inequalities
geometry
trigonometry
inequalities unsolved

[quote="pohoatza"]If a,b,c are three pozitive number...it is true the inequality:

       (a+b+c)^3 (-a+b+c)(a-b+c)(a+b-c) >= 27a^2b^2c^2[/quote]

He meant  for $a,b,c\ge0$ prove that:

$(a+b+c)^3 (-a+b+c)(a-b+c)(a+b-c)\ge27a^2b^2c^2$.

My conclusion:
$a=3,b=c=1$ it doesn't hold.

my mistake...sorry....a,b,c are 3 numbers that respect : a>=b>=c with b+c>a :) sorry for my mistake...now the inequality holds :)

[quote="pohoatza"]a,b,c are 3 numbers that respect : a>=b>=c with b+c>a  [/quote]
But I think, it still doesn't hold.
Just choose $a=2,b=1.1,c=1$

Yes, this problem is false.
Easy to see that $a,b,c$ are side-lengths of a triangle. If $F$ is the area of this triangle.
$(a+b+c)^3(a+b-c)(b+c-a)(c+a-b) \ge 27a^2b^2c^2 \newline \iff 8(a+b+c)^2 \cdot F^2 \ge 27a^2b^2c^2 \newline \iff 8(a+b+c)^2 \cdot \frac{a^2b^2c^2}{16R^2} \ge 27a^2b^2c^2 \newline \iff (a+b+c)^2 \ge \frac{27}{8}R^2 \newline \iff \sin A+\sin B+\sin C \ge \frac{3\sqrt{3}}{2\sqrt{2}} \newline \iff ?$

it holds...it can be writen like 64 p^2 S^2 > 27a^b^c^ were S = the surface of triangle ABC with sides a,b,c and p = a+b+c/2
...Now you divide the inequality by 16S^2 to obtain R^2=a^2b^2c^2/16S^2
<=> 4p^2>27 R^2
<=>2p>3 sqrt 3 R
<=> p/R > 3sqrt 3 / 2

But a/sin A = b/sin B=c/sin C = 2R...so...sin A = a/2R ...sin B=b/2R...sin C = c/2R...
so...sin A + sin B + sin C = a/2R + b/2R + c/2R = 2p/2R = p/R

Now we replace the result in our inequality...

sin A + sin B + sin C = p/R so...sin A +sin B + sin C > 3 sqrt 3 / 2 witch is true...the inequality holds for a,b,c triangle numbers


geometry
perimeter
inequalities
inequalities proposed

[quote="fuzzylogic"]Let $a, b, c$ be the sides of a triangle, $m_a,m_b,m_c$ be the corresponding medians. Let $s$ be the semi-perimeter and $R$ be the circumradius. Prove that

$m_aa+m_bb + m_cc \leq 3sR$.[/quote]

I found an inequality which is stronger than fuzzylogic 's one

$m_aa+m_bb + m_cc \leq \frac{\sqrt{3}}{2} (a^2+b^2+c^2) \leq 3sR $

But my solution is very long, so I wait for another one.

[quote="tranminhhoang"]I found an inequality which is stronger than fuzzylogic 's one

$m_aa+m_bb + m_cc \leq \frac{\sqrt{3}}{2} (a^2+b^2+c^2) \leq 3sR $

But my solution is very long, so I wait for another one.[/quote]

Actually, I have an inequality stronger than this one and I have a short proof. So I will leave others to try.


Fie triunghiul oarecare $\ ABC$ si $\ A'\in (BC),B'\in (CA),C'\ (AB)$.Notam $\ m(<AA'B')=\alpha,m(<AA'C')=\beta,m(<CA'B')=\gamma,m(<BA'C')=\pi$.Demonstrati ca dreptele $\ AA',BB',CC'$ sunt concurente daca si numai daca $\sin\alpha\cdot\sin\pi=\sin\beta\cdot\sin\gamma$.
$\text{ Calin Burdusel}$

hai ca dau si eu <refresh>...nimeni nu stie sa o faca?? :(

Aplici teorema sinusului in cele 4 triunghiuri. Nu m-ar mira sa dam chiar peste Ceva.


trigonometry
function
calculus
calculus computations

Solve this differential equation
$ (2x\ cosy\plus{}3x^2y)dx\plus{}(x^3\minus{}x^2\ siny \minus{}y)dy\equal{}0$

Is the given equation exact? If so how do you proceed?

From: $ (2x\cos y\plus{}3x^{2}y)dx\plus{}(x^{3}\minus{}x^{2}\sin y \minus{}y)dy\equal{}0$, write the following:

$ M(x,y) \equal{} 2x\cos y\plus{}3x^{2}y$ and $ N(x,y) \equal{} x^{3}\minus{}x^{2}\sin y \minus{}y$

$ M_{y}(x,y) \equal{} 3x^2 \minus{} 2x \sin y$ and $ N_{x}(x,y) \equal{} 3x^2 \minus{} 2x \sin y$

Since $ M_{y}(x,y) \equal{} N_{x}(x,y)$ there exists a function $ f(x,y)$ such that the following hold:

(1) $ f_{x} \equal{} 2x \cos y \plus{} 3x^2y$

(2) $ f_{y} \equal{} x^3 \minus{} x^2 \sin y \minus{} y$

Integrating (1) with respect x, we get $ f(x,y) \equal{} x^2 \cos y \plus{} x^3 y \plus{} g(y)$ *

Now differentiate the * with respect to y and you get $ f_{y} \equal{} x^3 \minus{} x^2 \sin y \plus{} g'(y)$**

Now the ** must equal (2) and so $ g'(y) \equal{} \minus{}y$, which means $ g(y) \equal{} \minus{}\frac{1}{2}y^2$.

Hence, $ f(x,y) \equal{} x^2 \cos y \plus{} x^3y \minus{} \frac{1}{2}y^2$. The desired solution is $ x^2 \cos y \plus{} x^3y \minus{} \frac{1}{2}y^2 \equal{} c$

By some arrangement, the given DE can be rewritten as $ (x^2\cos{y} \plus{} x^3y \minus{} \frac {1}{2}y^2)' \equal{} 0$.

So, the general solution is $ 2x^2\cos{y} \plus{} 2x^3y \minus{} y^2 \plus{} C \equal{} 0$.


In this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70613]thread[/url] Nameless makes an assertion:

[quote="Nameless"]Maybe he misunderstand in $\sum\frac{1}{\sqrt{b+c}}\ge\frac{3}{2}$
We have $\sum\frac{1}{\sqrt{b+c}}\ge\frac{9}{\sum\sqrt{b+c}}\ge\frac{9}{\sqrt{6(a+b+c)}}$[/quote]

This last step implies $\sum\sqrt{b+c} \le \sqrt{6(a+b+c)}$ and Nameless just spouts it off like it is a trivial inequality**. So could we have some nice and insightful proofs for it, like are there some well-known inequalities that have been hiding from me that turn this into a one-liner. If anyone knows these "secret" Ineq's, please post a solution to this one, thanks :lol:.

**The only way I could prove it was with THE trivial inequality, funny, haha.

? It's just Cauchy...

$\left(\sum (b+c) \right)^2 \leq \left(\sum (b+c) \right) \left(\sum 1 \right) = 6\sum a$

Jensen's on the concave function $f(x) = \sqrt{x}$ with equal weights yields:

$\frac{1}{3}(f(a+b)+f(b+c)+f(c+a)) \le f\left(\frac{2(a+b+c)}{3}\right)$

$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a} \le 3\sqrt{\frac{2(a+b+c)}{3}} = \sqrt{6(a+b+c)}$

as desired.


Consider the triangle $ ABC$. Denote by $ M$ and $ N$ the midpoints of $ AB,AC$, respectively. Find the locus of the centroid of the triangle $ \triangle MNP$ while $ P$ moves on $ [BC]$.

Hello , I called L the middle of MN and G' the centroid of $ \Delta MNP$ . The locus is a parallel line to MN or parallel to BC , because PG' = 2G'L.

 [geogebra]1a1c526d3aed3a4186ea492dabd13391ca8a2f64[/geogebra]


I don't quite understand what the roots of unity filter does/means. I read
http://www.mathlinks.ro/weblog_entry.php?t=245909

[quote]The aim of using this method is to isolate the coefficients of some polynomial and sum them up. It is particularly effective whenever the desired indexes of that polynomial are periodic modulo some $ n$. Essentially, it states that
\[ \frac {P(1) \plus{} P(\zeta) \plus{} ... \plus{} P(\zeta^{n \minus{} 1})}{n}\]
is the sum of all indexes of the polynomial divisible by $ n$ where $ \zeta$ is an $ n^\text{th}$ root of unity. (This is a common notation)[/quote]

Look at some simple examples.

[hide="n is 2"]$ n \equal{} 2 \implies \zeta \equal{} \minus{} 1$

$ P(1) \equal{} a_0 \plus{} a_1 \plus{} a_2 \plus{} a_3 \plus{} \ldots$
$ P( \minus{} 1) \equal{} a_0 \minus{} a_1 \plus{} a_2 \minus{} a_3 \plus{} \ldots$

Add and divide by two to get:
$ a_0 \plus{} a_2 \plus{} a_4 \plus{} \ldots$
[/hide]
[hide="n is 3 in detail"]Consider $ n \equal{} 3$:

$ P(1) \equal{} a_0 \plus{} a_1 \plus{} a_2 \plus{} a_3 \plus{} \ldots$
$ P(\zeta) \equal{} a_0 \plus{} a_1 \zeta \plus{} a_2 \zeta ^2 \plus{} a_3 \plus{} \ldots$
$ P(\zeta ^2) \equal{} a_0 \plus{} a_1 \zeta ^2 \plus{} a_2 \zeta \plus{} a_3 \plus{} \ldots$

Where I've used the fact that $ \zeta^3 \equal{} 1$. Now, note that $ \zeta ^3 \equal{} 1 \implies (\zeta \minus{} 1) (\zeta ^2 \plus{} \zeta \plus{} 1) \equal{} 0$, and since $ \zeta \neq 1$, we have $ \zeta ^2 \plus{} \zeta \plus{} 1 \equal{} 0$. Thus, adding and grouping gives:

$ 3 a_0 \plus{} 0 a_1 \plus{} 0 a_2 \plus{} 3 a_3 \plus{} 0 a_4 \plus{} 0 a_5 \plus{} \ldots$

Dividing by $ n \equal{} 3$ gives:
$ a_0 \plus{} a_3 \plus{} a_6 \plus{} \ldots$[/hide]

This generalizes to higher $ n$.

How would you modify the above procedure if you wanted to find, say, $ a_1 \plus{} a_4 \plus{} a_7$ in the polynomial $ P(x) \equal{} (1\plus{}x)^8$.

One specific use for this is to simplify problems involving binomial coefficient sums. For example, something like $ \sum\binom{2010}{3k}$ can be reduced with a roots of unity filter like so:

[hide]The sum is the sum of every third coefficient of $ (x\plus{}1)^{2010}$.

We know that this sum is equal to $ \frac{(1\plus{}1)^{2010}\plus{}(\zeta\plus{}1)^{2010}\plus{}(\zeta^2\plus{}1)^{2010}}{3}$ by the roots of unity filter method.

Now we express $ \zeta$ as $ \cos\frac{2\pi}{3}\plus{}i\sin\frac{2\pi}{3}\equal{}\minus{}\frac 12\plus{}\frac{\sqrt{3}}2i$

And by DeMoivre's, $ \zeta^2\equal{}\cos\frac{4\pi}3\plus{}i\sin\frac{4\pi}{3}\equal{}\minus{}\frac 12\minus{}\frac{\sqrt{3}}2i$

Adding $ 1$ to both of these gives $ \zeta\plus{}1\equal{}\frac 12\plus{}\frac{\sqrt 3}{2}i\equal{}\cos\frac{\pi}{3}\plus{}i\sin\frac{\pi}{3}$ and $ \zeta^2\plus{}1\equal{}\frac 12\minus{}\frac{\sqrt{3}}2i\equal{}\cos\frac{5\pi}3\plus{}i\sin\frac{5\pi}{3}$.

By DeMoivre again, $ \text{cis}^{2010}\frac{\pi}3\equal{}\text{cis}\ 670\pi\equiv \text{cis}\ 0\equal{}1$, and $ \text{cis}^{2010}\frac{4\pi}{3}\equal{}1$ as well.

So we end up with $ \boxed{\frac{2\plus{}2^{2010}}{3}}$.[/hide]

Pretty cool stuff...

Thanks for the replies. That cleared up my confusion pretty well.


prove or disprove:
if $ P_1,P_2,. . .,P_k$ are all primes, then

$ \sqrt(P_1) \plus{} \sqrt(P_2) \plus{} . . . \plus{} \sqrt(P_k)$ is irrational for all k, regardless of the choices for the $ P_i$.

This is true for the set of all non-square positive integers, not just primes.

yes, i suspected that this holds for a more general form, but how would you prove such a statement

(you may prove your generalized statement if that is easier :P )


algebra
polynomial
inequalities
integration
calculus
triangle inequality
complex analysis

I have two questions that I cannot solve at this time  :( .

(1). Prove there exists $ \alpha \in \mathbb C$ such that $ |\alpha| \equal{} 1$ and $ |p(\alpha)| \geq 1$ where $ p(z) \equal{} z^n \plus{} a_{n \minus{} 1}z^{n \minus{} 1} \plus{} \cdots \plus{} a_1 z \plus{} a_0$. I got a hint that we need to count zeros of $ a_{n \minus{} 1}z^{n \minus{} 1} \plus{} \cdots \plus{} a_1 z \plus{} a_0$.

(2). If $ p(z) \equal{} a_n z^n \plus{} a_{n \minus{} 1}z^{n \minus{} 1} \plus{} \cdots \plus{} a_1 z \plus{} a_0$ is such that $ |p(z)| \leq 1$ for all $ z$ satisfying $ |z| \equal{} 1$. Prove that $ a_k \leq 1$ for every $ k \equal{} 0,1,...,n$.

Any idea? Thank you.

# 2, You can assume one of $ a_{k} > 1$, use the triangle inequality for $ |p(z)|$ to derive a contradiction that $ |p(z)| \leq 1$ .

