limit integration inequalities calculus function logarithms

here is the hardest limit problem i have ever seen.try to crack it :wink:

$ \lim_{n\rightarrow\infty}\sqrt[n]{\int_{1}^{3}x^{n/x} dx}$

EDIT:
wrong post..
I had just posted a wrong solution...

But I guess the answer is e^(1/e)...

Nice Problem  :)
What I did is the following:  Could someone help me with sharpen the left side of the inequality:

Now, if m and M (attained at e) be the global minimum and maximum of f(x)= the integrand of the integral, then
m(3-1) <= the integral <=  (3-1) e^(n/e)

Taking nth roots throughout, and allowing n to tend to infinity,
we get
1<== the required limit <= e^ (1/e)
Could some one tell me what i cud do abt the left side to finish up the problem using a sandwich argument  :?:

I think the answer is three..
My method..
$ L \equal{} (\lim_{x \to \infty}(\int_{1}^{3} x^(\frac {n}{x})dx))^\frac {1} {n} \Longrightarrow log L \equal{} \frac {(\lim_{x \to \infty}(\int_{1}^{3} x^(\frac{n}{x})dx))} {n} Using L \minus{} Hospital's rule, \Longrightarrow log L \equal{} \frac {(\lim_{x \to \infty}(\frac {\int_{1}^{3} log x x^(\frac{n}{x}) dx} {x}))} {(\lim_{x \to \infty}(\int_{1}^{3} x^(\frac{n}{x})dx))} \equal{} \frac {L log 3} {3} Comparing \frac {log 3}{3} \equal{} \frac {log L}{L}, L \equal{} 3.$

[quote="Euclidean Geometer"]Illa da chapps,
U ve terribly misunderstood da problem,
n tends to infinity not x  :P[/quote]

Hey you know tamil??? And I was just about to post that..

The answer is $ e^{\frac {1}{e}}$
[b]Claim[/b]
Let $ f(x)$ be a positive continious function in an closed interval $ [a,b]$ such that it has a maximum at $ x_{0}$ then we have
$ \boxed{\lim_{n \to \infty} (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \equal{} f(x_{0})}$

[b]Proof[/b]
clearly we have
$ (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \leq f(x_{0})(b \minus{} a)^{\frac {1}{n}}$
now by the continuity property we have for every $ \epsilon \in (0,1)$ we have a $ \delta < \frac {1}{2}$ such that for all $ x$ satisfying $ |x \minus{} x_{0}| < \delta$ we have $ f(x) > f(x_{0}) \minus{} \epsilon$
hence we have since $ f(x)$ is positive in $ [a,b]$ and using the above bound we have
$ (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \geq (2\delta(b \minus{} a))^{\frac {1}{n}} f(x_{0})$
now observe that since $ 0 < \delta < 1$ and if $ u > v$ then
$ \delta^{v} > \delta^{u}$ taking power $ \frac {1}{uv}$ we have
$ \delta^{\frac {1}{u}} > \delta^{\frac {1}{v}}$
now since $ 1 \minus{} \epsilon$ is postive we  choose a $ N \equal{} \frac {\ln(2 \delta)}{\ln (1 \minus{} \epsilon)}$
we have for all $ n > N$
$ (2\delta)^\frac {1}{N} > 1 \minus{} \epsilon$
thus we have
$ (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \geq (2\delta(b \minus{} a))^{\frac {1}{n}} f(x_{0}) > (1 \minus{} \epsilon)(f(x_{0}) \minus{} \epsilon)(b \minus{} a)^{\frac {1}{n}}$
thus we have for $ \epsilon \in (0,1)$ the above inequality [i]eventually[/i] holds
thus we claim
$ (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \geq f(x_{0})(b \minus{} a)^{\frac {1}{n}}$
because consider the function $ \Psi(\epsilon) \equal{} (1 \minus{} \epsilon)(f(x_{0}) \minus{} \epsilon),\epsilon \in [0,1]$
it is continious in $ \epsilon$ also $ \Psi(0) \equal{} f(x_{0}),\Psi(1) \equal{} 0$
so we have by intermediate value theorem [b]if the integral is less than $ f(x_{0})$[/b] then there exists some $ \epsilon^{'} \in (0,1)$ for which
$ (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \equal{} \Psi(\epsilon^{'})$
a contradiction so we have

$ f(x_{0})(b \minus{} a)^{\frac {1}{n}} \geq (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \geq f(x_{0})(b \minus{} a)^{\frac {1}{n}}$ taking $ n \to \infty$
we get
$ \boxed{\lim_{n \to \infty} (\int_{a}^{b} f(x)^{n} dx)^{\frac {1}{n}} \equal{} f(x_{0})}$

Here $ f(x) \equal{} x^{\frac {1}{x}}$ so the limit is $ e^{\frac {1}{e}}$

[quote="pardesi"]
[b]Claim[/b]
Let $ f(x)$ be a positive continious function in an closed interval $ [a,b]$ such that it has a maximum at $ x_{0}$ then we have
$ \boxed{\lim_{n \to \infty} (\int f(x)^{n} dx)^{\frac {1}{n}} \equal{} f(x_{0})}$

[/quote]

Wasn't this a sum that you gave before??? Or was it in the calculus forum?

Anyhow great job. The power of small age rules!!  :10:  :10:  :clap:

yes it was the sum posted not so properly  by [b]alter ego[/b] not me :) ...had tried it partially the other day...just had to finish it now

[quote="pardesi"]

[b]Proof[/b]
clearly we have
$ (\int f(x)^{n} dx)^{\frac {1}{n}} \leq f(x_{0})$[/quote]

Hmm..Is this extremely obvious for any given f(x)???

actually i missed out the limits were $ a$ to $ b$..
the actual integral shud be $ (\int_{a}^{b}f(x)^{n})^{\frac{1}{n}} \leq f(x_{0})(b\minus{}a)^{\frac{1}{n}}$
I AM EDITING THAT

I was breaking my head over that and then realized you forgot (b-a)

Im sorry, the answer is 3^(1/3) not e^(1/e)..
And I did n-> infinity not x...
L hospitals because of infinity/infinity form...

no i proved it....it's surely $ e^{\frac{1}{e}}$..... :D

do elucidate ur method...we wanna see it properly... :D

now by the continuity property we have for every $ \epsilon \in (0,1)$ we have a $ \delta < \frac {1}{2}$ such that for all $ x$ satisfying $ |x \minus{} x_{0}| < \delta$ we have $ f(x) > f(x_{0}) \minus{} \epsilon$
How da?? :?:  :(

because as $ f$ is continious the values of the function must smoothly tend towards $ f(x_{0})$ hence as we get closer to $ x \equal{} x_{0}$ the values must get closer to $ f(x_{0})$....further we can get as close to $ f(x_{0})$ as we can this follows by thee definition of continious function...
the definition says $ f$ is conitnious at $ x_{0}$ if $ \lim_{x \to x_{o}}f(x) \equal{} f(x_{0})$ this means at nearby $ x \equal{} x_{0}$ we have values close to $ f(x_{0})$ since they get arbitrarily close after some $ x$ on both the sides(that is right and left) i can always find an $ \delta$ to any $ \epsilon$...
otherwise we can argue on the basis of contradiction because if sucha thing doesn't happen then the function can't be continious as $ f(x_{0})$ is the max

[quote="SOUMYASHANT NAYAK"]Chhapli ..

its e^(1/e) and not 3^(1/3)..

Mathematica 5.1 rules... :rotfl:

I just put n = 10000,it gives 1.44....which is e^(1/e) approx...[/quote]

That is exactly what I would have done had this been an objective question in JEE or FIITJEE>  :rotfl:  :rotfl:

Shortcuts rule.....

[quote="Chapplikka"]Hey is the integral <= f(x0)?? Its less than or equal to (b-a)^(1/n) only na??[/quote]
i don't get u :maybe:

[quote="Chapplikka"]Hey is the integral <= f(x0)?? Its less than or equal to (b-a)^(1/n) only na??[/quote]
ok i edited that actually since $ \lim_{n \to \infty}(b\minus{}a)^{\frac{1}{n}}\equal{}1$ so i am forgetting that :D

[quote="SOUMYASHANT NAYAK"]Chhapli ..

its e^(1/e) and not 3^(1/3)..

Mathematica 5.1 rules... :rotfl:

I just put n = 10000,it gives 1.44....which is e^(1/e) approx...[/quote]

wel for d fact is even 3^(1/3) gives 1.44... :o
but den soumyashant could i know how u got 1.44.. by substitutin n=10000 :?:


Slove the equation
 $ x^3\minus{}8x^2\plus{}23x\minus{}78\equal{}6(1\plus{}\frac{5}{x})\sqrt{4x\minus{}1}$

hello, $ x\equal{}4\plus{}\sqrt{11}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, $ x \equal{} 4 \plus{} \sqrt {11}$.
Sonnhard.[/quote]
Dear Dr Sonnhard Graubner, why do you always just state the answer you get from maple or whatever?
You always do this, and most times with wrong answers, why? I'm pretty sure the poster wants a solution to the problem, and not just an answer.
Also, there should be at least two more complex roots to this..

hello, you have right, $ x_{2_3} \approx 0.4781 \plus{} \minus{} 4.0744i$ are also solutions.
Sonnhard.

[quote="Dr Sonnhard Graubner"]helleo, you have right, $ x_{2_3} \approx 0.4781 \plus{} \minus{} 4.0744i$ are also solutions.
Sonnhard.[/quote]
Now people might think u r using some software.
Its ok you are sending solutions. But few are interested in those. A proof would be more appretiated.


From a cube with side lengths of three inches, a small tetrahedron with side lengths of one inch is cut from each of the vertices. What is the maximum number of edges of the resulting polyhedron?

Add $ 3\cdot8$ edges for each vertice to get $ \boxed{36}$.

Hmm... I'm not all that sure if it's correct though.

It doesn't seem possible to cut regular tetrahedron's out of the corners of a cube in this way.  Am i wrong?

It seems that while a cube has angles of 90 degrees between edges, the angles between the edges of a regular tetrahedron is 60 degrees.  When the problem says "a tetrahedron with side lengths of one inch" it sounds like it's talking of regular tetrahedrons.  Should the problem say "a tetrahedron with side lengths of no more than one inch"?  If some edges are 1 and others are 3 or more, the problem can be rather confounding, though it seems the answer would still be 36.

This is not an area I'm good at.  I'm really asking.

I'm not really sure what the gamebot means when referring to the tetrahedron being cut out of the vertices of the cube. So, do like take the vertice out of the cube... ?


Umm, well...
Everyone talks about this guy called Arnav..
But, seriously, WHO IS HE?
BTW, visit alan's blog!  :D

Google "Arnav Tripathy." His AOPS sn is MysticTerminator

So basically, he is a guy in high school.

[quote="trigonometry1.01"]POOR POOR HELPLESS CHILD[/quote]

AMEN
your lack of knowledge is unhealthy.

Hmmmmm I think people should stop spamming.


If $ a,b,c > 0$ show that :
$ \sum \frac{a}{b^2\plus{}bc \plus{} c^2} \geq \frac{3}{a\plus{}b\plus{}c}$

good prob:
now by cs :
\[ \sum {a({b^2}}  \plus{} bc \plus{} {c^2}).{\sum {\frac{a}
{{{b^2} \plus{} bc \plus{} {c^2}}} \geqslant \left( {\sum a } \right)} ^2}\]
it's enough to show that:
\[ \frac{{{{\left( {\sum a } \right)}^2}}}
{{\sum {a({b^2}}  \plus{} bc \plus{} {c^2})}} \geqslant \frac{3}
{{\sum a }} \Leftrightarrow \sum {{a^3} \geqslant 3abc}\] true!

[quote="hendrata01"]If $ a,b,c > 0$ show that :
$ \sum \frac {a}{b^2 \plus{} bc \plus{} c^2} \geq \frac {3}{a \plus{} b \plus{} c}$[/quote]

It's trival by Cauchy-Schwarz or AM-GM,...

The following Inequality is stronger

$ \sum\frac a{2b^2\plus{}bc\plus{}2c^2}\ge\frac2{5}\left(\sum\frac1{b\plus{}c}\right)$

This one is more easier and i can say it comes from dduclam's :

$ \sum_{cyc}\frac{c}{7a^2\plus{}6ab\plus{}7b^2}\ge \frac{1}{10}\sum_{cyc}\frac{1}{a\plus{}b}$

[quote="hendrata01"]If $ a,b,c > 0$ show that :
$ \sum \frac {a}{b^2 \plus{} bc \plus{} c^2} \geq \frac {3}{a \plus{} b \plus{} c}$[/quote]
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ ate non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that:
\[ \frac{a}{4b^2\plus{}bc\plus{}4c^2}\plus{}\frac{b}{4a^2\plus{}ac\plus{}4c^2}\plus{}\frac{c}{4a^2\plus{}ab\plus{}4b^2}\geq\frac{1}{a\plus{}b\plus{}c}\]

[quote="arqady"][quote="hendrata01"]If $ a,b,c > 0$ show that :
$ \sum \frac {a}{b^2 \plus{} bc \plus{} c^2} \geq \frac {3}{a \plus{} b \plus{} c}$[/quote]
The following inequality is also true.
Let $ a,$ $ b$ and $ c$ ate non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a}{4b^2 \plus{} bc \plus{} 4c^2} \plus{} \frac {b}{4a^2 \plus{} ac \plus{} 4c^2} \plus{} \frac {c}{4a^2 \plus{} ab \plus{} 4b^2}\geq\frac {1}{a \plus{} b \plus{} c}
\]
[/quote]

It is obvious by Cauchy-Schwarz. (Problem 451.3 in my ebook).
It is also consequence of a problem which proposed by me, hungkhtn and can_hang:
\[ \sum\frac a{\sqrt{4b^2\plus{}bc\plus{}4c^2}}\ge1\]

[quote="dduclam"]
It is also consequence of a problem which proposed by me, hungkhtn and can_hang:
\[ \sum\frac a{\sqrt {4b^2 \plus{} bc \plus{} 4c^2}}\ge1
\]
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183367

[quote="arqady"]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183367[/quote]
It makes me remember my old result:  :)
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ \frac {a}{\sqrt {4b \plus{} 4c^2 \plus{} 1}} \plus{} \frac {b}{\sqrt {4c \plus{} 4a^2 \plus{} 1}} \plus{} \frac {c}{\sqrt {4a \plus{} 4b^2 \plus{} 1}} \geq\ 1$
[quote="nguoivn"][quote="dduclam"]
The following Inequality is stronger

$ \sum\frac a{2b^2 \plus{} bc \plus{} 2c^2}\ge\frac2{5}\left(\sum\frac1{b \plus{} c}\right)$[/quote]
It's obviously trues by CS and Schur  :)[/quote]
And here is my proof:  :)


We will mass expand and simplify and then nicely factor the expression into:
$(y-z)(y-x)(z-x)(x+y+z)=0$
Because $(x+y+z)$ is not $0$, one of $(y-z),(y-x),(z-x)$ is $0$ so have $x=y$,$y=x$,or $x=z$, and the result follows.
To prove the other part, assume wolog $x=y$ and subst $x$ instead of $y$.
The expression becomes:
$\frac{x^2-xz}{x+z}+\frac{xz-x^2}{x+z}+\frac{z(x-x)}{2x}$ and that is $0$.
QED

[quote="perfect_radio"]Let $x,y,z$ be real numbers whose sum is $\neq 0$. Prove that
\[ \frac{x(y-z)}{y+z} + \frac{y(z-x)}{z+x} + \frac{z(x-y)}{x+y} = 0  \]
holds if and only if two of the numbers are equal.

