote]

There is a problem with that solution. If there are five professors, the fifth one can memorize the last number, he can figure out what the first professor made.

Prof#1 writes any arbitary salary figure and passes that to Prof # 2. Prof # 2 adds his arbitary salary to the received figure; forwards the total to the 3rd Prof after tearing off the value received from the 1st professor. This process continues

and then they start again and again and finally after n times they will find the mean...

[quote="mustafa"]There is a problem with that solution. If there are five professors, the fifth one can memorize the last number, he can figure out what the first professor made.[/quote]

No, he can't.  It goes around to all professors.  Then the first professor takes the result back, subtracts the arbitrary number he began with, and adds his own salary.  (He has to not let the second professor see the final total before he does this.  In fact, there must be strictly more than 2 professors, since knowing the mean of 2 numbers and one of the numbers tells you the other number.)  Then he divides by the number of professors, and the result is the mean.  The arbitrary initial value is highly relevant.

I have a feeling Shamik has seen this type of puzzle before and therefore has the expected and correct solution. Or he's just a genius with great insight :lol:

Similar: Ten people has to select one from two candidates. However they do not want to make it openly and there is no poll. Assume that they will not cheat.

Make it so that the number the first professor wrote can be negative


[quote="DonkeyKong"]um.. as far as I know, it uses the d=rt.

I think your equation looks right, but just remeber to plug it back in for passenger train.[/quote]Ya I agree with DonkeyKong it should use the d=rt

I agree w/ DK also, it does use d=r x t
when it says they both travel the same distance, you should use d=r x t because you already know the rate and you can just substitute r x t for distance.
Your equation looks fine to me!  :)

[quote="epatjn"]A passenger's train speed is 60 mph, and freight train speed is 40 mph.  The passenger train travels the same distance in 1.5 h less time than the freight train.  how long does each train take to make the trip?

Does this problem use d = rt?

if i use (60 * t-1.5)=40t     would that be right?[/quote]

Uh, technically, it should be $ 60\times (t\minus{}1.5)\equal{}40t$.  The parentheses are very important.

And then make sure you answer the question.  :wink:

you guys are all correct
i mean it looks right

 :play_ball:
 :coolspeak:


geometry
calculus
integration
real analysis
real analysis theorems

I was wondering how the Jacobian was derived.  It seems that dealing with projections would lead me there (similar to the surface area integral), but I want to see the full thing.

So far I have:

area(F1)=area(F2)cos(theta)

Then I can replace the left side of the equation with dxdy, but I'm not sure what to do from there.

Any ideas?

Thanks.

Actually, I asked the same question a while ago, here is the discussion: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32350[/url]


Find $ x \in R$ of this equation: $ 4\sqrt { 1\plus{} x} \minus{} 1 \equal{} 3x \plus{} 2\sqrt {1 \minus{} x} \plus{} \sqrt {1 \minus{} x^2}$

We can rewrite the equation as
$ (\sqrt{1\minus{}x}\plus{}\sqrt{1\plus{}x}\minus{}2)(\sqrt{1\minus{}x}\minus{}2\sqrt{1\plus{}x})\equal{}0$
The solution is $ 0$ or $ \minus{}\frac35$.


Let $m$ be a 4-digit number such that $m^2-n^2$ is a square number, where $n$ is the 4-digit number formed by reversing the digits of $m$. Find $m$.

This is similar to one of the 2005 AMC 12B problems, except that $m$ is a 4-digit number, not 2-digit...

[hide="my work so far"](EDITED) Let $m=1000a+100b+10c+d$ and $n=a+10b+100c+1000d$.

$m^2-n^2=(m+n)(m-n)$
$=(1001a+110b+110c+1001d)(999a+90b-90c-999d)$
$=99(91(a+d)+10(b+c))(111(a-d)+10(b-c))=k^2$

$(91(a+d)+10(b+c))(111(a-d)+10(b-c))$ must be in the form of $11p^2$.
But then, it's a little hard to check the values, and I don't know if this is the right approach :(  [/hide]

[quote="frt"]This is similar to one of the 2005 AMC 12B problems, except that $m$ is a 4-digit number, not 2-digit...

[hide="my work so far"](EDITED) Let $m=1000a+100b+10c+d$ and $n=a+10b+100c+1000d$.

$m^2-n^2=(m+n)(m-n)$
$=(1001a+110b+110c+1001d)(999a+90b+90c+999d)$
$=99(91(a+d)+10(b+c))(111(a+d)+10(b+c))=k^2$

Let $a+d=x$ and $b+c=y$, then $(91x+10y)(111x+10y)$ must be in the form of $11p^2$.
But then, it's a little hard to check the values for $x$ and $y$, and I don't know if this is the right approach :(  [/hide][/quote]

I thought it should be
$m^2-n^2=(m+n)(m-n)$
$=(1001a+110b+110c+1001d)(999a+90b-90c-999d)$.....