For #2, can I use the following argument: since \[ {a_k} \equal{} \frac{1}
{{2\pi i}}\int\limits_{\left| z \right| \equal{} 1} {\frac{{f\left( z \right)}}
{{{z^{k \plus{} 1}}}}dz} \]we have
\[ \left| {{a_k}} \right| \equal{} \frac{1}
{{2\pi }}\int\limits_{\left| z \right| \equal{} 1} {\frac{{\left| {f\left( z \right)} \right|}}
{{{{\left| z \right|}^{k \plus{} 1}}}}dz}  \leqslant 1.\]
 :oops:

For number 1, you can use Cauchy's Integral formula as follows:
\begin{eqnarray*}1 & = & \left|\frac {p^{(n)}(0)}{n!}\right| = \left|\frac {1}{2\pi i}\int_{S^1}\frac {p(z)}{z^{n + 1}}dz\right| \\
& \leq & \frac {1}{2\pi}\int_{S^1}\frac {|p(z)|}{|z|^{n + 1}}|dz|\leq \sup_{z\in S^1}|p(z)|\end{eqnarray*}
Since $ S^1\subset \mathbb{C}$ is compact and $ p$ is continuous, this supremum is achieved, as required. $ \Box$


Who can introduce some problem?I want to get some and we can discuss together,okay?  :lol:

Trịnh Lưu Ly, i think you can post this topic in Vietnam forum, and join to solve problems on Mathlinks forum  :(

I don't known  :rotfl: . But i think she is a character from Fantasy - a cartoon  :D

Yeah...I think it is Rinoa from final fantasy 8...brilliant game...


Points A and B are selected on the graph of $y=-\frac{1}{2}x^{2}$ so that triangle ABO (where O is the origin) is equilateral. What is the length of a side of triangle ABO?

If you're just looking for a solution, you can easily find several instances of this problem in the MATHCOUNTS forum, like [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120939[/url]

Thanks :)
[color=white]The message is too small. Please make the message longer before submitting.[/color]


Prove that f(x) is not bijection: (bijection=song ánh) satisfy the condition:
$f: N^{*}\rightarrow N$
$f(mn)=f(m)+f(n)+3.f(m).f(n) (\forall m,n \in Z^{+})$

Let $f(1)=x$, then $x+3x^{2}=0\Longrightarrow x=0\not \in N.$

Ok! Let $f: N\to Z_{+}$. We have f(1)=0. Let $g(n)=3f(n)+1$, then $g(mn)=g(m)g(n)$.
If f(n) is bijective, then for all number form $3k+1$ exist n, suth that g(n)=3k+1. Let $g(m)=22,g(n)=40, g(k)=16,g(l)=55$. We have $g(mn)=g(kl)$, therefore mn=kl. $22\not=ab$ for any $a,b=1(mod 3)$, therefore m is prime.  It give contradition.


function
integration
calculus
geometry
real analysis
real analysis unsolved

Sorry, I know almost nothing in diff. equations and I have some problems in solving the following one:
   Let $f$ be a $C^{1}$  class function defined for nonnegative reals. Then any solution of the differential equation $ y\"+fy=0$ is bounded if and only if $f$ has a strictly positive limit at infinity. IMHO, I think we should add some more condition, but I can be of course wrong. Anyway, could someone elucidate this? Thanks.

Here a classical version $f$:$)0,+\infty( \rightarrow R$ class $C^{1}$

such that there exist A>0, for any $x\geq A$, $f(x)>0, f'(x)\geq 0$

Then any solution $g$ of $y\"+f(x)y=0$ is bounded

[quote="harazi"]
   Let $f$ be a $C^{1}$  class function defined for nonnegative reals. Then any solution of the differential equation $ y\"+fy=0$ is bounded if and only if $f$ has a strictly positive limit at infinity.[/quote]
This condition is neither necessary, nor sufficient. It is not necessary, because, to have all solutions bounded, it suffices to have $\int_0^{+\infty}|f(t)-1|\,dt<+\infty$. Indeed, then for $E=y^2+y'^2$, we have $E'=2(yy'+y'y'')=2(1-f)yy'\leqslant |f-1| E$, so $E(t)\leqslant E(0)\exp\left\{\int_0^t |f(s)-1|\,ds\right\}$. But there are plenty of $C^1$ functions that satisfy this integral condition and do not have a limit. On the other hand, the condition is not sufficient either. Indeed, we can choose $f$ to be the identically $1$ function perturbed by small negative humps. The humps may be chosen inductively so that $yy'>0.1 E$ for the duration of each hump (because the point $(y,y')$ essentially moves on a circle with slowly changing radius and  with almost constant angular speed). Then, if the total area of humps is infinite, we shall get an unbounded solution. I don't know whether some reasonable necessary and sufficient condition can be given, but it is definitely not this one.

In the problem mentioned by Moubinool we can replace the condition that $f$ is $C^{1}$ function just considering a positive increasing continious  function $q:[a,+\infty){\rightarrow}{\mathbb{R}}$ and an diff. equation:
$y^{''}+qy=0$.
Talking about this problem we can also say that every solution of that diff. equation has a countable set of isolated zeroes with decreasing distance between them.It can also be proved that the oscillation between two neighbor zeroes is decreasing as zeroes go to infinity.

Thank you very much, fedja, so my "conjecture"  was true and my friend who had this problem in oral examination was not lucky at all to receive a completely false problem. Thanks again.

[quote="harazi"] my friend who had this problem in oral examination was not lucky at all to receive a completely false problem. [/quote] In my student years we considered such a situation quite lucky because if you could demonstrate that the problem given to you was false, you would get a higher grade than formally deserved (if possible). I should confess though that the people who were capable of doing such thing usually got the highest grade anyway :lol:.


Okay, so here we go:

If I can go from one floor of a building to another in 20 sec on an escalator (standing still), and there is a set of steps which I can walk up at a steady pace in 15 sec, how long will it take me to get from one floor to another if I walk up the escalator at a steady pace?

I liked this problem a lot.

The sum of the recipricals of two numbers is twice the product of their recipricals, what is the sum of the two numbers?

Have fun.
lyra

Source: MML

If the height of the storey is $h$, then your speed on the escalator is ${h\over t_{1}}$ and on the steps it's ${h\over t_{2}}$. Your combined speed is ${h\over t_{1}}+{h\over t_{2}}=\frac{h(t_{1}+t_{2})}{t_{1}t_{2}}$, hence you'll reach the next floor in $t_{0}=\frac{t_{1}t_{2}}{t_{1}+t_{2}}={60\over 7}$ seconds.

Thats exactly how I solved it, nice problems don't you think?

In one second, you travel $\frac{1}{20}+\frac{1}{15}=\frac{7}{60}$ of the way up, so you can get up in $\frac{60}{7}$ seconds.

$\frac{1}{x}+\frac{1}{y}=\frac{2}{xy}\rightarrow \frac{xy}{x}+\frac{xy}{y}=x+y=2$

[hide="My Solution"]

[hide="The first one"]

Say your constant speed is x, and the escalator's speed is y.

If x=15, then your speed up the stairs will be $\frac{1}{15}$ of the distance per second.  Meanwhile, if the escalator's speed is y, and y=20, then your speed with the escalator will be $\frac{1}{20}$ of the distance per second.  Your total speed per second will then be $\frac{1}{15}+\frac{1}{20}=\frac{4}{60}+\frac{3}{60}=\frac{7}{60}$ then $1\div\frac{7}{60}=\frac{60}{7}$ seconds.[/hide]

[hide="The second one"]

Say the numbers are x and y.  Then $\frac{1}{x}+\frac{1}{y}=2\left(\frac{1}{x}\times\frac{1}{y}\right)$.

Then,

$\frac{x+y}{xy}=\frac{2}{xy}$.

Then, multiply by xy.

$x+y=2$, thus the sum is 2.[/hide]

[/hide]


If the digits of X don't include zero,and X is divisible by S(X)   (S(X) is the sum of digits of X),then we call X a good number.
Prove that for every positive integer K,there is a good K-digit number.

See problem 7 of the ShortList IMO 1998. :)


algebra
polynomial
calculus
integration
Pascal\u0027s Triangle

I read something about constructing a polynomial using the elements in the "zero diagonal" of the finite differences table. Can somebody explain?

I'm too lazy to retype this cause I did it for homework last year, so I've attached it in the PDF.

There's an element of non-rigor to it: namely, it uses Pascal's Triangle, and the demonstration is thus limited to integer inputs. However, if you extend the binomial coefficient to be meaningful for non-integers, it gives the general polynomial interpolation. This is because if two polynomials have equal values at all integral inputs, they must be equal at all inputs.

For all it's worth though, I bet you already know it/have used it.


inequalities
function
inequalities unsolved

Show  that  $a^ab^bc^c \geq \displaystyle \left( \frac {a+b+c}3 \right)^{a+b+c}$.

This is just Jensen's inequality for the convex function $f(x)=x \cdot ln(x)$

Pierre.


quadratics
real analysis
real analysis unsolved

This is about flash scripting, actually. There's that thing called quadratic bezier spline,

$p_{0}(1-t)^{2}+2p_{1}t(1-t)+p_{2}t^{2}$,

right, and flash plugin has that "curveTo" command that apparently uses above expression for x(t) and y(t) to draw a curve between two points on screen. So, I was wondering how do I make it so that N points could be connected by smooth curve using N-1 such segments.

What I did originally was solving $x'_{i-1}(1) = x'_{i}(0)$ for subsequent control points ($p_{1,i}$) and same for y'-s:

$p_{1,i}= 2p_{2,i-1}-p_{1,i-1}$,

but this did not worked out well, because control points tend to jump around like crazy this way.

So, my next step was getting dx/dy to be same. I concluded that the curve will be still smooth if you multiply x' and y' by any positive number, and so my solution is now $p_{1,i}= p_{2,i-1}+a ((2p_{2,i-1}-p_{1,i-1})-p_{2,i-1}), 0 < a < 1$ (see example for a = 0.6 [url=http://img504.imageshack.us/my.php?image=58649373hf5.swf][u]here[/u][/url]; click a few times in white rectangle).

My problem is that I don't like the way "a" hangs in it as free parameter. I need some idea how to calculate it based on real curve data (e.g., $x_{i}, y_{i}$). The objective can be vaguely described as getting curve to look as nice as [url=http://img509.imageshack.us/my.php?image=49492299vs2.swf][u]Catmull-Rom spline[/u][/url] does.

Thanks for reading.

I've suddenly realized that my pitfall was attempting to solve 2D problem in 1D... so, get ready for another update.


if $S=\frac{(2^3-1)(3^3-1)(4^3-1)....(100^3-1)}{(2^3+1)(3^3+1)(4^3+1)....(100^3+1)}$. then which of the following values is closest to $S$?
$A)0.6,   B)0.67  C)0.667   D)0.6667   E)0.66667$
by the way this is a problem from a really old IRMO, it's pretty easy,
don't use computers, etc ... :lol:

[hide]
Since $(a^3-1)$ and $(a^3+1)$ can be factored into $(a-1)(a^2+a+1)$ and $(a+1)(a^2+a+1)$, we can factor both the numerator and the denominator.

$\frac{(2-1)(2^2+2+1)(3-1)(3^2+3+1) \ldots (100-1)(100^2+100+1)}{(2+1)(2^2-2+1)(3+1)(3^2-3+1) \ldots (100+1)(100^2-100+1)}$

It's easy to notice that $a+1=(a+2)-1$ so that the $(2+1)$ in the denominator will cancel with the $(4-1)$ in the numerator.  This means that everything will cancel except the $k \cdot \frac{(2-1)(3-1)}{(99+1)(100+1)}$ with $k$ being the rest of the stuff in the form of $(a^2+a+1)$  or $(a^2+a+1)$.

$a^2+a+1=(a+1)^2-(a+1)+1$ so that $(2^2+2+1)$ will cancel with $(3^2-3+1)$ leaving $k=\frac{(100^2+100+1)}{(2^2-2+1)}$ giving us the final answer of $\frac{3367}{5050}$ which is closest to $0.6667$.

To find an approximation by say that $\frac{3367}{5050}=\frac{2}{3}+\frac{1}{3 \cdot 5050}$.  From there we can say that $\frac{1}{15150} \approx \frac{1}{15*1000} \approx .000066$
[/hide]

[quote="amirhtlusa"]your answer is correct,
i solved it in a diffren way... :D[/quote]

which is...?


I'm playing a Chopin nocturine: Opus 9 No. 2
and a Mozart Concerto in B-flat K.595

I am currently working on Concerto by Nikolai Rimsky-Korsakov (Its a piece for trombone), and is written with a piano accompaniment as well.

Rimsky-Kosakov is a great composer! ;D
I like his Dance of the Tumblers for orchestra...

I'm playing a bach prelude and fugue (#21 in WTC 1), haydn piano concerto in d major, and a chopin prelude, maybe then 1 mentioned above, haven't decided yet

[quote="istos"]I love that nocturne! ;]
I'm playing a couple of different pieces right now.
My favorite is the Fantasie-Impromptu by Chopin.[/quote]
You know there's a reason why composers choose not to publish things.
Fantasie Impromptu is what I'm referring to.

[quote="myyellowducky82"]but my favorite piece by mozart is his last fantasy[/quote]
Was that a reply to my comment? Because that would make no sense.

GO MENDELSON!!! WHOOT! (i just spelled is name wrong again, didnt i?)

lol, I think it's Mendelssohn. I like him too.
  YAY MENDELSSOHN!

i don't know the spelling, I remember playing some of his songs without words a like 2 yrs ago, haven't played anything since, his name has an h right?

Ever since I came to university, it's really hard to find time to play again.  My project for the summer is Chopin's Nocturne in E minor...No. 19 i think?
Oh, and Chopin's Nocturne no. 9 op 2 is a beautiful piece!

I'm playing Abegg Variations by Schumann and Mozart Concert No. 23 both on the piano.

i'm playing alot of pieces including etude op 10 no 8 by chopin 4th movt of some sonata by prokofiev tijuca by milhaud a prelude and fugue by shostokavich( i dunno how to spell) and 3 polonaises by chopin

I don't know if this is considered a piece but... What Does The Fox Say By: Ylvis... Someone made me do it... Other than that I'm playing fur elise.

I'm playing Seitz Student Sonata 2nd one 1st movement.

Haydn Sonata Hob. XVI: 32
Beethoven Concerto No. 4
Beethoven Sonata Op. 2 No. 3
Chopin Scherzo No. 1
Chopin Etude Op. 25 No. 6
Chopin Barcarolle Op. 60
Prokofiev Sonata No. 7
these are all piano solo pieces btw (except for the concerto obviously.)

Right now:

Liszt Concert Etude No. 2 ("La leggierezza," or "lightness")
Prokofiev Sonata No. 1 [it's just one movement]
Ravel Concerto [starting with first movement]

Hm @above I am also working on Ravel concerto 1st movement!