[i]Virgil Nicula[/i][/quote]
Solution:
---------
The proof uses the following well known trick to simplify the expression on the left hand side $(LHS)$ of the given equation: $x-y = (x-z) + (z-y)$

$LHS$
$= x\cdot\frac{y-z}{y+z} + y\cdot\frac{z-x}{z+x} + z\cdot\frac{(x-z)+(z-y)}{x+y}$
$= (y-z)(\frac{x}{y+z}-\frac{z}{x+y}) + (z-x)(\frac{y}{z+x}-\frac{z}{x+y})$
$= (y-z)\frac{(x^2+xy-yz-z^2)}{(x+y)(y+z)} + (z-x)\frac{(xy+y^2-z^2-zx)}{(z+x)(x+y)}$
$= (y-z)\biggl(\frac{(x-z)(x+z)+y(x-z)}{(x+y)(y+z)}\biggr) + (z-x)\biggl(\frac{(y-z)(y+z)+x(y-z)}{(z+x)(x+y)}\biggr)$
$= (y-z)(x-z)\frac{(x+y+z)}{(x+y)(y+z)} + (z-x)(y-z)\frac{(x+y+z)}{(z+x)(x+y)}$
$= \frac{(y-z)(x-z)(x+y+z)}{(x+y)}\cdot\Bigl(\frac{1}{(y+z)}-\frac{1}{(z+x)}\Bigr)$
$= -\frac{(x-y)(y-z)(z-x)(x+y+z)}{(x+y)(y+z)(z+x)}$

Now, we know that $x+y+z\neq 0.$
So, $LHS = 0$
$\Longleftrightarrow \frac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)} = 0$
$\Longleftrightarrow (x-y)(y-z)(z-x) = 0$
$\Longleftrightarrow$ if any two of $x,y,z$ are equal.

[quote="perfect_radio"]Let $x,y,z$ be real numbers whose sum is $\neq 0$. Prove that
\[ \frac{x(y-z)}{y+z} + \frac{y(z-x)}{z+x} + \frac{z(x-y)}{x+y} = 0  \]
holds if and only if two of the numbers are equal.

[i]Virgil Nicula[/i][/quote]

$f(x)=(x+y)(y+z)(z+x)\left(\frac{x(y-z)}{y+z} + \frac{y(z-x)}{z+x} + \frac{z(x-y)}{x+y}\right)$

say $f(x)$ is a polynomial about $x$, where $y,z$ are constants, then

through substitution, $f(y)=f(-x-y)=0$, then by factor theorem

$f(x)=(x-y)(x+y+z)q(x)$, ($q(x)$ is a polynomial) we note that $f(x)$ is 3rd degree, so $q(x)$ is degree 1, also by symmetry, we have the other variables as factors, so

$f(x)=(x-y)(y-z)(z-x)(x+y+z)q_2(x)$

then $q_2(x)$ is constant due to degrees

then we can plug in arbitrary values of x,y,z to find $q_2(x)=1$

the rest follows by zero product property

[quote="matt276eagles"][quote="rem"]We will mass expand and simplify and then nicely factor the expression into:
$(y-z)(y-x)(z-x)(x+y+z)=0$
[/quote]

Could you explain that step more?  :)[/quote]
Mass expand to get:
$x^3y- x^3z- x^2z^2 + x^2y^2 + x^2yz -x^2zy+ y^2xz- z^2xy- y^3x+ y^3z+ y^2z^2- y^2x^2+ y^2xz- y^2zx- x^2yz+ z^2xy+ z^3x- z^3y+ z^2x^2- y^2x^2+ z^2xy- z^2yx+ x^2zy- y^2zx$
$=x^3y- y^3x+ y^3z- z^3y+ z^3x- x^3z$ (*)
$=xy(x-y)(x+y)+z(x-y)(x^2+xy+x^2)+z^3(x-y)$
Because the expression (*) is symmetrical(ie we can replace $y$ with $z$ and get the same result), it is divisible by $(x-y),(x-z),(y-z)$ and then, dividing the polynomial by $(x-y),(x-z),(y-z)$, get $(x+y+z)$. therefore:
$x^3y- y^3x+ y^3z- z^3y+ z^3x- x^3z=(y-z)(y-x)(z-x) (x+y+z)=0$.


Let $ a$, $ b$, $ c$ be positive real numbers.

Prove that:

$ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{\sqrt[3]{abc}\cdot\left(\sqrt[3]{abc}+1\right)}$.

I mean, this is not extremely hard:
Since
$ \frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca}$
$ =(1+abc)(\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)})+3$,
we can write the desired inequality in the form $ \frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca}\geq \frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}$. Now, break $ \frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca}$ into
 $ \frac{1+a}{ab+a}+\frac{b+1}{bc+b}+\frac{c+1}{ca+c}\geq\frac{3}{\sqrt[3]{abc}}$ and $ \frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1}\geq 3\sqrt[3]{abc}$, and it is proven.

Oh my..
I understand now,thanks a lot!!

But I dont think its "easy" to notice
that we can write another form like this... ;)
BTW, this solution is so nice :lol:
I will never think about this~~
thanks your help

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 for a weaker inequality:

$ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{abc+1}$.

Note that some proofs of this weaker inequality can be easily modified to prove the above inequality

$ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{\sqrt[3]{abc}\left(\sqrt[3]{abc}+1\right)}$.

  Darij

The keypoint of my solution is to set $ a=t \frac{q}{p}, b=t\frac{r}{q}, c=t\frac{p}{r}$ where $ t^{3}=abc$.

The solution I know use only AM-GM .
We have $(1+abc)(\sum\frac{1}{a(1+b)})+3=\sum\frac{1+a}{a(1+b)}+\sum\frac{b(c+1)}{1+b}\geq \frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}$

I think we can also kill it by a kind of natural method.

Take $a=\lambda x/y,\ b=\lambda y/z, c=\lambda z/x\ (x,y,z,\lambda>0)$ we just need to prove
$\frac{yz}{\lambda xy+zx}+\frac{zx}{\lambda yz+xy}+\frac{xy}{\lambda zx+yz}\geqslant\frac{3}{\lambda+1}$
Take $u=yz,\ v=zx,\ w=xy$ and it comes down to
$\frac{u}{\lambda w+v}+\frac{v}{\lambda u+w}+\frac{w}{\lambda v+u}\geqslant\frac{3}{\lambda+1}$
Use Cauchy-Schwarz we have
$\left(\frac{u}{\lambda w+v}+\frac{v}{\lambda u+w}+\frac{w}{\lambda v+u}\right)[u(\lambda w+v)+v(\lambda u+w)+w(\lambda v+u)]\\ \geqslant(u+v+w)^{2}$
And notice that
$u(\lambda w+v)+v(\lambda u+w)+w(\lambda v+u)=(\lambda+1)(uv+vw+wu)\leqslant\frac{\lambda+1}{3}(u+v+w)^{2}$
So we get what we need.


function
calculus
derivative
inequalities
inequalities proposed

Let $ x,y,z$ be positive real numbers belonging to the interval $ [a,b]$. Find the best M (which does not depend on $ x,y,z$) such that

$ x\plus{}y\plus{}z\le 3M\sqrt[3] {xyz}.$

The formula for the best $ M$ is nice. A nice M (which is not the best) is

$ M \equal{} \left(\frac {\sqrt\frac a{b} \plus{} \sqrt\frac b{a}}{2}\right)^\frac 4{3}$.

[quote="Vasc"]Let $ x,y,z$ be positive real numbers belonging to the interval $ [a,b]$. Find the best M (which does not depend on $ x,y,z$) such that

$ x \plus{} y \plus{} z\le 3M\sqrt [3] {xyz}.$[/quote]
If I amnot wrong, then
$ M_{\min} \equal{} \frac{2a\plus{}b}{3\sqrt[3]{a^2b}}.$
Am I right, Vasc?

[quote="Allnames"][quote="Vasc"]Let $ x,y,z$ be positive real numbers belonging to the interval $ [a,b]$. Find the best M (which does not depend on $ x,y,z$) such that

$ x \plus{} y \plus{} z\le 3M\sqrt [3] {xyz}.$[/quote]
Let $ f(x;y;z) \equal{} \frac {x \plus{} y \plus{} z}{\sqrt [3] {xyz}}$
 We will find maximum value of $ f(x;y;z)$ in interval
$ [a;b]$.
Now fix $ y;z$ our $ f(x;y;z)$ is a function with $ x$- variable.
Notice that $ g(t) \equal{} \frac {t^3 \plus{} k}{mt}$ (where $ k;m$ are positive constants ) is  concave ( $ g''(x) > 0$) thus its maximum value will hold when $ x \equal{} a$ or $ x \equal{} b$.
Repeat that argument for $ y$; $ z$; respectively we obtain $ f(x;y;z)$ get its maximum value when $ (x;y;z)$ is either $ (a;a;a)$ or $ (a;a;b)$ or $ (a;b;b)$ or $ (b;b;b)$.
The last is only checking what is the greatest value of above triples.
That is enough.[/quote]
Your proof is correct. Also, we can prove it without convexity (and derivative g").
In addition, the inequality holds:

$ a_1 \plus{} a_2 \plus{} ... \plus{} a_n\le nM\sqrt [n]{a_1a_2...a_n}$,

where $ M \equal{} \left(\frac {\sqrt\frac a{b} \plus{} \sqrt\frac b{a}}{2}\right)^{2 \minus{} \frac 2{n}}$.

for the collection
\[ \sqrt{\sum_{i\equal{}1}^na_i^2}\sqrt{\sum\limits_{i\equal{}1}^nb_i^2}\leq \frac{1}{2}\left(\sqrt{\frac{M_1M_2}{m_1m_2}}\plus{}\sqrt{\frac{m_1m_2}{M_1M_2}}\right)\sum\limits_{i\equal{}1}^na_ib_i\]
where $ 0<m_1\leq a_i\leq M_1, 0<m_2\leq b_i\leq M_2$


If $ a,b,c$ are non-negative real numbers such that $ a\plus{}b\plus{}c\equal{}3$,prove that \[ \frac{a^2}{b^2\plus{}2}\plus{}\frac{b^2}{c^2\plus{}2}\plus{}\frac{c^2}{a^2\plus{}2} \geq 1\]

It reminds me the inequality of Vasc which was

$ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3$ prove that $ \sum\frac{a}{b\plus{}2}\leq 1$

but I can't find a link for this one  :(  (could anyone find ?)

silouan, I can't find a link too, but I know the Vasc's proof:
Let $ b$ is a middle number of $ \{a,b,c\}.$ Then
$ \sum_{cyc}\frac {a}{b \plus{} 2}\leq1\Leftrightarrow(b \minus{} 1)^2(b \plus{} 2) \minus{} a(b \minus{} a)(b \minus{} c)\geq0.$ :lol:

I knew also this proof from Vasc's book ,but I am wondering if there exist another proof ...


[b]Five balls are numbered 1 to 5. Three boxes are numbered 1 to 3. How many ways can the balls be put in the boxes if two boxes have two balls each and the other box has the remaining ball?[/b]

[hide] I thought there are ${5\choose 2}{3\choose 2}\prod uct 3!=180$ ways. But it's not the exact answer in the text book...:([/hide]

5x4/2 x 3x2/2 = 30 ways

i dunno i might be wrong.

[quote="fizsle"]You can choose 2 boxes in 3 ways
You can select 2 balls for the first box in 5C2 ways
you can select 2 balls for the next box in 3 ways

so answer = 3 x 10 x 3 = 90 ways....

Am i right?  :?:[/quote]

That's the answer in the text book!

[hide=" :D "]Okay, first, there are $\binom{3}{2}=3$ ways to choose 2 boxes with 2 balls.  Next, you have to choose 2 of the 5 to put in the first box, which is $\binom{5}{2}=10$  Finally, you have to choose 2 of the 3 balls left to put in the second box.  $\binom{3}{2}=3$  The final answer is 3*10*3=90.[/hide]


For every $M>0$, prove that there exist integers $a>b>M$ such that the set of prime divisors of $a^2+1$ coincides with that of $b^2+1$.

Enjoy!  :D

--Vesselin

well, any success on it?

vess, any hints?

You've just shown that the set of prime divisors of $a^2+1$ is a subset of the set of prime divisors of $b^2+1$ for some $b$. The two sets must be precisely the same. Am I misunderstanding you?

grobber, of course your'e right...everyone can make stupid things :)

ok, this time will be better :)

Let's proove that there are infinitely many positive integers $(a,b)$ , such that $a^2+1=5(b^2+1)$,
and $5|b^2+1$, what will give a positive answer for this problem.