[quote="seamusoboyle"]Notice the three numbers form a pythagorean triple....[/quote]
but I cant understand how that would help.

It puts severe limits on what the numbers can be.

I cant think how you can do this using logical hit and trial way or using frt's way.

[quote="gunes"]i have two numbers
for       $m=6565$
$6565^2-5656^2=3333^2$
for       $m=5625$
$5625^2-5265^2=1980^2$

i found them with microsoft excel[/quote]
Wow... power of technology :lol:
I should have thought of $6565^2-5656^2=3333^2$ since I knew $65^2-56^2=33^2$...

[quote="jelly119"]What about a 3-digit number? Think about it![/quote]
I don't think there's any 3-digit number that satisfies the given condition.
Here's what I did.
[hide]Let $m=100a+10b+c$
Then $(100a+10b+c)^2-(a+10b+100c)^2$
$=(101a+20b+101c)(99a-99c)=99(101a+20b+101c)(a-c)=k^2$
This means $(101a+20b+101c)(a-c)$ has to be in the form of $11s^2$.
Note that $a-c\ne 11$
We then have to think about some cases.
[hide="case 1"]$a-c=p^2$ and $101(a+c)+20b=11q^2$

From the second equation, we obtain $b=\frac{11q^2-101(a+c)}{20}$ ($2\le a+c\le 18$ and $0\le b\le 9$), but there doesn't exist any solution.[/hide]
[hide="case 2"]$a-c=p$ and $101(a+c)+20b=11p$

Combine two equations to get $101(a+c)+20b=11(a-c)$ or $90a+20b+112c=0$, but there's no solution for $a,c\ge 1$ and $b\ge 0$.[/hide]
[hide="case 3"]$a-c=p$ and $101(a+c)+20b=11p^3$

Again, we obtain $b=\frac{11p^3-101(a+c)}{20}$ from the second equation, but there's no solution for $2\le a+c\le 18$ and $0\le b\le 9$.[/hide]
There are still some minor cases like $a-c=p$ and $101(a+c)+20b=11p^5$ etc., so I'm not completely sure, but probably there is no such 3-digit integer.[/hide]
Since it's late at night, I hope I didn't do my arithmetic wrong...

[quote="frt"][quote="jelly119"]What about a 3-digit number? Think about it![/quote]
I don't think there's any 3-digit number that satisfies the given condition.
Here's what I did.
[hide]Let $m=100a+10b+c$
Then $(100a+10b+c)^2-(a+10b+100c)^2$
$=(101a+20b+101c)(99a-99c)=99(101a+20b+101c)(a-c)=k^2$
This means $(101a+20b+101c)(a-c)$ has to be in the form of $11s^2$.
Note that $a-c\ne 11$
We then have to think about some cases.
[hide="case 1"]$a-c=p^2$ and $101(a+c)+20b=11q^2$

From the second equation, we obtain $b=\frac{11q^2-101(a+c)}{20}$ ($2\le a+c\le 18$ and $0\le b\le 9$), but there doesn't exist any solution.[/hide]
[hide="case 2"]$a-c=p$ and $101(a+c)+20b=11p$

You've got the correct answer!! :)
Combine two equations to get $101(a+c)+20b=11(a-c)$ or $90a+20b+112c=0$, but there's no solution for $a,c\ge 1$ and $b\ge 0$.[/hide]
[hide="case 3"]$a-c=p$ and $101(a+c)+20b=11p^3$

Again, we obtain $b=\frac{11p^3-101(a+c)}{20}$ from the second equation, but there's no solution for $2\le a+c\le 18$ and $0\le b\le 9$.[/hide]
There are still some minor cases like $a-c=p$ and $101(a+c)+20b=11p^5$ etc., so I'm not completely sure, but probably there is no such 3-digit integer.[/hide]
Since it's late at night, I hope I didn't do my arithmetic wrong...[/quote]


inequalities
real analysis
real analysis solved

[quote="liyi"]$f(x)\geq 0$ is Riemann-integrable on $[a,b]$.
Show that $\sqrt{f(x)}$ is also Riemann-integrable on $[a,b]$ using $\sum_i \omega_i \Delta x_i<\varepsilon$.

========================================

My puzzle is: I don't know how to use $\sum_i \omega_i \Delta x_i<\varepsilon$ directly. I use $\sum_{\omega_i\geq \epsilon} \Delta x_i<\sigma$.[/quote]

 We have $\sum_i{\omega_i}\Delta x_i <\varepsilon$
 A fundamental inequality gives : \[\sum_i \left( \sqrt{\omega_i}\Delta x_i \right)^2  \leq \left(\sum_i{\omega_i}\Delta x_i \right) \cdot \left( \sum_i \Delta x_i \right ) <\varepsilon \cdot (b-a) \] so \[\sum_i \left( \sqrt{\omega_i}\Delta x_i \right) < \sqrt{(b-a)\varepsilon} \]

I worked it out later. My proof is exactly the same. Thank you all the same.
The inequality is called Cauchy's Inequality.