So piano:

Debussy: L'Isle Joyeuse
Chopin: 3rd and 1st Ballades (polishing 3rd, starting 1st)
Rachmaninoff: Prelude Op 23 No 5 (G minor)
Beethoven: Sonata Op 31 No 2 ("Tempest" -- working on 1st and 3rd movements right now)\
Ravel: Piano Concerto 1st movement (the G major one -- not the left hand one XD)
Re-learning Beethoven: Moonlight Sonata

Violin:

Kreisler: Sicilienne and Rigaudon (that's it -- I haven't had a lesson in a while)

Sicilienne and Rigaudon is (according to my teacher) something I play well so she keeps making me audition with it for everything :/

Besides this I'm playing Scene de Ballet by de Beriot and the 3rd Mozart concerto.

Im doing the art of finger dexterity no. 14
(I think)
And mozart's k333. (Sonate of course)

Currently doing Scott Joplin's [i]Maple Leaf Rag[/i].

I play piano btw. Doing Chopin's Valse Opus 69 #2 which is above my level (whatever that is... I play for fun so I don't care)

I'm trying to make a piece (Its an assignment).

[quote=Dojo][quote="Quickster94"]Cat and Mouse is a lot of fun to play.  It's something I think most people either really connect with or struggle through.

I personally had planned to play it, but after looking at the sheetmusic lost enthusiasm.

It's definitely a very fun piece, actually not as difficult as it seems[/quote]
Gah, seeing this 10 year old prodigy play it made me want to cry.[/quote]

I just noticed this. I'm nearly certain I know this prodigy who you speak of in person. She goes to my music school and I know her decently well- and I'm pretty sure she's the correct age.
Actually no, she's thirteen now, but look up "Avery Gagliano - From the Top" on YouTube. :)

It's a piano piece by Aaron Copland -- rather fun "scherzo" type piece.


Mrs. Smith can knit one pair of mittens with a ball of yarn six inches in diameter. How many pairs of mittens can she knit with a ball of yarn twelve inches in diameter?

[hide]8 pairs of mittens?

$4/3\times\pi\times r^{3}$

$4/3\times\pi\times3^{3}$=

$36\pi$ amt. of yarn needed for one pair.

$4/3\times\pi\times6^{3}$=

$4/3\times\pi\times216$=

$288\pi$

divide by 36 you get 8.
btw how do you do pi in latex?[/hide]

[size=75][color=darkred]$\pi$ is done by "\pi" and also, try to use "\cdot", $\cdot$, or "\times", $\times$, for multiplication instead. Also, don't forget hide tags.
-nebula42[/color][/size]

[quote="sonny"]8 pairs of mittens?

$4/3*pi*r^{3}$

$4/3*pi*3^{3}$=

$36pi$ amt. of yarn needed for one pair.

$4/3*pi*6^{3}$=

$4/3*pi*216$=

$288pi$

divide by 36 you get 8.
btw how do you do pi in latex?[/quote]
That is correct, please hide your answer, and $\pi$ is \pi. :D

[quote="ragnarok23"][quote="jhredsox"]i thought a sphere that is double in diameter is 4x bigger (or is that just circles?)?
-jorian[/quote]
This is only for circles. The ratio between the two diamters for a circle is 2, therefore the ratio between the areas is 4.
However, this is not the case with spheres, or any other 3-D figure. When the ratio between the diamters is 2, the ratio between the volumes is $2^{3}=8$.[/quote]
THink about it, when finding volume you cube the edge then multiply it by other stuffs. So the ratio of the edges is 2, which is cubed to be 8... (as ragnorak said).
[size=75][color=orange]
[C*p can always be replaced with some other word, unless c*p is in the word crapulent. No bad language, and c*p counts. Remember, 6th graders do MC too.-Billy][/color][/size]

[hide]
Scan knit a pair of mittens with $3^{3}\times\frac{4}{3}\pi$ inches of string.
With the other one, $6^{3}\times\frac{4}{3}\pi$
$\frac{6^{3}\times\frac{4}{3}\pi}{3^{3}\times\frac{4}{3}\pi}=\frac{216}{27}=8$
Since the bottom is one pair, the ball of yarn with 12 inches of diameter is enough for 8 pairs.[/hide]

[hide="solution"]We find the areas of the two:

$\frac{4}{3}\pi3^{3}$

and

$\frac{4}{3}\pi6^{3}=8 \times \frac{4}{3}\pi3^{3}$

There is 8 times of yarn in the bigger ball, so she can knit 8 mittons.[/hide]

[quote="nutz_for2.718281828"][hide="solution"]There is 8 times of yarn in the bigger ball, so she can knit 8 mittons.[/hide][/quote]
A couple things I wanted to point out. [hide="info about the problem"]Your solution doesn't show how there are 8 times the yarn in the bigger ball, which just so happens to be the whole point of the problem[/hide]
Secondly, it is ridiculous, the number of solutions that have been posted to some problems that are identical. Here is a suggestion. If a solution like yours has been posted $\geq 2$ times, than what is the point of posting yours. The forums are getting clogged by stuff like this and it is spam. If you want to solve the problem, do so and compare it to others' answer ok. (Note: I am not being specific when I say you).

[quote="13375P34K43V312"]Dude, you're not a moderator. If you want to complain about this, tell the Admins, then they'll make you one.[/quote]
I don't want to be a mod. I am just saying that this is clogging the forums and people are reposting solutions to problems to increase their post count like crazy.

:ninja: wow. im posting on aops at 2 in the morning. im officially a math ninja.  :ninja:

[hide]you dont need to really find the areas.  the only thing that differs are the radii.  and for the volume of a sphere, the radii are cubed.  so you just take (6^3)/(3^3)=8.

(its actually a proportion, but since its 1 for the first, you dont really have to do that  :) )[/hide]


calculus
integration
calculus computations

$ f$ is increasing on $ [0,a]$ and $ \forall_{0\leq x < a},0\leq f(x) < 1$, so $ I_n$ is a bounded and decreasing sequence, so $ In$ is convergent to a limit $ L$.
For all $ 0 < u < 1$ we can write $ 0 \leq I_n \equal{} \int_{0}^{a}f^n(t).dt \equal{} \int_{0}^uf^n(t).dt \plus{} \int_u^af^n(t).dt$$ \leq uf^n_{(u)} \plus{} (a \minus{} u)f^n_{(1)}$
It's easy to see that $ f^n_{(u)}\rightarrow 0$, so $ 0\leq L \leq a \minus{} u$ for all $ 0\leq u < a$ , and finally $ L \equal{} 0$
 :cool:

[quote="Diogene"]so $ 0\leq L \leq a \minus{} u$ for all $ 0\leq u < a$ ,[u] and finally[/u] $ L \equal{} 0$
[/quote]

Is this by taking $ u \to a$ ?


Can someone give me proofs for multiplicative arithmetic functions.  These are functions for which f(mn)=f(m)f(n) iff gcd(m,n)=1.  Some examples are :phi:(n),  :stddev:(n), d(n),  :mu:(n).  If you have any more functions that are multiplicative, or proofs concerning these functions (even if they are not about being multiplicative) please let me know.

Thanks


often i make a poll in the G&FF forum but it keeps disappearing
i try to create it again but it disappears again  :o
idk how to use screenshots to show this though...

What is the thing dissapearing?
a) the whole thread
b) only the poll

If a):
Are you sure it isn't spam?
If it isn't, then scroll down, or check both Games and Fun Factory forums.

If b):
... I don't know :lol:

That always happens with me, but whenever I try to remake the options, it shows the options doubled. THen I go to the forum and reclick my topic, and I don't see the poll either?


[quote="Virgil Nicula"]Let $ ABC$ be a triangle. Prove that $ b(a^2 \minus{} bc)(c^2 \minus{} b^2) \equal{} a(c^2 \minus{} ab)(a^2 \minus{} c^2)\ \Longleftrightarrow\ a \equal{} b \equal{} c$ .[/quote]
$ b(a^2 \minus{} bc)(c^2 \minus{} b^2) \equal{} a(c^2 \minus{} ab)(a^2 \minus{} c^2)$
$ \iff ac^2(c^2 \minus{} a^2) \plus{} ba^2(a^2 \minus{} b^2) \plus{} cb^2(b^2 \minus{} a^2) \equal{} 0$
I will prove that the inequality :
$ ac^2(c^2 \minus{} a^2) \plus{} ba^2(a^2 \minus{} b^2) \plus{} cb^2(b^2 \minus{} a^2) \geq 0$
I think the inequality is right because it' is similar the inequality in IMO 1981 :
$ ac^2(c \minus{} a) \plus{} ba^2(a \minus{} b) \plus{} cb^2(b \minus{} a) \geq 0$
I think let $ a \equal{} y \plus{} z, b \equal{} z \plus{} x, c \equal{} x \plus{} y$.

[quote="conan_naruto236"][quote="Virgil Nicula"]Let $ ABC$ be a triangle. Prove that $ b(a^2 \minus{} bc)(c^2 \minus{} b^2) \equal{} a(c^2 \minus{} ab)(a^2 \minus{} c^2)\ \Longleftrightarrow\ a \equal{} b \equal{} c$ .[/quote]
$ b(a^2 \minus{} bc)(c^2 \minus{} b^2) \equal{} a(c^2 \minus{} ab)(a^2 \minus{} c^2)$
$ \iff ac^2(c^2 \minus{} a^2) \plus{} ba^2(a^2 \minus{} b^2) \plus{} cb^2(b^2 \minus{} a^2) \equal{} 0$
I will prove that the inequality :
$ ac^2(c^2 \minus{} a^2) \plus{} ba^2(a^2 \minus{} b^2) \plus{} cb^2(b^2 \minus{} a^2) \geq 0$
[b]I think the inequality is right because it' is similar the inequality in IMO 1981 :[/b]
$ ac^2(c \minus{} a) \plus{} ba^2(a \minus{} b) \plus{} cb^2(b \minus{} a) \geq 0$
I think let $ a \equal{} y \plus{} z, b \equal{} z \plus{} x, c \equal{} x \plus{} y$.[/quote]
Why think you that it is right ? Only because it is similarly to .... ?!

[quote="Virgil Nicula"]Let $ ABC$ be a triangle. Prove that $ b(a^2 \minus{} bc)(c^2 \minus{} b^2) \equal{} a(c^2 \minus{} ab)(a^2 \minus{} c^2)\ \Longleftrightarrow\ a \equal{} b \equal{} c$ .[/quote]
My solution :)

I couldn't find your solution. Please, would you post it ?

Thanks


The inequality is not valid.
Let $z \rightarrow 0$, $y=4$, $x=1$. ;)


abstract algebra
calculus
integration
vector
algebra
function
domain

I need help in the following question:

Let M be a module over R, an integral domain. Suppose M has rank n (the maximal number of linearly independent vectors), and x1, ..., xn be such a set of linearly independent vectors of M. Let N be the submodule generated by these elements. Prove that N is isomorphic to R^n and that the quotient M/N is a torsion R-module.

Suddenly struck with an idea: Can I pass to the field of fractions of R (sorry, I should have clarified that by domain, it is commutative)? Let it be F.

Then we get N' = Fx1 + Fx2 + ... + Fxn which is a vector space over F, dimension n (a priori <= n. But if we get a dependence relation in F, clearing factors gives a dependence relation in R).

Then if we treat M as a vector space over F, then its dimension is no more than n. N' incl in M and dim (N') >= dim (M).

Now let y be any element in M. Then get y = sum (ai)(xi) = sum (pi/qi) (xi).

Then (product of qi's) y = sum (ri) (xi), each ri in R, i.e.,

M/N is a torsion module.

Please correct me if you find the argument dubious.


geometry
MATHCOUNTS
trigonometry
Pythagorean Theorem
trig identities
Law of Cosines

Here was a quite straightforward idea I had to prove Heron's forrmula. I didn't find it anywhere else on the internet. Has anyone else proved it this way before?

Let the sides of a triangle be a, b, and c. Draw the altitude to c, call the length h. Call the area of the triangle A. Since ch/2=A, h=2A/c. Use the Pythagorean theorem on both right triangles to obtain $ sqr(a^2\minus{}(4A^2/c^2))\plus{}sqr(b^2\minus{}(4A^2/c^2))\equal{}c$ Solving for A gives Heron's formula.

(I wasn't sure how to type square roots).

I'm fairly sure I've seen this proof in some geometry text book or other. Solving for $ A$ involves some nasty algebra, though.

Square roots are
[code]
\sqrt{stuff}
[/code]

[quote="dgreenb801"]Here was a quite straightforward idea I had to prove Heron's forrmula. I didn't find it anywhere else on the internet. Has anyone else proved it this way before?

Let the sides of a triangle be a, b, and c. Draw the altitude to c, call the length h. Call the area of the triangle A. Since ch/2=A, h=2A/c. Use the Pythagorean theorem on both right triangles to obtain $ sqr(a^2 \minus{} (4A^2/c^2)) \plus{} sqr(b^2 \minus{} (4A^2/c^2)) \equal{} c$ Solving for A gives Heron's formula.

(I wasn't sure how to type square roots).[/quote]
iv never seen this, but i like it. it's straightforward and easy, even a basic mathcounts student could come up with it. the proof i learned required using K=1/2*ab*sinC. sinC=sqrt{1-cos^2c} and than using law of cosines and a lot of redundant and straightforward algebra. :-)

It's probably so common that no one bothers to post it...


Έστω ακολουθία θετικών αριθμών $ a_n$ ώστε η σειρά $ \sum_{n\equal{}1}^\infty a_n$ να συγκλίνει.

Τί μπορείτε να πείτε για τη σύγκλιση της σειράς $ \sum_{n\equal{}1}^\infty a_n \log n$? Συγκλίνει πάντα ή υπάρχει περίπτωση να αποκλίνει?

Cheerio,

Durandal 1707

[quote="Demetres"][hide]$ 1/n, 1/(n\log{n}), 1/(n(\log{n})^2)$ και integral test[/hide][/quote]

Κάτι μάλλον δεν καταλαβαίνω... το $ \int_1^\infty \frac{dx}{x \log^k x}$ αποκλίνει πάντα,  ένα $ \log$ παραπάνω ή παρακάτω στον παρονομαστή δε θα αλλάξει κάτι :huh:

Τί χάνω?

Durandal 1707

[quote="Durandal"][quote="Demetres"][hide]$ 1/n, 1/(n\log{n}), 1/(n(\log{n})^2)$ και integral test[/hide][/quote]

Κάτι μάλλον δεν καταλαβαίνω... το $ \int_1^\infty \frac {dx}{x \log^k x}$ αποκλίνει πάντα,  ένα $ \log$ παραπάνω ή παρακάτω στον παρονομαστή δε θα αλλάξει κάτι :huh:

Τί χάνω?