$a^2+1=5(b^2+1)$ is equivalent to $a^2-5b^2=4$. Applying the method introduced at:
[url=http://www.mathlinks.ro/Forum/topic-40958.html]www.mathlinks.ro/Forum/topic-40958.html[/url], we
want to find anifinitely many solutions of $a^2-5b^2=4$. Let's take the first pair satysfying this equation
$(3,1)$. Now we find any solution of $a^2-5b^2=1$, for instance $(9,4)$, so if $(x,y)$ is our solution of
the equation $x^2-5y^2=4$ then $(9x+20y,4x+9y)$ is also a solution. Actually, if we consider $mod$ $5$ we can reduce our transormation $(x,y)->(9x+20y,4x+9y)$ to $(x,y)->(4x,4(x+y))=(-x,-(x+y))$ $(mod 5)$, so we have
$(x,y)->(-x,-y-x)->(x,y+2x)->(-x,-y-3x)-> ... ->
((-1)^nx,(-1)^n(y+nx))->...(\mod 5)$
But because of the fact that $(x_0,y_0)=(3,1)$ - our first solution, then $(3,1)->(-3,-4)->(3,7)->(-3,-10)->(3,13)->... $ $(\mod 5)$.
Now we take the solutions $(x,y)$ for which $y^2+1=0 (mod 5)$ what's equivalent to $y=2 $or $3 (mod 5)$.
There are infinitely many of those $y$ because there are infinitely many numbers $2$ or $3$ $(\mod 5)$ in the sequence $(-1)^n(1+3n)$, so there are infinitely many pairs $(x,y)$ for which $x^2+1=5(y^2+1)$
and $y^2+1=0$ $(\mod 5)$.

ps. What's interesting , looking at $(3,1)->(-3,-4)->(3,7)->(-3,-10)->(3,13)->...$ $(\mod 5)$ we can realise how many of these satisfying $(x,y)$ are in consecutive solutions, quite many...


http://cms.math.ca/MediaReleases/2007/cmo.html

u guys are awesome!!   :D

hey congratz Cynthia on winning CMO... lol i remember how after she wrote it she thought she did bad (so modest)..

Congratulations to all Winners and people in the Honourable Mention.

sorry for the double post..but you can also find it here:

[url]http://www.newswire.ca/en/releases/archive/May2007/25/c4925.html[/url]

Congratulations to Cynthia, Johnathan and other winners! :thumbup:  :clap2:

Congrats to the other winners!

I want to meet Jarno! Does anyone know if he's coming to the Waterloo Seminar?

but who is going to IMO? we have seven winners, but only 6 of them will go, right?

alex, you will go for sure, i'll see you in the awards ceremoney for the tournment of the towns :lol:
l

[quote="cynthia1991"]I was not trying to be modest. I just feelthat I am not really good compared to some people, so there must be somebody who did better than me.[/quote]

o..well you won..so congratz again..


MiTex netinstaller seems to have stopped not crashed, about 75% complete.

Is it safe to close and then restart without re-dl-ing all the files??

closed out the installer and it upon re-startup does a check to see what has been dl-ed before and even repairs files with bad chksums.  Which I had 2, and continues.


trigonometry
calculus
calculus computations

Let $f(x,y)=x^3\sin\left(\frac1x\right)+y^2$ when $x\ne 0$ and $f(0,y)=y^2$ when $x=0$.

Notation: $D_1f(0,0)=\frac{\partial f}{\partial x}(0,0)$

My question is...
I don't know how to show that $D_1f(0,0)$ is not continuous at $(0,0)$ because I think it is continuous. Can somebody clarify this for me?

Yes, it's continuous.

This is basically a one-dimensional problem- $y$ is uninteresting.

Thank you jmerry, because the book asked me to show that it is not continuous and I insisted that the $D_1$ is continuous.


analytic geometry
geometry
circumcircle
conics
perpendicular bisector
power of a point

points x(0,0), y(4,6), and z(10,0) are given. how many units are in the length of the radius of the circle, which passes through the points x,y, and z? express your answer to the nearest tenth. how do you do this?
thanks

Here's one solution, and I believe the fastest, although there are doubtless others.

We have a circle and an inscribed triangle.  There is a theorem, I believe, which says that the perpendicular bisectors of the sides of a triangle meet, and in the case of an inscribed triangle, the point where they meet is the center.

So drawing a quick sketch shows that the three perpendicular bisectors meet at (5,1), which means that the radius is sqrt(25+1) = sqrt(26) = 5.1

Not as fast as mathfanatic's solution, I suppose, but bear in mind that since the circle passes through all three points, they have to be equidistant from the center. That should enable you to come up with an equation that you can solve for the coordinates of the center. And yes, it's (5,1).

Here's an alternative solution:

We have three points on a circle, (0,0), (4,6), and (10,0).

The standard form of a circle's equation is (x-h):^2: + (y-k):^2: = r:^2:

Plug in the first and last points to get:
(0-h):^2: + (0-k):^2: = r:^2:
(10-h):^2: + (0-k):^2: = r:^2:

Rewrite as:
h:^2: + k:^2: = r:^2:
(10-h):^2: + k:^2: = r:^2:

The second equation can be expanded to:
h:^2: - 20h + 100 + k:^2: = r:^2:

Now, subracting the first equation from the rewritten second equation gives:
h:^2: - 20h + 100 + k:^2: - h:^2: - k:^2: = r:^2: - r:^2:
-20h + 100 = 0
20h = 100
h = 5

Now, plug in h = 5 to the equation h:^2: + k:^2: = r:^2:
25 + k:^2: = r:^2:

And plugging h = 5 and the point (4,6) into (x-h):^2: + (y-k):^2: = r:^2:,
(4-5):^2: + (6-k):^2: = r:^2:
1 + (6-k):^2: = r:^2:
1 + 36 - 12k + k:^2: = r:^2:

Now subracting the equation 25 + k:^2: = r:^2: from the above equation:
1 + 36 - 12k + k:^2: - 25 - k:^2: = r:^2: - r:^2:
12 - 12k = 0
k = 1

So the center point (h,k) is (5,1)

Using the distance formula, we know that the radius is
sqrt[(5-0):^2: + (1-0):^2:] = sqrt[25+1] = sqrt[26] = 5.1

Hope this helps!

- mathfanatic
  :D

How about something a little less computation-intensive, since we only want the radius and not the incenter.

K=abc/4R, where K is the area, a, b, c are the sides, and R is the circumradius.

You can prove that easily from the (extended) law of sines.

[quote="TripleM"][quote="mathfanatic"]
We have a circle and an inscribed triangle.  There is a theorem, I believe, which says that the perpendicular bisectors of the sides of a triangle meet, and in the case of an inscribed triangle, the point where they meet is the center.
[/quote]

Yep, that comes from the fact that the perpendicular bisector of a chord in a circle passes through the centre..[/quote]

Thanks, I wasn't sure why this was but I knew there was some reason...

And the second solution was just there to show the "book" method of doing it, if you didn't know the K=abc/4R, which I'd heard of but forgotten...

Since a circle is defined by three points, a triangle must have a circumcenter.

[quote="Nukular"]which of course assumes that a triangle has a circumcircle in the first place[quote] Yep, that comes from the fact that the perpendicular bisector of a chord in a circle passes through the centre.. [/quote][/quote]

Well, if you mean a circumscribed circle, that was in the given.  If you mean a circumcenter, all triangles have one.

HAHA... speaking of vicious circles....

how do you KNOW a circle is defined by three points?

See here is how u can prove a triangle has a circumcenter.

It is easy enough to show that all points on the perpendicular bisector of a line are equidistant from the two endpoints.
Consider triangle ABC.

Let m be the perp. bisector of AB. then all points on the line are equidistant from A,B.

Let n be the perp. bisector of BC. then all points on the line BC are equidistant from B, C. Since the triangle is non-degenerate, AB is not parallel to BC, => m is not parallel to n, but m is coplanar with n so they must intersect at a point O. O is equidistant from A, B, C, so it is the center of a circle that passes through all three points.

This is the circumcircle. Note also the third p. bisector must go through O as well, but this is not necessary to prove the result.

It is also a corollary that this circle is unique, since O is uniquely determined from A, B, and C, so OA has fixed length, and is the radius of the circle.

Thats how you prove three points determine a circle.

:)

You can prove a circle is defined by three points because the equation of a general circle is (x-h)^2 + (y-k)^2 = r^2, and we can solve for h, k, and r by solving a system of equations.

[color=cyan]But Simon, can you prove that that equation defines a circle?  You keep defending your statements without going far enough back.

Nukular, you didn't really prove the uniqueness of the circumcircle -- the intersection of the perpendicular bisectors of AB and AC could be a different point than the intersection of the perpendicular bisectors of AB and BC, according to your proof.  You needed to say that the perpendicular bisector represents [i]all[/i] points equidistant from the two endpoints of the segment, not just that all points on the perpendicular bisector are equidistant from the endpoints.  If you want to be really rigorous, that is :).[/color]

I believe that Descartes [i]defined[/i] a circle by that equation.

Is defining it by (x-h)^2 + (y-k)^2 = r^2 different from defining it by center and radius? The center is (h,k), and the radius is r.

[color=cyan]Descartes definitely did not have dibs on defining a circle, the Greeks (or quite possibly earlier peoples) did.  So they get their definition :).  Actually, I wasn't really thinking very hard about your definition when I wrote that, since yes, for a circle it does follow without really having to say very much.  Although you do have to say a little bit to make it truly formal, which I guess is what this whole little sidetrack is about.  I was really thinking about other conic sections, wehre it takes a bit of work to prove the equivalence of the standard and analytic defintions.[/color]

Sure, the Greeks defined a circle first, but Descartes updated the definition to make it more ``user-friendly.''

[color=cyan]That is definitely not more user-friendly.  And much less elegant than the definition in terms of points equidistant or in terms of a conic section, both of which at least tell you something about what you're looking at.[/color]

If all you want to do is visualize circles without drawing them, the Greek definition is terrific. If you want to solve challenging problems that involve circles, the Descartes definition is much more useful most of the time.

[color=cyan]That's so untrue . . . the things that make solving circle problems possible almost never have anything to do with analytic geometry.  All their nice properties make for terrible analytic expressions.  I mean, Ptolemy's Theorem, Power of a Point, cyclic quadrilaterals and all the nice angular stuff, arclength, [i]anything[/i] involving angles, even inversion, they all become nasty when you switch to coordinates.  I mean, go do the problem on (I think) the Looking For Answers forum that involves a circle and a right triangle.  I mean, it practically gives you axes, and it's still messy!  So, in summary, I disagree.[/color]


algebra
polynomial
integration
trigonometry
real analysis
real analysis solved

Hello,

I've been trying to solve this for a while now, but no success!  I mean, I have an expression for this but not a closed one. I derived something in terms of Tchebychev Polynomials :  cos(nx) = P_n(cos(x)). Any help is appreciated.

Compute the following primitive:
\[ \int \frac{x\sin\left(2004 x\right)}{\tan x}\ dx\]

Let's use the formula:

$\displaystyle \int fg' dx = fg - \int gf' dx.$

Where $f=x$ and $\displaystyle g' = \frac{\sin 2004x}{\tan x}$ so $ \displaystyle g= \int \frac{\sin 2004x}{\tan x} dx.$

So it looks like so:

$\displaystyle \int \frac{x \sin 2004 x}{\tan x} dx = x \int \frac{ \sin 2004x}{\tan x} dx - \int \left( \int \frac{\sin 2004 x}{\tan x} dx \right) dx.$

Now all we need it to compute: $\displaystyle \int \frac{\sin 2004x}{\tan x} dx.$

Thus: $F_{n}(x) = \int \frac{\sin 2nx}{\tan x} dx.$

For $n>1$ we have $\displaystyle F_{n}(x) - F_{n-1}(x) = \frac{1}{2n} \sin 2nx + \frac{1}{2n-2} \sin (2n-2)x.$

Then by adding these relations(n=1 to n you just have to compute $F_{2}(x) - F_{1}(x)$ which is easy) you get:

$\displaystyle F_{n}(x) = \frac{1}{2n} \sin nx +  \frac{4}{2n-2} \sin 2(n-1)x + \ldots + \frac{2}{4} \sin 4x + \frac{1}{2} \sin 2x.$ etc

I hope I am right! Is this okay?  :blush:

cheers! :D :D


Hi, im new to this so here is my inventory

1991 mathcounts nationals target and sprint
1994 mathcounts nationals target and sprint
1996 mathcounts nationals target
1999 mathcounts nationals answers sprint target team
2000 solutions
all in 2001 - 2008 nationals(not all have answers)
1994 state
2000- 2007, 2009 amc 10s
1996-2009 amc 12s (no 2008)
i have amc 8s from 1993-2005 2008 and answers from 1985-2005 and 2008
2002 aime

I want anything i dont have

email me if you are interested at deadbanannas@gmail.com

Um. Could a admin or mod please move this or just delete it?
I know you are new, but I'm going to teach you something. When you have already posted something, then please do not post the same thing. You have already posted in the Mathcounts Resources thread so please do not post again. Thnx.  :D

Well, its supposed to be in the other forum because that forum is all about trading tests. There is no need to start a new topic.


Let $ a,b,c>0$. Prove that

$ \frac{2a^2\minus{}bc}{a^2\plus{}2bc}\plus{}\frac{2b^2\minus{}ca}{b^2\plus{}2ca}\plus{}\frac{2c^2\minus{}ab}{c^2\plus{}2ab}\geq1$

[quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ \frac {2a^2 \minus{} bc}{a^2 \plus{} 2bc} \plus{} \frac {2b^2 \minus{} ca}{b^2 \plus{} 2ca} \plus{} \frac {2c^2 \minus{} ab}{c^2 \plus{} 2ab}\geq1$[/quote]
We have
\[ \sum \frac{2a^2\minus{}bc}{a^2\plus{}2bc}\geq \sum\frac{2a^2\minus{}bc}{a^2\plus{}b^2\plus{}c^2}\equal{}2\minus{}\frac{ab\plus{}cb\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\]
(from AM-GM)
$ \geq 2\minus{}1\equal{}1$, since $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ :)
Nice inequality. :)

[quote="Potla"]
\[ \frac {2a^2 \minus{} bc}{a^2 \plus{} 2bc}\geq \frac {2a^2 \minus{} bc}{a^2 \plus{} b^2 \plus{} c^2}\]
[/quote]

I think it isn't true if $ bc\ge 2a^2$ .