[quote="dickchimney"]
 A fundamental inequality gives : \[\sum_i \left( \sqrt{\omega_i}\Delta x_i \right)^2  \leq \left(\sum_i{\omega_i}\Delta x_i \right) \cdot \left( \sum_i \Delta x_i \right ) \][/quote]
A typo there.
It must be $\left(\sum  \sqrt{\omega_i}\Delta x_i \right)^2  \leq \left(\sum{\omega_i}\Delta x_i \right) \cdot \left( \sum \Delta x_i \right )$


If $ x \plus{} 5 < 8$ and $ x$ is a prime number, what is the value of $ x$?

$ x\plus{}5<8$
$ x<3$
x is prime. Only prime number under 3 is 2.


complex analysis
complex analysis unsolved

Interpret this geometrically. Draw the circle centered at the origin with radius $ r\equal{}|z_1|$, and some segments as appropriate. What are these angles in the picture?

Yes, I see. Perifirical angle is  $ 1/2$ of central.


What are some good books for introductory physics?

What level? [url=http://www.amazon.com/University-Physics-Modern-Mastering-MasteringPhysics/dp/080538684X]University Physics (Young & Freedman)[/url] is a widely used calculus based introduction, first year college. Another one is [url=http://www.amazon.com/Physics-Scientists-Engineers-Douglas-Giancoli/dp/0132431068/ref=pd_sim_b_36]Physics for Scientists and Engineers (Giancoli)[/url], and [url=http://www.amazon.com/Fundamentals-Physics-Extended-David-Halliday/dp/0471758019/ref=cm_cr_pr_pb_t]Funamentals of Physics (Halliday & Resnick)[/url] is also known.
All these books are very big, covering lots of material (all the basic stuff: mechanics, thermodyamics, electricity+magnetism, waves, optics, etc.), but of course also on a very basic level.

If you're self studying, and can follow your own plan, I think there are better ways to learn physics. For example, first get a good working knowledge of calculus/linear algebra (at a computational level), and then directly go to some deeper, advanced introductory texts, like [url=http://www.amazon.com/Introduction-Mechanics-Daniel-Kleppner/dp/0070350485/ref=sr_1_4?ie=UTF8&s=books&qid=1234976789&sr=1-4]An Introduction to Mechanics (Kleppner)[/url], [url=http://www.amazon.com/Introduction-Electrodynamics-3rd-David-Griffiths/dp/013805326X/ref=pd_sim_b_3]Introduction to Electrodynamics (Griffiths)[/url] and [url=http://www.amazon.com/Special-Relativity-M-I-T-Introductory-Physics/dp/0393097935/ref=pd_sim_b_10]Special Relativity (French)[/url]

I think Halliday & Resnick is the best at an introductory level. it explains things very clearly and there are thought-provoking questions and problems behind every chapter. It is great  :omighty: .

French's Speical Relativity is awesome!  I'd highly recommmend it.  Also Wheeler's and Taylor's [i]Spacetime Physics [/i]is pretty for starters.


inequalities
geometry
circumcircle
inradius
geometry solved

Let ABC be a triangle with circumradius R and inradius r. Let [tex]m_a, h_a[/tex] be the lengths of the median and altitude, respectively, through A. Prove that

[tex]\frac{R}{2r} \geq \frac{m_a}{h_a}[/tex]

This was definitely posted before. I don't know whether it was posted in the "Geometry" or "Inequalities" section, however..

[color=red][Moderator edit: Indeed. http://www.mathlinks.ro/Forum/viewtopic.php?t=2558 ][/color]


[b]Let [/b]$ A \in M_{n} \mathbb{(R)}$,[b]I is identity nxn matrix .[/b]
 [b]Such that :[/b] $ 19A^{4}-5A^{3}+1909A^{2}-5A+1890I=0$
[b]Prove that :[/b]
  $ det^{2}(A+xI)+det^{2}(A+yI) \geq 2det^{2}(A+\sqrt{xy}I)$, $ \forall x,y \in R^{+}$
[b]Here x, y are positive real number .[/b]
[b]"..Do math to eat very sweet ..."[/b]

[b]You can prove from inequality :[/b]
 $ (ax^{2}\plus{}bx\plus{}c)(ay^{2}\plus{}by\plus{}c) \geq (axy\plus{}b \sqrt {xy}\plus{}c)^{2}$ , $ a>0,b<0,c>0 ,b^{2}\minus{}4ac < 0$
 $ (x^{2}\plus{}1)(y^{2}\plus{}1) \geq (xy\plus{}1)^{2}$ ,$ x ,y \geq 0$
[b]" .. You can real Jean Marie Monier professor to find the nice problem of math ..."[/b]
[b]"..I can copy all books to understand but I will creat to find nice problem of math ..."[/b][/b]
A


analytic geometry
graphing lines
slope
calculus
derivative
function
symmetry

What do you mean by slope? Any non-linear curve has variable slope... it's differentiation in calculus. Unless I misunderstood your question.