Durandal 1707[/quote]

Αν $ f(x) \equal{} \minus{}\frac{1}{\log(x)}$ τότε $ f'(x) \equal{} \frac{1}{x (\log{x})^2}$. Αυτά που αποκλίνουν είναι τα $ \int \frac{1}{x \log{x} \log{(\log{x})}}$ κλπ.

[quote="Demetres"]Αν $ f(x) \equal{} \minus{} \frac {1}{\log(x)}$ τότε $ f'(x) \equal{} \frac {1}{x (\log{x})^2}$. Αυτά που αποκλίνουν είναι τα $ \int \frac {1}{x \log{x} \log{(\log{x})}}$ κλπ.[/quote]

Ωχ, έλεος, τα θυμόμουν ανάποδα και το είχα βάλει για τα nested! :rotfl: :rotfl:

Meeeeh, too bad... έτσι μου πέθανε το δεύτερο ερώτημα που ήθελα να βάλω (και που τώρα καταλαβαίνω γιατί μου φαινόταν παράξενο :P).

Cheerio,

Durandal 1707


analytic geometry
real analysis
real analysis unsolved

Is it possible to construct a continuous and differentiable  curve in the cartesian plane such that  both the co- ordinates can never be simultaneously rational  ?

take 0 < x,y <1

Unless, I'm missing something, $ f(x)\equal{}\pi x$.  If $ x\equal{}\frac{p}{q}$, then $ y\equal{}\frac{p\pi}{q}\notin\mathbb{Q}$ and if $ y\equal{}\frac{a}{b}$, then $ x\equal{}\frac{a}{b\pi}\notin\mathbb{Q}$.  Since you specified to work on $ (0,1)\times (0,1)$, $ (0,0)$ is not a point we have to deal with.

And it's not hard to modify this for all of $ \bf R$: for example, $ y \equal{} \sqrt{2} \plus{} \pi x$ works nicely.


function
floor function
logarithms
algebra unsolved
algebra

[color=blue][b]Prove the following:

If $n \in \mathbb{N},n\geq 2$
then $\left\lfloor {\sqrt n } \right\rfloor + \left\lfloor {\sqrt[3]{n}} \right\rfloor + \cdots + \left\lfloor {\sqrt[n]{n}} \right\rfloor = \left\lfloor {\log _2 n} \right\rfloor + \left\lfloor {\log _3 n} \right\rfloor + \cdots + \left\lfloor {\log _n n} \right\rfloor$
[/b][/color]

The LHS is:
"number of squares $\leq n$" + "number of third powers $\leq n$" ...
The RHS is:
"number of powers of $2$ $\leq n$" + "number of powers of $3$ $\leq n$" + ...

When considering that the LHS contains also $+1$ for every summand ($1$ is a perfect power of any degree), these $n-1$ extra-$1$'s equal the $n-1$ 'first' powers on the RHS, thus RHS=LHS (we have $n-1$ plus the number of perfect powers $p^d,d>1,1<p\leq n$ on both sides, where some powers are counted in multiple ways).


Trick or Treat Part 1

We learn the importance of distinguishability in counting problems by counting the number of ways to distribute candy to children at Halloween.

It was kinda hard to hear Mr. Rusczyk through the mask.

This video is AWESOME!! Why did they take it away?!

They just changed the name to distinguishability


He has defined a sequence

$a_1=a_2=1$
$\forall n\geq 3$ $a_n =a_{n-a_{n-1}} + a_{n-a_{n-2}}$

Tell us what did you discover about this sequence
-Is it well defined ?
-Comportement ?

[url=http://mathworld.wolfram.com/HofstadtersQ-Sequence.html]Here[/url]'s some info about this sequence. Apparently, lots of questions about this sequence are still unsolved.

Hmm, from what I know, it seems to be defined under the restriction Q(n) < n.
However, other variations of the sequence by altering the initial conditions yield a quasi-periodic sequence.

This has some interesting observations:
http://webhome.cs.uvic.ca/~ruskey/Publications/MetaFib/FiboHofs.pdf

Daniel


Right now, I'm using this page (the "Using Asymptote in LaTeX" part): http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration

I'm at the step where you refresh the database, but every time I do it, I get a Problem Report saying:

MiKTeX Problem Report
Message: Windows API error 32: The process cannot access the file because it is being used by another process.

Data: C:\Documents and Settings\chaic\Local Settings\Application Data\MiKTeX\2.6\miktex\config\76677b985678706e11dce4472d6544f4.fndb
Source: C:\work3\miktex-2.6\Libraries\MiKTeX\Core\win\winFile.cpp
Line: 520
MiKTeX: 2.6
OS: Microsoft Windows XP Professional Service Pack 2 (Build 2600)
SystemAdmin: yes
PowerUser: no
SharedSetup: unknown
BinDir: C:\Program Files\MiKTeX 2.6\miktex\bin
Root0: C:\Documents and Settings\chaic\Application Data\MiKTeX\2.6
Root1: C:\Documents and Settings\chaic\Local Settings\Application Data\MiKTeX\2.6
Root2: C:\Program Files\MiKTeX 2.6
Install: C:\Program Files\MiKTeX 2.6
UserConfig: C:\Documents and Settings\chaic\Application Data\MiKTeX\2.6
UserData: C:\Documents and Settings\chaic\Local Settings\Application Data\MiKTeX\2.6

and then I can't use pictures in my pdf's. How do I fix this, or is there an alternate way to incorporate pictures?

I would try refreshing the file database immediately after booting up to ensure that nothing else is accessing the file. You could also try the free programs [url=http://www.dr-hoiby.com/WhoLockMe/]Who Lock Me?[/url] and [url=http://ccollomb.free.fr/unlocker/]Unlocker[/url] to see what is preventing access to the database and perhaps free it up.

Otherwise you could try re-installing MikTeX over the top of the old version in case that helps.

Temporarily, you can put your tex file in the Asymptote directory and LaTeX will then find Asymptote or else create the Asymptote image using the command line (or via TexnicCenter) then use [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Pictures]includegraphics[/url] to include the picture as you would normally do for pictures.

Ok, I refreshed it after rebooting, and that worked fine (thanks). But now, when I try the example file, the pdf comes out without a picture. The text and math is there, but no picture.

There are 3 stages:
1. compile LaTeX=>PDF
2. run Asymptote
3. re-compile LaTeX=>PDF
Check that 1. has created an asy file and that 2. has created the pdf

Do I run Asymptote on the same .tex file, or do I open the .asy file that was created and run it on that?

1. did create an .asy file, but nothing happened on 2. (the Asymptote window opened momentarily, and then closed)

Check Problem 2 at http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Help
If that doesn't help then [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]Showing Asymptote error messages in TeXnicCenter[/url] will tell you what is going wrong with Asymptote.

Ok, the troubleshooting didn't help much, and I can't find any .log file.

What files have you got? If the original fle is called test.tex then
1. creates test.log, test.asy, test.pdf (and other unimportant files)
2. asymptote creates test_1.pdf with a smiley (and test_asy.log if you are using the batch file)
3. recreates test.pdf.

I never got any .log after 1, and then 2 doesn't create anything.

Should the argument for customizing Asymptote be
-batchView -tex "latex" %tc
or
-batchView -tex "pdflatex" %tc

[quote="le00327146"]I never got any .log after 1[/quote]
That is serious because it means that LaTeX isn't working at all, since it [i]always[/i] produces a log file which is reproduced in the bottom window of TexnicCenter. What error does TexnicCenter give when you compile the tex file in step 1?
You can't proceed until this is sorted, and a .asy file won't be produced without it.

[quote="le00327146"]Should the argument for customizing Asymptote be
-batchView -tex "latex" %tc
or
-batchView -tex "pdflatex" %tc[/quote]

You are producing pdfs so the instructions say:
[quote="PDF format and beyond"]PDF output configuration
Although gsview can configured as the pdf viewer, we recommend using the default pdf viewer, Acrobat Reader.

Now open TeXnicCenter. Go to the Tools menu and click Customize, then go to the Tools tab. Click to modify your Asymptote tool that you made in the Interactive Mode section. In the Arguments line, simply change latex to pdflatex:

 -batchView -tex "pdflatex" %tc[/quote]

Then maybe I'm not seeing the .log, since it does make an .asy file, and there aren't any red X's when I do ctrl-F7 on the .tex file in step 1. But when I look in the Asymptote folder, there's nothing that ends in .log

would ghostscript have anything to do with the fact that nothing happens in step 2?

The log is in the same directory as the tex file, not in the Asymptote folder. The asy file will be there too.

I originally had everything saved in the Asymptote folder. I tried saving them somewhere else, and there's still nothing ending in .log, but a .asy file was created.

Does the .log file look like a little legal pad? I opened that and it looks similar to the stuff in the bottom window of TexnicCenter.

And now, after all that toying around, Asymptote doesn't even work by itself normally anymore  :(

If the tex file is called test.tex then the log file is called test.log. You [b]do[/b] have the file extensions turned on don't you? Without it you can't see what files are which as test.tex looks the same as test.log or test.pdf and you won't know what type your asymptote files are saved as.
Tools -> Folder Options -> View -> Untick [i]Hide extensions for known file types[/i] -> OK

ohh.....ok now I have the .log files. Sorry

Maybe I'm wrong about this (I've noticed that I've been wrong a lot lately), but it seems like the "PDF Format and Beyond" section is entirely separate from the "Using Asymptote in LaTeX" section. Should it be
-batchView -tex "pdflatex" %tc if I'm just trying to put a picture in my LaTeX pdf?

YAAAY! Never mind I fixed it. One more question: how do I shift the picture to the right within the pdf?

To centre the figure use the center environment:
[code]\begin{figure}[h]
\begin{center} % add this line
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{center} %add this line
\end{figure}[/code]and to put it on the right
[code]\begin{figure}[h]
\begin{flushright} % add this line
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{flushright} %add this line
\end{figure}[/code]


In a class of 180 eighth-grade students, $ 80\%$ went to the Science Museum. Ninety percent of the boys and 90 girls went on this trip. How many girls are in the eighth-grade class?

I asked this question, and I did not understand it.  But now, I do, because of Kouichi Nakagawa's solution.  However, the problem asks for the number of girls in the class, and Kouichi Nakagawa found the number of boys instead.  Therefore, our answer is actually $ 180\minus{}60\equal{}\boxed{120}$.

Oh, no.
I calculated a boy and girl with having misunderstood a problem sentence.
Thank you for indication.


function
inequalities
inequalities unsolved

1. Let $ x,y,z$ be positive real numbers satisfying $ x\plus{}y\plus{}z\equal{}1$. Show that
$ \left( 1\plus{} \frac{1}{x} \right) \cdot \left( 1\plus{}\frac{1}{y} \right) \cdot \left( 1\plus{} \frac{1}{z} \right) \ge 64$.
When does equality hold?

It considers the function $ f(x) \equal{} ln(1\plus{} \frac{1}{x})$ and applies jensen inequality  :)

This is probably the easiest solution to the given problem:
\[ \left( 1 \plus{} \frac {1}{x} \right) \cdot \left( 1 \plus{} \frac {1}{y} \right) \cdot \left( 1 \plus{} \frac {1}{z} \right) \ge 64
\]

\[ \left( \frac {1 \plus{} x}{x} \right) \cdot \left(\frac {1 \plus{} y}{y} \right) \cdot \left(\frac {1 \plus{} z}{z} \right) \ge 64
\]

\[ \left( \frac {x \plus{} x \plus{} y \plus{} z}{x} \right) \cdot \left(\frac {y \plus{} y \plus{} x \plus{} z}{y} \right) \cdot \left(\frac {z \plus{} z \plus{} x \plus{} y}{z} \right) \ge 64
\]
So we should prove
\[ (x \plus{} x \plus{} y \plus{} z)\cdot (y \plus{} y \plus{} z \plus{} x) \cdot (z \plus{} z \plus{} x \plus{} y) \geq 64xyz
\]
But from AM-GM inequality, we have:

$ x \plus{} x \plus{} y \plus{} z \geq 4 \cdot \sqrt [4]{x^2yz}$

$ y \plus{} y \plus{} x \plus{} z \geq 4 \cdot \sqrt [4]{xy^2z}$

$ z \plus{} z \plus{} x \plus{} y \geq 4 \cdot \sqrt [4]{xyz^2}$

Multiplying last three inequalities will yield the result

It is pure Muirhead.
Multiply through by xyz, expand and simplify, you get
$ xy \plus{} yz \plus{} xz \plus{} 2 \ge 63xyz$
$ (x \plus{} y \plus{} z)(xy \plus{} yz \plus{} xz) \plus{} 2(x \plus{} y \plus{} z)^3 \ge 63xyz$
There are 63 terms on each side, but every term of the left side majorizes the right side and hence it is greater by Muirhead.
Of course, shaam did it most elegantly  :lol: .

here is my soln and most elegant one according to me

using holders we have the given expression>={(1 + [1/(xyz)^1/3] )^3}

but it is known that [1/(xyz)^1/3]>=3  (am,gm since x+y+z=1)

hence we have {(1 + [1/(xyz)^1/3] )^3}>=(4)^3=64

hence qed........ :)

$ \prod_{cyc} \left(1 \plus{}\frac{1}{x}\right) \ge 64 \iff p\plus{}q\minus{}63r\plus{}1 \ge 9\minus{}27q \ge 9\minus{}9p^2 \equal{} 0$ :)


If you are doing a maths problem which gives space for interpretaition how do you know what to write when the question ends saying EXPLAIN :?:

Write the answer first, then show your solution.


How do people do a science fair?

I doubt that I can come up with anything that has not already been discovered considering the fact that there are people with PhDs in science and math.

Also, where do you get the equipment and stuff for a science fair?

I want to participate in a science fair, but I am not sure if I have the ability and resources to do a successful one.

Your standard science fair does not require you to do original research.  For example, if you were submitting a mathematics research paper, you might study something well-known but advanced (darkquantum is doing a project on Pell Equations).

How would you be "researching" something like that?

The research has already been done for you by countless other mathematicians. Do you just open up a number theory book and summarize what other people did?

Thanks

It becomes more of a research paper in the sense of a history research paper.  As in history, however, you have some obligation to put a slant on it -- try to prove simpler cases of a larger theorem on your own (for example, solve certain Pell Equations without resorting to computers or continued fractions).  Explain the importance of something.  Relate two different things.  There are lots of things you can do with other people's work that isn't original but still interesting.  I know someone who did a project on the Game of Life, which has been extensively studied, by changing the rules slightly and seeing what happened.  There are lots of different ways to go.

If you want to do it in a non-math field, I can't really help you, though.

ever wondered why there are so many different textbooks which have the same material but are very very different in quality?

Yeah, but most of the, say, siemens semifinalists have projects that are original and about things that have not been delved into already by other people, right?