$ \sum_{cyc}\left(\frac {2a^2 \minus{} bc}{a^2 \plus{} 2bc}\plus{}1\right)\equal{}\sum_{cyc}\left(\frac{1}{2}\plus{}\frac{5}{2}.\frac{a^2}{a^2\plus{}2bc}\right) \ge \sum_{cyc}\left(\frac{1}{2}\plus{}\frac{5}{2}.\frac{a^2}{a^2\plus{}b^2\plus{}c^2}\right)\equal{}4$

[quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ \frac {2a^2 \minus{} bc}{a^2 \plus{} 2bc} \plus{} \frac {2b^2 \minus{} ca}{b^2 \plus{} 2ca} \plus{} \frac {2c^2 \minus{} ab}{c^2 \plus{} 2ab}\geq1$[/quote]

the inequality is equivalent  to
$ 5a^3b^3 \plus{} 5a^3c^3 \plus{} 5b^3c^3 \minus{} 15a^2b^2c^2\geq 0$

from AM-GM  the inequality holds


inequalities
geometry
circumcircle
geometry proposed

A triangle $ ABC$ $ D,E,F$ be points on $ BC,AC,AB$ respectively
$ R \equal{}$ circumradius
show that $ (DE\plus{}EF\plus{}FD)(\frac{1}{{AD}} \plus{} \frac{1}{{BE}} \plus{} \frac{1}{{CF}}
)   \ge \frac {AB\plus{}BC\plus{}CA}{R}$


analytic geometry
graphing lines
slope
search
geometry
3D geometry
calculus

We'll start easy.

A father and a son are driving through the Great Smoky Mountains when suddenly a bear jumps out. Swerving to miss the bear, they get into an accident and only the son survived. When they pull him into the ER, the doctor there says, "I can't operate on him,  he's my son." How could this have happened?

[quote]A father and a son are driving through the Great Smoky Mountains when suddenly a bear jumps out. Swerving to miss the bear, they get into an accident and only the son survived. When they pull him into the ER, the doctor there says, "I can't operate on him, he's my son." How could this have happened?[/quote]

Doctor was his mother.

There are 1 grandfather, 2 fathers, 2 sons and a grandson in a room. How many people are in the room.

[quote="NapoleonXIV"]
Doctor was his mother.

There are 1 grandfather, 2 fathers, 2 sons and a grandson in a room. How many people are in the room.[/quote]

It is possible he was a priest?

At least 3.

I'll add a puzzle in a second

Edit:

Two men went into a restaurant. They both ordered the same dish from the menu. After they tasted it one of the men immediately got up from the table, went outside of the restaurant and shot himself. Why?

[quote="Temperal"]The man had been on a deserted island with a few companions a while back. They had nothing to eat, and so were starving. One day, one of his companions went missing. That same day, his other companions and he ate what they told him was whatever dish was mentioned in the problem. Later they're rescued. However, when he eats the dish for real, he realized that the earlier dish wasn't the same and he had actually eaten his missing companion.

Hm, how about this one: five men are lying dead in the middle of the ocean with a boat next to them. What happened?[/quote]

Well, so far I've gathered that they must be on an island, so perhaps the boat is the observer who has spotted them?

Btw, nice post count :)

[quote="Temperal"]Hm, how about this one: five men are lying dead in the middle of the ocean with a boat next to them. What happened?[/quote]

They all jumped down to go snorkeling but they forgot to put the ladder down on their yacht ... so they couldn't get up and drowned.

Next one:
When they tried to throw Uncle John into the ocean, the wind simply blew him back onto the boat.

[quote="AiW"][quote="randomdragoon"]
Next one:
When they tried to throw Uncle John into the ocean, the wind simply blew him back onto the boat.[/quote]
I think, Uncle John was cremated.   :maybe:[/quote]

I agree with you. Riddle, please ^_^.

A man was outside taking a walk when it began to rain. He did not have an umbrella and he wasn't wearing a hat. His clothes were soaked, yet not a single hair on his head got wet. How could this happen?

He was bald.

A person begins asking himself questions one day. Then, he soon gains the ability to fly. How?

Does this have anything to do with Hitchhiker's guide to the Galaxy?

If it does, I'd say he figured out the key to flying was to jump and miss the ground.

If it doesn't, I'd say he asked himself, "does gravity exist?" and realized it didn't.

Maybe, the person asking himself questions was a wise philosophical old man who, one day, died?

Well, it was a noble attempt. How about that myth, I forgot the title to, but a man and his son were kept prisoner by a God, They wanted to escape and then picked up all of the eagle feathers that were dropped.

So then the man who was kept by the God asked himself why mankind couldn't fly. So he smothered wax over the eagle feathers and made two sets for him and his son. Eventually they started flying but then his son got too noble, flew up towards the sun, and the feathers melted and he dropped.

Is this it...?

Uh... he's male? Because males are more likely to be bald than females?

THe real reason why he is male is this:
[quote=happycupcake]
Someone is walking down the street. [b]He[/b]'s bald. What is the person's gender?[/quote]

I'm not sure if this has been used, but here goes. (This is a modified version of an original so the answer won't be wrong)
Riddle:
One day, the police heard a gunshot by a lake and arrived at the scene 3 minutes later. They found a man dead inside a hut. In his left hand, he's holding a gun. In his right hand, he has a recording. When the recording is played, it starts with the man talking about how horrible life is and how he wants it to end. The recording ends with a gunshot. The police are about to call it a suicide until you point out a very important clue. What is it?

Hint:
[hide]Nobody has touched the man yet. (Police, Doctors, etc.)[/hide]

That’s an old one. Who pressed stop on the recording?

[quote=ythomashu]what if he shot the gun without killing himself, stopped recording, and then killed himself?[/quote]

LOL

What is common, has a head and tail, is brown, but has no legs

Yeah, it was used in the first 100 posts of the marathon. :/

IDK where this is from, but
12 pears, hanging high
12 men, passing bye
Each took a pear
And left 11 hanging there

Idrk... maybe "Each" was the name of one of the men???

[quote=ythomashu]what if he shot the gun without killing himself, stopped recording, and then killed himself?[/quote]

*Vsauce music starts playing*


function
limit
real analysis
real analysis unsolved

Does the sequence of functions $ f_n(x)\equal{}nx^n$ converge in $ (\minus{}1,1)$? If so, does it converge uniformly?

I am trying to show that it does converge uniformly to the zero function via the epsilon definition of uniform convergence, but I am having trouble obtaining $ n_0$ for given epsilon.

Well, since the convergence is [i]not[/i] uniform, proving it is going to be a bit difficult!

Yes, the pointwise limit is the zero function. We can say that $ f_n$ converges uniformly to $ f$ on the set $ S$ if and only if
\[ \lim_{n\to\infty}\left(\sup_{x\in S}|f_n(x) \minus{} f(x)|\right) \equal{} 0.
\]
But in this case, $ \sup_{x\in( \minus{} 1,1)}|nx^{n \minus{} 1} \minus{} 0| \equal{} n$ and that doesn't go to zero.

It is true that for any given fixed $ \delta>0,$ $ nx^{n\minus{}1}$ converges uniformly to zero on $ [\minus{}1\plus{}\delta,1\minus{}\delta]$ Consider it a case of uniform convergence on compact subsets.


inequalities
function
algebra proposed
algebra

This problem has been posed with hint such as  N.T.TUAN did for entrance examination of Waseda University/Politics & Economics 1998.

In notes of  pbornsztein and other, has following : Find all continuous functions $f: \mathbb{R}\to \mathbb{R}$ such that $|f(x)-f(q)|\leq 7(x-q)^{2}\;\forall x\in\mathbb{R}\;\forall q\in\mathbb{Q}$. Solution is same as above.

It's easy to see that $f$ is continuous on $\mathbb Q$ so it can be extended uniquely (because $\mathbb Q$ is dense in $\mathbb R$) to a function $g: \mathbb R\to\mathbb R$ so that $g|_{\mathbb Q}\equiv f$. It is easy to see that $g$ satisfies the inequality $|g(x)-g(y)|\le(x-y)^{2},\forall x,y\in\mathbb R$ (just use the fact that $\mathbb Q$ is dense in $\mathbb R$ and that $g$ satisfies the ineq for rationals).

If we multiply the inequality with $|x-y|^{-1}$ and let $y\to x$ we get that $g'(x)=0,\forall x\in \mathbb R$, so $g(x)=c\in\mathbb R$ hence $f$ is constant.

Instead of $(x-y)^{2}$ we could have $(x-y)^{\alpha}, \alpha>1$ and the conclusion still holds.

[quote=mateivld]Let $f: \mathbb{Q}\rightarrow \mathbb{R}$ be a function such that \[|f(x)-f(y)|\leq (x-y)^{2}\] for all $x,y \in\mathbb{Q}$. Prove that $f$ is constant.[/quote]
I really enjoyed this problem :)
[hide=Solution]Let $P(x,y)$ be the asertion of the function $f$.
\\ $P(\frac{x+y}{2} ,y)\Rightarrow |f(\frac{x+y}{2}) -f(y)|\leq \frac{(x-y)^2}{4}$
\\ $P(x,\frac{x+y}{2})\Rightarrow |f(x)-f(\frac{x+y}{2})|\leq \frac{(x-y)^2}{4}$
\\ Summing these up and using $|x|+|y|\geq |x+y|$, we get $|f(x)-f(y)|\leq \frac{1}{2} (x-y)^2.$ And by induction, $|f(x)-f(y)|\leq \frac{1}{2^n} (x-y)^2$, for any natural number $n$, which means $|f(x)-f(y)|=0\iff f(x)=f(y)=c$, thus $f$ is constant.    [/hide]


Five friends have their pictures taken.  If they take pictures in
groups of three, how many different groups of people can have
their picture taken?

Solution 1: Use combinations, 5 C 3=10.
Solution 2: Say the 5 people are A,B,C,D, and E.
2 people must be "left out", so you could leave out
AB
AC
AD
AE
BC
BD
BE
CD
CE
DE
Thats 4+3+2+1=10.
Notice that AB is the same as BA since order doesnt matter.

[quote="FantasyLover"]$ \binom{5}{3} \equal{} \binom{5}{2} \equal{} 10$

answer : 10[/quote]

Wondering:  Noticed a pattern in $ \binom{x}{2}$.  Is it always equal to $ x\minus{}1\plus{}x\minus{}2\plus{}x\minus{}3\plus{}...\plus{}2\plus{}1$?

$ \binom{x}{2}\equal{}x(x\minus{}1)/2$ which is the sum of the numbers from $ 1$ to $ (x\minus{}1)$.


Solution please? I'm not exactly what you'd um...call...proficient at Basic and have no idea what that THEN 30 means either.

[quote="#H34N1"]Write the output of this BASIC program:

[code]
10 LET S=O
20 LET N=O
30 LET N=N+1
40 LET S=S-N
50 IF N<5 THEN 30
60 PRINT S
70 END
[/code]

(Yes, it was a nat'l sprint)[/quote]

Oh, umm..

[hide]
I made a list of N and S, starting with both having values of 0.

the list would look like this:

N = 0 1 2 3 4 5
S = -1 -3 -6 -10 -15

because line 50 means that if N < 5, go back to line 30 and repeat the processes. You stop with N = 5 because line 50 basically says that if N = 5, stop.

Then PRINT S.
[/hide]

[quote="indianamath"][quote="undefined117"][hide]
-15
[/hide]

Huh. That was a nationals? It's kinda easy..[/quote]

[b]1985[/b] nATS[/quote]

Oh so competitions in the past are "easier"..?

Read the problem more carefully... the answer is -10.

Oh yeah, this is the one where they were obsessed with BASIC problems. Wasn't 39 also BASIC?

Yes, but 39 was pretty easy.

Do they ask basic anymore?

@undefined: One of the questions was: If the event hasprobability 7/10, what is the probability it won't occur.

Not since the 80s, I don't think. Now, they're more careful about their required knowledge. They can't assume people know BASIC.


analytic geometry
invariant
geometry
rectangle

A 2007×2008 chessboard is covered by a few 2×2 and 1×4 dominoes without overlaps. One 2×2 domino of the set of dominoes used is replaced with a 1×4 domino. Prove that it is not possible to cover the chessboard with the modified set.

How does one prove this ?

[quote="iin77"]If you start with all 1x4 dominoes, then when you substitute one 2x2 domino, it's obvious to see you can't cover the board.[/quote]

ofcourse it's obvious, but i need a proof :).

[quote="#H34N1"][hide="Outline"]Since the checkerboard is covered by all 1x4 dominoes and you take one out, there will be a 1x4 blank space. In order to compensate for this, a 1x4 piece must be put down in replacement, as there is no other way to fit a single 2x2 piece.[/hide][/quote]

The checkerboard is not covered by all 1x4 dominoes ... ?

like... a mathematical proof? I thought obviousness was proof enough =-)

For each $ i$, number the squares in column $ 4i \plus{} 1$ with $ 1$, in column $ 4i \plus{} 3$ with $ \minus{} 1$ and in column $ 2i$ with $ 0$.  Then any $ 1 \times 4$ block covers either $ (0,0,0,0), (1,1,1,1), ( \minus{} 1, \minus{} 1, \minus{} 1, \minus{} 1)$ or $ (0,1,0, \minus{} 1)$, while any $ 2\times2$ block covers either $ (0,0,1,1)$ or $ (0,0, \minus{} 1, \minus{} 1)$.  Conclude.

[quote="JBL"]Number the squares in column $ 4i \plus{} 1$ with $ 1$, in $ 4i \plus{} 3$ with $ \minus{} 1$ and in $ 2i$ with $ 0$.  Then any $ 1 \times 4$ block covers either $ (0,0,0,0), (1,1,1,1), ( \minus{} 1, \minus{} 1, \minus{} 1, \minus{} 1)$ or $ (0,1,0, \minus{} 1)$, while any $ 2\times2$ block covers either $ (0,0,1,1)$ or $ (0,0, \minus{} 1, \minus{} 1)$.  Conclude.[/quote]

Very nice, i was searching somethling like that.

Thanks

[hide="Proof by Contradiction"]Assume that a 1x4 domino can replace a 2x2 domino. Therefore, it should be possible to replace a domino with coordinates on $ (0,0), (0,1), (0,4), (1,4)$. A 2x2 domino has coordinates on $ (0,0),(2,0),(0,2),(2,2)$. Since only one of these coordinates and not all match, the domino does not fit. Therefore, the original assumption is disproved and a 1x4 domino cannot replace a 2x2 domino inside the modified chess set.[/hide][/hide]

#H34N1, after changing one piece I can rearrange the remaining pieces in myriad ways, and your solution doesn't account for this.  This is a classic problem for an invariant of the type I suggested.