[quote="Flirrrt"]I recently begun study regression, for personal interest, but now i've hit the wall. So, I have a little problem, that i think should be fairly easy to you guys.

I need the slope, of a curvilinear (non-linear) regression. I will use this on a time series study, and I can get the value correct (intercept) but not the slope. I think you somehow shoul be able to extract the slope from the formula of non-linear regression, but i can't pull it off.

Anyone willing to show me what formula to use for the non- linear regressino slope, whit the dependent variable x = timeserie?

Thanks guys.[/quote]

You need calculus (differentiation to find the slope).
However, if you haven't studied calculus yet, here's an approach:

Let your regression function be $ f(x)$, and suppose you would like to find the "slope" at $ x_{0}$. Pick a very small number $ \Delta x > 0$, and calculate
$ m = \frac{f(x+\Delta x)-f(x)}{\Delta x}$

Of course, this is only an approximation. But as you decrease the size of $ \Delta x$, then you can get really close to the actual value.

Well, what power is the function closest to? 2? 3? 4? if greater than 2, I would second vish. Otherwise, calculate the distance from the point on the function $ p_{o}$ to the approximate focus. The extend another line from the focus, along the line of symmetry, that has that same length $ d$, and ends at point $ p_{f}$. The lines that connects these is tangent to point $ p_{o}$.

If I'm not mistaken this is a more rudimentary form of calculus differentiation.

EDIT: whew, gotta pay attention in english :(


Write the parametric equations of the line that goes through the point (6,-2,1) and its perpendicular to both:
[x,y,z]=[1,4,-2]+t[3,-1,1] and [x,y,z]=[9,5,-3]+s[1,-3,7]

And what have you thought to try?  (A hint: in what context have you heard the phrase "perpendicular to both" before?)


symmetry
geometry
circumcircle
incenter
ratio
trigonometry
number theory

Today, 5 November the IMAR contest was held at National College Sf.
  Sava between 10-13. There were given 3 problems. The results will be available on Tuesday.

 [b] Problem 1. [/b]

Consider the equation
\[\frac{xy-C}{x+y}= k, \]
where all symbols used are positive integers.

1. Show that, for any (fixed) values $C, k$, this equation has at least a solution $x, y$;

2. Show that, for any (fixed) values $C, k$, this equation has at most a finite number of solutions $x, y$;

3. Show that, for any $C, n$, there exists $k = k(C,n)$ such that the equation has more than $n$ solutions $x, y$.

 [hide="Solution"]

 The equation can be written $(x-k)(y-k) = k^{2}+C$. In order to have solution(s) we
need $x-k$ and $y-k$ of same sign. Now, if both $x \leq k$ and $y \leq k$, then
$(x-k)(y-k) = (k-x)(k-y) < k\cdot k = k^{2}< k^{2}+C$, therefore we need have
$x > k$ and $y > k$. But if either of them is greater than $k^{2}+k+C$, again there is no
solution, so the number of possible solutions is finite. As $k^{2}+k+C$ and $k+1$ are a solution, we have
answered affirmatively to the stated assertion of the problem. In fact, the number of distinct positive integer
solutions is equal to the number of positive divisors of $k^{2}+C$.  $\square$

In order to prove 3., we shall use the {\it lemma} that
$m^{3}+1 = (m+1)[(m+1)(m-2)+3]$, so a common divisor of the two factors can be at most 3, but this
cannot happen when $m$ is itself divisible by 3, therefore the two factors are relatively prime.
Moreover, for $m > 1$, both factors are greater than 1. By repeated application, for $m_{0}= m$, $m_{i+1}= m_{i}^{3}$,
it follows that $m_{n}+1$ has more than $n$ relatively prime factors.

Let us now take
\[p = C\cdot 3^{3^{n}}, \quad{ }m = C\cdot 3^{3^{n}+1}, \quad{ }\textrm{and}\quad{ }k = p^{\frac{3^{n}+1}{2}}, \]
therefore $k^{2}+C = C(m^{3^{n}}+1)$, where $m$ is seen to be divisible with 3. According to the above lemma,
this will have more than $n$ relatively prime divisors. \hfill $\blacksquare$

[/hide]

 [b] Problem 2 [/b]

 A number of $n > m \geq 1$ soccer teams play a full tournament, each team meeting (once) each other.
Points are awarded: 2 for a victory, 1 for a tie and 0 for a loss. At the end, each team has won half of its points
against the $m$ teams placed last (including each of these teams, who won half of its points against the other $m-1$).