Siemens-Westinghouse and Intel are very different.  They expect original research.  That is something you need to do over a long period of time -- usually at least the summer.  I worked on mine over the course of junior year and a good bit at the beginning of senior.  For normal science fairs, however, that is not a reasonable expectation.

http://www.sciserv.org is the homepage of the Science Service, who are in charge of a variety of competitions including the Intel Science Talent Search, the Intel International Science and Engineering Fair, and I believe several others.  There's also the Junior Science and Humanities Symposium and the Siemens-Westinghouse contests, which I believe are independant of Science Service.  If they're what you're interested in, look them up.

Joel, what project did you submit for siemens/intel? Did you do all the research by yourself, or did you have a helping professor?

I did a project on game theory -- I created a full analysis of a 2-player game.  There are references to it around here somewhere.  If you want, I could PM you something I wrote for a couple of other people explaining all the background of what I did.

I did all the research myself.  I had a mentor, Professor Joel Spencer at NYU, who initially got me started on the project and who I met with every week or two to check up and make sure everything was going okay and show him what I'd done.  All of the math was mine, though.

Maybe you aren't winning because the teachers know you're just doing the same project over and over again.  Or maybe there's something they don't like about the project.  In either case, resubmitting the project each year won't help you at all, since in the former case the teachers will just get tired of you turning in the same thing (they may think "he's submitting the same work and not actually spending any time on it since it's all ready), and in the latter one, no matter how many times you submit it the thing they consider wrong with it will still be there.

Oh, I think I misunderstood...

When you said "in 6th grade I did the same thing" and the same for 7th grade, I assumed that meant you did the [i]same project[/i] each of those years.  Now, I take what you said to mean you submitted projects each of those years but never placed.  Anyway, good luck this year.  Sequences are interesting.


Evaluate

\[\sum_{n=1}^\infty \frac{\phi(n)}{2^n-1}.\]

More generally, prove that
\[\sum_{n = 1}^\infty \frac{\phi(n) x^n}{1 - x^n} = \frac{x}{(1 - x)^2}.\]

The coefficient of $x^n$ in the series on the left is $\displaystyle\sum_{d|n}\phi(d)$, while in the series on the right it's $n$, and it's well-known that these are equal.

Could you explain why your statement about the series on the left is true? I understand the part on the right.

Let $\displaystyle S_d=\frac{x^d}{1-x^d}=x^d+x^{2d}+x^{3d}+\ldots$. $S_d$ contains $x^n$ iff $d|n$, and since the coefficient of $x^n$ in $\phi(d)S_d$ is $\phi(d)$, it means that when we sum up all the coefficients of $x^n$ in all $S_d,\ d|n$ we get $\displaystyle\sum_{d|n}\phi(d)$.


ratio
trigonometry
complex numbers
geometry proposed
geometry

let $P$ be a regular n-gone in $R^2$. Suppose that  3 of its vertices are lattice points, 2 of them being consecutive vertices. Prove that $P$ is a square !
nice isn't it  :)

Let $ABC$ be the points, s.t. $A,B$ are consecutive vertices of the polygon, and assume WLOG that $C$ is, in fact, $O$, the origin of the plane.

Let $a,b\in\mathbb Z[i]$ be the complex numbers corresponding to $a,b$. We get $\frac ba=\sqrt r\cdot e^{\frac{\pi i}n}$, where $r$ is some rational number (in fact, it's the ratio $\frac{|b|}{|a|}$). Take $\sqrt r$ to the left side and square everything to get $e^{\frac{2\pi i}n}=\frac {a^2}{rb^2}\in\mathbb Q[i]$. In other words, both $\cos\left(\frac{2\pi i}n\right)$ and $\sin\left(\frac{2\pi i}n\right)$ are rational, and it's easy enough from here.

This looks too easy to be ok. What am I doing wrong? :?

Note that $\left[\mathbb{Q}\left(\cos\frac{2\pi}{n}\right):\mathbb{Q}\right]=\frac{\phi (n)}{2},\quad \forall n>2.$


[img]http://www.bearsystems.com/PegLegBates/PegLegBates.jpg[/img]

Peg Leg Bates has peg legs that last three years and peg legs that last four years.

If his total peg leg collection can last him 17 years, how many peg legs does he have in his collection?

Peg Leg Bates has peg legs that last three years and peg legs that last four years.

If his total peg leg collection can last him 17 years, how many peg legs does he have in his collection?

[hide]
First, I'm going to write an equation that won't necessarily solve the problem, but will help us in solving it.

3x+4y=17
x=peg legs that last for 3 years
y=peg legs that last for 4 years

Now, I would just guess and check. I'd take away multiples of 3 from 17 until we get an answer that is a multiple of 4.

17-3=14. Not a multiple of four.
17-6=11. Not a multiple of four.
17-9=8. A multiple of four!

So we have three peg legs that last for 3 years and two peg legs that last for 4 years. To check, plug back into the equation.

3*3+4*2=9+8=17!

This checks, so our answer is correct. In total, Mr. Bates has five peg legs.
[/hide]

[hide]try seeing if it works with 4 4-years and then three and so on

you get 2 4-years and 3 3-years[/hide]


Source: Olympiad Math Madness: Kid's Stuff Problem 2

A king went around a 8x8 chessboard, stopping in each square exactly once and ending up in the beginning square. His path was drawn by connecting the centers of the squares he was in, and it did not cross itself.
a) Think of an example in which the king made exactly 28 horizontal and vertical moves.
b) Prove that the king couldn't have made less than 28 horizontal
and vertical moves.
c) What is the maximum and minimum distances the king could have
travelled, if each square has a side of 1 inch?

Do you mean 28 vertical moves and 28 horizontal moves or the total number of horizontal and vertical moves was 28?

Both.  That is, the sum of the number of both horizontal and vertical moves.

But I'm not sure about this...it doesn't make much sense.  This should teach me to do a problem before I post it... I'm going to delete this if no one minds?

Don't delete it. I believe that it is correct. At least, my basic work on it tells me that it is true.

Alright.  I deleted any referenced to deleting, but the problem stands.

[hide]a.  I found an example for which the king moves horizontally and vertically 28 times.

b.  I have yet to figure out how to prove that that is the least number of such moves possible.

c.  Minimum distance = 64 inches. (Every move is horizontal of vertical)

Maximum distance = 28 + 36:rt2: inches. (The least possible number of horizontal and vertical moves)[/hide]

Can you describe your path? I think it will help you solving part b.


heyy so ive been stuck  on this question. You are given two segments of a hypotenuse and the line that divides it up is perpendicular to the hypotnuse, the question is asking for the area, how would find it?

In the future, look at the Contests/Resources section to find the original thread for the problem. Keep all the discussion confined to one thread; that way, if someone else has the same question later, it doesn't need to be asked twice.


If f : R-> R  (R = set of real nos) such that

  f(x) - f(y) |  <= | x - y |^3 for all x,y belonging to  R, the set of real nos, prove that [b]f[/b]  is a constant function.

PS: It's a problem from an old IIT entrance exam (1988)

The first bar of the first modulus sign is missing..

read the condition as |f(x)-f(y)| <= |x-y|^3

You can edit your own posts by hitting the "edit" button that is found near the corner of the post.
All nicened up:

If $f: \mathbb{R} \to\mathbb{R}$ is a function satisfying $|f(x) - f(y)| \leq |x-y|^3$ for all real x, y, prove that $f$ is constant.

Solution :

 Let h be != 0 and x, any real no.

 Then | (f(x+h) - f(x))/h   -  0  |  = | (f(x+h) - f(x))|/ |h|

                                                 <= | x+ h - x|^3/|h| (by the conditions of the problem)
                                                 = |h|^2

Now for any +ve no. e however small,

| (f(x+h) - f(x))/h   -  0   | < e if  |h| ^2 < e    i.e.   if    |h|  < sqrt(e) = d (say)

So, by the definitions of limit and differentiability of a function ,

 we conclude that   f'(x) = 0 for any real no. x

So, f is a constant ( this could be further established by Lagrange's MVT ,based on the fact that f'(x) = 0)


Solve for $ x$:    $ \frac{x\plus{}1}{x\minus{}1} \equal{} \frac{x\minus{}2}{x\plus{}2}$

This equals $ x^2\plus{}3x\plus{}2\equal{}x^2\minus{}3x\plus{}2$. From this, you can cancel out $ x^2$ ans $ 2$, leaving you with $ 3x\equal{}\minus{}3x$ By moving $ \minus{}3x$ to the right side, you get $ 6x\equal{}0$, so  the answer $ x\equal{}0$.

The answer is 0. Imagine the x's disappeared, as they would with zero. 1/-1=-2/2 is true.

You could also cross muliply and solve.

x^2 - 3x + 2 = x^2 + 3x + 2

-3x = 3x

6x = 0

x = 0

EDIT: This was posted just to note the fast FTW way to complete the problem.

guys cant u see that for any integer besides 0 its impossible? u dont have to factor anything

hello, you will get
$\frac{x+1}{x-1}-\frac{x-2}{x+2}=\frac{6x}{(x-1)(x+2)}$.
Sonnhard.


inequalities
number theory proposed
number theory

Find all $ n$ such that $ [\sqrt[n]{100}]$ divides $ 111$

Since $ 111\equal{}3\cdot37$ we have to make some distinguishing:

a.) $ [\sqrt[n]{100}] \equal{} 1$
So we have
$ 1\leq\sqrt[n]{100}< 2$
$ 1\leq 100 < 2^{n}$
This inequality holds if $ n$ is bigger than or equal 7.

b.) $ [\sqrt[n]{100}] \equal{} 3$
So we have
$ 3\leq\sqrt[n]{100}< 4$
$ 3^{n}\leq 100 < 4^{n}$

Right inequality holds if $ n$ is bigger that or equal 4, left inequality holds if $ n$ is smaller than or equal 4, so $ n\equal{}4$.

c.) $ [\sqrt[n]{100}] \equal{} 37$
So
$ 37\leq\sqrt[n]{100}< 38$
$ 37^{n}\leq 100 < 38^{n}$
Clearly, no integer $ n$ satisfies this condition.

d.) $ [\sqrt[n]{100}] \equal{} 111$
Also here we can't have solutions since $ \sqrt[n]{100}\leq 100\leq 111$

So we have solutions for $ n\equal{}4$ and $ n\geq 7$


Solve that equation in Real Numbers: $ x^2  \minus{} 3x \plus{} 1 \equal{}  \minus{} \frac{{\sqrt 3 }}{3}\sqrt {x^4  \plus{} x^2  \plus{} 1}$

hello, writing your equation in the form $ \minus{} x^2 \plus{} 3x \minus{} 1 \equal{} \frac {\sqrt {3}}{3}\sqrt {x^4 \plus{} x^2 \plus{} 1}$ we have the interval $ \frac {1}{2}( 1\minus{}\sqrt {17}) \le x \le \frac {1}{2}(1 \plus{} \sqrt {17})$. Squaring both sides and factoring we get $ (x^2 \minus{} 7x \plus{} 1)(x \minus{} 1)^2 \equal{} 0$. Only $ x \equal{} 1$ is the solution of our equation.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, writing your equation in the form $ \minus{} x^2 \plus{} 3x \minus{} 1 \equal{} \frac {\sqrt {3}}{3}\sqrt {x^4 \plus{} x^2 \plus{} 1}$ we have the interval $ \frac {1}{2}( 1 \minus{} \sqrt {17}) \le x \le \frac {1}{2}(1 \plus{} \sqrt {17})$. Squaring both sides and factoring we get $ (x^2 \minus{} 7x \plus{} 1)(x \minus{} 1)^2 \equal{} 0$. Only $ x \equal{} 1$ is the solution of our equation.
Sonnhard.[/quote]

Ur solution is very great, but I have an other soltion and that's solution is "under" great! :D

What do you think we replace  $ \minus{} \frac{{\sqrt 3 }}{3}$ with $ m$, and we find $ m$ make that's equation has roots? ^^

We have:

$ x^4  \plus{} x^2  \plus{} 1 \equal{} \left( {x^2  \plus{} x \plus{} 1} \right)\left( {x^2  \minus{} x \plus{} 1} \right)$ and $ x^2  \minus{} 3x \plus{} 1 \equal{} 2\left( {x^2  \minus{} x \plus{} 1} \right) \minus{} \left( {x^2  \plus{} x \plus{} 1} \right)$ and you can to solve then.


calculus
integration
limit
real analysis
real analysis unsolved

Let $L_n: = \sqrt[n+1]{(n+1)!} - \sqrt[n]{n!} . \;$ Find the best constants  $m, M \in \left[-\frac{1}{2}, 1\right]$  such that
\[ \frac{m}{n} < L_n - \frac{1}{e} < \frac{M}{n}\; \; \; ,\; \; \forall n\in {\mathbb N },\; n>2 .  \]

[quote="eugene"]I've found that $M$ can be $M=\frac{C}{e}$, where $C=\frac{1}{2}ln(2\pi)$ and $m<\frac{C}{e}\frac{n}{n+1}$[/quote]
Please  post the solution ! It's elementary, or use differential/or integral calculus ?l

[quote="eugene"]It uses Striling's formula:$n!=n^{n+\frac{1}{2}}e^{-n+C+o(1)}$, where the constant $C$ could be found with well known Wallis representation for $\pi=\displaystyle \lim_{n\to\infty}\frac{1}{n} (\frac{(2n)!!}{(2n-1)!!})^2$ , exactly $C=\frac{1}{2}ln(2\pi)$.
From now i just wrote $(n+1)!$ and $n!$ using this formula[/quote]
Do you have an elementary prooof of Stirling or Wallis formulae [without differenetial or/and integral calculus] ?
However please post your  complete solution , i.e. with all steps.


Dar mosalase ABC az A be I (markaz mohati) vasl kardehim va dar I be AI amud karde ta BC ra dar A' ghat konad. A' ra nesbat be OI gharine kardim ta A" ijad shavad. be hamin surat B" , C". sabet konid AA", BB" , CC" hamresand. ( dar hadde soale 6 e marhaleye dovom! )  :blush:  :lol:


While I was sitting on a toilet seat this came into my head and I couldn't solve it, so I'm posting it here..

Solve the $a^b+1=b^a$ in natural numbers.

[quote="krustyteklown"]a=2 and b=3 popped into my head, working on a solution rather than random answers right now, heh[/quote]

Yeah, I found that one too, actually that's how I came to this problem.. Also, (0, 1) is a solution in (nonnegative) integers.. And I don't know how did I manage not to see this one before: (1, 2)..

For large x $\frac{\ln x}{x}$ is decreasing.  This is intuitive if you think in terms of the graph of ln x.  As a result of this fact, $a^b$ and $b^a$ will be very far apart for large(ish) a, b.