Yeah so this is my idea. We use a special type of coloring. If (i,j) are the coordinates of a square, we colour all squares for which i and j are both odd. Next we see that this means we coloured exactly 1004x1004 squares. Next we observe the fact that 1x4 rectangle covers either 2 or none of these squares. And 2x2 squares cover exactly one of them. So if we have n 1x4 which cover 2 squares and m 2x2 which cover one in the beginning we have that 1004x1004=nx2+m so m is even. On the other hand when we exchange 1x4 with a 2x2 we get that m+1(number of 2x2 squares) is odd. And in the second case if the rectangles 1x4 covering two squares are Y then 2xY+m+1=1004x1004. That is in fact impossible since the left side is odd and the right isn't.
         Using this we can obtain a generalization since by changing 2 rectangles we see that the left side in case 1 is odd and in case 2 i even (or the opposite) and since in the two cases it would equal the same number of coloured squares it means that the board cannot be covered in both cases.
Nice and easy. :D


I have #3 solved but this one will be pain in the butt to write in computer because triangle construction etc. So, is it OK if I scan the solution in pencil and put this as image? I'm not sure if the grader will be able to read clearly though. Is this OK or not?

Try it and see, but we strongly recommend writing it in [b]pen[/b] rather than pencil if you're going to scan it.

Anything that is legible is OK.  But examine it first and determine whether or not it's legible.


Maybe this is gonna get moved but I didnt see a forum with a trig subheading so sorry if it's in the wrong place.

Can't seem to work through this, can anyone give me a hand?

Show that:

Tan 3x = 3tanx - tan^3x / 1 - 3tan^2x

Thanks in advance for any help...

hello, do you mean $ \tan(3x)\equal{}\frac{3\tan(x)\minus{}\tan(x)^3}{1\minus{}3\tan(x)^2}$.
Sonnhard.

hello, you will need this formula
$ \tan(x \plus{} y) \equal{} \frac {\tan(x) \plus{} \tan(y)}{1 \minus{} \tan(x)\tan(y)}$
Sonnhard.

I got that far but then I used tan(x) = sin (x) / cos (x) and then eventually hit a dead end so I'm thinking that was the wrong way to think

Sorry, still no luck!! Usually when I work with these problems I only work on one side of the = sign till it becomes the same as the other side but in this case it feels that I need to work on both sides on the = sign. Some more info would be greatly appreciated!!

The correct expression is
\[ \tan 3x \equal{} \frac {3 \tan x \minus{} {\tan}^3 x}{1 \minus{} 3  {\tan}^2 x}
\]
an you'll get it  softly just using the expression given in the post above for the tangent  of a sum, putting 3x= x+2x ( also  for 2x you can use the same expression with 2x=x+x).


abstract algebra
invariant
ratio
IMO
IMO 2007

the photos of participants are available : http://imo2007.edu.vn/index.php?module=participant

the moroccan IMO team  :          http://imo2007.edu.vn/index.php?module=ViewInforTableForm.php&ct=MAR

omfg... what happened to the photos? ^^

Naphthalin

http://imo2007.edu.vn/index.php?module=participant

Just click on the name of the country, names of participants with their photos will be there for you.

The observers from the [url=http://imo2007.edu.vn/index.php?module=ViewInforTableForm.php&ct=ROM]Romanian Team[/url] have been heavily distorted  :oops:
Fortunately, the contestants did not experience these width to height ratio problems :D

Brian Lawrence is a member of the forum ? if yes who ?
BTW , it seems that I need a shave .  :P  (see my phtoto and you will understand)

me too ;)
what is your picture?

Naphthalin

[quote="rem"]I am pretty sure it's this one:
http://imo2007.edu.vn/index.php?module=ViewInforTableForm.php&ct=GRE[/quote]
Or more precisely this: http://imo2007.edu.vn/images/Participants/GRE705.jpg

[quote="darij grinberg"]Is it just my browser or do the webmasters of the IMO site believe that we are invariant under affine transformations?

  darij[/quote]

 :rotfl:

Italians are known for their very thin heads ---> http://imo2007.edu.vn/index.php?module=ViewInforTableForm.php&ct=ITA
(I am Andrea)

It's weird that there are no Vietnameze participants. :maybe:
http://www.imo2007.edu.vn/index.php?module=ViewInforTableForm.php&ct=VIE

to [b]silouan[/b]
Why Greece team consists of 5 (not 6)contestants?

I just thought that there will be 5 contestants from Greece Now it's OK

[quote="silouan"]I really don't know . The other person is Nikos Romanidis .Maybe he has not sent his photo yet or his photo is bigger than 10 KB .[/quote]

But [url=http://www.imo2007.edu.vn/index.php?module=participant]this[/url] also says that Greece has five participants...

You are right [b]kdano[/b]

[i]No picicture is available [/i] or something like that should be written if there is a problem with picture (look at malaysia team)

[quote]or his photo is bigger than 10 KB .
[/quote]
That is not problem I have also sent picture that is bigger than 10 kb It was accepted.

Hmmm... It is quite strange, but on the other hand: what's with the Montenegrin team? They have confirmed participation this year on the IMO, selected the team... but it doesn't look that way in the table...

[quote="delegat"]Pa polako bre jađo Previše im je i što su potvrdili da će doći Nek se malo odmore Valja to poslat a to je težak posao za koji će im trebat još mnogo vremena...[/quote]

English please...

OK Sorry That was just "a local" joke...

[quote="delegat"]Pa polako bre jađo Previše im je i što su potvrdili da će doći Nek se malo odmore Valja to poslat a to je težak posao za koji će im trebat još mnogo vremena...[/quote]
You're underestimating the young generations of Montenegrins...  :lol:
They don't fit in the old stereotypes today (just like we Bosnians don't fit in stereotypes anymore :) ). But I'd really like to hear what happened with "djetici" (Montenegro) team  :maybe: ...

Just to inform that the photos of the Montenegrin team for the IMO 2007 have been posted:) I've  made a fuss out of it few weeks go, so I felt the need to say that everytihnig is ok now  :lol:

A niste valjda mislili bez nas na ovo besplatno ljetovanje u Vijetnam!

Actually, we still don’t really know if we are going to Vietnam. First they told us>u might be going, then>there\s no money, then suddenly I see there are 4 contestants on the IMO site, and then more of the U R GOING, U R NOT GOING, and finally when I thought I can go on the coast for a nice little vacation (yes we Montenegrins LOVE vacations, all generations included), they said they will send an application after all. So now we are on the site lol
If, in the end, we get to Vietnam, I rely on my friends Nikola and Marica to do the best they can cuz ill need a big rest from tomorrows Rolling Stones concert !  :P    cya guys

[quote="darij grinberg"]Is it just my browser or do the webmasters of the IMO site believe that we are invariant under affine transformations?

  darij[/quote]
It's not just the website, the photos on the nametags are equally squished.


function
integration
real analysis
real analysis unsolved

Suppose that $ f: [a,b]\rightarrow \mathbb{R} $ be a monotonic function and for every $ x_1,x_2\in [a,b] $ that $ x_1<x_2 $ ,there exist $ c\in (a,b) $ such that $ \int _{x_1}^{x_2}f(x)dx=f(c)(x_1-x_2) $
a) Show that $ f $ be the continuous function on interval $ (a,b) $
b) Suppose that $ f $ is  integrable function on interval $ [a,b] $ but  $ f $ isn't a monotonic function then ,is it the result of  part a) right?

Suppose for exemple that $f(x)$ is an increasing function (take $-f$ if $f$ is decreasing).
First of all   : $f(x_1)(x_2-x_1) \leq \int_{x_1}^{x_2}f(x)dx=f(c)(x_1-x_2) \leq f(x_2)(x_2-x_1)$ , so there is a such $c\in ]x_1,x_2[$
Let $u,r,t$ be such that $u\in ]a,b[ , 0 < r < 1, 0 < t < \frac{\min (u-a,b-u)}2 $
We can write $f(c_{(rt)})*(r+1)t = \int _{u-t}^{u+rt}f(x)dx = $ $\int_{u-t}^{u}f(x)dx + \int_{u}^{u+rt}f(x)dx = f(p_{(t)})*t + f(q_{(rt)})*t$ , so
 $f(c_{(rt)})=\frac {f(p_{(t)})+rf(q_{(rt)})}{1+r}$, where $c_{(t)}\in ]u-t,u+t[ \ ,\  u-t < p_{(t)} < u < q_{(rt)} < u+rt$
If $t \longrightarrow 0$ then  $p_{(t)}\longrightarrow u \ ,\ q_{(rt)}\longrightarrow u$ , so  $f(p_{(t)}) \longrightarrow f_{u-} \ , \ f(q_{(rt)}) \longrightarrow  f_{u+}\ $ (left and right limits of $f$) , so
$f(c_{(rt)})\longrightarrow \frac {f_{u-}+rf_{u+}}{1+r}\  ,\  f_{u-}\leq f(u) \leq f_{u+}$ , .
But $\ c_{(rt)}\in ]u-t,u[\ \cup \  {u\}\  \cup \ ]u,u+rt[}$ $\Longrightarrow f(c_{(rt)})\in\  ]f(u-t),f(u)[\ \cup\ \{f(u)\}\  \cup \ ]f(u),f(u+rt)[$
So for the limit of $f(c_{(rt)})$ we have :$ \frac {f_{u-}+rf_{u+}}{1+r} \in \{f_{u-},f(u), f_{u+}\} , \forall 0<r<1$
It's easy to see that : $\frac {f_{u-}+rf_{u+}}{1+r}=f_{u-} \Longrightarrow f_{u-}=f_{u+} \ ;\ \frac {f_{u-}+f_{u+}}{1+r}=f_{u+} \Longrightarrow f_{u-}=f_{u+}$
Suppose that $f_{u-} \neq f_{u+}$ , so $\frac {f_{u-}+rf_{u+}}{1+r}= f(u), \forall 0<r<1\ $, so $f_{u-} = f_{u+}$

 :cool:  $f_{u-} = f_{u+} \Longrightarrow f_{u-} = f_{u+} =f(u) \Longrightarrow \ continuity $ :cool:

If $f$ is not monotonic, the result does'nt hold : take for exemple $f(0)=0 \ , \ f(x)=1 \forall x\neq 0$
  :cool:


floor function
combinatorics unsolved
combinatorics

Let $ p > 2007$ be a prime number, and let $ n$ be an integer exceeding $ 2006p$. Prove that
$ C^{2006p}_{n} - C^{2006}_{k}$ is a multiple of $ p$, where $ k = \left\lfloor\frac {n}{p}\right\rfloor$ is the largest integer which is not larger than $ \frac {n}{p}$.

[i]Remark.[/i] Here, $ C^a_b$ is just another notation for $ \binom{b}{a}$.

It from Math and Young-a magazine in Vietnam, Please don't reply when i find a mod to lock this topic :mad:

[color=red][[b]Moderator edit:[/b] Yes, it is indeed a problem from the M&Y Magazine. The deadline is 15 October 2007. This deadline is now over and the problem can be freely discussed.][/color]

Let $ e_{p}(m)\equal{}a$
Then
$ C_{mp\minus{}1}^{p\minus{}1}\equal{}1 (mod p^{3\plus{}a})$

$ n(n\minus{}1)...(n\plus{}1\minus{}lp)\equal{}((p\minus{}1)!)^l p^lk(k\minus{}1)...(k\plus{}1\minus{}l)(1\plus{}O(p))$, were $ k\equal{}[n/p]$, therefore $ C_n^{lp}\equal{}C_k^l(1\plus{}O(p))$ or $ p|C_n^{lp}\minus{}C_k^l$.

Your solution is very short and hard to understand! Can you do more clearly?
What is $ O(p)$? And is this problem true if p isn't a prime number?

The problem as written seems to be incorrect (try $ n\equal{}2006p\plus{}1$, for example).  Instead, I think you want to show $ \binom{n}{2006p}\equal{}\binom{k}{2006}$.

My suggestion is for you to take a look at some properties of the following definitions and map.

Given a subset $ S \in \{1, 2, \dots n\}$ with $ |S|\equal{}2006p$ and an integer $ m$ satisfying $ 1 \leq m \leq k$, let $ S_m$ denote the intersection of $ S$ with $ \{(m\minus{}1)p\plus{}1,(m\minus{}1)p\plus{}2, \dots mp \}$.

Let $ m(S)$ denote the smallest $ m$ such that $ 0<|S_m|<p$.  If no such $ m$ exists (if $ |S_1|, \dots |S_k|$ are each either 0 or $ p$), then $ m(S)$ is undefined.

Given these definitions, we define $ f(S)$ as follows:

-If $ m(S)$ is undefined, $ f(S)$ is also undefined.

-If $ m(S)\equal{}m$, then $ x \in f(S)$ if one of the following three conditions hold:

1. $ x \in S$ and $ x$ does NOT satisfy $ (m\minus{}1)p<x \leq mp$.
2. $ (m\minus{}1)p\plus{}2 \leq x \leq mp$ and $ x\minus{}1 \in S$
3. $ x\equal{}(m\minus{}1)p\plus{}1$ and $ mp \in S$.

Intuitively what we're doing here dividing our interval up into subintervals of length $ p$, and then looking for the first subinterval which intersects $ S$ only partially.  Once we find that subinterval, we shift $ S$ by 1 inside that subinterval, including wrapping around from the end of the subinterval to 1.  On the other hand, we leave everything outside that subinterval alone.

You should try and argue that $ f$ has the following properties

A: Everywhere it is defined, applying $ f(S)$ a total of $ p$ times gets you $ S$ again.
B:  On the other hand, applying $ f(S)$ fewer than $ p$ times gets you something different than $ S$.   (This property is where you will use that $ p$ is prime!)
C: The total number of places where $ f$ is defined is divisible by $ p$.
D: The total number of places where $ f$ is not defined is equal to $ \binom{k}{2006}$.

[quote="kevinatcausa"]The problem as written seems to be incorrect (try $ n \equal{} 2006p \plus{} 1$, for example).  Instead, I think you want to show $ \binom{n}{2006p} \equal{} \binom{k}{2006}$.[/quote]

You are right (at least, if your equal sign means congruence modulo $ p$): This is what the original M&Y source wanted us to show.
I have edited the first post accordingly now.

Anyway, how has a trivial particular case of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160610]Lucas theorem[/url] (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=165946 ) made it into M&Y?

  darij

I steel don't known your solutions! Hmm :(!

Re-read the post above you... It should be at least a good hint.

  darij


Να αποδείξετε ότι  ο αριθμός

     $ {{{{{6^{n}}^{n}}^{n}^{...}}^{n}}+{{{{{4^{n}}^{n}}^{n}^{...}}^{n}}-2$

 όπου $ n$ είναι αρτιος φυσικος αριθμός , ειναι πάντα πολλαπλασιο του 5.