Find all possible values for $n$ and $m$, supported with examples of such tournaments.

  [hide="Solution."]

 Let us call the last $m$ teams {\it the losers}, and the other $n-m$ teams {\it the winners}.
It is clear that the losers together have won 2(2${m}\choose{2}$) points, while
the winners together have won 2(2${n-m}\choose{2}$) points; but in all there are 2${n}\choose{2}$
points awarded, hence we have the equation
2(2${m}\choose{2}$) + 2(2${n-m}\choose{2}$) = 2${n}\choose{2}$
which yields $n^{2}-(4m+1)n+4m^{2}= 0$, or $(n-2m)^{2}= n$, so we need have $n = (k+1)^{2}$, with $k \geq 1$.
But at least one loser won not less than $2(m-1)$ points, while at least one winner has won at most $2(n-m-1)$
points; this implies $n \geq 2m$, therefore $n-2m = k+1$, so $m = k(k+1)/2$ and $n-m = (k+1)(k+2)/2$.

We will give an example of high symmetry: assume all losers tied the matches between themselves, and similarly,
all winners tied the matches between themselves; this gives that each loser won $2(m-1) = (k-1)(k+2)$ points,
while each winner won $2(n-m-1) = k(k+3)$ points. We shall now consider the possible cases:

{\bf k even}: arrange the winners in a rectangular array of $(k+1)$ rows and $(k+2)/2$ columns; there will be
${k+1}\choose{k-1}$ = ${k+1}\choose{2}$ = $k(k+1)/2 = m$ groups of $(k-1)$ rows. Establish a one-to-one
mapping between the losers and these groups, and have each loser tie the winners in the corresponding group, and lose to the others.

{\bf k odd}: arrange the winners in a rectangular array of $(k+1)/2$ rows and $(k+2)$ columns; there will be
${(k+1)/2}\choose{(k-1)/2}$ = ${(k+1)/2}\choose{1}$ = $(k+1)/2 = m/k$ groups of $(k-1)/2$ rows. Establish a one-to-one
mapping between the partition of the losers in $m/k$ classes of $k$ losers each and these groups, and have each loser tie
the winners in the corresponding group, and lose to the others.
Thus, each loser ties $(k-1)(k+2)/2$ and loses the other $(k+2)$ of these cross-matches, accounting for
$(k-1)(k+2)/2$ points, while each winner ties
$k(k-1)/2$ and wins the other $k$ of these cross-matches, accounting for $k(k+3)/2$ points, as needed. \hfill $\blacksquare$

 {\bf Remarks.} This problem is based on a Swiss problem, where $m = 10$ was given (therefore $k = 4$);
thus $n$ came out to be 25; but the examples given there lacked the grouping idea behind. Our implementation both
generalizes the problem and offers a generic way of constructing the required examples.

 [/hide]

  [b] Problem 3. [/b]

   Consider the isosceles triangle $ABC$ with $AB = AC$, and $M$ the midpoint of $BC$.
Find the locus of the points $P$ interior to the triangle, for which $\angle BPM+\angle CPA = \pi$.

  [hide="Solution"]

  When asked during the contest, the problem started with the following helping preliminary question
(omitted in this published version), which we will use as a {\it lemma:}

{\it Prove that if, for a point $P$ interior to the triangle, $\angle ABP = \angle BCP$, then $\angle BPM+\angle CPA = \pi$.}

The proof starts by noticing that the configuration is fully symmetrical:
\[\angle ABP = \angle BCP \quad{ }\textrm{iff}\quad{ }\angle ACP = \angle CBP\]
\[\angle BPM+\angle CPA = \pi \quad{ }\textrm{iff}\quad{ }\angle CPM+\angle BPA = \pi\]
and also that $P \in AM$ guarantees the result, therefore we may assume wlog (for construction's sake)
that $P$ is interior to the triangle $ABM$. The given angle equality is readily seen to be equivalent with $P \in \Gamma$,
where $\Gamma$ is the arc interior to $ABC$ of the circumcircle $\mathcal K$ of triangle $BCI$, with $I$ the incenter of $ABC$.
It is also immediately seen that $AB$ and $AC$ are tangent to $\mathcal K$.