Edit -- Sorry, I meant decreasing.

Actually $\frac{ln x}{x}$ has limit 0 as x approaches infinity...

[quote="shyong"][quote="Hexaditidom"][url=http://mathworld.wolfram.com/CatalansConjecture.html]Catalan's Conjecture[/url] states that $2^3$ and $3^2$ are the only positive powers with a difference of 1.[/quote]

However , example above also shows that $2^1$ and $1^2$ works ...

Anyway , when $a$ is odd and $b$ is even , for $b\geq 4$

we notice that $a^b+1\equiv 1+1 \mod 4$ which contradict with $b^a\equiv 0 \mod4$

This implies that $b$ can only be $2$ and $a=1$ . Which is $2^1$ and $1^2$

But for the case $a$ is even and $b$ is odd , I have no idea how to solve .... can anyone continue ? :?[/quote]

From link Hexaditidom provided:

[quote="http://mathworld.wolfram.com/CatalansConjecture.html"]Interestingly, more than 500 years before Catalan formulated his conjecture, Levi ben Gerson (1288-1344) had already noted that the only powers of 2 and 3 that apparently differed by 1 were  and  (Peterson 2002).

This conjecture had defied all attempts to prove it for more than 150 years, although Hyyro and Makowski proved that no three consecutive powers exist (Ribenboim 1996), and it was also known that 8 and 9 are the only consecutive cubic and Square Numbers (in either order). The conjecture was finally proved by Preda Mihailescu in a manuscript privately circulated on April 18, 2002.[/quote]

Care, da sam na tvom mestu nikad vise ne bih srao...

See

http://www.kalva.demon.co.uk/putnam/putn76.html

prob. A3.

[b]singular[/b], the problem A3 you mentioned is different from the problem proposed here  :!:
[hide="hint"]
clearly $a\not =b$ and you can prove that if $6<a<b$ then $a^b>2b^a$
you prove that $a^{a+1}>2{(a+1)}^a>{(a+1)}^a+1$ for $a>6$
because the second part is obvious and the first part becomes $a>2(1+\frac{1}{a})^a$
which is easy (actually  $(1+\frac{1}{n})^n\rightarrow e$ when $n\rightarrow \infty$)
proving that $(1+\frac{1}{a})^a<3$ using induction to prove first that
$(1+\frac{1}{a})^k<\frac{k^2}{a^2}+\frac{k}{a}+1$ then it easy to prove that this works also for $b-a=m$
and then to finish the solution but it's to boring to write  ;)
 [/hide]


What classes are recommended for preparing for AMC 8? I have taken Intro. to Algebra, Intro. to Geometry, Intro. to Number Theory, and Algebra 3. (I plan to take Algebra 2 as a review.) However, I am not able to take Advanced Mathcounts/AMC 8 during the fall because I have something going on during that time, and I believe that the Basics class will be too easy...what should I do?

The only class left is Intro Counting.  After that (having taken the other Intro classes), you should be all set for the AMC 8.


AMC
USA(J)MO
USAMO
USAMTS
AIME
AMC 10
AMC 12 A

Suppose that on USAMTS I get a 72, on AMC 12 a 120, and on AMC 10 a 130, thus qualifying for AIME by all three methods. (I guess the USAMO index for USAMTS is just doubled, like 72*2 = 144, right? If I'm right, please show me a link to where it says so.) Even if I have AMC, will USAMTS (if it is my highest score) be used to determine my USAMO index?

I *think* that all scores qualifying to AIME by USAMTS are counted as 100.

In past years, I'm almost sure that USAMTS to AIME would be counted as 100, though doubling would definitely seem more appropriate.  Even if you qualify in all three ways, they will only take the score that makes your USAMO index the highest.


Θανάση γειά σου .
  Η λύση μου είναι στο attachment .

ολοσωστη! :)

Εσύ την έφτιαξες την άσκηση Θανάση ??Κοίταξε στο νέο τόπικ που έβαλα για την ΑΜ-ΓΜ τις προτεινόμενες ασκήσεις.....Και πες μου πως σου φαίνονται.... ;)

οχι δε την κατασκευασα εγω.

Ειχα καποτε κανει μια ανισοτητα. Αν τη βρω σας τη λεω!  :)

για δείτε και αυτην.........νομιζω είναι σωστη........
ελέγξτε πρωτα!

αν $a,b,c,d>0$ νδο

$\frac{d+a^2}{d+bc}+\frac{d+b^2}{d+ac}+\frac{d+c^2}{d+ab}\geq 3$

 :D

Από AM-GM αρκεί νδο:

(d + a^2)(d + b^2)(d + c^2) >= (d + ab)(d + bc)(d + ca)

άρα αρκέι νδο, χωρίς περιορισμό της γενικότητας, ότι:
(d + a^2)(d + b^2) >= (d + ab)^2, που ισχύει! (πράξεις!!)

πολλ/ζοντας, και παίρνοντας τη ρίζα (όλα είναι > 0), βγαίνει αυτό που ζητάς.

ναι σωστα...
εγω την έλυσα με τσεμπισεβ.... :D

Ρε cmad τι πράξεις , BCS για το τελευταίο! Τελική φάση πληροφορική θα έρθω


Let be given a equilateral triangle ABC. Let us consider an arbitrary point M in its plane. Draw perpendiculars MH, MK, ML to BC, CA, AB respectively. Find the minimal and maximal value of

        $ P = \frac{MA+MB+MC}{MH+MK+ML} $

From what I know, if $M$ is in the interior of the triangle $ABC$, P is constant. ( if we assume that $ABC$ is equilateral.) ;)

I don't think that's true, since $MH + MK + ML$ is constant while $MA + MB + MC$ isn't.

Arne, I don't quite think so. Please, remember that $ABC$ is a equilateral triangle. ;)

Yes, and $MA + MB + MC$ is definitely not constant!

Anyway, I have proved that if $ABC$ is an equilateral triangle of side equal to $l$, then the following inequality hold:

   $PA+PB+PC <2l.$ ;)


modular arithmetic
number theory unsolved
number theory

The square of a whole number finishes in four equal digits.Find that digit.
Artiona

$0$ ?

Just look at all the possible residues for the squares modulo $1000$.

Pierre.

I suppose you mean 10000. I'd like to see you do that by hand though.. (a faster way would just be starting with 2 equal digits and working upwards with a bit of mods).

On that note.. what digit not including 0 would the answer be if it were 3 equal digits rather than 4?

It can't be an odd digit, because all odd squares has an even second digit.  ;)
So it is an even digit: either 2, 4, 6, 8 or 0.
1) 2 and 6 are wrong candidats (consider $\pmod{4}$).
2) 4 and 8 are wrong candidats too (consider $\pmod{16}$).

 :D  :D  :D


calculus
integration
function
complex analysis

I'm trying to integrate the following:

I = 1/sqrt(1 - x^2)

from -1 to 1 along the real axis.  I've drawn a branch cut between -1 and 1 and found a closed contour that circles the entire thing in the counterclockwise sense.  Reversing the direction, I find that the original integral, [i]I[/i], is just i2pi times the residue at infinite.  I'm stuck at correctly determining the residue at infinite while keeping track of the phase changes.  Here's what I've done so far:

1. F(z) = (1 - z^2)^(-1/2)
2. I chose to go to infinity along the positive real axis, so the factor of (1 - z) picks up a phase of -pi and the factor of (1 + z) remains unchanged.
3. I factored a -z^2 from the function and pulled out a 1/iz
4. The phase factor, after the square root, produces an additional i on the function
5. After this, F(z) = 1/z(1 - 1/z^2)^(-1/2)
6. Expanding this about zero for large values of z, I see that the a[size=59]-1[/size] term is 1, thus the residue at infinite is -1

This is obviously wrong, since it yields a complex value for the integral, but I've not been able to figure out my error.  I think I fundamentally misunderstand how to determine the residue at infinity for a function that has branch cuts.  I've been stuck on this for a while and I think I understand most of the pieces, but so far the fog hasn't lifted yet.  I should add that I'm not asking someone to tell me what the residue is - it's really more of a fundamental matter, so if someone could identify the flaw in my procedure and explain how to, in general, determine the residue of a function about a point in the presence of branch cuts, I'd really appreciate it.  Thanks.

~ George

There is no singularity at $ \infty$, and thus the residue is zero.

Your loop integrates to zero, because the opposite signs of the two sides cancel each other. Your current method cannot be used to find $ \int_{\minus{}1}^1 \frac1{\sqrt{1\minus{}x^2}}\,dx$. Fortunately, that integral is easy using real techniques.

No, jmerry; despite $ 1/z$ is not a singularity at $ 0$, the contour integral doesn't tend to $ 0$ ;). Also, the sides do not cancel each other and the method is applicable. Note that $ \frac 1{\sqrt{1\minus{}z^2}}\asymp \frac 1{iz}$ (or $ \minus{}\frac 1{iz}$, depending on which branch is chosen); so the residue is plus or minus $ i$. It cancels with $ i$ in the $ 2\pi i$ factor in the residue theorem. $ \sqrt{1\minus{}z^2}$ is real on both sides, so I do not see any problems unless there are some stupid arithmetic mistakes somewhere (I do not quite understand Step 4 in the original post, for example).

OK- I haven't messed with integrals around branch cuts enough, and I was silly to say the opposite sign cancels the integral.

[quote]There is no singularity at infinite, and thus the residue is zero. [/quote]

A function can have a residue at infinite without being singular there.

I think I've figured out what the problem is.  In order to expand the function for large values of z, i need to pick a direction to approach infinite.  Since I know that the phase of the function just above the origin is 0 (ie, F(0) = 1), then I can head off to infinite along the positive real axis and I get that z = x.  By doing so, it's clear that the phase of the (1 + z) factor is unchanged but the phase of the (1 - z) factor changes by -pi.  So, now that I know the phase at infinite, I just needed to find the magnitude of the expansion there.

A simpler way to do this would have been to let z = iy and go to infinite along the positive imaginary axis.  Since the phase changes of each branch point would have canceled each other out, I could have just let z = iy and expanded for large values of y and the substituted y = -iz back in and read off the residue.  In either case, the expansions are identical, as they should be.  If anyone cares, the residue at infinite is -i.

[quote]I do not quite understand Step 4 in the original post, for example[/quote]

Yeah, I apologize for that - it wasn't very clear.  What I meant was that, if the (1 - z) factor picked up a phase of -pi, then 1/sqrt(1 - z) would pick up a phase of i*pi/2.


geometry
parallelogram
circumcircle
geometry unsolved

In a paralleogram ABCD, the bisector of angle BAD meets BC at M and DC produced at N. Prove that the center of the circumcircle of triangle CMN lies on the circumcircle of CBD.


I have noticed that the mathematicians from former Yugoslavia are pretty active on MathLinks forums these days... I think that it would be nice for us to have our own local community (where we could write in our native language(s), share useful infos about competitions etc...

Still, there is a question: how would we call the community? For me, personally, the title "former Yugoslavia" isn't a problem, but it could produce some negative comments by some of the "natives"...

[quote="delegat"]You are absolutely right I agree with you and really it is time to have our own community.
I also agree with name of community Maybe it is the best choice at the moment or if somebody else has differnet idea please post it...

Hajde vi iz Hrvatske [b]mornik [/b] [b], djuro ... [/b], Makedonije [b]bodan [/b]i ostali (da ne nabrajamo) podrzite nas.[/quote]

Volim matematiku, razumem i govorim srpski, hrvatski, podrzavam inicijativu. Javicu i drugima koji nisu to procitali.

Pozdrav, Bodan

meni je apsostrumfno svejedno kako biste nazvali podforum, cudno mi je da sam bas danas ujutro i o ovom razmisljao, al ne znam kaj cemo nasega mi postat ovdje?
u svakom slucaju ne bi bilo lose da se pokoji put opustim od englishkog....
mozemo i slovence pozvat ako ih se nadje na forumu ;)

Former Yugoslavia je sasvim Ok, ali nije bas kreativno...

da.true. stalno je neka ex yu finta u igri...
mozda balkan forum??? :)
bar da zbunimo grke malo ;)
ionako se stalno hvalimo da smo balkanci. :D

Ma nece nam dati Bugari, Albanci, Rumuni... Mozda Zapadni Balkan? A Slovenci svakako imaju svoj podforum (koji vise niko ne posjecuje :lol: ), pa nam ne moraju ni dolaziti (kad su oni vec Europejci, a ne Balkanci)...

ako ce se ko bunit, mijenjat cemo u zapadni balkan, a do tad, ko ih ...:)
fantasticno je sta kad bi psovao na ovom forumu, ne bi se nista desilo, morat cu to jednom isprobat :D

Hajd ako je vec frka za ovo Yugoslavia i ja se slazem da forum nazovemo [b]zapadni balkan[/b]
Javite i ostalima koje znate

Pozdrav...

Slazem se, super ideja za taj podforum.

Sto se tice naziva zapadni balkan je ok, ali ja predlazem "Skoro Evropljani".

katastrofa...
nemoj nas gurat u europu. dok smo bili u ljubljani na IMOu nismo ni cevape nigdje mogli naci, a onda na izletu naprave rostilj sa slovenskim "nacionalnim jelom" cevapima. pusti europu sinko, rjesavaj nejednakosti ;)
ajde stvarno, mozda nas se i skupi desetak za taj forum...

Može, ja sam za. Kak god ju nazovete, meni svejedno. Zapadni Balkan mi se čini u redu. Makar, to uključuje i Albaniju. Može i Južnoslavenske zemlje (doduše, to uključuje i Bugarsku). A Bivša Jugoslavija mi je jedno onak dost glupo rješenje. Kao da kažemo, svojevremeno smo bili bar nešto, a sad više nismo apsolutno ništa.  :lol: Može i Tito's Boys, što se mene tiče. :D
Ozbiljno, Zapadni Balkan je OK.

[quote]mozemo i slovence pozvat ako ih se nadje na forumu[/quote]
Ma ne, pa moramo imati Sloveniju, i oni su "naši".

Evo, maloprije sam poslao PM (personal message, da ne bude zabune  :rotfl: ) Valentinu Vornicuu u vezi sa ovom inicijativom... Pa cemo vidjeti, vjerovatno ce on uskoro ovdje postati nesto...