Δεν ειναι δυσκολη oσο φαινεται!Μια μικρη γενικευση της ειναι και η εξης: $ 5/6^{nk}\plus{}4^{nk}\minus{}2$ για $ n$ αρτιο.
ΑΠΟΔΕΙΞΗ
Για $ n$ αρτιο εχουμε:
$ 6^{nk}\equal{} 1mod5$
$ 4^{nk}\equal{} 1m0d5$
Aρα $ 6^{nk}\plus{}4^{nk}\minus{}2 \equal{} 0mod5$

Γράφω τα ίδια με τον Αρχημήδη απλά χρησιμοποιώ την δικό μου τρόπο.

Ο αριθμός 6^ν έχει τελευταίο ψηφίο το 6.

Ο αριθμός 4^ν έχει σαν τελευταία ψηφία τους αριθμούς 6 ( αν ν άρτιος ) ή 4 ( αν ν περιττός )
Στην δικιά μας περίπτωση επειδή ο ν είναι άρτιος ο εκθέτης του 4 στην άσκηση είναι άρτιος.

Άρα το τελευταίο ψηφίο της παράστασης θα είναι  6 ( από το 6^ν) + 6 ( από το 4^ν) - 2 = 0
Ο αριθμός τελειώνει σε μηδέν. Άρα είναι άρτιος και διαιρείται με 5.


function
geometry
geometric transformation
reflection
algorithm
absolute value
calculus

FIND THE NON-DIFFERENTIAL POINTS OF THE FUNCTION:

f(x) = |||||||x| - 3| - 5| - 7| -2| - 4| - 6|

|.| --------> INDICATES ABSOLUTE VALUE SIGN.

$ f(x) \equal{} |x|$ is non-differentiable at $ x\equal{}0$, therefore...

What have you done to your teacher? (Incorrect graph deleted)

your graph not showing that it is an even  function.  :|

I forgot to put abs() around $ x$. Anyway, how do we solve the problem?
- $ f$ is not differentiable at $ 0$. As for other non-diff points, we only need to find them for $ x>0$ and then reflect them about $ 0$.
- $ f$ is not differentiable at $ 3$. As for other non-diff points, we only need to find them for $ x>3$ and then reflect them about $ 3$. (But ignore those that fall to the left of $ 0$ after reflection, see above).
- $ f$ is not differentiable at $ 3\plus{}5\equal{}8$. As for other non-diff points, we only need to find them for $ x>8$ and then reflect them about $ 8$. (But ignore those that fall to the left of $ 3$ after reflection, see above).
- ...
This boils down to the following algorithm: marks the points 0, 3, 8, 15, 17, 21, and 27 on the real axis (call them $ a_1<a_2<\dots < a_n$.) For $ i\equal{}n\minus{}1,n\minus{}2,\dots,1$ do the following: reflect all marked points to the right of $ a_i$ about $ a_{i}$, but only mark their reflections if they fall to the right of $ a_{i\minus{}1}$.
This process adds the points 13, 9, and 7, and a bunch of negative values. Thus, the answer is: $ 0, \pm 3, \pm7, \pm8, \pm9, \pm13, \pm15, \pm17, \pm21, \pm27$


Let $ ABC$ be a triangle and $ O,O_a,O_b,O_c$ are cicumcentre of $ (ABC),(OBC),(OCA),(OAB)$ respectively.
    a) Prove that $ AO_a,BO_b,CO_c$ are concurent at $ P$
    b) let $ Q$ be image of $ P$ pass inversion $ I_O^{R^2}$ where $ R$ is radius of $ (ABC)$ .$ Q_a,Q_b,Q_C$ are symetric with $ Q$ on $ BC,CA,AB$. prove that $ AQ_a,BQ_B,CQ_c$ are concurent at a point lie on $ (ABC)$.

[quote="hoangclub"]....
b) let $ Q$ be image of $ P$ pass inversion $ I_O^{R^2}$ where $ R$ is radius of $ (ABC)$ .$ Q_a,Q_b,Q_C$ are symetric with $ Q$ on $ BC,CA,AB$. prove that $ AQ_a,BQ_B,CQ_c$ are concurent at a point lie on $ (ABC)$.[/quote]
Do you mean dear [b][size=100]hoangclub[/size][/b], that the point $ Q$ lies on the line segment $ OP$ towards $ P,$ such that $ OP\cdot OQ \equal{} R^{2}\ ?$

[quote="¬[ƒ(Gabriel)³²¹º]¼"]a) $ \triangle O_aO_bO_c$ and $ \triangle ABC$ are Orthologic Triangles and they Orthology Centers are equal to O so the triangle are also perspective.[/quote]
Do you know dear [b][size=100]Gabriel[/size][/b] or somebody else, any reference about the proof of the perspectivity of two orthologic triangles, whose the othologic centers are coincided?

I have in mind a rather long proof of this result by [b][size=100]Menelaus theorem[/size][/b] and I am wondering about another simpler and sorter one.

Thank you, Kostas Vittas.


i got either a 116 or 119.5

i'm going to take the B though :)

oh and Press alt f4, what does Press alt f4 do?

[quote="Press alt f4"]Press alt f4[/quote]
ha, ha, ha.  very funny

 :mad:

(nobody press alt f4, it closes the web site)
[color=darkred]
Wrong. Pressing Alt F4 speeds up your connection by piggybacking your ISP onto my proxy server.[/color]

You didn't know that tell now :rotfl:  :rotfl:  :rotfl:  :rotfl:

Alt+F4 speeds up your internet connection (see above)
Alt+Tab(hold)  allows you to switch programs without using program bar.
Cntl+Esc  opens start menu (for those with keyboards lacking the Windows sign)

I think I got somthing in the 110 range...
I'l hopefully do the AMC 10 B

WE'll have o wait and see for the Chapter.

my math teacher only gives AMC 12 cuz shes weird and says why do AMC 10 when u can do AMC 12?  so i got a 99.5!!!!!!!!!! :mad: pooey cuz 100 is the score needed to pass... well, i'm not at "the age" yet so who cares?

Do a different topic for keyboard shortcuts, ok? It's not that I don't like them, altF4 is great, but it is way offtopic


There are some black and white marbles in a row. We can do three actions:
1. One black marble which is neighbor with to adjacent white marbles, can jump through both of them and delete one of them.
2. An arbitrary(black or white) marble which is neighbor with three adjacent white marbles can jump through all of them and delete all of them.
3. An arbitrary marble which is neighbor with two black marbles can jump through both of them and delete the both.
At the beginnig we have 2007 adjacent black marbles, then 2008 adjacent white marbles and then 2007 adjacent black marbles in a row. Can we reach with these three actions to a state in which there is no white marble in the row?

[hide="Hint"]Consider action of $ S_3$.[/hide]


Hi,

i'm not sure how to put my equations in the text, so i attach a file with a summation formula i'm trying to turn into a simpler expression.

i also put there a solution that i'm aware of for a simpler version of the problem, that might be of help to make the extra step i don't manage to.

thanks.

This is the way to do this:

First of all:

$ {(x\plus{}y)^n\equal{}\sum^{n}_{k\equal{}0}{\binom{n}{k}x^k y^{n\minus{}k}}\equal{}\sum^{n}_{k\equal{}0}{\binom{n}{k}y^k x^{n\minus{}k}}}$

Now, let us try reducing your expression to one similar to this:

$ {\sum^{s\minus{}1}_{k\equal{}0} {\binom{s\minus{}1}{k}u^k (1\minus{}u)^{s\minus{}k\minus{}1}\frac{p^{k\plus{}1}}{k\plus{}1}}}$

First we can use:

$ {\binom{n}{k}\equal{}\frac{n!}{k! (n\minus{}k)!}}$

and

$ {n(n\minus{}1)!\equal{}n!}$

This way we get:

$ {\binom{s\minus{}1}{k}\frac{1}{k\plus{}1}\equal{}\frac{(s\minus{}1)!}{k! (k\plus{}1) (s\minus{}k\minus{}1)!}\equal{}\frac{\frac{s!}{s}}{(k\plus{}1)!(s\minus{}k\minus{}1)!}\equal{}\frac{1}{s}\binom{s}{k\plus{}1}}$

Now we have:

$ {\frac{p}{s}\sum^{s\minus{}1}_{k\equal{}0} {\binom{s}{k\plus{}1} (u p)^k (1\minus{}u)^{s\minus{}k\minus{}1}}\equal{}\frac{p}{s}\sum^{s}_{k\equal{}1} {\binom{s}{k} (u p)^{k\minus{}1} (1\minus{}u)^{s\minus{}k}}\equal{}\frac{1}{s u}\sum^{s}_{k\equal{}1} {\binom{s}{k} (u p)^k (1\minus{}u)^{s\minus{}k}}}$

$ {\equal{}\frac{1}{s u}\sum^{s}_{k\equal{}0} {\binom{s}{k} (u p)^k (1\minus{}u)^{s\minus{}k}}\minus{}\frac{1}{s u}\binom{s}{0} (u p)^0 (1\minus{}u)^{s\minus{}0}\equal{}\frac{1}{s u} (up\plus{}1\minus{}u)^s\minus{}\frac{1}{su} (1\minus{}u)^s}$

great, this does look simpler than i expected. thanks.


Stiind ca $a$,$b$,$c$ sunt numere reale, calculati partea intreaga a numarului $A$, unde :

$A=(a^{2}+bc)/(a-b)(a-c)+(b^{2}+ac)/(b-c)(b-a)+(c^{2}+ab)/(c-a)(c-b)$.

P.S.: poate cineva sa imi explice ce vrea de fapt enuntul problemei? evident ca partea intreaga a lui $A$ nu este constanta. atunci ce trebuie sa aflam? partea intreaga ca un interval de numere intregi sau sa o exprimam in fuctie de $a$,$b$,$c$? orice parere este bine venita.

S-ar putea sa fie si constanta sau sa fie in functie de a,b,c.Tu va trebui sa stabilesti si sa demonstrezi asta.

$[A]=2$ pt ca $\sum_{cyc}\frac{a^{2}+bc}{(a-b)(a-c)}=2,\forall a,b,c\in \mathbb R$ $a\neq b\neq c\neq a$

multumesc mult. demonstrasem ca $A\geq 2$ dar nu eram sigur ca $[A]=2$  pentru orice $a$, $b$, $c$ reale.

Pai da' $\frac{a^{2}+bc}{(a-b)(a-c)}=\frac{a^{2}+bc}{a^{2}+bc-ab-ac}\ ge 1$ daca a,b,c>0.Si daca adunam relatiile ne da ca e mai mare ca 3. Deci?

nu neaparat
daca $0<b\leq a\leq c$, atunci numitorul ala e negativ, si numaratorul e pozitiv, deci fractia e numar negativ, care nu poate sa fie mai mare sau egal cu $1$.
n`am stat sa verific calculele, dar si eu cred ca expresia aia e $2$.

Ups sorry...
Da tot nu m-am prins d c e 2

Fie
\[P\left( x\right) =\sum \frac{\left( x-b\right) \left( x-c\right) }{\left( a-b\right) \left( a-c\right) }. \]
Evident, $P$ este un polinom de grad cel mult 2, şi cum $P\left( a\right) =P\left( b\right) =P\left( c\right) =1,$ rezultă că $P\left( x\right) =1,$ pentru orice $x.$ Atunci
\[4=P\left( a+b+c\right)+3=\sum \left( \frac{\left( a+c\right) \left( a+b\right) }{\left( a-b\right) \left( a-c\right) }+1\right) =2\sum \frac{ a^{2}+bc}{\left( a-b\right) \left( a-c\right) }. \]

A se remarca totusi deosebita originalitate a propunatorului...  :rotfl:

[quote="mychrom"]harazi, eu n-am zis ca asta e un exercitiu nemaiintalnit, doar mi s-a parut interesant pt clasele 7-8.[/quote]
Nu era vorba de cine a propus topicul, ci de cel care a propus problema asta într-un concurs sau într-o revistă.... :)


analytic geometry
real analysis
real analysis unsolved

It's possible this goes in Computations and Tutorials, but I thought it was a decently hard problem. I am bad at this type of problems. I can solve it, but it is sort of messy. I'd like to see more elegant solutions.

A wine glass with the 'cup' $z = x^2 + y^2, z \le 1$ is filled to the height $h$. If the glass is tilted by $45$ degrees then the wine just starts spilling out of the glass. What is $h$?

you said you have a solution

is it $\frac{\sqrt{10}-\sqrt{6}}{4}$ ?
if so I would explain but i don't think it will be much shorter
the main point is that I don't change my axes and I use polar coordinates

No, it is $h = 1/4$ (I also have the answers). What my solution does is to find an ON-basis where the body that the wine describes in its tilted state is easy to describe. I can calculate the volume of the wine by using Fubini's theorem and a change of variables. Then I just find the volume of the wine in its normal state for any h, and then match up the h so the volume will be correct.


abstract algebra
group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

1) If $ N \triangleleft G , [G : N]$ finite , $ H<G , |H|$ finite , and $ [G : N]$ and $ |H|$ are relatively prime , then $ H<N$. (< means subgroup).

[hide]At first I thought , I solved it but a question came . Consider the projection $ \pi : G \longrightarrow G/N$. If I could show that $ |\pi(H)| \mid |H|$ and $ |\pi(H)| \mid |\frac{G}{N}|$, then $ |\pi(H)| \mid gcd(|H|,\frac{G}{N}|)\equal{}1$ and $ H<N$. Is it true$ |\pi(H)| \mid |H|$ ? how?[/hide]

2)If $ N \triangleleft G , |N|$ finite , $ H<G , [G : H]$ finite, and $ [G : H]$ and $ |N|$ are relatively prime, the $ N<H$.

1: If $ \pi : A \to B$ is any group homomorphism, then $ \mathrm{im}(\pi) \cong A/\mathrm{ker}(\pi)$. Then $ |A| \equal{} | \mathrm{im}(\pi) | \cdot | \mathrm{ker}(\pi) |$ is immediate.

2: Play a similar game. Consider the same projection and look at the kernel (using the formula above).

For any two subgroups $ A,B\le G$ with $ AB\le G$ we get
a well-defined bijection between the sets of cosets
$ A/A\cap B\to AB/B, a(A\cap B)\mapsto aB$, which gives $ |A: A\cap B| \equal{} |AB: B|$.

So
1) $ |H: H\cap N| \equal{} |HN: N|$ divides $ \gcd(|H|,|G: N| ) \equal{} 1$, which gives $ H \equal{} H\cap N\le N$
and
2) $ |N: N\cap H| \equal{} |NH: H|$ divides $ \gcd(|N|,|G: H| ) \equal{} 1$, which gives $ N \equal{} N\cap H\le H$.