Consider the Apollonius circle $\mathcal A$ for points $A$, $M$ and ratio $BA/BM$;
denote by $U$ and $V$ the points where $AM$ meets $\mathcal A$. Clearly, $B, C \in \mathcal A$. Then $BA/BM = UA/UM = VA/VM$,
from which follows $BA^{2}/BM^{2}= UA\cdot VA/UM\cdot VM = UA\cdot VA/BM^{2}$, by symmetry and the power of a point relation, so
$BA^{2}= UA\cdot VA$, hence $AB$ is tangent to $\mathcal A$, again by the power of a point relation. Therefore the two
circles $\mathcal K$ and $\mathcal A$ coincide. Now, prolong $AP$ until it meets $\mathcal K$ again at $Q$, and prolong
$PM$ until it meets $\mathcal K$ again at $R$. Then $BA/BM = PA/PM = QA/QM$, therefore $BA^{2}/BM^{2}= PA\cdot QA/PM\cdot QM = BA^{2}/PM\cdot QM$, by the power of a point relation, so $BM^{2}= PM\cdot QM$. But $BM^{2}= PM\cdot RM$, again by the power
of a point relation, so $QM = RM$, hence $BQ = CR$. It follows that $\angle BPQ = \angle CPR$, and this is enough
to yield $\angle BPM+\angle CPA = \pi$. \hfill $\square$

Alternatively, one can calculate ratios using the symmetrical points $P'$ and $Q'$ with respect to $AM$, and
denoting by $N$ the meeting point of $PQ'$ and $P'Q$. It can be obtained that $AM = AN$, therefore $M \equiv N$
and the rest easily follows as above.

\bigskip

We claim the locus is the arc $\Gamma$ (defined in the above), together with the (open) segment $(AM)$.
Clearly $P \in (AM)$ fills the bill, so from now on we will assume $P \notin (AM)$. Also, as above, we will
assume $P$ interior to the triangle $ABM$ (otherwise we work with the symmetrical relations).

Assume $P \notin \Gamma$, equivalent to $\angle BPC \neq \pi-\angle ABC$; then $AB$ and $AC$ are not tangent to $\mathcal K$.
Take $B'$ and $C'$ to be the tangency points on $\mathcal K$ from $A$, and $M'$ to be the midpoint of $B'C'$.
We are now under the conditions from the lemma (for triangle $AB'C'$), therefore $\angle B'PM'+\angle C'PA = \pi$.
But $\angle B'PM' = \angle BPM+\delta(\angle B'PB+\angle MPM')$ and $\angle C'PA = \angle CPA-\delta\angle C'PC$,
where $\delta = 1$ if $\angle BPC < \pi-\angle ABC$, respectively $\delta =-1$ if $\angle BPC > \pi-\angle ABC$.
We have $\angle B'PB = \angle C'PC$ from the symmetry of the configuration, and $\angle BPM+\angle CPA = \pi$ given;
these relations therefore imply $\angle MPM' = 0$, which can only happen when $M$, $M'$ and $P$ are collinear,
i.e. $P \in AM$, which was ruled out from the start in this part of the proof. The contradiction thus reached
confirms our claim. $\blacksquare$

Alternatively, when $\angle BPC > \pi/2$, construct the point $A'$ as the apex of the isosceles triangle of base $BC$ and
equal angles of measure $\angle PBC+\angle PCB < \pi/2$, on the same side of $BC$ with $A$;
then $P$ is like in the lemma (for triangle $A'BC$),
therefore $\angle BPM+\angle CPA' = \pi$, hence $\angle CPA = \angle CPA'$ and so $A \equiv A'$, as $P$, $A$ and $A'$
are not collinear. But this is clearly indicating that $P \in \Gamma$.

Showing that if $\angle BPC \leq \pi/2$, then $\angle BPM+\angle CPA < \pi$, will finish the proof; this follows
from noticing that, considering $\Gamma_{M}$ the circumcircle of triangle $BPM$, $\Gamma_{M}$ will meet $(AM)$ at a point $X$
(in fact we even have $X \in (AH)$, with $H$ the orthocenter of $ABC$).
Also, considering the (open) hemicircle $\gamma$ of diameter $BC$ and on the same side of $BC$ with $A$,
$\Gamma_{M}$ will meet $\gamma$ at a point $Y$ ($X$ may even coincide with $Y$, when $CX$ is tangent to $\Gamma_{M}$).

But the points $C$, $X$ and $Y$ are collinear, so the point $P$, being on $\Gamma_{M}$ and interior to $ABM$, will be such that
$X$ is within the triangle $CPA$, therefore $\angle CXA > \angle CPA$. But $X$ and $P$ being on $\Gamma_{M}$
yields $\angle BPM = \angle BXM$, while $X$ being on $(AM)$ yields $\angle BXM+\angle CXA = \pi$. Together
these add up to $\angle BPM+\angle CPA < \pi$, as needed.

[b] Remarks. [/b] This problem is based on a Moldavian problem (the preliminary question), to which the locus
question has been added (as kind of a converse proposition).