Hvala, fakat si borac za "našu stvar".  :D A s obzirom na to da smo ipak Balkanci, mogao si mu poslati i drugi PM...  :D Just kidding.  :D

Cini mi se da smo postigli dogovor za ime: Zapadni Balkan (West Balcan Countries). Eh sada nam ostaje jos jedno (malo problematicno :) ) pitanje: pored imena naseg podforuma bi trebala stajati neka zastava, ili nesto nalik zastavi. Vjerovatno postoji mogucnost da idemo i bez toga (kao sto je recimo africki podforum), ali eto, mozemo malo razmisliti o tome... Evo dva prijedloga od mene, pa ako se kome svide (meni se licno ne svidjaju ali nisam imao sta pametnije jutros da radim nego da crtam u Fireworksu:)), neka kaze, ili neka predlozi nesto novo...
Ipak, najrealnije zasada izgleda da idemo bez ikakvog obiljezja, zar ne?

E hvala puno. Balkan se piše s [b]Balkans[/b] i u engleskom, tak da obrati pozornost na to... a Zapadni Balkan je [b]Western Balkans[/b]... bar mislim. A što se tiče zastave, može ništa, može i ona tvoja prva slika (ala EU)... meni svejedno, makar mi je ona prva slika baš  :cool: . Neka ostali odluče.

We cannot have a forum called "Former Yugoslavia" unfortunately (we have slovenia already separate, and as far as I understand they want to be separate). I have made one for Serbia and Montenegro instead: http://mathlinks.ro/Forum/index.php?f=272

Hmh... I don't know if Slovenia would like to stay separate as none of their people posted on this topic, but I'm afraid Serbia and Montenegro isn't right. I didn't see anyone from Serbia or Montenegro even post in this topic while there were posters from Macedonia, Croatia and Bosnia and Herzegovina. Furthermore, Serbia and Montenegro are not one country anymore (for at least a year now). Also, I'd like to point out that Slovenian national forum has not had any activity from Dec 31, 2006 and that two posts weren't made by someone from Slovenia. The last Slovenian that posted in their own forum was on Aug 11, 2006. Now, I don't see where the problem is with Western Balkans / Former Yugoslavia as Slovenia has little activity on Mathlinks and, if anything, a new subforum with many more of their local posters will maybe even increase their activity. I doubt that anyone from Slovenia will have any objections to this. Thanks.

I've merged the forums into one (the link is edited above) called (Former) Yugoslavia with all the countries listed.

neda nam balkan...nasi snovi ce brzo nestati...:(

Thank you Valentin, you're our man! We won't bother you with the flag (in part because we could reach a consensus ourselves  :blush: )  :D . It's great, thank you!

@goc: pogledaj ponovno - poslao sam čovjeku PM i sve je riješeno...  :D Samo se zovemo (Bivša) Jugoslavija...

Imamo forum!

pa vidim to seljak jedan, al to je tocno ono sto nismo htjeli... ajde neka...
bas me zanima ko ce lupit prvi post ;)

Stop posting in other languages than english in this forum please!

Sorry. I think you can lock the topic, though.


As ksekinisoume ena topic afierwmeno stis anisotites gia na min tis vazoume stis askiseis tis Algevras.

[color=red]ASKHSH 1[/color]

Ean $a,b,c\in R^{\star}$ na deiksete oti

$\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+b^{2}}+\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}}\geq \frac{3}{5}$

Alexandros

Μάλιστα , μάλιστα....Πριν από κάποιες μέρες κάναμε κάποιες ανισότητες στην κατεύθυνση και κάποιος λέει στον καθηγητή ότι εγώ είμαι καλός στις ανισότητες και δεν πρόκειται να βρει ανισότητα που να μην τη λύσω .Ε λοιπόν αυτός ορμόμενος από τη δύσκολη λύση που ήξερε για την παραπάνω ανισότητα μου την πρότεινε και του έδωσα την παρακάτω λύση .

Αρκεί να αποδείξουμε την ανισότητα για θετικούς πραγματικούς αριθμούς (γιατί?) και λόγω του ότι η ανισότητα είναι ομοιογενής υποθέτουμε ότι $a+b+c=3$ (θα μπορούσαμε να υποθέσουμε οτιδήποτε άλλο ας πούμε $a+b+c=m$ ) .
Τότε η προς απόδειξη γίνεται
$\sum\frac{(3-2a)^{2}}{(3-a)^{2}+a^{2}}\geq \frac{3}{5}$ ή ισοδύναμα
$\sum\frac{(3-2a)^{2}}{(3-a)^{2}+a^{2}}-\frac{1}{5}+\frac{18(x-1)}{25}\geq 0$

ή ισοδύναμα
$\sum\frac{18(a-1)^{2}(2a+1)}{25(2a^{2}-6a+9)}\geq 0$

που προφανώς ισχύει .
(Πάντως ο καθηγητής ακόμα ψάχνει πως σκέφτηκα τη διάσπαση στην δεύτερη γραμμή  :rotfl: )

ΑΣΚΗΣΗ 1 -ΛΥΣΗ

Οι παρονομαστές είναι θετικοί άρα μπορούμε να εφαρμόσουμε την Ανισότητα Andreescu.

Θέτω x=b+c/a , y = c+a/b , w =a+b/c και η ανισότητα γίνεται

(x-1)^2/x^2 +1  +   (y-1)^2/y^2 +1   +  (w-1)^2/w^2 +1 >= 5/3

Επειδή οι παρονομαστές είναι θετικοί μπορούμε να εφαρμόσουμε την ανισότητα Andreescu.Οπότε αρκεί να ισχύει

(x+y+w-3)^2[b]/[/b]x^2 + y^2 + w^2 + 3 >= 3/5

Μετά από πράξεις παίρνουμε

2(x^2+y^2+w^2) - 15(x+y+w) + 5(xy+yw+wx) +18>=0 :arrow:
(x+y+w)^2  - 15(x+y+w) + 3(xy+yw+wx) >= 0.

Από την ανισότητα του Scur  θα έχουμε (xy + yw + wx)>= 2(x+y+w) (αφού πρώτα αντικταστήσουμε τα x,y,z).

Επιπρόσθετα έχουμε x+y+z>=6 αφού
x + y + z = b+c/a + c+a/b + a+b/c >= 3[(a+b)(b+c)(c+a)/abc]^1/3   (από την AM-GM)

Και επειδή (a+b)(b+c)(c+a)>=8abc θα έχουμε
x+y+z>=6

Τελικά θα έχουμε

(x+y+z)^2 -15(x+y+x) + 3(xy+yz+zy) >= (x+y+z)^2 -9(x+y+z) + 18 = (x+y+z-3)(x+y+z-6)>=0  που ισχύει αφού x+y+x>=6
Η ισότητα ισχύει για a=b=c.

ΥΣ1.Πολύ ωραία άσκηση.Την έχω ξαναδεί και τη θυμήθηκα γιατί μου είχε πάρει αρκετές μέρες να τη σκεφτώ.Νομίζω ότι υπάρχει και στο βιβλίο του κ.Στεργίου.

ΥΣ2.Δεν ξέρω πώς να πληκτρολογώ τα κλάσματα και τις δυνάμεις για αυτό η λύση είναι λίγο κακογραμμένη.

Αν χ , ψ ,z >= 0 και χ + ψ + z + χψz = 4 , να δείξετε οτι
                                 χ + ψ + z >= xψ + ψz + zχ .

[quote="fotis"][b][color=red]ΑΣΚΗΣΗ 1 -ΛΥΣΗ[/color][/b]

Οι παρονομαστές είναι θετικοί άρα μπορούμε να εφαρμόσουμε την Ανισότητα Andreescu.

Θέτω $x=\frac{b+c}{a}, y = \frac{c+a}{b}, w =\frac{a+b}{c}$ και η ανισότητα γίνεται

$\frac{(x-1)^{2}}{x^{2}+1}+\frac{(y-1)^{2}}{y^{2}+1}+\frac{(w-1)^{2}}{w^{2}+1}\geq \frac{5}{3}$

Επειδή οι παρονομαστές είναι θετικοί μπορούμε να εφαρμόσουμε την ανισότητα Andreescu.Οπότε αρκεί να ισχύει

$\frac{(x+y+w-3)^{2}}{x^{2}+y^{2}+w^{2}+3}\geq \frac{5}{3}$

Μετά από πράξεις παίρνουμε

$2(x^{2}+y^{2}+w^{2})-15(x+y+w)+5(xy+yw+wx)+18 \geq 0$ ή  $(x+y+w)^{2}-15(x+y+w)+3(xy+yw+wx) \geq 0$.

Από την ανισότητα του Scur  θα έχουμε $(xy+yw+wx) \geq 2(x+y+w)$ (αφού πρώτα αντικταστήσουμε τα $x,y,z$).

Επιπρόσθετα έχουμε $x+y+z \geq 6$,  αφού
$x+y+z = \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 3 \frac{[(a+b)(b+c)(c+a)}{abc}^\frac{1}{3}$   (από την $AM-GM$)

Και επειδή $(a+b)(b+c)(c+a) \geq 8abc$  θα έχουμε
$x+y+z \geq 6$

Τελικά θα έχουμε

$(x+y+z)^{2}-15(x+y+x)+3(xy+yz+zy) \geq (x+y+z)^{2}-9(x+y+z)+18 = (x+y+z-3)(x+y+z-6) \geq 0$

  που ισχύει ,  αφού  $x+y+x \geq 6$
Η ισότητα ισχύει για $a=b=c$.

[b]ΥΣ1.[/b][i]Πολύ ωραία άσκηση.Την έχω ξαναδεί και τη θυμήθηκα γιατί μου είχε πάρει αρκετές μέρες να τη σκεφτώ.Νομίζω ότι υπάρχει και στο βιβλίο του κ.Στεργίου.[/i]
[b]ΥΣ2.[/b][i]Δεν ξέρω πώς να πληκτρολογώ τα κλάσματα και τις δυνάμεις για αυτό η λύση είναι λίγο κακογραμμένη[/i].[/quote]

[color=blue] Φαντάζομαι τώρα να μπορείτε να μελετήσετε τη λύση πιο ξεκούραστα.  [/color]

                         [color=red] Fotis , WELCOME !!! [/color]

Μπάμπης

μηπως η πρωτη ανισοτητα ηταν σε καποιο Αρχιμηδη ?

Άσκηση 2 Λύση
Απο προφανή ανισότητα ξέρουμε ότι
χψ+ψζ+ζχ=<(χ+ψ+ζ)^2 /3 (1)
Από την δοσμένη ανισότητα υψώνοντας στο τετράγωνο κι χρησιμοποιώντας αυτήν (1) έχουμε ότι αρκεί νδο
χ+ψ+ζ>=3
Έστω χ+ψ+ζ<3 (2)
προσθέτοντας χψζ  κι εφαρμόζοντας την δοσμένη ισότητα έχουμε ότι γιοα να ισχύει η (2) πρέπει χψζ<1
Αλλά να στην ισότητα εφαρμόσουμε Α-ΓΜ βγάζουμε ότι
χψζ>=1
Άρα άτοπο άρα ισχύει η ζητούμενη

Fisika,

Htan ston Arximidi tou 1998 kai se diagwnismo tis Iapwnias to 1997 kai epeidi i lisi pou edine i epitropi tou diagwnismou tis EME to 1998 itan aparadekti (giati eixe mono prakseis) gi'auto tin evala edw! Gia na graftei me pio omorfo tropo pou ap'oti eida egine idi!!! Mpravo loipon!

Silouan to $x$ ti akrivws einai sti lisi sou?

Kalwsirthes Fwti. (O tropos me ton opoio grafoume ta mathimatika keimena edw, stirizetai se mia "mathimatiki glwssa" pou legetai LaTeX, i opoia mas epitrepei na grafoume poli eukola kai apla mathimatikes ekfraseis arkei na grapsoume tin antistoixi entoli pou theloume kai na tin kleisoume anamesa se dio dollaria $\$$. Gia na deis tis entoles pou iparxoun kai na pareis mia prwti geusi mporeis na deis stin selida http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php h opoia periexei ta vasika simvola kai tis entoles me tis opoies ta emfanizeis kai episis des stin selida http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php gia tis prwtes aples entoles emfanisis mathimatikwn ennoiwn (klasmata, athroismata, oloklirwmata, ktl ktl... Sigoura tha diaskedaseis grafontas mathimatika me auto ton tropo. Stin arxi tha diskoleuteis ligaki alla meta tha ginetai olo kai pio elkistiko.)

Etsi loipon gia na grapseis to (b+c)^2/a^3+(a+b)^2 >= 3 tha patiseis

[code] $\frac{(b+c)^2}{a^3+(a+b)^2} \geq 3$ [/code] kai tha emfanistei $\frac{(b+c)^{2}}{a^{3}+(a+b)^{2}}\geq 3$. Kali arxi!!

Alexandros

καλα θυμαμαι λοιπον.πραγματι εκεινη η λυση ηταν αλησμονητη.... :lol:

Foti ean theleis na dokimaseis tis entoles kai na "paikseis" prin steileis post mporeis na pas stin selida http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php kai ekei mesa na pliktrologiseis (opws leei kai stis odigies) tin entoli pou theleis alla xwris dollaria, dioti etsi ki alliws einai mono gia dokimes mathimatikwn tipwn.

(Vevaia mporeis prin steileis opoiodipote post, na patiseis "preview" kai na deis pws emfanizetai. An einai ok patas "Submit" kai etoimos. H parapanw lisi einai mono kai mono gia na dokimaseis merikes vasikes entoles kai na ksekiniseis na grafeis.)

Alexandros

[b][color=red]ΑΣΚΗΣΗ 3[/color][/b]

 [i]Ας προσπαθήσουμε και  μία έξυπνη άσκηση , επιπέδου Αρχιμήδη για Junior ![/i]

 Αν $a , b >0$  και $a \leq 1 , a+b \leq 10$ , να αποδείξετε ότι :

      $\sqrt a+\sqrt b \leq 4$

Η λύση μου είναι λίγο βιαστική, οπότε συγχωρείστε με αν μου ξεφύγει κάτι. Έχουμε λοιπόν:

$\sqrt{a}+\sqrt{b}\leq4\quad(1)$

και επίσης $a\leq1, b\leq9$

Για $a_{max}=1$ και $b_{max}=9$ έχουμε $1+3=4$ (ισότητα).
Έτσι για όσο μικρότερες τιμές των $a,b$ θα ισχύει η ανίσωση.

b max=9???
a=0,5 b=9.5????.......

gia tin 1...
ontos ine 8ema iaponias....(eki tin ixa protodi..)
kai ontos poli palio 8ema arximidi...
vsk ego tin elisa,8etontas a+b+c=1(logo omogenias)kai meta jensen gia mia sinartisi pou prokipti...