Of course we also need $ |G: B|\equal{}|G: A|\cdot |A: B|$ for any $ G\ge A\ge B$.
For that we need, that if $ S$ is a set of representatives of $ G/A$,
then we get the well-defined bijection $ S\times (A/B)\to G/B, (s,aB)\mapsto saB$.


A student is asked his age, and he responds, "My age is nine less than half my professor's age." The professor is asked her age, and she responds, "My age is three times my student's age." What is the age of the student?

[hide=" Cheesy Way"]
Students have professors in college, and typically begin college at age $ \boxed{18}$.
[/hide]

[hide="Algebraic Way"]
Let the student's age be $ s$. We can express the professor's age as $ 3s$, and write the equation $ s \equal{} \frac{3s}{2} \minus{} 9 \implies 2s \equal{} 3s \minus{} 18 \implies s \equal{} \boxed{18}$.
[/hide]


$ 1$ Show that if $ k,n \in N$ and $ n \ge 2$, then $ (n \minus{} 1)^2 |n^k  \minus{} 1 \Leftrightarrow (n \minus{} 1)|k$

$ 2$ If $ n$ is an odd number and $ n \in N$, then $ \frac{{n(n \plus{} 1)}}{2}|n!$

[quote="Nea"]$ 1$ Show that if $ k,n \in N$ and $ n \ge 2$, then $ (n \minus{} 1)^2 |n^k \minus{} 1 \Leftrightarrow (n \minus{} 1)|k$

$ 2$ If $ n$ is an odd number and $ n \in N$, then $ \frac {{n(n \plus{} 1)}}{2}|n!$[/quote]

[hide="Question 2"]
Let $ n\equal{}2m\plus{}1$
Hence $ \frac {n(n \plus{} 1)}{2}\equal{}(2m\plus{}1)(m\plus{}1)$, clearly this divides $ (2m\plus{}1)!$
[/hide]

[quote="Nea"]$ 1$ Show that if $ k,n \in N$ and $ n \ge 2$, then $ (n \minus{} 1)^2 |n^k \minus{} 1 \Leftrightarrow (n \minus{} 1)|k$

$ 2$ If $ n$ is an odd number and $ n \in N$, then $ \frac {{n(n \plus{} 1)}}{2}|n!$[/quote]

[hide="Question 1"]
It is well-known that $ n^k\minus{}1 \equal{} (n\minus{}1)\sum_{i\equal{}0}^{k\minus{}1}n^i.$ So we can reduce the question to when $ n\minus{}1$ divides $ \sum_{i\equal{}0}^{k\minus{}1}n^i\equiv \sum_{i\equal{}0}^{k\minus{}1} 1 \equiv k \pmod{n\minus{}1}$ which proves the statement.[/hide]


$\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}}\geq 1 \iff$
$\frac a{a+3b}+\frac b{b+3a}+2\sqrt{\frac{ab}{(a+3b)(b+3a)}}\geq 1\iff$

$\frac{a(b+3a)+b(a+3b)+2\sqrt{ab\cdot (a+3b)(b+3a)}}{(a+3b)(b+3a)}$ $\geq 1 \iff$ NUMARATOR$\geq$ NUMITOR.Notam $a+b=s,a\cdot b=p$ si trebuie sa demonstram:

$ab+3a^{2}+ab+3b^{2}+2\sqrt{ab(a+3b)(b+3a)}\geq (a+3b)(b+3a)$

$a^{2}+b^{2}=s^{2}-2p,(a+3b)(b+3a)=3s^{2}+4p$ rezulta  ca inegalitatea este echivalenta cu:

$2p+3(s^{2}-2p)+2\sqrt{p(3s^{2}+4p)}\geq 3s^{2}+4p$,ceea ce este echivalent cu a arata ca $\sqrt{p(3s^{2}+4p)}\geq 4p$,echivalent ,prin ridicare la patrat,cu

$3ps^{2}+4p^{2}\geq 16p^{2}\iff3ps^{2}\geq 12p^{2}\iff s^{2}\geq 4p$,ultima din urma fiind evident adevarata,caci este echivalenta cu $(a+b)^{2}\geq 4ab$.Si fiindca am pornit cu relatzii de echivalentza,rezulta ca inegalitatea din enunt este demonstrata!!

End of story

$\sqrt{\frac{1}{1+3\frac{b}{a}}}+\sqrt{\frac{1}{1+3\frac{a}{b}}}\geq1$
Notand $\frac{b}{a}=x$ inegalitatea devine :
$\frac{1}{\sqrt{1+3x}}+\sqrt{\frac{x}{x+3}}\geq1$
Aducem la acelasi numitor, ridicam la patrat de doua ori si obtinem
$3x(x-1)^{2}\geq0$

[quote=Mircea Lascu]Fie $a,b$ numere reale strict pozitive. Aratati ca: \[\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}}\geq 1.\][/quote]
Fie $a,b,c$ strict pozitive.[url=https://artofproblemsolving.com/community/c275h100451p567530 ]Aratati ca[/url]\[\sqrt{\frac{a}{3a+b}}+\sqrt{\frac{b}{3b+c}}+\sqrt{\frac{c}{3c+a}}\leq\frac{3}{2}\]
[url]https://artofproblemsolving.com/community/c6h526120p2981935[/url]

[quote=Mircea Lascu]Fie $a,b$ numere reale strict pozitive. Aratati ca: \[\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}}\geq 1.\][/quote]
Let $a,b>0. $ [url=https://artofproblemsolving.com/community/c6h487873p27085382]Prove that[/url]
$$\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}}\le\frac{3}{2 \sqrt 2} $$


I have a few questions on this achievement:

1. Why did Perelman decline the Fields Medal (something about ethical issues)?
2. Who should actually take credit for the proof?
3. What are some implications of the Poincare Conjecture/Theorem?

I might ask more when I think of them.

As I recall, Perelman declined the Fields Medal because he wasn't interested in the publicity.  Where did you hear about ethical issues?  As far as I can tell, Perelman's proof is genuine.

I read on the internet about some issues with people taking credits for his proof or plagiarism I am not really sure but something along those lines.

An somewhat lengthy article in which all of these issues are discussed can be found here: [url=http://www.newyorker.com/archive/2006/08/28/060828fa_fact2]New Yorker Magazine, Aug. 2006[/url].

There is also a lot of information and juicy gossip at the [url=http://science.slashdot.org/article.pl?sid=06/09/21/1442239]Slashdot[/url] website.

You can read what Fedja had to say about the New Yorker article [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=881934#881934]here[/url].

Fedja is actually quoted in the New Yorker article.

Well I guess their will always be some controversy over who actually proved what. But the important thing is that it is proved. Future generations shall benefit from the techniques created by these Mathematicians and continue to formulate new conjectures.


quadratics
number theory
Diophantine equation
algebra
number theory unsolved

This problem was given in a slightly different form in Problem Solving Strategies (by Engel) as follows:

a).  Prove that if $\displaystyle a, b, \frac{a^2+b^2}{ab-1} = q$ are integers, then q = 5

b).  prove that the diophantine equation $a^2+b^2-5ab+5=0$ has infinitely many solutions

[edited to fix the equation in b)]

[quote="zscool"]a).  Prove that if $\displaystyle a, b, \frac{a^2+b^2}{ab-1} = q$ are integers, then q = 5[/quote]
Just write $a^2-qb\cdot a+(b^2+q)=0$. This quadratic equation has two integer roots $a$ and $a'=\frac{b^2+q}{a}$. If we assume $a>b$ then $a'=\frac{b^2+q}{a}<a$ (check it using $a +a'=bq$), so we have a descend algorithm.
Finally, we will come to $b=1$. So $a^2-aq+1+q=0$, i.e. $D=(q-2)^2-8$ is a perfect square. Thus $q=5$.

[quote]b).  prove that the diophantine equation $a^2+b^2-5ab+5=0$ has infinitely many solutions[/quote]
Just apply inverse step for descend algorithm.

[Edited: there was a typo, I have replaced "5" with "q" is some places :blush:]

[quote="Myth"]
Just write $a^2-qb\cdot a+(b^2+5)=0$.
[/quote]
I don't think this step is completely right :D

I don't know hot it appears in my proof  :blush:
Anyway, I don't use it.

[quote="Myth"]If we assume $a>b$ and  then $\frac{b^2+5}{a}<a$ (check it! :) ), so we have descend algorithm.
[/quote]

It does appear in your proof and it makes your solution wrong.

And what?
What do you want to say?  :mad:

I am sorry Myth, don't want to offend you :) certainly not!
I am just saying that your solution needs to be fixed.

[quote="zscool"]dont you just replace that 5 with q, and it works out?[/quote]
Of course, it works.

[quote][quote]b).  prove that the diophantine equation $ a^2 \plus{} b^2 \minus{} 5ab \plus{} 5 \equal{} 0$ has infinitely many solutions[/quote]
Just apply inverse step for descend algorithm.[/quote]

Can you explain it in detail please?

[quote="Myth"]
 $ a' \equal{} \frac {b^2 \plus{} q}{a} < a$ (check it using $ a \plus{} a' \equal{} bq$)[/quote]

why

[quote="Myth"] $ a' \equal{} \frac {b^2 \plus{} q}{a} < a$ (check it using $ a \plus{} a' \equal{} bq$)[/quote]
Why???  :)

[quote="Myth"] $ a' \equal{} \frac {b^2 \plus{} q}{a} < a$ (check it using $ a \plus{} a' \equal{} bq$)[/quote]
Why???  :D

[hide=Solution][quote=zscool]a).  Prove that if $\displaystyle a, b, \frac{a^2+b^2}{ab-1} = q$ are integers, then q = 5
[/quote] Are [b]positive[/b] integers. Оtherwise, if $a>0,b<0$  then $q$ can be any negative square (see IMO 1988, problem 6)

Use Vieta jumping
1. $q \ge 3\; (a^2+b^2 \ge 2ab)$

2. For every fixed $q$, let $a_1 \ge b$ be a [b]minimal[/b] solution of equation $a^2-qab+b^2+q=0$ (if solutions exists)

$a_2 \in \mathbb{Z}$ because $a_2=qab-a_1$
$a_2 \in \mathbb{N}$ because $a_1a_2=b^2+q>0$
So, $a_2 \ge a_1 \ge b$ Or:

$(a_1-b)(a_2-b) \ge 0$
$a_1a_2-b(a_1+a_2)+b^2 \ge 0$

$b^2+q-b(qb)+b^2 \ge 0$

$(q-2)b^2 \le q \; \Rightarrow b^2 \le 1 +\frac{2}{q-2}$

So, a minimum solution exists only if $b=1$. Then $q=5,a_1=2,a_2=3$

and all solutions are two consecutive terms of the sequence $c_n=5c_{n-1}-c_{n-2}$ with first terms
$c_1=1,c_2=2\;\;(1,2,9,43,206,\cdots)$
and
$c_1=1,c_2=3\;\;(1,3,14,67,321,\cdots)$

[/hide]


calculus
integration
function
derivative
Euler
complex analysis
geometric series

[url]http://mathworld.wolfram.com/PiFormulas.html[/url], about 1/5 down the page.

That doesn't really qualify as a solution . . .

Well, since no elementary solution is available, at least we can satisfy ourselves with the knowledge that :pi:2 / 6 = Integral (0 to :inf:) u / (eu - 1) du.  :roll:

Haha.  I don't have the faintest idea what either of these problems are doing on the intermediate forum, but I guess I'll have to wait until Mr. McGeek gets back to find out.  Not that they aren't cool, I just don't know why they're [i]here[/i].

I agree, this isn't intermediate, although this sequence is so well known that you would never need to prove it anyway, you just state it.

There are ways of summing this that aren't incredibly hard (to understand; to come up with them is another story). This one (Euler's original solution) is my favorite:

Call the sum S. Consider the function sin(x)/x. Let's also define sin(0)/0 to be 1 since that's its limit anyway. Its Taylor series is

1-x2/3!+x4/5!-x6/7!+...

We know the roots of sin(x)/x: it's all the roots of sin(x) except 0, namely all integral multiples of pi. It makes sense, therefore, that we can write

sin(x)/x=(1-x/pi)(1+x/pi)(1-x/2pi)(1+x/2pi)...

(This isn't actually as harmless as it looks: multiplying by e^x doesn't give us any new roots, but clearly the product can't go to both of them. Nonetheless, it is correct.)

Let's combine consecutive terms to give us

sin(x)/x=(1-x2/pi2)(1-x2/4pi2)(1-x2/9pi2)...

Now expand out the x2 term, which is -x2/pi2*S. However, this must be equal to the x2 term of the Taylor series, which is -x2/3!. Thus S=pi2/6.

That proof isn't even remotely close to elementary, Simon.  (Although I do think it's one of the coolest proofs I've ever seen.)  And as far as I can gather, the proof that that infinite produce is equal to sin(x) must be very tough.  (The only reason I say that is because I have seen several different individuals present that particular proof of this theorem, and every single one of them has hand-waved away the problem of having some sort of multiplier in front.  This leads me to believe that it must be very difficult to establish that there is not such term.)

Well, what does elementary mean for you? In the mathematical sense that I have heard most often used, it means that it doesn't use complex analysis. Except for the product thing (possibly; I don't know how to prove it), there is no complex analysis here.

[quote]Well, what does elementary mean for you? [/quote]

Elementary means that I could come up with it.  :D

But more seriously, isn't elementary anything under "higher math", which is calculus and above and not in most well-known high school competitions?

I generally take elementary to mean no calculus, including real analysis.  I've never really thought very hard about it, though.  I bet a rigorous proof of that product-sum identity does require complex analysis, though.

That proof seemed easy for me to understand, and I'm not in calculus yet (I will be next year).  I haven't seen an easier proof than what Simon showed.

But you had to take not only the product definition for granted, but also the Taylor Series. When he wrote
sin(x)/x = 1 - x2/3! + x4/5! - x6/7! + ...
you didn't question how he got that?

But you had to take not only the product definition for granted, but also the Taylor Series.  When he wrote
sin(x)/x = 1-x2/3!+x4/5!-x6/7!+...
you didn't question how he got that?

Sorry I put this in Intermediate...

I didn't really know where this should go :oops:

Posted by MrMcGeek

Anything that uses calculus beyond the level of perhaps basic derivatives or limits should go to Advanced or to Other Problem Solving Topics.