 [/hide]

i`ll give now another solution to problem 3 :)

so..let`s take $\angle BPM=\alpha$ and $\angle CPM = \beta$ so $\angle APC=180-\alpha$ and $\angle APB=180-\beta$

we consider $\angle CBP = u$ and $\angle BCP = v$

applying the law of sin in triangles $ABP$ and $ACP$ we have:

$\frac{l}{\sin({180-\beta})}=\frac{AP}{\sin({B-u})}$ and

$\frac{l}{\sin({180-\alpha})}=\frac{AP}{\sin({C-v})}$ , where $l=AB=AC$

by dividing the two relations we obtain that :

$\frac{\sin{\alpha}}{\sin{\beta}}=\frac{\sin({C-v})}{\sin({B-u})}$ $(\star)$ (we used the fact that $\sin({180-x})=\sin x$)

now let`s apply the law of sin in triangles $BPM$ and $CPM$:

$\frac{PM}{\sin{u}}=\frac{BM}{\sin{\alpha}}$

$\frac{PM}{\sin{v}}=\frac{CM}{\sin{\beta}}$

by dividing the two relations we obtain that:

$\frac{\sin{\alpha}}{\sin{\beta}}=\frac{\sin{u}}{\sin{v}}$ $(\star\star)$

by $(\star)$ and $(\star\star)$ we have that : $\frac{\sin{u}}{\sin{v}}=\frac{\sin({C-v})}{\sin({B-u})}$

$\Rightarrow \sin{u}\cdot \sin(B-u) = \sin{v}\cdot \sin(C-v)$

it is well known that we have the following relation : $\sin{x}\sin{y}=\frac{\cos(x-y)-\cos(x+y)}{2}$

so : $\cos(B-2u)-\cos{B}= \cos(C-2v)-\cos{C}$, but $B=C$

$\Rightarrow \cos(B-2u) = \cos(B-2v)$

$1.$ $B-2u = B-2v$ $\Rightarrow u=v$ so $P \in (AM)$

$2.$ $B-2u = 2v-B$ $\Rightarrow B=u+v$ so $v=\angle ABP \Rightarrow \angle ABP+\angle ACP = \angle PBC+\angle PCB \Rightarrow P$ is on arc $BIC$

the end  :oops:


algebra
binomial theorem
number theory unsolved
number theory

Assume that the equation x^2 - d*y^2 = -1 is solvable, with least positive solution x_0 + y_o*sqrt(d) > 1.

Let also x_1 + y_1*sqrt(d) be a solution to x^2 - d*y^2 = 1.

Show that (x_1 + y_1*sqrt(d)) = (x_0 + y_0*sqrt(d))^2.

Ok, let me rephrase the question.
Suppose $d$ is not a perfect square. Let $(x, y) = (x_1, y_1)$ be the pair of smallest positive integers satisfying $x^2 - dy^2 = -1$. Let $(a,b) = (a_1, b_1)$ be the pair of smallest positive integers satisfying $a^2 - db^2 = 1$. Show that $a_1 + b_1 \sqrt{d} = {(x_1 + y_1 \sqrt{d})}^2$.

Let $x_2 + y_2 \sqrt{d} = {(x_1 + y_1 \sqrt{d})}^2$. Note that by Binomial Theorem and by grouping the rational and irrational terms together, we also have $x_2 - y_2 \sqrt{d} = {(x_1 - y_1 \sqrt{d})}^2$. Thus ${x_2}^2 - d{y_2}^2 = {(x_1 + y_1 \sqrt{d})}^2 {(x_1 - y_1 \sqrt{d})}^2 = ({x_1}^2 - d{y_1}^2)^2 = 1$. Thus $(a,b) = (x_2, y_2)$ is a solution to $a^2 - db^2 = 1$.

Now note that ${a_1}^2 - d{b_1}^2 = 1$ by definition, thus $a_1 \leq x_2, b_1 \leq y_2$. Suppose $a_1 < x_2, b_1 < y_2$, then since $a_1 + b_1 \sqrt{d} > 1$, $\exists$ some $m \in \mathbb{Z}^+$ such that \[ {(a_1 + b_1 \sqrt{d})}^m \leq x_2 + y_2 \sqrt{d} < {(a_1 + b_1 \sqrt{d})}^{m+1}. \] Multiplying throughout by ${(a_1 - b_1 \sqrt{d})}^m$, we get \[ 1 \leq (x_2 + y_2 \sqrt{d}){(a_1 - b_1 \sqrt{d})}^m < a_1 + b_1 \sqrt{d}. \] Let $(x_2 + y_2 \sqrt{d}){(a_1 - b_1 \sqrt{d})}^m = a_0 + b_0 \sqrt{d}$ $\Rightarrow (x_2 - y_2 \sqrt{d}){(a_1 + b_1 \sqrt{d})}^m = a_0 - b_0 \sqrt{d}$. Thus we have ${a_0}^2 - d{b_0}^2 = ({x_2}^2 - d{y_2}^2){({a_1}^2 - d{b_1}^2)}^m = 1$.