 :D

[quote="RAKAR"]b max=9???
a=0,5 b=9.5????.......
[/quote]

Έχεις δίκιο, όποιος βιάζεται σκοντάφτει  :oops:

lol, για να πω την αλήθεια δεν κάθισα να τη check, αλλά με το σκεπτικό ότι το πρότεινε ο RAKAR θεώρησα ότι είναι σωστή  :P

orea mostel... :coolspeak:
+mu doses tin idea gia mia akomi lisi(afu afti me jensen den piani..)
8eto $x=\frac{b+c}{a}$ $y=\frac{c+a}{b}$ $z=\frac{a+b}{c}$(klasikos metasximatismos...)
 opote

$\frac{(x+2)^{2}}{x^{2}+2}+\frac{(y+2)^{2}}{y^{2}+2}+\frac{(z+2)^{2}}{z^{2}+2}\leq 8$

$\frac{2x+1}{x^{2}+2}+\frac{2y+1}{y^{2}+2}+\frac{2z+1}{z^{2}+2}\leq \frac{5}{2}$

$\frac{(x-1)^{2}}{x^{2}+2}+\frac{(y-1)^{2}}{y^{2}+2}+\frac{(z-1)^{2}}{z^{2}+2}\geq \frac{1}{2}$

vsk olo ayto to ekana gt i8ela nte +kala na efarmoso C-B-S :cursing:  ala ixa anti8eti fora anisotitas...eno stin teleytea sxesi..tin exo feri sta metra mu...opote apo C-B-S arki ndo $\frac{(x+y+z-3)^{2}}{x^{2}+y^{2}+z^{2}+6}\geq \frac{1}{2}$

 i isodinama oti $x^{2}+y^{2}+z^{2}+4(xy+yz+zx)-12(x+y+z)+12\geq 0$

diladi  $(x+y+z)^{2}-12(x+y+z)+36+2(xy+yz+zx)-24\geq 0$

 $(x+y+z-6)^{2}+2(xy+yz+zx)-24\geq 0$

opote arki ndo $xy+yz+zx\geq 12$...pu isxii apo AM-GM afu $xyz\geq 8$....

 :D

"Καλήν ημέραν άρχοντες...!"
Καλές γιορτές σε όλους με υγεία, αγάπη, ευτυχία!

Ωραία RAKAR  :lol: . Να διευκρινίσω πως η άσκηση ήταν από USAMO το 2003. Θα τη βρείτε στο "μεγάλο" PDF που ετοίμασε ο Αλέξανδρος (τον ευχαριστούμε πολύ) όπου είναι ripped όλο το site του kalva! Λεπτομέρειες στις σελίδες 3901-3909! Αν θέλετε μπορώ να βάλλω και τη λύση που προτείνει εκεί, άλλα πιστεύω πως αυτές που ειπώθηκαν εδώ είναι πολύ καλλίτερες  :)

Πολύ ωραίες λύσεις... :lol:
Ράλλη την 5 την έχω χωρίς απαλοιφή  :wink:  . Αλλά και από εκεί που έφτασες υπάρχει τρόπος αν ξέρεις μια μαγική ταυτότητα  :)

[quote="silouan"]ΑΣΚΗΣΗ -5
Έστω $a,b,c>0$ . Δείξτε ότι

$\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\geq\frac{12}{(a+b+c)^{2}}$[/quote]

Μια βοήθεια Υποθέτουμε ότι $a+b+c=1$ .
Γράψτε την ισοδύναμα στη μορφή
$\sum\frac{1-a^{2}-bc}{a^{2}+bc}\geq 9$

Μπορεί κάποιος να την τελειώσει τώρα?

***
Έχω 5 αστέρια πλέον .  :D

[b][color=red]Άσκηση 8 [/color][/b]

[img]http://img512.imageshack.us/img512/8263/9038rq9.png[/img]

Κλείνω τον κύκλο των ανισώσεων και το 2006 (συγκίνηση) με την παρακάτω:

Αποδείξτε ότι για $a$, $b$, $c$ $\in [0,1]$,

$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)\leq1$

[i][b][color=red]ΚΑΛΗ ΧΡΟΝΙΑ ΣΕ ΟΛΟΥΣ!!![/color][/b][/i]

Καλή χρονιά . Πάλι με ανισότητες με βρήκε να ασχολούμε ο καινούργιος χρόνος  :mad:
Αλλα βγήκε κάτι ωραίο η άσκηση 9 ας πούμε  :)  (προσπαθείστε την αξίζει τον κόπο . Α μην υψώστε στην 9η  :D )

[color=red]ΑΣΚΗΣΗ 10-ΛΥΣΗ [/color]
Αφού έχουμε κλειστό διάστημα [0,1] η άσκηση τελειώνει πριν αρχίσει  :lol:

Έστω $f(a,b,c)$ το δεξί μέλος της ανισότητας .Η $f$ είναι κυρτή σε κάθε μεταβλητή της (απλά παραγωγίζουμε μερικά δύο φορές ) .Άρα η $f$ επιτυγχάνει μέγιστο στα άκρα του διαστήματος άρα αρκεί να εξετάσουμε τις τιμές $f(0,0,0)$,....,$f(1,1,1)$ (όλες οι τιμές δίνουν 1)
Άρα $f(a,b,c)\leq 1$

ΧΡΟΝΙΑ ΠΟΛΛΑ ευτηχησμενο το 2007 (αληθεια ειναι πρωτος αριθμος το 2007 ?) πολυ ωραια λυση.αυτο το θεμα ειχε πεσει USAMO και η γενικη περιπτωση ν μεταβλητων ειναι συμφωνα με ενα βιβλιο η  ανισοτητα Giroux η κατι τετοιο.

καλή χρονιά σε όλους κι εγώ όπως ο σιλουανός λίγο πρίν αλλάξει ο χρόνος με ανισότητες ασχολιόμουν
λοιπόν 2 ερωτήσεις
η 10 δεν λύνεται διαφορετικά χωρίς συνναρτήσεις ??
η 9 ειναι τρομερή όποιος την λυσει να βάλει αναλυτικά λύση

Nai Αμβρόσιε ξέρω μία λύση για την 10 με AM-GM (πολύ ωραία λύση αλλά δύσκολη στο να τη σκεφτείς )

[quote="ambros"]
η 10 δεν λύνεται διαφορετικά χωρίς συνναρτήσεις ??
[/quote]

Εχω βρει στοιχειώδη λύση για τη 10 , αλλά είναι μεγάλη και καθόλου κομψή.....Καλύτερα είναι με συναρτήσεις! :wink:

Ωραία Silouan, τελικά μάλλον ήταν πολύ εύκολη  :oops:


I was wondering if the following expressions are capable of being represented as a product of  sine and cosine;

cos(x) + a*sin(x) = B*cos(d) *sin(f) ?

or

cos(x) - a * sin(x)= B*cos(d) *sin(f) ?

[hide="We can do better!"]  In fact, $ A \cos x \plus{} B \sin x$ can be represented as a [b]single[/b] cosine (or sine).  Here's how.  Let $ a \equal{} \frac{A}{ \sqrt{A^2 \plus{} B^2}} \equal{} \cos \theta, b \equal{} \frac{B}{ \sqrt{A^2 \plus{} B^2}} \equal{} \sin \theta$.  Then

$ A \cos x \minus{} B \sin x \equal{} \sqrt{A^2 \plus{} B^2} \left( \cos \theta \cos x \plus{} \sin \theta \sin x \right) \equal{} \boxed{ \sqrt{A^2 \plus{} B^2} \cos (x \minus{} \theta) }$

by the cosine angle addition formula.  In your case,

$ \cos x \plus{} a \sin x \equal{} \sqrt{1 \plus{} a^2} \cos (x \minus{} \arctan a)$
$ \cos x \minus{} a \sin x \equal{} \sqrt{1 \plus{} a^2} \cos (x \plus{} \arctan a)$. [/hide]


trigonometry
calculus
integration
function
real analysis
real analysis theorems

In a recent topic, LideR_M made the following claim:

[quote="LideR_M"]$ \int_{0}^\pi x f(\sin x)dx=\frac{\pi}{2}\int_{0}^\pi f(\sin x)dx$[/quote]
Prove that this claim is correct (given reasonable hypotheses as to the nature of $ f.$)

Function $ f$ must be continous on $ [0;1]$.

And to prove the claim, we should for example take $ x=\pi-t$

Continuity of $ f$ will do for a sufficient condition; we could certainly weaken that some if we needed to.

The substitution is a very strong hint, but I was hoping someone would post the complete argument, with explanation.

As $ \sin(\pi\minus{}x)\equal{}\sin x$, we have that

$ \int_{0}^{\pi}xf(\sin x)dx\equal{}\int_{0}^{\pi}(\pi\minus{}x)f\left(\sin(\pi\minus{}x)\right)dx\equal{}\pi\int_{0}^{\pi}f\left(\sin x\right)dx\minus{}\int_{0}^{\pi}xf\left(\sin x\right)dx$,

and then,

$ \int_{0}^{\pi}xf(\sin x)dx\equal{}\frac{\pi}{2}\int_{0}^{\pi}f(\sin x)dx$.


do any of you participate in MAO write-in
from National Assessment and testing
like Ciphering Time Trial or Startup?

wow.. that was an amazingly helpful post. Seriously, I must say, u really deserve to be put on a pedestal for that one.

Jese man... give some kind of informative conversation to get this topic going.

Although my school doesnt actually participate in the mao one, I do know some worthwhile people who do. The top Mao school in the country (Stoneman Douglas) does it, and the all say that its really not worht the time. I know from the first round, their top scores were int he 14-16 range. They said that its all about computation, so blah.

hahaha....so yea, we do it...i did pretty bad [i missed like 3 or 4 questions] on round 1(out of 3), which i thought was pretty hard...
that time trial thing was extremely hard...mostly because there was a lot of time pressure, questions were pretty easy though...i got around 70/100 on that, i think
round 2 was last week, did better than the first time, got 18/20...

our school does the MAO write-in.. it really isn't anything special, but it's something to do.

like jelyman said, it's pretty basic computational stuff

found out today that i placed in the top 10 in that time-ciphering thing a couple weeks ago...got a nice plaque...pretty cool.


a,b,c are the sides of the triangle and r=inradius R=circumradius
prove that
$(1-\frac{2a}{b+c})^{2}<1-\frac{2r}{R}$
this one's kinda cooool (more so coz it is the 1st mathematical problem i've solved since 3 months :lol: )
cheers

Not being in town right now, makes me hard to make computations, so.. did u solve this one like so:

you replaced:

$\displaystyle \frac{R}{r}=\frac{abc}{4(p-a)(p-b)(p-c)}$ and then made some computations?

well turn it into it's trigonometrical equivalent and proceed (this is the kinda problem which solves itself,no new ideas in here assuming my proof is correct)


Two shaded identical rectangular decorative tiles are first placed (one each) at the top and at the base of a door frame for a hobbit's house, as shown in Figure 1. The distance from W to H is 45 inches. Then the same two tiles are rearranged at the top
and at the base of the door frame, as shown in Figure 2. The distance from Y to Z is 37 inches. What is the height of the door frame?

[asy]draw((0,0)--(34,0));
draw((1,0)--(1,30)--(11,30)--(11,0));
draw((21,0)--(21,30)--(31,30)--(31,0));
fill((11,0)--(11,2)--(15,2)--(15,0)--cycle,gray(.7));
fill((9,30)--(9,34)--(11,34)--(11,30)--cycle,gray(.7));
fill((31,0)--(31,4)--(33,4)--(33,0)--cycle,gray(.7));
fill((27,30)--(27,32)--(31,32)--(31,30)--cycle,gray(.7));
dot((11,2));
label("H",(11,2),NE);

dot((11,34));
label("W",(11,34),NW);

dot((31,32));
label("Y",(31,32),NW);

dot((31,4));
label("Z",(31,4),NW);

label("Figure 1",(6,0),S);
label("Figure 2",(26,0),S);[/asy]

You can average the given numbers and get 41 the right answer, but why does that work?

Let $ h$ represent the height of the door, $ l$ represent the length of one of the blocks (longer side), and $ w$ represent the width of one of the blocks (shorter side). Then we have
$ h\plus{}l\minus{}w\equal{}45$
$ h\minus{}l\plus{}w\equal{}37$
Adding, we have $ 2h\equal{}82\implies h\equal{}\boxed{41}$.


modular arithmetic
Diophantine Equations

Suppose that $a, b$, and $p$ are integers such that $b \equiv 1 \; \pmod{4}$, $p \equiv 3 \; \pmod{4}$, $p$ is prime, and if $q$ is any prime divisor of $a$ such that $q \equiv 3 \; \pmod{4}$, then $q^{p}\vert a^{2}$ and $p$ does not divide $q-1$ (if $q=p$, then also $q \vert b$). Show that the equation \[x^{2}+4a^{2}= y^{p}-b^{p}\] has no solutions in integers.

Conditions of this problem are a little bit weird.

Note that $ y \equiv 1 \pmod{4}$, since otherwise the right hand side would be congruent to $ 2$ or $ 3$ mod $ 4$ while the left is congruent to $ 0$ or $ 1$. Let us rewrite the equation in the form:
\[ x^2 \plus{} 4a^2 \equal{} (y\minus{}b)\frac{y^p\minus{}b^p}{y\minus{}b} \equal{} (y\minus{}b)(y^{p\minus{}1} \plus{} y^{p\minus{}2}b \plus{} ... \plus{} yb^{p\minus{}2} \plus{} b^{p\minus{}1}).\]
Since $ y \equiv b \equiv 1 \pmod{4}$ we obtain that $ \frac{y^p\minus{}b^p}{y\minus{}b} \equiv 1\plus{}1...\plus{}1 \equiv p \equiv 3 \pmod{4}$.
Let $ d \equal{} (y,b)$ and write $ y \equal{} dy_0$, $ b\equal{}db_0$. Then $ \frac{y^p\minus{}b^p}{y\minus{}b} \equal{} d^{p\minus{}1} \frac{y_0^p\minus{}b_0^p}{y_0\minus{}b_0}$ and since $ d^{p\minus{}1} \equiv 1 \pmod{4}$ we conclude that $ \frac{y_0^p\minus{}b_0^p}{y_0\minus{}b_0} \equiv 3 \pmod{4}$. Therefore this number has a prime divisor of the form $ 4k\plus{}3$, call it $ q$. Since $ y_0, b_0$ are coprime $ q$ is not a divisor of $ y_0, b_0$ and we can easily prove that order of $ \frac{y_0}{b_0}$ mod $ q$ is $ p$, so $ p|q\minus{}1$. But at the same time we have $ q|x^2 \plus{} (2a)^2$ and since $ q\equal{}4k\plus{}3$ it is well known that we must have $ q|x$ and $ q|a$. That's a contradiction with the given condition.

I haven't used all conditions, is it wrong?