The proof of the product expansion of sin x (and other analytic functions) is a typically a topic covered in complex analysis courses. It's possible to prove it by other means, but it is by no means "elementary." Euler actually had a particular habit of manipulating infinite series and products to his advantage without justifying them. This was because rigorous analysis and the rules of how to manipulate infinite series hadn't yet been ironed out yet.

So certain things like the fact that the geometric series 1 - 1 + 1 - 1 +... should equal 1/(1+1) = 1/2 thouroughly confused him and many other people of that time. Some of them tried to justify it in all kinds of ways, like saying that it was half the time 1 and half the time 0 so therefore it was a half.

Anyway. As for the original problem, the simplest proof of the sum of that series probably comes from using the Fourier Series of the "saw-tooth" function f(x) = x for 0 :le: x < 2pi (and periodic before and after that), and applying Plancherel's Identity to it. There are many other ways of proving it (and other interesting series that you wouldn't think you could sum) by using Fourier Series as well.


expand

LHS =$4\sum a^4bc+2\sum a^3b^3+9a^2b^2c^2$
RHS =$\sum a^3b^3 + 3\sum a^3b^2c+6a^2b^2c^2$

LHS - RHS = $4\sum a^4bc+\sum a^3b^3-3\sum a^3b^2c+3a^2b^2c^2$

By Schur $\sum a^4bc+3a^2b^2c^2\geq \sum a^3b^2c$

So it remains to prove $3\sum a^4bc+\sum a^3b^3\geq 2\sum a^3b^2c$ which is true by Muirhead .

Do you have any solution without expanding ? Arqady ?

[quote="andyciup"]Setting $x=ab,y=ac,z=bc$ we need to prove that $(\frac{xy}z+2z)(\frac{xz}y+2y)(\frac{yz}x+2x) \geq (x+y+z)^3$
$\iff (xy+2z^2)(xz+2y^2)(yz+2x^2) \geq (x+y+z)^3 \cdot xyz$.But by cauchy:
$(xy+2z^2)(xz+2y^2) \geq(x\sqrt{yz}+2yz)^2=yz(x+2\sqrt{yz})^2$.We do the same for $(xy+2z^2)(yz+2x^2)$ and $(xz+2y^2)(yz+2x^2)$ and by multiplying these last 3 ineqs and squaring,we get that:
$(xy+2z^2)(xz+2y^2)(yz+2x^2) \geq xyz(x+2\sqrt{yz})(y+2\sqrt{xz})(z+2\sqrt{xy})$ and now if we prove that
$(x+2\sqrt{yz})(y+2\sqrt{xz})(z+2\sqrt{xy}) \geq (x+y+z)^3$,then the problem is solved..I really hope the last one is true :D[/quote]

it is wrong , pick $x=y=0$ then LHS = $0$ but RHS = $z^3$

[quote="shyong"]
Do you have any solution without expanding ? Arqady ?[/quote]
My proof is similar, but LHS - RHS = $4\sum_{cyc} a^4bc+\sum_{cyc} a^3b^3-3\sum_{sym} a^3b^2c+3a^2b^2c^2=$
$=\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+2\cdot\sum_{cyc}(2a^4bc-a^3b^2c-a^3c^2b)\geq0.$
Schur and Miurhead too! ;)


Given a sequence $ x_{n}$ such that : $ x_{0}\equal{}a ; x_{n\plus{}1}\equal{}\frac{1\minus{}6x_n}{1\minus{}4x_n} \forall n\geq 0$
Find the general formula of $ x_{n}$
Hope see your answer soon :)

[hide="Outline"]Let $ x_n\equal{}\frac{b_n}{a_n}$ with $ a_0\equal{}1, b_0\equal{}a$. Plugging into the original recursion, we see $ a_n\equal{}a_{n\minus{}1}\minus{}4b_{n\minus{}1}$ (1) and $ b_n\equal{}a_{n\minus{}1}\minus{}6b_{n\minus{}1}$ (2). Subtracting (2) from (1), we have $ a_n\minus{}b_n\equal{}2b_{n\minus{}1}$, or $ a_{n\minus{}1}\equal{}b_{n\minus{}1}\plus{}2b_{n\minus{}2}$ (3). Plugging (3) into (2), we have $ b_n\equal{}\minus{}5b_{n\minus{}1}\plus{}2b_{n\minus{}2}$. Similarly, we have $ a_n\equal{}\minus{}5a_{n\minus{}1}\plus{}2a_{n\minus{}2}$. Now, it's easy to use the initial condition to solve the two recursions.[/hide]

用不动点法可以了吧?

结果是

$ x_n \equal{} \frac{\left(\left(\sqrt{33}\minus{}7\right) a\plus{}2\right) \left(5\minus{}\sqrt{33}\right)^n\plus{}\left(\left(7\plus{}\sqrt{33}\right) a\minus{}2\right) \left(5\plus{}\sqrt{33}\right)^n}{\left(\minus{}8 a\plus{}\sqrt{33}\plus{}7\right) \left(5\minus{}\sqrt{33}\right)^n\plus{}\left(8 a\plus{}\sqrt{33}\minus{}7\right) \left(5\plus{}\sqrt{33}\right)^n}$


I need to find the solution of this: http://mathzone.net.tc

Please help!

You have at least posted the same question for six times in five days:

http://www.mathlinks.ro/Forum/viewtopic.php?t=38212

http://www.mathlinks.ro/Forum/viewtopic.php?t=38202

http://www.mathlinks.ro/Forum/viewtopic.php?t=38194

http://www.mathlinks.ro/Forum/viewtopic.php?t=37946

http://www.mathlinks.ro/Forum/viewtopic.php?t=37756

http://www.mathlinks.ro/Forum/viewtopic.php?t=37735

First, please do not post the same thing in so many topics.

Sencond, I have tried the links, but failed; maybe this is the reason why you receive no replies.

At last, you can type the problems yourself in one of the topics mentioned above, this may be helpful to let us know your question.


Circumcircles -
Suppose you have ABC, and points, D,E, and F on BC, AC, and AB respectively. Then, prove that the circumcircles of AFE, BFD, and EDC are concurrent.

Now, [another problem] let ABCD be a quadrilateral, such that AB and CD [extended] meet at E and AD and BC meet at F. Show that the circumcircles of FAB, FCD, EAD, EBC are concurrent.

Sequences -
Show that for an integer $ n$, there exist $ n$ consecutive composite positive integers

Thanks!  :)

[quote="firestar58"]
Sequences -
Show that for an integer $ n$, there exist $ n$ consecutive composite positive integers
[/quote]

[hide="sneaky solution"]
use
$ (n\plus{}1)!\plus{}2, (n\plus{}1)!\plus{}3, \ldots (n\plus{}1)!\plus{}n\plus{}1$
It is easy to check that this sequence has n terms.

Each term is composite because if we express the number as $ (n\plus{}1)!\plus{}k$ for $ 2 \le k \le n\plus{}1$
Then k is a factor that is neither 1 nor $ (n\plus{}1)!\plus{}k$ that divides both terms and therefore the sum.
[/hide]

i'll try the other problems

[hide="second"]
It is the well-known [b]miquel point[/b] of the complete quadrilateral :wink: .
If no-one else post solution i will popst mine....[/hide]

wait... so would the proof for the miquel point thing be the same as this?

what is the proof?

[hide="First problem"]In the first problem, the point of concurrency is also a miquel point, although I know no proof of the existence of this point.[/hide]

[quote="Dimitris X"]Ok i will post it.

Let $ M$ be the point where the circles $ C_{AED},C_{FAB}$ intersects.

Then we only need to prove that:
$ MECB$ and $ MFCD$ are incribed.

I will prove this for the first.
First of all $ \angle MBF \equal{} \angle MAF$.
But $ \angle MAF \equal{} \angle MED$ (because $ MAED$ is inscribed).
But also $ \angle MAF \equal{} \angle MBF$ (because $ MABF$ is inscribed).

So $ \angle MEC \equal{} \angle MBF$ so the $ MBCE$ is incribed indeed :wink:

I hope this helps

Dimitris[/quote]

Wait... I'm not quite sure I follow your explanation.  :read:  :(  :fool: According to my drawing... MECB isn't a polygon. Do you mean MEBC? Also, how does showing that those two are inscribed show that the all the circles coincide?

And is there a solution to the first one? The one with the triangles? Thanks

[quote="firestar58"]
 how does showing that those two are inscribed show that the all the circles coincide?
[/quote]

We have already the intersection point of FAB and EAD at M.
So if we prove that:
MFCD and MBCE are inscribed then ALL the circles will go through M.(the miquel point).

I cant understan what you dont understan :) ??

Ok... I think I understand. I just didn't see the connection, but I do now.

So, would you use this similar approach for the triangle problem?

Dear Mathlinkers,
the first question is the pivot theorem coming from Miquel.

Sincerely
Jean-Louis


algebra
polynomial
Functional Analysis
advanced fields
advanced fields unsolved

Let's consider the Fourier-Plancherel transfrom  $ f \to \hat{f}$ from $ L^2(\mathbb{R})$ to itself. The Plancherel theorem states that it is a unitary operator.
What is the spectrum of this operator ?

The fourth power of the operator is the identity. Its spectrum consists of precisely the fourth roots of unity: $ \{1,i, \minus{} 1, \minus{} i\}.$

With appropriate scaling, the $ k$th degree Hermite polynomial multiplied by the appropriate Gaussian is an eigenfunction with eigenvalue $ i^k.$

(Some details of this depend on precisely where you put the $ 2\pi$ in your definition of the Fourier transform. But when eugene called it a unitary operator, he was narrowing the conventional choices to cases in which the details are as I stated them.)

So, i understand that the point spectrum i.e. the eigenvalues of the Fourier transform are fourth roots of unity.
Is it obvious that the Fourier operator (let's call it $ F$), defined above, doesn't have continuous spectrum, i.e. those $ \lambda$ such that $ F\minus{}\lambda I$ is injective, but its range is not the whole $ L^2(\mathbb{R})$ ?

Consider the consequences of the polynomial identity $ F^4\minus{}I\equal{}0.$

Let's suppose $ \lambda^4\ne 1.$ Then we can write

$ x^4\minus{}1\equal{}(x\minus{}\lambda)Q(x)\plus{}(\lambda^4\minus{}1)$ for some cubic polynomial $ Q.$

$ 0\equal{}F^4\minus{}I\equal{}(F\minus{}\lambda I)Q(F)\plus{}(\lambda^4\minus{}1)I.$ Hence,

$ (F\minus{}\lambda I)\cdot\left[\frac{\minus{}1}{\lambda^4\minus{}1}Q(F)\right]\equal{}I.$

Thus, $ F\minus{}\lambda I$ is invertible whenever $ \lambda^4\ne 1.$

When $ \lambda^4\equal{}1,$ then $ F$ has (by the Hermite construction) eigenfunctions, and $ F\minus{}\lambda I$ is not injective.

In fact, $ L^2$ is simply the orthogonal direct sum of the four point eigenspaces.

[quote="eugene"]
Is it obvious that the Fourier operator (let's call it $ F$), defined above, doesn't have continuous spectrum, i.e. those $ \lambda$ such that $ F \minus{} \lambda I$ is injective, but its range is not the whole $ L^2(\mathbb{R})$ ?[/quote]

That's out of the question, for reasons which have nothing to do with Fourier analysis :): an isolated point of the spectrum of a normal operator $ A$ on a complex Hilbert space $ \mathcal H$ is an eigenvalue. Given this, it should be easy to show that the sum of the eigenspaces of a normal operator with finite spectrum is the whole space.

I believe the result invoked above has been discussed before around here. Anyway, you might want to give it a shot; it makes a nice exercise :).


inequalities
induction
trigonometry
limit
LaTeX
algebra unsolved
algebra

The sequences $ a_0, a_1, \ldots$ and $ b_0, b_1, \ldots$ are defined for $ n \equal{} 0, 1, 2, \ldots$ by the equalities
\[ a_0 \equal{} \frac {\sqrt {2}}{2}, \quad a_{n \plus{} 1} \equal{} \frac {\sqrt {2}}{2} \cdot \sqrt {1 \minus{} \sqrt {1 \minus{} a^2_n}}
\]
and
\[ b_0 \equal{} 1, \quad b_{n \plus{} 1} \equal{} \frac {\sqrt {1 \plus{} b^2_n} \minus{} 1}{b_n}
\]
Prove the inequalities for every $ n \equal{} 0, 1, 2, \ldots$
\[ 2^{n \plus{} 2} a_n < \pi < 2^{n \plus{} 2} b_n.
\]

Proceed by induction to prove that

[b]i)[/b] $ a_n \equal{} \sin \frac {\pi}{2^{n \plus{} 2}}$
[b]ii)[/b] $ b_n \equal{} \tan \frac {\pi}{2^{n \plus{} 2}}$

From which the inequality follows simply

I'm not sure about the induction proof SimonM mentioned, but to derive $ a_n\equal{}\sin \frac{\pi}{2^{n\plus{}2}}$, start from the half angle identity $ \sin x \equal{} \sqrt{\frac{1\minus{}\cos 2x}2} \equal{} \frac{\sqrt{2\minus{}2\cos 2x}}2$ and then replaced $ x$ by $ \pi/2^{n\plus{}2}$ to get the desired result.  Using similar technique, one can prove this $ \pi$ identity:
$ \pi \equal{} \lim_{n \to \infty} 2^{n\plus{}1} \sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2\plus{}...}}}$ where the underbrace below the nested square roots should say $ n$ times; I'm not sure about the latex code for that.


topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Problem 1: Let f: S^2 to T^2 be continuous. Show that f is homotopically trivial.

Attempted Solution: I want to lift f to the plane which is the universal cover of the torus. But to do that I need show that $ f_{*}(\pi_1(S^2,s_0)) \subset p_*(\pi_1(R^2,0))$, right? Any ideas how to do that?

Problem 2: Show that any map from S^2 to the Klein bottle is homotopically trivial.

Can I solve this problem similarly? Does the Klein bottle have a universal cover?

Yes. The plane covers the Klein bottle; we identify points $ (x,y)$ and $ (x\plus{}1,y)$, and also $ (x,y)$ and $ (\minus{}x,y\plus{}1)$. Extend that to a group of transformations, and you have the universal cover.

As for the fundamental group piece- note that the fundamental group of the sphere is trivial, so monodromy works automatically.
The final step is the easy one: just construct an explicit homotopy $ g(t,x)\equal{}(1\minus{}t)g(x)$, where $ g$ is the lifted map.