So $(a,b) = (a_0, b_0)$ is a solution of $a^2 - db^2 = 1$ such that $1 \leq a_0 + b_0 \sqrt{d} < a_1 + b_1 \sqrt{d}$, which contradicts the minimality of $(a_1, b_1)$.

$\therefore a_1 + b_1 \sqrt{d} = {(x_1 + y_1 \sqrt{d})}^2$.


[b]Find  roots of equation ![/b]
x^4 -5x^3+4x^2-2x+2=0

By the rational roots theorem, we see that the possible roots are $ \pm1, \pm2$.

By plugging in $ 1$ in synthetic division, we see that it's a root. So we can factor it to:

$ (x\minus{}1)(x^3 \minus{}4x^2 \minus{}2) \equal{} 0$

Now for $ x^3 \minus{}4x^2 \minus{}2$, there is 1 real root and 2 complex roots. I don't know if you are solving for reals or complex...

$ x^3 \minus{} 4x^{^2 } \minus{} 2 \equal{} 0$
Maybe, we will find the real root with formula [b]Cadano.[/b]/

It's irrational and somewhere between 4 and 5. You can use a graphing calculator or Wolfram Alpha, but not by hand.


[quote="most-wanted"]since long that i haven't been here.
just had exams.
here is the question.
A man have head 40 Cows and birds.
he counts the legs of them 148 totallly.
how much cows and birds does he have ?[/quote]
I used to do questions like this all the time. :)
Let x be the no. of cows, and y be the no. of birds. So,
$4x+2y=148$-->$2x+y=74$
$x+y=40$
So, sub the first in the second to get $x=34$
And then $y=6$.
So, 6 birds and 34 cows. :)

[quote="most-wanted"]i am not sure how you got that.
can you please explain it in details.
how you got 2x + y = 74 ?[/quote]
Well, a cow has 4 legs, and a bird 2 (unless they are mutated...), so we get the equation
4x+2y=148, where x is the no. of cows and y the no. of bids. The common factor is 2, so we divide both sides of the equation by it, to get
2x+y=74. Which, is the simplfied form. :)

thanks i would remember that always by now :lol:

[hide]using elimination, $\boxed{34}$ cows and $\boxed{6}$ birds[/hide]

[hide]Call cows $x$ and birds $y$. We can set up the equations $4x+2y=148$ and $x+y=40$. Solving we find $(x,y)=(34,6)$.[/hide]

[quote="most-wanted"]since long that i haven't been here.
just had exams.

here is the question.
A man have head 40 Cows and birds.
he counts the legs of them 148 totallly.

how much cows and birds does he have ?[/quote]
[hide="for those who do not get systems"]If they all are cows there would be 160 legs. There are two less legs per member. Twelve legs means six birds and 40-6=34 cows.[/hide]

[hide]
using systems of eqations and then using elimination i get 34 and 6[/hide]

[quote="most-wanted"]since long that i haven't been here.
just had exams.

here is the question.
A man have head 40 Cows and birds.
he counts the legs of them 148 totallly.

how much cows and birds does he have ?[/quote]

[hide]
c+b=40
4c+2b=148

4c+4b=160
4c+2b=148

2b=12
b=6
c=34[/hide]


When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole number quotient with a remainder of 1.  How many distinct values are possible for this three-digit dividend?

[hide="Hint"]
This is the same as if the units digit was 1 and the remainder was 0. Then immediately, you know that the divisors cannot be some values, like 2, 4, 5, 6, 8.[/hide]

[hide]there are 30 3-digit with lats digit as 1 divisible by 3
7*23=161, ...7*133=931
so there are 12 divisible by 7

30+12=42

bu the answer should be 38

what did I miss?[/hide]

how did you conclude that their are 30 three digit numbers that are divisible by 3? Also since there are numbers like 231 which are divisible by three and seven their could be some overlaps.


inequalities
trigonometry
inequalities proposed

Consider $ a_{1}$, $ a_{2}$, $ \ldots$, $ a_{10}$ pozitive real numbers such that $ a_{1}\plus{}a_{2}\plus{}\ldots\plus{}a_{10}\equal{}1$. Prove that
\[ \sqrt{a_{1}a_{2}}\plus{}\sqrt{a_{2}a_{3}}\plus{}\ldots\plus{}\sqrt{a_{9}a_{10}} \leq \cos{\frac{\pi}{11}}.\]

I think if you use mixing variables, you can get maximum occurs when $ a_1 \equal{} a_3 \equal{} ... \equal{} a_9$ and $ a_2 \equal{} a_4 \equal{} ... \equal{} a_10$. From this I think the maximum occurs when all the variables are equal, so the value is 0.9 < cos pi/11... this looks strange. Is there any case where the max is > 0.9?